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Calculus Answers - Section 1.1 1 then x 1 1(a f(−a = =−...

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Unformatted text preview: Section 1.1 1 , then: x 1 1 (a) f (−a) = =− ; −a a 1 (b) f (a−1 ) = −1 = a; a √ 1 1 (c) f ( a ) = √ = 1/2 = a−1/2 ; a a C01S01.001: If f (x) = (d) f (a2 ) = 1 = a−2 . a2 C01S01.002: If f (x) = x2 + 5, then: (a) f (−a) = (−a)2 + 5 = a2 + 5; (b) f (a−1 ) = (a−1 )2 + 5 = a−2 + 5 = 1 1 + 5a2 +5= ; 2 a a2 √ √2 (c) f ( a ) = ( a ) + 5 = a + 5; (d) f (a2 ) = (a2 )2 + 5 = a4 + 5. 1 , then: x2 + 5 1 1 (a) f (−a) = =2 ; (−a)2 + 5 a +5 C01S01.003: If f (x) = (b) f (a−1 ) = 1 a2 1 1 · a2 = ; = −2 = −2 2 (a−1 )2 + 5 a +5 a · a + 5 · a2 1 + 5a2 √ 1 1 (c) f ( a ) = √ 2 = ; a+5 ( a) + 5 (d) f (a2 ) = 1 (a2 )2 +5 C01S01.004: If f (x) = (a) f (−a) = (b) f (a−1 ) = = a4 1 . +5 √ 1 + x2 + x4 , then: √ 1 + (−a)2 + (−a)4 = 1 + a2 + a4 ; 1 + (a−1 )2 + (a−1 )4 = √ 1 + a−2 + a−4 = √ a4 + a2 + 1 a4 + a2 + 1 √ = = ; a2 a4 √ √2 √4√ (c) f ( a ) = 1 + ( a ) + ( a ) = 1 + a + a2 ; √ (d) f (a2 ) = 1 + (a2 )2 + (a4 )2 = 1 + a4 + a8 . a4 + a2 + 1 = a4 √ (a4 ) · (1 + a−2 + a−4 ) a4 C01S01.005: If g (x) = 3x + 4 and g (a) = 5, then 3a + 4 = 5, so 3a = 1; therefore a = 1 . 3 C01S01.006: If g (x) = 1 and g (a) = 5, then: 2x − 1 1 1 = 5; 2a − 1 1 = 5 · (2a − 1); 1 = 10a − 5; 10a = 6; a= C01S01.007: If g (x) = √ 3 . 5 x2 + 16 and g (a) = 5, then: a2 + 16 = 5; a2 + 16 = 25; a2 = 9; a = 3 or a = −3. C01S01.008: If g (x) = x3 − 3 and g (a) = 5, then a3 − 3 = 5, so a3 = 8. Hence a = 2. C01S01.009: If g (x) = √ 3 x + 25 = (x + 25)1/3 and g (a) = 5, then (a + 25)1/3 = 5; a + 25 = 53 = 125; a = 100. C01S01.010: If g (x) = 2x2 − x + 4 and g (a) = 5, then: 2a2 − a + 4 = 5; 2a2 − a − 1 = 0; (2a + 1)(a − 1) = 0; 2a + 1 = 0 or a − 1 = 0; a=− 1 or a = 1. 2 C01S01.011: If f (x) = 3x − 2, then f (a + h) − f (a) = [3(a + h) − 2] − [3a − 2] = 3a + 3h − 2 − 3a + 2 = 3h. C01S01.012: If f (x) = 1 − 2x, then 2 f (a + h) − f (a) = [1 − 2(a + h)] − [1 − 2a] = 1 − 2a − 2h − 1 + 2a = −2h. C01S01.013: If f (x) = x2 , then f (a + h) − f (a) = (a + h)2 − a2 = a2 + 2ah + h2 − a2 = 2ah + h2 = h · (2a + h). C01S01.014: If f (x) = x2 + 2x, then f (a + h) − f (a) = [(a + h)2 + 2(a + h)] − [a2 + 2a] = a2 + 2ah + h2 + 2a + 2h − a2 − 2a = 2ah + h2 + 2h = h · (2a + h + 2). C01S01.015: If f (x) = 1 , then x f (a + h) − f (a) = = C01S01.016: If f (x) = 1 1 a a+h −= − a+h a a(a + h) a(a + h) a − (a + h) −h = . a(a + h) a(a + h) 2 , then x+1 f (a + h) − f (a) = 2 2 2(a + 1) 2(a + h + 1) − = − a+h+1 a+1 (a + h + 1)(a + 1) (a + h + 1)(a + 1) = 2a + 2 2a + 2h + 2 (2a + 2) − (2a + 2h + 2) − = (a + h + 1)(a + 1) (a + h + 1)(a + 1) (a + h + 1)(a + 1) = 2a + 2 − 2a − 2h − 2 −2h = . (a + h + 1)(a + 1) (a + h + 1)(a + 1) C01S01.017: If x > 0 then f (x) = x x = = 1. |x| x If x < 0 then f (x) = x x = = −1. |x| −x We are given f (0) = 0, so the range of f is {−1, 0, 1}. That is, the range of f is the set consisting of the three real numbers −1, 0, and 1. C01S01.018: Given f (x) = [[3x ]], we see that 3 f (x) = 0 if 0 x < 1, 3 f (x) = 1 if 1 3 x < 2, 3 f (2) = 2 if 2 3 x < 1; moreover, f (x) = −3 if −1 f (x) = −2 if − 2 3 x < − 1, 3 f (x) = −1 if − 1 3 x < 0. f (x) = 3m if m x < m + 1, 3 f (x) = 3m + 1 if m+ 1 3 x < m + 2, 3 f (x) = 3m + 2 if m+ 2 3 x < m + 1. x < − 2, 3 In general, if m is any integer, then Because every integer is equal to 3m or to 3m + 1 or to 3m + 2 for some integer m, we see that the range of f includes the set Z of all integers. Because f can assume no values other than integers, we can conclude that the range of f is exactly Z. C01S01.019: Given f (x) = (−1)[[x]] , we first note that the values of the exponent [[ x ]] consist of all the integers and no other numbers. So all that matters about the exponent is whether it is an even integer or an odd integer, for if even then f (x) = 1 and if odd then f (x) = −1. No other values of f (x) are possible, so the range of f is the set consisting of the two numbers −1 and 1. C01S01.020: If 0 < x 1, then f (x) = 34. If 1 < x 2 then f (x) = 34 + 21 = 55. If 2 < x 3 then f (x) = 34 + 2 · 21 = 76. We continue in this way and conclude with the observation that if 11 < x < 12 then f (x) = 34 + 11 · 21 = 265. So the range of f is the set {34, 55, 76, 97, 118, 139, 160, 181, 202, 223, 244, 265}. C01S01.021: Given f (x) = 10 − x2 , note that for every real number x, x2 is defined, and for every such real number x2 , 10 − x2 is also defined. Therefore the domain of f is the set R of all real numbers. C01S01.022: Given f (x) = x3 + 5, we note that for each real number x, x3 is defined; moreover, for each such real number x3 , x3 + 5 is also defined. Thus the domain of f is the set R of all real numbers. √ C01S01.023: √ Given f (t) = t2 , we observe that for every real number t, t2 is defined and nonnegative, and hence that t2 is defined as well. Therefore the domain of f is the set R of all real numbers. √2 √ √ C01S01.024: Given g (t) = t , we observe that t is defined exactly when t 0. In this case, t is also defined, and hence the domain of g is the set [0, + ∞) of all nonnegative real numbers. 2 √ C01S01.025: Given f (x) = 3x − 5, we note that 3x − 5 is defined for all real numbers x, but that its square root will be defined when and only when 3x − 5 is nonnegative; that is, when 3x − 5 0, so that 5 5 x 3 . So the domain of f consists of all those real numbers x in the interval 3 , + ∞ . 4 √ C01S01.026: Given g (t) = 3 t + 4 = (t + 4)1/3 , we note that t + 4 is defined for every real number t and the cube root of t + 4 is defined for every possible resulting value of t + 4. Therefore the domain of g is the set R of all real numbers. √ C01S01.027: Given f (t) = 1 − 2t, we observe that 1 − 2t is defined for every real number t, but that its square root is defined only when 1 − 2t is nonnegative. We solve the inequality 1 − 2t 0 to find that f (t) is defined exactly when t 1 . Hence the domain of f is the interval − ∞, 1 . 2 2 C01S01.028: Given g (x) = 1 , (x + 2)2 we see that (x + 2)2 is defined for every real number x, but that g (x), its reciprocal, will be defined only when (x + 2)2 = 0; that is, when x + 2 = 0. So the domain of g consists of those real numbers x = −2. C01S01.029: Given f (x) = 2 , 3−x we see that 3 − x is defined for all real values of x, but that f (x), double its reciprocal, is defined only when 3 − x = 0. So the domain of f consists of those real numbers x = 3. C01S01.030: Given 2 , 3−t g (t) = it is necessary that 3 − t be both nonzero (so that its reciprocal is defined) and nonnegative (so that the square root is defined). Thus 3 − t > 0, and therefore the domain of g consists of those real numbers t < 3. √ C01S01.031: Given f (x) = x2 + 9, observe that for each real number x, x2 + 9 is defined and, moreover, is positive. So its square root is defined for every real number x. Hence the domain of f is the set R of all real numbers. C01S01.032: Given h(z ) = √ 1 , 4 − z2 we note that 4 − z 2 is defined for every real number z , but that its square root will be defined only if 4 − z2 0. Moreover, the square root cannot be zero, else its reciprocal will be undefined, so we need to solve the inequality 4 − z 2 > 0; that is, z 2 < 4. The solution is −2 < z < 2, so the domain of h is the open interval (−2, 2). √ √ C01S01.033: Given f (x) = 4 − x , note first that we require x 0 in order that x be defined. In √ addition, we require 4 − x 0 so that its square root will be defined as well. So we solve [simultaneously] √ x 0 and x 4 to find that 0 x 16. So the domain of f is the closed interval [0, 16]. C01S01.034: Given x+1 , x−1 f (x) = 5 we require that x = 1 so that the fraction is defined. In addition we require that the fraction be nonnegative so that its square root will be defined. These conditions imply that both numerator and denominator be positive or that both be negative; moreover, the numerator may also be zero. But if the denominator is positive then the [larger] numerator will be positive as well; if the numerator is nonpositive then the [smaller] denominator will be negative. So the domain of f consists of those real numbers for which either x − 1 > 0 or x + 1 0; that is, either x > 1 or x −1. So the domain of f is the union of the two intervals (−∞, −1] and (1, + ∞). Alternatively, it consists of those real numbers x not in the interval (−1, 1]. C01S01.035: Given: g (t) = t . |t| This fraction will be defined whenever its denominator is nonzero, thus for all real numbers t = 0. So the domain of g consists of the nonzero real numbers; that is, the union of the two intervals (−∞, 0) and (0, + ∞). C01S01.036: If a square has edge length x, then its area A is given by A = x2 and its perimeter P is given by P = 4x. To express A in terms of P : x = 1P; 4 A = x2 = 2 1 4P = 2 1 16 P . Thus to express A as a function of P , we write A(P ) = 0 2 1 16 P , P < + ∞. (It will be convenient later in the course to allow the possibility that P , x, and A are zero. If this produces an answer that fails to meet real-world criteria for a solution, then that possibility can simply be eliminated when the answer to the problem is stated.) C01S01.037: If a circle has radius r, then its circumference C is given by C = 2π r and its area A by A = π r2 . To express C in terms of A, we first express r in terms of A, then substitute in the formula for C : A = πr2 ; r= C = 2π r = 2π A ; π √ π2 A = 2 π A. π A =2 π Therefore to express C as a function of A, we write √ C (A) = 2 π A, 0 A < + ∞. √ It is also permissible simply to write C (A) = 2 π A without mentioning the domain, because the “default” domain is correct. In the first displayed equation we do not write r = ± A/π because we know that r is never negative. C01S01.38: If r denotes the radius of the sphere, then its volume is given by V = 4 π r3 and its surface 3 area by S = 4π r2 . Hence 6 r= 1 2 S ; π V= Answer: V (S ) = 1 π 6 S π 43 4 1 πr = π · 3 3 8 0 3 /2 3 /2 S < + ∞. S π = 1 π 6 S π . 3 /2 , C01S01.039: To avoid decimals, we note that a change of 5◦ C is the same as a change of 9◦ F, so when the temperature is 10◦ C it is 32 + 18 = 50◦ F; when the temperature is 20◦ C then it is 32 + 2 · 18 = 68◦ F. 1 In general we get the Fahrenheit temperature F by adding 32 to the product of 10 C and 18, where C is the Celsius temperature. That is, 9 F = 32 + C, 5 and therefore C = 5 (F − 32). Answer: 9 C (F ) = 5 (F − 32), 9 F > −459.67. C01S01.040: Suppose that a rectangle has base length x and perimeter 100. Let h denote the height of such a rectangle. Then 2x + 2h = 100, so that h = 50 − x. Because x 0 and h 0, we see that 0 x 50. The area A of the rectangle is xh, so that A(x) = x(50 − x), 0 50. x C01S01.041: Let y denote the height of such a rectangle. The rectangle is inscribed in a circle of diameter 4, so the bottom side x √ the left side y√ the two legs of a right triangle with hypotenuse 4. Consequently and are x2 + y 2 = 16, so y = 16 − x2 (not − 16 − x2 because y 0). Because x 0 and y 0, we see that 0 x 4. The rectangle has area A = xy , so A(x) = x 16 − x2 , 0 x 4. C01S042.042: We take the problem to mean that current production is 200 barrels per day per well, that if one new well is drilled then the 21 wells will produce 195 barrels per day per well; in general, that if x new wells are drilled then the 20 + x wells will produce 200 − 5x barrels per day per well. So total production would be p = (20 + x)(200 − 5x) barrels per day. But because 200 − 5x 0, we see that x 40. Because x 0 as well (you don’t “undrill” wells), here’s the answer: p(x) = 4000 + 100x − 5x2 , 0 x 40, x an integer. C01S01.043: The square base of the box measures x by x centimeters; let y denote its height (in centimeters). Because the volume of the box is 324 cm3 , we see that x2 y = 324. The base of the box costs 2x2 cents, each of its four sides costs xy cents, and its top costs x2 cents. So the total cost of the box is C = 2x2 + 4xy + x2 = 3x2 + 4xy. 7 (1) Because x > 0 and y > 0 (the box has positive volume), but because y can be arbitrarily close to zero (as well as x), we see also that 0 < x < + ∞. We use the equation x2 y = 324 to eliminate y from Eq. (1) and thereby find that C (x) = 3x2 + 1296 , x 0 < x < + ∞. C01S01.044: If the rectangle is rotated around its side S of length x to produce a cylinder, then x will also be the height of the cylinder. Let y denote the length of the two sides perpendicular to S ; then y will be the radius of the cylinder; moreover, the perimeter of the original rectangle is 2x + 2y = 36. Hence y = 18 − x. Note also that x 0 and that x 18 (because y 0). The volume of the cylinder is V = π y 2 x, and so V (x) = π x(18 − x)2 , 0 18. x C01S01.045: Let h denote the height of the cylinder. Its radius is r, so its volume is π r2 h = 1000. The total surface area of the cylinder is A = 2π r2 + 2π rh h= 1000 , πr2 (look inside the front cover of the book); so A = 2π r2 + 2π r · 1000 2000 = 2π r2 + . 2 πr r Now r cannot be negative; r cannot be zero, else π r2 h = 1000. But r can be arbitrarily small positive as well as arbitrarily large positive (by making h sufficiently close to zero). Answer: A(r) = 2π r2 + 2000 , r 0 < r < + ∞. C01S01.046: Let y denote the height of the box (in centimeters). Then 2x2 + 4xy = 600, so that y= 600 − 2x2 . 4x (1) The volume of the box is V = x2 y = (600 − 2x2 ) · x2 1 1 = (600x − 2x3 ) = (300x − x3 ) 4x 4 2 by Eq. (1). Also x > 0 by Eq. √ but the maximum value of x is attained when Eq. (1) forces y to be zero, (1), √ at which point x = 300 = 10 3. Answer: V (x) = 300x − x3 , 2 √ 10 3. 0<x C01S01.047: The base of the box will be a square measuring 50 − 2x in. on each side, so the open-topped box will have that square as its base and four rectangular sides each measuring 50 − 2x by x (the height of the box). Clearly 0 x and 2x 50. So the volume of the box will be V (x) = x(50 − 2x)2 , 8 0 x 25. C01S01.048: Recall that A(x) = x(50 − x), 0 at some special numbers in its domain: x0 A0 5 225 10 400 15 525 50. Here is a table of a few values of the function A x 20 600 25 625 30 600 35 525 40 400 45 225 50 0 It appears that when x = 25 (so the rectangle is a square), the rectangle has maximum area 625. C01S01.049: Recall that the total daily production of the oil field is p(x) = (20 + x)(200 − 5x) if x new wells are drilled (where x is an integer satisfying 0 x 40). Here is a table of all of the values of the production function p: x 0 p 4000 1 4095 2 4180 3 4 5 6 7 4255 4320 4375 4420 4455 x 8 p 4480 9 4495 10 4500 11 12 13 14 15 4495 4480 4455 4420 4375 x 16 p 4320 17 4255 18 4180 19 20 21 22 23 4095 4000 3895 3780 3655 x 24 p 3520 25 3375 26 3220 27 28 29 30 31 3055 2880 2695 2500 2295 x 32 p 2080 33 1855 34 1620 35 36 1375 1120 37 855 38 580 39 295 and, finally, p(40) = 0. Answer: Drill ten new wells. C01S01.050: The surface area A of the box of Example 8 was A(x) = 2x2 + The restrictions x 1 and y 1 imply that 1 three places, are given in the following table. x 1 A 502 2 258 3 185 4 157 500 , x 0 < x < ∞. √ x 125. A small number of values of A, rounded to 5 6 7 8 9 10 11 150 155 169 191 218 250 287 It appears that A is minimized when x = y = 5. C01S01.051: If x is an integer, then Ceiling(x) = x and −Floor(−x) = −(−x) = x. If x is not an integer, then choose the integer n so that n < x < n + 1. Then Ceiling(x) = n + 1, −(n + 1) < −x < −n, and −Floor(−x) = −[−(n + 1)] = n + 1. In both cases we see that Ceiling(x) = −Floor(−x). C01S01.052: The range of Round(x) is the set Z of all integers. If k is a nonzero constant, then as x varies through all real number values, so does kx. Hence the range of Round(kx) is Z if k = 0. If k = 0 then the range of Round(kx) consists of the single number zero. 9 C01S01.053: By the result of Problem 52, the range of Round(10x) is the set of all integers, so the range 1 1 of g (x) = 10 Round(10x) is the set of all integral multiple of 10 . 1 C01S01.054: What works for π will work for every real number; let Round2(x) = 100 Round(100x). To be certain that this is correct (we will verify it only for positive numbers), write the [positive] real number x in the form x = k+ t h m + + + r, 10 100 1000 where k is a nonnnegative integer, t (the “tenths” digit) is a nonnegative integer between 0 and 9, h (the “hundredths” digit) is a nonnegative integer between 0 and 9, as is m, and 0 r < 0.001. Then Round2(x) = m + 0.5) = If 0 1 100 Floor(100x 1 100 Floor(100k + 10t + h + 1 10 (m + 5) + 100r). 4, the last expression becomes 1 t h (100k + 10t + h) = k + + , 100 10 100 which is the correct two-digit rounding of x. If 5 m 9, it becomes 1 t h+1 (100k + 10t + h + 1) = k + + , 100 10 100 also the correct two-digit rounding of x in this case. 1 C01S01.055: Let Round4(x) = 10000 Round(10000x). To verify that Round4 has the desired property for [say] positive values of x, write such a number x in the form x=k+ d1 d2 d3 d4 d5 + + + + + r, 10 100 1000 10000 100,000 where k is a nonnegative integer, each di is an integer between 0 and 9, and 0 of Round4 to x then produces 1 Floor(10000k + 1000d1 + 100d2 + 10d3 + d4 + 10000 Then consideration of the two cases 0 d5 four-place rounding of x in both cases. C01S01.056: Let Chop4(x) = 4 and 5 1 10000 Floor(10000x). x=k+ d5 1 10 (d5 r < 0.00001. Application + 5) + 10000r). 9 will show that Round4 produces the correct Suppose that x > 0. Write x in the form d1 d2 d3 d4 + + + + r, 10 100 1000 10000 where k is a nonnegative integer, each of the di is an integer between 0 and 9, and 0 Chop4(x) produces 1 10000 Floor(10000k = + 1000d1 + 100d2 + 10d3 + d4 + 10000r) 1 10000 (10000k + 1000d1 + 100d2 + 10d3 + d4 ) 10 r < 0.0001. Then because 10000r < 1. It follows that Chop4 has the desired effect. C01S01.057: x y 0.0 1.0 0.2 0.44 0.4 −0.04 0.6 −0.44 The sign change occurs between x = 0.2 and x = 0.4. x y 0.20 0.44 0.25 0.3125 0.30 0.19 0.35 0.0725 The sign change occurs between x = 0.35 and x = 0.40. x y 0.35 0.0725 0.36 0.0496 0.37 0.0269 0.38 0.0044 From this point on, the data for y will be rounded. x y 0.380 0.0044 0.382 −0.0001 0.384 −0.0045 0.386 −0.0090 0.8 −0.76 1.0 −1.0 0.40 −0.04 0.39 −0.0179 0.40 −0.04 0.388 −0.0135 0.390 −0.0179 Answer (rounded to two places): 0.38. The quadratic formula yields the two roots of these is approximately 0.381966011250105151795. 1 2 3± √ 5 ; the smaller Problems 58 through 66 are worked in the same way as Problem 57. C01S01.058: The sign change intervals are [2, 3], [2.6, 2.8], [2.60, 2.64], and [2.616, 2.624]. Answer: √ 1 5 ≈ 2.62. 2 3+ C01S01.059: √ The sign change intervals are [1, 2], [1.2, 1.4], [1.20, 1.24], [1.232, 1.240], and [1.2352, 1.2368]. Answer: −1 + 5 ≈ 1.24. C01S01.060: The sign change intervals are [−4, −3], [−3.4, −3.2], [−3.24, −3.20], [−3.240, −3.232], and √ [−3.2368, −3.2352]. Answer: −1 − 5 ≈ −3.24. C01S01.061: The sign change intervals are [0, 1], [0.6, 0.8], [0.68, 0.72], [0.712, 0.720], and [0.7184, 0.7200]. √ Answer: 1 7 − 17 ≈ 0.72. 4 C01S01.062: The sign change intervals are [2, 3], [2.6, 2.8], [2.76, 2.80], [2.776, 2.784], and [2.7792, 2.7808]. √ Answer: 1 7 + 17 ≈ 2.78. 4 C01S01.063: The sign change intervals are [3, 4], [3.2, 3.4], [3.20, 3.24], [3.208, 3.216], and [3.2080, 3.2096]. √ Answer: 1 11 − 21 ≈ 3.21. 2 C01S01.064: The sign change intervals are [7, 8], [7.6, 7.8], [7.76, 7.80], [7.784, 7.792], and [7.7904, 7.7920]. √ Answer: 1 11 + 21 ≈ 7.79. 2 C01S01.065: The sign change intervals √ [1, 2], [1.6, 1.8], [1.60, 1.64], [1.608, 1.616], [1.6144, 1.6160], are and [1.61568, 1.61600]. Answer: 1 −23 + 1069 ≈ 1.62. 6 C01S01.066: The sign change intervals are [−10, −9], [−9.4, −9.2], [−9.32, −9.28], [−9.288, −9.280], and √ [−9.2832, −9.2816]. Answer: 1 −23 − 1069 ≈ −9.28. 6 11 Section 1.2 C01S02.001: The slope of L is m = (3 − 0)/(2 − 0) = 3 , so L has equation 2 y−0= 3 (x − 0); 2 that is, 2y = 3x. C01S02.002: Because L is vertical and (7, 0) lies on L, every point on L has Cartesian coordinates (7, y ) for some number y (and every such point lies on L). Hence an equation of L is x = 7. C01S02.003: Because L is horizontal, it has slope zero. Hence an equation of L is y − (−5) = 0 · (x − 3); that is, y = −5. C01S02.004: Because (2, 0) and (0, −3) lie on L, it has slope (0 + 3)/(2 − 0) = 3 . Hence an equation of 2 L is y − 0 = 3 (x − 2); 2 that is, y = 3 x − 3. 2 C01S02.005: The slope of L is (3 − (−3))/(5 − 2) = 2, so an equation of L is y − 3 = 2(x − 5); that is, y = 2x − 7. C01S02.006: An equation of L is y − (−4) = 1 (x − (−1)); that is, 2y + 7 = x. 2 C01S02.007: The slope of L is tan(135◦ ) = −1, so L has equation y − 2 = −1 · (x − 4); that is, x + y = 6. C01S02.008: Equation: y − 7 = 6(x − 0); that is, y = 6x + 7. C01S02.009: The second line’s equation can be written in the form y = −2x + 10 to show that it has slope −2. Because L is parallel to the second line, L also has slope −2 and thus equation y − 5 = −2(x − 1). C01S02.010: The equation of the second line can be rewritten as y = − 1 x + 17 to show that it has slope 2 2 − 1 . Because L is perpendicular to the second line, L has slope 2 and thus equation y − 4 = 2(x + 2). 2 C01S02.011: x2 − 4x + 4 + y 2 = 4: (x − 2)2 + (y − 0)2 = 22 . Center (2, 0), radius 2. C01S02.012: x2 + y 2 + 6y + 9 = 9: (x − 0)2 + (y + 3)2 = 32 . Center (0, −3), radius 3. C01S02.013: x2 + 2x + 1 + y 2 + 2y + 1 = 4: (x + 1)2 + (y + 1)2 = 22 . Center (−1, −1), radius 2. C01S02.014: x2 + 10x + 25 + y 2 − 20y + 100 = 25: (x + 5)2 + (y − 10)2 = 52 . Center (−5, 10), radius 5. C01S02.015: x2 + y 2 + x − y = 1 : x2 + x + 1 + y 2 − y + 1 = 1; (x + 1 )2 + (y − 1 )2 = 1. Center: (− 1 , 1 ), 2 4 4 2 2 22 radius 1. C01S02.016: x2 + y 2 − 2 x − 4 y = 3 3 ( 1 , 2 ), radius 4 . 33 3 11 9: x2 − 2 x + 1 + y 2 − 4 y + 4 = 3 9 3 9 1 16 9; (x − 1 )2 + (y − 2 )2 = ( 4 )2 . Center 3 3 3 C01S02.017: y = (x − 3)2 : Opens upward, vertex at (3, 0). C01S02.018: y − 16 = −x2 : Opens downward, vertex at (0, 16). C01S02.019: y − 3 = (x + 1)2 : Opens upward, vertex at (−1, 3). C01S02.020: 2y = x2 − 4x + 4 + 4: y − 2 = 1 (x − 2)2 . Opens upward, vertex at (2, 2). 2 C01S02.021: y = 5(x2 + 4x + 4) + 3 = 5(x + 2)2 + 3: Opens upward, vertex at (−2, 3). C01S02.022: y = −(x2 − x) = −(x2 − x + 1 ) + 1 : y − 4 4 1 4 = −(x − 1 )2 . Opens downward, vertex at ( 1 , 1 ). 2 24 C01S02.023: x2 − 6x + 9 + y 2 + 8y + 16 = 25: (x − 3)2 + (y + 4)2 = 55 . Circle, center (3, −4), radius 5. C01S02.024: (x − 1)2 + (y + 1)2 = 0: The graph consists of the single point (1, −1). C01S02.025: (x + 1)2 + (y + 3)2 = −10: There are no points on the graph. C01S02.026: x2 + y 2 − x + 3y + 2.5 = 0: x2 − x + 0.25 + y 2 + 3y + 2.25 = 0: (x − 0.5)2 + (y + 1.5)2 = 0. The graph consists of the single point (0.5, −1.5). C01S02.027: The graph is the straight line segment connecting the two points (−1, 7) and (1, −3) (including those two points). C01S02.028: The graph is the straight line segment connecting the two points (0, 2) and (2, −8), including the first of these two points but not the second. C01S02.029: The graph is √ parabola that opens downward, symmetric around the y -axis, with vertex the at (0, 10) and x-intercepts ± 10. C01S02.030: The graph of y = 1 + 2x2 is a parabola that opens upwards, is symmetric around the y -axis, and has vertex at (0, 1). C01S02.031: The graph of y = x3 can be visualized by modifying the familiar graph of the parabola with equation y = x2 : The former may be obtained by multiplying the y -coordinate of the latter’s point (x, x2 ) by x. Thus both have flat spots at the origin. For 0 < x < 1, the graph of y = x3 is below that of y = x2 . They cross at (1, 1), and for x > 1 the graph of y = x2 is below that of y = x3 , with the difference becoming arbitrarily large as x increases without bound. If the graph of y = x3 for x 0 is rotated 180◦ around the point (0, 0), the graph of y = x3 for x < 0 is the result. C01S02.032: The graph of f (x) = x4 can be visualized by first visualizing the graph of y = x2 . If the y -coordinate of each point on this graph is replaced with its square (x4 ), the result is the graph of f . The effect on the graph of y = x2 is to multiply the y -coordinate by x2 , which is between 0 and 1 for 0 < | x | < 1 and which is larger than 1 for | x | > 1. Thus the graph of f superficially resembles that of y = x2 , but is much closer to the x-axis for | x | < 1 and much farther away for | x | > 1. The two graphs cross at (0, 0) (where each has a flat spot) and at (±1, 1), but the graph of f is much steeper at the latter two points. √ C01S02.033: To graph y = f (x) = 4 − x2 , note that y 0 and that y 2 = 4 − x2 ; that is, x2 + y 2 = 4. Hence the graph of f is the upper half of the circle with center (0, 0) and radius 2. √ C01S02.034: To graph y = f (x) = − 9 − x2 , note that y 0 and that y 2 = 9 − x2 ; that is, that 2 2 x + y = 9. Hence the graph of f is the lower half of the circle with center (0, 0) and radius 3. 2 √ C01S02.035: To graph f (x) = x2 − 9, note that there is no graph for −3 < √ < 3, that f (±3) = 0, and x √ that f (x) > 0 for x < −3 and for x > 3. If x is large positive, then x2 − 9 ≈ x2 = x, so the graph of f has x-intercept (3, 0) and rises as x increases, nearly coinciding with the graph of y = √ for x large positive. x √ The case x < −3 is trickier. In this case, if x is a large negative number, then f (x) = x2 − 9 ≈ x2 = −x (Note the minus sign!). So for x −3, the graph of f has x-intercept (−3, 0) and, for x large negative, almost coincides with the graph of y = −x. Later we will see that the graph of f becomes arbitrarily steep as x gets closer and closer to ±3. C01S02.036: As x increases without bound—either positively or negatively—f (x) gets arbitrarily close to zero. Moreover, if x is large positive then f (x) is negative and close to zero, so the graph of f lies just below the x-axis for such x. Similarly, the graph of f lies just above the x-axis for x large negative. If x is slightly less than 1 but very close to 1, then f (x) is the reciprocal of a tiny positive number, hence is a large positive number. So the graph of f just to the left of the vertical line x = 1 almost coincides with the top half of that line. Similarly, just to the right of the line x = 1, then graph of f almost coincides with the bottom half of that line. There is no graph where x = 1, so the graph resembles the one in the next figure. The only intercept is the y -intercept (0, 1). The graph correctly shows that the graph of f is increasing for x < 1 and for x > 1. 4 2 -4 -2 2 4 6 -2 -4 C01S02.037: Note that f (x) is positive and close to zero for x large positive, so that the graph of f is just above the x-axis—and nearly coincides with it—for such x. Similarly, the graph of f is just below the x-axis and nearly coincides with it for x large negative. There is no graph where x = −2, but if x is slightly greater than −2 then f (x) is the reciprocal of a very small positive number, so f (x) is large and nearly coincides with the upper half of the vertical line x = −2. Similarly, if x is slightly less than −2, then the graph of f (x) is large negative and nearly coincides with the the lower half of the line x = −3. The graph of f is decreasing for x < −2 and for x > −2 and its only intercept is the y -intercept 0, 1 . 2 C01S02.038: Note that f (x) is very small but positive if x is either large positive or large negative. There is no graph for x = 0, but if x is very close to zero, then f (x) is the reciprocal of a very small positive number, and hence is large positive. So the graph of f is just above the x-axis and almost coincides with it if | x | is large, whereas the graph of f almost coincides with the positive y -axis for x near zero. There are no intercepts; the graph of f is increasing for x < 0 and is decreasing for x > 0. 3 C01S02.039: Note that f (x) > 0 for all x other than x = 1, where f is not defined. If | x | is large, then f (x) is near zero, so the graph of f almost coincides with the x-axis for such x. If x is very close to 1, then f (x) is the reciprocal of a very small positive number, hence f (x) is large positive. So for such x, the graph of f (x) almost coincides with the upper half of the vertical line x = 1. The only intercept is (0, 1). C01S02.040: Note first that f (x) is undefined at x = 0. To handle the absolute value symbol, we look at two cases: If x > 0, then f (x) = 1; if x < 0, then f (x) = −1. So the graph of f consists of the part of the horizontal line y = 1 for which x > 0, together with the part of the horizontal line y = −1 for which x < 0. C01S02.041: Note that f (x) is undefined when 2x + 3 = 0; that is, when x = − 3 . If x is large positive, 2 then f (x) is positive and close to zero, so the graph of f is slightly above the x-axis and almost coincides with the x-axis. If x is large negative, then f (x) is negative and close to zero, so the graph of f is slightly below the x-axis and almost coincides with the x-axis. If x is slightly greater than − 3 then f (x) is very 2 large positive, so the graph of f almost coincides with the upper half of the vertical line x = − 3 . If x is 2 slightly less than − 3 then f (x) is very large negative, so the graph of f almost coincides with the lower half 2 of that vertical line. The graph of f is decreasing for x < − 3 and also decreasing for x > − 3 . The only 2 2 intercept is at 0, 1 . 3 C01S02.042: Note that f (x) is undefined when 2x + 3 = 0; that is, when x = − 3 . If x is large positive 2 or large negative, then f (x) is positive and close to zero, so the graph of f is slightly above the x-axis and almost coincides with the x-axis for | x | large. If x is close to − 3 then f (x) is very large positive, so the 2 graph of f almost coincides with the upper half of the vertical line x = − 3 . The graph of f is increasing for 2 x < − 3 and decreasing for x > − 3 . The only intercept is at 0, 1 . 2 2 9 √ C01S02.043: Given y = f (x) = 1 − x, note that y 0 and that y 2 = 1 − x; that is, x = 1 − y 2 . So the graph is the part of the parabola x = 1 − y 2 for which y 0. This parabola has horizontal axis of symmetry the y -axis, opens to the left (because the coefficient of y 2 is negative), and has vertex (1, 0). Therefore the graph of f is the upper half of this parabola. C01S02.044: Note that the interval x < 1 is the domain of f , so there is no graph for x 1. If x is a large negative number, then the denominator is large positive, so that its reciprocal f (x) is very small positive. As x gets closer and closer to 1 (while x < 1), the denominator approaches zero, so its reciprocal f (x) takes on arbitrarily large positive values. So the graph of f is slightly above the x-axis and almost coincides with that axis for x large negative; the graph of f almost coincides with the upper half of the vertical line x = 1 for x near (and less than) 1. The graph of f is increasing for all x < 1 and (0, 1) is the only intercept. C01S02.045: Note that f (x) is defined only if 2x + 3 > 0; that is, if x > − 3 . Note also that f (x) > 0 2 for all such x. If x is large positive, then f (x) is positive but near zero, so the graph of f is just above the x-axis and almost coincides with it. If x is very close to − 3 (but larger), then the denominator in f (x) is 2 very tiny positive, so the graph of f almost coincides with the upper half of the vertical line x = − 3 for 2 such x. The graph of f is decreasing for all x > − 3 . 2 C01S02.046: Given: f (x) = | 2x − 2 |. Case 1: x 1. Then 2x − 2 0, so that f (x) = 2x − 2. Because f (1) = 0, the graph of f for x 1 consists of the part of the straight line through (1, 0) with slope 2. Case 2: x < 1. Then 2x − 2 < 0, so that f (x) = −2x + 2. The line y = −2x + 2 passes through (1, 0), so the graph of f for x < 1 consists of the part of the straight line through (1, 0) with slope −2. C01S02.047: Given: f (x) = | x | + x. If x 0 then f (x) = x + x = 2x, so if x 0 then the graph of f is the part of the straight line through (0, 0) with slope 2 for which x 0. If x < 0 then f (x) = −x + x = 0, so the rest of the graph of f coincides with the negative x-axis. 4 C01S02.048: Given: f (x) = | x − 3 |. If x 0 then f (x) = x − 3, so the graph of f consists of the straight line through (3, 0) with slope 1 for x 3. If x < 0 then f (x) = −x + 3, so the graph of f consists of that part of the straight line with slope −3 and y -intercept (0, 3). These two line segments fit together perfectly at the point (3, 0); there is no break or gap or discontinuity in the graph of f . C01S02.049: Given: f (x) = | 2x + 5 |. The two cases are determined by the point where 2x + 5 changes sign, which is where x = − 5 . If x − 5 , then f (x) = 2x + 5, so the graph of f consists of the part of the 2 2 line with slope 2 and y -intercept 5 for which x − 5 . If x < − 5 , then the graph of f is the part of the 2 2 straight line y = −2x − 5 for which x < − 5 . 2 C01S02.050: The graph consists of the part of the line y = −x for which x < 0 together with the part of the parabola y = x2 for which x 0. The two graphs fit together perfectly at the point (0, 0); there is no break, gap, jump, or discontinuity there. The graph is shown next. 4 3 2 1 -2 -1 1 2 C01S02.051: The graph consists of the horizontal line y = 0 for x < 0 together with the horizontal line y = 1 for x 0. As x moves from left to right through the value zero, there is an abrupt and unavoidable “jump” in the value of f from 0 to 1. That is, f is discontinuous at x = 0. To see part of the graph of f , enter the Mathematica commands f [ x ] := If [ x < 0, 0, 1] Plot [ f [ x ], {x, −3.5, 3.5 }, AspectRatio −> Automatic, PlotRange −> { { −3.5, 3.5 }, {−1.5, 2.5 }} ]; C01S02.052: The graph of f consists of the open intervals . . . , (−2, −1), (−1, 0), (0, 1), (1, 2), (2, 3), . . . on the x-axis together with the isolated points . . . , (−1, 1), (0, 1), (1, 1), (2, 1), (3, 1), . . . . There is a discontinuity at every integral value of x. A Mathematica plot of f [ x ] := If [IntegerQ[x], 1, 0 ] will produce a graph that’s completely different because Mathematica, like most plotting programs, “connects the dots,” in effect assuming that every function is continuous at every point in its domain. C01S02.053: Because the graph of the greatest integer function changes at each integral value of x, the graph of f (x) = [[2x ]] changes twice as often—at each integral multiple of 1 . So as x moves from left to 2 right through such points, the graph jumps upward one unit. Thus there is a discontinuity at each integral multiple of 1 . Because f is constant otherwise, these are the only discontinuities. To see something like the 2 graph of f , enter the Mathematica commands 5 f [ x ] := Floor [ 2∗x ]; Plot [ f [ x ], {x, −3.5, 3.5 }, AspectRatio −> Automatic, PlotRange −> { { −3.5, 3.5 }, {−4.5, 4.5 }} ]; Mathematica will draw vertical lines connecting points that it shouldn’t, making the graph look like treads and risers of a staircase, whereas only the treads are on the graph. C01S02.054: The function f is undefined at x = 1. The graph consists of the horizontal line y = 1 for x > 1 together with the horizontal line y = −1 for x < 1. There is a discontinuity at x = 1. C01S02.055: Given: f (x) = [[ x ]]. If n is an integer and n x < n + 1, then express x as x = n + ((x)) where ((x)) = x − [[ x ]] is the fractional part of x. Then f (x) = n − x = n − [n + ((x))] = − ((x)). So f (x) is the negative of the fractional part of x. So as x ranges from n up to (but not including) n + 1, f (x) begins at 0 and drops linearly down not quite to −1. That is, on the interval (n, n + 1), the graph of f is the straight line segment connecting the two points (n, 0) and (n + 1, −1) with the first of these points included and the second excluded. There is a discontinuity at each integral value of x. C01S02.056: Given: f (x) = [[ x ]] + [[ − x ]] + 1. If x is an integer, then f (x) = x + (−x) + 1 = 1. If x is not an integer, then choose the integer n such that n < x < n + 1. Then −(n + 1) < −x < −n, so f (x) = [[ x ]] + [[ − x ]] + 1 = n − (n + 1) + 1 = 0. So f is the same function as the one defined in Problem 52 and has the same discontinuities: one at each integral value of x. C01S02.057: Because y = 2x2 − 6x + 7 = 2(x2 − 3x + 3.5) = 2(x2 − 3x + 2.25 + 1.25) = 2(x − 1.5)2 + 2.5, the vertex of the parabola is at (1.5, 2.5). C01S02.058: Because y = 2x2 − 10x + 11 = 2(x2 − 5x + 5.5) = 2(x2 − 5x + 6.25 − 0.75) = 2(x − 2.5)2 − 1.5, the vertex of the parabola is at (2.5, −1.5). C01S02.059: Because y = 4x2 − 18x + 22 = 4(x2 − (4.5)x + 5.5) = 4(x2 − (4.5)x + 5.0625 + 0.4375) 2 = 4(x − 2.25) + 1.75, the vertex of the parabola is at (2.25, 1.75). C01S02.060: Because y = 5x2 − 32x + 49 = 5(x2 − (6.4)x + 9.8) = 5(x2 − (6.4)x + 10.24 − 0.44) = 5(x − 3.2)2 − 2.2, the vertex of the parabola is at (3.2, −2.2). C01S02.061: Because y = −8x2 + 36x − 32 = −8(x2 − (4.5)x + 4) = −8(x2 − (4.5)x + 5.0625 − 1.0625) = −8(x − 2.25)2 + 8.5, the vertex of the parabola is at (2.25, 8.5). C01S02.062: Because y = −5x2 − 34x − 53 = −5(x2 + (6.8)x + 10.6) = −5(x2 + (6.8)x + 11.56 − 0.96) = −5(x + 3.4)2 + 4.8, the vertex of the parabola is at (−3.4, 4.8). C01S02.063: Because y = −3x2 −8x+3 = −3 x2 + 8 x − 1 = −3 x2 + 8 x + 3 3 the vertex of the parabola is at − 4 , 25 . 3 3 C01S02.064: = −9 x − 17 2 9 + Because y = −9x2 + 34x − 28 = −9 x2 − 37 9, the vertex of the parabola is at 17 37 9, 9 . 34 9x + 28 9 16 9 − 25 9 = −9 x2 − = −3 x + 34 9x + 289 81 − 4 2 25 3 +3, 37 81 C01S02.065: To find the maximum height y = −16t2 + 96t of the ball, we find the vertex of the parabola: y = −16(t2 − 6t) = −16(t2 − 6t + 9 − 9) = −16(t − 3)2 + 144. The vertex of the parabola is at (3, 144) and therefore the maximum height of the ball is 144 ft. 6 C01S02.066: Recall that the area of the rectangle is given by y = A(x) = x(50 − x). To maximize A(x) we find the vertex of the parabola: y = 50x − x2 = −(x2 − 50x) = −(x2 − 50x + 625 − 625) = −(x − 25)2 + 625. Because the vertex of the parabola is at (25, 625) and x = 25 is in the domain of the function A, the maximum value of A(x) occurs at x = 25 and is A(25) = 625 (ft2 ). C01S02.067: If two positive numbers x and y have sum 50, then y = 50 − x and x < 50 (because y > 0). To maximize their product p(x) we find the vertex of the parabola y = p(x) = x(50 − x) = −(x2 − 50x) = −(x2 − 50x + 625 − 625) = −(x − 25)2 + 625, which is at (25, 625). Because 0 < 25 < 50, x = 25 is in the domain of the product function p(x) = x(50 − x), and hence the maximum value of the product of x and y is p(25) = 625. C01S02.068: Recall that if x new wells are drilled, then the resulting total production p is given by p(x) = 4000 + 100x − 5x2 . To maximize p(x) we find the vertex of the parabola y = p(x) = −5x2 + 100x + 4000 = −5(x2 − 20x − 800) = −5(x2 − 20x + 100 − 900) = −5(x − 10)2 + 4500. The vertex of the parabola y = p(x) is therefore at (10, 4500). Because x = 10 is in the domain of p (it is an integer between 0 and 40) and because the parabola opens downward (the coefficient of x2 is negative), x = 10 indeed maximizes p(x). C01S02.069: The graph looks like the graph of y = | x | because the slope of the left-hand part is −1 and that of the right-hand part is 1; but the vertex is shifted to (−1, 0), so—using the translation principle—the graph in Fig. 1.2.29 must be the graph of f (x) = | x + 1 |, −2 x 2. C01S02.070: Because the graph in Fig. 1.2.30 is composed described most easily using a “three-part” function: if −3 2x + 6 f (x) = 2 if −2 1 (10 − 2x) if 2 3 of three straight-line segments, it can be x < −2; x < 2; x 5. C01S02.071: The graph in Fig. 1.2.31 is much like the graph of the greatest integer function—it takes on only integral values—but the “jumps” occur twice as often, so this must be very like—indeed, it is exactly—the graph of f (x) = [[2x , −1 x < 2. C01S02.072: The graph in Fig. 1.2.32 resembles the graph of the greatest integer function in that it takes on all integral values and only those, but it is decreasing rather than increasing and the “jumps” occur only at the even integers. Thus it must be the graph of something similar to f (x) = − [[ 1 x]], −4 x < 4. 2 Comparing values of f at x = −4, −3, −2.1, −2, −1, −0.1, 0, 1, 1.9, 2, 3, and 3.9 with points on the graph is sufficient evidence that the graph of f is indeed that shown in the figure. C01S02.073: Clearly x(t) = 45t for the first hour; that is, for 0 t 1. In the second hour the graph of x(t) must be a straight line (because of constant speed) of slope 75, thus with equation x(t) = 75t + C for some constant C . The constant C is determined by the fact that 45t and 75t + C must be equal at time 7 t = 1, as the automobile cannot suddenly jump from one position to a completely different position in an instant. Hence 45 = 75 + C , so that C = −30. Therefore x(t) = 45t 75t − 30 if 0 t if 1 < t 1; 2. To see the graph of x(t), plot in Mathematica x[t ] := If[t < 1, 45∗t, 75∗t − 30 ] on the interval 0 t 2. C01S02.074: The graph of x(t) will consist of three straight-line segments (because of the constant speeds), the first of slope 60 for 0 t 1, the second of slope zero for 1 t 1.5, and the third of slope 60 for 1.5 t 2.5. The first pair must coincide when t = 1 and the second pair must coincide when t = 1.5 because the graph of x(t) can have no discontinuities. So if we write x(t) = 60 for 0 t 1, we must have x(t) = 60 for 1 t 1.5. Finally, x(t) = 60t + C for some constant C if 1.5 t 2.5, but the latter must equal 60 when t = 1.5, so that C = −30. Hence if 0 t 1, 60t x(t) = 60 if 1 < t 1.5, 60t − 30 if 1.5 < t 2.5. The graph of x(t) is shown next. 120 100 80 60 40 20 0.5 1 1.5 2 2.5 C01S02.075: The graph must consist of two straight-line segments (because of the constant speeds). The first must have slope 60, so we have x(t) = 60t for 0 t 1. The second must have slope −30, negative because you’re driving in the reverse direction, so x(t) = −30t + C for some constant C if 1 t 3. The two segments must coincide when t = 1, so that 60 = −30 + C . Thus C = 90 and thus a formula for x(t) is x(t) = 60t 90 − 30t if 0 t if 1 < t 1, 3. C01S02.076: We need three straight line segments, the first of slope 60 for 0 t 0.5, the second of slope −60 for 0.5 t 1, and the third of slope 60 for 1 t 3. Clearly the first must be x(t) = 60t for 0 t 0.5. The second must have the form x(t) = −60t + C for some constant C , and the first and second must coincide when t = 0.5, so that 30 = −30 + C , and thus C = 60. The third segment must have the form x(t) = 60t + K for some constant K , and the second and third must coincide when t = 1, so that 0 = 60 + K , and so K = −60. Therefore a formula for x(t) is 8 The graph of x(t) is shown next. 60t x(t) = 60 − 60t 60t − 60 if 0 t 0.5, if 0.5 < t 1, if 1 < t 3. 120 100 80 60 40 20 0.5 1 1.5 2 2.5 3 C01S02.077: Initially we work in units of pages and cents (to avoid decimals and fractions). The graph of C , as a function of p, must be a straight line segment, and its slope is (by information given) C (79) − C (34) 305 − 170 135 = = = 3. 79 − 34 79 − 34 45 Thus C (p) = 3p + K for some constant K . So 3 · 34+ K = 170, and it follows that K = 68. So C (p) = 3p +68, 1 p 100, if C is to be expressed in cents. If C is to be expressed in dollars, we have C (p) = (0.03)p + 0.68, 1 p 100. The “fixed cost” is incurred regardless of the number of pamphlets printed; it is $0.68. The “marginal cost” of printing each additional page of the pamphlet is the coefficient $0.03 of p. C01S02.078: We are given C (x) = a + bx where a and b are constants; we are also given 99.45 = C (207) = a + 207b and 79.15 = C (149) = a + 149b. Subtraction of the second equation from the first yields 20.3 = 58b, so that b = 0.35. Substitution of this datum in the first of the preceding equations then yields 99.45 = a + 207 · 0.35 = a + 72.45, so that a = 27. Therefore C (x) = 27 + (0.35)x, 0 x < +∞. Thus if you drive 175 miles on the third day, the cost for that day will be C (175) = 88.25 (in dollars). The slope b = 0.35 represents a cost of $0.35 per mile. The C -intercept a = 27 represents the daily base cost of renting the car. In civil engineering and in some branches of applied mathematics, the intercept a = 27 is sometimes called the offset, representing the vertical amount by which C (0) is “offset” from zero. C01S02.079: Suppose that the letter weighs x ounces, 0 < x 16. If x 8, then the cost is simply 8 (dollars). If 8 < x 9, add $0.80; if 9 < x 10, add $1.60, and so on. Very roughly, one adds $0.80 if 9 [[ x − 8]] = 1, $1.60 if [[ x − 8]] = 2, and so on. But this isn’t quite right—we are using the Floor function of Section 1.1, whereas we should really be using the Ceiling function. By the result of Problem 51 of that section, we see that instead of cost C (x) = 8 + (0.8)[[ x − 8]] for 8 < x 16, we should instead write C (x) = 8 if 0 < x 8, 8 − (0.8)[[ −(x − 8)]] if 8 < x 16. 12.5 15 The graph of the cost function is shown next. 15 12.5 10 7.5 5 2.5 2.5 5 7.5 10 C01S02.080: Solve this problem like Problem 79 (but it is more complicated). Result: if 0 < x 2; 3 C (x) = 3 − 0.5[[ −2(x − 2)]] if 2 < x 10; 11 − 0.5[[−(x − 10)]] if 10 < x 20. The graph of C is shown below. 15 12.5 10 7.5 5 2.5 5 10 15 20 C01S02.081: Boyle’s law states that under conditions of constant temperature, the product of the pressure p and the volume V of a fixed mass of gas remains constant. If we assume that pV = c, a constant, for the given data, we find that the given five data points yield the values c = 1.68, 1.68, 1.675, 1.68, and 1.62. The 10 average of these is 1.65 (to two places) and should be a good estimate of the true value of c. Alternatively, you can use a computer algebra program to find c; in Mathematica, for example, the command Fit will fit given data points to a sum of constant multiples of functions you specify. We used the commands data = {{0.25, 6.72}, {1.0, 1.68}, {2.5, 0.67}, {4.0, 0.42}, {6.0, 0.27}}; Fit [ data, { 1/p }, p ] to find that V (p) = 1.67986 p yields the best least-squares fit of the given data to a function of the form V (p) = c/p. We rounded the numerator to 1.68 to find the estimates V (0.5) ≈ 3.36 and V (5) ≈ 0.336 (L). The graph of V (p) is shown next. 8 6 4 2 1 2 3 4 5 6 C01S02.082: It seems reasonable to assume that the maximum average temperature occurs on July 15 and the minimum on January 15, so that a multiple of a cosine function should fit the given data if we take t = 0 on July 15. So we assume a solution of the form T (t) = c1 + c2 cos 2π t 365 . Also assuming that the average year-round daily temperature is the average of the minimum and the maximum, we find that c1 = 61.25, so we could find c2 by the averaging method of Problem 81. Alternatively, we could use the Fit command in Mathematica to find both c1 and c2 simultaneously as follows: data = {{0, 79.1}, {62, 70.2}, {123, 52.3}, {184, 43.4}, {224, 52.2}, {285, 70.1}}; Fit [ data, { 1, Cos[2∗Pi∗t/365] }, t ] The result is the formula T (t) = 62.9602 + (17.437) cos 2π t 365 . The values predicted by this function at the six dates in question are [approximately] 80.4, 71.4, 53.9, 45.5, 49.8, and 66.3. Not bad, considering we are dealing with weather, a most unpredictable phenomenon. The graph of T (t) is shown next. Units on the horizontal axis are days, measured from July 15. Units on 11 the vertical axis are degrees Fahrenheit. Remember that these are average daily temperatures; it is not uncommon for a winter low in Athens to be below 28◦ F and for a summer high to be as much as 92◦ F. 80 60 40 20 50 100 150 200 12 250 300 350 Section 1.3 C01S03.001: The domain of f is R, the set of all real numbers; so is the domain of g , but g (x) = 0 when x = 1 and when x = −3. So the domain of f + g and f · g is the set R and the domain of f /g is the set of all real numbers other than 1 and −3. Their formulas are (f + g )(x) = x2 + 3x − 2, (f · g )(x) = (x + 1)(x2 + 2x − 3) = x3 + 3x2 − x − 3, and f x+1 (x) = 2 . g x + 2x − 3 C01S03.002: The domain of f consists of all real numbers other than 1 and the domain of g consists of all real numbers other than − 1 . Hence the domain of f + g , f · g , and f /g consists of all real numbers other 2 than − 1 and 1. For such x, 2 (f + g )(x) = (f · g )(x) = 1 1 3x + = , x − 1 2x + 1 (x − 1)(2x + 1) 1 , (x − 1)(2x + 1) and f 2x + 1 (x) = . g x−1 Note that, in spite of the last equation, the domain of f /g does not include the number − 1 . 2 C01S03.003: The domain of f is the interval [0, + ∞) and the domain of g is the interval [2, + ∞). Hence the domain of f + g and f · g is the interval [2, + ∞), but because g (2) = 0, the domain of f /g is the open interval (2, + ∞). The formulas for these combinations are (f + g )(x) = √ x+ √ x − 2, √√ x x − 2 = x2 − 2x, √ f x x (x) = √ . = g x−2 x−2 (f · g )(x) = and C01S03.004: The domain of f is the interval [−1, + ∞) and the domain of g is the interval (−∞, 5]. Hence the domain of f + g and f · g is the closed interval [−1, 5], but because g (5) = 0, the domain of f /g is the half-open interval [−1, 5). Their formulas are √ √ x + 1 + 5 − x, √ f x+1 x+1 . (x) = √ = g 5−x 5−x (f + g )(x) = (f · g )(x) = √ x+1 √ 5−x= 5 + 4x − x2 , and C01S03.005: The domain of f is the set R of all real numbers; the domain of g is the open interval (−2, 2). Hence the domain of f + g and f · g is the open interval (−2, 2); because g (x) is never zero, the domain of f /g is the same. Their formulas are 1 (f + g )(x) = f (x) = g x2 1 +1+ √ , 4 − x2 x2 + 1 4 − x2 = √ x2 + 1 (f · g )(x) = √ , 4 − x2 and 4 + 3x2 − x4 . C01S03.006: The domain of f is the set of all real numbers other than 2 and the domain of g is the set of all real numbers other than −2. Hence the domain of f + g and f · g is the set of all real numbers other than ± 2. But because g (−1) = 0, −1 does not belong to the domain of f /g , which therefore consists of all real numbers other than −2, −1, and 2. The formulas of these combinations are (f + g )(x) = x−1 x+1 2x2 − 4 + =2 , x−2 x+2 x −4 (f · g )(x) = f x−1 x+2 x2 + x − 2 (x) = · =2 . g x−2 x+1 x −x−2 x−1 x+1 x2 − 1 · =2 , x−2 x+2 x −4 and C01S03.007: f (x) = x3 − 3x + 1 has 1, 2, or 3 zeros, approaches + ∞ as x does, and approaches −∞ as x does. Because f (0) = 0, the graph does not match Fig. 1.3.26, so it must match Fig. 1.3.30. C01S03.008: f (x) = 1 + 4x − x3 has one, two, or three zeros, approaches −∞ as x → + ∞ and approaches + ∞ as x → −∞. Hence its graph must be the one shown in Fig. 1.3.28. C01S03.009: f (x) = x4 − 5x3 + 13x + 1 has four or fewer zeros and approaches + ∞ as x approaches either + ∞ or −∞. Hence its graph must be the one shown in Fig. 1.3.31. C01S03.010: f (x) = 2x5 − 10x3 + 6x − 1 has between one and five zeros, approaches + ∞ as x does, and approaches −∞ as x does. So its graph might be the one shown in Fig. 1.3.26, the one in Fig. 1.3.29, or the one in Fig. 1.3.30. But f (0) = 0, so Fig. 1.3.26 is ruled out, and we have already found that the graph in Fig. 1.3.30 matches the function in Problem 7. Therefore the graph of f must be the one shown in Fig. 1.3.29. Alternatively, the observation that f (x) changes sign on the five intervals [−3, −2], [−1, 0], [0, 0.5], [0.5, 1], and [2, 3] shows that f (x) has five zeros; therefore the graph must be the one shown in Fig. 1.3.29. C01S03.011: f (x) = 16 + 2x2 − x4 approaches −∞ as x approaches either + ∞ or −∞, so its graph must be the one shown in Fig. 1.3.27. C01S03.012: f (x) = x5 + x approaches + ∞ as x does and approaches −∞ as x does. Moreover, f (x) > 0 if x > 0 and f (x) < 0 if x < 0, which rules out every graph except for the one shown in Fig. 1.3.26. C01S03.013: The graph of f has vertical asymptotes at x = −1 and at x = 2, so its graph must be the one shown in Fig. 1.3.34. C01S03.014: The graph of f (x) has vertical asymptotes at x = ± 3, so its graph must be the one shown in Fig. 1.3.32. C01S03.015: The graph of f has no vertical asymptotes and has maximum value 3 when x = 0. Hence its graph must be the one shown in Fig. 1.3.33. C01S03.016: The denominator x3 − 1 = (x − 1)(x2 + x + 1) of f (x) is zero only when x = 1 (because 2 0 for all x), so its graph must be the one shown in Fig. 1.3.35. x2 + x + 1 > x2 + x + 1 = x + 1 4 2 2 √ C01S03.017: The domain of f (x) = x x + 2 is the interval [−2, + ∞), so its graph must be the one shown in Fig. 1.3.38. √ C01S03.018: The domain of f (x) = 2x − x2 consists of those numbers for which 2x − x2 0; that is, x(2 − x) 0. This occurs when x and 2 − x have the same sign and also when either is zero. If x > 0 and 2 − x > 0, then 0 < x < 2. If x < 0 and 2 − x < 0, then x < 0 and x > 2, which is impossible. Hence the domain of f is the closed interval [0, 2]. So the graph of f must be the one shown in Fig. 1.3.36. √ C01S03.019: The domain of f (x) = x2 − 2x consists of those numbers x for which x2 − 2x 0; that is, x(x − 2) 0. This occurs when x and x − 2 have the same sign and also when either is zero. If x > 0 and x − 2 > 0, then x > 2; if x < 0 and x − 2 < 0, then x < 0. So the domain of f is the union of the two intervals (−∞, 0] and [2, +∞). So the graph of f must be the one shown in Fig. 1.3.39. C01S03.020: The domain of f (x) = 2(x2 − 2x)1/3 is the set R of all real numbers because every real number has a [unique] cube root. By the analysis in the solution of Problem 19, x2 − 2x < 0 if 0 < x < 2 and x2 − 2x 0 otherwise. Hence f (x) < 0 if 0 < x < 2 and f (x) 0 otherwise. This makes it certain that the graph of f is the one shown in Fig. 1.3.37. C01S03.021: Good viewing window: −2.5 C01S03.022: Good viewing window: −3 C01S03.023: Good viewing window: −3.5 2.5. Three zeros, approximately −1.88, 0.35, and 1.53. x 3. Two zeros: −2 and 1. x 2.5. One zero, approximately −2.10. x C01S03.024: Good viewing window: −1.6 2.20. x 2.8. Four zeros, approximately −1.28, 0.61, 1.46, and C01S03.025: Good viewing window: −1.6 approximately 2.30. x 2.8. Three zeros: approximately −1.30, exactly 1, and C01S03.026: Good viewing window: −1.6 x C01S03.027: 7.91. Good viewing window: −7.5 2.8. Two zeros, approximately −1.33 and 2.37. x 8.5. Three zeros: Approximately −5.70, −2.22, and C01S03.028: Good viewing window: None; it takes three: −22 x 8 shows that there is a zero near −20 and that the graph crosses the x-axis somewhere in the vicinity of x = 0. The window −3 x 3 shows that something interesting happens near x = −1 and that there is a zero near 1.8. The window −1.4 x 0.4 shows that there are zeros near −1.1 and −0.8. Closer approximations to these four zeros are −19.88, −1.09, −0.79, and 1.76. C01S03.029: The viewing window −11 x 8 shows that there are five zeros, although the two near 2.5 may be only one. The window 1.5 x 3.5 shows that there are in fact two zeros near 2.5. Approximate values of the five zeros are −10.20, −7.31, 1.98, 3.25, and 7.28. C01S03.030: The viewing window −16 x 16 shows that there are zeros near ± 15 and perhaps a few more near x = 0. The window −4 x 4 shows that there are in fact four zeros near x = 0. Approximate values of the six are ± 15.48, ± 3.04, and ± 1.06. C01S03.031: Every time c increases by 1, the graph is raised 1 unit (in the positive y -direction), but there is no other change. 3 C01S03.032: The graph starts with two “bends” when c = −5. As c increases the bends become narrower and narrower and disappear when c = 0. Then the graph gets steeper and steeper. See the following figure. 15 c=5 c = –5 10 5 -4 -2 2 4 -5 -10 C01S03.033: The graph always passes through (0, 0) and is tangent to the x-axis there. When c = −5 there is another zero at x = 5. As c increases this zero shifts to the left until it coincides with the one at x = 0 when c = 0. At this point the “bend” in the graph disappears. As c increases from 1 to 5, the bend reappears to the left of the x-axis and the second zero reappears at −c. C01S03.034: The graph is always tangent to the x-axis at x = 0 and is always symmetric around the y -axis. When c = −5 there is another pair of zeros near ± 2.2. As c increases these zeros move closer to x = 0 and the bends in the graph get smaller and smaller. They disappear when c = 0 and, at the same time, the zeros merge with the one at x = 0. Thereafter the graph simply becomes steeper and steeper. See the following figure. c=5 15 10 5 -3 -2 c = –5 -1 1 2 3 -5 C01S03.035: The graph is always symmetric around the origin (and, consequently, always passes through the origin). When c = −5 there is another pair of zeros near ± 2.2. As c increases the graph develops positive slope at x = 0, two more bends, and two more zeros on either side of the origin. They move outward and, when c = −2, they coincide with the outer pair of zeros, which have also been moving toward the origin. They reach the origin when c = 0 and thereafter the graph simply becomes steeper and steeper. C01S03.036: As c increases the “mountain” around the y -axis gets narrower and steeper. See the following 4 figure. 1 0.8 0.6 c = –5 0.4 0.2 c=5 -3 -2 -1 1 2 3 C01S03.037: As c increases the graph becomes wider and taller; its shape does not seem to change very much. C01S03.038: The length of the airfoil is approximately 1.0089 and its width is approximately 0.200057. 5 Section 1.4 C01S04.001: Because g (x) = 2x increases—first slowly, then rapidly—on the set of all real numbers, with values in the range (0, + ∞), the given function f (x) = 2x − 1 must increase in the same way, but with values in the range (−1, + ∞). Therefore its graph is the one shown in Fig. 1.4.29. C01S04.002: Given: f (x) = 2 − 3−x . The graph of g (x) = 3x increases, first slowly, then rapidly, on its domain the set R of all real numbers. Hence h(x) = 3−x decreases, first rapidly, then slowly, on R, with values in the interval (0, + ∞). Hence j (x) = −3−x increases, first rapidly, then slowly, on R, with values in the interval (−∞, 0). Therefore f (x) = 2 − 3−x increases, first rapidly, then slowly, on R, with values in the interval (−∞, 2). Therefore its graph must be the one shown in Fig. 1.4.33. C01S04.003: The graph of f (x) = 1 + cos x is simply the graph of the ordinary cosine function raised 1 unit—moved upward 1 unit in the positive y -direction. Hence its graph is the one shown in Fig. 1.4.27. C01S04.004: The graph of g (x) = 2 sin x resembles the graph of the ordinary sine function, but with values ranging from −2 to 2. The graph of h(x) = −2 sin x is the same, but turned “upside down.” Add 2 to get f (x) = 2 − 2 sin x and the graph of h is raised 2 units, thus taking values in the range [0, 4]. So the graph of f is the one shown in Fig. 1.4.32. C01S04.005: The graph of g (x) = 2 cos x resembles the graph of the cosine function, but with all values doubled, so that its range is the interval [−2, 2]. Add 1 to get f (x) = 1 + 2 cos x and the range is now the interval [−1, 3]. So the graph of f is the one shown in Fig. 1.4.35. C01S04.006: Turn the graph of the sine function upside down, then add 2 to get f (x) = 2 − sin x, with range the interval [1, 3]. Hence the graph of f is the one shown in Fig. 1.4.28. C01S04.007: The graph of g (x) = 2x increases, first slowly, then rapidly, on the set of all real numbers, with range the interval (0, + ∞). So its reciprocal h(x) = 2−x decreases, first rapidly, then slowly, with the same domain and range. Multiply by x to obtain f (x) = x · 2−x . The effect of multiplication by x is to change large positive values into large negative values for x < 0, to cause f (0) to be zero, and to multiply very small positive values (of 2−x ) by somewhat large positive values (of x) for x > 0, resulting in values that are still small and positive, even when x is quite large. So the graph of f must increase rapidly through negative values, pass through (0, 0), rise to a maximum, then decrease rapidly through positive values toward zero. Hence the graph of f must be the one shown in Fig. 1.4.31. C01S04.008: The graph of g (x) = log x has domain the set (0, + ∞) of all positive real numbers; it rises, first rapidly, then more slowly, with range the set of all real numbers, and its graph passes through the point (1, 0). Division by x > 0 will have little effect if x is near zero, as this will merely multiply large negative values of log x by large positive numbers. But when x is large positive, it will be much larger than log x, and thus the graph of f (x) will rise to a maximum somewhere to the right of x = 1, then decreases fairly rapidly toward zero. So the graph of f is the one shown in Fig. 1.4.36. C01S04.009: The graph of g (x) = 1 + cos 6x will resemble the graph of the cosine function, but raised 1 unit (so that its range is the interval [0, 2]) and with much more “activity” on the x-axis (because of the factor 6). Division by 1 + x2 will have little effect until x is no longer close to zero, and then the effect will be to divide values of g (x) by larger and larger positive numbers, so that the cosine oscillations have a much smaller range that 0 x 2; they will range from 0 to smaller and smaller positive values as | x | increases. So the graph of f is the one shown in Fig. 1.4.34. 1 C01S04.010: The graph of g (x) = sin 10x resembles that of the sine function, but with much more “activity” because of the factor 10. Multiply by the rapidly decreasing positive numbers 2−x and you will see the sine oscillations decreasing from the range [−1, 1] when x is near zero to very small oscillations—near zero—as x increases. So the graph of f is the one shown in Fig. 1.4.30. C01S04.011: Given f (x) = 1 − x2 and g (x) = 2x + 3, f (g (x)) = 1 − (g (x))2 = 1 − (2x + 3)2 = −4x2 − 12x − 8 and g (f (x)) = 2f (x) + 3 = 2(1 − x2 ) + 3 = −2x2 + 5. C01S04.012: Given f (x) = −17 and g (x) = | x |, f (g (x)) = −17 and g (f (x)) = | f (x) | = | − 17 | = 17. The first result is a little puzzling until one realizes that to obtain f (g (x)), one substitutes g (x) for x for every occurrence of x in the formula for f . No x there means there’s no place to put g (x). Indeed, f (h(x)) = −17 no matter what the formula of h. C01S04.013: If f (x) = √ x2 − 3 and g (x) = x2 + 3, then f (g (x)) = (g (x))2 − 3 = g (f (x)) = (f (x))2 + 3 = (x2 + 3)2 − 3 = x2 − 3 2 x4 + 6x2 + 6 and + 3 = x2 − 3 + 3 = x2 . The domain of f (g ) is the set R of all real numbers, but the domain of g (f ) is the same as the domain of f , the set of all real numbers x such that x2 3. C01S04.014: If f (x) = x2 + 1 and g (x) = 1 , then x2 + 1 f (g (x)) = (g (x))2 + 1 = g (f (x)) = 1 x4 + 2x2 + 2 +1= 4 (x2 + 1)2 x + 2x2 + 1 and 1 1 1 =2 =4 . 2+1 2+1 (f (x)) (x + 1) x + 2x2 + 2 C01S04.015: If f (x) = x3 − 4 and g (x) = (x + 4)1/3 , then f (g (x)) = (g (x))3 − 4 = (x + 4)1/3 1/3 g (f (x)) = (f (x) + 4) 3 = x3 − 4 + 4 −4=x+4−4=x 1/3 = x3 1 /3 and = x. The domain of both f (g ) and g (f ) is the set R of all real numbers, so here is an example of the highly unusual case in which f (g ) and g (f ) are the same function. C01S04.016: If f (x) = √ x and g (x) = cos x, then 2 √ f (g (x)) = f (cos x) = cos x and √ √ g (f (x)) = g x = cos x . C01S04.017: If f (x) = sin x and g (x) = x3 , then f (g (x)) = f x3 = sin x3 = sin x3 and 3 g (f (x)) = g (sin x) = (sin x) = sin3 x. We note in passing that sin x3 and sin3 x don’t mean the same thing! C01S04.018: If f (x) = sin x and g (x) = cos x, then f (g (x)) = f (cos x) = sin(cos x) and g (f (x)) = g (sin x) = cos(sin x). C01S04.019: If f (x) = 1 + x2 and g (x) = tan x, then f (g (x)) = f (tan x) = 1 + (tan x)2 = 1 + tan2 x and g (f (x)) = g (1 + x2 ) = tan(1 + x2 ). C01S04.020: If f (x) = 1 − x2 and g (x) = sin x, then f (g (x)) = f (sin x) = 1 − (sin x)2 = 1 − sin2 x = cos2 x and g (f (x)) = g (1 − x2 ) = sin(1 − x2 ). Note: The answers to Problems 21 through 30 are not unique. We have generally chosen the simplest and most natural answer. C01S04.021: h(x) = (2 + 3x)2 = (g (x))k = f (g (x)) where f (x) = xk , k = 2, and g (x) = 2 + 3x. C01S04.022: h(x) = (4 − x)3 = (g (x))3 = f (g (x)) where f (x) = xk , k = 3, and g (x) = 4 − x. C01S04.023: h(x) = (2x − x2 )1/2 = (g (x))1/2 = f (g (x)) where f (x) = xk , k = 1 , and g (x) = 2x − x2 . 2 C01S04.024: h(x) = (1 + x4 )17 = (g (x))17 = f (g (x)) where f (x) = xk , k = 17, and g (x) = 1 + x4 . C01S04.025: h(x) = (5 − x2 )3/2 = (g (x))3/2 = f (g (x)) where f (x) = xk , k = 3 , and g (x) = 5 − x2 . 2 4 C01S04.026: h(x) = (4x − 6)1/3 = (4x − 6)4/3 = (g (x))4/3 = f (g (x)) where f (x) = xk , k = 4 , and 3 g (x) = 4x − 6. Alternatively, h(x) = (g (x))4 = f (g (x)) where f (x) = xk , k = 4, and g (x) = (4x − 6)1/3 . C01S04.027: h(x) = (x + 1)−1 = (g (x))−1 = f (g (x)) where f (x) = xk , k = −1, and g (x) = x + 1. C01S04.028: h(x) = (1 + x2 )−1 = (g (x))−1 = f (g (x)) where f (x) = xk , k = −1, and g (x) = 1 + x2 . C01S04.029: h(x) = (x + 10)−1/2 = (g (x))−1/2 = f (g (x)) where f (x) = xk , k = − 1 , and g (x) = x + 10. 2 C01S04.030: h(x) = (1+ x + x2 )−3 = (g (x))−3 = f (g (x)) where f (x) = −xk , k = −3, and g (x) = 1+ x + x2 . C01S04.031: Recommended window: −2 x exactly one solution (approximately 0.641186). 2. The graph makes it evident that the equation has 3 C01S04.032: Recommended window: −5 x 5. The graph makes it evident that the equation has exactly three solutions (approximately −3.63796, −1.86236, and 0.88947). C01S04.033: Recommended window: −5 x exactly one solution (approximately 1.42773). 5. The graph makes it evident that the equation has C01S04.034: Recommended window: −6 x 6. The graph makes it evident that the equation has exactly three solutions (approximately −3.83747, −1.97738, and 1.30644). C01S04.035: Recommended window: −8 x 8. The graph makes it evident that the equation has exactly five solutions (approximately −4.08863, −1.83622, 1.37333, 5.65222, and 6.61597). C01S04.036: Recommended window: 0.1 x exactly one solution (approximately 1.32432). 20. The graph makes it evident that the equation has C01S04.037: Recommended window: 0.1 x 20. The graph makes it evident that the equation has exactly three solutions (approximately 1.41841, 5.55211, and 6.86308). C01S04.038: Recommended window: −4 x exactly two solutions (approximately ± 1.37936). 4. The graph makes it evident that the equation has C01S04.039: Recommended window: −11 x 11. The graph makes it evident that the equation has exactly six solutions (approximately −5.92454, −3.24723, 3.04852, 6.75738, 8.59387, and [exactly] 0). C01S04.040: Recommended window: 0.1 x 20. The graph makes it evident that the equation has exactly six solutions (approximately 0.372968, 1.68831, 4.29331, 8.05637, 11.1288, and 13.6582). C01S04.041: Graphical methods show that the solution of 10 · 2t = 100 is slightly less than 3.322. We began with the viewing window 0 t 6 and gradually narrowed it to 3.321 t 3.323. C01S04.042: Under the assumption that the interest is compounded continuously at a rate of 7.696% (for an annual yield of 8%), we solved the equation 5000 · (1.07696)t = 15000 for t ≈ 14.8176. We began with the viewing window 10 t 20 and gradually narrowed it to 14.81762 y 14.81763. Under the assumption that the interest is compounded yearly at an annual rate of 8%, we solved the equation A(t) = 5000 · (1.08)t = 15000 by evaluating A(14) ≈ 14686 and A(15) ≈ 15861. Thus in this case you’d have to wait a full 15 years for your money to triple. C01S04.043: Graphical methods show that the solution of (67.4)·(1.026)t = 134.8 is approximately 27.0046. We began with the viewing window 20 t 30 and gradually narrowed it to 27.0045 t 27.0047. C01S04.044: Graphical methods show that the solution of A(t) = (0.9975)t = 0.5 is approximately 276.912. We began with the viewing window 200 t 300 and gradually narrowed it to 276.910 t 276.914. C01S04.045: Graphical methods show that the solution of A(t) = 12 · (0.975)t = 1 is approximately 98.149. We began with the viewing window 50 t 250 and gradually narrowed it to 98.148 t 98.150. C01S04.046: Graphical methods show that the negative solution of x2 = 2x is approximately −0.76666. We began with the viewing window −1 x 0 and gradually narrowed it to −0.7667 t −0.7666. C01S04.047: We plotted y = log10 x and y = 1 x1/5 simultaneously. We began with the viewing window 2 1 x 10 and gradually narrowed it to 4.84890 x 4.84892. Answer: x ≈ 4.84891. 4 C01S04.048: We began with the viewing window −2 x 2, which showed the two smaller solutions but not the larger solution. We first narrowed this window to −0.9054 x −0.9052 to get the first solution, x ≈ −0.9053. We returned to the original window and narrowed it to 1.1324 x 1.1326 to get the second solution, x ≈ 1.1325. We looked for a solution in the window 20 x 30 but there was none. But the exponential graph was still below the polynomial graph, so we checked the window 30 x 32. A solution was evident, and we gradually narrowed this window to 31.3636 x 31.3638 to discover the third solution, x ≈ 31.3637. 5 Chapter 1 Miscellaneous Problems C01S0M.001: The domain of f (x) = interval [4, + ∞). √ x − 4 is the set of real numbers x for which x − 4 0; that is, the C01S0M.002: The domain of f consists of those real numbers x for which 2 − x = 0; that is, the set of all real numbers other than 2. C01S0M.003: The domain of f consists of those real numbers for which the denominator is nonzero; that is, the set of all real numbers other than ± 3. C01S0M.004: Because x2 + 1 is never zero, the domain of f is the set R of all real numbers. √ √ C01S0M.005: If x 0, then x exists; there is no obstruction to adding 1 to x nor to cubing the sum. Hence the domain of f is the set [0, + ∞) of all nonnegative real numbers. C01S0M.006: Given: f (x) = x+1 . x2 − 2x The only obstruction to computing the number f (x) is the possibility that the denominator is zero. Thus we must eliminate from the set of all real numbers those for which x2 − 2x = 0; that is, x(x − 2) = 0. Therefore the domain of f is the set of all real numbers other than 0 and 2. C01S0M.07: whenever The function f (x) = √ 2 − 3x is defined whenever the radicand is nonnegative; that is, 2 − 3x 3x 2 3 2; x Hence the domain of f is the interval − ∞, 0; 2 3. . C01S0M.008: In order that the square root is defined, we require 9 − x2 0; we also need the denominator in f (x) to be nonzero, so we further require that 9 − x2 = 0. Hence 9 − x2 > 0; that is, x2 < 9, so that −3 < x < 3. Hence the domain of f is the open interval (−3, 3). C01S0M.009: Regardless of the value of x, it’s always possible to subtract 2 from x, to subtract x from 4, and to multiply the results. Hence the domain of f is the set R of all real numbers. C01S0M.010: The domain of f consists of those real numbers x for which (x − 2)(4 − x) is nonnegative. That is, x − 2 and 4 − x are both positive, or x − 2 and 4 − x are both negative, or either is zero. First case: x − 2 > 0 and 4 − x > 0. Then 2 < x < 4, so the interval (2, 4) is part of the domain of f . Second case: x − 2 < 0 and 4 − x < 0. These inequalities imply that x < 2 and 4 < x. No real numbers satisfy both these inequalities. So the second case contributes no numbers to the domain of f . Third case: x − 2 = 0 or 4 − x = 0. That is, x = 2 or x = 4. Therefore the domain of f is the closed interval [2, 4]. C01S0M.011: Because 100 V 200 and p > 0, it follows that 100p we see that 100p 800 200p, so that p 8 2p. That is, p 8 and 4 range of possible values of p. 1 pV 200p. Because pV = 800, p, so that 4 p 8. This is the C01S0M.012: If 70 F 90, then 70 32 + 9 C 5 70 − 32 90. Hence 90 − 32; 9 5C 38 9 5C 58; 190 9C 290; 190 9 C 290 9. Answer: The Celsius temperature ranged from a low of about 21.1◦ C to a high of about 32.2◦ C. C01S0M.013: Because 25 < R < 50, 25I < IR < 50I , so that 25I < E < 50I ; 25I < 100 < 50I ; I < 4 < 2I ; I<4 and 2 < I. Therefore the current I lies in the range 2 < I < 4. C01S0M.014: Because 3 < L < 4, we see that 3 L 4 < < ; 32 32 32 3 < 32 3 < 2π 32 2π L < 2π 32 3 <T <π 2 π 2 1 ; 8 L < 32 1 ; 8 1 . 2 In approximate terms, 1.923825 < T < 2.221441. C01S0M.015: If a cube has edge length x, then its volume is V = x3 and its total surface area is S = 6x2 (because each of its six faces has area x2 ). Hence x = S/6 , and therefore V (S ) = S 6 3 = S 6 3/2 , 0 < S < + ∞. Under certain circumstances it would be both permissible and desirable to let the domain of V be the interval [0, + ∞). C01S0M.016: Let r denote the radius, and h the height, of the cylinder. Then its volume V and total surface area A are given by V = πr2 h and A = 2π rh + 2π r2 (look inside the front cover of the textbook). In this problem we are given h = r, so that V = π r3 and A = 4π r2 . Therefore 2 Answer: V π A(V ) = 4π 1 /3 V π r= and so A = 4π V π 2/3 . 2 /3 , 0 < V < + ∞. It is permissible, and sometimes desirable, to use instead the domain 0 V < + ∞. C01S0M.017: The following figure shows an equilateral triangle with sides of length 2x and an altitude of length h. Because T is a right triangle, we see that x2 + h2 = (2x)2 , √ h = x 3. so that The area of this triangle is A = hx and its perimeter is P = 6x. So √ A = x2 3 √ P2 3 Therefore A(P ) = , 36 and x= P . 6 0 < P < ∞. 2x h T x C01S0M.018: The square has perimeter x and thus edge length y = 1 x. The circle has circumference 4 100 − x. Thus if z is the radius of the circle, then 2π z = 100 − x, so that z = (100 − x)/(2π ). The area of the square is y 2 and the area of the circle is π z 2 , so that the sum of the areas of the square and the circle is given by A(x) = x2 +π 16 100 − x 2π 2 , 0 < x < 100. Looking ahead to Chapter 3, it will be advantageous to use the closed interval [0, 100] for the domain of the function A. C01S0M.019: The slope of L is 13 − 5 = 2, so an equation of L is 1 − (−3) y − 5 = 2(x + 3); that is, y = 2x + 11. C01S0M.020: An equation of L is y − (−1) = −3(x − 4); that is, 3x + y = 11. C01S0M.021: The point (0, −5) lies on L, so an equation of L is 3 y − (−5) = 1 (x − 0); 2 alternatively, 2y + 10 = x. C01S0M.022: The equation 3x − 2y = 4 of the other line may be written in the form y = 3 x − 2, revealing 2 that it and L have slope 3 . Hence an equation of L is 2 y − (−3) = 3 (x − 2); 2 that is, y= 3 x − 6. 2 C01S0M.023: The equation y − 2x = 10 may be written in the form y = 2x + 10, showing that it has slope 2. Hence the perpendicular line L has slope − 1 . Therefore an equation of L is 2 1 y − 7 = − (x − (−3)); 2 that is, x + 2y = 11. C01S0M.024: The segment S joining (1, −5) and (3, −1) has slope (−1 − (−5))/(3 − 1) = 2 and midpoint (2, −3), and hence L has slope − 1 and passes through (2, −3). So an equation of L is 2 1 y − (−3) = − (x − 2); 2 that is, x + 2y = −4. C01S0M.025: The graph of y = f (x) = 2 − 2x − x2 is a parabola opening downward. The only such graph is shown in Fig. 1.MP.6. C01S0M.026: Given: f (x) = x3 − 4x2 + 5. Because f (−1) = 0, f (1) = 2 > 0 > −3 = f (2), and f (3) = −4 < 0 < 5 = f (4), the graph of f crosses the x-axis at x = −1, between x = 1 and x = 2, and between x = 3 and x = 4. Hence the graph of f is the one shown in Fig. 1.MP.9. C01S0M.027: Given: f (x) = x4 − 4x3 + 5. Because the graph of f has no vertical asymptotes and because f (x) approaches + ∞ as x approaches either + ∞ or −∞, the graph of f must be the one shown in Fig. 1.MP.4. C01S0M.028: Given: f (x) = 5 5 = . x2 − x − 6 (x − 3)(x + 2) The denominator in f (x) is zero when x = 3 and when x = −2 (and the numerator is not zero), so the graph of y = f (x) has vertical asymptotes at x = −2 and at x = 3. Also f (x) approaches zero as x approaches either + ∞ or −∞. Therefore the graph of y = f (x) must be the one shown in Fig. 1.MP.11. C01S0M.029: Given: f (x) = x2 5 20 20 =2 = . −x+6 4x − 4x + 1 + 23 (2x − 1)2 + 23 The algebra displayed here shows that the denominator in f (x) is never zero, so there are no vertical asymptotes. It also shows that the maximum value of f (x) occurs when the denominator is minimal; that is, when x = 1 . Finally, f (x) approaches zero as x approaches either + ∞ or −∞. So the graph of y = f (x) 2 must be the one shown in Fig. 1.MP.3. C01S0M.030: If y = f (x) = √ 8 + 2x − x2 , then 4 y 2 = 8 + 2x − x2 ; x2 − 2x + 1 + y 2 = 9; (x − 1)2 + (y − 0)2 = 32 . The last is the equation of a circle with center (1, 0) and radius 3. But y half of that circle, and it is shown in Fig. 1.MP.10. 0, so the graph of f is the upper C01S0M.031: Given: f (x) = 2−x − 1. The graph of y = 2x is an increasing exponential function, so the graph of y = 2−x is a decreasing exponential function, approaching 0 as x approaches + ∞. So the graph of f approaches −1 as x approaches + ∞. Moreover, f (0) = 0. Therefore the graph of f is the one shown in Fig. 1.MP.7. C01S0M.032: The graph of f (x) = log10 (x + 1) is obtained from the graph of g (x) = log10 x by translation one unit to the left; note also that f (0) = 0. Therefore the graph of f is the one shown in Fig. 1.MP.2. C01S0M.033: The graph of y = 3 sin x oscillates between its minimum value −3 and its maximum value 3, so the graph of f (x) = 1 + 3 sin x oscillates between −2 and 4. This graph is shown in Fig. 1.MP.8. C01S0M.034: The graph of f (x) = x + 3 sin x viewed at a great distance resembles the graph of y = x. A closer view shows oscillations, due to the sine function, superposed on the graph of y = x. Thus the graph of f is the one shown in Fig. 1.MP.5. C01S0M.035: The graph of 2x − 5y = 7 is the straight line with x-intercept 7 2 and y -intercept − 7 . 5 C01S0M.036: If | x − y | = 1, then x − y = 1 or x − y = −1. The graph of the first of these is the straight line y = x − 1 with slope 1 and y -intercept −1; the graph of the second is the straight line y = x + 1 with slope 1 and y -intercept 1. So the graph of | x − y | = 1 consists of these two parallel lines. C01S0M.037: We complete the square: x2 − 2x + 1 + y 2 = 1, so that (x − 1)2 + (y − 0)2 = 12 . Thus the graph of the given equation is the circle with center (1, 0) and radius 1. C01S0M.038: We complete the square in x and in y to obtain x2 + 6x + 9 + y 2 − 4y + 4 = 16; (x + 3)2 + (y − 2)2 = 42 . Therefore the graph of the given equation is the circle with center (−3, 2) and radius 4. C01S0M.039: The graph is a parabola opening upward. To find its vertex, we complete the square: y = 2 x2 − 2x − 1 2 = 2 x2 − 2x + 1 − 3 2 = 2(x − 1)2 − 3. So the vertex of this parabola is at the point (1, −3). C01S0M.040: The graph is a parabola opening downward. To find its vertex, we complete the square: y = 4x − x2 = −(x2 − 4x) = −(x2 − 4x + 4 − 4) = 4 − (x − 2)2 . 5 Thus the vertex of this parabola is at the point (2, 4). C01S0M.041: The graph has a vertical asymptote at x = −5 and is shown next. 20 10 -8 -6 -4 -2 -10 -20 C01S0M.042: The graph has vertical asymptotes at x = ± 2 and is shown next. 6 4 2 -4 -2 2 4 -2 -4 -6 C01S0M.043: The graph of f is obtained by shifing the graph of g (x) = | x | three units to the right, so that the graph of f has its “vertex” at the point (3, 0). C01S0M.044: Given: f (x) = | x − 3 | + | x + 2 |. If x 3 then f (x) = x − 3 + x + 2 = 2x − 1, so the graph is the unbounded line segment with slope 2 and endpoint (3, 5) for x 3. If −2 x 3 then f (x) = 3 − x + x + 2 = 5, so another part of the graph is the horizontal line segment joining (−2, 5) with (3, 5). If x −2 then f (x) = 3 − x − x − 2 = −2x + 1, so the rest of the graph is the unbounded line segment with slope −2 and endpoint (−2, 5) for x −2. The graph is shown next. 8 6 4 2 -4 -2 2 6 4 C01S0M.045: Suppose that a, b, and c are arbitrary real numbers. Then |a + b + c| = |(a + b) + c| |a + b| + |c| |a| + |b| + |c| . . C01S0M.046: Suppose that a and b are arbitrary real numbers. Then |a| = |(a − b) + b| Therefore |a| −| b| |a − b|. |a − b| + |b|. C01S0M.047: If x − 3 > 0 and x + 2 > 0, then x > 3 and x > −2, so x > 3. If x − 3 < 0 and x + 2 < 0, then x < 3 and x < −2, so x < −2. Answer: (−∞, −2) ∪ (3, ∞). C01S0M.048: (x − 1)(x − 2) < 0: x − 1 and x − 2 have opposite signs, so either x < 1 and x > 2 (which leads to no values of x) or x > 1 and x < 2. Answer: (1, 2). C01S0M.049: (x − 4)(x + 2) > 0: Either x > 4 and x > −2 (so that x > 4) or x < 4 and x < −2 (so that x < −2). Answer: (−∞, −2) ∪ (4, + ∞). C01S0M.050: 2x 15 − x2 : x2 + 2x − 15 0, so (x − 3)(x + 5) x + 5 0. Thus x 3 or x −5. Answer: (−∞, −5 ] ∪ [ 3, + ∞). 0. Now x + 5 > x − 3, so x − 3 0 or C01S0M.051: The viewing window −3 x 8 shows a solution near −1 and another near 5. Gradual magnification of the region near −1 shows a solution between −1.1405 and −1.1395. Similarly, the other solution is between 6.1395 and 6.1405. So the solutions are approximately −1.140 and 6.140. C01S0M.052: The viewing window −2 x 5 shows a solution near −1 and another near 4. To approximate the first more closely, we used the method of repeated tabulation on [−1.0, −0.8], then on [−0.88, −0, 86], then on [−0.872, −0.870]. To approximate the second, we used the interval [4.1, 4.3], then [4.20, 4.22], then [4.204, 4.206]. To three places, the solutions are −0.872 and 4.205. C01S0M.053: The viewing window 0.5 x 3 shows one solution near 1.2 and another near 2.3. The method of repeated tabulation with successive intervals [1.1, 1.3], [1.18, 1.20], and [1.190, 1.192] yields the approximation 1.191 to the first solution. The successive intervals [2.2, 2.4], [2.30, 2.32], and [2.308, 2.310] yield the approximation 2.309 to the second solution. C01S0M.054: The viewing window −7 x 2 shows one solution near −6 and another near 1. The method of repeated tabulation with successive intervals [−6.1, −5.9], [−5.98, −5.96], and [−5.974, −5.970] yield the approximation −5.972 to the first solution. Simlarly, we find the second solution to be approximately 1.172. C01S0M.055: The viewing window −6 x 2 shows one solution near −5 and another near 1. The method of repeated tabulation with successive intervals [−5.1, −4.9], [−5.04, −5.02], and [−5.022, −5.020], then with the intervals [0.8, 1.0], [0.88, 0.90], and [0.896, 0.898], yields the two approximations −5.021 and 0.896 to the two solutions. C01S0M.056: The viewing window −11 x 3 shows one solution near −10 and another near 1.7. The method of repeated tabulation, first with the intervals [−10.0, −9.9], [−9.97, −9.96], and [−9.963, −9.962], then with [1.7, 1.8], [1.73, 1.74], and [1.739, 1.741], yields the two approximations −9.962 and 1.740 to the two solutions. 7 C01S0M.057: The viewing window 2 x 3 shows the low point with x-coordinate near 2.5. The method of repeated tabulation, using the successive intervals [2.4, 2.6], [2.48, 2.52], and [2.496, 2.504], indicates that the low point is very close to (2.5, 0.75). C01S0M.058: The viewing window −1 x 4 shows the low point with x-coordinate near 1.7. The method of repeated tabulation, with the successive intervals [1.6, 1.8], [1.64, 1.68], and [1.664, 1.672], shows that the low point is quite close to (1.66, 2.67). C01S0M.059: The viewing window −0.5 x 4 shows that the low point has x-coordinate near 1.8. The method of repeated tabulation, with the successive intervals [1.7, 1.9], [1.72, 1.78], and [1.744, 1.756], shows that the low point is very close to (1.75, −1.25). C01S0M.060: The viewing window −5 x 1 shows that the low point has x-coordinate near −2.5. The method of repeated tabulation indicates that the low point is very close to (−2.4, 6.2). C01S0M.061: The viewing window −5 x 1 show that the x-coordinate of the low point is close to −2. The method of repeated tabulation shows that the low point is very close to (−2.0625, 0.96875). C01S0M.062: The viewing window −7 x 1 shows that the x-coordinate of the low point is close to −4. The method of repeated tabulation indicates that the low point is very close to (−4.111, 3.889). C01S0M.063: The small rectangle has dimensions 10 − 4x by 7 − 2x; (7)(10) − (10 − 4x)(7 − 2x) = 20, which leads to the quadratic equation 8x2 − 48x + 20 = 0. One solution of this equation is approximately 5.5495, which must be rejected; it is too large. The value of x is the other solution: x ≈ 0.4505. C01S0M.064: After shrinking, the tablecloth has dimensions 60 − x by 35 − x. The area of this rectangle is 93% of the area of the original tablecloth, so (60 − x)(35 − x) = (0.93)(35)(60). The larger solution of this quadratic equation is approximately 93.43, which we reject as too large. Answer: x ≈ 1.573. C01S0M.065: The viewing window −4 x C01S0M.066: The viewing window −3 x4 > | − 3x2 + 4x − 5 | if | x | > 3. x 4 shows three solutions (and there can be no more). 3 shows two solutions, and there can be no more because C01S0M.067: We plotted y = sin x and y = x3 − 3x + 1 simultaneously to see where they crossed. The viewing window −2.2 x 2.2 shows three solutions, and there can be no more because | x3 − 3x + 1 | > 1 if | x | > 2.2. C01S0M.068: We plotted y = cos x and y = x4 − x simultaneously to see where they crossed. The viewing window −2 x 2 shows two solutions, and there can be no more because x4 − x > 1 if | x | > 2. C01S0M.069: We plotted y = cos x and y = log10 x simultaneously to see where they crossed. The viewing window 0.1 x 14 shows three solutions, and there can be no more because log10 x < −1 if 0 < x < 0.1 and log10 x > 1 if x > 14. C01S0M.070: We plotted y = 10−x and y = log10 x simultaneously to see where they crossed. The viewing window 0.1 x 3 shows one solution, and there can be no more because the exponential function is decreasing for all x and the logarithm function is increasing for all x > 0. 8 Section 2.1 C02S01.001: f (x) = 0 · x2 + 0 · x + 5, so m(a) = 0 · 2 · a + 0 ≡ 0. In particular, m(2) = 0, so the tangent line has equation y − 5 = 0 · (x − 0); that is, y ≡ 5. C02S01.002: f (x) = 0 · x2 + 1 · x + 0, so m(a) = 0 · 2 · a + 1 ≡ 1. In particular, m(2) = 1, so the tangent line has equation y − 2 = 1 · (x − 2); that is, y = x. C02S01.003: Because f (x) = 1 · x2 + 0 · x + 0, the slope-predictor is m(a) = 2 · 1 · a + 0 = 2a. Hence the line L tangent to the graph of f at (2, f (2)) has slope m(2) = 4. So an equation of L is y − f (2) = 4(x − 2); that is, y = 4x − 4. C02S01.004: Because f (x) = −2x2 +0 · x +1, the slope-predictor for f is m(a) = 2 · (−2)a +0 = −4a. Thus the line L tangent to the graph of f at (2, f (2)) has slope m(2) = −8 and equation y − f (2) = −8(x − 2); that is, y = −8x + 9. C02S01.005: Because f (x) = 0 · x2 + 4x − 5, the slope-predictor for f is m(a) = 2 · 0 · a + 4 = 4. So the line tangent to the graph of f at (2, f (2)) has slope 4 and therefore equation y − 3 = 4(x − 2); that is, y = 4x − 5. C02S01.006: Because f (x) = 0 · x2 − 3x + 7, the slope-predictor for f is m(a) = 2 · 0 · a − 3 = −3. So the line tangent to the graph of f at (2, f (2)) has slope −3 and therefore equation y − 1 = −3(x − 2); that is, y = −3x + 7. C02S01.007: Because f (x) = 2x2 − 3x + 4, the slope-predictor for f is m(a) = 2 · 2 · a − 3 = 4a − 3. So the line tangent to the graph of f at (2, f (2)) has slope 5 and therefore equation y − 6 = 5(x − 2); that is, y = 5x − 4. C02S01.008: Because f (x) = (−1) · x2 − 3x +5, the slope-predictor for f is m(a) = 2 · (−1) · a − 3 = −2a − 3. So the line tangent to the graph of f at (2, f (2)) has slope −7 and therefore equation y + 5 = −7(x − 2); that is, y = −7x + 9. C02S01.009: Because f (x) = 2x2 + 6x, the slope-predictor for f is m(a) = 4a + 6. So the line tangent to the graph of f at (2, f (2)) has slope m(2) = 14 and therefore equation y − 20 = 14(x − 2); that is, y = 14x − 8. C02S01.010: Because f (x) = −3x2 + 15x, the slope-predictor for f is m(a) = −6a + 15. So the line tangent to the graph of f at (2, f (2)) has slope m(2) = 3 and therefore equation y − 18 = 3(x − 2); that is, y = 3x + 12. 1 2 C02S01.011: Because f (x) = − 100 x2 + 2x, the slope-predictor for f is m(a) = − 100 a + 2. So the line 1 tangent to the graph of f at (2, f (2)) has slope m(2) = − 25 +2 = 49 and therefore equation y − 99 = 49 (x−2); 25 25 25 that is, 25y = 49x + 1. C02S01.012: Because f (x) = −9x2 − 12x, the slope-predictor for f is m(a) = −18a − 12. So the line tangent to the graph of f at (2, f (2)) has slope m(2) = −48 and therefore equation y + 60 = −48(x − 2); that is, y = −48x + 36. C02S01.013: Because f (x) = 4x2 + 1, the slope-predictor for f is m(a) = 8a. So the line tangent to the graph of f at (2, f (2)) has slope m(2) = 16 and therefore equation y − 17 = 16(x − 2); that is, y = 16x − 15. C02S01.014: Because f (x) = 24x, the slope-predictor for f is m(a) = 24. So the line tangent to the graph of f at (2, f (2)) has slope m(2) = 24 and therefore equation y − 48 = 24(x − 2); that is, y = 24x. 1 C02S01.015: If f (x) = −x2 + 10, then the slope-predictor for f is m(a) = −2a. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 0. So the tangent line is horizontal at the point (0, 10) and at no other point of the graph of f . C02S01.016: If f (x) = −x2 + 10x, then the slope-predictor for f is m(a) = −2a + 10. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 5. So the tangent line is horizontal at the point (5, 25) and at no other point of the graph of f . C02S01.017: If f (x) = x2 − 2x + 1, then the slope-predictor for f is m(a) = 2a − 2. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 1. So the tangent line is horizontal at the point (1, 0) and at no other point of the graph of f . C02S01.018: If f (x) = x2 + x − 2, then the slope-predictor for f is m(a) = 2a + 1. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = − 1 . So the tangent line is horizontal at the 2 point − 1 , − 9 and at no other point of the graph of f . 2 4 1 1 C02S01.019: If f (x) = − 100 x2 + x, then the slope-predictor for f is m(a) = − 50 a + 1. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 50. So the tangent line is horizontal at the point (50, 25) and at no other point of the graph of f . C02S01.020: If f (x) = −x2 + 100x, then the slope-predictor for f is m(a) = −2a + 100. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 50. So the tangent line is horizontal at the point (50, 2500) and at no other point of the graph of f . C02S01.021: If f (x) = x2 − 2x − 15, then the slope-predictor for f is m(a) = 2a − 2. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 1. So the tangent line is horizontal at the point (1, −16) and at no other point of the graph of f . C02S01.022: If f (x) = x2 − 10x + 25, then the slope-predictor for f is m(a) = 2a − 10. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 5. So the tangent line is horizontal at the point (5, 0) and at no other point of the graph of f . C02S01.023: If f (x) = −x2 + 70x, then the slope-predictor for f is m(a) = −2a + 70. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 35. So the tangent line is horizontal at the point (35, 1225) and at no other point of the graph of f . C02S01.024: If f (x) = x2 − 20x + 100, then the slope-predictor for f is m(a) = 2a − 20. A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 10. So the tangent line is horizontal at the point (10, 0) and at no other point of the graph of f . C02S01.025: If f (x) = x2 , then the slope-predictor for f is m(a) = 2a. So the line tangent to the graph of f at the point P (−2, 4) has slope m(−2) = −4 and the normal line at P has slope 1 . Hence an equation 4 for the line tangent to the graph of f at P is y − 4 = −4(x + 2); that is, y = −4x − 4. An equation for the line normal to the graph of f at P is y − 4 = 1 (x + 2); that is, 4y = x + 18. 4 C02S01.026: If f (x) = −2x2 − x + 5, then the slope-predictor for f is m(a) = −4a − 1. So the line tangent to the graph of f at the point P (−1, 4) has slope m(−1) = 3 and the normal line at P has slope − 1 . Hence 3 an equation for the line tangent to the graph of f at P is y − 4 = 3(x + 1); that is, y = 3x + 7. An equation for the line normal to the graph of f at P is y − 4 = − 1 (x + 1); that is, x + 3y = 11. 3 C02S01.027: If f (x) = 2x2 + 3x − 5, then the slope-predictor for f is m(a) = 4a + 3. So the line tangent 1 to the graph of f at the point P (2, 9) has slope m(2) = 11 and the normal line at P has slope − 11 . Hence 2 an equation for the line tangent to the graph of f at P is y − 9 = 11(x − 2); that is, y = 11x − 13. An 1 equation for the line normal to the graph of f at P is y − 9 = − 11 (x − 2); that is, x + 11y = 101. C02S01.028: If f (x) = x2 , then the slope-predictor for f is m(a) = 2a. Hence the line L tangent to the graph of f at the point (x0 , y0 ) has slope m(x0 ) = 2x0 . Because y0 = x2 , an equation of L is 0 y − x2 = 2x0 (x − x0 ). To find where L meets the x-axis, we substitute y = 0 in the equation of L and solve 0 for x: 0 − x2 = 2x0 (x − x0 ); 0 x − x0 = − 1 x0 2 (if x0 = 0); x = x0 − 1 x0 = 1 x0 . 2 2 Therefore if x0 = 0, L meets the x-axis at the point 1 x0 , 0 . If x0 = 0, then L is the x-axis and therefore 2 meets the x-axis at 1 x0 , 0 = (0, 0) as well as at every other point. 2 C02S01.029: If the ball has height y (t) = −16t2 + 96t (feet) at time t (s), then the slope-predictor for y is m(a) = −32a + 96. Assuming that the maximum height of the ball occurs at the point on the graph of y where the tangent line is horizontal, we find that point by solving m(a) = 0 and find that a = 3. So the highest point on the graph of y is the point (3, y (3)) = (3, 144). Therefore the ball reaches a maximum height of 144 (ft). C02S01.030: The slope-predictor for A(x) = −x2 + 50x is m(a) = −2a + 50. The highest point on the graph of A occurs where the tangent line is horizontal; that is, when 2a = 50, so that a = 25. (We know it’s the high point rather than the low point because the graph of y = A(x) is a parabola that opens downward.) So the highest point on the graph of A is the point (25, A(25)) = (25, 625). Because a = 25 is in the domain [0, 50] of the function A, the maximum possible area of the rectangle is 625 (ft2 ). C02S01.031: If the two positive numbers x and y have sum 50, then y = 50 − x, x > 0, and x < 50 (because y > 0). So the product of two such numbers is given by p(x) = x(50 − x), 0 < x < 50. The graph of p(x) = −x2 + 50x has a highest point because the graph of y = p(x) is a parabola that opens downward. The slope-predictor for the function p is m(a) = −2a + 50. The highest point on the graph of p will occur when the tangent line is horizontal, so that m(a) = 0. This leads to a = 25, which does lie in the domain of p. Therefore the highest point on the graph of p is (25, p(25)) = (25, 625). Hence the maximum possible value of p(x) is 625. So the maximum possible product of two positive numbers with sum 50 is 625. 1 2 C02S01.032: If y = f (x) = − 625 x2 + x, then the slope predictor for f is m(a) = − 625 a + 1. (a) The projectile hits the ground at that point x for which f (x) = 0; that is, x2 = 625x, so that x = 0 (which we reject; this is where the projectile leaves the ground) or x = 625. Because the projectile travels from x = 0 to x = 625, the horizontal distance it travels is 625 (ft). (b) To find the maximum height of the projectile, we find where the line tangent to the graph of f is horizontal. This occurs when m(a) = 0, so that a = 312.5. So the maximum height of the projectile is f (312.5) = 156.25 (ft). (It’s a maximum rather than a minimum because the graph of y = f (x) is a parabola that opens downwards and x = 312.5 does lie in the domain [0, 625] of the function f .) C02S01.033: Suppose that the “other” line L is tangent to the parabola at the point (a, a2 ). The slopepredictor for y = f (x) = x2 is m(a) = 2a, so the line L has slope m(a) = 2a. (Note that a changes from a variable to a constant in the last sentence. This is dangerous but the notation has forced this situation 3 upon us.) Using the two-point formula for slope, we can compute the slope of L in another way and equate our two results: a2 − 0 = 2a; a−3 a2 = 2a(a − 3); a = 2a − 6; (because a = 0); a = 6. Therefore L has slope m(6) = 12. Because L passes through (3, 0), an equation of L is y − 0 = 12(x − 3); that is, y = 12x − 36. C02S01.034: If y = f (x) = −x2 + 4x, then the slope-predictor for f is m(a) = −2a + 4. Suppose that the line L passes through the point P (2, 5) and is tangent to the graph of f . Let Q(c, f (c)) = (c, 4c − c2 ) be the point of tangency. We can use the two points P and Q to compute the slope of L. We can also use the slope-predictor. We do so and equate the results: 4c − c2 − 5 = −2c + 4; c−2 4c − c2 − 5 = (c − 2)(−2c + 4) = −2c2 + 8c − 8; c2 − 4c + 3 = 0; (c − 1)(c − 3) = 0. Therefore c = 1 or c = 3. We have discovered that there are two points at which L may be tangent to the graph of f : (1, f (1)) = (1, 3) and (3, f (3)) = (3, 3). Thus one tangent line has slope 2 and the other has slope −2; their equations may be written as y − 5 = 2(x − 2) and y − 5 = −2(x − 2). C02S01.035: Suppose that (a, a2 ) is the point on the graph of y = x2 closest to (3, 0). Let L be the line segment from (3, 0) to (a, a2 ). Under the plausible assumption that L is normal to the tangent line at (a, a2 ), we infer that the slope m of L is − 1/(2a) because the slope of the tangent line is 2a. Because we can also compute m by using the two points known to lie on it, we find that m=− 1 a2 − 0 = . 2a a−3 This leads to the equation 0 = 2a3 + a − 3 = (a − 1)(2a2 + 2a + 3), which has a = 1 as its only real solution (note that the discriminant of 2a2 + 2a + 3 is negative). Intuitively, it’s clear that there is a point on the graph nearest (3 , 0), so we have found it: That point is (1, 1). Alternatively, if (x, x2 ) is an arbitrary point on the given parabola, then the distance from (x, x2 ) to (3, 0) is the square root of f (x) = (x2 − 0)2 + (x − 3)2 = x4 + x2 − 6x + 9. A positive quantity is minimized when its square is minimized, so we minimize the distance from (x, x2 ) to (3, 0) by minimizing f (x). The slope-predictor for f is m(a) = 4a3 + 2a − 6 = 2(a − 1)(2a2 + 2x + 3), and (as before) the equation m(a) = 0 has only one real solution, a = 1. Again appealing to intuition for the existence of a point on the parabola nearest to (3, 0), we see that it can only be the point (1, 1). In Chapter 3 we will see how the existence of the closest point can be established without an appeal to the intuition. C02S01.036: Given: f (x) = x2 and a = −1. We computed 4 f (a + h) − f (a − h) 2h (1) for h = 10−1 , 10−2 , . . . , and 10−10 . The values of the expression in (1) were all −2.00000000000000000000 (to twenty places). The numerical evidence overwhelmingly suggests that the slope of the tangent line is exactly −2 and thus that it has equation y = −2x − 1. The graph of this line and y = f (x) are shown next. 4 3 2 1 -2 -1.5 -1 -0.5 -1 C02S01.037: Given: f (x) = x3 and a = 2. We computed f (a + h) − f (a − h) 2h (1) for h = 10−1 , 10−2 , . . . , 10−10 . The values of the expression in (1) were 12.01, 12.0001, 12.000001, . . . , 12.00000000000000000001. The numerical evidence overwhelmingly suggests that the slope of the tangent line is 12 and thus that it has equation y = 12x − 16. The graph of this line and y = f (x) are shown next. 25 20 15 10 5 1.5 2 2.5 3 C02S01.038: Using the techniques in the previous solution produced strong evidence that the slope of the tangent line is 3, so that its equation is y = 3x + 2. The graphs of f (x) = x3 and the tangent line are shown 5 next. 2 -2 -1.5 -1 -0.5 -2 -4 -6 -8 C02S01.039: The numerical evidence suggests that the slope of the tangent line is 1 , so that its equation 2 √ is y = 1 (x + 1). The graph of the tangent line and the graph of f (x) = x are shown next. 2 1.4 1.2 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 C02S01.040: The numerical evidence suggests that the slope of the tangent line is 1 , so that its equation 4 √ is y = 1 (x + 4). The graph of the tangent line and the graph of f (x) = x are shown next. 4 2.2 2.1 3.5 4 4.5 5 1.9 1.8 C02S01.041: The numerical evidence suggests that the slope of the tangent line is −1, so that its equation 6 is y = −x + 2. The graph of the tangent line and the graph of f (x) = 1/x are shown next. 2 1.8 1.6 1.4 1.2 0.6 0.8 1.2 1.4 0.8 0.6 C02S01.042: The numerical evidence suggests that the slope of the tangent line is −4, so that its equation is y = −4x − 4. The graph of the tangent line and the graph of f (x) = 1/x are shown next. -1 -1.5 -2 -2.5 -3 -3.5 -0.6 -0.5 -0.4 -0.3 C02S01.043: The numerical evidence suggests that the slope of the tangent line is 0, so that its equation is y = 1. The graph of the tangent line and the graph of f (x) = cos x are shown next. -0.4 -0.2 0.2 0.4 0.99 0.98 0.97 0.96 0.95 C02S01.044: The numerical evidence suggests that the slope of the tangent line is 10π , so that its equation is y = 10π x. The graph of the tangent line and the graph of f (x) = 10π x are shown next. 7 3 2 1 -0.1 -0.05 0.05 0.1 -1 -2 -3 C02S01.045: equation is √ The numerical evidence suggests that the slope of the tangent line is −1/ 2, so that its √ √ 2 2 y− =− 2 2 x− π . 4 The graph of the tangent line and the graph of f (x) = cos x are shown next. 1.25 1 0.75 0.5 0.25 0.5 1 1.5 -0.25 C02S01.046: The numerical evidence suggests that the tangent line is horizontal, so that its equation is y ≡ 1. The graph of the tangent line and the graph of f (x) = sin 10π x are shown next. 8 1 0.8 0.6 0.4 0.2 0.02 0.04 0.06 0.08 0.1 C02S01.047: The numerical evidence suggests that the tangent line is horizontal, so that its equation is √ y ≡ 5. The graph of the tangent line and the graph of f (x) = 25 − x2 are shown next. -0.4 -0.2 0.2 0.4 4.998 4.996 4.994 4.992 4.99 C02S01.048: The numerical evidence suggests that the tangent line√ slope − 3 , so that its equation is has 4 3x + 4y = 25. The graph of the tangent line and the graph of f (x) = 25 − x2 are shown next. 5 4.5 4 3.5 1.5 2.5 3 3.5 2.5 9 4 4.5 Section 2.2 C02S02.001: lim (3x2 + 7x − 12) = 3 lim x x→3 C02S02.002: x→3 lim lim C02S02.006: x7 + 7x − 4 = lim x2 − 1 · lim x7 + 7x − 4 = 0 · 4 = 0. x→1 x→1 x→−2 x3 − 3x + 3 · lim x→−2 x2 + 2x + 5 = 1 · 5 = 5. x→1 t→−2 x→3 x→−2 lim (x + 1) x+1 2 x→1 = =. 2 2+x+1 x 3 lim (x + x + 1) lim C02S02.007: lim x→3 x→−2 x3 − 3x + 3 x2 + 2x + 5 = lim x→−2 x→1 x→3 x→−2 x→1 C02S02.005: lim + 7 lim x − lim 12 = 3 · 32 + 7 · 3 − 12 = 36. x3 − 3x2 + 5 = lim x3 − 3 lim x2 + lim 5 = −15. x→−2 C02S02.003: lim x2 − 1 C02S02.004: 2 lim (t + 2) t+2 0 t→−2 = = = 0. t2 + 4 8 lim (t2 + 4) t→−2 x2 + 1 3 3 (x3 25) = 3 lim x2 + 1 x→3 lim x3 − 25 x→3 = 3 3 lim x2 + 1 3 lim x3 − 25 x→3 = 103 1000 = = 125. 23 8 x→3 10 C02S02.008: 3z 2 + 2z + 1 lim x→1 (z 3 + 5) √ C02S02.010: lim 4x + 5 = 27 − y →4 √ C02S02.011: lim x2 − 1 3 /2 t→−4 3z + 2z + 1 z →−1 lim lim (4x + 5) = x→1 = lim (27 − y →4 √ lim x2 − 1 √ z3 + 5 lim t→−4 lim z 2/3 z 2 /3 z →8 √ C02S02.013: lim = √ z →8 z − 2z lim z − 2z = = √ 25 = 5. √ 4 2 =√ = . 3 9 4 = 1. 4 z →8 C02S02.014: lim 3 t→2 3t3 + 4t − 5 = 4 C02S02.015: lim C02S02.016: lim t→−4 3 6 (t + 1) = lim (3t3 + 4t − 5) = 3. t→2 4 (w − 2) = w→0 3 lim (w − 2) = w →0 3 z →−1 √ = 83/2 = 16 2. 3 /2 25 − t2 5 lim 9 = 3. y) = lim (t + 8) t→−4 10 3z 2 + 2z + 1 lim z →−1 x→3 t+8 = 25 − t2 lim lim 2 z →−1 y= x→3 C02S02.012: = 5 z →−1 C02S02.009: lim 10 6 lim (t + 1) = 9. t→−4 1 (−2)4 = 4. 5 z3 + 5 = 210 = 1. 45 C02S02.017: lim x→−2 2y 2 + 2y + 4 6y − 3 C02S02.018: lim y →5 C02S02.019: x+2 = (x − 2)2 3 lim x→−1 x2 3 lim x→−2 1 /3 = (x + 2) = 0. (x − 2)2 64 27 1/3 = 4 . 3 x+1 x+1 1 1 = lim = lim =− . − x − 2 x→−1 (x + 1)(x − 2) x→−1 x − 2 3 C02S02.020: lim t2 − 9 (t − 3)(t + 3) = lim = lim (t + 3) = 6. t→3 t→3 t−3 t−3 C02S02.021: lim x2 + x − 2 (x + 2)(x − 1) x+2 3 = lim = lim =− . x2 − 4x + 3 x→1 (x − 3)(x − 1) x→1 x − 3 2 t→3 x→1 C02S02.022: C02S02.023: 4y 2 − 1 (2y − 1)(2y + 1) 2y − 1 2 = lim = lim = − = −1. + 8y + 3 y→−1/2 (2y + 3)(2y + 1) y→−1/2 2y + 3 2 lim y →−1/2 4y 2 lim t→−3 t2 + 6t + 9 (t + 3)(t + 3) t+3 = lim = lim = 0. t→−3 (t + 3)(t − 3) t→−3 t − 3 t2 − 9 x2 − 4 (x − 2)(x + 2) x+2 2 = lim = lim =. − 2x − 8 x→2 (x − 2)(3x + 4) x→2 3x + 4 5 C02S02.024: lim x→2 3x2 C02S02.025: lim z →−2 (z + 2)2 (z + 2)(z + 2) z+2 = lim = lim = 0. z →−2 (z + 2)(z − 2)(z 2 + 4) z →−2 (z − 2)(z 2 + 4) z 4 − 16 C02S02.026: lim t3 − 9t t(t2 − 9) = lim 2 = 3. 2−9 t→3 t − 9 t C02S02.027: lim x3 − 1 (x − 1)(x2 + x + 1) x2 + x + 1 3 = lim = lim =. 4−1 x→1 (x − 1)(x + 1)(x2 + 1) x→1 (x + 1)(x2 + 1) x 4 t→3 x→1 C02S02.028: lim y →−3 y 3 + 27 (y + 3)(y 2 − 3y + 9) y 2 − 3y + 9 27 9 = lim = lim =− =− . 2−9 y →−3 y →−3 y (y + 3)(y − 3) y−3 6 2 11 − x 3 = lim C02S02.029: lim x→3 x − 3 x→3 3−x 3x 1 1 − 2 + t 2 = lim C02S02.030: lim t→0 t→0 t 1 x−3 2 − (2 + t) 2(2 + t) = lim x→3 1 t −1 1 =− . 3x 9 = lim t→0 2 − 2 − t) 2(2 + t) √ √ √ x−4 ( x − 2)( x + 2) √ C02S02.031: lim √ = lim = lim x + 2 = 4. x→4 x→4 x − 2 x→4 x−2 C02S02.032: lim x→9 √ √ 3− x 3− x 1 1 √ √ √=. = lim = lim x→9 (3 − 9−x 6 x )(3 + x ) x→9 3 + x 2 1 t = lim t→0 −1 1 =− . 2(2 + t) 4 C02S02.033: lim √ t→0 1 C02S02.034: lim h→0 h √ √ t+4 −2 t+4 +2 ·√ t t+4 +2 t+4−4 = lim √ t→0 t t+4 +2 t 1 1 = lim √ = lim √ =. t→0 t t→0 4 t+4 +2 t+4 +2 t+4 −2 = lim t→0 t √ 3− 9+h √ = lim h→0 3h 9 + h √ √ 3− 9+h 3+ 9+h √ √ = lim · h→0 3h 9 + h 3+ 9+h 9 − (9 + h) −1 √ √ =√ = lim √ h→0 3 9 + h 3 + 3h 9 + h 3 + 9 + h 9+h 1 1 √ − 3 9+h =− 1 . 54 √ √ √ x2 − 16 (x + 4)( x − 2)( x + 2) √ = lim √ = lim −(x + 4)( x + 2) = −32. x→4 2 − x→4 x→4 x 2− x C02S02.035: lim C02S02.036: lim x→0 √ 1+x − x √ 1−x √ √ √ √ 1+x − 1−x 1+x + 1−x √ √ = lim · x→0 x 1+x + 1−x (1 + x) − (1 − x) √ √ = lim x→0 x 1+x + 1−x 2 √ = lim √ = 1. x→0 1+x + 1−x f (x + h) − f (x) (x + h)3 − x3 x3 + 3x2 h + 3xh2 + h3 − x3 = = = 3x2 + 3xh + h2 → 3x2 as h h h h → 0. When x = 2, y = f (2) = x3 = 8 and the slope of the tangent line to this curve at x = 2 is 3x2 = 12, so an equation of this tangent line is y = 12x − 16. C02S02.037: 1 1 − f (x + h) − f (x) x − (x + h) −1 1 x+h x C02S02.038: = = = → − 2 as h → 0. When h h hx(x + h) x(x + h) x x = 2, y = f (2) = 1 and the slope of the line tangent to this curve at x = 2 is − 1 , so an equation of this 2 4 tangent line is y − 1 = − 1 (x − 2); that is, y = − 1 (x − 4). 2 4 4 1 1 −2 2 f (x + h) − f (x) −2x − h 2 x2 − (x + h)2 (x + h) x C02S02.039: =2 → − 3 as h → 0. When = = h h hx2 (x + h)2 x (x + h)2 x x = 2, y = f (2) = 1 and the slope of the line tangent to this curve at x = 2 is − 1 , so an equation of this 4 4 tangent line is y − 1 = − 1 (x − 2); that is, y = − 1 (x − 3). 4 4 4 1 1 − f (x + h) − f (x) x+1−x−h−1 −1 x+h+1 x+1 C02S02.040: = = = . h h h(x + 1)(x + h + 1) (x + 1)(x + h + 1) 1 This approaches − as h approaches 0. When x = 2, y = f (2) = 1 and the slope of the line tangent 3 (x + 1)2 1 1 to this curve at x = 2 is − 9 , so an equation of this tangent line is y − 3 = − 1 (x − 2); that is, y = − 1 (x − 5). 9 9 C02S02.041: f (x + h) − f (x) = h 2 x+h−1 − h 2 x−1 3 = 2(x − 1 − x − h + 1) −2 = . h(x − 1)(x + h − 1) (x − 1)(x + h − 1) −2 as h approaches 0. When x = 2, y = f (2) = 2 and the slope of the line tangent to (x − 1)2 this curve at x = 2 is −2, so an equation of this tangent line is y − 2 = −2(x − 2); alternatively, y = −2(x − 3). This approaches f (x + h) − f (x) C02S02.042: = h x+h x+h−1 x x−1 − = (x − 1)(x + h) − x2 − xh + x h(x − 1)(x + h − 1) h −1 −1 = → as h → 0. (x − 1)(x + h − 1) (x − 1)2 When x = 2, y = f (2) = 2 and the slope of the line tangent to this curve at x = 2 is −1, so an equation of this tangent line is y − 2 = −1(x − 2); that is, y = −x + 4. C02S02.043: f (x + h) − f (x) = h √ 1 x+h+2 √ − 1 x+2 h √ √ √ √ x+2 − x+h+2 x+2 + x+h+2 √ √ √ = ·√ h x+2 x+h+2 x+2 + x+h+2 −h −1 √ √ √ √ =√ → (x + 2) 2 x + 2 h x+2 x+h+2 x+2 + x+h+2 1 as h → 0. When x = 2, y = f (2) = 1 and the slope of the line tangent to this curve at x = 2 is − 16 , so an 2 1 1 1 equation of this tangent line is y − 2 = − 16 (x − 2); that is, y = − 16 (x − 10). 3 3 (x + h)2 + − x2 − f (x + h) − f (x) x+h x C02S02.044: = h h −3 3 = (2x + h) + → 2x − 2 as h → 0. x(x + h) x When x = 2, y = f (2) = 11 and the slope of the line tangent to this curve at x = 2 is 2 this tangent line is y − 11 = 13 (x − 2). 2 4 C02S02.045: f (x + h) − f (x) = h = = 2(x + h) + 5 − h √ 2(x + h) + 5 − h 2 2(x + h) + 5 + 2x + 5 2x + 5 · →√ 2(x + h) + 5 + 2(x + h) + 5 + 2 C02S02.046: − x2 x+1 √ √ 2x + 5 2x + 5 1 as h → 0. 2x + 5 When x = 2, y = f (2) = 3 and the slope of the line tangent to this curve at x = 2 is this tangent line is y − 3 = 1 (x − 2); if you prefer, y = 1 (x + 7). 3 3 (x + h) x+h+1 so an equation of 2x + 5 √ √ 13 4, 1 3, so an equation of 2 (x + 1) (x + h) − (x + h + 1) x2 h h(x + 1)(x + h + 1) 2 2 x + xh + 2x + h x + 2x = as h → 0. → (x + 1)(x + h + 1) (x + 1)2 f (x + h) − f (x) = h = When x = 2, y = f (2) = 4 and the slope of the line tangent to this curve at x = 2 is 8 , so an equation of 3 9 this tangent line is y − 4 = 8 (x − 2); that is, 9y = 8x − 4. 3 9 4 C02S02.047: x 10−2 10−4 10−6 10−8 10−10 2. f (x) 2.01 2.001 2. 2. x −10−2 −10−4 −10−6 −10−8 f (x) 1.99 1.9999 2. 2. −10−10 2. The limit appears to be 2. C02S02.048: x 1 + 10−2 1 + 10−4 1 + 10−6 1 + 10−8 f (x) 4.0604 4.0006 4.00001 4. x 1 − 10−2 1 − 10−4 1 − 10−6 1 − 10−8 f (x) 3.9404 3.9994 3.99999 1 + 10−10 4. 1 − 10−10 4. 4. The limit appears to be 4. C02S02.049: x 10−2 10−4 10−6 10−8 10−10 f (x) 0.16662 0.166666 0.166667 0.166667 0.166667 x −10−2 −10−4 −10−6 −10−8 −10−10 f (x) 0.166713 0.166667 0.166667 0.166667 0.166667 The limit appears to be 1 . 6 C02S02.050: x 4 + 10−2 4 + 10−4 4 + 10−6 4 + 10−8 f (x) 3.00187 3.00002 3. 3. x 4 − 10−2 4 − 10−4 4 − 10−6 4 − 10−8 f (x) 2.99812 2.99998 3. 4 + 10−10 3. 4 − 10−10 3. 3. The limit appears to be 3. C02S02.051: x 10−2 10−4 10−6 10−8 10−10 f (x) −0.37128 −0.374963 −0.375 −0.375 −0.375 x −10−2 −10−4 −10−6 −10−8 −10−10 f (x) −0.378781 −0.375038 −0.375 −0.375 −0.375 5 The limit appears to be − 3 . 8 C02S02.052: 10−2 x 10−4 10−6 10−8 10−10 f (x) −0.222225 −0.222222 −0.222222 −0.222225 x −10−2 −10−4 −10−6 −10−8 f (x) −0.222225 −0.222222 −0.222222 −0.222222 −0.222222 −10−10 −0.222222 Limit: − 2 . 9 C02S02.053: 10−2 x 10−4 10−6 10−8 10−10 1. f (x) 0.999983 1. 1. 1. x −10−2 −10−4 −10−6 −10−8 10−4 10−6 f (x) 0.999983 1. 1. −10−10 1. 1. The limit appears to be 1. C02S02.054: 10−2 x 10−8 f (x) 0.49996 0.5 0.5 0.499817 x −10−2 −10−4 −10−6 10−10 −10−8 f (x) 0.499996 0.5 0.5 0 −10−10 0.499817 0 Beware of round-off errors. The limit is 0.5. C02S02.055: 10−2 x 10−3 10−4 10−5 10−6 f (x) 0.166666 0.166667 0.166667 0.166667 x −10−2 −10−3 −10−4 −10−5 f (x) 0.166666 0.166667 0.166667 0.166667 −10−6 0.166667 0.166667 The limit appears to be 1 . 6 C02S02.056: 10−2 10−4 10−6 f (x) 1.04723 1.00092 x −10−2 −10−4 x f (x) .954898 .999079 10−8 10−10 1.00001 1. 1. −10−6 −10−8 .999986 1. −10−10 1. C02S02.057: x 2−1 2−5 2−10 2−15 2−20 (1 + x)1/x 2.25 2.67699 2.71696 2.71824 2.71828 x −2−1 −2−5 −2−10 −2−15 −2−20 (1 + x)1/x 4. 2.76210 2.71961 2.71832 2.71828 6 The limit appears to be 1. C02S02.058: The graph of y = (sin x)/x on the interval [−0.01, 0.01] is next. -0.01 -0.005 0.005 0.01 0.999997 0.999995 0.999992 0.99999 0.999987 0.999985 C02S02.059: lim x→0 x − tan x 1 =− . 3 x 3 Answer: −0.3333. sin 2x 2 =. x→0 tan 5x 5 C02S02.060: lim π π = 0 for every positive integer n. Therefore lim sin , if it x→0 x π were to exist, would be 0. Notice however that sin 3n · alternates between +1 and −1 for n = 1, 2, 3, . . . . 2 π does not exist. Therefore lim sin x→0 x C02S02.061: sin 2−n = sin 2π · 2(n−1) C02S02.062: The graph of f (x) = sin x + 10−5 cos x on the interval [−0.00001, 0.00001] is shown next. The graph makes it clear that the limit is certainly not zero and almost certainly is 10−5 . 0.00002 0.000015 0.00001 -0.00001 0.00001 −1/32 C02S02.063: The graph of f (x) = (log10 (1/| x |) ) is shown next, as well as a table of values of f (x) for x very close to zero. The table was generated by Mathematica, version 3.0, but virtually any computer algebra system will produce similar results. 7 0.95 0.948 0.946 0.944 0.942 -0.00001 0.00001 0.938 f (x) x f (x) x 10 −6 1.0000 0.9455 0.9786 10 −7 10 10 0.9663 10 −8 10 0.9576 10 −9 10 0.9509 10 −10 x 10 −1 −2 −3 −4 −5 f (x) x 10 −11 0.9278 0.9410 10 −12 0.9371 10 −13 0.9336 10 −14 0.9306 10 −15 f (x) 10 −16 0.9170 0.9253 10 −17 0.9153 0.9230 10 −18 0.9136 0.9208 10 −19 0.9121 0.9189 10 −20 0.9106 C02S02.064: The slope of the line tangent to the graph of y = 10x at the point (0, 1) is L = lim h→0 100+h − 100 10h − 1 = lim . h→0 h h With h = 0.1, 0.01, 0.001, . . . , 0.000001, a calculator reports that the corresponding values of (10h − 1)/h are (approximately) 2.58925, 2.32930, 2.30524, . . . , and 2.30259. This is fair evidence that L = ln 10 ≈ 2.302585. The slope-predictor for y = 10x is m(x) = lim h→0 10x+h − 10x = 10x · h lim h→0 10h − 1 h = L · 10x . The line tangent to the graph of y = 10x at the point P (a, 10a ) has predicted equation y − 10a = L · 10a · (x − a). To see the graph of y = 10x near P and the line predicted to be tangent to that graph at P , enter the Mathematica commands a = 2; (* or any other value you please *) Plot[ { 10∧x, 10∧a + (10∧a)∗Log[ 10 ]∗(x − a) }, { x, a − 1, a + 1 } ]; 8 Section 2.3 C02S03.001: θ · C02S03.002: θ → 0 · 1 = 0 as θ → 0. sin θ sin θ sin θ · → 1 · 1 = 1 as θ → 0. θ θ C02S03.003: Multiply numerator and denominator by 1 + cos θ (the conjugate of the numerator) to obtain lim θ →0 C02S03.004: 1 − cos2 θ sin θ sin θ 1 1 1 = lim · · =1·1· = . θ2 (1 + cos θ) θ→0 θ θ 1 + cos θ 2 2 tan θ sin θ sin θ 1 1 = = · → 1 · = 1 as θ → 0. θ θ cos θ θ cos θ 1 C02S03.005: Divide each term in numerator and denominator by t. Then it’s clear that the denominator is approaching zero whereas the numerator is not, so the limit does not exist. Because the numerator is positive and the denominator is approaching zero through negative values, the answer −∞ is also correct. C02S03.006: As θ → 0, so does ω = θ2 , and sin 2ω 2 sin ω cos ω sin ω = = · 2 cos ω → 1 · 2 · 1 = 2. ω ω ω C02S03.007: Let z = 5x. Then z → 0 as x → 0, and sin 5x 5 sin z = → 5 · 1 = 5. x z sin 2z 2 sin z cos z 2 cos z sin z 1 = = · → 2 · · 1 = 2 as z → 0. z cos 3z z cos 3z cos 3z z 1 √ C02S03.009: This limit does not exist because x is not defined for x near 0 if x < 0. But C02S03.008: lim x→0+ √ sin x sin x √ = lim x· + x x x→0 = 0 · 1 = 0. C02S03.010: Using the identity sin2 x = 1 (1 − cos 2x) (inside the front cover), we obtain 2 lim x→0 1 − cos 2x 2(1 − cos 2x) 2 sin2 x sin x = lim = lim = lim (2 sin x) = 2 · 0 · 1 = 0. x→0 x→0 x→0 x 2x x x Alternatively, you could multiply numerator and denominator by 1+cos 2x (the conjugate of the numerator). C02S03.011: Let x = 3z . Then x → 0 is equivalent to z → 0, and therefore lim x→0 1 1 1 sin z x 1 1 sin = lim sin z = lim · = ·1= . z →0 3 x 3 z→0 3z z 3 3 C02S03.012: Let x = 3θ. Then θ = 1 x and θ → 0 is equivalent to x → 0. Hence 3 lim θ →0 (sin 3θ)2 = lim x→0 θ2 cos θ sin x sin x 1 (sin x)2 1 = lim 9 · = 9 · 1 · 1 · = 9. · · x→0 x x 1 cos 1 x cos 1 x 3 3 12 9x C02S03.013: Multiply numerator and denominator by 1 + cos x (the conjugate of the numerator) to obtain 1 lim x→0 (1 − cos x)(1 + cos x) 1 − cos2 x sin2 x sin x 0 = lim = lim = lim = = 0. x→0 (sin x)(1 + cos x) x→0 (sin x)(1 + cos x) x→0 1 + cos x (sin x)(1 + cos x) 1+1 C02S03.014: By Problem 4, (tan x)/x → 1 as x → 0. This observation implies that lim x→0 tan kx =1 kx and lim x→0 kx =1 tan kx for any nonzero constant k . Hence lim x→0 tan 3x 3 tan 3x 5x 3 3 = lim · · = ·1·1= . tan 5x x→0 5 3x tan 5x 5 5 1 1 and csc x = . Hence cos x sin x 1 x 1 lim x sec x csc x = lim · = · 1 = 1. x→0 x→0 cos x sin x 1 C02S03.015: Recall that sec x = We also used the fact that lim x→0 C02S03.016: lim θ →0 x 1 1 = = 1. = lim x→0 sin x sin x 1 x sin 2θ 2 sin θ cos θ sin θ = lim = lim (2 cos θ) · = 2 · 1 · 1 = 2. θ →0 θ →0 θ θ θ C02S03.017: Multiply numerator and denominator by 1 + cos θ (the conjugate of the numerator) to obtain lim θ →0 (1 − cos θ)(1 + cos θ) sin2 θ sin θ = lim = lim θ →0 (θ sin θ )(1 + cos θ ) θ →0 θ (1 + cos θ ) (θ sin θ)(1 + cos θ) = lim θ →0 sin θ 1 1 1 · =1· =. θ 1 + cos θ 1+1 2 C02S03.018: lim sin2 θ sin θ = lim · sin θ = 1 · 0 = 0. θ →0 θ θ C02S03.019: lim tan z sin z 1 1 1 =. = lim = lim = sin 2z z→0 (cos z )(2 sin z cos z ) z→0 2 cos2 z 2 · 12 2 C02S03.020: lim tan 2x 2 tan 2x 2 2 = lim · = ·1= x→0 3 3x 2x 3 3 θ →0 z →0 x→0 C02S03.021: lim x cot 3x = lim x→0 x→0 (with the aid of Problem 4). x cos 3x 3x cos 3x 1 1 = lim · = 1· = x→0 sin 3x sin 3x 3 3 3 x − tan x C02S03.022: lim = lim x→0 x→0 sin x x tan x − x x sin x x = 1−1 =0 1 C02S03.023: Let x = 1 t. Then x → 0 is equivalent to t → 0, so 2 2 (see Problem 15, last line). (with the aid of Problem 4). t 2 sin lim t→0 t 2 = lim x→0 sin x = 1. x Therefore lim t→0 1 sin2 t2 C02S03.024: Because t 2 14 · sin2 4 t2 = lim t→0 2 t sin 1 2 = 1 · 12 = 1 . = lim · t t→0 4 4 4 2 t 2 sin x → 1 as x → 0, it follows that x lim x→0 sin kx =1 kx and lim x→0 kx =1 sin kx for any nonzero constant k . Hence lim x→0 sin 2x 2 sin 2x 5x 2 2 = lim · · = ·1·1= . x→0 5 sin 5x 2x sin 5x 5 5 C02S03.025: Because −1 cos 10x 1 for all x, −x2 x2 cos 10x x2 for all x. But both −x2 and x2 approach zero as x → 0. Therefore lim x2 cos 10x = 0. The second inequality is illustrated next. x→0 2 1 -1.5 -1 -0.5 0.5 1 1.5 -1 -2 1 x2 for all x = 0. Because both −x2 x 1 and x2 approach zero as x → 0, it follows from the squeeze law that lim x2 sin = 0. The second inequality x→0 x is illustrated next. C02S03.026: Because −1 sin x 1 for all x, also −x2 3 x2 sin 0.075 0.05 0.025 -0.6 -0.4 -0.2 -0.025 0.2 0.4 0.6 -0.05 -0.075 C02S03.027: First, −1 cos x 1 for all x. Therefore 1 x2 cos √ 3 x −x2 x2 for all x = 0. Finally, both x2 and −x2 approach zero as x → 0. Therefore, by the squeeze law, 1 lim x2 cos √ = 0. 3 x x→0 C02S03.028: Because −1 sin x 1 for all x, − √ 3 √ 3 x x sin 1 x √ 3 x √ √ for all x = 0. Because both − | 3 x | and | 3 x | approach zero as x → 0, it follows from the squeeze law that √ 3 lim x→0 C02S03.029: lim x→0+ 3− √ x =3− x sin 1 = 0. x lim x = 3 − 0 = 3. x→0+ 3 /2 C02S03.030: C02S03.031: lim x→0+ lim x→1− 4 + 3x3/2 = 4 + 3 · √ lim x + = 4 − 3 · 0 = 4. x→0 x − 1 does not exist because if x < 1, then x − 1 < 0. C02S03.032: Because x → 4− , x < 4, so that √ √ and lim− 4 − x = 4 − 4 = 0. √ 4 − x is defined for all such x. Therefore the limit exists x→4 C02S03.033: Because x → 2+ , x > 2, so that x2 > 4. Hence √ lim+ x2 − 4 = 4 − 4 = 0. √ x2 − 4 is defined for all such x and x→2 C02S03.034: Because x → 3+ , x > 3, so that 9 − x2 < 0 for all such x. Thus the given limit does not exist. 4 C02S03.035: Because x → 5− , x < 5 and x > 0 for x sufficiently close to 5. Therefore x(5 − x) > 0 for such x, so that x(5 − x) exists for such x. Therefore lim− x(5 − x) = x→5 lim− x x→5 5 − lim− x x→5 = √ 5 · 0 = 0. C02S03.036: As x → 2− , x < 2, and x > −2 for x su√ ciently close to 2. For such x, 4 − x2 > 0, so that ffi √ 4 − x2 exists. Therefore lim 4 − x2 = 4 − 22 = 0 = 0. x→2− C02S03.037: As x → 4+ , x > 4, so that both 4x and x − 4 are positive. Hence the radicand is positive and the square root exists. But the denominator in the radicand is approaching zero through positive values while the numerator is approaching 16. So the fraction is approaching + ∞. Therefore 4x = +∞. x−4 lim x→4+ It is also correct to say that this limit does not exist. C02S03.038: First, 6 − x − x2 = (3 + x)(2 − x), so that as x → −3+ , x > −3, and thus 3 + x > 3 − 3 = 0. Also x < 2 if x is sufficiently close to −3, so that 2 − x > 0. Therefore (3 + x)(2 − x) > 0, and so the square root is defined. Finally, lim x→−3+ 6 − x − x2 = C02S03.039: If x < 5, then x − 5 < 0, so lim x→−3+ (3 + x)(2 − x) = √ 0·5= √ 0 = 0. x−5 x−5 = = −1. Therefore the limit is −1. |x − 5| −(x − 5) √ 16 − x2 C02S03.040: If −4 < x < 4, then 16 − x2 > 0, so √ = 16 − x2 → 0 as x → −4+ . 2 16 − x C02S03.041: If x > 3, then x −6x+9 = (x−3) > 0 and x−3 > 0, so 2 2 as x → 3+ . √ x2 − 6x + 9 |x − 3| x−3 = = →1 x−3 x−3 x−3 x−2 x−2 1 = = → −1 as x → 2+ . Indeed, the two-sided limit exists x2 − 5x + 6 (x − 2)(x − 3) x−3 and is equal to −1. C02S03.042: C02S03.043: If x > 2 then x − 2 > 0, so 2−x 2−x = = −1. Therefore the limit is also −1. |x − 2| x−2 C02S03.044: If x < 7 then x − 7 < 0, so 7−x 7−x = = 1. So the limit is 1. |x − 7| −(x − 7) C02S03.045: 1 − x2 (1 + x)(1 − x) = = 1 + x, so the limit is 2. 1−x 1−x C02S03.046: As x → 0− , x < 0, so that x − | x | = x − (−x) = 2x. Therefore lim x→0− x x 1 1 = lim− = lim− = . x − |x| 2 x→0 2x x→0 2 5 √ C02S03.047: Recall first that z 2 = | z | for every real number z . Because x → 5+ , x > 5, so 5 − x < 0. Therefore (5 − x)2 = | 5 − x | = −(5 − x) = x − 5. Therefore (5 − x)2 x−5 = lim+ = lim+ (−1) = −1. 5−x x→5 5 − x x→5 lim x→5+ √ C02S03.048: Recall that z 2 = | z | for every real number z . Because x → −4− , we know that x < −4. Hence 4 + x < 4 + (−4) = 0. Therefore lim x→−4− 4+x (4 + x)2 = lim x→−4− 4+x 4+x = lim = lim (−1) = −1. − −(4 + x) |4 + x| x→−4 x→−4− C02S03.049: The right-hand and left-hand limits both fail to exist at a = 1. The behavior of f near a is best described by observing that lim x→1+ 1 = +∞ x−1 and lim x→1− 1 = −∞. x−1 C02S03.050: The right-hand and left-hand limits both fail to exist at a = 3. The behavior of f near a is best described by observing that lim x→3+ 2 = −∞ 3−x and lim− x→3 2 = + ∞. 3−x C02S03.051: The right-hand and left-hand limits both fail to exist at a = −1. The behavior of f near a is best described by observing that lim x→−1+ x−1 = −∞ x+1 and lim x→−1− x−1 = + ∞. x+1 C02S03.052: The right-hand and left-hand limits both fail to exist at a = 5. The behavior of f near a is best described by observing that lim x→5+ 2x − 5 = −∞ 5−x and lim x→5− 2x − 5 = + ∞. 5−x C02S03.053: The right-hand and left-hand limits both fail to exist at a = −2. If x is slightly greater than −2, then 1 − x2 is close to 1 − 4 = −3, while x + 2 is a positive number close to zero. In this case f (x) is a large negative number. Similarly, if x is slightly less than −2, then 1 − x2 is close to −3, while x + 2 is a negative number close to zero. In this case f (x) is a large positive number. The behavior of f near −2 is best described by observing that lim x→−2+ 1 − x2 = −∞ x+2 and lim − x→−2 1 − x2 = + ∞. x+2 C02S03.054: The right-hand and left-hand limits fail to exist at a = 5. If x is close to 5 but x = 5, then x − 5 is close to zero, so that (x − 5)2 is a positive number still very close to zero. Its reciprocal is therefore a very large positive number. That is, lim x→5+ 1 1 = lim− = + ∞. (x − 5)2 x→5 (x − 5)2 6 (1) Unlike the previous problems of this sort, we may in this case also write lim x→5 1 = + ∞. (x − 5)2 Nevertheless, Eq. (1) implies that neither the left-hand nor the right-hand limit of f (x) exists (is a real number) at x = 5. C02S03.055: The left-hand and right-hand limits both fail to exist at x = 1. To simplify f (x), observe that |1 − x| |1 − x| 1 = = . (1 − x)2 | 1 − x |2 |1 − x| f (x) = Therefore we can describe the behavior of f (x) near a = 1 in this way: lim f (x) = lim x→1 x→1 1 = + ∞. |1 − x| C02S03.056: Because x2 + 6x + 9 = (x + 3)2 , the denominator in f (x) is zero when x = −3, and so the left-hand and right-hand limits fail to exist at a = −3. When x is close to −3 but x = −3, (x + 3)2 is a positive number very close to zero, while the numerator x + 1 is close to −2. Therefore f (x) is a very large negative number. That is, lim x→−3 x+1 = −∞. x2 + 6x + 9 C02S03.057: First simplify f (x): If x2 = 4 (that is, if x = ± 2), then f (x) = x−2 x−2 −1 = = . 2 4−x (2 + x)(2 − x) 2+x So even though f (2) does not exist, there is no real problem with the limit of f (x) as x → 2: lim f (x) = lim x→2 x→2 −1 1 =− . 2+x 4 But the left-hand and right-hand limits of f (x) fail to exist at x = −2, because lim x→−2+ f (x) = lim x→−2+ −1 = −∞ 2+x and lim x→−2− f (x) = lim x→−2− −1 = + ∞. 2+x C02S03.058: First simplify: f (x) = x2 x−1 x−1 1 = = − 3x + 2 (x − 1)(x − 2) x−2 if x = 1 and x = 2. But even though f (1) is undefined, lim f (x) = lim x→1 x→1 But the one-sided limits fail to exist at x = 2: lim f (x) = lim x→2+ x→2+ 1 = +∞ x−2 1 1 = = −1. x−2 1−2 and 7 lim f (x) = lim− x→2− x→2 1 = −∞. x−2 C02S03.059: shown next. lim+ x→2 x2 − 4 x2 − 4 = 4 and lim− = −4. The two-sided limit does not exist. The graph is |x − 2| x→2 |x − 2| 4 2 1.5 2 2.5 3 -2 -4 x4 − 8x + 16 x4 − 8x + 16 = +∞ and lim− = +∞, the two-sided limit also |x − 2| |x − 2| x→2+ x→2 fails to exist. The graph is shown next. C02S03.060: Because lim 2000 1500 1000 500 1.5 2 2.5 3 C02S03.061: If x is an even integer then f (x) = 3, if x is an odd integer then f (x) = 1, and lim f (x) = 2 x→a for all real number values of a. The graph of f is shown next. 4 3 2 1 -1 1 2 3 4 C02S03.062: If n is any integer then f (x) → n as x → n. Note: lim f (x) = a for all real number values x→a 8 of a. The graph of f is shown next. 4 3 2 1 -1 1 2 3 4 -1 C02S03.063: For any integer n, lim f (x) = 10n − 1 and lim f (x) = 10n. Note: lim f (x) exists if and − + x→n only if 10a is not an integer. The graph is shown next. x→a x→n 4 3 2 1 -0.2 -0.1 0.1 0.2 0.3 0.4 -1 -2 -3 C02S03.064: If n is an odd integer, then f (x) = n − 1, an even integer, for n − 1 an odd integer, for n x < n + 1. Therefore lim f (x) = 1 x→n− and x < n and f (x) = n, lim f (x) = −1. x→n+ Similarly, if n is an even integer, then lim f (x) = −1 x→n− and lim f (x) = 1. x→n+ Finally, lim f (x) exists if and only if a is not an integer. The graph of f is shown next. x→a 9 1 0.5 -1 1 2 3 4 -0.5 -1 C02S03.065: If n is an integer and n < x < n + 1, then write x = n + t where 0 < t < 1. Then f (x) = n + t − n − 1 = t − 1 . Moreover, x → n+ is equivalent to t → 0+ . Therefore 2 2 lim f (x) = lim t − + + x→n x→n 1 2 = lim t − + t→0 1 2 1 =− . 2 Similar reasoning, with n − 1 < x < n, shows that if n is an integer, then lim f (x) = − x→n 1 . 2 Finally, if a is a real number other than an integer, then lim f (x) exists. The graph of f is next. x→a 0.4 0.2 -1 1 2 3 4 -0.2 -0.4 C02S03.066: Given the real number x, there is a [unique] integer n such that 2n x < 2n + 2. Thus 1 n 2 x < n + 1, and in this case f (x) = n. So if m = 2n is an even integer, then f (x) → m as x → m+ and x → a for every real number a strictly between 2n and 2n + 2. But if 2n − 2 < x < 2n, then n − 1 < 1 x < n, 2 so that f (x) = n − 1; in this case f (x) → n − 1 as x → m− . Therefore: If k is an odd integer, then lim f (x) = 1 (k − 1). 2 x→k If k is an even integer, then lim+ f (x) = 1 k and lim− f (x) = 1 (k − 2). 2 2 x→k x→k Finally, lim f (x) exists if and only if a is not an even integer. The graph of f is next. x→a 10 3 2 1 -2 2 4 6 -1 -2 C02S03.067: If x is an integer, then f (x) = x − x = 0. If x is not an integer, choose the [unique] integer n such that n < x < n + 1. Then −(n + 1) < −x < −n, so f (x) = n − (n + 1) = −1. Therefore lim f (x) = lim (−1) = −1 x→a x→a for every real number a. In particular, for every integer n, lim f (x) = −1 x→n− and lim f (x) = −1. x→n+ The graph of f is shown next. -1 1 2 3 4 -0.5 -1 -1.5 -2 C02S03.068: If n is a positive integer, then lim f (x) = x→n− n−1 n and lim f (x) = 1. x→n+ For any integer n < 0, lim f (x) = x→n− n+1 n and lim+ f (x) = 1. x→n Also, lim f (x) = + ∞ x→0− and 11 lim f (x) = 0. x→0+ Note: lim f (x) exists if and only if a is not an integer. The graph of f is next. x→a 4 3 2 1 -1 1 2 3 4 C02S03.069: The values of a for which lim g (x) exists are those real numbers not integral multiples of 1 10 . If b is an integral multiple of 1 10 , then x→a lim g (x) = b − x→b− 1 10 and lim g (x) = b. x→b+ The graph of g is shown next. 0.4 0.3 0.2 0.1 -0.2 -0.1 0.1 0.2 0.3 0.4 -0.1 -0.2 -0.3 C02S03.070: Let f (x) = sgn(x) and g (x) = − sgn(x). Clearly neither f (x) nor g (x) has a limit as x → 0 (for example, f (x) → 1 as x → 1+ but f (x) → −1 as x → 1− ). But if x > 0, 1−1 f (x) + g (x) = −1 + 1 if x < 0, 0+0 if x = 0, so that f (x) + g (x) ≡ 0, and therefore f (x) + g (x) → 0 as x → 0. Also, f (x) · g (x) = −1 0 if x = 0, if x = 0. Therefore lim f (x) · g (x) = −1. x→0 C02S03.071: Because −x2 f (x) x2 for all x and because −x2 → 0 and x2 → 0 as x → 0, it follows from the squeeze law for limits that lim f (x) = 0 = f (0). x→0 12 C02S03.072: As x → 0+ , 1/x → + ∞, so 1 + 21/x → + ∞ as well. Therefore lim x→0+ 1 = 0. 1 + 21/x As x → 0− , 1/x → −∞, so 21/x → 0. Consequently, lim x→0− 1 = 1. 1 + 21/x Therefore lim f (x) does not exist. This function approaches its one-sided limits at x = 0 very rapidly. For example, x→0 f (0.01) ≈ 7.888609052 × 10−31 and f (−0.01) ≈ 1 − 7.888601052 × 10−31 . The graph of f for x near zero is next. 1 0.8 0.6 0.4 0.2 -1 1 2 3 4 C02S03.073: Given: f (x) = x · [[1/x ]]. Let’s first study the right-hand limit of f (x) at x = 0. We need consider only values of x in the interval (0, 1), and if 0 < x < 1 then 1< 1 , x so that n 1 <n+1 x for some [unique] positive integer n. Moreover, if so then 1 <x n+1 1 . n Therefore f (x) = x · n, so that n < f (x) n+1 n = 1. n (1) As x → 0+ , n → ∞, so the bounds on f (x) in (1) both approach 1. Therefore lim f (x) = 1. x→0+ A similar (but slightly more delicate) argument shows that f (x) → 1 as x → 0− as well. Therefore lim f (x) x→0 exists and is equal to 1. The graph of f is next. 13 1.6 1.4 1.2 -1.5 -1 -0.5 0.5 1 1.5 0.8 0.6 0.4 C02S03.074: Here, f (x) is obtained from the function in Problem 73 by multiplication by x. Therefore, because the function in Problem 73 had limit 1 as x → 0, the product rule for limits implies that f (x) → 0 · 1 = 0 as x → 0. The graph of f near zero is next. 0.4 0.2 -0.4 -0.2 0.2 0.4 -0.2 -0.4 -0.6 C02S03.075: Given > 0, let δ = /7. Suppose that 0 < | x − (−3) | < δ. Then |x + 3| < 7 ; | 7x + 21 | < ; | 7x − 9 + 30 | < ; | (7x − 9) − (−30) | < . Therefore, by definition, lim (7x − 9) = −30. x→−3 C02S03.076: Given > 0, let δ = /17. Suppose that 0 < | x − 5 | < δ . Then 14 | 17x − 85 | < 17δ ; | (17x − 35) − 50 | < . Therefore, by definition, lim (17x − 35) = 50. x→5 C02S03.077: Definition: We say that the number L is the right-hand limit of the function f at x = a provided that, for every > 0, there exists δ > 0 such that, if 0 < | x − a | < δ and x > a, then | f (x) − L | < . √ To prove that lim x = 0, suppose that > 0 is given. Let δ = 2 . Suppose that | x − 0 | < δ and x→0+ √ that x > 0. Then 0 < x < δ = 2 . Hence x < , and therefore So, by definition, lim x→0+ √ √ | x − 0| < . x = 0. C02S03.078: Let > 0 be given. Let δ = Hence | x2 − 0 | < . Therefore, by definition, √ . Suppose that 0 < | x − 0 | < δ . Then 0 < x2 < δ 2 = . lim x2 = 0. x→0 C02S03.079: Suppose that > 0 is given. Let δ be the minimum of the two numbers 1 and /5 and suppose that 0 < | x − 2 | < δ . Then | x − 2 | < 1; − 1 < x − 2 < 1; 3 < x + 2 < 5; | x + 2 | < 5. Therefore | x2 − 4 | = | x + 2 | · | x − 2 | < 5 · δ 5· 5 =. Hence, by definition, lim x2 = 4. x→2 C02S03.080: Given > 0, choose δ to be the minimum of 1 and /10. Suppose that 0 < | x − 7 | < δ . Then | x − 7 | < 1; − 1 < x − 7 < 1; 8 < x + 2 < 10; | x + 2 | < 10. Therefore | (x2 − 5x − 4) − 10 | = | x + 2 | · | x − 7 | < 10 · δ 15 10 · 10 =. Thus, by definition, lim (x2 − 5x − 4) = 10. x→7 C02S03.081: Given > 0, let δ be the minimum of 1 and /29. Suppose that 0 < | x − 10 | < δ . Then 0 < | x − 10 | < 1; − 1 < x − 10 < 1; − 2 < 2x − 20 < 2; 25 < 2x + 7 < 29; | 2x + 7 | < 29. Thus | (2x2 − 13x − 25) − 45 | = | 2x + 7 | · | x − 10 | < 29 · δ 29 · 29 =. Therefore, by definition, lim (2x2 − 13x − 25) = 45. x→10 C02S03.082: Given > 0, choose δ to be the minimum of 1 and /19. Suppose that 0 < | x − 2 | < δ . Then 0 < | x − 2 | < 1; − 1 < x − 2 < 1; 1 < x < 3; 1 < x2 < 9 and 2 < 2x < 6; 3 < x2 + 2x < 15; 7 < x2 + 2x + 4 < 19; | x2 + 2x + 4 | < 19. Consequently, | x3 − 8 | = | x2 + 2x + 4 | · | x − 2 | < 19 · δ 19 · 19 =. Therefore, by definition, lim x3 = 8. x→2 C02S03.083: In Problem 78 we showed that if a = 0, then lim x2 = lim x2 = 0 = 02 = a2 , x→a x→0 so the result we are to prove here holds when a = 0. Next case: Suppose that a > 0. Let > 0 be given. Choose δ to be the minimum of the numbers 1 and /(2a +1). Note that δ > 0. Suppose that 0 < | x − a | < δ . Then | x − a | < 1; − 1 < x − a < 1; 16 2a − 1 < x + a < 2a + 1; | x + a | < 2a + 1. Thus | x2 − a2 | = | x + a | · | x − a | < (2a + 1) · =. 2a + 1 Therefore, by definition, lim x2 = a2 if a > 0. x→a Final case: a < 0. Given > 0, let δ = min 1, . | 2a − 1 | Note that δ > 0. Suppose that 0 < | x − a | < δ . Then | x − a | < 1; − 1 < x − a < 1; 2a − 1 < x + a < 2a + 1; | x + a | < | 2a − 1 | (because | 2a − 1 | > | 2a + 1 | if a < 0). It follows that | x2 − a2 | = | x + a | · | x − a | < | 2a − 1 | · | 2a − 1 | =. Therefore, by definition, lim x2 = a2 if a < 0. x→a C02S03.084: Suppose that > 0 is given. Case (1): a = 0. Let δ = solution of Problem 78. Case (2): a > 0. Let δ = min a4 , 2 19a2 . Note that δ > 0. Suppose that 0 < | x − a | < δ . Then: |x − a| < − a ; 2 a a < x−a < ; 2 2 a 3a <x< ; 2 2 a2 9a2 < x2 < 4 4 (because x > 0); a2 3a2 < ax < ; 2 2 3a2 15a2 < x2 + ax < ; 4 4 17 √ 3 and proceed much as in the 7a2 19a2 < x2 + ax + a2 < ; 4 4 | x2 + ax + a2 | < 19a2 . 4 Therefore | x3 − a2 | = | x2 + ax + a2 | · | x − a | < 4 19a2 · =. 4 19a2 Thus, by definition, lim x2 = a2 if a > 0. Case (3), in which a < 0, is similar. x→a 18 Section 2.4 C02S04.001: Suppose that a is a real number. Then lim f (x) = lim (2x5 − 7x2 + 13) = x→a x→a = lim 2 lim x x→a 5 x→a lim 2x5 − lim 7x2 + lim 13 x→a − lim 7 x→a lim x x→a 2 x→a x→a + 13 = 2a5 − 7a2 + 13 = f (a). Therefore f is continuous at x for every real number x. C02S04.002: Suppose that a is a real number. Then lim f (x) = x→a lim 7x3 − lim (2x + 1)5 = x→a lim 7 x→a = 7a3 − lim 2 x→a lim x + lim 1 x→a x→a lim x 5 x→a x→a 3 − lim (2x + 1) 5 x→a = 7a3 − (2a + 1)5 = f (a). Therefore f is continuous at x for every real number x. C02S04.003: Suppose that a is a real number. Then lim g (x) = lim x→a x→a = lim (2x − 1) 2x − 1 = x→a 2 4x2 + 1 lim (4x + 1) x→a lim 2 lim x − lim 1 x→a x→a lim 4 lim x x→a 2 x→a x→a = + lim 1 2a − 1 = g (a). 4a2 + 1 x→a Therefore g is continuous at x for every real number x. C02S04.004: Suppose that a is a fixed real number. Then lim g (x) = x→a limx→a limx→a x3 x2 + 2 limx→a x + 5 3 = (limx→a x) 2 (limx→a x) + 2 limx→a x + 5 = a2 a3 = g (a). + 2a + 5 Therefore g is continuous at x for all real x. C02S04.005: Suppose that a is a fixed real number. Then a2 + 4a + 5 = (a + 2)2 + 1 > 0, so h(a) is defined. Moreover, lim h(x) = lim x→a x→a = x2 + 4x + 5 = lim x x→a 2 lim (x2 + 4x + 5) 1 /2 x→a 1 /2 + lim 4 x→a lim x + lim 5 x→a x→a = a2 + 4a + 5 = h(a). Therefore, by definition, h is continuous at x = a. Because a is arbitrary, h is continuous at x for every real number x. 1 C02S04.006: Suppose that a is a real number. Then h(a) exists because x1/3 is defined for every real number x. Moreover, lim h(x) = lim (1 − 5x)1/3 = x→a x→a = x→a lim 1 − lim 5 x→a 1 /3 lim (1 − 5x) 1 /3 lim x x→a x→a = (1 − 5a)1/3 = h(a). Therefore h is continuous at x = a, and thus continuous at every real number x. C02S04.007: Suppose that a is a real number. Then 1 + cos2 a = 0, so that f (a) is defined. Note also that lim sin x = sin a and x→a lim cos x = cos a x→a because the sine and cosine functions are continuous at every real number (Theorem 1). Moreover, lim f (x) = lim x→a x→a lim (1 − sin x) 1 − sin x = x→a 1 + cos2 x lim (1 + cos2 x) x→a lim 1 − lim sin x x→a = x→a lim 1 + lim cos x x→a = 2 1 − sin a = f (a). 1 + cos2 a x→a Therefore f is continuous at a. Because a is arbitrary, f is continuous at x for every real number x. C02S04.008: Suppose that a is a real number. Then 0 sin2 a 1, so that 1 − sin2 a 0, and thus 2 1 /4 g (a) = (1 − sin a) exists. Because the sine function is continuous on the set of all real numbers (Theorem 1), we know also that sin x → sin a as x → a. Therefore lim g (x) = lim (1 − sin2 x)1/4 = x→a x→a = lim (1 − sin2 x) x→a lim 1 − lim sin x x→a 1 /4 2 1/4 x→a = 1 − sin2 a 1 /4 = g (a). Therefore g is continuous at a for every real number a. C02S04.009: If a > −1, then f (a) exists because a = −1. Moreover, lim f (x) = lim x→a x→a lim 1 1 = x+1 x→a lim x + lim 1 x→a = 1 = f (a). a+1 x→a Therefore f is continuous on the interval x > −1. C02S04.010: If −2 < a < 2, then f (a) exists because a2 − 4 = 0. Moreover, lim f (x) = lim x→a x→a x−1 = x2 − 4 lim x − lim 1 x→a lim x x→a Therefore f is continuous at x if −2 < x < 2. 2 2 x→a − lim 4 x→a = a−1 = f (a). a2 − 4 C02S04.011: Because − 3 t 3 , 0 4t2 9, so that the radicand in g (t) is never negative. Therefore 2 2 g (a) is defined for every real number a in the interval − 3 , 3 , and 22 lim g (t) = lim (9 − 4t2 )1/2 = lim (9 − 4t2 ) t→a t→a = t→a lim 9 − 4 lim t t→a 2 1 /2 t→a = (9 − 4a2 )1/2 = g (a). Therefore g is continuous at a for every real number a in − 3 , 2 C02S04.012: If 1 z 3, then 0 z − 1 2, −3 −z is nonnegative for such values of z , and therefore if 1 a 1 /2 lim h(z ) = lim [(z − 1)(3 − z ) ] z →a z →a = 3 2 . −1, and 0 3 then = lim (z − 1)(3 − z ) 3−z 2. So the radicand in h(z ) 1/2 z →a lim z − lim 1 z →a 1 /2 lim 3 − lim z z →a z →a 1 /2 z →a 1 /2 = [ (a − 1)(3 − a) ] = h(a). Therefore h is continuous at a for all real numbers a in the interval [1, 3]. C02S04.013: If − 1 π < x < 1 π , then cos x = 0, so f (x) is defined for all such x. In addition, cos x → cos a 2 2 as x → a because the cosine function is continuous everywhere (Theorem 1). Therefore lim f (x) = lim x→a x→a lim x x a = x→a = = f (a). cos x lim cos x cos a x→a Therefore f is continuous at x if − 1 π < x < 1 π . 2 2 C02S04.014: If − 1 π < t < 1 π , then − 1 < sin t < 1 , so that 1 − 2 sin t > 0 for such values of t. Therefore 6 6 2 2 g (a) is defined if − 1 π < a < 1 π . Moreover, for such values of a, we have 6 6 lim g (t) = lim (1 − 2 sin t)1/2 = lim (1 − 2 sin t) t→a t→a = lim 1 − lim 2 t→a 1/2 t→a t→a lim sin t t→a 1/2 = (1 − 2 sin a)1/2 = g (a). Therefore g is continuous at a for each real number a in the interval − 1 π , 1 π . 6 6 √ C02S04.015: The root law of Section 2.2 implies that g (x) = 3 x is continuous on the set R of all real numbers. We know that the polynomial h(x) = 2x is continuous on R (Section 2.4, page 88). Hence the sum f (x) = h(x) + g (x) is continuous on R. C02S04.016: The polynomial f (x) = x2 is continuous on R (the set of all real numbers) and the quotient h(x) = 1 x of continuous functions is continuous where its denominator is not zero. Hence the sum g (x) = f (x) + h(x) is continuous on its domain, the set of all nonzero real numbers. C02S04.017: Because f (x) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), f is continuous wherever its denominator is nonzero. Therefore f is continuous on its domain, the set of all real numbers other than −3. 3 C02S04.018: Because f (t) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), f is continuous wherever its denominator is nonzero. Therefore f is continuous on its domain, the set of all real numbers other than 5. C02S04.019: Because f (x) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), f is continuous wherever its denominator is nonzero. Therefore f is continuous on its domain, the set of all real numbers. C02S04.020: Because g (z ) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), g is continuous wherever its denominator is nonzero. Therefore g is continuous on its domain, the set of all real numbers other than ± 1. C02S04.021: Note that f (x) is not defined at x = 5, so it is not continuous there. Because f (x) = 1 for x > 5 and f (x) = −1 for x < 5, f is a polynomial on the interval (5, + ∞) and a [another] polynomial on the interval (−∞, 5). Therefore f is continuous on its domain, the set of all real numbers other than 5. C02S04.022: Because h(x) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), h is continuous wherever its denominator is nonzero. Therefore h is continuous on its domain, the set of all real numbers. C02S04.023: Because f (x) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), f is continuous wherever its denominator is nonzero. Therefore f is continuous on its domain, the set of all real numbers other than 2. C02S04.024: Because g (t) = 4 + t4 is a polynomial, it is continuous everywhere (and never negative). √ Because h(t) = 4 t is a root function, it is continuous wherever t 0. Therefore (by Theorem 2) the composition f (t) = h(g (t)) is continuous everywhere. C02S04.025: Let h(x) = x+1 . x−1 Because h(x) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), h is continuous wherever its denominator is nonzero. Therefore h is continuous on √ its domain, the set of all real numbers other than 1. Now let g (x) = 3 x. By the root rule of Section 2.2, g is continuous everywhere. Therefore the composition f (x) = g (h(x)) is continuous on the set of all real numbers other than 1. √ C02S04.026: Here, F (u) = g (h(u)) where g (u) = 3 u and h(u) = 3 − u3 . Now g is continuous everywhere by the root rule of Section 2.2; h is continuous everywhere because h(u) is a polynomial. Therefore the composition F (u) = g (h(u)) of continuous functions is continuous where defined; namely, on the set R of all real numbers. C02S04.027: Because f (x) is the quotient of continuous functions (the numerator and denominator are polynomials, continuous everywhere), f is continuous wherever its denominator is nonzero. Therefore f is continuous on its domain, the set of all real numbers other than 0 and 1. C02S04.028: The domain of f is the interval −3 z 3, and on that domain f (z ) is the composition of continuous functions, thus f is continuous there. Because f (z ) is not defined if | z | > 3, it is not continuous for z < −3 nor for z > 3. But it is still correct to say simply that “f is continuous” (see the definition of continuous on page 88). 4 C02S04.029: Let h(x) = 4 − x2 . Then h is continuous everywhere because h(x) is a polynomial. The root √ √ function g (x) = x is continuous for x 0 by the root rule of Section 2.2. Hence g (h(x)) = 4 − x2 is continuous wherever x2 4; that is, on the interval [−2, 2]. The quotient f (x) = √ x x = 2 g (h(x)) 4−x is continuous wherever the numerator is continuous (that’s everywhere) and the denominator is both continuous and nonzero (that’s the open interval (−2, 2)). Therefore f is continuous on the open interval (−2, 2). That is, f is continuous on its domain. C02S04.030: Because f (x) is formed by the composition and quotient of continuous functions (polynomials and root functions), it will be continuous wherever the denominator in the fraction is nonzero and the fraction is nonnegative. So continuity of f will occur when both 4 − x2 = 0 and 1 − x2 4 − x2 0. The first inequality is equivalent to x = ±2 and the second will hold when 1 − x2 and 4 − x2 have the same sign (both positive or both negative) or the numerator is zero. If both are positive, then x2 < 1 and x2 < 4, so that −1 < x < 1. If both are negative, then x2 > 1 and x2 > 4, so that x2 > 4; that is, x < −2 or x > 2. Finally, 1 − x2 = 0 when x = ±1. Therefore f is continuous on its domain, (−∞, −2) ∪ [−1, 1] ∪ (2, + ∞). C02S04.031: Because f (x) is the quotient of continuous functions, it is continuous where its denominator is nonzero; that is, if x = 0. Thus f is continuous on its domain and not continuous at x = 0 (because it is undefined there). C02S04.032: Given: g (θ) = θ . cos θ Because g (θ) is the quotient of continuous functions, it is continuous wherever its denominator is nonzero; that is, at every real number x not an odd integral multiple of π /2. That is, g is discontinuous (because it is undefined) at ... , − 5π 3π π π 3π 5π 7π ,− ,− , , , , , ... . 2 2 222 2 2 Therefore g is continuous on its domain, the set · · · ∪ − 5 π, − 3 π ∪ − 3 π, − 1 π ∪ − 1 π, 1 π ∪ 2 2 2 2 2 2 1 3 2 π, 2 π ∪ 3 5 2 π, 2 π ∪ 5 7 2 π, 2 π ∪ ··· . C02S04.033: Given: f (x) = 1 . sin 2x The numerator in f (x) is a polynomial, thus continuous everywhere. The denominator is the composition of a function continuous on the set of all real numbers (the sine function) with another continuous function (a polynomial), hence is also continuous everywhere. Thus because f (x) is the quotient of continuous functions, it is continuous wherever its denominator is nonzero; that is, its only discontinuities occur when sin 2x = 0. Thus f is continuous at every real number other than an integral multiple of π /2. 5 √ C02S04.034: Because f (x) = sin x is the composition of continuous functions, it is continuous wherever it is defined; that is, wherever sin x 0. Hence f is continuous on its domain, the set · · · ∪ [−4π , −3π ] ∪ [−2π , −π ] ∪ [0, π ] ∪ [2π , 3π ] ∪ [4π , 5π ] ∪ · · · . C02S04.035: Given: f (x) = sin | x |. The sine function is continuous on the set of all real numbers, as is the absolute value function. Therefore their composition f is continuous on the set R of all real numbers. C02S04.036: Given: G(u) = √ 1 . 1 + cos u Because G(u) is the sum, composition, and quotient of continuous functions, it is continuous where it is √ defined. There is no obstruction to computing 1 + cos u because 1 + cos u 0 for every real number u. Hence G will be undefined, and thus not continuous, exactly when its denominator is zero, which is exactly when 1 + cos u = 0. Therefore G is continuous except at the odd integral multiples of π . Put another way, G is continuous on the union of open intervals of the form ([2n − 1]π , [2n + 1]π ) where n runs through all integral values. C02S04.037: The function x (x + 3)3 f (x) = is not continuous when x = −3. This discontinuity is not removable because f (x) → −∞ as x → −3+ , so that the limit of f (x) at x = −3 does not exist. C02S04.038: The function f (t) = t2 t −1 is not continuous when t = ± 1 because t2 − 1 = 0 then. These discontinuities are not removable because f (t) → + ∞ as t → 1+ and as t → −1+ , so that f has no limit at either t = 1 or t = −1. C02S04.039: First simplify f (x): f (x) = x−2 x−2 1 = = x2 − 4 (x + 2)(x − 2) x+2 if x = 2. Now f (x) is not defined at x = ± 2 because x2 − 4 = 0 for such x. The discontinuity at −2 is not removable because f (x) → + ∞ as x → −2+ . But f (x) → 1 as x → 2, so the discontinuity at x = 2 is removable; f 4 can be made continuous at x = 2 by defining its value there to be its limit there, 1 . 4 C02S04.040: First try to simplify the formula of G: G(u) = u+1 u+1 = . u2 − u − 6 (u − 3)(u + 2) This computation shows that G is not continuous at u = 3 and at u = −2. It also shows that these discontinuities are not removable because G(u) → + ∞ as u → 3+ and as u → −2+ , so that G has no limit at either of its discontinuities. 6 C02S04.041: Given: f (x) = 1 . 1 − |x| The function f is not continuous at ± 1 because its denominator is zero if x = −1 and if x = 1. Because f (x) → + ∞ as x → 1− and as x → −1+ (consider separately the cases x > 0 and x < 0), these discontinuities are not removable; f (x) has no limit at −1 or at 1. C02S04.042: If x > 1, then h(x) = |x − 1| x−1 1 = = . 3 3 (x − 1) (x − 1) (x − 1)2 Therefore h is discontinuous at x = 1 and, because h(x) → + ∞ as x → 1+ , this discontinuity is not removable. C02S04.043: If x > 17, then x − 17 > 0, so that f (x) = x − 17 x − 17 = = 1. | x − 17 | x − 17 But if x < 17, then x − 17 < 0, and thus f (x) = x − 17 x − 17 = = −1. | x − 17 | −(x − 17) Therefore h(x) has no limit as x → 17 because its left-hand and right-hand limits there are unequal. Thus the discontinuity at x = 17 is not removable. C02S04.044: First simplify: g (x) = x2 + 5x + 6 (x + 2)(x + 3) = =x+3 x+2 x+2 if x = −2. Therefore, although g is discontinuous at x = −2 (because it is not defined there), this discontinuity is removable; simply define g (−2) to be 1, the limit of g (x) as x → −2. C02S04.045: Although f (x) is not continuous at x = 0 (because it is not defined there), this discontinuity is removable. For it is clear that f (x) → 0 as x → 0+ and as x → 0− , so defining f (0) to be 0, the limit of f (x) at x = 0, will make f continuous there. C02S04.046: Although f (x) is not continuous at x = 1 (because it is not defined there), this discontinuity is removable. For it is clear that f (x) → 2 as x → 1+ and as x → 1− , so defining f (1) to be 2, the limit of f (x) at x = 1, will make f continuous there. C02S04.047: Although f (x) is not continuous at x = 0 (because it is not defined there), this discontinuity is removable. For it is clear that f (x) → 1 as x → 0+ and as x → 0− , so defining f (0) to be 1, the limit of f (x) at x = 0, will make f continuous there. C02S04.048: Although f (x) is not continuous at x = 0 (because it is not defined there), this discontinuity is removable. For lim f (x) = lim − − x→0 x→0 1 − cos x sin2 x sin x sin x 0 = lim− = lim− · =1· = 0, x x 1 + cos x 1+1 x→0 x(1 + cos x) x→0 7 and it is clear that f (x) → 0 as x → 0+ . So defining f (0) to be 0, the limit of f (x) at x = 0, will make f continuous there. C02S04.049: The given function is clearly continuous for all x except possibly for x = 0. For continuity at x = 0, the left-hand and right-hand limits of f (x) must be the same there. But lim f (x) = lim (x + c) = c x→0− x→0− and f (x) → 4 as x → 0+ . So continuity of f at x = 0 can occur only if c = 4. Moreover, if c = 4, then (as we have seen) f (x) → 4 as x → 0 and f (0) = 4, so f will be continuous at x = 0 if and only if c = 4. Answer: c = 4. C02S04.050: Clearly f is continuous if x = 3, for if x < 3 or if x > 3, then f (x) is a polynomial, regardless of the value of c. For continuity at x = 3, we require that the one-sided limits of f (x) at x = 3 be equal. But f (x) → 6 + c as x → 3− and f (x) → 2c − 3 as x → 3+ . Equality of the one-sided limits is equivalent to 6 + c = 2c − 3; that is, c = 9. Finally, if c = 9, then the two-sided limit of f (x) at x = 3 is 15 and f (3) = 2 · 3 + 9 = 15, so f will be continuous at x = 3 if c = 9. Answer: c = 9. C02S04.051: Note that f is continuous at x if x = 0, because f (x) is a polynomial for x < 0 and for x > 0 regardless of the value of c. To be continuous at x = 0, it’s necessary that the left-hand and right-hand limits exist and are equal there. Now lim f (x) = lim (c2 − x2 ) = c2 x→0− x→0− and lim f (x) = lim 2(x − c)2 = 2c2 , x→0+ x→0+ and therefore continuity at x = 0 will hold if and only if c2 = 2c2 ; that is, if c = 0. And if so, then f (0) = lim f (x) as well, so f will be continuous at x = 0. Answer: c = 0. x→0 C02S04.052: Note that f is continuous if x < π because f (x) is a polynomial for such x; also, f is continuous for x > π because (regardless of the value of c) f (x) is a constant multiple of a continuous function. For continuity of f at x = π , the left-hand and right-hand limits must be equal there. But lim f (x) = lim (c3 − x3 ) = c3 − π 3 x→π − x→π − and lim f (x) = lim c sin x = lim c sin π = 0. x→π + x→π + x→π + So continuity of f at x = π requires c3 −π 3 = 0; that is, c = π . And if so, then f (π ) = π 3 −π 3 = 0 = lim f (x), x→π so f will be continuous at x = π . Answer: c = π . C02S04.053: Let f (x) = x2 − 5. Then f is continuous everywhere because f (x) is a polynomial. So f has the intermediate value property on the interval [2, 3]. Also f (2) = −1 < 0 < 4 = f (3), so f (c) = 0 for some number c in [2, 3]. That is, c2 − 5 = 0. Hence the equation x2 − 5 = 0 has a solution in [2, 3]. C02S04.054: Let f (x) = x3 + x + 1. Then f is continuous everywhere because f (x) is a polynomial. So f has the intermediate value property on the interval [−1, 0]. Also f (−1) = −1 < 0 < 1 = f (0), so f (c) = 0 for some number c in [−1, 0]. That is, c3 + c + 1 = 0. Hence the equation x3 + x + 1 = 0 has a solution in [−1, 0]. C02S04.055: Let f (x) = x3 − 3x2 + 1. Then f is continuous everywhere because f (x) is a polynomial. So f has the intermediate value property on the interval [0, 1]. Also f (0) = 1 > 0 > −1 = f (1), so f (c) = 0 for 8 some number c in [0, 1]. That is, c3 − 3c2 + 1 = 0. Hence the equation x3 − 3x2 + 1 = 0 has a solution in [0, 1]. C02S04.056: Let f (x) = x3 − 5. Then f is continuous everywhere because f (x) is a polynomial. So f has the intermediate value property on the interval [1, 2]. Also f (1) = −4 < 0 < 3 = f (2), so f (c) = 0 for some number c in [1, 2]. That is, c3 − 5 = 0. Hence the equation x3 = 5 has a solution in [1, 2]. C02S04.057: Let f (x) = x4 + 2x − 1. Then f is continuous everywhere because f (x) is a polynomial. So f has the intermediate value property on the interval [0, 1]. Also f (0) = −1 < 0 < 2 = f (1), so f (c) = 0 for some number c in [0, 1]. That is, c4 + 2c − 1 = 0. Hence the equation x4 + 2x − 1 = 0 has a solution in [0, 1]. C02S04.058: Let f (x) = x5 − 5x3 + 3. Then f is continuous everywhere because f (x) is a polynomial. So f has the intermediate value property on the interval [−3, −2]. Also f (−3) = −105 < 0 < 11 = f (−2), so f (c) = 0 for some number c in [−3, −2]. That is, c5 − 5c3 + 3 = 0. Hence the equation x5 − 5x3 + 3 = 0 has a solution in [−3, −2]. C02S04.059: Given: f (x) = x3 − 4x + 1. Values of f (x): x f (x) −3 −14 −2 1 −1 4 0 1 1 −2 2 1 3 16 So f (xi ) = 0 for x1 in (−3, −2), x2 in (0, 1), and x3 in (1, 2). Because these intervals do not overlap, the equation f (x) = 0 has at least three real solutions. Because f (x) is a polynomial of degree 3, that equation also has at most three real solutions. Therefore the equation x3 − 4x + 1 = 0 has exactly three real solutions. C02S04.060: Given: f (x) = x3 − 3x2 + 1. Values of f (x): x f (x) −3 −2 −53 −19 −1 −3 0 1 1 −1 2 −3 3 1 So f (xi ) = 0 for x1 in (−1, 0), x2 in (0, 1), and x3 in (2, 3). Because these intervals do not overlap, the equation f (x) = 0 has at least three real solutions. Because f (x) is a polynomial of degree 3, that equation also has at most three real solutions. Therefore it has exactly three real solutions. C02S04.061: At time t, [[t]] years have elapsed, and at that point your starting salary has been multiplied by 1.06 exactly t times. Thus it is S (t) = 25 · (1.06)[[t]] . Of course S is discontinuous exactly when t is an integer between 1 and 5. The graph is next. 34 32 30 28 26 24 22 1 2 3 4 9 5 C02S04.062: The salary function is P (t) = 25 · (1.015)[[4t]] . It is discontinuous at the end of every threemonth period; that is, at integral multiples of 1 . You will accumulate more money with three-month raises 4 of 1.5% than with yearly raises of 6%; the total salary received by the end of the first five years with yearly raises would be $140,930 but with quarterly raises it would be $144,520. The graphs of the function S (t) of Problem 61 and P (t) are shown next. Although the graphs do not make it perfectly clear, it turns out that P (t) > S (t) if t 1 , so that the quarterly raise is better for you financially after the first three months and 4 continues to outpace the yearly raise as long as you keep the job. 34 32 30 28 26 24 22 1 2 3 4 5 C02S04.063: The next figure shows the graphs of two such functions f and g , with [a, b] = [1, 3], p = 2, and q = 5. Because f and g are continuous on [a, b], so is h = f − g . Because p = q , h(a) = p − q and h(b) = q − p have opposite signs, so that 0 is an intermediate value of the continuous function h. Therefore h(c) = 0 for some number c in (a, b). That is, f (c) = g (c). This concludes the proof. To construct the figure, we used (the given coefficients are approximate) f (x) = 1.53045 + (0.172739)ex and g (x) = 2.27857 + (2.05)x + (1.36429)x2 − (0.692857)x3 . 7 6 g a b 5 q 4 f 3 p 2 1 -1 1 2 3 4 C02S04.064: Let f (t) denote your distance from Estes Park during your trip today, with f measured in kilometers and t in hours, t = 0 corresponding to time 1 p.m. Let g (t) denote your distance from Estes Park during your trip tomorrow, with g in kilometers, t in hours, and t = 0 corresponding to 1 p.m. tomorrow. Assuming that both f and g are continuous, we use the facts that f (0) = 0, f (1) = M (where M is the distance from Estes Park to Grand Lake), g (0) = M , and g (1) = 0 and apply the result of Problem 63 to conclude that f (c) = g (c) for some number c in (0, 1). That is, at time t = c tomorrow you will be at exactly the same spot (at distance g (c) from Estes Park) as you will be at the same time t = c today at distance f (c) = g (c) from Estes Park. 10 The 1999 National Geographic Road Atlas indicates that M ≈ 101 (km). Making this trip in a single hour is unforgivable given the magnificent scenery (and probably impossible as well given the dozens of tight turns on the highway). C02S04.065: Given a > 0, let f (x) = x2 − a. Then f is continuous on [0, a + 1] because f (x) is a polynomial. Also f (a + 1) > 0 because f (a + 1) = (a + 1)2 − a = a2 + a + 1 > 1 > 0. So f (0) = −a < 0 < f (a + 1). Therefore, because f has the intermediate value property on the interval [0, a + 1], there exists a number r in (0, a + 1) such that f (r) = 0. That is, r2 − a = 0, so that r2 = a. Therefore a has a square root. Our proof shows that a has a positive square root. Can you modify it to show that a also has a negative square root? Do you see why we used the interval [0, a + 1] rather than the simpler [0, a]? C02S04.066: Clearly a = 0 has a cube root. Suppose first that a > 0. Let f (x) = x3 − a. Then f has the intermediate value property on [0, a + 1] because f (x) is a polynomial. Moreover, f (a + 1) = (a + 1)3 − a = a3 + 3a2 + 2a + 1 > 1 > 0 > −a = f (0). Therefore there exist a number c in [0, a + 1] such that f (c) = 0. That is, c3 − a = 0, so that c3 = a. Thus the positive real number a has a cube root. Moreover, (−c)3 = −(c3 ) = −a, so that every negative real number has a cube root as well. C02S04.067: Given the real number a, we need to show that lim cos x = cos a. x→a Let h = x − a, so that x = a + h. Then x → a is equivalent to h → 0; also, cos x = cos(a + h). Thus lim cos x = lim cos(a + h) = lim (cos a cos h − sin a sin h) = (cos a) · 1 − (sin a) · 0 = cos a. x→a h→0 h→0 Therefore the cosine function is continuous at x = a for every real number a. C02S04.068: If x is not an integer, choose that [unique] integer n = [[x]] such that n < x < n + 1. Then f (x) = x + n on the interval (n, n + 1), thus f (x) is effectively a polynomial on that interval. So f is continuous at x. But if m is an integer, then lim f (x) = lim (x + [[x]]) = lim (x + m − 1) = m + m − 1 = 2m − 1, − − x→m− x→m x→m whereas lim f (x) = lim (x + [[x]]) = lim (x + m) = m + m = 2m. + + x→m+ x→m x→m Because the left-hand and right-hand limits of f (x) differ at m, f is not continuous there. Thus f is discontinuous at each integer and continuous at every other real number. C02S04.069: Suppose that a is a real number. We appeal to the formal definition of the limit in Section 2.2 (page 74) to show that f (x) has no limit as x → a. Suppose by way of contradiction that f (x) → L as x → a. Then, for every > 0, there exists a number δ > 0 such that | f (x) − L | < for every number x such that 0 < | x − a | < δ . So this statement must hold if = 1 . 4 11 Case 1: L = 0. Then there must exist a number δ > 0 such that | f (x) − 0 | < 1 4 if 0 < | x − a | < δ . But, regardless of the value of δ , there exist irrational values of x satisfying this last inequality. (We’ll explain why in a moment.) Choose such a number x. Then | f (x) − 0 | = | 1 − 0 | = 1 < 1 . 4 This is impossible. So L = 0. Case 2: L = 1. Proceed exactly as in Case 1, except choose a rational value of x such that 0 < | x − a | < δ . Then | f (x) − 1 | = | 0 − 1 | = 1 < 1 . 4 This, too, is impossible. So L = 1. Case 3: L is neither 0 nor 1. Let that, if = 1 | L |. Note that 3 0 < | x − a | < δ, then > 0. Then suppose that there exists δ > 0 such | f (x) − L | < . (1) Choose a rational number x satisfying the left-hand inequality. Then | f (x) − L | = | 0 − L | = | L | = 3 . It follows from (1) that 3 < , which is impossible because > 0. In summary, L cannot be 0, nor can it be 1, nor can it be any other real number. Therefore f (x) has no limit as x → a. Consequently f is not continuous at x = a. In this proof we relied heavily on the fact that if a is any real number, then we can find both rational and irrational numbers arbitrarily close to a. Rather than providing a formal proof, we illustrate how to do this in the case that a = 1.23456789101112131415 · · · . (It doesn’t matter whether a is rational or irrational.) To produce rational numbers arbitrarily close to a, use 1.2, 1.23, 1.234, 1.2345, 1.23456, 1.234567, . . . . (2) The numbers in (2) are all rational because they all have terminating decimal expansions, and the n th number in (2) differs from a by less than 10−n , so there are rational numbers arbitrarily close to a. To get irrational numbers with the same properties, use 1.20100100010000100000100 · · · , 1.230100100010000100000100 · · · , 1.2340100100010000100000100 · · · , 1.23450100100010000100000100 · · · , . . . . These numbers are irrational because every one of them has a nonrepeating decimal expansion. C02S04.070: You can modify the argument in the solution of Problem 69 to show that f (x) has no limit at x = a if a = 0. Simply use a2 in place of 1 in that argument. Because 0 f (x) x2 for all x, and 2 because 0 → 0 and x → 0 as x → 0, it follows from the squeeze theorem that 12 lim f (x) = 0 = f (0). x→0 Therefore f is continuous at x = 0. C02S04.071: Let g (x) = x − cos x. Then g (x) is the sum of continuous functions, thus continuous everywhere, and in particular on the interval 0, 1 π . So g has the intermediate value property there. Also 2 g (0) = −1 < 0 and g (1) = 1 > 0. Therefore there exists a number c in 0, 1 π such that g (c) = 0. Thus 2 c − cos x = 0, so that c = cos c. Thus the equation x = cos x has a solution in 0, 1 π . (This solution is 2 approximately 0.7390851.) C02S04.072: Let h(x) = x + 5 cos x. Then h(x) is the sum and product of continuous functions, thus is continuous everywhere, including the intervals [−π , 0], [0, π ], and [π , 2π ]. Moreover, h(−π ) = −π − 5 < 0 < 5 = h(0), and h(0) = 5 > 0 > π − 5 = h(π ), h(π ) = π − 5 < 0 < 2π + 5 = h(2π ). By the intermediate value property of continuous functions, h(a) = 0 for some number a in I = (−π , 0), h(b) = 0 for some number b in J = (0, π ), and h(c) = 0 for some number c in K = (π , 2π ). Because no two of I , J , and K have any points in common, the numbers a, b, and c are distinct. Therefore the equation h(x) = 0 three distinct solutions; in other words, the equation x = −5 cos x has three distinct solutions. (We have not shown that there are no additional solutions, but this was not required.) Finally, a ≈ −1.30644, b ≈ 1.97738, and c ≈ 3.83747. C02S04.073: Because lim 21/x = + ∞, x→0+ f is not right continuous at x = 0. Because lim 21/x = lim 2u = lim u→−∞ x→0− z →+∞ 1 = 0 = f (0), 2z f is left continuous at x = 0. C02S04.074: Because 2 lim 2−1/x = 0 = f (0), x→0 the function f is both left and right continuous—thus continuous—at x = 0. C02S04.075: Because lim x→0+ 1 = 0 = f (0), 1 + 21/x f is not right continuous at x = 0. But lim x→0− 1 1 = = 1 = f (0), 1+0 1 + 21/x f is left continuous at x = 0. 13 C02S04.076: Because 1 1 = = 1 = f (0), −1/x2 1+0 1+2 lim x→0 the function f is both left and right continuous at x = 0—thus it is continuous there. C02S04.077: We consider only the discontinuity at x = a = π /2; the behavior of f is the same near all of its discontinuities (the odd integral multiples of a). Because lim x→a+ 1 1 = = 1 = f (1), 1 + 2tan x 1+0 the function f is right continuous at x = a. But 1 = 0 = f (0), 1 + 2tan x lim x→a− so f is not left continuous at x = a. C02S04.078: We consider only the discontinuities at x = 0 and x = π , because the behavior of f at every even integral multiple of π is the same as its behavior at x = 0, and its behavior at every odd integral multiple of π is the same as its behavior at x = π . We first note that lim x→0+ 1 = 0 = f (0), 1 + 21/ sin x so that f is right continuous at x = 0. But lim x→0− 1 1+ 21/ sin x = 1 = f (0), and thus f is not left continuous at x = 0. The situation is reversed at π , as one might gather from examining the graph of the sine function: lim x→π − 1 = 0 = f (π ), 1 + 21/ sin x so that f is left continuous at x = π , but lim x→π + 1 = 1 = f (π ). 1 + 21/ sin x Therefore f is not right continuous at x = π . 14 Chapter 2 Miscellaneous Problems 2 C02S0M.001: lim (x2 − 3x + 4) = lim x − 3 · lim x + 4 = 02 − 3 · 0 + 4 = 4. x→0 x→0 x→0 3 C02S0M.002: lim (3 − x + x3 ) = 3 − lim x + x→−1 x→−1 C02S0M.003: lim (4 − x2 )10 = 4 − lim x x→2 2 10 x→2 10 = 010 = 0. 17 x→1 1 + lim x 1 + x2 x→2 C02S0M.005: lim = x→2 1 − x2 1 − lim x = (12 + 1 − 1)17 = 1. x→1 2 2 = x→2 x→3 = 4 − 22 lim x + lim x − 1 x→1 2x = x2 − x − 3 = 3 − (−1) + (−1)3 = 3. x→−1 2 C02S0M.004: lim (x2 + x − 1)17 = C02S0M.006: lim lim x 1 + 22 1+4 5 = =− . 2 1−2 1−4 3 2 · lim x 2 x→3 = lim x − lim x − 3 x→3 x→3 2·3 6 = = 2. 32 − 3 − 3 3 C02S0M.007: x2 − 1 (x + 1)(x − 1) =− = −(x + 1) → −2 as x → 1. 1−x x−1 C02S0M.008: x+2 x+2 1 1 = = → − as x → −2. x2 + x − 2 (x + 2)(x − 1) x−1 3 C02S0M.009: t 2 + 6t + 9 (t + 3)2 t+3 −3 + 3 =− →− = 0 as t → −3. =− 9 − t2 (t + 3)(t − 3) t−3 −3 − 3 C02S0M.010: 4x − x3 4 − x2 4−0 4 x(4 − x2 ) = → = as x → 0. = 2 3x + x x(3 + x) 3+x 3+0 3 lim x − 1 x→3 C02S0M.012: lim x→2 C02S0M.013: lim x→3 C02S0M.014: 2x + 1 2x 2 5x + 1 x2 − 8 = (32 − 1)2/3 = 82/3 = 81/3 x→3 1/2 = 3 /4 2 /3 2 C02S0M.011: lim (x2 − 1)2/3 = = 2 2 · lim x + 1 x→2 /2 1 2 · lim x 5 · lim x + 1 /4 3 x→2 x→3 2 lim x − 8 x→3 = 2·4+1 2·2 = 161/4 3 1 /2 = 9 4 2 = 22 = 4. 1/2 = 3 . 2 = 8. x4 − 1 (x2 + 1)(x + 1)(x − 1) (x2 + 1)(x + 1) 2·2 = = → = 1 as x → 1. 2 + 2x − 3 x (x + 3)(x − 1) x+3 4 C02S0M.015: First multiply numerator and denominator by to obtain 1 √ x + 2 + 3 (the conjugate of the numerator) lim √ x→7 x+2−3 x+2−9 x−7 √ √ = lim = lim x→7 (x − 7) x→7 (x − 7) x−7 x+2+3 x+2+3 = lim √ x→7 lim 1 1 x √→7 = = x+2+3 lim x+2+3 x→7 1 2 + lim x x→7 C02S0M.016: Note that x > 1 as x → 1+ , so that √ lim x→1+ x− −1 = lim x − x→1+ lim x x→1+ = + lim 3 1 1 =. 3+3 6 x→7 x2 − 1 is defined for such x. Therefore 1 /2 2 x2 1 /2 − lim 1 x→1+ =1− 12 − 1 = 1 − 0 = 1. C02S0M.017: First simplify: 1 1 √ √ − 1 3 − 13 + x 3 − 13 + x 13 + x 3 √ = ·√ = . x+4 x+4 3 13 + x 3(x + 4) 13 + x √ Then multiply numerator and denominator by 3 + 13 + x, the conjugate of the numerator, to obtain √ 3(x + 4) 9 − (13 + x) √ √ 13 + x 3 + 13 + x = Now let x → −4 to obtain the limit − 3(x + 4) √ −(x + 4) √ 13 + x 3 + 13 + x =− 3 √ 13 + x 1 √ . 3 + 13 + x 1 1 =− . 3 · 3 · (3 + 3) 54 C02S0M.018: Because x → 1+ , x > 1, so that 1 − x < 0. Therefore lim x→1+ 1−x 1−x = lim = lim (−1) = −1. + −(1 − x) |1 − x| x→1 x→1+ C02S0M.019: First, 4 − 4x + x2 = (2 − x)2 = (x − 2)2 . Because x → 2+ , x > 2, so that x − 2 > 0. Hence √ 4 − 4x + x2 = (x − 2)2 = | x − 2 | = x − 2. Therefore lim √ x→2+ 2−x 2−x = lim+ = lim+ (−1) = −1. 2 x→2 x − 2 x→2 4 − 4x + x C02S0M.020: As x → −2− , x < −2, so that x + 2 < 0. Hence | x + 2 | = −(x + 2). Thus lim x→−2− x+2 x+2 = lim − = lim − (−1) = −1. |x + 2| x→−2 −(x + 2) x→−2 C02S0M.021: As x → 4+ , x > 4, so that x − 4 > 0. Therefore | x − 4 | = x − 4, and thus lim x→4+ x−4 x−4 = lim+ = lim+ 1 = 1. |x − 4| x→4 x − 4 x→4 C02S0M.022: As x → 3− , x < 3, so that x2 − 9 < 0 for −3 < x < 3. Therefore such x, and consequently lim x2 − 9 does not exist. x→3− 2 √ x2 − 9 is undefined for C02S0M.023: As x → 2+ , x > 2, so that 4 − x2 < 0. Therefore consequently lim 4 − x2 does not exist. √ 4 − x2 is undefined for all such x, and x→2+ C02S0M.024: As x → −3, (x + 3)2 → 0 whereas the numerator x approaches −3. Therefore this limit does not exist. Because (x + 3)2 is approaching zero through positive values, it is also correct to write x = −∞. (x + 3)2 lim x→−3 C02S0M.025: As x → 2, the denominator (x − 2)2 is approaching zero, while the numerator x + 2 is approaching 4. So this limit does not exist. Because the denominator is approaching zero through positive values, it is also correct (and more informative) to write lim x→2 x+2 = + ∞. (x − 2)2 C02S0M.026: As x → 1− , the denominator x − 1 is approaching zero, but the numerator x is not. Therefore this limit does not exist. Because the numerator is approaching 1 and the denominator is approaching zero through negative values, it is also correct to write lim x→1− x = −∞. x−1 C02S0M.027: Because x → 3+ , the denominator x − 3 is approaching zero, but the numerator x is not. Therefore this limit does not exist. Because the denominator is approaching zero through positive values while the numerator is approaching 3, it is also correct to write lim x→3+ x = + ∞. x−3 C02S0M.028: Because x2 x−2 x−2 1 = = − 3x + 2 (x − 1)(x − 2) x−1 if x = 2, the limit of this fraction as x → 1− does not exist: The numerator is approaching 1 while the denominator is approaching zero. Because the denominator is approaching zero through negative values, it is also correct to write lim x→1− x−2 = −∞. x2 − 3x + 2 C02S0M.029: As x → 1− , the numerator of the fraction is approaching 2, but the denominator is approaching zero. Therefore this limit does not exist. Because the denominator is approaching zero through negative values, it is also correct to write lim x→1− x+1 = −∞. (x − 1)3 C02S0M.030: Note first that 3 25 − x2 (5 + x)(5 − x) 5+x (5 + x)(5 x) = = = . 2 − 10x + 25 2 2 x (x − 5) (5 − x) 5−x Thus as x → 5+ , the numerator approaches 10 while the denominator is approaching zero through negative values. Therefore this limit does not exist. It is also correct to write lim x→5+ x2 25 − x2 = −∞. − 10x + 25 C02S0M.031: Let u = 3x. Then x = 1 u; also, x → 0 is equivalent to u → 0. Thus 3 lim x→0 sin 3x sin u 3 sin u = lim 1 = lim = u→0 u→0 x u u 3 lim 3 · lim u →0 u →0 sin u u = 3 · 1 = 3. C02S0M.032: Let u = 5x; then x = 1 u; moreover, x → 0 is equivalent to u → 0. Therefore 5 lim x→0 tan 5x tan u 5 sin u = lim = lim 1 =5· u→0 u→0 u cos u x u 5 lim u→0 sin u 1 · lim u→0 cos u u =5·1· 1 = 5. 1 C02S0M.033: The substitution u = kx shows that if k = 0, then lim x→0 It also follows that lim x→0 lim x→0 sin kx = 1. kx kx = 1. Therefore sin kx sin 3x sin 3x sin 3x 2x 3x 2x 3 3 3 = lim · · = lim · · =1·1· = . x→0 x→0 sin 2x 3x sin 2x 2x 3x sin 2x 2 2 2 C02S0M.034: We saw in the solution of Problem C02S03.014 that if k is a nonzero constant, then lim x→0 tan kx kx = 1 = lim . x→0 tan kx kx Therefore lim x→0 tan 2x tan 2x 3x 2x tan 2x 3x 2 2 2 = lim · · = lim · · =1·1· = . tan 3x x→0 2x tan 3x 3x x→0 2x tan 3x 3 3 3 C02S0M.035: Let x = u2 where u > 0. Then x → 0+ is equivalent to u → 0+ . Hence lim x→0+ x u2 u √ = lim = lim u · = 0 · 1 = 0. sin u sin x u→0+ sin u u→0+ C02S0M.036: First multiply numerator and denominator by the conjugate 1 + cos 3x of the numerator: 1 − cos 3x (1 − cos 3x)(1 + cos 3x) 1 − cos2 3x sin2 3x = = = 2x 2x(1 + cos 3x) 2x(1 + cos 3x) 2x(1 + cos 3x) = sin 3x sin 3x sin 3x 3x sin 3x sin 3x 3 sin 3x · = · · = ·· . 2x 1 + cos 3x 3x 2x 1 + cos 3x 3x 2 1 + cos 3x 4 Now let x → 0 to obtain the limit 1 · 3 0 · = 0. 2 1+1 C02S0M.037: First multiply numerator and denominator by the conjugate 1 + cos 3x of the numerator: 1 − cos 3x (1 − cos 3x)(1 + cos 3x) 1 − cos2 3x sin2 3x = =2 =2 2x2 2x2 (1 + cos 3x) 2x (1 + cos 3x) 2x (1 + cos 3x) = sin 3x sin 3x 1 sin 3x 3x sin 3x 3x 1 · · = · · · · 2x x 1 + cos 3x 3x 2x 3x x 1 + cos 3x = sin 3x 3 sin 3x 3 1 ·· ·· . 3x 2 3x 1 1 + cos 3x Now let x → 0 to obtain the limit 1 · 3 3 1 9 ·1· · =. 2 1 1+1 4 C02S0M.038: Express the cotangent and cosecant functions in terms of the sine and cosine functions to obtain lim x3 cot x csc x = lim (x3 ) · x→0 x→0 cos x 1 x x · = lim x · · (cos x) · = 0 · 1 · 1 · 1 = 0. sin x sin x x→0 sin x sin x C02S0M.039: Let u = 2x; then x = 1 u, and x → 0 is then equivalent to u → 0. Also express the secant 2 and tangent functions in terms of the sine and cosine functions. Result: lim x→0 sec 2x tan 2x sec u tan u 2 sin u 2 sin u 2 = lim = lim = lim · = · 1 = 2. 1 2u 2u u→0 u→0 u cos u→0 cos x u 1 2u C02S0M.040: Let u = 3x; then x = 1 u, and x → 0 is then equivalent to u → 0. Also express the 3 cotangent function in terms of sines and cosines. Result: lim x2 cot2 3x = lim x→0 x→0 x2 cos2 3x = lim u→0 sin2 3x cos2 u cos2 u u u 1 1 · · = ·1·1= . = lim 2 u →0 9 sin u sin u 9 9 sin u 12 9u C02S0M.041: Given f (x) = 2x2 + 3, a slope-predictor for f is m(x) = 4x. The slope of the line tangent to the graph of f at (1, f (1)) = (1, 5) is therefore m(1) = 4. So an equation of that line is y − 5 = 4(x − 1); that is, y = 4x + 1. C02S0M.042: Given f (x) = −5x2 + x, a slope-predictor for f is m(x) = −10x + 1. The slope of the line tangent to the graph of f at (1, f (1)) = (1, −4) is therefore m(1) = −9. So an equation of that line is y + 4 = −9(x − 1); that is, y = −9x + 5. C02S0M.043: Given f (x) = 3x2 + 4x − 5, a slope-predictor for f is m(x) = 6x + 4. The slope of the line tangent to the graph of f at (1, f (1)) = (1, 2) is therefore m(1) = 10. So an equation of that line is y − 2 = 10(x − 1); that is, y = 10x − 8. C02S0M.044: Given f (x) = −3x2 − 2x + 1, a slope-predictor for f is m(x) = −6x − 2. The slope of the line tangent to the graph of f at (1, f (1)) = (1, −4) is therefore m(1) = −8. So an equation of that line is y + 4 = −8(x − 1); that is, y = −8x + 4. 5 C02S0M.045: Given f (x) = (x − 1)(2x − 1) = 2x2 − 3x + 1, a slope-predictor for f is m(x) = 4x − 3. The slope of the line tangent to the graph of f at (1, f (1)) = (1, 0) is therefore m(1) = 1. So an equation of that line is y = x − 1. 2 1 C02S0M.046: Given f (x) = 1 x − 1 x = − 16 x2 + 1 x, a slope-predictor for f is m(x) = − 1 x + 1 . The 3 4 3 8 3 5 slope of the line tangent to the graph of f at (1, f (1)) = 1, 13 is therefore m(1) = − 1 + 1 = 24 . So an 48 8 3 5 equation of that line is y − 13 = 24 (x − 1); that is, 48y = 10x + 3. 48 C02S0M.047: If f (x) = 2x2 + 3x, then the slope-predicting function for f is m(x) = lim h→0 = lim h→0 f (x + h) − f (x) 2(x + h)2 + 3(x + h) − (2x2 + 3x) = lim h→0 h h 2x2 + 4xh + 2h2 + 3x + 3h − 2x2 − 3x 4xh + 2h2 + 3h h(4x + 2h + 3) = lim = lim h→0 h→0 h h h = lim (4x + 2h + 3) = 4x + 3. h→0 C02S0M.048: If f (x) = x − x3 , then the slope-predicting function for f is m(x) = lim h→0 f (x + h) − f (x) (x + h) − (x + h)3 − (x − x3 ) = lim h→0 h h = lim x + h − (x3 + 3x2 h + 3xh2 + h3 ) − x + x3 x + h − x3 − 3x2 h − 3xh2 − h3 − x + x3 = lim h→0 h h = lim h − 3x2 h − 3xh2 − h3 h(1 − 3x2 − 3xh − h2 ) = lim h→0 h h h→0 h→0 = lim (1 − 3x2 − 3xh − h2 ) = 1 − 3x2 . h→0 C02S0M.049: If f (x) = 1 , then the slope-predicting function for f is 3−x 1 1 − f (x + h) − f (x) 1 (3 − x) − (3 − x − h) 3 − (x + h) 3 − x m(x) = lim = lim = lim · h→0 h→0 h→0 h h h (3 − x − h)(3 − x) = lim 1 3−x−3+x+h h · = lim h (3 − x − h)(3 − x) h→0 h(3 − x − h)(3 − x) = lim 1 1 = . (3 − x − h)(3 − x) (3 − x)2 h→0 h→0 C02S0M.050: If f (x) = 1 , then the slope-predicting function for f is 2x + 1 6 1 1 − f (x + h) − f (x) 1 (2x + 1) − (2x + 2h + 1) 2(x + h) + 1 2x + 1 m(x) = lim = lim = lim · h→0 h→0 h→0 h h h (2x + 2h + 1)(2x + 1) = lim 2x + 1 − 2x − 2h − 1 −2h = lim h(2x + 2h + 1)(2x + 1) h→0 h(2x + 2h + 1)(2x + 1) = lim −2 2 =− . (2x + 2h + 1)(2x + 1) (2x + 1)2 h→0 h→0 C02S0M.051: If f (x) = x − 1 , then the slope-predicting function for f is x f (x + h) − f (x) m(x) = lim = lim h→0 h→0 h (x + h) − 1 1 − x− x+h x h = lim 1 1 1 · x+h− −x+ h x+h x = lim = lim 1 x+h−x · h+ h x(x + h) 1+ h→0 h→0 = lim h→0 C02S0M.052: If f (x) = m(x) = lim h→0 1+ 1 x(x + h) = lim h→0 =1+ h→0 1 1 1 · h+ − h x x+h h hx(x + h) 1 x2 + 1 = . x2 x2 x , then the slope-predicting function for f is x+1 f (x + h) − f (x) 1 x+h x = lim · − h→0 h h x+h+1 x+1 = lim 1 (x + h)(x + 1) − (x + h + 1)(x) 1 (x2 + x + hx + h) − (x2 + xh + x) · = lim · h→0 h h (x + h + 1)(x + 1) (x + h + 1)(x + 1) = lim x2 + x + hx + h − x2 − xh − x h = lim h→0 h(x + h + 1)(x + 1) h(x + h + 1)(x + 1) = lim 1 1 . = (x + h + 1)(x + 1) (x + 1)2 h→0 h→0 h→0 C02S0M.053: If f (x) = x+1 , then the slope-predicting function for f is x−1 7 m(x) = lim h→0 f (x + h) − f (x) 1 = lim · h→0 h h x+h+1 x+1 − x+h−1 x−1 = lim 1 (x + h + 1)(x − 1) − (x + h − 1)(x + 1) · h (x + h − 1)(x − 1) = lim (x2 + hx + x − x − h − 1) − (x2 + hx − x + x + h − 1) h(x + h − 1)(x − 1) = lim x2 + hx − h − 1 − x2 − hx − h + 1 −2h = lim h→0 h(x + h − 1)(x − 1) h(x + h − 1)(x − 1) = lim −2 2 =− . (x + h − 1)(x − 1) (x − 1)2 h→0 h→0 h→0 h→0 C02S0M.054: We must deal with | 2x + 3 |, and to do so we need to know when 2x + 3 changes sign: When 2x + 3 = 0; that is, when x = − 3 . If x > − 3 , then 2x + 3 > 0, so that 2 2 f (x) = 3x − x2 + (2x + 3) = −x2 + 5x + 3 if x > − 3. 2 By the theorem on page 58 (Section 2.1), the slope-predicting function for f will be m1 (x) = −2x + 5 if x > − 3 . But if x < − 3 , then 2x + 3 < 0, so that 2 2 f (x) = 3x − x2 − (2x + 3) = −x2 + x − 3 if x < − 3. 2 By the theorem just cited, the slope-predicting function for f will be m2 (x) = −2x + 1 if x < − 3 . Therefore 2 the general slope-predicting function for f will be m(x) = −2x + 5 −2x + 1 if x > − 3 , 2 if x < − 3 . 2 There will be no tangent line at x = − 3 . The reason is that 2 lim x→−1.5− f (x + h) − f (x) =4 h whereas lim x→−1.5+ f (x + h) − f (x) = 8. h Therefore there is no tangent line at the point − 3 , − 27 . But f is continuous at that point; indeed, f is 2 4 continuous on the set R of all real numbers. The graph of f is shown next on the left; the part of the graph near the corner point at − 3 , − 27 is shown magnified on the right. 2 4 -3 5 -4 -5 -3 -2 -1 1 -6 -5 -7 -10 -8 -15 -1.8 8 -1.6 -1.4 -1.2 -1 C02S0M.055: Following the suggestion, the line tangent to the graph of y = x2 at (a, a2 ) has slope 2a (because the slope-predicting function for f (x) = x2 is m(x) = 2x). But using the two-point formula for slope, this line also has slope a2 − 4 = 2a, a−3 √ so that a2 − 4 = 2a2 − 6a; that is, a2 − 6a + 4 = 0. The quadratic formula yields the two √ solutions a = 3 ± 5, √ so one of the two lines in question has slope 2 3 + 5 and the other has slope 2 3 − 5 . Both lines pass through (3, 4), so their equations are y−4=2 3+ √ 5 (x − 3) and y−4=2 3− √ 5 (x − 3). C02S0M.056: The given line has equation y = −x − 3, so its slope is −1. The radius of the circle from its center (2, 3) to the point (a, b) of tangency is perpendicular to that line, so has slope 1. So the radius lies on the line y − 3 = x − 2; that is, y = x + 1. We solve y = x + 1 and y = −x − 3 simultaneously to find √ the point of tangency (a, b) to be (−2, −1). The distance from the center of the circle to this point is 4 2. Therefore an equation of the circle is (x − 2)2 + (y − 3)2 = 32. C02S0M.057: First simplify f (x): f (x) = 1−x 1−x 1 = = 1 − x2 (1 + x)(1 − x) 1+x (1) if x = 1. Every rational function is continuous wherever it is defined, so f is continuous except at ± 1. The computations in (1) show that f (x) has no limit as x → −1, so f cannot be made continuous at x = −1. But the discontinuity at x = 1 is removable; if we redefine f at x = 1 to be its limit 1 there, then f will be 2 continuous there as well. C02S0M.058: Every rational function is continuous where it is defined; that is, where its denominator is nonzero. So f (x) = 1−x (2 − x)2 is continuous except at x = 2. This discontinuity is not removable because f (x) has no limit at x = 2. C02S0M.059: First simplify f (x): f (x) = x2 + x − 2 (x − 1)(x + 2) x+2 = = x2 + 2x − 3 (x − 1)(x + 3) x+3 (1) provided that x = 1. Note that f is a rational function, so f is continuous wherever it is defined: at every number other than 1 and −3. The computations in (1) show that f (x) has no limit at x = −3, so it cannot be redefined in such a way to be continuous there. But the discontinuity at x = 1 is removable; if we redefine f at x = 1 to be its limit 3 there, then f will be continuous everywhere except at x = −3. 4 C02S0M.060: Note that f (x) = 1 if x2 > 1; that is, if x > 1 or x < −1. But if x2 < 1, so that −1 < x < 1, then f (x) = −1. Hence f is continuous on (−∞, −1) ∪ (−1, 1) ∪ (1, + ∞). But f cannot be made continuous at either x = 1 or x = −1, because its left-hand and right-hand limits are unequal at each of these points. In any case, f is continuous wherever it is defined. 9 C02S0M.061: Let f (x) = x5 + x − 1. Then f (0) = −1 < 0 < 1 = f (1). Because f (x) is a polynomial, it is continuous on [0, 1], so f has the intermediate value property there. Hence there exists a number c in (0, 1) such that f (c) = 0. Thus c5 + c − 1 = 0, and so the equation x5 + x − 1 = 0 has a solution. (The value of c is approximately 0.754877666.) C02S0M.062: Let f (x) = x5 − 4x2 + 1. Here are some values of f (x): x f (x) −1 −4 0 1 1 −2 2 17 Because f (x) is a polynomial, it is continuous everywhere. Therefore f (x1 ) = 0 for some number x1 in (−1, 0), f (x2 ) = 0 for some number x2 in (0, 1), and f (x3 ) = 0 for some number x3 in (1, 2). The numbers x1 , x2 , and x3 are distinct because they lie in nonoverlapping intervals. Therefore the equation x5 − 4x2 + 1 = 0 has at least three real solutions. (The actual values are x1 ≈ 0.50842209, x2 ≈ 1.52864292, and x3 ≈ −0.49268877.) C02S0M.063: Let g (x) = x − cos x. Then g (0) = −1 < 0 < π /2 = g (π /2). Because g is continuous, g (c) = 0 for some number c in (0, π /2). That is, c − cos c = 0, so that c = cos c. C02S0M.064: Let h(x) = x + tan x. Then h(π ) = π > 0 and h(x) → −∞ as x approaches π /2 from above (from the right). This implies that h(r) < 0 for some number r slightly larger than π /2. Because h is continuous on the interval [r, π ], h has the intermediate value property there, so h(c) = 0 for some number c between r and π , and thus between π /2 and π . That is, c + tan c = 0, so that tan c = −c, and c does lie in the required interval 1 π , π . 2 C02S0M.065: Suppose that L is a straight line through 12, 15 that is normal to the graph of y = x2 at 2 the point (a, a2 ). The line tangent to the graph of y = x2 at that point has slope 2a, and the slope of L is then −1/(2a). We can equate this to the slope of L found by using the two-point formula: a2 − 15 1 2 =− ; a − 12 2a 2a a2 − 15 2 = −(a − 12); 2a3 − 15a = −a + 12; 2a3 − 14a − 12 = 0; a3 − 7a − 6 = 0. By inspection, one solution of the last equation is a = −1. By the factor theorem of algebra, we know that a − (−1) = a + 1 is a factor of the polynomial a3 − 7a − 6, and division of the former into the latter yields a3 − 7a − 6 = (a + 1)(a2 − a − 6) = (a + 1)(a − 3)(a + 2). So the equation a3 − 7a − 6 = 0 has the three solutions a = −1, a = 3, and a = −2. Therefore there are three lines through 12, 15 that are normal to the graph of y = x2 , and their slopes are 1 , 1 , and − 1 . 2 42 6 C02S0M.066: Let (0, c) be the center of a too-big circle, r its radius, and (a, a2 ) the point in the first quadrant where the too-big circle and the parabola are tangent. The idea is to solve for a in terms of c and (possibly) r, then to impose the condition that there is exactly one solution for a! This means that the circle just reaches to the bottom of the parabola and not beyond. 10 Consider the radius of the circle connecting (0, c) with (a, a2 ). The circle and the parabola are mutually tangent at (a, a2 ), so this radius must be normal not only to the circle, but also to the parabola at the point (a, a2 ). We compute the slope of this radius in two ways to find that a2 − c 1 =− ; a−0 2a 1 a2 − c = − ; 2 1 a2 = c − . 2 Now we impose the condition that there is only one point at which the circle and the parabola meet. The last equation will have exactly one solution when c = 1 , and in this case the radius of the circle—because it 2 touches the parabola only at (0, 0)—will also be r = 1 . Answer: 1 . 2 2 11 Section 3.1 C03S01.001: Given f (x) = 4x − 5, we have a = 0, b = 4, and c = −5, so f (x) = 2ax + b = 4. C03S01.002: Given g (t) = −16t2 + 100, we have a = −16, b = 0, and c = 100, so g (t) = 2at + b = −32t. C03S01.003: If h(z ) = z (25 − z ) = −z 2 +25z , then a = −1, b = 25, and c = 0, so h (z ) = 2az + b = −2z +25. C03S01.004: If f (x) = −49x + 16, then a = 0, b = −49, and c = 16, so f (x) = −49. C03S01.005: If y = 2x2 + 3x − 17, then a = 2, b = 3, and c = −17, so dy = 2ax + b = 4x + 3. dx C03S01.006: If x = −100t2 + 16t, then a = −100, b = 16, and c = 0, so C03S01.007: If z = 5u2 − 3u, then a = 5, b = −3, and c = 0, so dx = 2at + b = −200t + 16. dt dz = 2au + b = 10u − 3. du C03S01.008: If v = −5y 2 + 500y , then a = −5, b = 500, and c = 0, so dv = 2ay + b = −10y + 500. dy C03S01.009: If x = −5y 2 + 17y + 300, then a = −5, b = 17, and c = 300, so C03S01.010: If u = 7t2 + 13t, then a = 7, b = 13, and c = 0, so C03S01.011: f (x) = lim h→0 = lim h→0 2h = lim 2 = 2. h→0 h C03S01.012: f (x) = lim h→0 = lim h→0 h→0 h→0 f (x + h) − f (x) 2 − 3(x + h) − (2 − 3x) 2 − 3x − 3h − 2 + 3x = lim = lim h→0 h→0 h h h f (x + h) − f (x) (x + h)2 + 5 − (x2 + 5) = lim h→0 h h x + 2xh + h2 + 5 − x2 − 5 2xh + h2 = lim = lim (2x + h) = 2x. h→0 h→0 h h h→0 h→0 f (x + h) − f (x) 2(x + h) − 1 − (2x − 1) 2x + 2h − 1 − 2x + 1 = lim = lim h→0 h→0 h h h 2 C03S01.014: f (x) = lim = lim du = 2at + b = 14t + 13. dt −3h = lim (−3) = −3. h→0 h C03S01.013: f (x) = lim = lim f (x + h) − f (x) 3 − 2(x + h)2 − (3 − 2x2 ) = lim h→0 h h 3 − 2x − 4xh − 2h2 − 3 + 2x2 −4xh − 2h2 = lim = lim (−4x − 2h) = −4x. h→0 h→0 h h 2 1 1 − f (x + h) − f (x) 2(x + h) + 1 2x + 1 C03S01.015: f (x) = lim = lim h→0 h→0 h h = lim h→0 dx = 2ay + b = −10y + 17. dy 2x + 1 − (2x + 2h + 1) 2x + 1 − 2x − 2h − 1 −2h = lim = lim h(2x + 2h + 1)(2x + 1) h→0 h(2x + 2h + 1)(2x + 1) h→0 h(2x + 2h + 1)(2x + 1) 1 = lim h→0 −2 −2 . = (2x + 2h + 1)(2x + 1) (2x + 1)2 1 1 − f (x + h) − f (x) 3 − (x + h) 3 − x C03S01.016: f (x) = lim = lim h→0 h→0 h h = lim h→0 = lim h→0 (3 − x) − (3 − x − h) 3−x−3+x+h h = lim = lim h→0 h(3 − x − h)(3 − x) h→0 h(3 − x − h)(3 − x) h(3 − x − h)(3 − x) 1 1 = . (3 − x − h)(3 − x) (3 − x)2 √ 2(x + h) + 1 − 2x + 1 f (x + h) − f (x) C03S01.017: f (x) = lim = lim h→0 h→0 h h √ √ √ √ 2x + 2h + 1 − 2x + 1 2x + 2h + 1 + 2x + 1 (2h + 2h + 1) − (2x + 1) √ √ √ √ = lim = lim h→0 h→0 h h 2x + 2h + 1 + 2x + 1 2x + 2h + 1 + 2x + 1 = lim h→0 h √ 2h √ 2x + 2h + 1 + 2x + 1 = lim √ h→0 2 2 1 √ =√ =√ . 2 2x + 1 2x + 1 2x + 2h + 1 + 2x + 1 1 1 1 1 √ (f (x + h) − f (x)) = lim −√ h→0 h h→0 h x+1 x+h+1 √ √ √ √ √ √ x+1− x+h+1 x+1+ x+h+1 x+1− x+h+1 √ √ √ √ √ = lim √ = lim h→0 h x + h + 1 x + 1 h→0 h x+h+1 x+1 x+1+ x+h+1 C03S01.018: f (x) = lim = lim h→0 = lim h→0 = lim h→0 h √ (x + 1) − (x + h + 1) √ √ √ x+h+1 x+1 x+1+ x+h+1 h √ −h √ √ √ x+h+1 x+1 x+1+ x+h+1 √ −1 √ √ x+h+1 x+1 x+1+ x+h+1 √ C03S01.019: f (x) = lim h→0 = lim h→0 = lim h→0 = lim h→0 = lim h→0 2 √ 2 x+1 =− x+h x − 1 − 2(x + h) 1 − 2x x − 2x2 + h − 2xh − x + 2x2 + 2xh h = lim h→0 h(1 − 2x − 2h)(1 − 2x) h(1 − 2x − 2h)(1 − 2x) 1 1 = . (1 − 2x − 2h)(1 − 2x) (1 − 2x)2 h→0 h→0 1 1 (f (x + h) − f (x)) = lim h→0 h h x+1 −1 1 . 2(x + 1)3/2 1 (x + h)(1 − 2x) − (1 − 2x − 2h)(x) (x − 2x2 + h − 2xh) − (x − 2x2 − 2xh) · = lim h→0 h (1 − 2x − 2h)(1 − 2x) h(1 − 2x − 2h)(1 − 2x) C03S01.020: f (x) = lim = lim =√ f (x + h) − f (x) 1 = lim h→0 h h 1 (x + h + 1)(x − 1) − (x + h − 1)(x + 1) · h (x + h − 1)(x − 1) x+h+1 x+1 − x+h−1 x−1 (x2 − x + hx − h + x − 1) − (x2 + x + hx + h − x − 1) h(x + h − 1)(x − 1) 2 = lim h→0 = lim h→0 x2 − x + hx − h + x − 1 − x2 − x − hx − h + x + 1 −2h = lim h→0 h(x + h − 1)(x − 1) h(x + h − 1)(x − 1) −2 2 . =− (x + h − 1)(x − 1) (x − 1)2 C03S01.021: The velocity of the particle at time t is position of the particle then is x(0) = 100. C03S01.022: The velocity of the particle at time t is The position of the particle then is x(5) = 425. C03S01.023: The velocity of the particle at time t is The position of the particle then is x(2.5) = 99. C03S01.024: The velocity of the particle at time t is position of the particle then is x(0) = 50. C03S01.025: The velocity of the particle at time t is The position of the particle then is x(−2) = 120. dx = v (t) = −32t, so v (t) = 0 when t = 0. The dt dx = v (t) = −32t + 160, so v (t) = 0 when t = 5. dt dx = v (t) = −32t + 80, so v (t) = 0 when t = 2.5. dt dx = v (t) = 200t, so v (t) = 0 when t = 0. The dt dx = v (t) = −20 − 10t, so v (t) = 0 when t = −2. dt C03S01.026: The ball reaches its maximum height when its velocity v (t) = v (t) = 0 when t = 5. The height of the ball then is y (5) = 400 (ft). C03S01.027: The ball reaches its maximum height when its velocity v (t) = v (t) = 0 when t = 2. The height of the ball then is y (2) = 64 (ft). C03S01.028: The ball reaches its maximum height when its velocity v (t) = v (t) = 0 when t = 4. The height of the ball then is y (4) = 281 (ft). C03S01.029: The ball reaches its maximum height when its velocity v (t) = v (t) = 0 when t = 3. The height of the ball then is y (3) = 194 (ft). dy = −32t + 160 is zero, and dt dy = −32t + 64 is zero, and dt dy = −32t + 128 is zero, and dt dy = −32t + 96 is zero, and dt C03S01.030: Figure 3.1.22 shows a graph first increasing, then with a horizontal tangent at x = 0, then decreasing. Hence its derivative must be first positive, then zero when x = 0, then negative. This matches Fig. 3.1.28(c). C03S01.031: Figure 3.1.23 shows a graph first decreasing, then with a horizontal tangent where x = 1, then increasing thereafter. So its derivative must be negative for x < 1, zero when x = 1, and positive for x > 1. This matches Fig. 3.1.28(e). C03S01.032: Figure 3.1.24 shows a graph increasing for x < −1.5, decreasing for −1.5 < x < 1.5, and increasing for 1.5 < x. So its derivative must be positive for x < −1.5, negative for −1.5 < x < 1.5, and positive for 1.5 < x. This matches Fig. 3.1.28(b). C03S01.033: Figure 3.1.25 shows a graph decreasing for x < −1.5, increasing for −1.5 < x < 0, decreasing for 0 < 1 < 1.5, and increasing for 1.5 < x. Hence its derivative is negative for x < −1.5, positive for 3 −1.5 < x < 0, negative for 0 < x < 1.5, and positive for 1.5 < x. Only the graph in Fig. 3.1.28(f) shows these characteristics. C03S01.034: Figure 3.1.26 shows a graph with horizontal tangents near where x = −3, x = 0, and x = 3. So the graph of the derivative must be zero near these three points, and this behavior is matched by Fig. 3.1.28(a). C03S01.035: Figure 3.1.27 shows a graph that increases, first slowly, then rapidly. So its derivative must exhibit the same behavior, and thus its graph is the one shown in Fig. 3.1.28(d). C03S01.036: Note that C (F ) = 5 160 F− 9 9 and so F (C ) = 9 C + 32. 5 So the rate of change of C with respect to F is dC 5 = dF 9 C (F ) = and the rate of change of F with respect to C is F (C ) = dF 9 =. dC 5 C03S01.037: Let r note the radius of the circle. Then A = π r2 and C = 2π r. Thus r= C , 2π and so A(C ) = 12 C, 4π C > 0. Therefore the rate of change of A with respect to C is A (C ) = dA 1 = C. dC 2π C03S01.038: Let r denote the radius of the circular ripple in feet at time t (seconds). Then r = 5t, and the area within the ripple at time t is A = π r2 = 25π t2 . The rate at which this area is increasing at time t is A (t) = 50π t, so at time t = 10 the area is increasing at the rate of A (10) = 50π · 10 = 500π (ft2 /s). C03S01.039: The velocity of the car (in feet per second) at time t (seconds) is v (t) = x (t) = 100 − 10t. The car comes to a stop when v (t) = 0; that is, when t = 10. At that time the car has traveled a distance x(10) = 500 (ft). So the car skids for 10 seconds and skids a distance of 500 ft. 1 1 C03S01.040: Because V (t) = 10 − 1 t + 1000 t2 , V (t) = − 1 + 500 t and so the rate at which the water is 5 5 2 leaking out one minute later (t = 60) is V (60) = − 25 (gal/s); that is, −4.8 gal/min. The average rate of change of V from t = 0 until t = 100 is V (100) − V (0) 0 − 10 1 = =− . 100 − 0 100 10 1 The instantaneous rate of change of V will have this value when V (t) = − 10 , which we easily solve for t = 50. C03S01.041: First, P (t) = 100 + 30t + 4t2 . The initial population is 100, so doubling occurs when P (t) = 200; that is, when 4t2 +30t−100 = 0. The quadratic formula yields t = 2.5 as the only positive solution 4 of this equation, so the population will take two and one-half months to double. Because P (t) = 30 + 8t, the rate of growth of the population when P = 200 will be P (2.5) = 50 (chipmunks per month). C03S01.042: In our construction, the tangent line at 1989 passes through the points (1984, 259) and (1994, 423), and so has slope 16.4; this yields a rate of growth of approximately 16.4 thousand per year in 1989. Alternatively, using the Mathematica function Fit to fit the given data to a sixth-degree polynomial, we find that P (in thousands) is given in terms of t (as a four-digit year) by P (t) ≈(1.0453588 × 10−14 )x6 − (4.057899 × 10−11 )x5 + (3.9377735 × 10−8 )x4 + (5.93932556 × 10−11 )x3 + (5.972176 × 10−14 )x2 + (5.939427 × 10−17 )x + (3.734218 × 10−12 ), and that P (1989) ≈ 16.4214. Of course neither method is exact. C03S01.043: On our graph, the tangent line at the point (20, 810) has slope m1 ≈ 0.6 and the tangent line at (40, 2686) has slope m2 ≈ 0.9. A line of slope 1 on our graph corresponds to a velocity of 125 ft/s (because the line through (0, 0) and (10, 1250) has slope 1), and thus we estimate the velocity of the car at time t = 20 to be about (0.6)(125) = 75 ft/s, and at time t = 40 it is traveling at about (0.9)(125) = 112.5 ft/s. The method is crude; the answer in the back of the textbook is quite different simply because it was obtained by someone else. When we used the Mathematica function Fit to fit the data to a sixth-degree polynomial, we obtained x(t) ≈ 0.0000175721 + (6.500002)x + (1.112083)x2 + (0.074188)x3 − (0.00309375)x4 + (0.0000481250)x5 − (0.000000270834)x6 , which yields x (20) ≈ 74.3083 and x (40) ≈ 109.167. Of course neither method is exact. C03S01.044: With volume V and edge x, the volume of the cube is given by V (x) = x3 . Now which is indeed half the total surface area 6x2 of the cube. dV = 3x2 , dx dV C03S01.045: With volume V and radius r, the volume of the sphere is V (r) = 4 π r3 . Then = 4π r2 , 3 dr and this is indeed the surface area of the sphere. C03S01.046: A right circular cylinder of radius r and height h has volume V = π r2 h and total surface area S obtained by adding the areas of its top, bottom, and curved side: S = 2π r2 + 2π rh. We are given h = 2r, so V (r) = 2π r3 and S (r) = 6π r2 . Also dV /dr = 6π r2 = S (r), so the rate of change of volume with respect to radius is indeed equal to total surface area. C03S01.047: We must compute dV /dt when t = 30; V (r) = 4 π r3 is the volume of the balloon when its 3 60 − t radius is r. We are given r = , and thus 12 V (t) = 4 π 3 60 − t 12 3 = π (216000 − 10800t + 180t2 − t3 ). 1296 Therefore dV π = (−10800 + 360t − 3t2 ), dt 1296 and so V (t) = − 25π in.3 /s; that is, air is leaking out at approximately 6.545 in.3 /s. 12 5 1.68 1.68 we derive V (p) = − 2 . The rate of change of V with respect to p p p when p = 2 (atm) is then V (2) = −0.42 (liters/atm). C03S01.048: From V (p) = C03S01.049: Let V (t) denote the volume (in cm3 ) of the snowball at time t (in hours) and let r(t) denote its radius then. From the data given in the problem, r = 12 − t. The volume of the snowball is V= 43 4 4 π r = π (12 − t)3 = π 1728 − 432t + 36t2 − t3 , 3 3 3 so its instantaneous rate of change is V (t) = 4 π −432 + 72t − 3t2 . 3 Hence its rate of change of volume when t = 6 is V (6) = −144π cm3 /h. Its average rate of change of volume from t = 3 to t = 9 in cm3 /h is V (9) − V (3) 36π − 972π = = −156π (cm3 /h). 9−3 6 dy = −32t+96, which is zero when t = 3. So the maximum dt height of the ball is y (3) = 256 (ft). It hits the ground when y (t) = 0; that is, when −16t2 + 96t + 112 = 0. The only positive solution of this equation is t = 7, so the impact speed of the ball is | y (7) | = 128 (ft/s). C03S01.050: The velocity of the ball at time t is C03S01.051: The spaceship hits the ground when 25t2 − 100t + 100 = 0, which has solution t = 2. The velocity of the spaceship at time t is y (t) = 50t − 100, so the speed of the spaceship at impact is (fortunately) zero. 3 C03S01.052: Because P (t) = 100 + 4t + 10 t2 , we have P (t) = 4 + 3 t. The year 1986 corresponds to t = 6, 5 so the rate of change of P then was P (6) = 7.6 (thousands per year). The average rate of change of P from 1983 (t = 3) to 1988 (t = 8) was P (8) − P (3) 151.2 − 114.7 = = 7.3 (thousands per year). 8−3 5 C03S01.053: The average rate of change of the population from January 1, 1990 to January 1, 2000 was P (10) − P (0) 6 = = 0.6 10 − 0 10 (thousands per year). The instantaneous rate of change of the population (in thousands per year, again) at time t was P (t) = 1 − (0.2)t + (0.018)t2 . Using the quadratic formula to solve the equation P (t) = 0.6, we find two solutions: √ √ 50 − 10 7 50 + 10 7 t= ≈ 2.6158318766 and t = ≈ 8.4952792345. 9 9 These values of t correspond to August 12, 1992 and June 30, 1998, respectively. C03S01.054: (a) If f (x) = | x |, then 6 f− (0) = lim h→0− similarly, f+ (0) = 1. derivative there is |h| − 0 −h = lim = −1; − h h h→0 (b) The function f (x) = | 2x − 10 | is not differentiable at x = 5. Its right-hand f+ (5) = lim + h→0 | 2 · (5 + h) − 10 | − 0 10 + 2h − 10 = lim+ = 2. h h h→0 Similarly, f− (5) = −2. C03S01.055: The graphs of the function of part (a) is shown next, on the left; the graph of the function of part (b) is on the right. 1.4 2 1.2 1 1.5 0.8 1 0.6 0.4 0.5 0.2 -1 -0.5 0.5 1 -1 -0.5 0.5 1 -0.5 -1 (a) f− (0) = 1 while f+ (0) = 2. Hence f is not differentiable at x = 0. (b) In contrast, f− (0) = lim h→0− (0 + h)2 − 2 · 02 =0 h and f+ (0) = lim + h→0 2 · (0 + h)2 − 2 · 02 = 0; h therefore f is differentiable at x = 0 and f (0) = 0. C03S01.056: The function f is clearly differentiable except possibly at x = 1. But f− (1) = lim h→0− 2 · (1 + h) + 1 − 3 =2 h and f+ (1) = lim h→0+ 4 · (1 + h) − (1 + h)2 − 3 4 + 4h − 1 − 2h − h2 − 3 = lim+ = lim+ (2 − h) = 2. h h h→0 h→0 7 Therefore f is differentiable at x = 1 as well. C03S01.057: Clearly f is differentiable except possibly at x = 3. Moreover, f− (3) = lim − h→0 11 + 6 · (3 + h) − (3 + h)2 − 20 11 + 18 + 6h − 9 − 6h − h2 − 20 −h2 = lim− = lim− =0 h h h h→0 h→0 and f+ (3) = lim h→0+ (3 + h)2 − 6 · (3 + h) + 29 − 20 9 + 6h + h2 − 18 − 6h + 29 − 20 h2 = lim = lim = 0. h h h→0+ h→0+ h Therefore the function f is also differentiable at x = 3; moreover, f (3) = 0. C03S01.058: The graph of f (x) = x · | x | is shown next. 0.2 0.1 -1 -0.5 0.5 1 -0.1 -0.2 Because f (x) = x2 if x > 0 and f (x) = −x2 if x < 0, f is differentiable except possibly at x = 0. But f− (0) = lim − (0 + h) · | 0 + h | −h2 = lim− =0 h h h→0 f+ (0) = lim (0 + h) · | 0 + h | h2 = lim+ = 0. h h h→0 h→0 and h→0+ Therefore f is differentiable at x = 0 and f (0) = 0. Because 2x f (x) = 0 −2x we see that f (x) = 2| x | for all x. 8 if x > 0, if x = 0, if x < 0, C03S01.059: The graph of f (x) = x + | x | is shown next. 2 1.5 1 0.5 -1 -0.5 0.5 1 Because f (x) = 0 if x < 0 and f (x) = 2x if x > 0, clearly f is differentiable except possibly at x = 0. Next, f− (0) = lim − 0 + h + |0 + h| h−h = lim− = 0, h h h→0 f+ (0) = lim + 0 + h + |0 + h| 2h = lim+ = 2 = 0. h h h→0 h→0 whereas h→0 Therefore f is not differentiable at x = 0. In summary, f (x) = 2 if x > 0 and f (x) = 0 if x < 0. For a “single-formula” version of the derivative, consider f (x) = 1 + |x| . x C03S01.060: The graph of f (x) = x · (x + | x |), is shown next. 0.8 0.6 0.4 0.2 -1 -0.5 0.5 1 Because f (x) = 2x2 if x > 0 and f (x) = 0 if x < 0, f (x) exists except possibly at x = 0. But f− (0) = lim h→0− (0 + h) · (0 + h + | 0 + h |) h · (h − h) = lim− =0 h h h→0 and 9 f+ (0) = lim h→0+ (0 + h) · (0 + h + | 0 + h |) h · (h + h) = lim = 0. + h h h→0 Therefore f is differentiable at x = 0 and f (0) = 0. Finally, f (x) = 4x if x > 0 and f (x) = 0 if x < 0. For a “single-formula” version of the derivative, consider f (x) = 2 · (x + | x |). 10 Section 3.2 C03S02.001: Given: f (x) = 3x2 − x + 5. We apply the rule for differentiating a linear combination and the power rule to obtain f (x) = 3 Dx (x2 ) − Dx (x) + Dx (5) = 3 · 2x − 1 + 0 = 6x − 1. C03S02.002: Given: g (t) = 1 − 3t2 − 2t4 . We apply the rule for differentiating a linear combination and the power rule to obtain g (t) = Dt (1) − 3 Dt (t2 ) − 2 Dt (t4 ) = 0 − 3 · 2t − 2 · 4t3 = −6t − 8t3 . C03S02.003: Given: f (x) = (2x + 3)(3x − 2). We apply the product rule to obtain f (x) = (2x + 3) Dx (3x − 2) + (3x − 2) Dx (2x + 3) = (2x + 3) · 3 + (3x − 2) · 2 = 12x + 5. C03S02.004: Given: g (x) = (2x2 − 1)(x3 + 2). We apply the product rule, the rule for differentiating a linear combination, and the power rule to obtain g (x) = (2x2 − 1) Dx (x3 + 2) + (x3 + 2) Dx (2x2 − 1) = (2x2 − 1)(3x2 ) + (x3 + 2)(4x) = 10x4 − 3x2 + 8x. C03S02.005: Given: h(x) = (x + 1)3 . We rewrite h(x) in the form h(x) = (x + 1)(x + 1)(x + 1) and then apply the extended product rule in Eq. (16) to obtain h (x) = (x + 1)(x + 1) Dx (x + 1) + (x + 1)(x + 1) Dx (x + 1) + (x + 1)(x + 1) Dx (x + 1) = (x + 1)(x + 1)(1) + (x + 1)(x + 1)(1) + (x + 1)(x + 1)(1) = 3(x + 1)2 . Alternatively, we could rewrite h(x) in the form h(x) = x3 + 3x2 + 3x + 1 and then apply the power rule and the rule for differentiating a linear combination to obtain h (x) = Dx (x3 ) + 3 Dx (x2 ) + 3 Dx (x) + Dx (1) = 3x2 + 6x + 3. The first method gives the answer in a more useful form because it is easier to determine where h (x) is positive, where negative, and where zero. Zeugma! C03S02.006: Given: g (t) = (4t − 7)2 = (4t − 7) · (4t − 7). We apply the product rule and the rule for differentiating a linear combination to obtain g (t) = (4t − 7) Dt (4t − 7) + (4t − 7) Dt (4t − 7) = 4 · (4t − 7) + 4 · (4t − 7) = 8 · (4t − 7) = 32t − 56. C03S02.007: Given: f (y ) = y (2y − 1)(2y + 1). We apply the extended product rule in Eq. (16) to obtain 1 f (y ) = (2y − 1)(2y + 1) Dy (y ) + y (2y + 1) Dy (2y − 1) + y (2y − 1) Dy (2y + 1) = (2y − 1)(2y + 1) · 1 + y (2y + 1) · 2 + y (2y − 1) · 2 = 4y 2 − 1 + 4y 2 + 2y + 4y 2 − 2y = 12y 2 − 1. Alternatively, we could first expand: f (y ) = 4y 3 − y . Then we could apply the rule for differentiating a linear combination and the power rule to obtain f (y ) = 4 Dy (y 3 ) − Dy (y ) = 12y 2 − 1. C03S02.008: Given: f (x) = 4x4 − 1 . We apply various rules, including the reciprocal rule, to obtain x2 f (x) = 4 Dx (x4 ) − − Dx (x2 ) (x2 )2 = 4 · 4x3 + 2x 2 = 16x3 + 3 . 4 x x Alternatively, we could rewrite: f (x) = 4x4 − x−2 . Then we could apply the rule for differentiating a linear combination and the power rule (both for positive and for negative integral exponents) to obtain f (x) = 4 Dx (x4 ) − Dx (x−2 ) = 4 · 4x3 − (−2)x−3 = 16x3 + 2x−3 = 16x3 + 2 . x3 C03S02.009: We apply the rule for differentiating a linear combination and the reciprocal rule (twice) to obtain g (x) = Dx =− 1 x+1 − Dx 1 x−1 Dx (x + 1) Dx (x − 1) 1 1 + =− + . (x + 1)2 (x − 1)2 (x + 1)2 (x − 1)2 Looking ahead to later sections and chapters—in which we will want to find where g (x) is positive, negative, or zero—it would be good practice to simplify g (x) to g (x) = (x + 1)2 − (x − 1)2 4x = . (x + 1)2 (x − 1)2 (x + 1)2 (x − 1)2 C03S02.010: We apply the reciprocal rule to f (t) = f (t) = − 1 to obtain 4 − t2 Dt (4 − t2 ) −2t 2t =− = . (4 − t2 )2 (4 − t2 )2 (4 − t2 )2 C03S02.011: First write (or think of) h(x) as h(x) = 3 · 1 , x2 + x + 1 then apply the rule for differentiating a linear combination and the reciprocal rule to obtain h (x) = 3 · − Dx (x2 + x + 1) (x2 + x + 1)2 = −3 · (2x + 1) . (x2 + x + 1)2 Alternatively apply the quotient rule directly to obtain h (x) = (x2 + x + 1)Dx (3) − 3 Dx (x2 + x + 1) −3 · (2x + 1) =2 . (x2 + x + 1)2 (x + x + 1)2 2 C03S02.012: Multiply numerator and denominator in f (x) by x to obtain f (x) = 1 2 1− x = x . x−2 Then apply the quotient rule to obtain f (x) = (x − 2) Dx (x) − xDx (x − 2) (x − 2) · 1 − x · 1 −2 = = . 2 2 (x − 2) (x − 2) (x − 2)2 C03S02.013: Given g (t) = (t2 + 1)(t3 + t2 + 1), apply the product rule, the rule for differentiating a linear combination, and the power rule to obtain g (t) = (t2 + 1) Dt (t3 + t2 + 1) + (t3 + t2 + 1) Dt (t2 + 1) = (t2 + 1)(3t2 + 2t + 0) + (t3 + t2 + 1)(2t + 0) = (3t4 + 2t3 + 3t2 + 2t) + (2t4 + 2t3 + 2t) = 5t4 + 4t3 + 3t2 + 4t. Alternatively, first expand: g (t) = t5 + t4 + t3 + 2t2 + 1, then apply the rule for differentiating a linear combination and the power rule. C03S02.014: Given f (x) = (2x3 − 3)(17x4 − 6x + 2), apply the product rule, the rule for differentiating a linear combination, and the power rule to obtain f (x) = (2x3 − 3)(68x3 − 6) + (6x2 )(17x4 − 6x + 2) = (136x6 − 216x3 + 18) + (102x6 − 36x3 + 12x2 ) = 238x6 − 252x3 + 12x2 + 18. Alternatively, first expand f (x), then apply the linear combination rule and the power rule. C03S02.015: The easiest way to find g (z ) is first to rewrite g (z ): g (z ) = 1 1 1 1 − 2 = z −1 − z −2 . 2z 3z 2 3 Then apply the linear combination rule and the power rule (for negative integral exponents) to obtain g (z ) = 1 1 1 2 4 − 3z . (−1)z −2 − (−2)z −3 = − 2 + 3 = 2 3 2z 3z 6z 3 The last step is advisable should it be necessary to find where g (z ) is positive, where it is negative, and where it is zero. Hypozeuxis! C03S02.016: The quotient rule yields f (x) = x2 Dx (2x3 − 3x2 + 4x − 5) − (2x3 − 3x2 + 4x − 5) Dx (x2 ) (x2 )2 = (x2 )(6x2 − 6x + 4) − (2x3 − 3x2 + 4x − 5)(2x) (6x4 − 6x3 + 4x2 ) − (4x4 − 6x3 + 8x2 − 10x) = x4 x4 = 6x4 − 6x3 + 4x2 − 4x4 + 6x3 − 8x2 + 10x 2x4 − 4x2 + 10x 2x3 − 4x + 10 = = . x4 x4 x3 3 But if the objective is to obtain the correct answer as quickly as possible, regardless of its appearance, you could proceed as follows (using the linear combination rule and the power rule for negative integral exponents): f (x) = 2x − 3 + 4x−1 − 5x−2 , so f (x) = 2 − 4x−2 + 10x−3 . C03S02.017: Apply the extended product rule in Eq. (16) to obtain g (y ) = (3y 2 − 1)(y 2 + 2y + 3) Dy (2y ) + (2y )(y 2 + 2y + 3) Dy (3y 2 − 1) + (2y )(3y 2 − 1) Dy (y 2 + 2y + 3) = (3y 2 − 1)(y 2 + 2y + 3)(2) + (2y )(y 2 + 2y + 3)(6y ) + (2y )(3y 2 − 1)(2y + 2) = (6y 4 + 12y 3 + 18y 2 − 2y 2 − 4y − 6) + (12y 4 + 24y 3 + 36y 2 ) + (12y 4 − 4y 2 + 12y 3 − 4y ) = 30y 4 + 48y 3 + 48y 2 − 8y − 6. Or if you prefer, first expand g (y ), then apply the linear combination rule and the power rule to obtain g (y ) = (6y 3 − 2y )(y 2 + 2y + 3) = 6y 5 + 12y 4 + 16y 3 − 4y 2 − 6y, so g (y ) = 30y 4 + 48y 3 + 48y 2 − 8y − 6. C03S02.018: By the quotient rule, f (x) = (x2 + 4) Dx (x2 − 4) − (x2 − 4) Dx (x2 + 4) (x2 + 4)(2x) − (x2 − 4)(2x) 16x = =2 . 2 + 4)2 2 + 4)2 (x (x (x + 4)2 C03S02.019: Apply the quotient rule to obtain g (t) = = (t2 + 2t + 1) Dt (t − 1) − (t − 1) Dt (t2 + 2t + 1) (t2 + 2t + 1)(1) − (t − 1)(2t + 2) = 2 2 + 2t + 1)2 (t [ (t + 1)2 ] (t2 + 2t + 1) − (2t2 − 2) 3 + 2 t − t2 (t + 1)(t − 3) 3−t = =− = . 4 (t + 1) (t + 1)4 (t + 1)4 (t + 1)3 C03S02.020: Apply the reciprocal rule to obtain u (x) = − Dx (x2 + 4x + 4) 2x + 4 2 =− =− . (x + 2)4 (x + 2)4 (x + 2)3 C03S02.021: Apply the reciprocal rule to obtain v (t) = − Dt (t3 − 3t2 + 3t − 1) 3t2 − 6t + 3 3(t − 1)2 3 =− =− =− . (t − 1)6 (t − 1)6 (t − 1)6 (t − 1)4 C03S02.022: The quotient rule yields 4 h(x) = (2x − 5) Dx (2x3 + x2 − 3x + 17) − (2x3 + x2 − 3x + 17) Dx (2x − 5) (2x − 5)2 = (2x − 5)(6x2 + 2x − 3) − (2x3 + x2 − 3x + 17)(2) (2x − 5)2 = (12x3 − 26x2 − 16x + 15) − (4x3 + 2x2 − 6x + 34) (2x − 5)2 = 12x3 − 26x2 − 16x + 15 − 4x3 − 2x2 + 6x − 34 8x3 − 28x2 − 10x − 19 = . (2x − 5)2 (2x − 5)2 C03S02.023: The quotient rule yields g (x) = (x3 + 7x − 5)(3) − (3x)(3x2 + 7) 3x3 + 21x − 15 − 9x3 − 21x 6x3 + 15 = =− 3 . (x3 + 7x − 5)2 (x3 + 7x − 5)2 (x + 7x − 5)2 C03S02.024: First expand the denominator, then multiply numerator and denominator by t2 , to obtain 1 f (t) = t+ 1 t 2 = 1 t2 + 2 + 1 t2 = t2 . t4 + 2t2 + 1 Then apply the quotient rule to obtain f (t) = (t4 + 2t2 + 1)(2t) − (t2 )(4t3 + 4t) 2 [ (t2 + 1)2 ] = 2t5 + 4t3 + 2t − 4t5 − 4t3 2t − 2t5 =2 . (t2 + 1)4 (t + 1)4 A modest simplification is possible: f (t) = − 2t(t4 − 1) 2t(t2 + 1)(t2 − 1) 2t(t2 − 1) =− =− 2 . 2 + 1)4 2 + 1)4 (t (t (t + 1)3 C03S02.025: First multiply each term in numerator and denominator by x4 to obtain g (x) = x3 − 2x2 . 2x − 3 Then apply the quotient rule to obtain g (x) = (2x − 3)(3x2 − 4x) − (x3 − 2x2 )(2) (6x3 − 17x2 + 12x) − (2x3 − 4x2 ) 4x3 − 13x2 + 12x = = . 2 2 (2x − 3) (2x − 3) (2x − 3)2 It is usually wise to simplify an expression before differentiating it. C03S02.026: First multiply each term in numerator and denominator by x2 + 1 to obtain f (x) = x3 (x2 + 1) − 1 x5 + x3 − 1 =6 . x4 (x2 + 1) + 1 x + x4 + 1 Then apply the quotient rule to obtain 5 f (x) = (x6 + x4 + 1)(5x4 + 3x2 ) − (x5 + x3 − 1)(6x5 + 4x3 ) (x6 + x4 + 1)2 = (5x10 + 8x8 + 3x6 + 5x4 + 3x2 ) − (6x10 + 10x8 + 4x6 − 6x5 − 4x3 ) (x6 + x4 + 1)2 = 5x10 + 8x8 + 3x6 + 5x4 + 3x2 − 6x10 − 10x8 − 4x6 + 6x5 + 4x3 (x6 + x4 + 1)2 = −x10 − 2x8 − x6 + 6x5 + 5x4 + 4x3 + 3x2 . (x6 + x4 + 1)2 C03S02.027: If y (x) = x3 − 6x5 + 3 x−4 + 12, then the linear combination rule and the power rules yield 2 h (x) = 3x2 − 30x4 − 6x−5 . C03S02.028: Given: x(t) = 3 4 − 2 − 5 = 3t−1 − 4t−2 − 5, t t it follows from the linear combination rule and the power rule for negative integral exponents that x (t) = −3t−2 + 8t−3 = 8 3 8 − 3t − 2= . 3 t t t3 C03S02.029: Given: y (x) = 5 − 4x2 + x5 5 4x2 x5 = 3 − 3 + 3 = 5x−3 − 4x−1 + x2 , x3 x x x it follows from the linear combination rule and the power rules that y (x) = −15x−4 + 4x−2 + 2x = 2x + 4 15 2x5 + 4x2 − 15 − 4= . 2 x x x4 C03S02.030: Given u(x) = 2x − 3x2 + 2x4 2 32 = x−1 − + x2 , 2 5x 5 55 it follows from the linear combination rule and the power rules that 2 4 4x3 − 2 u (x) = − x−2 + x = . 5 5 5x2 C03S02.031: Because y (x) can be written in the form y (x) = 3x − 1 x−2 , the linear combination rule and 4 the power rules yield y (x) = 3 + 1 x−3 . 2 C03S02.032: We use the reciprocal rule, the linear combination rule, and the power rule for positive integral exponents: f (z ) = − Dz (z 3 + 2z 2 + 2z ) 3z 2 + 4z + 2 =− 2 2 . 2 (z 2 + 2z + 2)2 z z (z + 2z + 2)2 6 C03S02.033: If we first combine the two fractions, we will need to use the quotient rule only once: y (x) = x x+1 3x2 + x2 − 1 4x2 − 1 + = =2 , x−1 3x 3x(x − 1) 3x − 3x and therefore y (x) = (3x2 − 3x)(8x) − (4x2 − 1)(6x − 3) 24x3 − 24x2 − 24x3 + 12x2 + 6x − 3 −12x2 + 6x − 3 = = . 2 − 3x)2 2 − 3x)2 (3x (3x (3x2 − 3x)2 C03S02.034: First multiply each term in numerator and denominator by t2 to obtain u(t) = then apply the quotient rule: u (t) = 1 t2 =2 , −2 1 − 4t t −4 (t2 − 4)(2t) − (t2 )(2t) 8t =− 2 . 2 − 4)2 (t (t − 4)2 C03S02.035: The quotient rule (and other rules, such as the linear combination rule and the power rule) yield y (x) = = (x2 + 9)(3x2 − 4) − (x3 − 4x + 5)(2x) (x2 + 9)2 3x4 + 23x2 − 36 − 2x4 + 8x2 − 10x x4 + 31x2 − 10x − 36 = . (x2 + 9)2 (x2 + 9)2 C03S02.036: Expand w(z ) and take advantage of negative exponents: w(z ) = z 2 2z 3 − 3 4z 4 3 = 2z 5 − z −2 , 4 and so 3 3 20z 7 + 3 w (z ) = 10z 4 + z −3 = 10z 4 + 3 = . 2 2z 2z 3 C03S02.037: First multiply each term in numerator and denominator by 5x4 to obtain y (x) = 10x6 . 15x5 − 4 Then apply the quotient rule (among others): y (x) = (15x5 − 4)(60x5 ) − (10x6 )(75x4 ) 900x10 − 240x5 − 750x10 150x10 − 240x5 30x5 (5x5 − 8) = = = . 5 − 4)2 5 − 4)2 5 − 4)2 (15x (15x (15x (15x5 − 4)2 C03S02.038: First rewrite z (t) = 4 · 1 , t4 − 6t2 + 9 7 then apply the linear combination rule and the reciprocal rule to obtain z (t) = −4 · (t4 4t3 − 12t 48t − 16t3 16t(t2 − 3) 16t =2 =− 2 =− 2 . 2 + 9)2 4 − 6t (t − 3) (t − 3)4 (t − 3)3 C03S02.039: The quotient rule yields y (x) = (x + 1)(2x) − (x2 )(1) 2x2 + 2x − x2 x(x + 2) = = . 2 2 (x + 1) (x + 1) (x + 1)2 C03S02.040: Use the quotient rule, or if you prefer write h(w) = w−1 + 10w−2 , so that h (w) = −w−2 − 20w−3 = − 1 20 +3 2 w w =− w + 20 . w3 C03S02.041: Given f (x) = x3 and P (2, 8) on its graph, f (x) = 3x2 , so that f (2) = 12 is the slope of the line L tangent to the graph of f at P . So L has equation y − 8 = 12(x − 2); that is, 12x − y = 16. C03S02.042: Given f (x) = 3x2 − 4 and P (1, −1) on its graph, f (x) = 6x, so that f (1) = 6 is the slope of the line L tangent to the graph of f at P . So L has equation y + 1 = 6(x − 1); that is, 6x − y = 7. C03S02.043: Given f (x) = 1/(x − 1) and P (2, 1) on its graph, f (x) = − Dx (x − 1) 1 =− , (x − 1)2 (x − 1)2 so that f (2) = −1 is the slope of the line L tangent to the graph of f at P . So L has equation y − 1 = −(x − 2); that is, x + y = 3. C03S02.044: Given f (x) = 2x − x−1 and P (0.5, −1) on its graph, f (x) = 2 + x−2 , so that f (0.5) = 6 is the slope of the line L tangent to the graph of f at P . So L has equation y + 1 = 6 x − 1 ; that is, 2 6x − y = 4. C03S02.045: Given f (x) = x3 + 3x2 − 4x − 5 and P (1, −5) on its graph, f (x) = 3x2 + 6x − 4, so that f (1) = 5 is the slope of the line L tangent to the graph of f at P . So L has equation y + 5 = 5(x − 1); that is, 5x − y = 10. C03S02.046: Given f (x) = 1 1 − x x2 −1 = x−1 x2 −1 = x2 , x−1 and P (2, 4) on its graph, f (x) = (x − 1)(2x) − x2 x2 − 2x = , 2 (x − 1) (x − 1)2 so that f (2) = 0 is the slope of the line L tangent to the graph of f at P . So L has equation y − 4 = 0 · (x − 2); that is, y = 4. C03S02.047: Given f (x) = 3x−2 − 4x−3 and P (−1, 7) on its graph, f (x) = 12x−4 − 6x−3 , so that f (−1) = 18 is the slope of the line L tangent to the graph of f at P . So L has equation y − 7 = 18(x + 1); that is, 18x − y = −25. 8 C03S02.048: Given f (x) = 3x − 2 3x + 2 and P (2, 0.5) on its graph, f (x) = (3x + 2)(3) − (3x − 2)(3) 12 = , 2 (3x + 2) (3x + 2)2 3 so that f (2) = 16 is the slope of the line L tangent to the graph of f at P . So L has equation y − 1 = 2 that is, 3x − 16y = −2. 3 16 (x − 2); C03S02.049: Given f (x) = x2 3x2 +x+1 and P (−1, 3) on its graph, f (x) = (x2 + x + 1)(6x) − (3x2 )(2x + 1) 3x2 + 6x =2 , 2 + x + 1)2 (x (x + x + 1)2 so that f (−1) = −3 is the slope of the line L tangent to the graph of f at P . So an equation of the line L is y − 3 = −3(x + 1); that is, 3x + y = 0. C03S02.050: Given f (x) = 6 1 − x2 and P (2, −2) on its graph, f (x) = −6 · −2x 12x = , 2 )2 (1 − x (1 − x2 )2 so that f (2) = 8 is the slope of the line L tangent to the graph of f at P . So L has equation y +2 = 8 (x − 2); 3 3 that is, 8x − 3y = 22. C03S02.051: V = V0 (1 + αT + β T 2 + γ T 3 ) where α ≈ −0.06427 × 10−3 , β ≈ 8.5053 × 10−6 , and γ ≈ −6.79 × 10−8 . Now dV /dt = V0 (α +2β T +3γ T 2 ); V = V0 = 1000 when T = 0. Because V (0) = αV0 < 0, the water contracts when it is first heated. The rate of change of volume at that point is V (0) ≈ −0.06427 cm3 per ◦ C. 2 × 109 dW dW 4 × 109 62500 = (2 × 109 )R−2 , so ; when R = 3960, =− =− 2 R dR R3 dR 970299 (lb/mi). Thus W decreases initially at about 1.03 ounces per mile. C03S02.052: W= C03S02.053: Draw a cross section of the tank through its axis of symmetry. Let r denote the radius of the (circular) water surface when the height of water in the tank is h. Draw a typical radius, label it r, and label the height h. From similar triangles in your figure, deduce that h/r = 800/160 = 5, so r = h/5. The volume 1 of water in a cone of height h and radius r is V = 1 π r2 h, so in this case we have V = V (h) = 75 π h3 . The rate 3 1 of change of V with respect to h is dV /dh = 25 π h2 , and therefore when h = 600, we have V (600) = 14400π ; that is, approximately 45239 cm3 per cm. 9 C03S02.054: Because y (x) = 3x2 +2x +1, the slope of the tangent line at (1, 3) is y (1) = 6. The equation of the tangent line at (1, 3) is y − 3 = 6(x − 1); that is, y = 6x − 3. The intercepts of the tangent line are (0, −3) and 1 , 0 . 2 C03S02.055: The slope of the tangent line can be computed using dy/dx at x = a and also by using the two points known to lie on the line. We thereby find that 3a2 = a3 − 5 . a−1 This leads to the equation (a + 1)(2a2 − 5a + 5) = 0. The quadratic factor has negative discriminant, so the only real solution of the cubic equation is a = −1. The point of tangency is (−1, −1), the slope there is 3, and the equation of the line in question is y = 3x + 2. C03S02.056: Let (a, a3 ) be a point of tangency. The tangent line therefore has slope 3a2 and, because it passes through (2, 8), we have 3a2 = a3 − 8 ; a−2 that is, 3a2 (a − 2) = a3 − 8. This leads to the equation 2a2 − 2a − 4 = 0, so that a = −1 or a = 2. The solution a = 2 yields the line tangent at (2, 8) with slope 12. The solution a = −1 gives the line tangent at (−1, −1) with slope 3. The two lines have equations y − 8 = 12(x − 2) and y + 1 = 3(x + 1); that is, y = 12x − 16 and y = 3x + 2. C03S02.057: Suppose that some straight line L is tangent to the graph of f (x) = x2 at the points (a, a2 ) and (b, b2 ). Our plan is to show that a = b, and we may conclude that L cannot be tangent to the graph of f at two different points. Because f (x) = 2x and because (a, a2 ) and (b, b2 ) both lie on L, the slope of L is equal to both f (a) and f (b); that is, 2a = 2b. Hence a = b, so that (a, a2 ) and (b, b2 ) are the same point. Conclusion: No straight line can be tangent to the graph of y = x2 at two different points. C03S02.058: Let (a, 1/a) be a point of tangency. The slope of the tangent there is −1/a2 , so −1/a2 = −2. √ √ Thus there are two possible values for a: ± 1 2. These lead to the equations of the two lines: y = −2x +2 2 2 √ and y = −2x − 2 2. C03S02.059: Given f (x) = xn , we have f (x) = nxn−1 . The line tangent to the graph of f at the point P (x0 , y0 ) has slope that we compute in two ways and then equate: y − (x0 )n = n(x0 )n−1 . x − x0 To find the x-intercept of this line, substitute y = 0 into this equation and solve for x. It follows that the n−1 x-intercept is x = x0 . n C03S02.060: Because dy/dx = 5x4 + 2 2 > 0 for all x, the curve has no horizontal tangent line. The minimal slope occurs when dy/dx is minimal, and this occurs when x = 0. So the smallest slope that a line tangent to this graph can have is 2. C03S02.061: Dx [f (x)]3 = f (x)f (x)f (x) + f (x)f (x)f (x) + f (x)f (x)f (x) = 3[f (x)]2 f (x). C03S02.062: Suppose that u1 , u2 , u3 , u4 , and u5 are differentiable functions of x. Let primes denote derivatives with respect to x. Then 10 Dx [u1 u2 u3 u4 ] = Dx [(u1 u2 u3 )u4 ] = (u1 u2 u3 ) u4 + (u1 u2 u3 )u4 = (u1 u2 u3 + u1 u2 u3 + u1 u2 u3 )u4 + (u1 u2 u3 )u4 = u1 u2 u3 u4 + u1 u2 u3 u4 + u1 u2 u3 u4 + u1 u2 u3 u4 . Next, using this result, Dx [u1 u2 u3 u4 u5 ] = Dx [(u1 u2 u3 u4 )u5 ] = (u1 u2 u3 u4 ) u5 + (u1 u2 u3 u4 )u5 = (u1 u2 u3 u4 + u1 u2 u3 u4 + u1 u2 u3 u4 + u1 u2 u3 u4 )u5 + (u1 u2 u3 u4 )u5 = u1 u2 u3 u4 u5 + u1 u2 u3 u4 u5 + u1 u2 u3 u4 u5 + u1 u2 u3 u4 u5 + u1 u2 u3 u4 u5 . C03S02.063: Let u1 (x) = u2 (x) = u3 (x) = · · · = un−1 (x) = un (x) = f (x). Then the left-hand side of Eq. (16) is Dx [ (f (x))n ] and the right-hand side is f (x)[f (x)]n−1 + f (x)f (x)[f (x)]n−2 + [f (x)]2 f (x)[f (x)]n−3 + · · · + [f (x)]n−1 f (x) = n[f (x)]n−1 · f (x). Therefore if n is a positive integer and f (x) exists, then Dx [ (f (x))n ] = n(f (x))n−1 · f (x). C03S02.064: Substitution of f (x) = x2 + x + 1 and n = 100 in the result of Problem 63 yields Dx [ (x2 + x + 1)100 ] = Dx [ (f (x))n ] = n(f (x))n−1 · f (x) = 100(x2 + x + 1)99 · (2x + 1). C03S02.065: Let f (x) = x3 − 17x + 35 and let n = 17. Then g (x) = (f (x))n . Hence, by the result in Problem 63, g (x) = Dx [ (f (x))n ] = n(f (x))n−1 · f (x) = 17(x3 − 17x + 35)16 · (3x2 − 17). C03S02.066: We begin with f (x) = ax3 + bx2 + cx + d. Then f (x) = 3ax2 + 2bx + c. The conditions in the problem require that (simultaneously) 1 = f (0) = d, 0 = f (0) = c, 0 = f (1) = a + b + c + d, and 0 = f (1) = 3a + 2b + c. These equations have the unique solution a = 2, b = −3, c = 0, and d = 1. Therefore f (x) = 2x3 − 3x2 + 1 is the only possible solution. It is easy to verify that f (x) satisfies the conditions required in the problem. C03S02.067: If n is a positive integer and f (x) = xn , 1 + x2 then f (x) = (1 + x2 )(nxn−1 ) − (2x)(xn ) nxn−1 + nxn+1 − 2xn+1 xn−1 [n + (n − 2)x2 ] = = . 2 )2 2 )2 (1 + x (1 + x (1 + x2 )2 11 (1) If n = 0, then (by the reciprocal rule) f (x) = − 2x . (1 + x2 )2 If n = 2, then by Eq. (1) f (x) = 2x . (1 + x2 )2 In each case there can be but one solution of f (x) = 0, so there is only one horizontal tangent line. If n = 0 it is tangent to the graph of f at the point (0, 1); if n = 2 it is tangent to the graph of f at the point (0, 0). C03S02.068: If n = 1, then Eq. (1) of the solution of Problem 67 yields f (x) = 1 − x2 . (1 + x2 )2 The equation f (x) = 0 has the two solutions x = ± 1, so there are two points on the graph of f where the tangent line is horizontal: −1, − 1 and 1, 1 . 2 2 C03S02.069: If n is a positive integer and n 3, f (x) = 0 only when the numerator is zero in Eq. (1) of the solution of Problem 67; that is, when xn−1 (n + [n − 2]x2 ) = 0. But this implies that x = 0 (because n 3) or that n + [n − 2]x2 = 0. The latter is impossible because n > 0 and [n − 2]x2 0. Therefore the only horizontal tangent to the graph of f is at the point (0, 0). C03S02.070: By Eq. (1) in the solution of Problem 67, if f (x) = x3 , 1 + x2 then f (x) = x2 (3 + x2 ) . (1 + x2 )2 So f (x) = 1 when x2 (3 + x2 ) = 1; (1 + x2 )2 x2 (3 + x2 ) = (1 + x2 )2 ; x4 + 3x2 = x4 + 2x2 + 1; x2 = 1. Therefore there are two points where the line tangent to the graph of f has slope 1; they are −1, − 1 and 2 1, 1 . 2 C03S02.071: If f (x) = x3 , 1 + x2 then f (x) = x2 (3 + x2 ) x4 + 3x2 = , (1 + x2 )2 (1 + x2 )2 by Eq. (1) in the solution of Problem 67. A line tangent to the graph of y = f (x) will be horizontal when the derivative f (x) of f (x) is zero. But 12 Dx [f (x) ] = f (x) = = (1 + x2 )2 (4x3 + 6x) − (x4 + 3x2 )(4x3 + 4x) (1 + x2 )4 (1 + x2 )2 (4x3 + 6x) − (x4 + 3x2 )(4x)(x2 + 1) (1 + x2 )(4x3 + 6x) − (x4 + 3x2 )(4x) = (1 + x2 )4 (1 + x2 )3 4x3 + 6x + 4x5 + 6x3 − 4x5 − 12x3 6x − 2x3 2x(3 − x2 ) = = . 2 )3 2 )3 (1 + x (1 + x (1 + x2 )3 √ So f (x) = 0 when x = 0 and when x = ± 3. Therefore there√ three points on the graph of y = f (x) are √ at which the tangent line is horizontal: (0, 0), − 3, 9 , and 3, 9 . 8 8 = C03S02.072: (a) Using the quadratic formula, V (T ) = 0 when T = Tm √ 170100 − 20 59243226 = ≈ 3.96680349529363770572 4074 (in ◦ C) and substitution in the formula for V (T ) (Example 5) yields Vm = V (Tm ) ≈ 999.87464592037071155281 (cm3 ). (b) The Mathematica command Solve[ V(T) == 1000, T ] yielded three solutions, the only one of which is close to T = 8 was √ 85050 − 10 54879293 T= ≈ 8.07764394099814733845 1358 (in ◦ C). C03S02.073: The graph of f (x) = | x3 | is shown next. 0.8 0.6 0.4 0.2 -1.5 -1 -0.5 0.5 1 1.5 Clearly f is differentiable at x if x = 0. Moreover, f− (0) = lim h→0− | (0 + h)3 | − 0 −h3 = lim− =0 h h h→0 and f+ (0) = 0 by a similar computation. Therefore f is differentiable everywhere. 13 C03S02.074: The graph of f (x) = x3 + | x3 | is shown next. 0.5 0.4 0.3 0.2 0.1 -1.5 -1 -0.5 0.5 1 1.5 Clearly f is differentiable except possibly at x = 0. Moreover, f− (0) = lim h→0− (0 + h)3 + | (0 + h)3 | h3 − h3 = lim− =0 h h h→0 and f+ (0) = lim h→0+ (0 + h)3 + | (0 + h)3 | 2h 3 = lim+ = 0. h h h→0 Therefore f is differentiable at x for all x in R. C03S02.075: The graph of f (x) = 2 + 3x2 if x < 1, 3 + 2x3 if x 1, is shown next. 17.5 15 12.5 10 7.5 5 2.5 0.5 1 1.5 2 Clearly f is differentiable except possibly at x = 1. But f− (1) = lim − h→0 2 + 3(1 + h)2 − 5 6h + 3h2 = lim− =6 h h h→0 14 and f+ (1) = lim h→0+ 3 + 2(1 + h)3 − 5 6h + 6h2 + 2h3 = lim+ = 6. h h h→0 Therefore f (1) exists (and f (1) = 6), and hence f (x) exists for every real number x. C03S02.076: The graph of x4 if x < 1, 1 2− 4 x f (x) = if x 1 is shown next. 1.75 1.5 1.25 1 0.75 0.5 0.25 0.25 0.5 0.75 1 1.25 1.5 1.75 Clearly f is differentiable except possibly at x = 1. But here we have f− (1) = lim − h→0 (1 + h)4 − 1 4h + 6h2 + 4h3 + h4 = lim− =4 h h h→0 and f+ (1) = lim h→0+ = lim h→0+ 1 1 · 2− −1 h (1 + h)4 = lim h→0+ 1 1 · 1− h (1 + h)4 4h + 6h2 + 4h3 + h4 4 + 6 h + 4h 2 + h 3 4 = lim+ = = 4. 4 4 h(1 + h) (1 + h) 1 h→0 Therefore f is differentiable at x = 1 as well. C03S02.077: The graph of 1 2−x f (x) = x 15 if x < 1, if x 1 is shown next. 1.6 1.4 1.2 0.25 0.5 0.75 1 1.25 1.5 1.75 0.8 0.6 Clearly f is differentiable except possibly at x = 1. But f− (1) = lim − h→0 1 · h and 1 −1 2 − (1 + h) = lim+ h→0 1 · h 1 −1 1−h f+ (1) = lim + h→0 = lim− h→0 h 1 = lim− =1 h(1 − h) h→0 1 − h 1+h−1 = 1. h Thus f is differentiable at x = 1 as well (and f (1) = 1). C03S02.078: The graph of the function 12 (5 − x)2 f (x) = 2 x − 3x + 3 is next. if x < 3, if x 3 7 6 5 4 3 2.5 3 3.5 4 Clearly f is differentiable except possibly at x = 3. But f− (3) = lim − h→0 1 · h 12 −3 (5 − 3 − h)2 = lim− h→0 12 − 3(2 − h)2 12h − 3h2 12 − 3h = lim− = lim− =3 2 h(2 − h) h(2 − h)2 h→0 h→0 (2 − h)2 16 and f+ (3) = lim h→0+ (3 + h)2 − 3(3 + h) + 3 − 3 6h + h2 − 3h = lim+ = 3. h h h→0 Thus f is differentiable at x = 3 as well. 17 Section 3.3 C03S03.001: Given y = (3x + 4)5 , the chain rule yields dy = 5 · (3x + 4)4 · Dx (3x + 4) = 5 · (3x + 4)4 · 3 = 15(3x + 4)4 . dx C03S03.002: Given y = (2 − 5x)3 , the chain rule yields dy = 3 · (2 − 5x)2 · Dx (2 − 5x) = 3 · (2 − 5x)2 · (−5) = −15(2 − 5x)2 . dx C03S03.003: Rewrite the given function in the form y = (3x − 2)−1 in order to apply the chain rule. The result is dy 3 . = (−1)(3x − 2)−2 · Dx (3x − 2) = (−1)(3x − 2)−2 · 3 = −3(3x − 2)−2 = − dx (3x − 2)2 C03S03.004: Rewrite the given function in the form y = (2x + 1)−3 in order to apply the chain rule. The result is dy 6 . = (−3)(2x + 1)−4 · Dx (2x + 1) = (−3)(2x + 1)−4 · 2 = −6(2x + 1)−4 = − dx (2x + 1)4 C03S03.005: Given y = (x2 + 3x + 4)3 , the chain rule yields dy = 3(x2 + 3x + 4)2 · Dx (x2 + 3x + 4) = 3(x2 + 3x + 4)2 (2x + 3). dx C03S03.006: dy 24x2 = −4 · (7 − 2x3 )−5 · Dx (7 − 2x3 ) = −4 · (7 − 2x3 )−5 · (−6x2 ) = 24x2 (7 − 2x3 )−5 = . dx (7 − 2x3 )5 C03S03.007: We use the product rule, and in the process of doing so must use the chain rule twice: Given y = (2 − x)4 (3 + x)7 , dy = (2 − x)4 · Dx (3 + x)7 + (3 + x)7 · Dx (2 − x)4 dx = (2 − x)4 · 7 · (3 + x)6 · Dx (3 + x) + (3 + x)7 · 4(2 − x)3 · Dx (2 − x) = (2 − x)4 · 7 · (3 + x)6 · 1 + (3 + x)7 · 4(2 − x)3 · (−1) = 7(2 − x)4 (3 + x)6 − 4(2 − x)3 (3 + x)7 = (2 − x)3 (3 + x)6 (14 − 7x − 12 − 4x) = (2 − x)3 (3 + x)6 (2 − 11x). The last simplifications would be necessary only if you needed to find where y (x) is positive, where negative, and where zero. C03S03.008: Given y = (x + x2 )5 (1 + x3 )2 , the product rule—followed by two applications of the chain rule—yields 1 dy = (x + x2 )5 · Dx (1 + x3 )2 + (1 + x3 )2 · Dx (x + x2 )5 dx = (x + x2 )5 · 2 · (1 + x3 ) · Dx (1 + x3 ) + (1 + x3 )2 · 5 · (x + x2 )4 · Dx (x + x2 ) = (x + x2 )5 · 2 · (1 + x3 ) · 3x2 + (1 + x3 )2 · 5 · (x + x2 )4 · (1 + 2x) = · · · = x4 (x + 1)6 (x2 − x + 1)(16x3 − 5x2 + 5x + 5). Sometimes you have to factor an expression as much as you can to determine where it is positive, negative, or zero. C03S03.009: We will use the quotient rule, which will require use of the chain rule to find the derivative of the denominator: dy (3x − 4)3 Dx (x + 2) − (x + 2)Dx (3x − 4)3 = 2 dx [ (3x − 4)3 ] = (3x − 4)3 · 1 − (x + 2) · 3 · (3x − 4)2 · Dx (3x − 4) (3x − 4)6 = (3x − 4)3 − 3(x + 2)(3x − 4)2 · 3 (3x − 4) − 9(x + 2) 6x + 22 = =− . 6 4 (3x − 4) (3x − 4) (3x − 4)4 C03S03.010: We use the quotient rule, and need to use the chain rule twice along the way: dy (4 + 5x + 6x2 )2 · Dx (1 − x2 )3 − (1 − x2 )3 · Dx (4 + 5x + 6x2 )2 = dx (4 + 5x + 6x2 )4 = (4 + 5x + 6x2 )2 · 3 · (1 − x2 )2 · Dx (1 − x2 ) − (1 − x2 )3 · 2 · (4 + 5x + 6x2 ) · Dx (4 + 5x + 6x2 ) (4 + 5x + 6x2 )4 = (4 + 5x + 6x2 )2 · 3 · (1 − x2 )2 · (−2x) − (1 − x2 )3 · 2 · (4 + 5x + 6x2 ) · (5 + 12x) (4 + 5x + 6x2 )4 = (4 + 5x + 6x2 ) · 3 · (1 − x2 )2 · (−2x) − (1 − x2 )3 · 2 · (5 + 12x) 2(x2 − 1)2 (6x3 + 10x2 + 24x + 5) =− . (4 + 5x + 6x2 )3 (4 + 5x + 6x2 )3 C03S03.011: Here is a problem in which use of the chain rule contains another use of the chain rule. Given y = [1 + (1 + x)3 ]4 , dy = 4[1 + (1 + x)3 ]3 · Dx [1 + (1 + x)3 ] = 4[1 + (1 + x)3 ]3 · [0 + Dx (1 + x)3 ] dx = 4[1 + (1 + x)3 ]3 · 3 · (1 + x)2 · Dx (1 + x) = 12[1 + (1 + x)3 ]3 (1 + x)2 . C03S03.012: Again a “nested chain rule” problem: 2 dy = −5 · x + (x + x2 )−3 dx −6 = −5 · x + (x + x2 )−3 −6 1 + (−3) · (x + x2 )−4 · Dx (x + x2 ) = −5 · x + (x + x2 )−3 −6 1 + (−3) · (x + x2 )−4 · (1 + 2x) . C03S03.013: Given: y = (u + 1)3 and u = · Dx x + (x + x2 )−3 1 . The chain rule yields x2 dy −2 6 dy du = · = 3(u + 1)2 · 3 = − 3 dx du dx x x 1 +1 x2 2 =− 6(x2 + 1)2 . x7 C03S03.014: Write y = 1 u−1 − 1 u−2 . Then, with u = 2x + 1, the chain rule yields 2 3 dy dy du = · = dx du dx = −2 2 1 − u−2 + u−3 2 3 1 2 − 2(2x + 1)2 3(2x + 1)3 · 2 = −2 1 2 −3 2u 2 3u = ··· = 1 − 6x . 3(1 + 2x)3 C03S03.015: Given y = (1 + u2 )3 and u = (4x − 1)2 , the chain rule yields dy dy du = · = 6u(1 + u2 )2 · 8 · (4x − 1) dx du dx = 48 · (4x − 1)2 (1 + (4x − 1)4 )2 (4x − 1) = 48(4x − 1)3 (1 + (4x − 1)4 )2 . Without the chain rule, our only way to differentiate y (x) would be first to expand it: y (x) = 8 − 192x + 2688x2 − 25600x3 + 181248x4 − 983040x5 + 4128768x6 − 13369344x7 + 32636928x8 − 57671680x9 + 69206016x10 − 50331648x11 + 16777216x12 . Then we could differentiate y (x) using the linear combination and power rules: y (x) = −192 + 5376x − 76800x2 + 724992x3 − 4915200x4 + 24772608x5 − 93585408x6 + 261095424x7 − 519045120x8 + 692060160x9 − 553648128x10 + 201326592x11 . Fortunately, the chain rule is available—and even if not, we still have Maple, Derive, Mathematica, and MATLAB. 1 , then the chain rule yields 3x − 2 dy 5 Dx (3x − 2) 3 15 dy du =− · =− . = · = 5u4 · − dx du dx (3x − 2)2 (3x − 2)4 (3x − 2)2 (3x − 2)6 C03S03.016: If y = u5 and u = C03S03.017: If y = u(1 − u)3 and u = 1 , then the chain rule yields x4 dy dy du = · = (1 − u)3 − 3u(1 − u)2 · −4x−5 = (1 − x−4 )3 − 3x−4 (1 − x−4 )2 · −4x−5 , dx du dx 3 which a very patient person can simplify to dy 16 − 36x4 + 24x8 − 4x12 = . dx x17 C03S03.018: If y = u x and u = , then u+1 x+1 dy 1 1 dy du (u + 1) · 1 − u · 1 (x + 1) · 1 − x · 1 · = · = · = dx du dx (u + 1)2 (x + 1)2 (u + 1)2 (x + 1)2 = 1 x +1 x+1 2 · 1 = (x + 1)2 1 2x + 1 x+1 2 · 1 1 = . (x + 1)2 (2x + 1)2 C03S03.019: If y = u2 (u − u4 )3 and u = x−2 , then dy dy du = · = 2u(u − u4 )3 + 3u2 (u − u4 )2 (1 − 4u3 ) · (−2x−3 ) dx du dx = 2x−2 (x−2 − x−8 )3 + 3x−4 (x−2 − x−8 )2 (1 − 4x−6 ) · (−2x−3 ) = · · · = C03S03.020: If y = 28 − 66x6 + 48x12 − 10x18 . x29 u and u = x − 2x−1 , then (2u + 1)4 dy dy du (2u + 1)4 − 8u(2u + 1)3 2u + 1 − 8u = · = · (1 + 2x−2 ) = · (1 + 2x−2 ) 8 dx du dx (2u + 1) (2u + 1)5 1 − 6u x2 + 2 = · = (2u + 1)5 x2 1 − 6x + 2x − 12 x 4 +1 x 5 · x2 + 2 = x2 x − 6x2 + 12 x 2x2 − 4 + x x 5 = x4 (12 + x − 6x2 ) x2 + 2 x2 (12 + x − 6x2 )(x2 + 2) · = 2 + x − 4)5 2 (2x x (2x2 + x − 4)5 = x2 (12x2 + x3 − 6x4 + 24 + 2x − 12x2 ) x2 (24 + 2x + x3 − 6x4 ) = . 2 + x − 4)5 (2x (2x2 + x − 4)5 C03S03.021: Let u(x) = 2x − x2 and n = 3. Then f (x) = un , so that f (x) = nun−1 · du = 3u2 · (2 − 2x) = 3(2x − x2 )2 (2 − 2x). dx C03S03.022: Let u(x) = 2 + 5x3 and n = −1. Then f (x) = un , so that f (x) = nun−1 · du 15x2 = (−1)u−2 · 15x2 = − . dx (2 + 5x3 )2 C03S03.023: Let u(x) = 1 − x2 and n = −4. Then f (x) = un , so that 4 · x2 + 2 x2 f (x) = nun−1 · du 8x = (−4)u−5 · (−2x) = . dx (1 − x2 )5 C03S03.024: Let u(x) = x2 − 4x + 1 and n = 3. Then f (x) = un , so du = 3u2 · (2x − 4) = 3(x2 − 4x + 1)2 (2x − 4). dx f (x) = nun−1 · C03S03.025: Let u(x) = f (x) = nun−1 · x+1 and n = 7. Then f (x) = un , and therefore x−1 du (x − 1) − (x + 1) = 7u6 · =7 dx (x − 1)2 C03S03.026: Let n = 4 and u(x) = f (x) = nun−1 · = du =4 dx x2 + x + 1 x+1 x+1 x−1 6 · −2 (x + 1)6 = − 14 · . (x − 1)2 (x − 1)8 x2 + x + 1 . Thus x+1 3 · (x + 1)(2x + 1) − (x2 + x + 1) (x + 1)2 4(x2 + x + 1)3 2x2 + 3x + 1 − x2 − x − 1 4(x2 + x + 1)3 x2 + 2x 4(x2 + x + 1)3 (x2 + 2x) · = · = . (x + 1)3 (x + 1)2 (x + 1)3 (x + 1)2 (x + 1)5 C03S03.027: g (y ) = 1 + 5(2y − 3)4 · 2 = 1 + 10(2y − 3)4 . C03S03.028: h (z ) = 2z (z 2 + 4)3 + 3z 2 (z 2 + 4)2 · 2z = (2z 3 + 8z + 6z 3 )(z 2 + 4)2 = 8z (z 2 + 1)(z 2 + 4)2 . C03S03.029: If F (s) = (s − s−2 )3 , then F (s) = 3(s − s−2 )2 (1 + 2s−3 ) = 3 s − =3· C03S03.030: If G(t) = G (t) = 2 2t − 1 s2 2 · 1+ 2 s3 =3 s3 − 1 s2 2 · s3 + 2 s3 (s3 − 1)2 (s3 + 2) (s6 − 2s3 + 1)(s3 + 2) 3(s9 − 3s3 + 2) =3· = . s7 s7 s7 t2 + 1 + 1 t 2 , then 1 1 · t2 + 1 + 2 t t = 4t3 + 4t + 2 − 2 2 4t6 + 4t4 + 2t3 − 2t − 2 − 3= . t2 t t3 C03S03.031: If f (u) = (1 + u)3 (1 + u2 )4 , then f (u) = 3(1 + u)2 (1 + u2 )4 + 8u(1 + u)3 (1 + u2 )3 = (1 + u)2 (1 + u2 )3 (11u2 + 8u + 3). C03S03.032: If g (w) = (w2 − 3w + 4)(w + 4)5 , then g (w) = 5(w + 4)4 (w2 − 3w + 4) + (w + 4)5 (2w − 3) = (w + 4)4 (7w2 − 10w + 8). 5 v− 1− h (v ) = (−2) v − 1 − −1 −2 1 v 1 v C03S03.033: If h(v ) = C03S03.034: If p(t) = p (t) = 4 =4 −1 −3 −4 1 1 1 + 2+ 3 t t t 1 1 1 + 2+ 3 t t t t3 t2 + t + 1 , then 1 v −2 1 v2 = 2(v − 1)(v 2 − 2v + 2) . v 3 (2 − v )3 , then −5 5 1+ 1− 1 2 3 + 3+ 4 2 t t t t 2 + 2t + 3 t4 = =4 t2 + t + 1 t3 −5 t2 + 2t + 3 t4 4t15 (t2 + 2t + 3) 4t11 (t2 + 2t + 3) = . (t2 + t + 1)5 t4 (t2 + t + 1)5 C03S03.035: If F (z ) = (5z 5 − 4z + 3)−10 , then F (z ) = −10(5z 5 − 4z + 3)−11 (25z 4 − 4) = 40 − 250z 4 . (5z 5 − 4z + 3)11 C03S03.036: Given G(x) = (1 + [x + (x2 + x3 )4 ]5 )6 , G (x) = 6(1 + [x + (x2 + x3 )4 ]5 )5 · 5[x + (x2 + x3 )4 ]4 · [1 + 4(x2 + x3 )3 (2x + 3x2 )]. When G (x) is expanded completely (written in polynomial form), it has degree 359 and the term with largest coefficient is 74313942135996360069651059069038417440x287 . C03S03.037: Chain rule: dy dy = 4(x3 )3 · 3x2 . Power rule: = 12x11 . dx dx C03S03.038: Chain rule: dy = (−1) dx C03S03.039: Chain rule: dy dy = 2(x2 − 1) · 2x. Without chain rule: = 4x3 − 4x. dx dx C03S03.040: Chain rule: dy dy = −3(1 − x)2 . Without chain rule: = −3 + 6x − 3x2 . dx dx C03S03.041: Chain rule: dy dy = 4(x + 1)3 . Without chain rule: = 4x3 + 12x2 + 12x + 4. dx dx C03S03.042: Chain rule: dy dy 2x + 2 = −2(x + 1)−3 . Reciprocal rule: =− 2 . dx dx (x + 2x + 1)2 C03S03.043: Chain rule: dy dy 2x . = −2x(x2 + 1)−2 . Reciprocal rule: =− 2 dx dx (x + 1)2 C03S03.044: Chain rule: dy dy = 2(x2 + 1) · 2x. Product rule: = 2x(x2 + 1) + 2x(x2 + 1). dx dx 1 x −2 − 1 dy = 1. . Power rule: 2 x dx 6 C03S03.045: If f (x) = sin(x3 ), then f (x) = cos(x3 ) · Dx (x3 ) = 3x2 cos(x3 ) = 3x2 cos x3 . C03S03.046: If g (t) = (sin t)3 , then g (t) = 3(sin t)2 · Dt sin t = (3 sin2 t)(cos t) = 3 sin2 t cos t. C03S03.047: If g (z ) = (sin 2z )3 , then g (z ) = 3(sin 2z )2 · Dz (sin 2z ) = 3(sin 2z )2 (cos 2z ) · Dz (2z ) = 6 sin2 2z cos 2z. C03S03.048: If k (u) = sin(1 + sin u), then k (u) = [ cos(1 + sin u) ] · Du (1 + sin u) = [ cos(1 + sin u) ] · cos u. C03S03.049: The radius of the circular ripple is r(t) = 2t and its area is a(t) = π (2t)2 ; thus a (t) = 8π t. When r = 10, t = 5, and at that time the rate of change of area with respect to time is a (5) = 40π (in.2 /s). C03S03.050: If the circle has area A and radius r, then A = π r2 , so that r = seconds, then the rate of change of the radius of the circle is A/π . If t denotes time in dr dr dA 1 dA · = · =√ . dt dA dt 2 π A dt (1) We are given the values A = 75π and dA/dt = −2π ; when we substitute these values into the last expression √ √ dr 1 1 in Eq. (1), we find that = − 15 3. Hence the radius of the circle is decreasing at the rate of − 15 3 dt (cm/s) at the time in question. C03S03.051: Let A denote the area of the square and x the length of each edge. Then A = x2 , so dA/dx = 2x. If t denotes time (in seconds), then dA dA dx dx = · = 2x . dt dx dt dt All that remains is to substitute the given data x = 10 and dx/dt = 2 to find that the area of the square is increasing at the rate of 40 in.2 /s at the time in question. √ C03S03.052: √ x denote the length of each side of the triangle. Then its altitude is 1 x 3, and so its Let 2 area is A = 1 x2 3. Therefore the rate of change of its area with respect to time t (in seconds) is 4 dA = dt 1√ dx x3· . 2 dt √ We are given x = 10 and dx/dt = 2, so at that point the area is increasing at 10 3 (in.2 /s). dV dx = 3x2 . We dt dt are given dx/dt = −2, so when x = 10 the volume of the block is decreasing at 600 in.3 /h. C03S03.053: The volume of the block is V = x3 where x is the length of each edge. So C03S03.054: By the chain rule, f (y ) = h (g (y )) · g (y ). Then substitution of the data given in the problem yields f (−1) = h (g (−1)) · g (−1) = h (2) · g (−1) = −1 · 7 = −7. C03S03.055: G (t) = f (h(t)) · h (t). Now h(1) = 4, h (1) = −6, and f (4) = 3, so G (1) = 3 · (−6) = −18. C03S03.056: The derivative of f (f (f (x))) is the product of the three expressions f (f (f (x))), f (f (x)), and f (x). When x = 0, f (x) = 0 and f (x) = 1. Thus when x = 0, each of those three expressions has value 1, so the answer is 1. 7 C03S03.057: The volume of the balloon is given by V = 4 π r3 , so 3 dV dr dV dr = · = 4π r2 . dt dr dt dt Answer: When r = 10, dV /dt = 4π · 102 · 1 = 400π ≈ 1256.64 (cm3 /s). C03S03.058: Let V denote the volume of the balloon and r its radius at time t (in seconds). We are given dV /dt = 200π . Now dV dr dV dr = · = 4π r2 . dt dr dt dt When r = 5, we have 200π = 4π · 25 · (dr/dt), so dr/dt = 2. Answer: When r = 5 (cm), the radius of the balloon is increasing at 2 cm/s. dr dV dr C03S03.059: Given: = −3. Now = −300π = 4π r2 · ( ). So 4π r2 = 100π , and thus r = 5 (cm) at dt dt dt the time in question. C03S03.060: Let x denote the radius of the hailstone and let V denote its volume. Then V= 43 πx , 3 and so dV dx = 4π x2 . dt dt dV dx 1 dx 1 = −0.1, and therefore − = 4π · 22 · . So =− . Answer: At the time in dt 10 dt dt 160π 1 question, the radius of the hailstone is decreasing at cm/s —that is, at about 0.002 cm/s. 160π When x = 2, C03S03.061: Let V denote the volume of the snowball and A its surface area at time t (in hours). Then dV = kA dt and A = cV 2/3 (the latter because A is proportional to r2 , whereas V is proportional to r3 ). Therefore dV = α V 2 /3 dt and thus dt = β V −2/3 dV (α and β are constants). From the last equation we may conclude that t = γ V 1/3 + δ for some constants γ and δ , so that V = V (t) = (P t + Q)3 for some constants P and Q. From the information 500 = V (0) = Q3 √ √ √ and 250 = V (1) = (P + Q)3 , we find that Q = 5 3 4 and that P = −5 · ( 3 4 − 3 2). Now V (t) = 0 when P T + Q = 0; it turns out that T=√ 3 √ 3 2 ≈ 4.8473. 2−1 Therefore the snowball finishes melting at about 2:50:50 P.M. on the same day. C03S03.062: Let V denote the volume of the block, x the length of each of its edges. Then V = x3 . In 8 hours x decreases from 20 to 8, and dx/dt is steady, so t hours after 8:00 a.m. have 3 x = 20 − t. 2 Also 8 2 dV dV dx 3 9 3 = · = 3x2 · (− ) = − · 20 − t dt dx dt 2 2 2 At 12 noon we have t = 4, so at noon in.3 /h then. . dV = − 9 (20 − 6)2 = −882. Answer: The volume is decreasing at 882 2 dt C03S03.063: By the chain rule, dv dv dw = · , dx dw dx du du dv du dv dw = · = · · . dx dv dx dv dw dx and therefore n C03S03.064: Given: n is a fixed integer, f is differentiable, f (1) = 1, F (x) = f (xn ), and G(x) = [ f (x) ] . Then n F (1) = f (1n ) = f (1) = 1 = 1n = [ f (1) ] = G(1). Next, F (x) = Dx f (xn ) = f (xn ) · nxn−1 n−1 n and G (x) = Dx [ f (x) ] = n [ f (x) ] · f (x). Therefore n−1 F (1) = f (1n ) · n · 1 = nf (1) = n · 1n−1 · f (1) = n · [ f (1) ] C03S03.065: If h(x) = √ · f (1) = G (1). x + 4 , then 1 1 1 √ · Dx (x + 4) = √ ·1 = √ . 2 x+4 2 x+4 2 x+4 h (x) = C03S03.066: If h(x) = x3/2 = x · h (x) = 1 · √ √ x , then x + x · Dx √ x = √ √ x 1√ 3√ x+ √ = x+ x= x. 2 2 2x √ C03S03.067: If h(x) = (x2 + 4)3/2 = (x2 + 4) x2 + 4 , then 1 h (x) = 2x x2 + 4 + (x2 + 4) · √ · 2x = 2x x2 + 4 + x x2 + 4 = 3x x2 + 4 . 2+4 2x C03S03.068: If h(x) = | x | = √ x2 , then h (x) = 2x 1 x √ · Dx x2 = √ = . 2 2 |x| 2x 2x 9 Section 3.4 C03S04.001: Write f (x) = 4x5/2 + 2x−1/2 to find f (x) = 10x3/2 − x−3/2 = 10x3/2 − 1 x3/2 = 10x3 − 1 . x3/2 C03S04.002: Write g (t) = 9t4/3 − 3t−1/3 to find g (t) = 12t1/3 + t−4/3 = 12t1/3 + 1 t4 / 3 = 12t5/3 + 1 . t4 / 3 C03S04.003: Write f (x) = (2x + 1)1/2 to find f (x) = 1 1 (2x + 1)−1/2 · 2 = √ . 2 2x + 1 C03S04.004: Write h(z ) = (7 − 6z )−1/3 to find that 1 2 h (z ) = − (7 − 6z )−4/3 · (−6) = . 3 (7 − 6z )4/3 3 3(x2 + 2) C03S04.005: Write f (x) = 6x−1/2 − x3/2 to find that f (x) = −3x−3/2 − x1/2 = − . 2 2x3/2 C03S04.006: Write φ(u) = 7u−2/3 + 2u1/3 − 3u10/3 to find that φ (u) = − 14 −5/3 2 −2/3 2(15u4 − u + 7) +u − 10u7/3 = − . u 3 3 3u5/3 C03S04.007: Dx (2x + 3)3/2 = √ 3 (2x + 3)1/2 · 2 = 3 2x + 3. 2 C03S04.008: Dx (3x + 4)4/3 = √ 4 (3x + 4)1/3 · 3 = 4(3x + 4)1/3 = 4 3 3x + 4. 3 3 6x C03S04.009: Dx (3 − 2x2 )−3/2 = − (3 − 2x2 )−5/2 · (−4x) = . 2 (3 − 2x2 )5/2 2 6y 2 C03S04.010: Dy (4 − 3y 3 )−2/3 = − (4 − 3y 3 )−5/3 · (−9y 2 ) = . 3 (4 − 3y 3 )5/3 C03S04.011: Dx (x3 + 1)1/2 = 13 3x2 (x + 1)−1/2 · (3x2 ) = √ . 2 2 x3 + 1 C03S04.012: Dz (z 4 + 3)−2 = −2(z 4 + 3)−3 · 4z 3 = − C03S04.013: Dx (2x2 + 1)1/2 = 8z 3 . (z 4 + 3)3 1 2x . (2x2 + 1)−1/2 · 4x = √ 2 2x2 + 1 1 1 2t 4 C03S04.014: Dt t(1 + t4 )−1/2 = (1 + t4 )−1/2 − t(1 + t4 )−3/2 · 4t3 = − 4 )1/2 2 (1 + t (1 + t4 )3/2 1 = 1 − t4 . (1 + t4 )3/2 C03S04.015: Dt t C03S04.016: Dt 3 /2 √ √ 3 1 /2 √ 3t =t 2= √ . 2 2 2 1 √ · t−5/2 3 5 5 √. = − √ · t−7/2 = − 7/2 3 23 2t C03S04.017: Dx (2x2 − x + 7)3/2 = 3 3 (2x2 − x + 7)1/2 · (4x − 1) = (4x − 1) 2x2 − x + 7. 2 2 C03S04.018: Dz (3z 2 − 4)97 = 97(3z 2 − 4)96 · 6z = 582z (3z 2 − 4)96 . 4 4(6x2 − 1) C03S04.019: Dx (x − 2x3 )−4/3 = − (x − 2x3 )−7/3 · (1 − 6x2 ) = . 3 3(x − 2x3 )7/3 C03S04.020: Dt t2 + (1 + t)4 = 5 t2 + (1 + t)4 4 5 = 5 t2 + (1 + t)4 4 · Dt t2 + (1 + t)4 4 · 2t + 4(1 + t)3 · 1 = 5 t2 + (1 + t)4 · 2t + 4(1 + t)3 . C03S04.021: If f (x) = x(1 − x2 )1/2 , then (by the product rule and the chain rule, among others) 1 f (x) = 1 · (1 − x2 )1/2 + x · (1 − x2 )−1/2 · Dx (1 − x2 ) 2 1 = (1 − x2 )1/2 + x · (1 − x2 )−1/2 · (−2x) = 2 C03S04.022: Write g (x) = g (x) = 1 − x2 − √ x2 1 − 2x2 =√ . 1 − x2 1 − x2 (2x + 1)1/2 to find (x − 1)1/2 (x − 1)1/2 · 1 (2x + 1)−1/2 · 2 − 1 (x − 1)−1/2 · (2x + 1)1/2 2 2 (x − 1)1/2 2 = 2(x − 1)1/2 (2x + 1)−1/2 − (x − 1)−1/2 (2x + 1)1/2 2(x − 1) = 2(x − 1) − (2x + 1) 3 √ =− . 3 / 2 2x + 1 2(x − 1)(x − 1)1/2 (2x + 1)1/2 2(x − 1) t2 + 1 = t2 − 1 C03S04.023: If f (t) = f (t) = 1 2 t2 + 1 t2 − 1 −1/2 · C03S04.024: If h(y ) = h (y ) = 17 y+1 y−1 t2 + 1 t2 − 1 1 /2 , then (t2 − 1)(2t) − (t2 + 1)(2t) 1 = (t2 − 1)2 2 y+1 y−1 16 · t2 − 1 t2 + 1 1 /2 · −4t 2t √ . =− (t2 − 1)2 (t2 − 1)3/2 t2 + 1 17 , then (y − 1) · 1 − (y + 1) · 1 = 17 (y − 1)2 2 y+1 y−1 16 · −2 34(y + 1)16 =− . 2 (y − 1) (y − 1)18 C03S04.025: Dx x − 1 x 3 1 x =3 x− 2 1+ 1 x2 =3 x2 − 1 x 2 · x2 + 1 3(x2 − 1)2 (x2 + 1) = . x2 x4 C03S04.026: Write g (z ) = z 2 (1 + z 2 )−1/2 , then apply the product rule and the chain rule to obtain g (z ) = 2z (1 + z 2 )−1/2 + z 2 · − C03S04.027: Write f (v ) = f (v ) = 1 2 (1 + z 2 )−3/2 · 2z = 2z z3 z 3 + 2z − = . 2 )1/2 2 )3/2 (1 + z (1 + z (1 + z 2 )3/2 (v + 1)1/2 . Then v v · 1 (v + 1)−1/2 − 1 · (v + 1)1/2 v · (v + 1)−1/2 − 2(v + 1)1/2 v − 2(v + 1) v+2 2 = =2 =− 2 . 2 2 1 /2 v 2v 2v (v + 1) 2v (v + 1)1/2 C03S04.028: h (x) = 5 x 3 1 + x2 C03S04.029: Dx (1 − x2 )1/3 = 2 /3 · (1 + x2 ) · 1 − x · 2x 5 x = (1 + x2 )2 3 1 + x2 2 /3 · 1 − x2 . (1 + x2 )2 1 2x . (1 − x2 )−2/3 · (−2x) = − 3 3(1 − x2 )2/3 C03S04.030: Dx (x + x1/2 )1/2 = 1 1 (x + x1/2 )−1/2 1 + x−1/2 2 2 √ 1+2 x =√ √. 4 x x+ x C03S04.031: If f (x) = x(3 − 4x)1/2 , then (with the aid of the product rule and the chain rule) 1 2x 3(1 − 2x) f (x) = 1 · (3 − 4x)1/2 + x · (3 − 4x)−1/2 · (−4) = (3 − 4x)1/2 − =√ . 1 /2 2 (3 − 4x) 3 − 4x C03S04.032: Given g (t) = g (t) = = t − (1 + t2 )1/2 , t2 t2 1 − 1 (1 + t2 )−1/2 · 2t − 2t t − (1 + t2 )1/2 t 1 − t(1 + t2 )−1/2 − 2 t − (1 + t2 )1/2 2 = (t2 )2 t3 t − t2 (1 + t2 )−1/2 − 2t + 2(1 + t2 )1/2 −t(1 + t2 )1/2 − t2 + 2(1 + t2 ) t2 + 2 − t(1 + t2 )1/2 = = . 3 3 (1 + t2 )1/2 t t t3 (1 + t2 )1/2 C03S04.033: If f (x) = (1 − x2 )(2x + 4)1/3 , then the product rule (among others) yields 2 f (x) = −2x(2x + 4)1/3 + (1 − x2 ) · (2x + 4)−2/3 3 = −6x(2x + 4) + 2(1 − x2 ) −12x2 − 24x + 2 − 2x2 2 − 24x − 14x2 = = . 2 /3 2 /3 3(2x + 4) 3(2x + 4) 3(2x + 4)2/3 C03S04.034: If f (x) = (1 − x)1/2 (2 − x)1/3 , then f (x) = 1 (1 − x)−1/2 (−1) · (2 − x)1/3 + 1 (2 − x)−2/3 (−1) · (1 − x)1/2 = − 2 3 =− 3(2 − x) + 2(1 − x) 5x − 8 = . 6(2 − x)2/3 (1 − x)1/2 6(2 − x)2/3 (1 − x)1/2 3 (2 − x)1/3 (1 − x)1/2 + 2(1 − x)1/2 3(2 − x)2/3 C03S04.035: If g (t) = g (t) = 1+ = 3t · 1 t 2 1+ 1 t 2 (3t2 + 1)1/2 , then 1 1 · (3t2 + 1)−1/2 (6t) + 2 · 1 + 2 t − 1 (3t2 + 1)1/2 t2 (t + 1)2 2 t+1 2 3t2 (t + 1)2 2(t + 1)(3t2 + 1) 3t4 − 3t2 − 2t − 2 √ − 2· (3t + 1)1/2 = 3 2 −32 = . t t t2 (3t2 + 1)1/2 t (3t + 1)1/2 t (3t + 1)1/2 t3 3t2 + 1 C03S04.036: If f (x) = x(1 + 2x + 3x2 )10 , then f (x) = (1 + 2x + 3x2 )10 + 10x(1 + 2x + 3x2 )9 (2 + 6x) = (3x2 + 2x + 1)9 (63x2 + 22x + 1). C03S04.037: If f (x) = f (x) = 2x − 1 , then (3x + 4)5 2(3x + 4)5 − (2x − 1) · 5(3x + 4)4 · 3 2(3x + 4) − 15(2x − 1) 23 − 24x = = . 10 6 (3x + 4) (3x + 4) (3x + 4)6 C03S04.038: If h(z ) = (z − 1)4 (z + 1)6 , then h (z ) = 4(z − 1)3 (z + 1)6 + 6(z + 1)5 (z − 1)4 = (z − 1)3 (z + 1)5 (4(z + 1) + 6(z − 1)) = (z − 1)3 (z + 1)5 (10z − 2). C03S04.039: If f (x) = (2x + 1)1/2 , then (3x + 4)1/3 f (x) = = (3x + 4)1/3 (2x + 1)−1/2 − (2x + 1)1/2 (3x + 4)−2/3 (3x + 4)2/3 (3x + 4) − (2x + 1) x+3 = . 4/3 (2x + 1)1/2 4/3 (2x + 1)1/2 (3x + 4) (3x + 4) C03S04.040: If f (x) = (1 − 3x4 )5 (4 − x)1/3 , then 1 f (x) = 5(1 − 3x4 )4 (−12x3 )(4 − x)1/3 + (1 − 3x4 )5 · (4 − x)−2/3 (−1) 3 = −60x3 (1 − 3x4 )4 (4 − x)1/3 − = (1 − 3x4 )5 −180x3 (1 − 3x4 )4 (4 − x) (1 − 3x4 )5 = − 3(4 − x)2/3 3(4 − x)2/3 3(4 − x)2/3 (180x4 − 720x3 ) − (1 − 3x4 ) (1 − 3x4 )4 (183x4 − 720x3 − 1)(1 − 3x4 )4 = . 3(4 − x)2/3 3(4 − x)2/3 C03S04.041: If h(y ) = (1 + y )1/2 + (1 − y )1/2 , then y 5/3 4 h (y ) = = = y 5 /3 1 2 (1 1 2 (1 y + y )−1/2 − 1 (1 − y )−1/2 − 5 y 2/3 (1 + y )1/2 + (1 − y )1/2 2 3 y 10/3 + y )−1/2 − 1 (1 − y )−1/2 − 2 y 8 /3 (1 + y )1/2 + (1 − y )1/2 5 3 y 3(1 + y )−1/2 − 3(1 − y )−1/2 − 10 (1 + y )1/2 + (1 − y )1/2 6y 8/3 y 3(1 − y )1/2 − 3(1 + y )1/2 − 10 (1 + y )(1 − y )1/2 + (1 − y )(1 + y )1/2 6y 8/3 (1 − y )1/2 (1 + y )1/2 √ √ (7y − 10) 1 + y − (7y + 10) 1 − y √ √ = . 6y 8/3 1 − y 1 + y = C03S04.042: If f (x) = (1 − x1/3 )1/2 , then f (x) = 1 1 (1 − x1/3 )−1/2 − x−2/3 2 3 C03S04.043: If g (t) = t + (t + t1/2 )1/2 g (t) = 1 /2 1 t + (t + t1/2 )1/2 2 =− 6x2/3 1 √ . 1 − x1/3 , then −1/2 1 1 · 1 + (t + t1/2 )−1/2 1 + t−1/2 2 2 . It is possible to write the derivative without negative exponents. The symbolic algebra program Mathematica yields g (t) = − (t + (t + t1/2 )1/2 )1/2 1 − 4t3/2 − 4t2 + 3t1/2 1 + (t + t1/2 )1/2 + 2t 1 + (t + t1/2 )1/2 8t(1 + t1/2 )(t3/2 − t1/2 − 1) But the first answer that Mathematica gives is 1 1+ √ 2t 1+ √ 2 t+ t g (t) = √. 2 t+ t+ t C03S04.044: If f (x) = x3 1− f (x) = 3x2 1 , then x2 + 1 1− 1 1 1 + x3 1 − 2 2+1 x 2 x +1 −1/2 The symbolic algebra program Mathematica simplifies this to f (x) = (3x2 + 4) C03S04.045: Because 5 x2 x2 + 1 3 /2 . · (x2 2x . + 1)2 . y (x) = 2 dy = 1 /3 dx 3x is never zero, there are no horizontal tangents. Because y (x) is continuous at x = 0 and | y (x) | → +∞ as x → 0, there is a vertical tangent at (0, 0). √ C03S04.046: If f (x) = x 4 − x2 , then x2 2(2 − x2 ) =√ . 4 − x2 4 − x2 √ √ Hence there are horizontal tangents at − 2, −2 and at 2, 2 . Because f is continuous at ± 2 and f (x) = lim x→−2+ 4 − x2 − √ | f (x) | = + ∞ = lim | f (x) |, x→2− there are vertical tangents at (−2, 0) and (2, 0). C03S04.047: If g (x) = x1/2 − x3/2 , then √ 1 −1/2 3 1/2 1 3x 1 − 3x g (x) = x −x =√− =√. 2 2 2 2x 2x √ Thus there is a horizontal tangent at 1 , 2 3 . Also, because g is continuous at x = 0 and 39 lim | g (x) | = lim x→0+ x→0+ 1 − 3x √ = + ∞, 2x the graph of g has a vertical tangent at (0, 0). C03S04.048: If h(x) = (9 − x2 )−1/2 , then 1 x h (x) = − (9 − x2 )−3/2 · (−2x) = . 2 (9 − x2 )3/2 So the graph of h has a horizontal tangent at 0, 1 . There are no vertical tangents because, even though 3 | h (x) | → + ∞ as x → 3− and as x → −3+ , h is not continuous at 3 or at −3, and there are no other values of x at which | h (x) | → + ∞. C03S04.049: If y (x) = x(1 − x2 )−1/2 , then y (x) = dy 1 1 x2 1 = (1 − x2 )−1/2 − x(1 − x2 )−3/2 · (−2x) = + = . 2 )1/2 2 )3/2 dx 2 (1 − x (1 − x (1 − x2 )3/2 Thus the graph of y (x) has no horizontal tangents because y (x) is never zero. The only candidates for vertical tangents are at x = ± 1, but there are none because y (x) is not continuous at either of those two values of x. C03S04.050: If f (x) = (1 − x2 )(4 − x2 ) = (x4 − 5x2 + 4)1/2 , then f (x) = 14 (x − 5x2 + 4)−1/2 · (4x3 − 10x) = 2 x(2x2 − 5) (1 − x2 )(4 − x2 ) . There are no horizontal tangents where 2x2 = 5 because the two corresponding values of x are not in the domain (−∞, −2] ∪ [−1, 1] ∪ [2, + ∞) of f . There is a horizontal tangent at (0, 2). There are vertical 6 tangents at (−2, 0), (−1, 0), (1, 0), and (2, 0) because the appropriate one-sided limits of | f (x) | are all + ∞. √ C03S04.051: Let f (x) = 2 x. Then f (x) = x−1/2 , so an equation of the required tangent line is y − f (4) = f (4)(x − 4); that is, y = 1 (x + 4). The graph of f and this tangent line are shown next. 2 6 5 4 3 2 1 2 4 6 8 C03S04.052: If f (x) = 3x1/3 then f (x) = x−2/3 , so an equation of the required tangent line is y − f (8) = f (8)(x − 8); that is, y = 1 (x + 16). A graph of f and this tangent line are shown next. 4 7 6 5 4 3 2 1 2 4 6 8 10 12 C03S04.053: If f (x) = 3x2/3 , then f (x) = 2x−1/3 . Therefore an equation of the required tangent line is y − f (−1) = f (−1)(x + 1); that is, y = −2x + 1. A graph of f and this tangent line are shown next. 5 4 3 2 1 -2 -1.5 -1 -0.5 0.5 -1 7 1 C03S04.054: If f (x) = 2(1 − x)1/2 , then f (x) = −(1 − x)−1/2 , and therefore an equation of the required tangent line is y − f 3 = f 3 x − 3 ; that is, y = −2x + 5 . The graph of f and this tangent line are 4 4 4 2 shown next. 3 2 1 -0.5 0.5 1 1.5 C03S04.055: If f (x) = x(4 − x)1/2 , then 1 x 8 − 3x f (x) = (4 − x)1/2 − x(4 − x)−1/2 = (4 − x)1/2 − =√ . 2 2(4 − x)1/2 2 4−x So an equation of the required tangent line is y − f (0) = f (0)(x − 0); that is, y = 2x. A graph of f and this tangent line are shown next. 8 6 4 2 -2 -1 1 2 3 4 -2 -4 C03S04.056: If f (x) = x1/2 − x3/2 , then (as in the solution of Problem 47) f (x) = 1 − 3x √. 2x Therefore an equation of the required tangent line is y − f (4) = f (4)(x − 4); that is, y = − 11 x + 5. A graph 4 8 of f and this tangent line are shown next. 5 1 2 3 4 5 6 7 -5 -10 -15 C03S04.057: If x < 0 then f (x) < 0; as x → 0− , f (x) appears to approach −∞. If x > 0 then f (x) > 0; as x → 0+ , f (x) appears to approach + ∞. So the graph of f must be the one shown in Fig. 3.4.13(d). C03S04.058: If x = 0, then f (x) > 0; moreover, f (x) appears to be approaching zero as | x | increases without bound. In contrast, f (x) appears to approach + ∞ as x → 0. Hence the graph of f must be the one shown in Fig. 3.4.13(f). C03S04.059: Note that f (x) > 0 if x < 0 whereas f (x) < 0 if x > 0. Moreover, as x → 0, | f (x) | appears to approach + ∞. So the graph of f must be the one shown in Fig. 3.4.13(b). C03S04.060: We see that f (x) > 0 for x < 1.4 (approximately), that f (x) = 0 when x ≈ 1.4, and that f (x) < 0 for 1.4 < x < 2; moreover, f (x) → −∞ as x → 2− . So the graph of f must be the one shown in Fig. 3.4.13(a). C03S04.061: We see that f (x) < 0 for −2 < x < −1.4 (approximately), that f (x) > 0 for −1.4 < x < 1.4 (approximately), and that f (x) < 0 for 1.4 < x < 2. Also f (x) = 0 when x ≈ ± 1.4. Therefore the graph of f must be the one shown in Fig. 3.4.13(e). C03S04.062: Figure 3.4.12 shows a graph whose derivative is negative for x < −1, positive for −1 < x < −0.3 (approximately), negative for −0.3 < x < 0, positive for 0 < x < 0.3 (approximately), negative for 0.3 < x < 1, and positive for 1 < x. Moreover, f (x) = 0 when x = ± 1 and when x ≈ ± 0.3. Finally, f (x) → −∞ as x → 0− whereas f (x) → ∞ as x → 0+ . Therefore the graph of f must be the one shown in Fig. 3.4.13(c). P 2g dL dP Pg 2π 2 , so , and hence = = . Given g = 32 and P = 2, we find the value 4π 2 dP 2π 2 dL Pg 12 of the latter to be 32 π ≈ 0.308 (seconds per foot). C03S04.063: L = C03S04.064: dV /dA = 1 A/π , and A = 400π when the radius of the sphere is 10, so the answer is 5 (in 4 appropriate units, such as cubic meters per square meter). √ √ C03S04.065: Whether y = + 1 − x2 or y = − 1 − x2 , it follows easily that dy/dx = −x/y . The slope of √ the tangent is −2 when x = 2y , so from the equation x2 + y 2 = 1 we see that x2 = 4/5, so that x = ± 2 5. 5 √ √ √ √ Because y = 1 x, the two points we are to find are − 2 5, − 1 5 and 2 5, 1 5 . 2 5 5 5 5 9 C03S04.066: Using some of the results in the preceding solution, we find that the √ slope of the tan√ 1 3 1 gent is 3 when x = −3y , so that y 2 = 10 . So the two points of tangency are − 10 10, 10 10 and √ √ 3 1 10 10, − 10 10 . C03S04.067: The line tangent to the parabola y = x2 at the point Q(a, a2 ) has slope 2a, so the normal to the parabola at Q has slope −1/(2a). The normal also passes through P (18, 0), so we can find its slope another way—by using the two-point formula. Thus − 1 a2 − 0 = ; 2a a − 18 18 − a = 2a3 ; 2a3 + a − 18 = 0. By inspection, a = 2 is a solution of the last equation. Thus a − 2 is a factor of the cubic, and division yields 2a3 + a − 18 = (a − 2)(2a2 + 4a + 9). The quadratic factor has negative discriminant, so a = 2 is the only real solution of 2a3 + a − 18 = 0. Therefore the normal line has slope − 1 and equation x + 4y = 18. 4 C03S04.068: Let Q(a, a2 ) be a point on the parabola y = x2 at which some line through P (3, 10) is normal to the parabola. Then, as in the solution of Problem 67, we find that a2 − 10 1 =− . a−3 2a This yields the cubic equation 2a3 − 19a − 3 = 0, and after a little computation we find one of its small integral roots to be r = −3. So a + 3 is a factor of the cubic; by division, the other factor is 2a2 − 6a − 1, √ which is zero when a = 1 3 ± 11 . So the three lines have slopes 2 1 , 6 − 3− 1 √ , and 11 (x − 3), 11 − 3+ 1 √ 11 . Their equations are y − 10 = 1 (x − 3), 6 y − 10 = − 3− 1 √ and y − 10 = − 3+ 1 √ 11 (x − 3). C03S04.069: If a line through P 0, 5 is normal to y = x2/3 at Q(a, a2/3 ), then it has slope − 3 a1/3 . As 2 2 in the two previous solutions, we find that a2/3 − a 5 2 3 = − a1/3 , 2 which yields 3a4/3 + 2a2/3 − 5 = 0. Put u = a2/3 ; we obtain 3u2 + 2u − 5 = 0, so that (3u + 5)(u − 1) = 0. Because u = a2/3 > 0, u = 1 is the only solution, so a = 1 and a = −1 yield the two possibilities for the point Q, and therefore the equations of the two lines are y− 5 3 =− x 2 2 and 10 y− 5 3 = x. 2 2 C03S04.070: Suppose that P = P (u, v ), so that u2 + v 2 = a2 . Then √ slope of the radius OP is the √ mr = v/u if u = 0; if u = 0 then OP lies on the y -axis. Also, whether y = + a2 − x2 or y = − a2 − x2 , it follows that dy −x −x x = ±√ =± =− . dx ±y y a2 − x2 (1) Thus if u = 0 and v = 0, then the slope of the line tangent L to the circle at P (u, v ) is mt = −u/v . In this case mr · mt = v u ·− = −1, u v so that OP is perpendicular to L if u = 0 and v = 0. If u = 0 then Eq. (1) shows that L has slope 0, so that L and OP are also perpendicular in this case. Finally, if v = 0 then OP lies on the x-axis and L is vertical, so the two are also perpendicular in this case. In every case we see that L and OP are perpendicular. C03S04.071: Equation (3) is an identity, and if two functions have identical graphs on an interval, then their derivatives will also be identically equal to each other on that interval. (That is, if f (x) ≡ g (x) on an interval I , then f (x) ≡ g (x) there.) There is no point in differentiating both sides of an algebraic equation. C03S04.072: If f (x) = x1/2 and a > 0, then f (a) = lim x→a Therefore Dx x1/2 = x1/2 − a1/2 x1/2 − a1/2 1 1 = lim = lim 1/2 = . 1/2 − a1/2 )(x1/2 + a1/2 ) x→a (x x→a x x−a + a1/2 2a1/2 1 −1/2 x if x > 0. 2 C03S04.073: If f (x) = x1/3 and a > 0, then f (a) = lim x→a = lim x→a x1/3 − a1/3 x1/3 − a1/3 = lim x→a (x1/3 − a1/3 )(x2/3 + x1/3 a1/3 + a2/3 ) x−a 1 1 1 = 2 /3 = . x2/3 + x1/3 a1/3 + a2/3 a + a2/3 + a2/3 3a2/3 1 −2/3 x if x > 0. 3 This formula is of course valid for x < 0 as well. To show this, observe that the previous argument is valid if a < 0, or—if you prefer—you can use the chain rule, laws of exponents, and the preceding result, as follows. Suppose that x < 0. Then −x > 0; also, x1/3 = −(−x)1/3 . So Therefore Dx x1/3 = Dx (x1/3 ) = Dx −(−x)1/3 Therefore Dx (x1/3 ) = = − Dx (−x)1/3 = − 1 (−x)−2/3 · (−1) 3 = 1 1 (−x)2/3 = x−2/3 . 3 3 1 −2/3 x if x = 0. 3 C03S04.074: If f (x) = x1/5 and a > 0, then f (a) = lim x→a = lim xa x1/5 − a1/5 x1/5 − a1/5 = lim x→a (x1/5 − a1/5 )(x4/5 + x3/5 a1/5 + x2/5 a2/5 + x1/5 a3/5 + a4/5 ) x−a 1 1 1 = 4 /5 = 4/5 . x4/5 + x3/5 a1/5 + x2/5 a2/5 + x1/5 a3/5 + a4/5 a + a4/5 + a4/5 + a4/5 + a4/5 5a 11 1 −4/5 if x > 0. x 5 As in the concluding paragraph in the previous solution, it is easy to show that this formula holds for all x = 0. Therefore Dx (x1/5 ) = C03S04.075: The preamble to Problems 72 through 75 implies that if q is a positive integer and x and a are positive real numbers, then x − a = (x1/q − a1/q )(x(q−1)/q + x(q−2)/q a1/q + x(q−3)/q a2/q + · · · + x1/q a(q−2)/q + a(q−1)/q ). Thus if f (x) = x1/q and a > 0, then f (a) = lim x→a x1/q − a1/q x−a = lim x→a (x1/q = lim x→a = = − a1/q )(x(q−1)/q + x1/q − a1/q + x(q−3)/q a2/q + · · · + x1/q a(q−2)/q + a(q−1)/q ) x(q−2)/q a1/q 1 x(q−1)/q + x(q−2)/q a1/q + x(q−3)/q a2/q + · · · + a(q−1)/q 1 a(q−1)/q + a(q−1)/q + a(q−1)/q + · · · + a(q−1)/q 1 qa(q−1)/q = (q terms in the denominator) (still q terms in the denominator) 1 −(q−1)/q . a q 1 −(q−1)/q if x > 0 and q is a positive integer. This result is easy to extend to the x q case x < 0. Therefore if q is a positive integer and x = 0, then Therefore Dx (x1/q ) = Dx (x1/q ) = 12 1 (1/q)−1 x . q Section 3.5 C03S05.001: Because f (x) = 1 − x is decreasing everywhere, it can attain a maximum only at a left-hand endpoint of its domain and a minimum only at a right-hand endpoint of its domain. Its domain [−1, 1) has no right-hand endpoint, so f has no minimum value. Its maximum value occurs at −1, is f (−1) = 2, and is the global maximum value of f on its domain. C03S05.002: Because f (x) = 2x + 1 is increasing everywhere, it can have a minimum only at a left-hand endpoint of its domain and a maximum only at a right-hand endpoint of its domain. But its domain [ −1, 1) has no right-hand endpoint, so f has no maximum. It has the global minimum value f (−1) = −1 at the left-hand endpoint of its domain. C03S05.003: Because f (x) = | x | is decreasing for x < 0 and increasing for x > 0, it can have a maximum only at a left-hand or a right-hand endpoint of its domain (−1, 1). But its domain has no endpoints, so f has no maximum value. It has the global minimum value f (0) = 0. C03S05.004: Because g (x) = But √ √ x is increasing on (0, 1], its reciprocal f (x) = 1/ x is decreasing there. 1 lim √ = + ∞, + x x→0 so f has no maximum value. It has the global minimum value f (1) = 1 at the right-hand endpoint of its domain. C03S05.005: Given: f (x) = | x − 2 | on (1, 4]. If x > 2 then f (x) = x − 2, which is increasing for x > 2; if x < 2 then f (x) = 2 − x, which is decreasing for x < 2. So f can have a maximum only at an endpoint of its domain; the only endpoint is at x = 4, where f (x) has the maximum value f (4) = 2. Because f (x) → 1 as x → 1+ , the extremum at x = 4 is in fact a global maximum. Finally, f (2) = 0 is the global minimum value of f . C03S05.006: If f (x) = 5 − x2 , then f (x) = −2x, so x = 0 is the only critical point of f . We note that f is increasing for x < 0 and decreasing for x > 0, so f (0) = 5 is the global maximum value of f . Because f (−1) = 4 and f (x) → 1 as x → 2− , the minimum at x = −1 is local but not global. C03S05.007: Given: f (x) = x3 + 1 on [ −1, 1]. The only critical point of f occurs where f (x) = 3x2 is zero; that is, at x = 0. But f (x) < 1 = f (0) if x < 0 whereas f (x) > 1 = f (0) if x > 0, so there is no extremum at x = 0. By Theorem 1 (page 142), f must have a global maximum and a global minimum. The only possible locations are at the endpoints of the domain of f , and therefore f (−1) = 0 is the global minimum value of f and f (1) = 2 is its global maximum value. C03S05.008: If f (x) = 1 , x2 + 1 then f (x) = − 2x , (x2 + 1)2 so x = 0 is the only critical point of f . Because g (x) = x2 + 1 is increasing for x > 0 and decreasing for x < 0, we may conclude that f is decreasing for x > 0 and increasing for x < 0. Therefore f has the global maximum value f (0) = 1 at x = 0 and no other extrema of any kind. C03S05.009: If 1 f (x) = 1 , x(1 − x) then f (x) = 2x − 1 , x2 (1 − x)2 which does not exist at x = 0 or at x = 1 and is zero when x = 1 . But none of these points lies in the domain 2 [2, 3] of f , so there are no extrema at those three points. By Theorem 1 f must have a global maximum and a global minimum, which therefore must occur at the endpoints of its domain. Because f (2) = − 1 and 2 f (3) = − 1 , the former is the global minimum value of f and the latter is its global maximum value. 6 C03S05.010: If f (x) = 1 , x(1 − x) then f (x) = 2x − 1 , − x)2 x2 (1 which does not exist at x = 0 or at x = 1 and is zero when x = 1 . But the domain (0, 1) of f includes only 2 the last of these three points. We note that f 1 2 =4 and that lim f (x) = + ∞ = lim f (x), − x→0+ x→1 and therefore f has no global maximum value. The reciprocal of f (x) is g (x) = x − x2 = −(x2 − x) = − x2 − x + which has the global maximum value 1 4 1 4 + 1 4 = 1 4 − x− 12 2 , at x = 1 . Therefore f (x) has the global minimum value 4 at x = 1 . 2 2 C03S05.011: f (x) = 3 is never zero and always exists. Therefore f (−2) = −8 is the global minimum value of f and f (3) = 7 is its global maximum value. C03S05.012: f (x) = −3 always exists and is never zero. Therefore f (5) = −11 is the global minimum value of f and f (−1) = 7 is its global maximum value. C03S05.013: h (x) = −2x always exists and is zero only at x = 0, which is not in the domain of h. Therefore h(1) = 3 is the global maximum value of h and h(3) = −5 is its global minimum value. C03S05.014: f (x) = 2x always exists and is zero only at x = 0, an endpoint of the domain of f . Therefore f (0) = 3 is the global minimum value of f and f (5) = 28 is its global maximum value. C03S05.015: g (x) = 2(x − 1) always exists and is zero only at x = 1. Because g (−1) = 4, g (1) = 0, and g (4) = 9, the global minimum value of g is 0 and the global maximum is 9. If −1 < x < 0 then g (x) = (x − 1)2 < 4, so the extremum at x = −1 is a local maximum. C03S05.016: h (x) = 2x + 4 always exists and is zero only at x = −2. Because h(−3) = 4, h(−2) = 3, and h(0) = 7, the global minimum value of h is 3 and its global maximum is 7. Because the graph of h is a parabola opening upward, h(−3) = 4 is a local (but not global) maximum value of h. C03S05.017: f (x) = 3x2 − 3 = 3(x + 1)(x − 1) always exists and is zero when x = −1 and when x = 1. Because f (−2) = −2, f (−1) = 2, f (1) = −2, and f (4) = 52, the latter is the global maximum value of f and −2 is its global minimum value—note that the minimum occurs at two different points on the graph. Because f is continuous on [ −2, 2], it must have a global maximum there, and our work shows that it occurs at x = −1. But because f (4) = 52 > 2 = f (−1), f (−1) = 2 is only a local maximum for f on its domain [ −2, 4]. Summary: Global minimum value −2, local maximum value 2, global maximum value 52. 2 C03S05.018: g (x) = 6x2 − 18x + 12 = 6(x − 1)(x − 2) always exists and is zero when x = 1 and when x = 2. Because g (0) = 0, g (1) = 5, g (2) = 4, and g (4) = 32, the global minimum value of g is g (0) = 0 and its global maximum is g (4) = 32. Because g is continuous on [0, 2], it must have a global maximum there, so g (1) = 5 is a local maximum for g on [0, 4]. Because g is continuous on [1, 4], it must have a global minimum there, so g (2) = 4 is a local minimum for g on [0, 4]. C03S05.019: If h(x) = x + 4 , x then h (x) = 1 − 4 x2 − 4 = . 2 x x2 Therefore h is continuous on [1, 4] and x = 2 is the only critical point of h in its domain. Because h(1) = 5, h(2) = 4, and h(4) = 5, the global maximum value of h is 5 and its global minimum value is 4. 16 16 2x3 − 16 2(x − 2)(x2 + 2x + 4) = . So f is , then f (x) = 2x − 2 = x x x2 x2 continuous on its domain [1, 3] and its only critical point is x = 2. Because f (1) = 17, f (2) = 12, and f (3) = 43 ≈ 14.333, the global maximum value of f is 17 and its global minimum value is 12. Because f 3 is continuous on [2, 3], it must have a global maximum there, and therefore f (3) = 43 is a local maximum 3 value of f on [1, 3]. C03S05.020: If f (x) = x2 + C03S05.021: f (x) = −2 always exists and is never zero, so f (1) = 1 is the global minimum value of f and f (−1) = 5 is its global maximum value. C03S05.022: f (x) = 2x − 4 always exists and is zero when x = 2, which is an endpoint of the domain of f . Hence f (2) = −1 is the global minimum value of f and f (0) = 3 is its global maximum value. C03S05.023: f (x) = −12 − 18x always exists and is zero when x = − 2 . Because f (−1) = 8, f − 2 = 9, 3 3 and f (1) = −16, the global maximum value of f is 9 and its global minimum value is −16. Consideration of the interval −1, − 2 shows that f (−1) = 8 is a local minimum of f . 3 C03S05.024: f (x) = 4x − 4 always exists and is zero when x = 1. Because f (0) = 7, f (1) = 5, and f (2) = 7, the global maximum value of f is 7 and its global minimum value is 5. C03S05.025: f (x) = 3x2 − 6x − 9 = 3(x + 1)(x − 3) always exists and is zero when x = −1 and when x = 3. Because f (−2) = 3, f (−1) = 10, f (3) = −22, and f (4) = −15, the global minimum value of f is −22 and its global maximum is 10. Consideration of the interval [ −2, −1] shows that f (−2) = 3 is a local minimum of f ; consideration of the interval [3, 4] shows that f (4) = −15 is a local maximum of f . C03S05.026: f (x) = 3x2 + 1 always exists and is never zero, so f (−1) = −2 is the global minimum value of f and f (2) = 10 is its global maximum value. C03S05.027: f (x) = 15x4 − 15x2 = 15x2 (x + 1)(x − 1) always exists and is zero at x = −1, at x = 0, and at x = 1. We note that f (−2) = −56, f (−1) = 2, f (0) = 0, f (1) = −2, and f (2) = 56. So the global minimum value of f is −56 and its global maximum value is 56. Consideration of the interval [ −2, 0] shows that f (−1) = 2 is a local maximum of f on its domain [ −2, 2]. Similarly, f (1) = −2 is a local minimum of f there. Suppose that x is near, but not equal, to zero. Then f (x) = x3 (3x2 − 5) is negative if x > 0 and positive if x < 0. Therefore there is no extremum at x = 0. C03S05.028: Given: f (x) = | 2x − 3 | on [1, 2]. If x > 3 then f (x) = 2x − 3, so that f (x) = 2. If x < 3 2 2 then f (x) = 3 − 2x, so that f (x) = −2. Therefore f (x) is never zero. But it fails to exist at x = 3 . Because 2 f (1) = 1, f 3 = 0, and f (2) = 1, the global maximum value of f is 1 and its global minimum value is 0. 2 3 C03S05.029: Given: f (x) = 5 + | 7 − 3x | on [1, 5]. If x < 7 , then −3x > −7, so that 7 − 3x > 0; in this 3 case, f (x) = 12 − 3x and so f (x) = −3. Similarly, if x > 7 , then f (x) = 3x − 2 and so f (x) = 3. Hence 3 f (x) is never zero, but it fails to exist at x = 7 . Now f (1) = 9, f 7 = 5, and f (5) = 13, so 13 is the global 3 3 maximum value of f and 5 is its global minimum value. Consideration of the continuous function f on the interval 1, 7 shows that f (1) = 9 is a local maximum of f on its domain. 3 C03S05.030: Given: f (x) = | x + 1 | + | x − 1 | on [ −2, 2]. If x < −1 then f (x) = −(x + 1) − (x − 1) = −2x, so that f (x) = −2. If x > 1 then f (x) = x + 1 + x − 1 = 2x, so that f (x) = 2. If −1 x 1 then f (x) = x + 1 − (x − 1) = 2, so that f (x) = 0. But f (x) does not exist at x = −1 or at x = 1. We note that f (−2) = 4, f (x) = 2 for all x such that −1 x 1, and that f (2) = 4. So 4 is the global maximum value of f and 2 is its global minimum value. Observe that f has infinitely many critical points: every number in the interval [ −1, 1]. C03S05.031: f (x) = 150x2 − 210x + 72 = 6(5x − 3)(5x − 4) always exists and is zero at x = 3 and at 5 x = 4 . Now f (0) = 0, f 3 = 16.2, f 4 = 16, and f (1) = 17. Hence 17 is the global maximum value of 5 5 5 f and 0 is its global minimum value. Consideration of the intervals 0, 4 and 3 , 1 shows that 16.2 is a 5 5 local maximum value of f on [0, 1] and that 16 is a local minimum value of f there. 1 1 4x2 − 1 . Therefore f (x) exists for all x in , then f (x) = 2 − 2 = 2x 2x 2x2 the domain [1, 4] of f and there are no points in the domain of f at which f (x) = 0. Thus the global minimum value of f is f (1) = 2.5 and its global maximum value is f (4) = 8.125. C03S05.032: If f (x) = 2x + C03S05.033: If f (x) = x , x+1 then f (x) = 1 , (x + 1)2 so f (x) exists for all x in the domain [0, 3] of f and is never zero there. Hence f (0) = 0 is the global minimum value of f and f (3) = 3 is its global maximum value. 4 x 1 − x2 , then f (x) = , so f (x) exists for all x; the only point in the x2 + 1 (1 + x2 )2 3 domain of f at which f (x) = 0 is x = 1. Now f (0) = 0, f (1) = 1 , and f (3) = 10 , so 0 is the global minimum 2 value of f and 1 is its global maximum value. By the usual argument, there is a local minimum at x = 3. 2 C03S05.034: If f (x) = C03S05.035: If f (x) = 1−x , x2 + 3 f (x) = then (x + 1)(x − 3) , (x2 + 3)2 so f (x) always exists and is zero when x = −1 and when x = 3. Now f (−2) = 3 , f (−1) = 1 , f (3) = − 1 , 7 2 6 and f (5) = − 1 . So the global minimum value of f is − 1 and its global maximum value is 1 . Consideration 7 6 2 of the interval [ −2, −1] shows that 3 is a local minimum value of f ; consideration of the interval [3, 5] shows 7 that − 1 is a local maximum value of f . 7 C03S05.036: If f (x) = 2 − x1/3 , then f (x) = − 1 , 3x2/3 so f (x) is never zero and f (x) does not exist when x = 0. Nevertheless, f is continuous on its domain [ −1, 8]. And f (−1) = 3, f (0) = 2, and f (8) = 0, so the global maximum value of f is 3 and its global 4 minimum value is 0. Because g (x) = x1/3 is an increasing function, f (x) = 2 − x1/3 is decreasing on its domain, and therefore there is no extremum at x = 0. C03S05.037: Given: f (x) = x(1 − x2 )1/2 on [ −1, 1]. First, 1 x2 1 − 2x2 f (x) = (1 − x2 )1/2 + x · (1 − x2 )−1/2 · (−2x) = (1 − x2 )1/2 − =√ . 2 )1/2 2 (1 − x 1 − x2 Hence f (x) exists for −1 < x < 1 and not otherwise, but we will check the endpoints ±1 of the domain of f √ √ √ separately. Also f (x) = 0 when x = ± 1 2. Now f (−1) = 0, f − 1 2 = − 1 , f 1 2 = 1 , and f (1) = 0. 2 2 2 2 2 Therefore the global minimum value of f is − 1 and its global maximum value is 1 . Consideration of the 2 2 √ interval −1, − 1 2 shows that f (−1) = 0 is a local maximum value of f on [ −1, 1]; similarly, f (1) = 0 is 2 a local minimum value of f there. C03S05.038: Given: f (x) = x(4 − x2 )1/2 on [0, 2]. Then 1 x2 4 − 2x2 f (x) = (4 − x2 )1/2 + x · (4 − x2 )−1/2 · (−2x) = (4 − x2 )1/2 − =√ , 2 (4 − x2 )1/2 4 − x2 √ √ so f (x) exists if 0 x < 2 and is zero when x = 2. Now f (0) = 0 = f (2) and f 2 = 2, so the former is the global minimum value of f on [0, 2] and the latter is its global maximum value there. C03S05.039: Given: f (x) = x(2 − x)1/3 on [1, 3]. Then 1 x 6 − 4x f (x) = (2 − x)1/3 + x · (2 − x)−2/3 · (−1) = (2 − x)1/3 − = . 2 /3 3 3(2 − x) 3(2 − x)2/3 Then f (2) does not exist and f (x) = 0 when x = 3 . Also f is continuous everywhere, and f (1) = 1, 2 f 3 ≈ 1.19, and f (3) = −3. Hence the global minimum value of f is −3 and its global maximum value is 2 f 3 = 3 · 2−4/3 ≈ 1.190551. Consideration of the interval 1, 3 shows that f (1) = 1 is a local minimum 2 2 value of f . C03S05.040: Given: f (x) = x1/2 − x3/2 on [0, 4]. Then f (x) = 1 −1/2 3 1/2 1 3x1/2 1 − 3x x −x = 1/2 − =√. 2 2 2 2x 2x Then f (x) does not exist when x = 0, although f is continuous on its domain; also, f (x) = 0 when x = 1 . 3 √ Now f (0) √ 0, f 1 = 2 3, and f (4) = −6. So −6 is the global minimum value of f and its global maximum = 3 9 value is 2 3. Consideration of the interval 0, 1 shows that f (0) = 0 is a local minimum value of f . 9 3 C03S05.041: If A = 0, then f (x) ≡ A is never zero, but because f is continuous it must have global extrema. Therefore they occur at the endpoints. If A = 0, then f is a constant function, and its maximum and minimum value B occurs at every point of the interval, including the two endpoints. C03S05.042: The hypotheses imply that f has no critical points in (a, b), but f must have global extrema. Therefore they occur at the endpoints. C03S05.043: f (x) = 0 if x is not an integer; f (x) does not exist if x is an integer (we saw in Chapter 2 that f (x) = [[x]] is discontinuous at each integer). C03S05.044: If f (x) = ax2 + bx + c and a = 0, then f (x) = 2ax + b. Clearly f (x) exists for all x, and f (x) = 0 has the unique solution x = −b/(2a). Therefore f has exactly one critical point on the real number line. 5 C03S05.045: If f (x) = ax3 + bx2 + cx + d and a = 0, then f (x) = 3ax2 + 2bx + c exists for all x, but the quadratic equation 3ax2 + 2bx + c = 0 has two solutions if the discriminant ∆ = 4b2 − 12ac is positive, one solution if ∆ = 0, and no [real] solutions if ∆ < 0. Therefore f has either no critical points, exactly one critical point, or exactly two. Examples: f (x) = x3 + x f (x) = x3 f (x) = x − 3x 3 has no critical points, has exactly one critical point, and has exactly two critical points. C03S05.046: A formula for f is f (x) = min{ x − [[x]], 1 + [[x]] − x }. (1) If you are not comfortable with the idea that “min” is a “function,” an equivalent way of defining f is this: f (x) = 1 1 − 2x − 1 − 2[[x]] 2 . To verify that f performs as advertised, suppose that x is a real number and that n = [[x]], so that n n + 1. Case (1): n x n + 1 . Then 2 1 2 x − [[x]] = x − n and 1 + [[x]] − x = 1 + n − x = 1 − (x − n) x< 1 , 2 so that Eq. (1) yields f (x) = x − n, which is indeed the distance from x to the nearest integer, because in Case (1) the nearest integer is n. Case (2), in which n + 1 < x < n + 1, is handled similarly. 2 The graph of f is shown next. It should be clear that f (x) fails to exist at every integral multiple of 1 2 and that its derivative is either +1 or −1 otherwise. Hence its critical points are the integral multiples of 1 . 2 1 0.5 -1 1 2 3 -0.5 C03S05.047: The derivative is positive on (−∞, −1.3), negative on (−1.3, 1.3), and positive on (1.3, + ∞). So its graph must be the one in Fig. 3.5.15(c). (Numbers with decimal points are approximations.) C03S05.048: The derivative is negative on (−∞, −1.0), positive on (−1.0, 1.0), negative on (1.0, 3.0), and positive on (3.0, + ∞). So its graph must be the one shown in Fig. 3.5.15(f). (Numbers with decimal points are approximations.) C03S05.049: The derivative is positive on (−∞, 0.0), negative on (0.0, 2.0), and positive on (2.0, + ∞). So its graph must be the one shown in Fig. 3.5.15(d). (Numbers with decimal points are approximations.) 6 C03S05.050: The derivative is positive on (−∞, −2.0), negative on (−2.0, 0.0), positive on (0.0, 2.0), and negative on (2.0, + ∞). So its graph must be the one shown in Fig. 3.5.15(b). (Numbers with decimal points are approximations.) C03S05.051: The derivative is negative on (−∞, −2.0), positive on (−2.0, 1.0), and negative on (1.0, + ∞). Therefore its graph must be the one shown in Fig. 3.5.15(a). (Numbers with decimal points are approximations.) C03S05.052: The derivative is negative on (−∞, −2.2), positive on (−2.2, 2.2), and negative again on (2.2, + ∞). So its graph must be the one shown in Fig. 3.5.15(e). (Numbers with decimal points are approximations.) Note: In Problems 53 through 60, we used Mathematica 3.0 and Newton’s method (when necessary), carrying 40 decimal digits throughout all computations. Answers are correct or correctly rounded to the number of digits shown. Your answers may differ in the last (or last few) digits beause of differences in hardware or software. Using a graphing calculator or computer to zoom in on solutions has more limited accuracy when using certain machines. C03S05.053: Global maximum value 28 at the left endpoint x = −2, global minimum value approximately √ 6.828387610996 at the critical point where x = −1 + 1 30 ≈ 0.825741858351, local maximum value 16 at 3 the right endpoint x = 2. C03S05.054: Local minimum value 22 at the left endpoint x = −4, global maximum value approximately √ 31.171612389004 at the critical point x = −1 − 1 30 ≈ −2.825741858351, global minimum value approxi3 √ mately 6.828387610996 at the critical point x = −1 + 1 30 ≈ 0.825741858351, local maximum value 16 at 3 the right endpoint x = 2. C03S05.055: Global maximum value 136 at the left endpoint x = −3, global minimum value approximately −8.669500829438 at the critical point x ≈ −0.762212740507, local maximum value 16 at the right endpont x = 3. C03S05.056: Global maximum value 160 at the left endpoint x = −3, global minimum value approximately −16.048632589199 at the critical point x ≈ −0.950838582066, local maximum value approximately 8.976226903748 at the critical point x ≈ 1.323417756580, local minimum value −8 at the right endpoint x = 3. C03S05.057: Global minimum value −5 at the left endpoint x = 0, global maximum value approximately 8.976226903748 at the critical point x ≈ 1.323417756580, local minimum value 5 at the right endpoint x = 2. C03S05.058: Local maximum value 3 at the left endpoint x = −1, global minimum value approximately −5.767229705222 at the critical point x ≈ −0.460141424682, global maximum value approximately 21.047667292488 at the critical point x ≈ 0.967947424014, local minimum value 21 at the right endpoint x = 1. C03S05.059: Local minimum value −159 at the left endpoint x = −3, global maximum value approxiately 30.643243080334 at the critical point x ≈ −1.911336401963, local minimum value approximately −5.767229705222 at the critical point x ≈ −0.460141424682, local maximum value approximately 21.047667292488 at the critical point x ≈ 0.967947424014, global minimum value −345 at the right endpoint x = 3. 7 C03S05.060: Local minimum value 0 at the left endpoint x = 0, local maximum value approximately 21.047667292488 at the critical point x ≈ 0.967947424014, global minimum value approximately −1401.923680667600 at the critical point x ≈ 5.403530402632, global maximum value 36930 at the right endpoint x = 10. 8 Section 3.6 C03S06.001: With x > 0, y > 0, and x + y = 50, we are to maximize the product P = xy . P = P (x) = x(50 − x) = 50x − x2 , 0 < x < 50 (x < 50 because y > 0.) The product is not maximal if we let x = 0 or x = 50, so we adjoin the endpoints to the domain of P ; thus the continuous function P (x) = 50x − x2 has a global maximum on the closed interval [0, 50], and the maximum does not occur at either endpoint. Because f is differentiable, the maximum must occur at a point where P (x) = 0: 50 − 2x = 0, and so x = 25. Because this is the only critical point of P , it follows that x = 25 maximizes P (x). When x = 25, y = 50 − 25 = 25, so the two positive real numbers with sum 50 and maximum possible product are 25 and 25. C03S06.002: If two parallel sides of the rectangle both have length x and the other two sides both have length y , then we are to maximize the area A = xy given that 2x + 2y = 200. So A = A(x) = x(100 − x), 0 100. x Clearly the maximum value of A occurs at a critical point of A in the interval (0, 100). But A (x) = 100 − 2x, so x = 50 is the location of the maximum. When x = 50, also y = 50, so the rectangle of maximal area is a square of area 502 = 2500 ft2 . C03S06.003: If the coordinates of the “fourth vertex” are (x, y ), then y = 100 − 2x and the area of the rectangle is A = xy . So we are to maximize A(x) = x(100 − 2x) 0 50. x By the usual argument the solution occurs where A (x) = 0, thus where x = 25, y = 50, and the maximum area is 1250. C03S06.004: If the side of the pen parallel to the wall has length x and the two perpendicular sides both have length y , then we are to maximize area A = xy given x + 2y = 600. Thus A = A(y ) = y (600 − 2y ), 0 300. y Adjoining the endpoints to the domain is allowed because the maximum we seek occurs at neither endpoint. Therefore the maximum occurs at an interior critical point. We have A (y ) = 600 − 4y , so the only critical point of A is y = 150. When y = 150, we have x = 300, so the maximum possible area that can be enclosed is 45000 m2 . C03S06.005: If x is the length of each edge of the base of the box and y denotes the height of the box, then its volume is given by V = x2 y . Its total surface area is the sum of the area x2 of its bottom and four times the area xy of each of its vertical sides, so x2 + 4xy = 300. Thus V = V (x) = x2 · 300 − x2 300x − x3 = , 4x 4 Hence V (x) = 300 − 3x2 , 4 1 1 x √ 10 3. so V (x) always exists and V (x) = 0 when x = 10 (we discard the solution x = −10; it’s not in the domain of V ). Then V (1) = 299 = 74.75, 4 V (10) = 500, and √ V 10 3 = 0, so the maximum possible volume of the box is 500 in.3 . C03S06.006: The excess of the number x over its square is f (x) = x − x2 . In this problem we also know that 0 x 1. Then f (x) = 0 at the endpoints of its domain, so the maximum value of f (x) must occur at an interior critical point. But f (x) = 1 − 2x, so the only critical point of f is x = 1 , which must yield a 2 maximum because f is continuous on [0, 1]. So the maximum value of x − x2 for 0 x 1 is 1 . 4 C03S06.007: If the two numbers are x and y , then we are to minimize S = x2 + y 2 given x > 0, y > 0, and x + y = 48. So S (x) = x2 + (48 − x)2 , 0 x 48. Here we adjoin the endpoints to the domain of S to ensure the existence of a maximum, but we must test the values of S at these endpoints because it is not immediately clear that neither S (0) nor S (48) yields the maximum value of S . Now S (x) = 2x − 2(48 − x); the only interior critical point of S is x = 24, and when x = 24, y = 24 as well. Finally, S (0) = (48)2 = 2304 = S (48) > 1152 = S (24), so the answer is 1152. C03S06.008: Let x be the length of the side around which the rectangle is rotated and let y be the length of each perpendicular side. Then 2x + 2y = 36. The radius of the cylinder is y and its height is x, so its volume is V = π y 2 x. So V = V (y ) = π y 2 (18 − y ) = π (18y 2 − y 3 ), with natural domain 0 < y < 18. We adjoin the endpoints to the domain because neither y = 0 nor y = 18 maximizes V (y ), and deduce the existence of a global maximum at an interior critical point. Now V (y ) = π (36y − 3y 2 ) = 3π y (12 − y ). So V (y ) = 0 when y = 0 and when y = 12. The former value of y minimizes V (y ), so the maximum possible volume of the cylinder is V (12) = 864π . C03S06.009: Let x and y be the two numbers. Then x + y = 10, x the sum of their cubes, S = x3 + y 3 : S (x) = x3 + (10 − x)3 , 0, and y 0 x 0. We are to minimize 10. Now S (x) = 3x2 − 3(10 − x)2 , so the values of x to be tested are x = 0, x = 5, and x = 10. At the endpoints, S = 1000; when x = 5, S = 250 (the minimum). C03S06.010: Draw a cross section of the cylindrical log—a circle of radius r. Inscribe in this circle a cross section of the beam—a rectangle of width w and height h. Draw a diagonal of the rectangle; the Pythagorean theorem yields x2 + h2 = 4r2 . The strength S of the beam is given by S = kwh2 where k is a positive constant. Because h2 = 4r2 − w2 , we have S = S (w) = kw(4r2 − w2 ) = k (4wr2 − w3 ) with natural domain 0 < w < 2r. We adjoin the endpoints to this domain; this is permissible because S = 0 at each, and so is not maximal. Next, S (w) = k (4r2 − 3w2 ); S (w) = 0 when 3w2 = 4r2 , and the corresponding (positive) value of w yields the maximum of S (we know that S (w) must have a maximum on [0, 2r ] because 2 of the continuity of S on this interval, and we also know that the maximum does not occur at either endpoint, √ so there is only one possible location for the maximum). At maximum, h2 = 4r2 − w2 = 3w2 − w2 , so h = w 2 describes the shape of the beam of greatest strength. C03S06.011: As in Fig. 3.6.18, let y denote the length of each of the internal dividers and of the two sides parallel to them; let x denote the length of each of the other two sides. The total length of all the fencing is 2x + 4y = 600 and the area of the corral is A = xy . Hence A = A(y ) = 600 − 4y · y = 300y − 2y 2 , 2 0 y 150. Now A (y ) = 0 only when y = 75, and A(0) = 0 = A(150), and therefore the maximum area of the corral is A(75) = 11250 yd2 . C03S06.012: Let r denote the radius of the cylinder and h its height. We are to maximize its volume V = π r2 h given the constraint that the total surface area is 150π : 2π r2 + 2π rh = 150π , so that h= 75 − r2 . r Thus V = V (r) = π r(75 − r2 ) = π (75r − r3 ), 0<r< √ 75. √ We may adjoin both endpoints to this domain without creating a spurious maximum, so we use 0, 5 3 as the domain of V . Next, V (r) = π (75 − 3r2 ). Hence V (r) always exists and its only zero in the domain of V occurs when r = 5 (and h = 10). But V is zero at the two endpoints of its domain, so V (5) = 250π is the maximum volume of such a cylinder. C03S06.013: If the rectangle has sides x and y , then x2 + y 2 = 162 by the Pythagorean theorem. The area of the rectangle is then A(x) = x 256 − x2 , 0 x 16. A positive quantity is maximized exactly when its square is maximized, so in place of A we maximize 2 f (x) = (A(x)) = 256x2 − x4 . √ The only solutions of f (x) = 0 in the domain of A are x = 0 and x = 8 2. But A(0) = 0 = A(16), so √ x = 8 2 yields the maximum value 128 of A. C03S06.014: If the far side of the rectangle has length 2x (this leads to simpler algebra than length x), and the sides perpendicular to the far side have length y , then by the Pythagorean theorem, x2 + y 2 = L2 . The area of the rectangle is A = 2xy , so we maximize A(x) = 2x L2 − x2 , 0 x L by maximizing 2 f (x) = (A(x)) = 4(L2 x2 − x4 ). √ Now f (x) = 4(2L2 x − 4x3 ) = 8x(L2 − 2x2 ) is zero when √ = 0 (rejected; A(0) = 0) and when x = 1 L 2. x 2 Note also that A(L) = 0. By the usual argument, x = 1 L 2 maximizes f (x) and thus A(x). The answer is 2 √ A 1 L 2 = L2 . 2 3 C03S06.015: V (T ) = −0.06426 + (0.0170086)T − (0.0002037)T 2 . The equation V (T ) = 0 is quadratic with the two (approximate) solutions T ≈ 79.532 and T ≈ 3.967. The formula for V (T ) is valid only in the range 0 T 30, so we reject the first solution. Finally, V (0) = 999.87, V (30) ≈ 1003.763, and V (3.967) ≈ 999.71. Thus the volume is minimized when T ≈ 3.967, and therefore water has its greatest density at about 3.967◦ C. C03S06.016: Let P (x, 0) be the lower right-hand corner point of the rectangle. The rectangle then has base 2x, height 4 − x2 , and thus area A(x) = 2x(4 − x2 ) = 8x − 2x3 , Now A (x) = 8 − 6x2 ; A (x) = 0 when √ = x √ maximum possible area is A 2 3 = 32 3. 3 9 2 3 √ 0 x 2. 3. Because A(0) = 0, A(2) = 0, and A 2 3 √ 3 > 0, the C03S06.017: Let x denote the length of each edge of the base and let y denote the height of the box. We are to maximize its volume V = x2 y given the constraint 2x2 + 4xy = 600. Solve the latter for y to write 1 V (x) = 150x − x3 , 2 1 x √ 10 3. The solution of V (x) = 0 in the domain of V is x = 10. Because V (10) = 1000 > V (1) = 149.5 > √ V 10 3 = 0, this shows that x = 10 maximizes V and that the maximum value of V is 1000 cm3 . C03S06.018: Let x denote the radius of the cylinder and y its height. Then its total surface area is π x2 + 2π xy = 300π , so x2 + 2xy = 300. We are to maximize its volume V = π x2 y . Because y= 300 − x2 , 2x it follows that V = V (x) = π (300x − x3 ), 2 0 x √ 10 3. It is then easy to show that x = 10 maximizes V (x), that y = x = 10 as well, and thus that the maximum possible volume of the can is 1000π in.3 C03S06.019: Let x be the length of the edge of each of the twelve small squares. Then each of the three cross-shaped pieces will form boxes with base length 1 − 2x and height x, so each of the three will have volume x(1 − 2x)2 . Both of the two cubical boxes will have edge x and thus volume x3 . So the total volume of all five boxes will be V (x) = 3x(1 − 2x)2 + 2x3 = 14x3 − 12x2 + 3x, 0 x 1 . 2 Now V (x) = 42x2 − √ x + 3; V (x) = 0 when 14x2 − 8x − 1 = 0. The quadratic formula gives the two 24 1 solutions x = 14 4 ± 2 . These are approximately 0.3867 and 0.1847, and both lie in the domain of V . Finally, V (0) = 0, V (0.1847) ≈ 0.2329, V (0.3867) ≈ 0.1752, and V (0.5) = 0.25. Therefore, to maximize V , one must cut each of the three large squares into four smaller squares of side length 1 each and form the 2 resulting twelve squares into two cubes. At maximum volume there will be only two boxes, not five. C03S06.020: Let x be the length of each edge of the square base of the box and let h denote its height. Then its volume is V = x2 h. The total cost of the box is $144, hence 4xh + x2 + 2x2 = 144 and thus Therefore 4 h= 144 − 3x2 . 4x x 3 144 − 3x2 = 36x − x3 . 4 4 √ The natural domain of V is the open interval 0, 4 3 , but we may adjoin the endpoints as usual to obtain a closed interval. Also V = V (x) = 9 V (x) = 36 − x2 , 4 so V (x) always exists and is zero only at x = 4 (reject the other root x = −4). Finally, V (x) = 0 at the endpoints of its domain, so V (4) = 96 (ft3 ) is the maximum volume of such a box. The dimensions of the largest box are 4 ft square on the base by 6 ft high. C03S06.021: Let x denote the edge length of one square and y that of the other. Then 4x + 4y = 80, so y = 20 − x. The total area of the two squares is A = x2 + y 2 , so A = A(x) = x2 + (20 − x)2 = 2x2 − 40x + 400, with domain (0, 20); adjoin the endpoints as usual. Then A (x) = 4x − 40, which always exists and which vanishes when x = 10. Now A(0) = 400 = A(20), whereas A(10) = 200. So to minimize the total area of the two squares, make two equal squares. To maximize it, make only one square. C03S06.022: Let r be the radius of the circle and x the edge of the square. We are to maximize total area A = π r2 + x2 given the side condition 2π r + 4x = 100. From the last equation we infer that x= 100 − 2π r 50 − π r = . 4 2 So 1 A = A(r) = π r2 + (50 − π r)2 = 4 for 0 r 50/π (because x 1 π + π 2 r2 − 25π r + 625 4 0). Now 1 A (r) = 2 π + π 2 r − 25π ; 4 A (r) = 0 when r = 25 50 π = π + 4; 2+ 2 that is, when r ≈ 7. Finally, A(0) = 625, A 50 π ≈ 795.77 and A 50 π+4 ≈ 350.06. Results: For minimum area, construct a circle of radius 50/(π +4) ≈ 7.00124 (cm) and a square of edge length 100/(π + 4) ≈ 14.00248 (cm). For maximum area, bend all the wire into a circle of radius 50/π ≈ 15.91549 (cm). C03S06.023: Let x be the length of each segment of fence perpendicular to the wall and let y be the length of each segment parallel to the wall. 5 Case 1: The internal fence is perpendicular to the wall. Then y = 600 − 3x and the enclosure will have area A(x) = 600x − 3x2 , 0 x 200. Then A (x) = 0 when x = 100; A(100) = 30000 (m2 ) is the maximum in Case 1. Case 2: The internal fence is parallel to the wall. Then y = 300 − x, and the area of the enclosure is given by A(x) = 300x − x2 , 0 x 300. Then A (x) = 0 when x = 150; A(150) = 22500 (m2 ) is the maximum in Case 2. Answer: The maximum possible area of the enclosure is 30000 m2 . The divider must be perpendicular to the wall and of length 100 m. The side parallel to the wall is to have length 300 m. C03S06.024: See Fig. 3.6.22 of the text. Suppose that the pen measures x (horizontal) by y (vertical). Then it has area A = xy . Case 1: x 10, y 5. Then x + (x − 10) + y + (y − 5) = 85, so x + y = 50. Therefore A = A(x) = x(50 − x) = 50x − x2 , 10 45. x Then A (x) = 0 when x = 25; A(25) = 625. Note that A(10) = 400 and that A(45) = 225. Case 2: 0 x 10, y 5. Then x + y + (y − 5) = 85, so x + 2y = 90. Therefore A = A(x) = x 90 − x 1 = (90x − x2 ), 2 2 0 10. x In this case, A (x) = 0 when x = 45, but 45 doesn’t lie in the domain of A. Note that A(0) = 0 and that A(10) = 400. Case 3: x 10, 0 y 5. Then x + (x − 10) + y = 85, so 2x + y = 95. Therefore A = A(x) = x(95 − 2x) = 95x − 2x2 , 45 47.5. x In this case A (x) = 0 when x = 23.75, not in the domain of A. Note that A(45) = 225 and that A(47.5) = 0. Conclusion: The area of the pen is maximized when the pen is square, 25 m on each side (the maximum from Case 1). C03S06.025: Let the dimensions of the box be x by x by y . We are to maximize V = x2 y subject to some conditions on x and y . According to the poster on the wall of the Bogart, Georgia Post Office, the length of the box is the larger of x and y , and the girth is measured around the box in a plane perpendicular to its length. Case 1: x < y . Then the length is y , the girth is 4x, and the mailing constraint is 4x + y clear that we take 4x + y = 100 to maximize V , so that V = V (x) = x2 (100 − 4x) = 100x2 − 4x3 , 6 0 x 25. 100. It is Then V (x) = 4x(50 − 3x); V (x) = 0 for x = 0 and for x = 50/3. But V (0) = 0, V (25) = 0, and V (50/3) = 250000/27 ≈ 9259 (in.3 ). The latter is the maximum in Case 1. Case 2: x y . Then the length is x and the girth is 2x + y , although you may get some argument from a postal worker who may insist that it’s 4x. So 3x + 2y = 100, and thus 100 − 3x 2 V = V (x) = x2 3 = 50x2 − x3 , 2 0 x 100/3. Then V (x) = 100x − 9 x2 ; V (x) = 0 when x = 0 and when x = 200/9. But V (0) = 0, V (100/3) = 0, and 2 V (200/9) = 2000000/243 ≈ 8230 (in.3 ). Case 3: You lose the argument in Case 2. Then the box has length x and girth 4x, so 5x = 100; thus x = 20. To maximize the total volume, no calculus is needed—let y = x. Then the box of maximum volume will have volume 203 = 8000 (in.3 ). Answer: The maximum is 250000 in.3 27 C03S06.026: In this problem the girth of the package is its circumference; no one would interpret “girth” in any other way. So suppose that the package has length x and radius r. Then it has volume V = π r2 x where x + 2π r = 100. We seek to maximize V = V (r) = π r2 (100 − 2π r) = π (100r2 − 2π r3 ), 0 r 50 . π Now V (r) = π (200r − 6π r2 ) = 2π r(100 − 3π r); V (r) = 0 when r = 0 and when r = 100/(3π ). But V (0) = 0, V 50 π = 0, and V 100 3π = 1000000 ≈ 11789 (in.3 ), 27π and the latter is clearly the maximum of V . C03S06.027: Suppose that n presses are used, 1 C (n) = 5n + (10 + 6n) 8. The total cost of the poster run would then be n 50000 3600n = 5n + 125 9 10 +6 n dollars. Temporarily assume that n can take on every real number value between 1 and 8. Then C (n) = 5 − 125 10 ; · 9 n2 √ C (n) = 0 when n = 5 10 ≈ 5.27 presses. But an integral number of presses must be used, so the actual 3 number that will minimize the cost is either 5 or 6, unless the minimum occurs at one of the two endpoints. The values in question are C (1) ≈ 227.2, C (5) ≈ 136.1, C (6) ≈ 136.5, and C (8) ≈ 140.7. So to minimize cost and thereby maximize the profit, five presses should be used. C03S06.028: Let x denote the number of workers hired. Each worker will pick 900/x bushels; each worker will spend 180/x hours picking beans. The supervisor cost will be 1800/x dollars, and the cost per worker will be 8 + (900/x) dollars. Thus the total cost will be C (x) = 8x + 900 + 7 1800 , x 1 x. It is clear that large values of x make C (x) large, so the global minimum of C (x) occurs either at x = 1 or where C (x) = 0. Assume for the moment that x can take on all real number values in [1, + ∞), not merely integral values, so that C is defined. Then C (x) = 8 − 1800 ; C (x) = 0 x2 when x2 = 225. Thus C (15) = 0. Now C (1) = 2708 and C (15) = 1140, so fifteen workers should be hired; the cost to pick each bushel will be approximately $1.27. C03S06.029: We are to minimize the total cost C over a ten-year period. This cost is the sum of the initial cost and ten times the annual cost: C (x) = 150x + 10 1000 2+x 0 , 10. x Next, C (x) = 150 − 10000 ; (2 + x)2 C (x) = 0 when 150 = 10000 , (2 + x)2 so that (2 + x)2 = 200 . One of the resulting values of x is negative, so we reject it. The other is x = 3 −2 + 200/3 ≈ 6.165 (in.). The problem itself suggests that x must be an integer, so we check x = 6 and x = 7 along with the endpoints of the domain of C . In dollars, C (0) = 5000, C (6) ≈ 2150, C (7) ≈ 2161, and C (10) ≈ 2333. Result: Install six inches of insulation. The annual savings over the situation with no insulation at all then will be one-tenth of 5000 − 2150, about $285 per year. C03S06.030: We assume that each one-cent increase in price reduces sales by 50 burritos per night. Let x be the amount, in cents, by which the price is increased. The resulting profit is P (x) = (50 + x)(5000 − 5x) − 25(5000 − 50x) − 100000 = (25 + x)(5000 − 50x) − 100000 = 25000 + 3750x − 50x2 , −50 x. Because P (x) < 0 for large values of x and for x = −50, P will be maximized where P (x) = 0: P (x) = 3750 − 100x; P (x) = 0 when x = 37.5. Now P (37) = 953, P (37.5) ≈ 953.13, and P (38) = 953. Therefore profit is maximized when the selling price is either 87/ or 88/, and the maximum profit will be $953. c c C03S06.031: Let x be the number of five-cent fare increases. The resulting revenue will be R(x) = (150 + 5x)(600 − 40x), −15 x 15 (the revenue is the product of the price and the number of passengers). Now R(x) = 90000 − 3000x − 200x2 ; R (x) = −3000 − 400x; R (x) = 0 8 when x = −7.5. Because the fare must be an integral number of cents, we check R(−7) = 1012 = R(−8) (dollars). Answer: The fare should be either $1.10 or $1.15; this is a reduction of 40 or 35 cents, respectively, and each results in the maximum possible revenue of $1012 per day. C03S06.032: The following figure shows a central cross section of the sphere and inscribed cylinder. The radius of the cylinder is r and its height is h; the radius of the sphere is R. From the Pythagorean theorem we see that 4r2 + h2 = 4R2 . The volume of the cylinder is V = π r2 h, and therefore we find that 1 V = V (h) = π R2 − h2 h 4 = π (4R2 h − h3 ), 4 0 h 2R. Then V (h) = π (4R2 − 3h2 ), 4 √ so V (h) = 0 when 3h2 = 4R2 , so that h = 2 R 3. This value of h maximizes V because V (0) = 0 and 3√ V (2R) √ 0. The corresponding value of r is 1 R 6, so the ratio of the height of the cylinder to its radius is = 3 √ h/r = 2. The volume of the maximal cylinder is 4 π R3 3 and the volume of√ sphere is 4 π R3 ; the ratio the 9 3 of the volume of the sphere to that of the maximal inscribed cylinder is thus 3. h 2R 2r C03S06.033: The following figure shows a cross section of the cone and inscribed cylinder. Let x be the radius of the cylinder and y its height. By similar triangles in the figure, H y = , R R−x so y= H (R − x). R We are to maximize the volume V = π x2 y of the cylinder, so we write V = V (x) = π x2 =π H (R − x) R H (Rx2 − x3 ), R 9 0 x R. Because V (0) = 0 = V (r), V is maximized when V (x) = 0; this leads to the equation 2xR = 3x2 and thus to the results x = 2 R and y = 1 H . 3 3 H y x R C03S06.034: Let the circle have equation x2 + y 2 = 1 and let (x, y ) denote the coordinates of the upper right-hand vertex of the trapezoid (Fig. 3.6.25). Then the area A of the trapezoid is the product of its altitude y and the average of the lengths of its two bases, so 1 y (2x + 2) where y 2 = 1 − x2 . 2 A positive quantity is maximized when its square is maximized, so we maximize instead A= f (x) = A2 = (x + 1)2 (1 − x2 ) = 1 + 2x − 2x3 − x4 , 0 1. x Because f (0) = 0 = f (1), f is maximized when f (x) = 0: 0 = 2 − 6x2 − 4x3 = 2(1 + x)2 (1 − 2x). But the only solution of f√x) = 0 in the domain of f is x = 1 . Finally, f 1 = 27 , so the maximum possible ( 2 2 16 area of the trapezoid is 3 3. This is just over 41% of the area of the circle, so the answer meets the test of 4 plausibility. C03S06.035: Draw a circle in the plane with center at the origin and with radius R. Inscribe a rectangle with vertical and horizontal sides and let (x, y ) be its vertex in the first quadrant. The base of the rectangle has length 2x and its height is 2y , so the perimeter of the rectangle is P = 4x + 4y . Also x2 + y 2 = R2 , so P = P (x) = 4x + 4 R2 − x2 , P (x) = 4 − √ P (x) = 0 4x ; − x2 R2 when 4 R2 − x2 = 4x; R2 − x2 = x2 ; x2 = 0 12 R. 2 10 x R. √ √ Because x > 0, x = 1 R 2. The corresponding value of P (x) is 4R 2, and P (0) = 4R = P (R). So the 2 former value of x maximizes the perimeter P . Because y 2 = R2 − x2 and because R2 − x2 = x2 at maximum, y = x at maximum. Therefore the rectangle of largest perimeter that can be inscribed in a circle is a square. C03S06.036: Let (x, y ) be the coordinates of the vertex of the rectangle in the first quadrant. Then, by symmetry, the area of the rectangle is A = (2x)(2y ) = 4xy . But from the equation of the ellipse we find that y= 3 5 25 − x2 , A = A(x) = 12 5x so 25 − x2 , 0 5. x We can simplify the algebra by maximizing instead f (x) = 2 25 144 A = 25x2 − x4 ; f (x) = 50x − 4x3 ; √ f (x) = 0 when x = 0 and when x = 5 2. 2 √ √ Now A(0) = 0 = A(5), whereas A 5 2 = 30. So the rectangle of maximum area has base 2x = 5 2 and 2 √ height 2y = 3 2. C03S06.037: We are to maximize volume V = 1 π r2 h given r2 + h2 = 100. The latter relation enables us 3 to write V = V (h) = 1 1 π (100 − h2 )h = π (100h − h3 ), 3 3 0 h 10. √ Now V (h) = 1 π (100 − 3h2 ), so V (h) = 0 when 3h2 = 100, thus when h = 10 3. But V (h) = 0 at the 3 3 √ endpoints of its domain, so the latter value of h maximizes V , and its maximum value is 2000 π 3. 27 C03S06.038: Put the bases of the poles on the x-axis, one at the origin and the other at x = 10. Let the rope touch the ground at the point x. Then the rope reaches straight from (0, 10) to (x, 0) and straight from (x, 0) to (10, 10). In terms of x, its length is L(x) = 100 + x2 + 100 + (10 − x)2 = 100 + x2 + 200 − 20x + x2 , 0 x 10. So L (x) = √ x x − 10 +√ ; 2 100 + x 200 − 20x + x2 L (x) = 0 when x 200 − 20x + x2 = (10 − x) x2 + 100 ; x2 (x2 − 20x + 200) = 100 − 20x + x2 x2 + 100 ; x4 − 20x3 + 200x2 = x4 − 20x3 + 200x2 − 2000x + 10000; 2000x = 10000; 11 and thus when x = 5. Now L(0) = L(10) = 10 1 + length of the shortest possible rope. √ √ 2 , which exceeds L(5) = 10 5. So the latter is the C03S06.039: Let x and y be the two numbers. Then x 0, y 0, and x + y = 16. We are to find both the maximum and minimum values of x1/3 + y 1/3 . Because y = 16 − x, we seek the extrema of Now f (x) = x1/3 + (16 − x)1/3 , f (x) = = 0 16. x 1 −2/3 1 − (16 − x)−2/3 x 3 3 1 1 − ; 3x2/3 3(16 − x)2/3 f (x) = 0 when (16 − x)2/3 = x2/3 , so when 16 − x = x, thus when x = 8. Now f (0) = f (16) = 161/3 ≈ 2.52, so f (8) = 4 maximizes f whereas f (0) and f (16) yield its minimum. C03S06.040: If the base of the L has length x, then the vertical part has length 60 − x. Place the L with its corner at the origin in the xy -plane, its base on the nonnegative x-axis, and the vertical part on the nonnegative y -axis. The two ends of the L have coordinates (0, 60 − x) and (x, 0), so they are at distance d = d(x) = x2 + (60 − x)2 , 0 60. x A positive quantity is minimized when its square is minimal, so we minimize f (x) = (d(x))2 = x2 + (60 − x)2 , 0 x 60. Then f (x) = 2x − 2(60 − x) = 4x − 120; f (x) = 0 when x = 30. Now f (0) = f (60) = 3600, whereas f (30) = 1800. So x = 30 minimizes f (√ and thus d(x). The minimum possible distance between the two x) ends of the wire is therefore d(30) = 30 2. C03S06.041: If (x, x2 ) is a point of the parabola, then its distance from (0, 1) is d(x) = x2 + (x2 − 1)2 . So we minimize f (x) = (d(x))2 = x4 − x2 + 1, where the domain of f is the set of all real numbers. But because f (x) is large positive when |x| is large, we will not exclude a minimum if we restrict the domain of f to be an interval of the form [ −a, a ] where a is a large positive number. On the interval [ −a, a ], f is continuous and thus has a global minimum, which does not occur at ±a because f (±a) is large positive. Because f (x) exists for all x, the minimum of f occurs at a point where f (x) = 0: 4x3 − 2x = 0; 2x(2x2 − 1) = 0. √ √ Hence x = 0 or x = ± 1 2. Now f (0) = 1 and f ± 1 √ = 3 . So x = 0 yields a local maximum value for 2 22 4 √ f (x), and the minimum possible distance is 0.75 = 1 3. 2 C03S06.042: It suffices to minimize x2 + y 2 given y = (3x − 4)1/3 . Let f (x) = x2 + (3x − 4)2/3 . Then 12 f (x) = 2x + 2(3x − 4)−1/3 . Then f (x) = 0 when 2x + 2 = 0; (3x − 4)1/3 2x(3x − 4)1/3 = −2; x(3x − 4)1/3 = −1; x3 (3x − 4) = −1; x3 (3x − 4) + 1 = 0; 3x4 − 4x3 + 1 = 0; (x − 1)2 (3x2 + 2x + 1) = 0. Now x = 1 is the only real solution of the last equation, f (x) does not exist when x = 4 , and f (1) = 2 > 3 16 4 4 9 = f 3 . So the point closest to the origin is 3 , 0 . C03S06.043: Examine the plank on the right on Fig. 3.6.10. Let its height be 2y and its width (in the x-direction) be z . The total area of the four small rectangles in the figure is then√ = 4 √ · 2y = 8yz . The A ·z circle has radius 1, and by Problem 35 the large inscribed square has dimensions 2 by 2. Thus 1√ 2+z 2 2 + y 2 = 1. This implies that y= √ 1 − z 2 − z2 . 2 Therefore A(z ) = 8z √ 1 − z 2 − z2 , 2 0 z 1− 1√ 2. 2 Now A(z ) = 0 at each endpoint of its domain and √ √ 4 2 1 − 3z z − 4z 2 A (z ) = . √ 1 − 2z z − 2z 2 √ √ So A (z ) = 0 when z = 1 −3 2 ± 34 ; we discard the negative solution, and find that when A(z ) is 8 maximized, √ √ −3 2 + 34 z= ≈ 0.198539, 8 √ 7 − 17 2y = ≈ 0.848071, and 2 √ 142 + 34 17 A(z ) = ≈ 0.673500. 2 13 The four small planks use just under 59% of the wood that remains after the large plank is cut, a very efficient use of what might be scrap lumber. C03S06.044: Place the base of the triangle on the x-axis and its upper vertex on the y -axis. Then its √ lower right vertex is at the point 1 , 0 and its upper vertex is at 0, 1 3 . It follows that the slope of the 2 2 √ side of the triangle joining these two vertices is − 3. So this side lies on the straight line with equation y= √ 1 −x . 2 3 Let (x, y ) be the coordinates of the upper right-hand vertex of the rectangle. Then the rectangle has area A = 2xy , so A(x) = √ 3 (x − 2x2 ), 0 1 . 2 x Now A (x) = 0 when x = 1 , and because A(√) = 0 at the endpoints of its domain, it follows that the x 4 maximum area of such a rectangle is A 1 = 1 3. 4 8 C03S06.045: Set up a coordinate system in which the island is located at (0, 2) and the village at (6, 0), and let (x, 0) be the point at which the boat lands. It is clear that 0 x 6. The trip involves the land distance 6 − x traveled at 20 km/h and the water distance (4 + x2 )1/2 traveled at 10 km/h. The total time of the trip is then given by T (x) = 1 10 4 + x2 + 1 (6 − x), 20 0 x 6. Now x 1 −. 2 20 10 4 + x √ 0, we find that x = 2 3. The value of T there is 3 T (x) = Thus T (x) = 0 when 3x2 = 4; because x √ √ 1 3+ 3 10 ≈ 0.473, whereas T (0) = 0.5 and T (6) ≈ 0.632. Therefore the boater should make landfall at the point on the shore closest to the island. 2 3 √ 3 ≈ 1.155 km from C03S06.046: Set up a coordinate system in which the factory is located at the origin and the power station at (L , W ) in the xy -plane—L = 4500, W = 2000. Part of the path of the power cable will be straight along the river bank and part will be a diagonal running under water. It makes no difference whether the straight part is adjacent to the factory or to the power station, so we assume the former. Thus we suppose that the power cable runs straight from (0, 0) to (x, 0), then straight from (x, 0) to (L, W ), where 0 x L. Let y be the length of the diagonal stretch of the cable. Then by the Pythagorean theorem, W 2 + (L − x)2 = y 2 , so y= W 2 + (L − x)2 . The cost C of the cable is C = kx + 3ky where k is the cost per unit distance of over-the-ground cable. Therefore the total cost of the cable is C (x) = kx + 3k W 2 + (L − x)2 , 0 x L. It will not change the solution if we assume that k = 1, and in this case we have 14 3(L − x) C (x) = 1 − W 2 + (L − x)2 . Next, C (x) = 0 when x2 + (L − x)2 = 9(L − x)2 , and this leads to the solution 1√ 3√ x = L − W 2 and y = W 2. 4 4 It is not difficult to verify that the latter value of x yields a value of C smaller than either C (0) or C (L). √ √ Answer: Lay the cable x = 4500 − 500 2 ≈ 3793 meters along the bank and y = 1500 2 ≈ 2121 meters diagonally across the river. C03S06.047: The distances involved are |AP | = |BP | = minimize √ x2 + 1 and |CP | = 3 − x. Therefore we are to f (x) = 2 x2 + 1 + 3 − x, 0 3. x Now f (x) = √ 2x − 1; x2 + 1 f (x) = 0 when √ 2x = 1. x2 + 1 √ This leads to the equation 3x2 = 1, so x = 1 3. Now f (0) = 5, f (3) ≈ 6.32, and at the√ critical point, 3 √ 1 f (x) = 3 + 3 ≈ 4.732. Answer: The distribution center should be located at the point P ( 3 3, 0). C03S06.048: 1√ 2 1 a + x2 + (s − x)2 + b2 . c v x s−x (b) T (x) = √ − . 2 + x2 ca v (s − x)2 + b2 (a) T = T (x) = 0 √ when s−x x √ = 2 + x2 ca v (s − x)2 + b2 ; (x − s)2 + b2 c =; s−x v c sin α csc β = v sin α c = = n. sin β v x · a2 + x2 C03S06.049: We are to minimize total cost C = c1 C (x) = √ a2 + x2 + c2 c1 x − + x2 a2 C (x) = 0 when √ (L − x)2 + b2 . c2 (L − x) (L − x)2 + b2 c1 x = + x2 a2 ; c2 (L − x) (L − x)2 + b2 . The result in Part (a) is equivalent to the last equation. For Part (b), assume that a = b = c1 = 1, c2 = 2, and L = 4. Then we obtain 15 √ 2(4 − x) x = 1 + x2 (4 − x)2 + 1 ; x2 4(16 − 8x + x2 ) = ; 1 + x2 16 − 8x + x2 + 1 x2 (17 − 8x + x2 ) = (4 + 4x2 )(16 − 8x + x2 ); 17x2 − 8x3 + x4 = 64 − 32x + 68x2 − 32x3 + 4x4 . Therefore we wish to solve f (x) = 0 where f (x) = 3x4 − 24x4 + 51x2 − 32x + 64. Now f (0) = 64, f (1) = 62, f (2) = 60, f (3) = 22, and f (4) = −16. Because f (3) > 0 > f (4), we interpolate to estimate the zero of f (x) between 3 and 4; it turns out that interpolation gives x ≈ 3.58. Subsequent interpolation yields the more accurate estimate x ≈ 3.45. (The equation f (x) = 0 has exactly two solutions, x ≈ 3.452462314 and x ≈ 4.559682567.) C03S06.050: Because x3 + y 3 = 2000, y = (2000 − x3 )1/3 . We want to maximize and minimize total surface area A = 6x2 + 6y 2 ; A = A(x) = 6x2 + 6(2000 − x3 )2/3 , A (x) = 0 x √ 3 10 2. −12[x2 − x(2000 − x3 )1/3 ] . (2000 − x3 )1/3 √ Now A (x) = 0 at x = 0 and at x = 10; A (x) does not exist at x = 10 3 2, the right-hand endpoint of the domain of A (at that point, the graph of A has a vertical tangent). Also A(0) = 600 · 22/3 ≈ 952.441 and √ A 10 3 2 is the same; A(10) = 1200. So the maximum surface area is attained when each cube has edge √ length 10 and the minimum is attained when there is only one cube, of edge length 10 3 2 ≈ 12.5992. C03S06.051: Let r be the radius of the sphere and x the edge length of the cube. We are to maximize and minimize total volume V= 43 π r + x3 3 given 4π r2 + 6x2 = 1000. The latter equation yields x= 1000 − 4π r2 , 6 so V = V (r) = 43 πr + 3 500 − 2π r2 3 3/2 , 0 r Next, V (r) = 4π r2 − 2π r 16 500 − 2π r2 , 3 r1 = 5 10 . π and V (r) = 0 when 500 − 2π r2 . 3 4π r2 = 2π r So r = 0 or 2r = 500 − 2π r2 . 3 The latter equation leads to r = r2 = 5 10 . π+6 Now V (0) ≈ 2151.66, V (r1 ) ≈ 2973.54, and V (r2 ) ≈ 1743.16. Therefore, to minimize the sum of the volumes, choose r = r2 ≈ 5.229 in. and x = 2r2 ≈ 10.459 in. To maximize the sum of their volumes, take r = r1 ≈ 8.921 in. and x = 0 in. C03S06.052: Let the horizontal piece of wood have length 2x and the vertical piece have length y + z where y is the length of the part above the horizontal piece and z the length of the part below it. Then y= Also the kite area is A = x(y + z ); 4 − x2 and z = 16 − x2 . dA = 0 implies that dx y+z = x2 x2 +. y z Multiply each side of the last equation by yz to obtain y 2 z + yz 2 = x2 z + x2 y, so that yz (y + z ) = x2 (y + z ); x2 = yz ; x4 = y 2 z 2 = (4 − x2 )(16 − x2 ); x4 = 64 − 20x2 + x4 ; 20x2 = 64; x= Therefore L1 = 8 5 √ 4√ 2√ 8√ 5, y = 5, z = 5. 5 5 5 √ 5 ≈ 3.5777 and L2 = 2 5 ≈ 4.47214 for maximum area. 17 C03S06.053: The graph of V (x) is shown next. The maximum volume seems to occur near the point (4, V (4)) ≈ (4, 95.406), so the maximum volume is approximately 95.406 cubic feet. 100 80 60 40 20 1 2 3 4 5 C03S06.054: The graph of V (x) is shown next. The maximum volume seems to occur near the point (8 , V (8)) ≈ (8, 269.848), so the maximum volume is approximately 269.848 cubic feet. 300 200 100 2 4 6 8 10 C03S06.055: Let V1 and V2 be the volume functions of problems 53 and 54, respectively. Then √ 20 5 4x − x2 √ V1 (x) = , 3 5−x which is zero at x = 0 and at x = 4, and √ 10 5 8x − x2 √ V2 (x) = , 3 10 − x which is zero at x = 0 and at x = 8, as expected. Finally, √ V2 (8) = 2 2. V1 (4) C03S06.056: Let x denote the length of each edge of the base of the box; let y denote its height. If the box has total surface area A, then 2x2 + 4xy = A, and hence y= A − 2x2 . 4x The box has volume V = x2 y , so its volume can be expressed as a function of x alone: 18 (1) V (x) = Ax − 2x3 , 4 0 A/2 . x Then V (x) = A − 6x2 ; 4 V (x) = 0 when x = A/6 . This critical point clearly lies in the interior of the domain of V , and (almost as clearly) V (x) is increasing to its left and decreasing to its right. Hence this critical point yields the box of maximal volume. Moreover, when x = A/6 , we have—by Eq. (1)— √ √ √ A − (A/3) 6 6√ A 2A y == ·√ · A=√ . = = 3 4A 6 6 4 A/6 Therefore the closed box with square base, fixed surface area, and maximal volume is a cube. C03S06.057: Let x denote the length of each edge of the square base of the box and let y denote its height. Given total surface area A, we have x2 + 4xy = A, and hence y= A − x2 . 4x (1) The volume of the box is V = x2 y , and therefore V (x) = Ax − x2 , 4 0 x √ A. Next, V (x) = A − 3x2 ; 4 V (x) = 0 when x = A/3 . Because V (x) > 0 to the left of this critical point and V (x) < 0 to the right, it yields the global maximum value of V (x). By Eq. (1), the corresponding height of the box is 1 A/3 . Therefore the open box with 2 square base and maximal volume has height equal to half the length of the edge of its base. C03S06.058: Let r denote the radius of the base of the closed cylindrical can, h its height, and A its total surface area. Then 2π r2 + 2π rh = A, and hence h= A − 2π r2 . 2π r (1) The volume of the can is V = π r2 h, and thus V (r) = Ar − 2π r3 , 2 0 r A 2π 1/2 . Next, V (r) = A − 6π r2 ; 2 V (r) = 0 when r= A 6π 1 /2 . Because V (r) > 0 to the left of this critical point and V (r) < 0 to the right, it determines the global maximum value of V (r). By Eq. (1), it follows that the can of maximum volume has equal height and diameter. 19 C03S06.059: Let r denote the radius of the base of the open cylindrical can and let h denote its height. Its total surface area A then satisfies the equation π r2 + 2π rh = A, and therefore h= A − πr2 . 2π r (1) Thus the volume of the can is given by V (r) = Ar − π r3 , 2 0 A π r 1 /2 . Next, V (r) = A − 3π r2 ; 2 V (r) = 0 when A 3π r= 1 /2 . Clearly V (r) > 0 to the left of this critical point and V (r) < 0 to the right, so it determines the global maximum value of V (r). By Eq. (1) the corresponding value of h is the same, so the open cylindrical can of maximum volume has height equal to its radius. C03S06.060: Let r denote the interior radius of the cylindrical can and h its interior height. Because the thickness t of the material of the can will be very small in comparison with r and h, the total amount of material M used to make the can will be very accurately approximated by multiplying the thickness of the bottom by its area, the thickness of the curved side by its area, and the thickness of the top by its area. That is, π r2 t + 2π rht + 3π r2 t = M, so that h= M − 4π r 2 t . 2π rt (1) Thus the volume of the can will be given by (the very accurate approximation) V (r) = M r − 4π r 3 t , 2t 0 r M 4π t 1/2 . Next, V (r) = M − 12π r2 t ; 2t V (r) = 0 when r= M 1 · 2 3π t 1 /2 . Because V (r) > 0 to the left of this critical point and V (r) < 0 to the right, it yields the global maximum value of V (r). By (1), the height of the corresponding can is four times as great, so the can of maximum volume has height twice its diameter (approximately, but quite accurately). To solve this problem exactly, first establish that 4π t(r + t)2 + π ht2 (2 ∗ r + t) = M, then that V (r) = M r2 + 4π tr2 (r + t)2 . t2 (2 ∗ r + t) Then show that V (r) = 0 when 12π tr4 + 24π t2 r3 + (16π t3 − M )r2 + (4π t4 − M t)r = 0, 20 and that the relevant critical point is r= −3π t2 + 3π t(M − π t3 ) . 6π t C03S06.061: Let f (t) = 1 , 1 + t2 so that f (t) = − 2t . (1 + t2 )2 The line tangent to the graph of y = f (t) at the point (t, f (t)) then has x-intercept and y -intercept 1 + 3 t2 2t and 1 + 3 t2 , (1 + t2 )2 respectively. The area of the triangle bounded by the part of the tangent line in the first quadrant and the coordinate axes is A(t) = 1 1 + 3 t2 1 + 3 t2 · · , 2 2t (1 + t2 )2 (1) and A (t) = −9t6 + 9t4 + t2 − 1 . 4t2 (1 + t2 )3 Next, A (t) = 0 when (t − 1)(t + 1)(3t2 − 1)(3t2 + 1) = 0, and the only two critical points of A in the interval [0.5, 2] are √ 3 t1 = 1 and t2 = ≈ 0.57735. 3 Significant values of A(t) are then A(0.5) = 0.98, A(0.57735) ≈ 0.97428, A(1) = 1, and A(2) = 0.845. Therefore A(t) has a local maximum at t = 0.5, a local minimum at t2 , its global maximum at t = 1, and its global minimum at t = 2. To answer the first question in Problem 61, Eq. (1) makes it clear that A(t) → + ∞ as t → 0+ and, in addition, that A(t) → 0 as t → + ∞. C03S06.062: If 0 x < 1, then the cost of the power line will be C (x) = 40x + 100 1 + (1 − x)2 (in thousand of dollars). If x = 1, then the cost will be 80 thousand dollars because there is no need to use underground cable. Next, √ 100x − 100 + 40 x2 − 2x + 2 √ C (x) = , x2 − 2x + 2 and C (x) = 0 when 21 √ 21 − 2 21 x = x0 = ≈ 0.563564219528. 21 The graph of C (x) (using a computer algebra system) establishes that x0 determines the global minimum for C (x) on the interval [0, 1), yielding the corresponding value C (x0 ) ≈ 131.651513899117. Hence the global minimum for C (x) on [0, 1] is C (1) = 80 (thousand dollars). It is neither necessary to cross the park nor to use underground cable. 22 Section 3.7 C03S07.001: If f (x) = 3 sin2 x = 3(sin x)2 , then f (x) = 6 sin x cos x. C03S07.002: If f (x) = 2 cos4 x = 2(cos x)4 , then f (x) = 8(cos x)3 (− sin x) = −8 cos3 x sin x. C03S07.003: If f (x) = x cos x, then f (x) = 1 · cos x + x · (− sin x) = cos x − x sin x. C03S07.004: If f (x) = x1/2 sin x, then f (x) = 1 x−1/2 sin x + x1/2 cos x = 2 sin x + 2x cos x √ . 2x C03S07.005: If f (x) = sin x x cos x − sin x , then f (x) = . x x2 C03S07.006: If f (x) = x1/2 (− sin x) − 1 x−1/2 cos x cos x 2x sin x + cos x 2 √ . , then f (x) = =− 1 /2 x 2x x x C03S07.007: If f (x) = sin x cos2 x, then f (x) = cos x cos2 x + (sin x)(2 cos x)(− sin x) = cos3 x − 2 sin2 x cos x. C03S07.008: If f (x) = cos3 x sin2 x, then f (x) = (3 cos2 x)(− sin x)(sin2 x) + (2 sin x cos x)(cos3 x) = −3 cos2 x sin3 x + 2 sin x cos4 x. C03S07.009: If g (t) = (1 + sin t)4 , then g (t) = 4(1 + sin t)3 · cos t. C03S07.010: If g (t) = (2 − cos2 t)3 , then g (t) = 3(2 − cos2 t)2 · (2 cos t sin t) = 6(2 − cos2 t)2 (sin t cos t). C03S07.011: If g (t) = 1 cos t − sin t sin t − cos t , then (by the reciprocal rule) g (t) = − = . sin t + cos t (sin t + cos t)2 (sin t + cos t)2 C03S07.012: If g (t) = sin t , then (by the quotient rule) 1 + cos t g (t) = (1 + cos t)(cos t) − (sin t)(− sin t) sin2 t + cos2 t + cos t 1 + cos t 1 = = = . (1 + cos t)2 (1 + cos t)2 (1 + cos t)2 1 + cos t C03S07.013: If f (x) = 2x sin x − 3x2 cos x, then (by the product rule) f (x) = 2 sin x + 2x cos x − 6x cos x + 3x2 sin x = 3x2 sin x − 4x cos x + 2 sin x. C03S07.014: If f (x) = x1/2 cos x − x−1/2 sin x, f (x) = 1 −1/2 1 (1 − 2x2 ) sin x − x cos x √ x cos x − x1/2 sin x + x−3/2 sin x − x−1/2 cos x = . 2 2 2x x C03S07.015: If f (x) = cos 2x sin 3x, then f (x) = −2 sin 2x sin 3x + 3 cos 2x cos 3x. C03S07.016: If f (x) = cos 5x sin 7x, then f (x) = −5 sin 5x sin 7x + 7 cos 5x cos 7x. 1 C03S07.017: If g (t) = t3 sin2 2t = t3 (sin 2t)2 , then g (t) = 3t2 (sin 2t)2 + t3 · (2 sin 2t) · (cos 2t) · 2 = 3t2 sin2 2t + 4t3 sin 2t cos 2t. C03S07.018: If g (t) = g (t) = √ t cos3 3t = t1/2 (cos 3t)3 , then √ 1 −1/2 cos3 3t (cos 3t)3 + t1/2 · 3(cos 3t)2 · (−3 sin 3t) = √ − 9 t cos2 3t sin 3t. t 2 2t C03S07.019: If g (t) = (cos 3t + cos 5t)5/2 , then g (t) = 5 (cos 3t + cos 5t)3/2 (−3 sin 3t − 5 sin 5t). 2 C03S07.020: If g (t) = 1 sin t + sin 3t 2 2 = (sin2 t + sin2 3t)−1/2 , then 1 sin t cos t + 3 sin 3t cos 3t g (t) = − (sin2 t + sin2 3t)−3/2 (2 sin t cos t + 6 sin 3t cos 3t) = − . 2 (sin2 2t + sin2 3t)3/2 C03S07.021: If y = y (x) = sin2 √ x = (sin x1/2 )2 , then √ √ dy 1 sin x cos x √ . = 2(sin x1/2 )(cos x1/2 ) · x−1/2 = dx 2 x C03S07.022: If y = y (x) = cos 2x 2x sin 2x + cos 2x dy x · (−2 sin 2x) − 1 · cos 2x =− . , then = 2 x dx x x2 C03S07.023: If y = y (x) = x2 cos(3x2 − 1), then dy = 2x cos(3x2 − 1) − x2 · 6x · sin(3x2 − 1) = 2x cos(3x2 − 1) − 6x3 sin(3x2 − 1). dx C03S07.024: If y = y (x) = sin3 x4 = (sin x4 )3 , then dy = 3(sin x4 )2 · Dx (sin x4 ) = 3(sin x4 )2 · (cos x4 ) · Dx (x4 ) = 12x3 sin2 x4 cos x4 . dx C03S07.025: If y = y (x) = sin 2x cos 3x, then dy = (sin 2x) · Dx (cos 3x) + (cos 3x) · Dx (sin 2x) = −3 sin 2x sin 3x + 2 cos 3x cos 2x. dx C03S07.026: If y = y (x) = x dy (sin 3x) · 1 − x · Dx (sin 3x) sin 3x − 3x cos 3x , then = = . sin 3x dx (sin 3x)2 sin2 3x C03S07.027: If y = y (x) = cos 3x , then sin 5x dy (sin 5x)(−3 sin 3x) − (cos 3x)(5 cos 5x) 3 sin 3x sin 5x + 5 cos 3x cos 5x = =− . dx (sin 5x)2 sin2 5x C03S07.028: If y = y (x) = cos √ x = (cos x1/2 )1/2 , then 2 √ dy 1 1 sin x = (cos x1/2 )−1/2 (− sin x1/2 ) · x−1/2 = − √ √. dx 2 2 4 x cos x C03S07.029: If y = y (x) = sin2 x2 = (sin x2 )2 , then dy = 2(sin x2 ) · Dx (sin x2 ) = 2(sin x2 ) · (cos x2 ) · Dx (x2 ) = 4x sin x2 cos x2 . dx C03S07.030: If y = y (x) = cos3 x3 = (cos x3 )3 , then dy = 3(cos x3 )2 · Dx (cos x3 ) = 3(cos x3 )2 · (− sin x3 ) · Dx (x3 ) = −9x2 cos2 x3 sin x3 . dx √ C03S07.031: If y = y (x) = sin 2 x = sin(2x1/2 ), then √ dy cos 2 x . = cos(2x1/2 ) · Dx (2x1/2 ) = x−1/2 cos(2x1/2 ) = √ dx x √ C03S07.032: If y = y (x) = cos 3 3 x = cos(3x1/3 ), then √ dy sin 3 3 x 1 /3 1 /3 1/3 −2/3 )=− √ . = − sin(3x ) · Dx (3x ) = (− sin 3x ) · (x 3 dx x2 C03S07.033: If y = y (x) = x sin x2 , then C03S07.034: If y = y (x) = x2 cos dy 1 = 2x cos dx x = 2x cos C03S07.035: If y = y (x) = √ 1 x dy = 1 · sin x2 + x · (cos x2 ) · 2x = sin x2 + 2x2 cos x2 . dx 1 , then x + x2 − sin − x2 sin 1 x 1 x · Dx ·− 1 x2 1 x = 2x cos 1 x + sin 1 x . √ x sin x = x1/2 sin x1/2 , then √ √ √ √ √ dy cos x 1 sin x 1 sin x + x cos x √ + . = x−1/2 sin x1/2 + x1/2 (cos x1/2 ) · x−1/2 = √ = dx 2 2 2 2x 2x C03S07.036: If y = y (x) = (sin x − cos x)2 , then dy = 2(sin x − cos x)(cos x + sin x) = 2(sin2 x − cos2 x) = −2 cos 2x. dx C03S07.037: If y = y (x) = √ x (x − cos x)3 = x1/2 (x − cos x)3 , then 3 dy 1 = x−1/2 (x − cos x)3 + 3x1/2 (x − cos x)2 (1 + sin x) dx 2 = √ (x − cos x)3 (x − cos x)3 + 6x(x − cos x)2 (1 + sin x) √ √ + 3 x (x − cos x)2 (1 + sin x) = . 2x 2x C03S07.038: If y = y (x) = √ x sin x+ √ x = x1/2 sin(x + x1/2 )1/2 , then 1 dy 1 1 = x−1/2 sin(x + x1/2 )1/2 + x1/2 cos(x + x1/2 )1/2 · (x + x1/2 )−1/2 1 + x−1/2 . dx 2 2 2 The symbolic algebra program Mathematica simplifies this to (2x + dy = dx √ x ) cos x+ C03S07.039: If y = y (x) = cos(sin x2 ), then C03S07.040: If y = y (x) = sin 1 + √ √ x +2 x+ √ √ 4 x x+ x √ x sin x+ √ x . dy = − sin(sin x2 ) · (cos x2 ) · 2x = −2x sin(sin x2 ) cos x2 . dx sin x , then √ dy = cos 1 + sin x dx √ (cos x) cos 1 + sin x 1 √ · (sin x)−1/2 · cos x = . 2 2 sin x dx = sec t7 dt C03S07.041: If x = x(t) = tan t7 = tan(t7 ), then C03S07.042: If x = x(t) = sec t7 = sec(t7 ), then 2 · Dt (t7 ) = 7t6 sec2 t7 . dx = (sec t7 tan t7 ) · Dt (t7 ) = 7t6 sec t7 tan t7 . dt C03S07.043: If x = x(t) = (tan t)7 = tan7 t, then dx = 7(tan t)6 · Dt tan t = 7(tan t)6 sec2 t = 7 tan6 t sec2 t. dt C03S07.044: If x = x(t) = (sec 2t)7 = sec7 2t, then dx = 7(sec 2t)6 · Dt (sec 2t) = 7(sec 2t)6 (sec 2t tan 2t) · Dt (2t) = 14 sec7 2t tan 2t. dt C03S07.045: If x = x(t) = t7 tan 5t, then C03S07.046: If x = x(t) = dx = 7t6 tan 5t + 5t7 sec2 5t. dt sec t5 , then t dx t · (sec t5 tan t5 ) · 5t4 − sec t5 5t5 sec t5 tan t5 − sec t5 = = . dt t2 t2 C03S07.047: If x = x(t) = √ t sec √ t = t1/2 sec(t1/2 ), then 4 dx 1 sec 1 = t−1/2 sec(t1/2 ) + t1/2 sec(t1/2 ) tan(t1/2 ) · t−1/2 = dt 2 2 C03S07.048: If x = x(t) = sec dx = sec t1/2 dt √ √ t+ √ t sec √ 2t √ √ t tan t . √ t tan t = sec t1/2 tan t1/2 , then 1 −1/2 t sec2 t1/2 2 + 1 −1/2 t sec t1/2 tan t1/2 2 1 , then t2 dx 1 1 = − csc 2 cot 2 dt t t tan t 1 /2 = sec3 √ 1 t2 √ √ t + sec t tan2 t √ . 2t . C03S07.049: If x = x(t) = csc C03S07.050: If x = x(t) = cot C03S07.051: If x = x(t) = 2 t3 = 1 2 csc 2 3 t t cot = cot t−1/2 , then 1 √ t dx = − csc t−1/2 dt ·− 2 · Dt t−1/2 = 1 1 −3/2 2 t csc2 t−1/2 = √ csc2 √ 2 tt t . sec 5t , then tan 3t dx 5 tan 3t sec 5t tan 5t − 3 sec 5t sec2 3t = 5 cot 3t sec 5t tan 5t − 3 csc2 3t sec 5t. = dt (tan 3t)2 C03S07.052: If x = x(t) = sec2 t − tan2 t, then dx = (2 sec t)(sec t tan t) − (2 tan t)(sec2 t) ≡ 0. dt C03S07.053: If x = x(t) = t sec t csc t, then dx = sec t csc t + t sec t tan t csc t − t sec t csc t cot t = t sec2 t + sec t csc t − t csc2 t. dt C03S07.054: If x = x(t) = t3 tan3 t3 = t3 (tan t3 )3 , then dx = 3t2 (tan t3 )3 + t3 · 3(tan t3 )2 (sec t3 )2 · 3t2 = 3t2 tan3 t3 + 9t5 sec2 t3 tan2 t3 . dt C03S07.055: If x = x(t) = sec(sin t), then C03S07.056: If x = x(t) = cot(sec 7t), then dx = [sec(sin t) tan(sin t)] · cos t. dt dx = − csc2 (sec 7t) · 7 sec 7t tan 7t. dt C03S07.057: If x = x(t) = sin t dx = sin t cos t, then = cos2 t − sin2 t = cos 2t. sec t dt C03S07.058: If x = x(t) = sec t , then 1 + tan t dx sec t tan t + sec t tan2 t − (1 + tan2 t) sec t sec t tan t − sec t (1 + tan t) sec t tan t − sec t sec2 t = = . = 2 dt (1 + tan t) (1 + tan t)2 (1 + tan t)2 5 C03S07.059: If x = x(t) = √ 1 + cot 5t = (1 + cot 5t)1/2 , then dx 1 5 csc2 5t = (1 + cot 5t)−1/2 (−5 csc2 5t) = − √ . dt 2 2 1 + cot 5t C03S07.060: If x = x(t) = csc √ t = (csc t1/2 )1/2 , then cot dx 1 1 = (csc t1/2 )−1/2 (− csc t1/2 cot t1/2 ) · t−1/2 = − dt 2 2 √ t √ csc 4t √ t =− csc √ t 3 /2 √ 4t cos √ t . C03S07.061: If f (x) = x cos x, then f (x) = −x sin x + cos x, so the slope of the tangent at x = π is f (π ) = −π sin π + cos π = −1. Because f (π ) = −π , an equation of the tangent line is y + π = −(x − π ); that is, y = −x. The graph of f and this tangent line are shown next. 6 4 2 -2 2 4 6 -2 -4 -6 C03S07.062: If f (x) = cos2 x then f (x) = −2 cos x sin x, so the slope of the tangent at x = π /4 is f (π /4) = −2 cos(π /4) sin(π /4) = −1. Because f (π /4) = 1 , an equation of the tangent line is y − 1 = 2 2 −(x − π /4); that is, 4y = −4x + 2 + π . The graph of f and this line are shown next. 2 1.5 1 0.5 -0.5 0.5 1 1.5 2 -0.5 -1 f (1) = sec2 4 πx πx tan , then f (x) = sec2 , so the slope of the tangent at x = 1 is π 4 4 4 4 = 2. Because f (1) = , an equation of the tangent line is y − = 2 (x − 1); that is, π π If f (x) = C03S07.063: π 4 6 y = 2x − 2 + 4 . The graph of f and this tangent line are shown next. π 3 2.5 2 1.5 1 0.5 0.6 0.8 1.2 1.4 3 πx πx πx sin2 , then f (x) = 2 sin cos , so the slope of the tangent π 3 3 3 √ 5π 5π 9 at x = 5 is f (5) = 2 sin cos = − 1 3. Because f (5) = , an equation of the tangent line is 2 3 3 4π √ √ √ 9 x 3 9 + 10π 3 y− = − 1 3 (x − 5); that is, y = − + . The graph of f and this tangent line are shown 2 4π 2 4π next. C03S07.064: If f (x) = 4.6 4.8 5.2 5.4 0.8 0.6 0.4 dy C03S07.065: = −2 sin 2x. This derivative is zero at all values of x for which sin 2x = 0; i.e., values of dx x for which 2x = 0, ±π , ±2π , ±3π , . . . . Therefore the tangent line is horizontal at points with x-coordinate an integral multiple of 1 π . These are points of the form (nπ , 1) for any integer n and 1 mπ , −1 for any 2 2 odd integer m. dy = 1 − 2 cos x, which is zero for x = 1 π + 2k π and for x = − 1 π + 2k π for any integer 3 3 dx 1 1 k . The tangent line is horizontal at all points of the form ± 3 π + 2k π , y ± 3 π + 2k π where k is an integer. C03S07.066: C03S07.067: If f (x) = sin x cos x, then f (x) = cos2 x − sin2 x. This derivative is zero at x = 1 π + nπ and 4 at x = 3 π + nπ for any integer n. The tangent line is horizontal at all points of the form nπ + 1 π , 1 and 4 4 2 at all points of the form nπ + 3 π , − 1 where n is an integer. 4 2 C03S07.068: If 7 f (x) = 1 , 3 sin2 x + 2 cos2 x then f (x) = − sin 2x 2 + sin2 x 2. This derivative is zero at all values of x for which sin 2x = 0; i.e., values of x for which 2x = 0, ±π , ±2π , ±3π , . . . . Therefore the tangent line is horizontal at points with x-coordinate an integral multiple of 1 π . 2 These are points of the form nπ , 1 for any integer n and 1 mπ , 1 for any odd integer m. 2 2 3 C03S07.069: Let f (x) = x − 2 cos x. Then f (x) = 1 + 2 sin x, so f (x) = 1 when 2 sin x = 0; that is, when x = nπ for some integer n. Moreover, if n is an integer then f (nπ ) = nπ − 2 cos nπ , so f (nπ ) = nπ + 2 if n is even and f (nπ ) = nπ − 2 if n is odd. In particular, f (0) = 2 and f (π ) = π − 2. So the two lines have equations y = x + 2 and y = x − 2, respectively. C03S07.070: If f (x) = 16 + sin x , 3 + sin x then f (x) = − 13 cos x , (3 + sin x)2 so that f (x) = 0 when cos x = 0; that is, when x is an odd integral multiple of 1 π . In particular, 2 f similarly, f 3 4π = 15 2. 1 2π = 16 + 1 17 = ; 3+1 4 Hence equations of the two lines are y ≡ 17 4 and y ≡ 15 2. C03S07.071: To derive the formulas for the derivatives of the cotangent, secant, and cosecant functions, express each in terms of sines and cosines and apply the quotient rule (or the reciprocal rule) and various trigonometric identities (see Appendix C). Thus Dx cot x = Dx cos x − sin2 x − cos2 x 1 = − − 2 = − csc2 x, 2 sin x sin x sin x Dx sec x = Dx 1 − sin x 1 sin x =− = · = sec x tan x, 2x cos x cos cos x cos x Dx csc x = Dx 1 cos x 1 cos x =− 2 =− · = − csc x cot x. sin x sin x sin x sin x C03S07.072: If g (x) = cos x, then g (x) = lim h→0 g (x + h) − g (x) h cos x cos h − sin x sin h − cos x h→0 h = lim = lim h→0 1 − cos h sin h (− cos x) − lim (sin x) h→0 h h = 0 · (− cos x) − 1 · sin x = − sin x. C03S07.073: Write R = R(α) = 12 v sin 2α. Then 32 12 R (α) = v cos 2α, 16 8 and which is zero when α = π /4 (we assume 0 α π /2). Because R is zero at the endpoints of its domain, we conclude that α = π /4 maximizes the range R. C03S07.074: Let h be the altitude of the balloon (in feet) at time t (in seconds) and let θ be its angle of elevation with respect to the observer. From the obvious figure, h = 300 tan θ, so dh dθ = (300 sec2 θ) . dt dt When θ = π /4 and dθ = π /180, we have dt dh π 10π = 300 · 2 · = ≈ 10.47 (ft/s) dt 180 3 as the rate of the balloon’s ascent then. C03S07.075: Let h be the altitude of the rocket (in miles) at time t (in seconds) and let α be its angle of elevation then. From the obvious figure, h = 2 tan α, so dh dα = (2 sec2 α) . dt dt When α = 5π /18 and dα/dt = 5π /180, we have dh/dt ≈ 0.4224 (mi/s; about 1521 mi/h). C03S07.076: Draw a figure in which the airplane is located at (0, 25000) and the fixed point on the ground is located at (x, 0). A line connecting the two produces a triangle with angle θ at (x, 0). This angle is also the angle of depression of the pilot’s line of sight, and when θ = 65◦ , dθ/dt = 1.5◦ /s. Now tan θ = 25000 cos θ , so x = 25000 , x sin θ thus dx 25000 =− 2 . dθ sin θ The speed of the airplane is − When θ = dx 25000 dθ ·. = dt sin2 θ dt 13 dθ π π, = . So the ground speed of the airplane is 36 dt 120 25000 π ≈ 796.81 (ft/s). · 13π 120 2 sin 36 Answer: About 543.28 mi/h. C03S07.077: Draw a figure in which the observer is located at the origin, the x-axis corresponds to the ground, and the airplane is located at (x, 20000). The observer’s line of sight corrects the origin to the point (x, 20000) and makes an angle θ with the ground. Then tan θ = so that x = 20000 cot θ. Thus 9 20000 , x dx dθ = (−20000 csc2 θ) . dt dt dθ dθ = 0.5◦ /s; that is, = π /360 radians per second when θ = π /3. We dt dt evaluate dx/dt at this time with these values to obtain When θ = 60◦ , we are given dx 1 = (−20000) 2π dt sin 3 · 2000π π =− , 360 27 approximately −232.71 ft/s. Answer: About 158.67 mi/h. C03S07.078: The area of the rectangle is A = 4xy , but x = cos θ and y = sin θ, so A = A(θ) = 4 sin θ cos θ, 0 π /2. θ Now A (θ) = 4(cos2 θ − sin2 θ) = 4 cos 2θ, so A (θ) = 0 when cos 2θ = 0. Because 0 2θ π , it follows that 2θ = π /2, so θ = π /4. But A(0) = 0 = A(π /2) and A(π /4) = 2, so the latter is the largest possible area of a rectangle inscribed in the unit circle. C03S07.079: The cross section of the trough is a trapezoid with short base 2, long base 2 + 4 cos θ, and height 2 sin θ. Thus its cross-sectional area is 2 + (2 + 4 cos θ) · 2 sin θ 2 = 4(sin θ + sin θ cos θ), 0 A(θ) = θ π /2 (the real upper bound on θ is 2π /3, but the maximum value of A clearly occurs in the interval [0, π /2]). A (θ) = 4(cos θ + cos2 θ − sin2 θ) = 4(2 cos2 θ + cos θ − 1) = 4(2 cos θ − 1)(cos θ + 1). The only solution of A (θ) = 0 in the given domain occurs when cos θ = verify that this value of θ maximizes the function A. 1 2, so that θ = 1 3 π. It is easy to C03S07.080: In the situation described in the problem, we have D = 20 sec θ. The illumination of the walkway is then I = I (θ) = k sin θ cos2 θ, 400 0 θ π /2. dI k cos θ = (cos2 θ − 2 sin2 θ); dθ 400 dI/dθ = 0 when θ = π /2 and when cos2 θ = 2 sin2 θ√The solution θ in the domain of I of the latter equation . √ has the property that sin θ = 3/3 and cos θ = 6/3. But I (0) = 0 and I (θ) → 0 as θ → (π /2)− , so √ the optimal height of the lamp post occurs when sin θ = 3/3. This implies that the optimal height is √ 10 2 ≈ 14.14 m. C03S07.081: The following figure shows a cross section of the sphere-with-cone through the axis of the cone and a diameter of the sphere. Note that h = r tan θ and that 10 cos θ = R . h−R Therefore h = R + R sec θ, and thus r= R + R sec θ . tan θ Now V = 1 π r2 h, so for θ in the interval (0, π /2), we have 3 V = V (θ) = 1 3 (1 + sec θ)3 . πR · 3 tan2 θ Therefore V (θ) = π R3 3(tan2 θ)(1 + sec θ)2 sec θ tan θ − (1 + sec θ)3 (2 tan θ sec2 θ) . 3 tan4 θ If V (θ) = 0 then either sec θ = −1 (so θ = π , which we reject), or sec θ = 0 (which has no solutions), or tan θ = 0 (so either θ = 0 or θ = π , which we also reject), or (after replacement of tan2 θ with sec2 θ − 1) sec2 θ − 2 sec θ − 3 = 0. It follows that sec θ = 3 or sec θ = −1. We reject the latter as before, and find that sec θ = 3, so θ ≈ 1.23095 (radians). The resulting minimum volume of the cone is 8 π R3 , twice the volume of the sphere! 3 θ h θ R r C03S07.082: Let L be the length of the crease. Then the right triangle of which L is the hypotenuse has sides L cos θ and L sin θ. Now 20 = L sin θ + L sin θ cos 2θ, so L = L(θ) = Next, 20 , (sin θ)(1 + cos 2θ) dL = 0 when dθ 11 0<θ π . 4 (cos θ)(1 + cos 2θ) = (sin θ)(2 sin 2θ); (cos θ)(2 cos2 θ) = 4 sin2 θ cos θ; so cos θ = 0 (which is impossible given the domain of L) or cos2 θ = 2 sin2 θ = 2 − 2 cos2 θ; cos2 θ = 2 . 3 √ √ This implies that cos θ = 1 6 and sin θ = 1 3. Because L → +∞ as θ → 0+ , we have a minimum either 3 3 √ at the horizontal tangent just found or at√ endpoint θ = 1 π . The value of L at 1 π is 20 2 ≈ 28.28 √ the and 4 4 1 at the horizontal tangent we have L = 15 3 ≈ 25.98. So the shortest crease is obtained when cos θ = 3 6; that is, for θ approximately 35◦ 15 52 . The bottom of the crease should be one-quarter of the way across the page from the lower left-hand corner. C03S07.083: Set up coordinates so the diameter is on the x-axis and the equation of the circle is x2 + y 2 = 1; let (x , y ) denote the northwest corner of the trapezoid. The chord from (1, 0) to (x, y ) forms a right triangle with hypotenuse 2, side z opposite angle θ, and side w; moreover, z = 2 sin θ and w = 2 cos θ. It follows that y = w sin θ = 2 sin θ cos θ and −x = 1 − w cos θ = − cos 2θ. Now A = y (1 − x) = (2 sin θ cos θ)(1 − cos 2θ) = 4 sin θ cos θ sin2 θ, and therefore A = A(θ) = 4 sin3 θ cos θ, π 4 θ π . 2 A (θ) = 12 sin2 θ cos2 θ − 4 sin4 θ = (4 sin2 θ)(3 cos2 θ − sin2 θ). To solve A (θ) = 0, we note that sin θ = 0, so we must have 3 cos2 θ = sin2 θ; that is tan2 θ = 3. It follows that θ = 1 π . The value of A here exceeds its value at the endpoints, so we have found the maximum value 3 √ of the area—it is 3 3. 4 C03S07.084: Let θ = α/2 (see Fig. 3.7.18 of the text) and denote the radius of the circular log by r. Using the technique of the solution of Problem 82, we find that the area of the hexagon is A = A(θ) = 8r2 sin3 θ cos θ, 0 θ π . 2 After some simplifications we also find that dA = 8r2 (sin2 θ)(4 cos2 θ − 1). dθ Now dA/dθ = 0 when sin θ = 0 and when cos θ = 1 . When sin θ = 0, A = 0; also, A(0) = 0 = A 1 π . 2 2 Therefore A is maximal when cos θ = 1 : θ = 1 π . When this happens, we find that α = 2 π and that 2 3 3 β = π − θ = 2 π . Therefore the figure of maximal area is a regular hexagon. 3 12 C03S07.085: The area in question is the area of the sector minus the area of the triangle in Fig. 3.7.19 and turns out to be A= = 12 θ θ r θ − r2 cos sin 2 2 2 12 s2 (θ − sin θ) r (θ − sin θ) = 2 2θ 2 because s = rθ. Now dA s2 (2 sin θ − θ cos θ − θ) , = dθ 2θ3 so dA/dθ = 0 when θ(1 + cos θ) = 2 sin θ. Let θ = 2x; note that 0 < x condition that dA/dθ = 0 becomes π because 0 < θ 2π . So the sin θ = tan x. 1 + cos θ x= But this equation has no solution in the interval (0, π ]. So the only possible maximum of A must occur at an endpoint of its domain, or where x is undefined because the denominator 1 + cos θ is zero—and this occurs when θ = π . Finally, A(2π ) = s2 4π and A(π ) = s2 , 2π so the maximum area is attained when the arc is a semicircle. C03S07.086: The length of the forest path is 2 csc θ. So the length of the part of the trip along the road is 3 − 2 csc θ cos θ. Thus the total time for the trip is given by 2 cos θ 3− 2 sin θ . T = T (θ) = + 3 sin θ 8 Note that the range of values of θ is determined by the condition √ 3 13 cos θ 0. 13 After simplifications, we find that T (θ) = 3 − 8 cos θ . 12 sin2 θ ) Now T (θ√ = 0 when cos θ = 3 ; that is, when θ is approximately 67◦ 58 32 . For this value of θ, we find that 8 sin θ = 1 55. There’s no problem in verifying that we have found the minimum. Answer: The distance to 8 walk down the road is √ cos θ 6 55 3−2 =3− ≈ 2.19096 (km). sin θ sin θ= 1 √55 55 8 C03S07.087: Following the Suggestion, we note that if n is a positive integer and h= 2 , (4n + 1)π 13 then (4n + 1)π sin 1 (4n + 1)π f (h) f (0) 2 = = 1, h (4n + 1)π and that if h= 2 , (4n − 1)π then (4n − 1)π sin 1 (4n − 1)π f (h) − f (0) 2 = = −1. h (4n − 1)π Therefore there are values of h arbitrarily close to zero for which f (0 + h) − f (0) = +1 h and values of h arbitrarily close to zero for which f (0 + h) − f (0) = − 1. h It follows that lim h→0 f (0 + h) − f (0) h does not exist; that is, f (0) does not exist, and so f is not differentiable at x = 0. 1 h2 sin f (0 + h) − f (0) h = lim h sin 1 . C03S07.088: f (0) = lim = lim h→0 h→0 h→0 h h h It now follows from the Squeeze Law of Section 2.3 (page 79) that f (0) = lim h sin h→0 because −| h | h sin 1 h | h | if h = 0. 14 1 =0 h Section 3.8 C03S08.001: If f (x) = e2x , then f (x) = e2x · Dx (2x) = 2e2x . C03S08.002: If f (x) = e3x−1 , then f (x) = e3x−1 · Dx (3x − 1) = 3e3x−1 . C03S08.003: If f (x) = exp(x2 ), then f (x) = exp(x2 ) · Dx (x2 ) = 2x exp(x2 ). 3 3 3 C03S08.004: If f (x) = e4−x , then f (x) = e4−x · Dx (4 − x3 ) = −3x2 e4−x . 2 2 C03S08.005: If f (x) = e1/x , then f (x) = e1/x · Dx (1/x2 ) = − 2 1/x2 e . x3 C03S08.006: If f (x) = x2 exp(x3 ), then f (x) = 2x exp(x3 ) + x2 · 3x2 exp(x3 ) = (2x + 3x4 ) exp(x3 ). C03S08.007: If g (t) = t exp(t 1 /2 ), then g (t) = exp(t 1/2 √ 1 −1/2 2+ t 1 /2 )+t· t exp(t ) = exp(t1/2 ). 2 2 C03S08.008: If g (t) = (e2t + e3t )7 , then g (t) = 7(e2t + e3t )6 (2e2t + 3e3t ). C03S08.009: If g (t) = (t2 − 1)e−t , then g (t) = 2te−t − (t2 − 1)e−t = (1 + 2t − t2 )e−t . C03S08.010: If g (t) = (et − e−t )1/2 , then g (t) = 1t (e − e−t )−1/2 (et + e−t ). 2 C03S08.011: If g (t) = ecos t = exp(cos t), then g (t) = (− sin t) exp(cos t). C03S08.012: If f (x) = xesin x = x exp(sin x), then f (x) = exp(sin x) + (x cos x) exp(sin x) = esin x (1 + x cos x). C03S08.013: If g (t) = 1 − e−t te−t − (1 − e−t ) te−t + e−t − 1 = . , then g (t) = 2 t t t2 C03S08.014: If f (x) = e−1/x , then f (x) = C03S08.015: If f (x) = 1 −1/x e . x2 1−x , then ex (−1)ex − (1 − x)ex −1 − 1 + x x−2 f (x) = = = . (ex )2 ex ex √ √ C03S08.016: If f (x) = exp ( x ) + exp (− x ), then √ √ √ √ 1 −1/2 1 −1/2 exp ( x ) − exp (− x ) √ f (x) = x exp x − x exp − x = . 2 2 2x C03S08.017: If f (x) = exp (ex ), then f (x) = ex exp (ex ). C03S08.018: If f (x) = e2x + e−2x f (x) = 1 /2 , then 1 2x e + e−2x 2 −1/2 e2x − e−2x 2e2x − 2e−2x = √ . e2x + e−2x 1 C03S08.019: If f (x) = sin (2ex ), then f (x) = 2ex cos (2ex ). C03S08.020: If f (x) = cos (ex + e−x ), then f (x) = (e−x − ex ) sin (ex + e−x ). C03S08.021: If f (x) = ln(3x − 1), then f (x) = 1 3 · Dx (3x − 1) = . 3x − 1 3x − 1 C03S08.022: If f (x) = ln(4 − x2 ), then f (x) = 2x . x2 − 4 C03S08.023: If f (x) = ln (1 + 2x)1/2 , then f (x) = C03S08.024: If f (x) = ln (1 + x)2 , then f (x) = C03S08.025: If f (x) = ln (x3 − x)1/3 = 1 2 · 2(1 + 2x)−1/2 1 = . 1 + 2x (1 + 2x)1/2 2(1 + x) 2 . = (1 + x)2 1+x 1 3x2 − 1 ln(x3 − x), then f (x) = . 3 3(x3 − x) C03S08.026: If f (x) = ln (sin x)2 = 2 ln(sin x), then f (x) = C03S08.027: If f (x) = cos(ln x), then f (x) = − C03S08.028: If f (x) = (ln x)3 , then f (x) = C03S08.029: If f (x) = 2 cos x = 2 cot x. sin x sin(ln x) . x 3(ln x)2 . x 1 1 . , then (by the reciprocal rule) f (x) = − ln x x(ln x)2 C03S08.030: If f (x) = ln(ln x), then f (x) = 1 . x ln x C03S08.031: If f (x) = ln x(x2 + 1)1/2 , then f (x) = (x2 + 1)1/2 + x2 (x2 + 1)−1/2 2x2 + 1 = . x(x2 + 1) x(x2 + 1)1/2 C03S08.032: If g (t) = t3/2 ln(t + 1), then g (t) = 3 1 /2 t3 / 2 t1/2 [2t + 3 ln(t + 1) + 3t ln(t + 1) ] t ln(t + 1) + = . 2 t+1 2(t + 1) C03S08.033: If f (x) = ln cos x, then f (x) = − sin x = − tan x. cos x C03S08.034: If f (x) = ln(2 sin x) = (ln 2) + ln(sin x), then f (x) = cos x = cot x. sin x C03S08.035: If f (t) = t2 ln(cos t), then f (t) = 2t ln(cos t) − t2 sin t = t [2 ln(cos t) − t tan t ]. cos t C03S08.036: If f (x) = sin(ln 2x), then f (x) = [cos(ln 2x) ] · 2 cos(ln 2x) = . 2x x 2 C03S08.037: If g (t) = t(ln t)2 , then g (t) = (ln t)2 + t · 2 ln t = (2 + ln t) ln t. t 2 C03S08.038: If g (t) = t1/2 [cos(ln t) ] , then g (t) = 1 −1/2 − sin(ln t) [cos(ln t) ] [cos(ln t) − 4 sin(ln t) ] 2 t [cos(ln t) ] + 2t1/2 [cos(ln t) ] · = . 2 t 2t 1 / 2 C03S08.039: Because f (x) = 3 ln(2x + 1) + 4 ln(x2 − 4), we have 6 8x 22x2 + 8x − 24 +2 = . 2x + 1 x − 4 (2x + 1)(x2 − 4) f (x) = C03S08.040: If 1−x 1+x f (x) = ln 1/2 = 1 1 ln(1 − x) − ln(1 + x), 2 2 then f (x) = − 1 1 1 − = . 2(1 − x) 2(1 + x) (x + 1)(x − 1) C03S08.041: If f (x) = ln 4 − x2 9 + x2 1/2 = 1 1 ln(4 − x2 ) − ln(9 + x2 ), 2 2 then f (x) = − x x 13x − =2 . 4 − x2 9 + x2 (x − 4)(x2 + 9) C03S08.042: If f (x) = ln √ 4x − 7 1 = ln(4x − 7) − 3 ln(3x − 2), (3x − 2)3 2 then f (x) = 2 9 59 − 30x − = . 4x − 7 3x − 2 (3x − 2)(4x − 7) C03S08.043: If f (x) = ln x+1 = ln(x + 1) − ln(x − 1), x−1 then f (x) = C03S08.044: If 3 1 1 2 − =− . x+1 x−1 (x − 1)(x + 1) f (x) = x2 ln 1 = −x2 ln(2x + 1), 2x + 1 then f (x) = − 2x2 − 2x ln(2x + 1). 2x + 1 C03S08.045: If g (t) = ln t2 = 2 ln t − ln(t2 + 1), t2 + 1 then g (t) = 2 2t 2 −2 = . 2 + 1) t t +1 t(t C03S08.046: If f (x) = ln √ x+1 1 = ln(x + 1) − 3 ln(x − 1), (x − 1)3 2 then f (x) = 1 3 5x + 7 − =− . 2(x + 1) x − 1 2(x − 1)(x + 1) C03S08.047: Given: y = 2x . Then ln y = ln (2x ) = x ln 2; 1 dy · = ln 2; y dx dy = y (x) · ln 2 = 2x ln 2. dx C03S08.048: Given: y = xx . Then ln y = ln (xx ) = x ln x; 1 dy · = 1 + ln x; y dx dy = y (x) · (1 + ln x) = xx (1 + ln x). dx C03S08.049: Given: y = xln x . Then 2 ln y = ln xln x = (ln x) · (ln x) = (ln x) ; 1 dy 2 ln x · = ; y dx x dy 2 ln x 2xln x ln x = y (x) · = . dx x x C03S08.050: Given: y = (1 + x)1/x . Then ln y = ln(1 + x)1/x = 1 ln(1 + x); x 1 dy x − ln(1 + x) − x ln(1 + x) 1 ln(1 + x) = · = − ; 2 y dx x(1 + x) x x2 (1 + x) dy x − ln(1 + x) − x ln(1 + x) [ x − ln(1 + x) − x ln(1 + x) ] · (1 + x)1/x = y (x) · = . 2 (1 + x) dx x x2 (1 + x) 4 √ C03S08.051: Given: y = (ln x) x . Then √ ln y = ln (ln x) x = x1/2 ln (ln x) ; 1 dy 1 x1/2 · = x−1/2 ln (ln x) + ; y dx 2 x ln x dy ln (ln x) 1 = y (x) · + 1 /2 ; dx 2x1/2 x ln x √ dy 2 + (ln x) ln (ln x) x = · (ln x) . 1/2 ln x dx 2x x C03S08.052: Given: y = (3 + 2x ) . Then x ln y = ln (3 + 2x ) = x ln (3 + 2x ) ; 1 dy x · 2x · ln 2 + ln (3 + 2x ) ; · = y dx 3 + 2x dy x · 2x · (ln 2) + 3 ln (3 + 2x ) + 2x ln (3 + 2x ) ; = y (x) · dx 3 + 2x dy x · 2x · (ln 2) + 3 ln (3 + 2x ) + 2x ln (3 + 2x ) x = · (3 + 2x ) . dx 3 + 2x C03S08.053: If y = (1 + x2 )3/2 (1 + x3 )−4/3 , then ln y = 3 4 ln(1 + x2 ) − ln(1 + x3 ); 2 3 1 dy 4x2 3x − 4x2 − x4 3x − = · = ; y dx 1 + x2 1 + x3 (1 + x2 )(1 + x3 ) dy 3x − 4x2 − x4 3x − 4x2 − x4 (1 + x2 )3/2 = y (x) · = · ; 2 )(1 + x3 ) 2 )(1 + x3 ) (1 + x3 )4/3 dx (1 + x (1 + x dy (3x − 4x2 − x4 )(1 + x2 )1/2 . = dx (1 + x3 )7/3 C03S84.054: If y = (x + 1)x , then ln y = ln(x + 1)x = x ln(x + 1); 1 dy x · = + ln(x + 1); y dx x+1 dy x = + ln(x + 1) · (x + 1)x . dx x+1 2 C03S08.055: If y = (x2 + 1)x , then 5 2 ln y = ln(x2 + 1)x = x2 ln(x2 + 1); 1 dy 2x3 · =2 + 2x ln(x2 + 1); y dx x +1 2 dy 2x3 2x3 = y (x) · + 2x ln(x2 + 1) = + 2x ln(x2 + 1) · (x2 + 1)x . 2+1 2+1 dx x x C03S08.056: If y = 1+ 1 x x , then ln y = ln 1 + 1 x x = x ln 1 + 1 x = x ln(x + 1) − x ln x; 1 dy x · = + ln(x + 1) − 1 − ln x; y dx x+1 dy x 1 = + ln(x + 1) − 1 − ln x · 1 + dx x+1 x x . √ √x C03S08.057: Given: y = ( x ) . Then ln y = ln √ x √ x = x1/2 ln x1/2 = 1 1 /2 x ln x; 2 1 dy ln x 2 + ln x 1 √; · = 1 / 2 + 1 /2 = y dx 4x 2x 4x √ √x dy (2 + ln x) ( x ) √ . = dx 4x C03S08.058: If y = xsin x , then ln y = (sin x) ln x; 1 dy sin x · = + (cos x) ln x; y dx x dy sin x + x (cos x) ln x = · xsin x . dx x C03S08.059: If f (x) = xe2x , then f (x) = e2x + 2xe2x , so the slope of the graph of y = f (x) at (1, e2 ) is f (1) = 3e2 . Hence an equation of the line tangent to the graph at that point is y − e2 = 3e2 (x − 1); that is, y = 3e2 x − 2e2 . C03S08.060: If f (x) = e2x cos x, then f (x) = 2e2x cos x − e2x sin x, so the slope of the graph of y = f (x) at the point (0, 1) is f (0) = 2. So an equation of the line tangent to the graph at that point is y − 1 = 2(x − 0); that is, y = 2x + 1. C03S08.061: If f (x) = x3 ln x, then f (x) = x2 +3x2 ln x, so the slope of the graph of y = f (x) at the point (1, 0) is f (1) = 1. Hence an equation of the line tangent to the graph at that point is y − 0 = 1 · (x − 1); that is, y = x − 1. 6 C03S08.062: If ln x , x2 f (x) = then f (x) = x − 2x ln x 1 − 2 ln x = . 4 x x3 Hence the slope of the graph of y = f (x) at the point e, e−2 is f (e) = −1/e3 . Therefore an equation of the line tangent to the graph at that point is y− 1 1 = − 3 (x − e); 2 e e that is, y= 2e − x . e3 C03S08.063: If f (x) = e2x , then f (x) = 2e2x , f (x) = 4e2x , f (x) = 8e2x , f (4) (x) = 16e2x , and f (5) (x) = 32e2x . It appears that f (n) (x) = 2n e2x . C03S08.064: If f (x) = xex , then f (x) = (x+1)ex , f (x) = (x+2)ex , f (x) = (x+3)ex , f (4) (x) = (x+4)ex , and f (5) (x) = (x+5)ex . It appears that f (n) (x) = (x + n)ex . C03S08.065: If f (x) = e−x/6 sin x, then 1 6 cos x − sin x f (x) = − e−x/6 sin x + e−x/6 cos x = . 6 6ex/6 Hence the first local maximum point for x > 0 occurs when x = arctan 6 and the first local minimum point occurs when x = π + arctan 6. The corresponding y -coordinates are, respectively, 6 e(arctan 6)/6 √ 37 and − 6 e(π+arctan 6)/6 √ 37 . C03S08.066: Given f (x) = e−x/6 sin x, let g (x) = e−x/6 and h(x) = −e−x/6 . We solve the equation f (x) = g (x) by hand; the x-coordinate of the first point of tangency is π /2. Similarly, the x-coordinate of the second point of tangency is 3π /2. These are not the same as arctan 6 and π + arctan 6. C03S08.067: The viewing window 1.11831 x 1.11834 shows the intersection of the two graphs near 1.11833 (see the figure that follows this solution). Thus, to three decimal places, the indicated solution of ex = x10 is 1.118. 1.11832 1.11832 1.11833 1.11833 1.11834 1.11834 3.0598 3.0596 3.0594 C03S08.068: The viewing rectangle with 35.771515 x 35.771525 reveals a solution of ex = x10 near x = 35.75152. Therefore this solution is approximately 3.58 × 101 . Newton’s method applied to f (x) = ex − x10 reveals the more accurate approximation 35.7715206396. 7 C03S08.069: We first let f (k ) = 1 1+ 10k 10k . Then Mathematics yields the following approximations: k f (k ) (rounded) 1 2.593742460100 2 2.704813829422 3 2.716923932236 4 2.718145926825 5 2.718268237174 6 2.718280469319 7 2.718281692545 8 2.718281814868 9 2.718281827100 10 2.718281828323 11 2.718281828445 12 2.718281828458 13 2.718281828459 14 2.718281828459 15 2.718281828459 16 2.718281828459045099 17 2.719291929459045222 18 2.718281828459045234 19 2.718281828459045235 20 2.718281828459045235 21 2.718281828459045235 C03S08.070: If y = uv where all are differentiable functions of x, then ln y = v ln u. With u (x) denoted simply by u , etc., we now have 1 vu y = v ln u + . y u uv vu = vuv−1 u + uv (ln u)v . u dy (a) If u is constant, this implies that = uv(x) (ln u)v (x). dx Thus y = uv v ln u + 8 (b) If v is constant, this implies that dy v −1 = v (u(x)) u (x). dx C03S08.071: Solution: ln y = ln u + ln v + ln w − ln p − ln q − ln r; 1 dy 1 du 1 dv 1 dw 1 dp 1 dq 1 dr · =· +· +· −· −· −· ; y dx u dx v dx w dx p dx q dx r dx dy =y· dx 1 du 1 dv 1 dw 1 dp 1 dq 1 dr · +· +· −· −· −· u dx v dx w dx p dx q dx r dx . The solution makes the generalization obvious. C03S08.072: Suppose by way of contradiction that log2 3 is a rational number. Then log2 3 = p/q where p and q are positive integers (both positive because log2 3 > 0). Thus 2p/q = 3, so that 2p = 3q . But if p and q are positive integers, then 2p is even and 3q is odd, so they cannot be equal. Therefore the assumption that log2 3 is rational leads to a contradiction, and thus log2 3 is irrational. C03S08.073: (a): If f (x) = log10 x, then the definition of the derivative yields f (1) = lim h→0 f (1 + h) − f (1) 1 = lim log10 (1 + h) = lim log10 (1 + h)1/h . h→0 h h→0 h (b): When h = 0.1 the value of log10 (1 + h)1/h is approximately 0.4139. With h = 0.01 we get 0.4321, with h = 0.001 we get 0.4341, and with h = ±0.0001 we get 0.4343. C03S08.074: Because exp(ln x) = x, we see first that 10x = exp(ln 10x ) = exp(x ln 10) = ex ln 10 . Hence Dx 10x = Dx ex ln 10 = ex ln 10 ln 10 = 10x ln 10. Thus, by the chain rule, if u is a differentiable function of x, then Dx 10u = (10u ln 10) du . dx Finally, if u(x) = log10 x, so that 10u ≡ x, then differentiation of this last identity yields (10u ln 10) du ≡ 1, dx so that du 1 0.4343 = Dx log10 x = ≈ . dx x ln 10 x 9 Section 3.9 C03S09.001: 2x − 2y √ dy dy x dy x x x = 0, so = . Also, y = ± x2 − 1 , so = ±√ =√ =. dx dx y dx y x2 − 1 ± x2 − 1 dy dy y dy x−1 + y = 0, so = − . By substituting y = x−1 here, we get =− = −x−2 , dx dx x dx x which is the result obtained by explicit differentiation. C03S09.002: x C03S09.003: 32x + 50y get dy = dx √ dy dy 16x = 0; =− . Substituting y = ± 1 400 − 16x2 into the derivative, we 5 dx dx 25y 16x , which is the result obtained by explicit differentiation. 5 400 − 16x2 √ √ dy dy x2 dy x2 = 0, so = − 2 . y = 3 1 − x3 , so substitution results in =− . 2 /3 dx dx y dx (1 − x3 ) Explicit differentiation yields the same answer. C03S09.004: 3x2 + 3y 2 C03S09.005: 1 −1/2 2x + 1 y −1/2 2 C03S09.006: 4x3 +2x2 y C03S09.007: 2 −1/3 3x + 2 y −1/3 3 y . x dy =− dx dy dy +2xy 2 +4y 3 = 0: dx dx C03S09.008: y 2 + 2(x − 1)y C03S09.009: Given: dy = 0: dx dy = 0: dx dy =− dx (2x2 y +4y 3 ) x y −1/3 dy = −(4x3 +2xy 2 ); dx =− y x 1 /3 . dy dy 1 − y2 = 1, so = . dx dx 2y (x − 1) x3 − x2 y = xy 2 + y 3 : 3x2 − x2 dy dy dy − 2xy = y 2 + 2xy + 3y 2 ; dx dx dx 3x2 − 2xy − y 2 = (2xy + 3y 2 + x2 ) dy ; dx dy 3x2 − 2xy − y 2 =2 . dx 3y + 2xy + x2 C03S09.010: Given: x5 + y 5 = 5x2 y 2 : 5x4 + 5y 4 dy dy = 10x2 y + 10xy 2 ; dx dx dy 10xy 2 − 5x4 =4 . dx 5y − 10x2 y C03S09.011: Given: x sin y + y sin x = 1: x cos y dy dy + sin y + y cos x + sin x = 0; dx dx dy sin y + y cos x =− . dx x cos y + sin x 1 dy 4x3 + 2xy 2 =− 2 . dx 2x y + 4y 3 C03S09.012: Given: cos(x + y ) = sin x sin y : − sin(x + y )(1 + dy dy ) = sin x cos y + sin y cos x; dx dx dy sin y cos x + sin(x + y ) =− . dx sin(x + y ) + sin x cos y C03S09.013: Given: 2x + 3ey = ex+y . Differentiation of both sides of this equation (actually, an identity ) with respect to x yields 2 + 3ey dy dy = ex+y 1 + dx dx , and so dy ex+y − 2 3ey + 2x − 2 =y = . dx 3e − ex+y (ex − 3)ey C03S09.014: Given: xy = e−xy . Differentiation of both sides with respect to x yields x dy dy + y = −exy x +y , dx dx and so (1 + e−xy )x dy = −(1 + e−xy )y. dx Because 1 + e−xy > 0 for all x and y , it follows that dy y =− . dx x Another way to solve this problem is to observe that the equation e−z = z has exactly one real solution a ≈ 0.5671432904. Hence if e−xy = xy , then xy = a, so that y = a/x. Hence dy xy y a =− 2 =− 2 =− . dx x x x C03S09.015: y+4= 3 4 (x − 3). C03S09.016: y+2= 1 2 (x 2x + 2y x − 4). dy = 0: dx dy + y = 0: dx dy x =− . dx y dy y =− . dx x At (3, −4) the tangent has slope 3 4 and thus equation At (4, −2) the tangent has slope 1 2 and thus equation dy dy 1 − 2xy C03S09.017: x2 . At (2, 1) the tangent has slope − 3 and thus equation + 2xy = 1, so = 4 dx dx x2 3x + 4y = 10. dy C03S09.018: 1 x−3/4 + 1 y −3/4 = 0: 4 4 dx equation x + y = 32. dy = − (y/x)3/4 . At (16, 16) the tangent has slope −1 and thus dx dy dy + 2xy + x2 = 0: dx dx equation of the tangent there is y = −2. C03S09.019: y 2 + 2xy dy 2xy + y 2 =− . dx 2xy + x2 At (1, −2) the slope is zero, so an 1 1 dy dy (y + 1)2 = 0, so =− − · . At (1, 1) the tangent line has slope 2 2 dx (x + 1) (y + 1) dx (x + 1)2 −1 and thus equation y − 1 = −(x − 1). C03S09.020: − 2 dy 25y − 24x = . At (3, 4) the tangent line has slope dx 24y − 25x dy dy = 25y + 25x : dx dx and thus equation 4x = 3y . C03S09.021: 24x + 24y dy dy + 2y = 0: dx dx thus equation y + 2 = 4(x − 3). C03S09.022: 2x + y + x dy 2x + y =− . At (3, −2) the tangent line has slope 4 and dx x + 2y C03S09.023: dy 3e2x + 2ey = , so the tangent line at (0, 0) has slope dx 3e2x + ex+2y 5 4 C03S09.024: dy 12e2x − ye3y = , so the tangent line at (3, 2) has slope dx 18e2x + xe3y 10 21 C03S09.025: equation y − 1 dy 2 −1/3 + 2 y −1/3 = 0: 3x 3 dx 1 = − 2 (x − 8); that is, x + and equation 4y = 5x. and equation 10x + 12 = 21y . dy y 1 /3 = − 1/3 . At (8, 1) the tangent line has slope − 1 and thus 2 dx x 2y = 10. dy dy dy y − 2x − y + 2y = 0: = . At (3, −2) the tangent line has slope dx dx dx 2y − x 8 equation y + 2 = 7 (x − 3); that is, 7y = 8x − 38. C03S09.026: 2x − x C03S09.027: 2 x2 + y 2 2x + 2y dy dx 4 3 = 50x 8 7 and thus dy + 50y : dx dy 2x3 − 25y + 2xy 2 . =− dx −25x + 2x2 y + 2y 3 At (2, 4) the tangent line has slope 2 11 and thus equation y − 4 = 2 11 (x − 2); that is, 11y = 2x + 40. dy 3x2 + 14x dy = 3x2 + 14x: = . At (−3, 6) the tangent line has slope − 5 and thus 4 dx dx 2y 5 equation y − 6 = − 4 (x + 3); alternatively, 4y = 9 − 5x. C03S09.028: 2y C03S09.029: 3x2 + 3y 2 dy dy = 9x + 9y : dx dx (a): At (2, 4) the tangent line has slope dy 3y − x2 =2 . dx y − 3x 4 5 and thus equation y − 4 = 4 (x − 2); that is, 5y = 4x + 12. 5 dy = −1, 3y − x2 = −y 2 − 3x and x3 + y 3 = 9xy . This pair of dx simultaneous equations has solutions x = 0, y = 0 and x = 9 , y = 9 , but the derivative does not exist at 2 2 the point (0, 0). Therefore the tangent line with slope −1 has equation y − 9 = − x − 9 . 2 2 (b): At a point on the curve at which C03S09.030: First, 2x2 − 5xy + 2y 2 = (y − 2x)(2y − x). (a): Hence if 2x2 − 5xy + 2y 2 = 0, then y − 2x = 0 or 2y − x = 0. This is a pair of lines through the origin; the first has slope 2 and the second has slope 1 . 2 (b): Differentiating implicitly, we obtain 4x − 5x is 2 if y = 2x and − 1 if y = − 1 x. 2 2 dy dy dy 5y − 4x − 5y + 4y = 0, which gives = , which dx dx dx 4y − 5x dy 2−x = , so horizontal tangents can occur only if x = 2 and y = 2. When x = 2, dx y−2 √ the original equation yields y 2 − 4√ − 4 = 0, so that y = 2 ± 8. Thus there are two points at which the y √ tangent line is horizontal: 2, 2 − 8 and 2, 2 + 8 . C03S09.031: Here 3 dy y − x2 dx y2 − x =2 and = . Horizontal tangents require y = x2 , and the equation dx y −x dy y − x2 √ x3 + y 3 = 3xy of the folium yields x3 (x3 − 2) = 0, so either x = 0 or x = 3 2. But dy/dx is not defined at √√ (0, 0), so only at 3 2,√3 4 √is there a horizontal tangent. By symmetry or by a similar argument, there is a vertical tangent at 3 4, 3 2 and nowhere else. C03S09.032: First, C03S09.033: By direct differentiation, dx/dy = (1 + y )ey . By implicit differentiation, and the results are equivalent. dy 1 , = dx (1 + y )ey (a): At (0, 0), dy/dx = 1, so an equation of the line tangent to the curve at (0, 0) is y = x. (b): At (e, 1), dy/dx = 1/(2e), so an equation of the line tangent to the curve at (e, 1) is x + e = 2ey . C03S09.034: (a): By direct differentiation, dx/dy = (1 + y )ey , so there is only one point on the curve where the tangent line is vertical (dx/dy = 0): (−1/e, −1). dy 1 (b): Because is never zero, the graph has no horizontal tangents. = dx (1 + y )ey C03S09.035: From 2(x2 + y 2 ) 2x + 2y dy dx = 2x − 2y dy it follows that dx dy x[1 − 2(x2 + y 2 )] = . dx y [1 + 2(x2 + y 2 )] So dy/dx = 0 when x2 + y 2 = 1 , but is undefined when x = 0, for then y = 0 as well. If x2 + y 2 = 1 , then 2 2 x2 − y 2 = 1 , so that x2 = 3 , and it follows that y 2 = 1 . Consequently there are horizontal tangents at all 4 8 8 √ √ four points where |x| = 1 6 and |y | = 1 2. 4 4 Also dx/dy = 0 only when y = 0, and if so, then x4 = x2 , so that x = ±1 (dx/dy is undefined when x = 0). So there are vertical tangents at the two points (−1, 0) and (1, 0). C03S09.036: Base edge of block: x. Height: y . Volume: V = x2 y . We are given dx/dt = −2 and dy/dt = −3. Implicit differentiation yields dV dy dx = x2 + 2xy . dt dt dt When x = 20 and y = 15, dV /dt = (400)(−3) + (600)(−2) = −2400. So the rate of flow at the time given is 2400 in.3 /h. C03S08.037: Suppose that the pile has height h = h(t) at time t (seconds) and radius r = r(t) then. We are given h = 2r and we know that the volume of the pile at time t is V = V (t) = π2 2 r h = πr3 . 3 3 When h = 5, r = 2.5; at that time dV dV dr = ·, dt dr dt Now so 10 = 2π r2 dr . dt dr 10 4 = = ≈ 0.25645 (ft/s). 2 dt 2π (2.5) 5π C03S09.038: Draw a vertical cross section through the center of the tank. Let r denote the radius of the (circular) water surface when the depth of water in the tank is y . From the drawing and the Pythagorean theorem derive the relationship r2 + (10 − y )2 = 100. Therefore 2r dr dy − 2(10 − y ) = 0, dt dt and so r 4 dr dy = (10 − y ) . dt dt √ We are to find dr/dt when y = 5, given dy/dt = −3. At that time, r2 = 100 − 25, so r = 5 3. Thus √ 5 = √ (−3) = − 3. 53 y =5 y =5 √ Answer: The radius of the top surface is decreasing at 3 ft/s then. dr dt = 10 − y dy · r dt C03S09.039: We assume that the oil slick forms a solid right circular cylinder of height (thickness) h and radius r. Then its volume is V = π r2 h, and we are given V = 1 (constant) and dh/dt = −0.001. Therefore dh 1 dr dr r dr r 1 0 = πr2 = + 2π rh . Consequently 2h = , and so = . When r = 8, h = . At 2 dt dt dt 1000 dt 2000h πr 64π dr 8 · 64π 32π that time, = = ≈ 0.80425 (m/h). dt 2000 125 C03S09.040: Let x be the distance from the ostrich to the street light and u the distance from the base of the light pole to the tip of the ostrich’s shadow. Draw a figure and so label it; by similar triangles you find u u−x that = , and it follows that u = 2x. We are to find du/dt and Dt (u − x) = du/dt − dx/dt. But 10 5 u = 2x, so du dx =2 = (2)(−4) = −8; dt dt Answers: (a): +8 ft/s; du dx − = −8 − (−4) = −4. dt dt (b): +4 ft/s. C03S09.041: Let x denote the width of the rectangle; then its length is 2x and its area is A = 2x2 . Thus dA dx = 4x . When x = 10 and dx/dt = 0.5, we have dt dt dA = (4)(10)(0.5) = 20 (cm2 /s). dt x=10 C03S09.042: Let x denote the length of each edge of the triangle. Then the triangle’s area is A(x) = √ √ dA dx dx 1 1 2 4 3 x , and therefore dt = 2 3 x dt . Given x = 10 and dt = 0.5, we find that √ √ dA 3 53 = · 10 · (0.5) = (cm2 /s). dt x=10 2 2 C03S089.043: Let r denote the radius of the balloon and V its volume at time t (in seconds). Then V= 43 πr , 3 dV dr = 4π r2 . dt dt so We are to find dr/dt when r = 10, and we are given the information that dV /dt = 100π . Therefore 100π = 4π (10)2 dr dt , r =10 and so at the time in question the radius is increasing at the rate of dr/dt = C03S09.044: Because pV = 1000, V = 10 when p = 100. Moreover, p V = 10, and dp/dt = 2, we find that dV dt p=100 =− V dp · p dt p=100 5 =− 1 4 = 0.25 (cm/s). dV dp +V = 0. With p = 100, dt dt 10 1 ·2=− . 100 5 Therefore the volume is decreasing at 0.2 in.3 /s. C03S09.045: Place the person at the origin and the kite in the first quadrant at (x, 400) at time t, where x = x(t) and we are given dx/dt = 10. Then the length L = L(t) of the string satisfies the equation dx dL L2 = x2 + 160000, and therefore 2L = 2x . Moreover, when L = 500, x = 300. So dt dt dL 1000 = 600 · 10, dt L=500 which implies that the string is being let out at 6 ft/s. C03S09.046: Locate the observer at the origin and the balloon in the first quadrant at (300, y ), where y = y (t) is the balloon’s altitude at time t. Let θ be the angle of elevation of the balloon (in radians) from the observer’s point of view. Then tan θ = y/300. We are given dθ/dt = π /180 rad/s. Hence we are to find dy/dt when θ = π /4. But y = 300 tan θ, so dy dθ = (300 sec2 θ) . dt dt Substitution of the given values of θ and dθ/dt yields the answer dy dt θ =45◦ = 300 · 2 · π 10π = ≈ 10.472 (ft/s). 180 3 C03S09.047: Locate the observer at the origin and the airplane at (x, 3), with x > 0. We are given dx/dt where the units are in miles, hours, and miles per hour. The distance z between the observer and the airplane satisfies the identity z 2 = x2 + 9, and because the airplane is traveling at 8 mi/min, we find that x = 4, and therefore that z = 5, at the time 30 seconds after the airplane has passed over the observer. Also dz dx dz 2z = 2x , so at the time in question, 10 = 8 · 480. Therefore the distance between the airplane and dt dt dt the observer is increasing at 384 mi/h at the time in question. dV dy C03S09.048: In this problem we have V = 1 π y 2 (15 − y ) and (−100)(0.1337) = = π (10y − y 2 ) . 3 dt dt dy 13 · 37 Therefore =− . Answers: (a): Approximately 0.2027 ft/min; (b): The same. dt π y (10 − y ) C03S09.049: We use a = 10 in the formula given in Problem 42. Then V= 12 π y (30 − y ). 3 dV dy dy 13 · 37 = π (20y − y 2 ) . Thus =− . Substitution of y = 7 and y = 3 dt dt dt π y (20 − y ) now yields the two answers: Hence (−100)(0.1337) = (a): − 191 ≈ −0.047 (ft/min); 1300π (b): − 1337 ≈ −0.083 (ft/min). 5100π C03S09.050: When the height of the water at the deep end of the pool is 10 ft, the length of the water surface is 50 ft. So by similar triangles, if the height of the water at the deep end is y feet (y 10), then the length of the water surface is x = 5y feet. A cross section of the water perpendicular to the width of the pool thus forms a right triangle of area 5y 2 /2. Hence the volume of the pool is V (y ) = 50y 2 . Now dV dy 133.7 = = 100y , so when y = 6 we have dt dt 6 dy dt = y =6 133.7 ≈ 0.2228 (ft/min). 600 C03S09.051: Let the positive y -axis represent the wall and the positive x-axis the ground, with the top of the ladder at (0, y ) and its lower end at (x, 0) at time t. Given: dx/dt = 4, with units in feet, seconds, and dy dx feet per second. Also x2 + y 2 = 412 , and it follows that y = −x . Finally, when y = 9, we have x = 40, dt dt dy so at that time 9 = −40 · 4. Therefore the top of the ladder is moving downward at 160 ≈ 17.78 ft/s. 9 dt C03S09.052: Let x be the length of the base of the rectangle and y its height. We are given dx/dt = +4 and dy/dt = −3, with units in centimeters and seconds. The area of the rectangle is A = xy , so dA dy dx =x +y = −3x + 4y. dt dt dt Therefore when x = 20 and y = 12, we have dA/dt = −12, so the area of the rectangle is decreasing at the rate of 12 cm2 /s then. C03S09.053: Let r be the radius of the cone, h its height. We are given dh/dt = −3 and dr/dt = +2, with units in centimeters and seconds. The volume of the cone at time t is V = 1 π r2 h, so 3 dV 2 dr 1 dh = π rh + π r2 . dt 3 dt 3 dt dV 2 1 = · 24π · 2 + · 16π · (−3) = 16π , so the volume of the cone is increasing at dt 3 3 the rate of 16π cm3 /s then. When r = 4 and h = 6, C03S09.054: Let x be the edge length of the square and A = x2 its area. Given: But dA/dt = 2x(dx/dt), so dx/dt = 6 when x = 10. Answer: At 6 in./s. dA = 120 when x = 10. dt C03S09.055: Locate the radar station at the origin and the rocket at (4, y ) in the first quadrant at time t, with y in miles and t in hours. The distance z between the station and the rocket satisfies the equation dy dz y 2 + 16 = z 2 , so 2y = 2z . When z = 5, we have y = 3, and because dz/dt = 3600 it follows that dt dt dy/dt = 6000 mi/h. C03S09.056: Locate the car at (x, 0), the truck at (0, y ) (x, y > 0). Then at 1 p.m. we have x = 90 and y = 80. We are given that data dx/dt = 30 and dy/dt = 40, with units in miles, hours, and miles per hour. The distance z between the vehicles satisfies the equation z 2 = x2 + y 2 , so dz dx dy =x +y . dt dt dt √ Finally, at 1 p.m. z 2 = 8100 + 6400 = 14500, so z = 10 145 then. So at 1 p.m. z dz 2700 + 3200 590 √ = =√ dt 10 145 145 mi/h—approximately 49 mi/h. C03S09.057: Put the floor on the nonnegative x-axis and the wall on the nonnegative y -axis. Let x denote the distance from the wall to the foot of the ladder (measured along the floor) and let y be the distance from 7 the floor to the top of the ladder (measured along the wall). By the Pythagorean theorem, x2 + y 2 = 100, and we are given dx/dt = 22 (because we will use units of feet and seconds rather than miles and hours). 15 From the Pythagorean relation we find that 2x dx dy + 2y = 0, dt dt dy x dx 22x =− · =− . dt y dt 15y √ √ (a): If y = 4, then x = 84 = 2 21. Hence when the top of the ladder is 4 feet above the ground, it is moving a a rate of so that dy dt y =4 √ √ 44 21 11 21 =− =− ≈ −3.36 60 15 feet per second, about 2.29 miles per hour downward. (b): If y = 1 12 (one inch), then x2 = 100 − 1 14399 = , 144 144 so that x ≈ 9.99965. In this case, dy dt y =1/12 =− 22 · (9.99965) 88 = − · (9.99965) ≈ −176 1 5 15 · 12 feet per second, about 120 miles per hour downward. (c): If y = 1 mm, then x ≈ 10 (ft), and so dy 22 ≈ − · (3048) ≈ 4470 dt 15 feet per second, about 3048 miles per hour. The results in parts (b) and (c) are not plausible. This shows that the assumption that the top of the ladder never leaves the wall is invalid C03S09.058: Let x be the distance between the Pinta and the island at time t and y the distance between the Ni˜a and the island then. We know that x2 + y 2 = z 2 where z = z (t) is the distance between the two n ships, so 2z dz dx dy = 2x + 2y . dt dt dt (1) When x = 30 and y = 40, z = 50. It follows from Eq. (1) that dz/dt = −25 then. Answer: drawing closer at 25 mi/h at the time in question. They are C03S09.059: Locate the military jet at (x, 0) with x < 0 and the other aircraft at (0, y ) with y 0. With units in miles, minutes, and miles per minute, we are given dx/dt = +12, dy/dt = +8, and when t = 0, x = −208 and y = 0. The distance z between the aircraft satisfies the equation x2 + y 2 = z 2 , so dz = dt 1 x2 + y2 x dx dy +y dt dt 8 = 12x + 8y x2 + y 2 . The closest approach will occur when dz/dt = 0: y = −3x/2. Now x(t) = 12t − 208 and y (t) = 8t. So at closest approach we have 3 3 8t = y (t) = − x(t) = − (12t − 208). 2 2 Hence √ closest approach, 16t = 624 − 36t, and thus t = 12. At this time, x = −64, y = 96, and at z = 32 13 ≈ 115.38 (mi). C03S09.060: Let x be the distance from the anchor to the point on the seabed directly beneath the hawsehole; let L be the amount of anchor chain out. We must find dx/dt when L = 13 (fathoms), given dL dx dx L dL dL/dt = −10. Now x2 + 144 = L2 , so 2L = 2x . Consequently, =· . At the time in question dt dt dt x dt 2 2 2 in the problem, x = 13 − 12 , so x = 5. It follows that dx/dt = −26 then. Thus the ship is moving at 26 fathoms per minute—about 1.77 mi/h. C03S09.061: Let x be the radius of the water surface at time t and y the height of the water remaining at time t. If Q is the amount of water remaining in the tank at time t, then (because the water forms a cone) x 3 3y Q = Q(t) = 1 π x2 y . But by similar triangles, = , so x = . So 3 y 5 5 19 3 Q(t) = π y 3 = πy3 . 3 25 25 dQ dy 9 We are given dQ/dt = −2 when y = 3. This implies that when y = 3, −2 = = π y 2 . So at the time dt 25 dt in question, dy dt y =3 =− 50 ≈ −0.1965 (ft/s). 81π C03S09.062: Given V = 1 π (30y 2 − y 3 ), find dy/dt given V , y , and dy/dt. First, 3 dV 1 dy dy = π (60y − 3y 2 ) = π (20y − y 2 ) . dt 3 dt dt So dy 1 dV = · . Therefore, when y = 5, we have 2) dt π (20y − y dt dy (200)(0.1337) ≈ 0.113488 (ft/min). = dt y=5 π (100 − 25) C03S09.063: Let r be the radius of the water surface at time t, h the depth of water in the bucket then. By similar triangles we find that r−6 1 h = , so r = 6 + . h 4 4 The volume of water in the bucket then is 1 π h(36 + 6r + r2 ) 3 1 3 1 = π 36 + 36 + h + 36 + 3h + h2 3 2 16 V= = 1 9 1 π h 108 + h + h2 . 3 2 16 9 Now dV = −10; we are to find dh/dt when h = 12. dt dV dh 1 3 = π (108 + 9h + h2 ) . dt 3 16 dt Therefore dh dt = h=12 3 · π −10 108 + 9 · 12 + 3 · 12 16 2 =− 10 ≈ −0.0393 (in./min). 81π C03S09.064: Let x denote the distance between the ship and A, y the distance between the ship and B , h the perpendicular distance from the position of the ship to the line AB , u the distance from A to the foot of this perpendicular, and v the distance from B to the foot of the perpendicular. At the time in question, we know that x = 10.4, dx/dt = 19.2, y = 5, and dy/dt = −0.6. From the right triangles involved, we see that u2 + h2 = x2 and (12.6 − u)2 + h2 = y 2 . Therefore x2 − u2 = y 2 − (12.6 − u)2 . (1) We take x = 10.4 and y = 5 in Eq. (1); it follows that u = 9.6 and that v = 12.6 − u = 3. From Eq. (1), we know that x dx du dy du −u =y + (12.6 − u) , dt dt dt dt so du 1 = dt 12.6 x dx dy −y dt dt From the data given, du/dt ≈ 16.0857. Also, because h = dh dx du Moreover, h =x − u , and therefore dt dt dt dh dt h=4 ≈ √ . x2 − u2 , h = 4 when x = 10.4 and y = 9.6. 1 [(10.4)(19.2) − (9.6)(16.0857)] ≈ 11.3143. 4 dh/dt ≈ 0.7034, so the ship is sailing a course about 35◦ 7 north or south of east at a speed of du/dt (du/dt)2 + (dh/dt)2 ≈ 19.67 mi/h. It is located 9.6 miles east and 4 miles north or south of A, or 10.4 miles from A at a bearing of either 67◦ 22 48 or 112◦ 37 12 . Finally, C03S09.065: Set up a coordinate system in which the radar station is at the origin, the plane passes over it at the point (0, 1) (so units on the axes are in miles), and the plane is moving along the graph of the equation y = x + 1. Let s be the distance from (0, 1) to the plane and let u be the distance from the radar station to the plane. We are given du/dt = +7 mi/min. We may deduce from the law of cosines that √ u2 = s2 + 1 + s 2. Let v denote the speed of the plane, so that v = ds/dt. Then √ √ du = 2sv + v 2 = v 2s + 2 , dt 2u du √· . 2s + 2 dt √ √ When u = 5, s2 + s 2 − 24 = 0. The quadratic formula yields the solution s = 3 2, and it follows that √ v = 5 2 mi/min; alternatively, v ≈ 424.26 mi/h. 2u and so v = C03S09.066: V (y ) = 1 π (30y 2 − y 3 ) where the depth is y . Now 3 10 dV dV dy √ = −k y = · , and therefore dt dy dt √ √ ky ky dy =− =− . dV dt π (20y − y 2 ) dy To minimize dy/dt, write F (y ) = dy/dt. It turns out (after simplifications) that F (y ) = k 20y − 3y 2 · √. 2π (20y − y 2 )2 y So F (y ) = 0 when y = 0 and when y = 20 . When y is near 20, F (y ) is very large; the same is true for y 3 near zero. So y = 20 minimizes dy/dt, and therefore the answer to part (b) is 6 ft 8 in. 3 C03S09.067: Place the pole at the origin in the plane, and let the horizontal strip 0 y 30 represent the road. Suppose that the person is located at (x, 30) with x > 0 and is walking to the right, so dx/dt = +5. √ Then the distance from the pole to the person will be x2 + 900 . Let z be the length of the person’s shadow. √ dz dx By similar triangles it follows that 2z = x2 + 900 , so 4z 2 = x2 + 900, and thus 8z = 2x . When dt dt x = 40, we find that z = 25, and therefore that 100 dz dt z =25 = 40 · 5 = 200. Therefore the person’s shadow is lengthening at 2 ft/s at the time in question. C03S09.068: Set up a coordinate system in which the officer is at the origin and the van is moving in the positive direction along the line y = 200 (so units on the coordinate axes are in feet). When the van is at dx dz position (x, 200), the distance from the officer to the van is z , where x2 + 2002 = z 2 , so that x =z . dt dt √ When the van reaches the call box, x = 200, z = 200 2, and dz/dt = 66. It follows that dx dt √ = 66 2, x=200 which translates to about 63.6 mi/h. 11 Section 3.10 Note: Your answers may differ from ours in the last one or two decimal places because of differences in hardware or in the way the problem was solved. We used Mathematica 3.0 and carried 40 decimal digits throughout all calculations, and our answers are correct or correctly rounded to the number of digits shown here. In most of the first 20 problems the initial guess x0 was obtained by linear interpolation. Finally, the equals mark is used in this section to mean “equal or approximately equal.” C03S10.001: With f (x) = x2 − 5, a = 2, b = 3, and x0 = a − (b − a)f (a) = 2.2, f (b) − f (a) we used the iterative formula xn+1 = xn − f (xn ) f (xn ) for n 0. Thus we obtained x1 = 2.236363636, x2 = 2.236067997, and x3 = x4 = 2.236067977. Answer: 2.2361. C03S10.002: x0 = 1.142857143; we use f (x) = x3 − 2. Then x1 = 1.272321429, x2 = 1.260041515, x3 = 1.259921061, and x4 = x5 = 1.259921050. Answer: 1.2599. C03S10.003: x0 = 2.322274882; we use f (x) = x5 − 100. Then x1 = 2.545482651, x2 = 2.512761634, x3 = 2.511887041, and x4 = x5 = 2.511886432. Answer: 2.5119. C03S10.004: Let f (x) = x3/2 − 10. Then x0 = 4.628863603. From the iterative formula x ←− x − x3/2 − 10 3 1 /2 2x we obtain x1 = 4.641597575, x2 = 4.641588834 = x3 . Answer: 4.6416. C03S10.005: 0.25, 0.3035714286, 0.3027758133, 0.3027756377. Answer: 0.3028. C03S10.006: 0.2, 0.2466019417, 0.2462661921, 0.2462661722. Answer: 0.2463. C03S10.007: x0 = −0.5, x1 = −0.8108695652, x2 = −0.7449619516, x3 = −0.7402438226, x4 = −0.7402217826 = x5 . Answer: 0.7402. C03S10.008: Let f (x) = x3 + 2x2 + 2x − 10. With initial guess x0 = 1.5 (the midpoint of the interval), we obtain x1 = 1.323943661972, x2 = 1.309010783652, x3 = 1.308907324710, x4 = 1.308907319765, and x5 = 1.308907319765. Answer: 1.3089. C03S10.009: With f (x) = x − cos x, f (x) = 1 + sin x, and calculator set in radian mode, we obtain x0 = 0.5854549279, x1 = 0.7451929664, x2 = 0.7390933178, x3 = 0.7390851332, and x4 = x3 . Answer: 0.7391. C03S10.010: Let f (x) = x2 − sin x. Then f (x) = 2x − cos x. The linear interpolation formula yields x0 = 0.7956861008, and the iterative formula x ←− x − x2 − sin x 2x − cos x 1 (with calculator in radian mode) yields the following results: x1 = 0.8867915207, x2 = 0.8768492470, x3 = 0.8767262342, and x4 = 0.8767262154 = x5 . Answer: 0.8767. C03S10.011: With f (x) = 4x − sin x − 4 and calculator in radian mode, we get the following results: x0 = 1.213996400, x1 = 1.236193029, x2 = 1.236129989 = x3 . Answer: 1.2361. C03S10.012: x0 = 0.8809986055, x1 = 0.8712142932, x3 = 0.8712215145, and x4 = x3 . Answer: 0.8712. C03S10.013: With x0 = 2.188405797 and the iterative formula x ←− x − x4 (x + 1) − 100 , x3 (5x + 4) we obtain x1 = 2.360000254, x2 = 2.339638357, x3 = 2.339301099, and x4 = 2.339301008 = x5 . Answer: 2.3393. C03S10.014: x0 = 0.7142857143, x1 = 0.8890827860, x2 = 0.8607185590, x3 = 0.8596255544, and x4 = 0.8596240119 = x5 . Answer: 0.8596. C03S10.015: The nearest discontinuities of f (x) = x − tan x are at π /2 and at 3π /2, approximately 1.571 and 4.712. Therefore the function f (x) = x − tan x has the intermediate value property on the interval [2, 3]. Results: x0 = 2.060818495, x1 = 2.027969226, x2 = 2.028752991, and x3 = 2.028757838 = x4 . Answer: 2.0288. C03S10.016: As 7 π ≈ 10.9956 and 9 pi ≈ 14.1372 are the nearest discontinuities of f (x) = x − tan x, this 2 2 function has the intermediate value property on the interval [11, 12]. Because f (11) ≈ −214.95 and f (12) ≈ 11.364, the equation f (x) = 0 has a solution in [11, 12]. We obtain x0 = 11.94978618 by interpolation, and the iteration x ←− x − x + tan x 1 + sec2 x of Newton’s method yields the successive approximations x1 = 7.457596948, x2 = 6.180210620, x3 = 3.157913273, x4 = 1.571006986; after many more iterations we arrive at the answer 2.028757838 of Problem 15. The difficulty is caused by the fact that f (x) is generally a very large number, so the iteration of Newton’s method tends to alter the value of x excessively. A little experimentation yields the fact that f (11.08) ≈ −0.736577 and f (11.09) ≈ 0.531158. We begin anew on the better interval [11.08, 11.09] and obtain x0 = 11.08581018, x1 = 11.08553759, x2 = 11.08553841, and x3 = x2 . Answer: 11.0855. C03S10.017: x − e−x = 0; [0, 1]: x0 = 0.5, x1 ≈ 0.5663, x2 ≈ 0.5671, x3 ≈ 0.5671. C03S10.018: x0 = 2.058823529, x1 = 2.095291459, x2 = 2.094551790, x3 = 2.094551482 = x4 . Answer: 2.0946. C03S10.019: ex + x − 2 = 0; [0, 1]: x0 = 0.5, x1 ≈ 0.4439, x2 ≈ 0.4429, x3 ≈ 0.4429. C03S10.020: e−x − ln x = 0; [1, 2]: x0 = 1.5, x1 ≈ 1.2951, x2 ≈ 1.3097, x3 ≈ 1.3098 ≈ x4 . C03S10.021: Let f (x) = x3 − a. Then the iteration of Newton’s method in Eq. (6) takes the form 2 xn+1 = xn − f (xn ) (xn )3 − a 2(xn )3 + a 1 = = = xn − 2 2 f (xn ) 3(xn ) 3(xn ) 3 2xn + a (xn )2 . √ Because 1 < 3 2 < 2, we begin with x0 = 1.5 and apply this formula with a = 2 to obtain x1 = 1.296296296, x2 = 1.260932225, x3 = 1.259921861, and x4 = 1.259921050 = x5 . Answer: 1.25992. C03S10.022: The formula in Eq. (6) of the text, with f (x) = xk − a, takes the form xn+1 = xn − f (xn ) (xn )k − a (k − 1)(xn )k + a 1 a . = xn − = = (k − 1)xn + f (xn ) k (xn )k−1 k (xn )k−1 k (xn )k−1 We take a = 100, k = 10, and x0 = 1.5 and obtain x1 = 1.610122949, x2 = 1.586599871, x3 = 1.584901430, x4 = 1.584893193, and x5 = 1.584893192 = x6 . Answer: 1.58489. C03S10.023: We get x0 = 0.5, x1 = 0.4387912809, x2 = 0.4526329217, x3 = 0.4496493762, . . . , x14 = 0.4501836113 = x15 . The method of repeated substitution tends to converge much more slowly than Newton’s method, has the advantage of not requiring that you compute a derivative or even that the functions involved are differentiable, and has the disadvantage of more frequent failure than Newton’s method when both are applicable (see Problems 24 and 25). C03S10.024: Our results using the first formula: x0 = 1.5, x1 = 1.257433430, x2 = 1.225755182, x3 = 1.221432153, . . . , x10 = 1.220745085 = x11 . When we use the second formula, we obtain x1 = 4.0625, x2 = 271.3789215, x3 = 5423829645, and x4 has 39 digits to the left of the decimal point. It frequently requires some ingenuity to find a suitable way to put the equation f (x) = 0 into the form x = G(x). C03S10.025: Beginning with x0 = 0.5, the first formula yields x0 = 0.5, x1 = −1, x2 = 2, x3 = 2.75, x4 = 2.867768595, . . . , x12 = 2.879385242 = x13 . Wrong root! At least the method converged. If your calculator or computer balks at computing the cube root of a negative number, then you can rewrite the second formula in Problem 25 in the form x = Sgn(3x2 − 1) · | 3x2 − 1 |1/3 . The results, again with x0 = 0.5, are x1 = −0.629960525, x2 = 0.575444686, x3 = −0.187485243, x4 = −0.963535808, . . . , x25 = 2.877296053, x26 = 2.877933902, . . . , and x62 = 2.879385240 = x63 . Not only is convergence extremely slow, the method of repeated substitution again leads to the wrong root. Finally, the given equation can also be written in the form x= √ 1 , 3−x and in this case, again with x0 = 0.5, we obtain x1 = 0.632455532, x2 = 0.649906570, x3 = 0.652315106, x4 = 0.652649632, . . . , and x12 = 0.652703645 = x13 . C03S10.026: If f (x) = 1 − a, then Newton’s method uses the iteration x 1 −a 1 x x ←− x − = x + x2 − a = 2x − ax2 . 1 x −2 x C03S10.027: Let f (x) = x5 + x − 1. Then f (x) is a polynomial, thus is continuous everywhere, and thus has the intermediate value property on every interval. Also f (0) = −1 and f (1) = 1, so f (x) must 3 assume the intermediate value 0 somewhere in the interval [0, 1]. Thus the equation f (x) = 0 has at least one solution. Next, f (x) = 5x4 + 1 is positive for all x, so f is an increasing function. Because f is continuous, its graph can therefore cross the x-axis at most once, and so the equation f (x) = 0 has at most one solution. Thus it has exactly one solution. Incidentally, Newton’s method yields the approximate solution 0.75487766624669276. To four places, 0.7549. C03S10.028: Let f (x) = x2 − cos x. The graph of f on [ −1, 1] shows that there are two solutions, one near −0.8 and the other near 0.8. With x0 = 0.8, Newton’s method yields x1 = 0.824470434, x2 = 0.824132377, and x3 = 0.824132312 = x4 . Because f (−x) = f (x), the other solution is −0.824132312. Answer: ± 0.8241. C03S10.029: Let f (x) = x − 2 sin x. The graph of f on [ −2, 2] shows that there are exactly three solutions, the largest of which is approximately x0 = 1.9. With Newton’s method we obtain x1 = 1.895505940, and x2 = 1.895494267 = x3 . Because f (−x) = −f (x), the other two solutions are 0 and −1.895494267. Answer: ± 1.8955 and 0. C03S10.030: Let f (x) = x + 5 cos x. The graph of f on the interval [ −5, 5] shows that there are exactly three solutions, approximately −1.3, 2.0, and 3.9. Newton’s method then yields n First xn Second xn Third xn 1 −1.306444739 1.977235450 3.839096917 −1.306440008 1.977383023 3.837468316 −1.306440008 1.977383029 3.837467106 −1.306440008 1.977383029 3.837467106 2 3 4 Answers: −1.3064, 1.9774, and 3.8375. C03S10.031: Let f (x) = x7 − 3x3 + 1. Then f (x) is a polynomial, so f is continuous on every interval of real numbers, including the intervals [ −2, −1], [0, 1], and [1, 2]. Also f (−2) = −103 < 0 < 3 = f (−1), f (0) = 1 > 0 > −1 = f (1), and f (1) = −1 < 0 < 105 = f (2). Therefore the equation f (x) = 0 has one solution in (−2, −1), another in (0, 1), and a third in (1, 2). (It has no other real solutions.) The graph of f shows that the first solution is near −1.4, the second is near 0.7, and the third is near 1.2. Then Newton’s method yields n First xn Second xn Third xn 1 −1.362661201 0.714876604 1.275651936 −1.357920265 0.714714327 1.258289744 −1.357849569 0.714714308 1.256999591 −1.357849553 0.714714308 1.256992779 −1.357849553 0.714714308 1.256992779 2 3 4 5 Answers: −1.3578, 0.7147, and 1.2570. C03S10.032: Let f (x) = x3 − 5. Use the iteration x ←− x − x3 − 5 . 3x2 With x0 = 2, we obtain the sequence of approximations 1.75, 1.710884354, 1.709976429, 1.709975947, and 1.709975947. Answer: 1.7100. C03S10.033: There is only one solution of x3 = cos x for the following reasons: x3 < −1 cos x if 3 3 x < −1, x < 0 < cos x if −1 < x < 0, x is increasing on [0, 1] whereas cos x is decreasing there (and their 4 graphs cross in this interval as a consequence of the intermediate value property of continuous functions), and x3 > 1 cos x for x > 1. The graph of f (x) = x3 − cos x crosses the x-axis near x0 = 0.9, and Newton’s method yields x1 = 0.866579799, x2 = 0.865475218, and x3 = 0.865474033 = x4 . Answer: Approximately 0.8654740331016145. C03S10.034: The graphs of y = x and y = tan x show that the smallest positive solution of the equation f (x) = x − tan x = 0 is between π and 3π /2. With initial guess x = 4.5 we obtain 4.493613903, 4.493409655, 4.493409458, and 4.493409458. Answer: Approximately 4.493409457909064. C03S10.035: With x0 = 3.5, we obtain the sequence x1 = 3.451450588, x2 = 3.452461938, and finally x3 = 3.452462314 = x4 . Answer: Approximately 3.452462314057969. C03S10.036: To find a zero of f (θ) = θ − sin θ − 1 2 θ ←− θ − The results, with θ0 = 1.5 (86◦ 56 37 ), are: θ3 = 1.568140. 17 50 π , we use the iteration θ − 1 sin θ − 17 π 2 50 . 1 − 1 cos θ 2 θ1 = 1.569342 (89◦ 55 00 ), θ2 = 1.568140 (89◦ 50 52 ), C03S10.037: If the plane cuts the sphere at distance x from its center, then the smaller spherical segment has height h = a − x = 1 − x and the larger has height h = a + x = 1 + x. So the smaller has volume V1 = 12 1 π h (3a − h) = π (1 − x)2 (2 + x) 3 3 and the larger has volume V2 = 12 1 π h (3a − h) = π (1 + x)2 (2 − x) = 2V1 . 3 3 These equations leads to (1 + x)2 (2 − x) = 2(1 − x)2 (2 + x); (x2 + 2x + 1)(x − 2) + 2(x2 − 2x + 1)(x + 2) = 0; x3 − 3x − 2 + 2x3 − 6x + 4 = 0; 3x3 − 9x + 2 = 0. The last of these equations has three solutions, one near −1.83 (out of range), one near 1.61 (also out of range), and one near x0 = 0.2. Newton’s method yields x1 = 0.225925926, x2 = 0.226073709, and x3 = 0.226073714 = x4 . Answer: 0.2261. C03S10.038: This table shows that the equation f (x) = 0 has solutions in each of the intervals (−3, −2), (0, 1), and (1, 2). x −3 −2 −1 0 1 2 3 f (x) −14 1 4 1 −2 1 16 The next table shows the results of the iteration of Newton’s method: 5 n xn xn xn 0 1 1.5 2.090909091 0.5 0.2307692308 2 1.895903734 0.2540002371 −2.5 −2.186440678 3 1.861832371 0.2541016863 4 5 6 1.860806773 1.860805853 1.860805853 0.2541016884 0.2541016884 −2.118117688 −2.114914461 −2.114907542 −2.114907541 −2.114907541 Answer: −2.1149, 0.2541, and 1.8608. C03S10.039: We iterate using the formula x + tan x . 1 + sec2 x Here is a sequence of simple Mathematica commands to find approximations to the four least positive solutions of the given equation, together with the results. (The command list=g[list] was executed repeatedly, but deleted from the output to save space.) x ←− x − list={2.0, 5.0, 8.0, 11.0} f[x ]:=x+Tan[x] g[x ]:=N[x−f[x]/f [x], 10] list=g[list] 2.027314579, 4.879393859, 7.975116372, 11.00421012 2.028754298, 4.907699753, 7.978566616, 11.01202429 2.028757838, 4.913038110, 7.978665635, 11.02548807 2.028757838, 4.913180344, 7.978665712, 11.04550306 2.028757838, 4.913180439, 7.978665712, 11.06778114 2.028757838, 4.913180439, 7.978665712, 11.08205766 2.028757838, 4.913180439, 7.978665712, 11.08540507 2.028757838, 4.913180439, 7.978665712, 11.08553821 2.028757838, 4.913180439, 7.978665712, 11.08553841 Answer: 2.029 and 4.913. C03S10.040: Plot the graph of f (x) = 4x3 − 42x2 − 19x − 28 on [ −3, 12] to see that the equation f (x) = 0 has exactly one real solution neear x = 11. The initial guess x0 = 0 yields the solution x = 10.9902 after 20 iterations. The initial guess x0 = 10 yields the solution after three iterations. The initial guess x0 = 100 yields the solution after ten iterations. C03S10.041: Similar triangles show that x 5 = u+v v and 6 y 5 =, u+v u so that x=5· u+v = 5(1 + t) v and y =5· u+v 1 =5 1+ u t . Next, w2 + y 2 = 400 and w2 + x2 = 225, so that: 400 − y 2 = 225 − x2 ; 175 + x2 = y 2 ; 175 + 25(1 + t)2 = 25 1 + 1 t 2 ; 175t2 + 25t2 (1 + t)2 = 25(1 + t)2 ; 7t2 + t4 + 2t3 + t2 = t2 + 2t + 1; t4 + 2t3 + 7t2 − 2t − 1 = 0. The graph of f (t) = t4 + 2t3 + 7t2 − 2t − 1 shows a solution of f (t) = 0 near x0 = 0.5. Newton’s method yields x1 = 0.491071429, x2 = 0.490936940, and x3 = 0.490936909 = x4 . It now follows that x = 7.454684547, that y = 15.184608052, that w = 13.016438772, that u = 4.286063469, and that v = 8.730375303. Answers: t = 0.4909 and w = 13.0164. C03S010.042: We let f (x) = 3 sin x − ln x. The graph of f on the interval [1, 22] does not make it clear whether there are no solutions of f (x) = 0 between 20 and 22, or one solution, or two. But the graph on [20, 21] makes it quite clear that there is no solution there: The maximum value of f (x) there is approximately −0.005 and occurs close to x = 20.4. The iteration of Newton’s method, xn+1 = xn − f (xn ) , f (xn ) beginning with the initial values x0 = 7, x0 = 9, x0 = 13.5, and x0 = 14.5, yielded the following (rounded) results: n xn xn xn xn 1 6.9881777136 8.6622012723 13.6118476398 14.6151381365 2 6.9882410659 8.6242919485 13.6226435579 14.6025252483 3 6.9882410677 8.6236121268 13.6227513693 14.6023754151 4 6.9882410677 8.6236119024 13.6227513801 14.6023753939 5 6.9882410677 8.6236119024 13.6227513801 14.6023753939 The last line in the table gives the other four solutions to ten-place accuracy. C03S010.043: Let f (θ) = (100 + θ) cos θ − 100. The iterative formula of Newton’s method is θi+1 = θi − 7 f (θi ) f (θi ) (1) where, of course, f (θ) = cos θ − (100 + θ) sin θ. Beginning with θ0 = 1, iteration of the formula in (1) yields 0.4620438212, 0.2325523723, 0.1211226155, 0.0659741863, 0.0388772442, 0.0261688780, 0.0211747166, 0.0200587600, 0.0199968594, 0.0199966678, 0.0199966678, 0.0199966678. We take the last value of θi to be sufficiently accurate. The corresponding radius of the asteriod is thus approximately 1000/θ12 ≈ 50008.3319 ft, about 9.47 mi. C03S010.044: The length of the circular arc is 2Rθ = 5281; the length of its chord is 2R sin θ = 5280 (units are radians and feet). Division of the second of these equations by the first yields sin θ 5280 = . θ 5281 To solve for θ by means of Newton’s method, we let f (θ) = 5281 sin θ − 5280θ. The iterative formula of Newton’s method is θi+1 = θi − 5281 sin θ − 5280θ . 5281 cos θ − 5280 (1) Beginning with the [poor] initial guess θ0 = 1, iteration of the formula in (1) yields these results: 0.655415, 0.434163, 0.289117, 0.193357, 0.130147, 0.0887267, 0.0621344, 0.0459270, 0.0373185, 0.0341721, 0.0337171, 0.0337078, 0.0337078, 0.0337078, 0.0337078. Hence the radius of the circular arc is R≈ 5281 ≈ 78335.1, 2θ15 and its height at its center is x = R(1 − cos θ) ≈ 44.4985. That is, the maximum height is about 44.5 feet! Surprising to almost everyone. 8 Chapter 3 Miscellaneous Problems C03S0M.001: If y = y (x) = x2 + 3x−2 , then dy 6 = 2x − 6x−3 = 2x − 3 . dx x C03S0M.002: Given y 2 = x2 , implicit differentiation with respect to x yields dy = 2x, dx 2y C03S0M.003: If y = y (x) = √ dy x =. dx y so that 1 x + √ = x1/2 + x−1/3 , then 3 x dy 1 1 1 1 3x5/6 − 2 = x−1/2 − x−4/3 = 1/2 − 4/3 = . dx 2 3 2x 3x 6x4/3 C03S0M.004: Given y = y (x) = (x2 + 4x)5/2 , the chain rule yields dy 5 = (x2 + 4x)3/2 (2x + 4). dx 2 C03S0M.005: Given y = y (x) = (x − 1)7 (3x + 2)9 , the product rule and the chain rule yield dy = 7(x − 1)6 (3x + 2)9 + 27(x − 1)7 (3x + 2)8 = (x − 1)6 (3x + 2)8 (48x − 13). dx C03S0M.006: Given y = y (x) = x4 + x2 , the quotient rule yields +x+1 x2 dy (x2 + x + 1)(4x3 + 2x) − (x4 + x2 )(2x + 1) 2x5 + 3x4 + 4x3 + x2 + 2x = = . 2 + x + 1)2 dx (x (x2 + x + 1)2 1 2x2 4 dy = 4 3x − 1 x−2 2 dx 3 C03S0M.007: If y = y (x) = 3x − 4 = 3x − 1 x−2 , then 2 · 3 + x−3 = 4 3x − 1 2x2 3 · 3+ 1 x3 . C03S0M.008: Given y = y (x) = x10 sin 10x, the product rule and the chain rule yield dy = 10x9 sin 10x + 10x10 cos 10x = 10x9 (sin 10x + x cos 10x). dx C03S0M.009: Given xy = 9, implicit differentiation with respect to x yields x Alternatively, y = y (x) = dy + y = 0, dx so that dy y =− . dx x 9 dy 9 , so that = − 2. x dx x C03S0M.010: y = y (x) = (5x ) 6 −1/2 C03S0M.011: Given y = y (x) = √ 1 dy 3 35 6 −3/2 5 = − (5x ) : (30x ) = − √ =− 4. dx 2 5x x 5x6 1 (x3 − x)3 = (x3 − x)−3/2 , 1 dy 3 3(3x2 − 1) = − (x3 − x)−5/2 (3x2 − 1) = − . dx 2 2(x3 − x)5/2 C03S0M.012: Given y = y (x) = (2x + 1)1/3 (3x − 2)1/5 , dy 2 3 = (2x + 1)−2/3 (3x − 2)1/5 + (3x − 2)−4/5 (2x + 1)1/3 dx 3 5 = C03S0M.013: So 10(3x − 2) + 9(2x + 1) 48x − 11 = . 15(2x + 1)2/3 (3x − 2)4/5 15(2x + 1)2/3 (3x − 2)4/5 dy dy du −2u −2x 1 x4 + 2x2 + 2 = · = · . Now 1 + u2 = 1 + = . 2 )2 (1 + x2 )2 2 )2 dx du dx (1 + u (1 + x (1 + x2 )2 dy −2 (1 + x2 )4 −2(1 + x2 )3 −2u = ·4 =4 . = du (1 + u2 )2 1 + x2 (x + 2x2 + 2)2 (x + 2x2 + 2)2 Therefore dy −2(1 + x2 )3 −2x 4x(1 + x2 ) =4 · =4 . dx (x + 2x2 + 2)2 (1 + x2 )2 (x + 2x2 + 2)2 C03S0M.014: 3x2 = 2 dy dy 3x2 sin y cos y , so = . dx dx 2 sin y cos y C03S0M.015: Given y = y (x) = x1/2 + 21/3 x1/3 dy 7 1 /2 = x + 21/3 x1/3 dx 3 C03S0M.016: Given y = y (x) = √ 7 /3 4 /3 , · 1 −1/2 21/3 −2/3 . x + x 2 3 3x5 − 4x2 = (3x5 − 4x2 )1/2 , dy 15x4 − 8x 1 . = (3x5 − 4x2 )−1/2 · (15x4 − 8x) = √ dx 2 2 3x5 − 4x2 C03S0M.017: If y = u+1 and u = (x + 1)1/2 , then u−1 dy 2 1 dy du (u − 1) − (u + 1) 1 1 · (x + 1)−1/2 = − ·√ = · = =− √ dx du dx (u − 1)2 2 (u − 1)2 2 x + 1 x+1−1 2 √ x+1 . C03S0M.018: Given y = y (x) = sin(2 cos 3x), dy = [cos(2 cos 3x) ] · (− 6 sin 3x) = −6(sin 3x) cos(2 cos 3x). dx C03S0M.019: Given x2 y 2 = x + y , we differentiate (both sides) implicitly with respect to x and obtain 2 2xy 2 + 2x2 y 2x2 y dy dy =1+ ; dx dx dy dy − = 1 − 2xy 2 ; dx dx dy 1 − 2xy 2 =2 ; dx 2x y − 1 dy y x − 2x2 y 2 = · 22 ; dx x 2x y − y dy y x − 2(x + y ) =· ; dx x 2(x + y ) − y dy y −x − 2y (x + 2y )y =· =− . dx x 2x + y (2x + y )x Of course you may stop with the third line, but to find horizontal and vertical lines tangent to the graph of x2 y 2 = x + y , the last line is probably the most convenient. C03S0M.020: Given y = y (x) = 1 + sin x1/2 dy 1 1 + sin x1/2 = dx 2 C03S0M.021: Given y = y (x) = x+ dy 1 = x + 2x + (3x)1/2 dx 2 −1/2 1 /2 , √ 1 cos x cos x1/2 · x−1/2 = √ √. 2 4 x 1 + sin x 2x + 1/2 −1/2 √ 3x = x + 2x + (3x)1/2 · 1+ 1 2x + (3x)1/2 2 −1/2 1 /2 1 /2 , 3 · 2 + (3x)−1/2 2 . The symbolic algebra program Mathematica writes this answer without exponents as follows: √ 3 2+ √ 2x 1+ √ dy 2 2 x + 3x = . √ dx 2 x + 2x + 3x C03S0M.022: dy 1 − x2 − x sin x + cos x + x2 cos x (x2 + cos x)(1 + cos x) − (x + sin x)(2x − sin x) = . = dx (x2 + cos x)2 (x2 + cos x)2 C03S0M.023: Given x1/3 + y 1/3 = 4, differentiate both sides with respect to x: 1 −2/3 1 −2/3 dy x +y = 0, 3 3 dx so dy y =− dx x 2 /3 . C03S0M.024: Given x3 + y 3 = xy , differentiate both sides with respect to x to obtain 3x2 + 3y 2 dy dy =x + y, dx dx so that C03S0M.025: Given y = (1 + 2u)3 where u = (1 + x)−3 : 3 dy y − 3x2 =2 . dx 3y − x dy 18(1 + 2u)2 18(1 + 2(1 + x)−3 )2 dy du =− = · = 6(1 + 2u)2 · (−3)(1 + x)−4 = − dx du dx (1 + x)4 (1 + x)4 =− C03S0M.026: 18(1 + x)6 (1 + 2(1 + x)−3 )2 18((1 + x)3 + 2)2 (x3 + 3x2 + 3x + 3)2 =− = −18 · . (1 + x)10 (1 + x)10 (x + 1)10 dy = − 2 cos(sin2 x) sin(sin2 x) · 2 sin x cos x . dx C03S0M.027: Given y = y (x) = dy 1 = dx 2 = 1 /2 sin2 x 1 + cos x sin2 x 1 + cos x 1 + cos x sin2 x , −1/2 1 /2 · · (1 + cos x)(2 sin x cos x) + sin3 x (1 + cos x)2 2 sin x cos x + 2 sin x cos2 x + sin3 x . 2(1 + cos x)2 √2 √4 √3 √3 dy 3 (1 + x ) √ C03S0M.028: = (1 − 2 3 x ) + 4 (1 − 2 3 x ) − 2 x−2/3 (1 + x ) . 3 dx 2x cos 2x C03S0M.029: Given: y = y (x) = √ = (cos 2x)(sin 3x)−1/2 , sin 3x dy 1 = (−2 sin 2x)(sin 3x)−1/2 + (cos 2x) − (sin 3x)−3/2 (3 cos 3x) dx 2 2 sin 2x 4 sin 2x sin 3x + 3 cos 2x cos 3x 3 cos 2x cos 3x = −√ =− . − 2(sin 3x)3/2 2(sin 3x)3/2 sin 3x C03S0M.030: 3x2 − x2 dy dy dy dy 3x2 − 2xy + y 2 − 2xy + y 2 + 2xy − 3y 2 = 0: =2 . dx dx dx dx 3y − 2xy + x2 C03S0M.031: dy = ex (cos x − sin x). dx C03S0M.032: dy = e−2x (3 cos 3x − 2 sin 3x). dx C03S0M.033: dy 3ex =− dx (2 + 3ex )5/2 1 + (2 + 3ex )−3/2 C03S0M.034: dy = 5(ex + e−x )4 (ex − e−x ). dx C03S0M.035: cos2 [1 + ln x]1/3 sin [1 + ln x]1/3 dy . =− dx x[1 + ln x]2/3 1/3 . C07S0M.036: If f (x) = cos(1 − e−x ), then f (x) = −e−x sin(1 − e−x ). 2 C03S0M.037: If f (x) = sin2 (e−x ) = [sin(e−x )] , then f (x) = −2e−x sin(e−x ) cos(e−x ). 4 C03S0M.038: If f (x) = ln(x + e−x ), then f (x) = 1 − e−x . x + e−x C03S0M.039: If f (x) = ex cos 2x, then f (x) = ex cos 2x − 2ex sin 2x. C03S0M.040: If f (x) = e−2x sin 3x, then f (x) = 3e−2x cos 3x − 2e−2x sin 3x. 2 C03S0M.041: If g (t) = ln(tet ) = (ln t) + t2 ln e = (ln t) + t2 , then g (t) = 1 1 + 2t2 + 2t = . t t C03S0M.042: If g (t) = 3(et − ln t)5 , then g (t) = 15(et − ln t)4 et − 1 . t C03S0M.043: If g (t) = sin(et ) cos(e−t ), then g (t) = et cos(et ) cos(e−t ) + e−t sin(et ) sin(e−t ). C03S0M.044: If f (x) = 2 + 3x , then e4x f (x) = C03S0M.045: If g (t) = 3e4x − 4(2 + 3x)e4x 3 − 8 − 12x 12x + 5 = =− . (e4x )2 e4x e4x 1 + et (1 − et )et + (1 + et )et 2et , then g (t) = = . 1 − et (1 − et )2 (1 − et )2 C03S0M.046: Given xey = y , we apply Dx to both sides and find that ey + xey dy dy = ; dx dx (1 − xey ) dy ey = ; dx 1 − xey dy = ey ; dx dy ey = . dx 1−y In the last step we used the fact that xey = y to simplify the denominator. C03S0M.047: Given sin (exy ) = x, we apply Dx to both sides and find that [cos (exy ) ] · exy · y + x dy dx = 1; xexy [cos (exy ) ] · dy = 1 − yexy cos (exy ) ; dx dy 1 − yexy cos (exy ) = . dx xexy cos (exy ) C03S0M.048: Given ex + ey = exy , we apply Dx to both sides and find that ex + ey dy dy = exy y + x dx dx ; (ey − xexy ) dy = yexy − ex ; dx dy yexy − ex . =y dx e − xexy 5 C03S0M.049: Given x = yey , we apply Dx to both sides and obtain 1 = ey dy 1 y y =y =y = . dx e + yey ye + y 2 ey x + xy dy dy + yey ; dx dx We used the fact that yey = x in the simplification in the last step. Here is an alternative approach to finding dy/dx. Beginning with x = yey , we differentiate with respect to y and find that dx = ey + yey , dy dy 1 =y . dx e + yey so that C03S0M.050: Given ex−y = xy , we apply Dx to both sides and find that ex−y 1 − dy dx =y+x dy ; dx x + ex−y dy = ex−y − y ; dx dy xy − y (x − 1)y = = . dx xy + x (y + 1)x dy ex−y − y = x−y ; dx e +x We used the fact that ex−y = xy to make the simplification in the last step. C03S0M.051: Given x ln y = x + y , we apply Dx to both sides and find that ln y + x dy dy · = 1+ ; y dx dx x dy −1 · = 1 − ln y ; y dx x − y dy · = 1 − ln y ; y dx dy y (1 − ln y ) = . dx x−y C03S0M.052: Given: y = √ (x2 − 4) 2x + 1 . Thus ln y = ln (x2 − 4)(2x + 1)1/2 1 /2 = 1 1 1 ln (x2 − 4)(2x + 1)1/2 = ln(x2 − 4) + ln(2x + 1) . 2 2 2 Therefore 1 dy x 1 5x2 + 2x − 4 · =2 + = , y dx x − 4 2(2x + 1) 2(x2 − 4)(2x + 1) and so √ (5x2 + 2x − 4) (x2 − 4) 2x + 1 dy 5x2 + 2x − 4 = y (x) · = . dx 2(x2 − 4)(2x + 1) 2(x2 − 4)(2x + 1) C03S0M.053: Given: y = (3 − x2 )1/2 (x4 + 1)−1/4 . Thus 6 ln y = 1 1 ln(3 − x2 ) − ln(x4 + 1), 2 4 and therefore 1 dy x x3 x(3x2 + 1) · =− −4 =2 . y dx 3 − x2 x +1 (x − 3)(x4 + 1) Thus dy x(3x2 + 1) x(3x2 + 1) x(3x2 + 1)(3 − x2 )1/2 =− . = y (x) · 2 =− 4 + 1) 2 )(x4 + 1)(x4 + 1)1/4 dx (x − 3)(x (3 − x (3 − x2 )1/2 (x4 + 1)5/4 C03S0M.054: Given: y = ln y = (x + 1)(x + 2) (x2 + 1)(x2 + 2) 1 /3 . Then 1 ln(x + 1) + ln(x + 2) − ln(x2 + 1) − ln(x2 + 2) ; 3 1 dy 1 · = y dx 3 1 1 2x 2x + − − x + 1 x + 2 x2 + 1 x2 + 2 ; dy 6 − 8x − 9x2 − 8x3 − 9x4 − 2x5 = y (x) · ; dx 3(x + 1)(x + 2)(x2 + 1)(x2 + 2) dy (x + 1)(x + 2) 6 − 8x − 9x2 − 8x3 − 9x4 − 2x5 = · dx 3(x + 1)(x + 2)(x2 + 1)(x2 + 2) (x2 + 1)(x2 + 2) 1 /3 ; dy 6 − 8x − 9x2 − 8x3 − 9x4 − 2x5 . = dx 3(x + 1)2/3 (x + 2)2/3 (x2 + 1)4/3 (x2 + 2)4/3 C03S0M.055: If y = (x + 1)1/2 (x + 2)1/3 (x + 3)1/4 , then ln y = 1 1 1 ln(x + 1) + ln(x + 2) + ln(x + 3); 2 3 4 1 dy 1 1 1 · = + + ; y dx 2(x + 1) 3(x + 2) 4(x + 3) dy 13x2 + 55x + 54 13x2 + 55x + 54 = y (x) · = . dx 12(x + 1)(x + 2)(x + 3) 12(x + 1)1/2 (x + 2)2/3 (x + 3)3/4 x C03S0M.056: If y = x(e ) , then ln y = ex ln x; 1 dy ex · = + ex ln x; y dx x dy (1 + x ln x)ex = y (x) · ; dx x x dy (1 + x ln x)ex = · x(e ) . dx x ln x C03S0M.057: Given: y = (ln x) , x > 1. Then 7 ln y = (ln x) ln (ln x) ; 1 dy 1 ln x 1 + ln (ln x) · = ln (ln x) + = ; y dx x x ln x x dy 1 + ln (ln x) ln x = · (ln x) . dx x dy 2 (x − 1) − (x + 1) = =− ; the slope of the line tangent at (0, −1) is −2; an dx (x − 1)2 (x − 1)2 equation of the tangent line is y + 1 = −2x; that is, 2x + y + 1 = 0. C03S0M.058: dy dy 1 dy , so = . Because is undefined at (1, π /4), there may well dx dx 2 cos 2y dx dx be a vertical tangent at that point. And indeed there is: = 0 at (1, π /4). So an equation of the tangent dy line is x = 1. C03S0M.059: 1 = (2 cos 2y ) C03S0M.060: that is, x = 2y . C03S0M.061: dy 3y − 2x = ; at (2, 1) the slope is 1 . So an equation of the tangent is y − 1 = 1 (x − 2); 2 2 dx 4y − 3x dy dx 2x + 1 ; at (0, 0), = = 0, so the tangent line is vertical. Its equation is x = 0. dx 3y 2 dy C03S0M.062: V (x) = 1 π (36x2 − x3 ): V (x) = π x(24 − x). 3 dx dx 36π = −108π , so = − 1 (in./s) when x = 6. 3 dt dt Now dV dV dx = · ; when x = 6, dt dx dt C03S0M.063: Let r be the radius of the sandpile, h its height, each a function of time t. We know that 2r = h, so the volume of the sandpile at time t is V= 12 2 πr h = πr3 . 3 3 So 25π = substitution of r = 5 yields the answer: dr/dt = 1 2 dV dr = 2π r2 ; dt dt (ft/min) when r = 5 (ft). C03S0M.064: Divide each term in the numerator and denominator by sin x to obtain lim x→0 C03S0M.065: x cot 3x = C03S0M.066: x 1 − lim = 1 − 1 = 0. sin x x→0 cos x 1 3x 1 1 · → ·1= as x → 0. 3 sin 3x 3 3 sin 2x 2 sin 2x 5x 2 =· · → as x → 0. sin 5x 5 2x sin 5x 5 C03S0M.067: x2 csc 2x cot 2x = 1 2x 2x 1 1 · · · cos 2x → · 1 · 1 · 1 = as x → 0. 4 sin 2x sin 2x 4 4 8 C03S0M.068: −1 sin u 1 for all u. So −x2 x2 sin 1 x2 x2 for all x = 0. But x2 → 0 as x → 0, so the limit of the expression caught in the squeeze is also zero. C03S0M.069: −1 sin u 1 for all u. So √ −x √ for all x > 0. But √ x sin 1 x √ x x → 0 as x → 0+ , so the limit is zero. C03S0M.070: h(x) = (x + x4 )1/3 = f (g (x)) where f (x) = x1/3 and g (x) = x + x4 . h (x) = f (g (x)) · g (x) = 1 (x + x4 )−2/3 · (1 + 4x3 ). 3 Therefore C03S0M.071: h(x) = (x2 + 25)−1/2 = f (g (x)) where f (x) = x−1/2 and g (x) = x2 + 25. Therefore h (x) = f (g (x)) · g (x) = − 1 (x2 + 25)−3/2 · 2x. 2 C03S0M.072: First, x = x2 + 1 h(x) = x x2 + 1 1 /2 = f (g (x)) where f (x) = x1/2 and g (x) = x . x2 + 1 Therefore h (x) = 1 x 2+1 2x −1/2 · x2 + 1 − 2x2 = (x2 + 1)2 x2 + 1 x 1/2 · 1 − x2 1 − x2 = 1 /2 2 . 2(x2 + 1)2 2x (x + 1)3/2 C03S0M.073: One solution: h(x) = (x − 1)5/3 = f (g (x)) where f (x) = x5/3 and g (x) = x − 1. Therefore h (x) = f (g (x)) · g (x) = 5 (x − 1)2/3 · 1 = 5 (x − 1)2/3 . You might alternatively choose f (x) = x1/3 3 3 and g (x) = (x − 1)5 . C03S0M.074: If h(x) = (x + 1)10 , (x − 1)10 then h(x) = f (g (x)) where f (x) = x10 and g (x) = x+1 . x−1 Hence h (x) = f (g (x)) · g (x) = 10 x+1 x−1 9 · (x + 1) − (x − 1) x+1 = 10 2 (x − 1) x−1 9 · 2 20(x + 1)9 = . (x − 1)2 (x − 1)11 C03S0M.075: h(x) = cos(x2 + 1) = f (g (x)) where f (x) = cos x and g (x) = x2 + 1. h (x) = f (g (x)) · g (x) = −2x sin(x2 + 1). L dT π 32 dT C03S0M.076: T = 2π ; = . So 32 dL 32 L dL at approximately 0.27768 seconds per foot. 9 L=4 Therefore √ π2 = . Hence when L = 4, T is changing 16 C03S0M.077: Of course r denotes the radius of the sphere. First, A = 4π r2 , so dV dV r 1 · 8π r = 4π r2 , and therefore == dA dA 2 4 dV dA dV 4 · = . Now V = π r3 and dA dr dr 3 A . π C03S0M.078: Let (a, b) denote the point of tangency; note that 1 b=a+ , a a > 0, and h (x) = 1 − 1 . x2 The slope of the tangent line can be computed using the two-point formula for slope and by using the derivative. We equate the results to obtain 1 −0 1 a2 − 1 a =1− 2 = . a−1 a a2 a+ √ It follows that a3 + a = (a − 1)(a2 − 1) = a3 − a2 − a + 1. Thus a2 + 2a − 1 = √ and so a = −1 + 2 (the 0, positive root because a > 0). Consequently the tangent line has slope −2 1 + 2 and thus equation y = −2 1 + √ 2 (x − 1). C03S0M.079: Let y = y (t) denote the altitude of the rocket at time t; let u = u(t) denote the angle of elevation of the observer’s line of sight at time t. Then tan u = y/3, so that y = 3 tan u and, therefore, dy du = (3 sec2 u) . dt dt π and find that the speed of the rocket is 3 dy 3 π 2 = · = π ≈ 1.2566 (mi/s), 2 (π /3) 30 dt u=60◦ cos 5 When u = 60◦ , we take du/dt = about 4524 mi/h, or about 6635 ft/s. C03S0M.080: Current production per well: 200 (bbl/day). Number of new wells: x (x per well: 200 − 5x. Total production: T = T (x) = (20 + x)(200 − 5x), 0 0). Production 40. x Now T (x) = 4000 + 100x − 5x2 , so T (x) = 100 − 10x. T (x) = 0 when x = 10. T (0) = 4000, T (40) = 0, and T (10) = 4500. So x = 10 maximizes T (x). Answer: Ten new wells should be drilled, thereby increasing total production from 4000 bbl/day to 4500 bbl/day. C03S0M.081: Let the circle be the one with equation x2 + y 2 = R2 and let the base of the triangle lie on the x-axis; denote the opposite vertex of the triangle by (x, y ). The area of the triangle A = Ry is clearly maximal when y is maximal; that is, when y = R. To solve this problem using calculus, let θ be the angle of the triangle at (−R, 0). Because the triangle has a right angle at (x, y ), its two short sides are 2R cos θ and 2R sin θ, so its area is A(θ) = 2R2 sin θ cos θ = R2 sin 2θ, 0 θ π . 2 Then A (θ) = 2R2 cos 2θ; A (θ) = 0 when cos 2θ = 0; because θ lies in the first quadrant, θ = 1 π . Finally, 4 A(0) = 0 = A(π /2), but A(π /4) = R2 > 0. Hence the maximum possible area of such a triangle is R2 . 10 C03S0M.082: Let x be the length of the edges of each of the 20 small squares. The first five boxes measure 210 − 2x by 336 − 2x by x. The total volume is then V (x) = 5x(210 − 2x)(336 − 2x) + 8x3 , 0 105. x Thus V (x) = 28x3 − 5460x2 + 352800x, and so V (x) = 84x2 − 10290x + 352800 = 84(x2 − 130x + 4200) = 84(x − 60)(x − 70). So V (x) = 0 when x = 60 and when x = 70. But V (0) = 0, V (60) = 7,560,000, V (70) = 7,546,000, and V (105) = 9,261,000. Answer: For maximal volume, make x as large as possible: 105 cm. This yields the maximum volume, 9,261,000 cm3 . Note that it is attained by constructing one large cubical box and that some material is wasted. C03S0M.083: Let one sphere have radius r; the other, s. We seek the extrema of A = 4π (r2 + s2 ) given 4 3 3 3 π (r + s ) = V, a constant. We illustrate here the method of auxiliary variables: dA ds = 4π 2r + 2s dr dr ; the condition dA/dr = 0 yields ds/dr = −r/s. But we also know that both sides of this identity with respect to r yields 4 ds π 3r2 + 3s2 3 dr = 0, 4 3 3 π (r + s3 ) = V ; differentiation of and so r = 0; s r2 − rs = 0. 3r2 + 3s2 − Therefore r = 0 or r = s. Also, ds/dr is undefined when s = 0. So we test these three critical points. If r = 0 or if s = 0, there is only one sphere, with radius (3V /4π )1/3 and surface area (36π V 2 )1/3 . If r = s, then there are two spheres of equal size, both with radius 1 (3V /π )1/3 and surface area (72π V 2 )1/3 . Therefore, 2 for maximum surface area, make two equal spheres. For minimum surface area, make only one sphere. C03S0M.084: Let x be the length of the edge of the rectangle on the side of length 4 and y the length of the adjacent edges. By similar triangles, 3/4 = (3 − y )/x, so x = 4 − 4 y . We are to maximize A = xy ; that 3 is, 4 A = A(y ) = 4y − y 2 , 3 0 y 3. Now dA/dy = 4 − 8 y ; dA/dy = 0 when y = 3 . Because A(0) = A(3) = 0, the maximum is A(2) = 3 (m2 ). 3 2 C03S0M.085: Let r be the radius of the cone; let its height be h = R + y where 0 y R. (Actually, −R y R, but the cone will have maximal volume if y 0.) A central vertical cross section of the figure (draw it! ) shows a right triangle from which we read the relation y 2 = R2 − r2 . We are to maximize V = 1 π r2 h, so we write 3 V = V (r) = 1 π r2 R + 3 R2 − r 2 , 0 r R. √ The condition V (r√= 0 leads to the equation r 2R2 − 3r2 + 2R R2 − r2 = 0, which has the two solutions ) r = 0 √ r = 2 R 2. Now V (0) = 0, V (R) = 1 π R3 (which is one-fourth the volume of the sphere), and and 3 3 2 V 3 R 2 = 32 π R3 (which is 8/27 of the volume of the sphere). Answer: The maximum volume is 32 π R3 . 81 81 11 C03S0M.086: Let x denote the length of the two sides of the corral that are perpendicular to the wall. There are two cases to consider. Case 1: Part of the wall is used. Let y be the length of the side of the corral parallel to the wall. Then y = 400 − 2x, and we are to maximize the area A = xy = x(400 − 2x), 150 200. x Then A (x) = 400 − 4x; A (x) = 0 when x = 100, but that value of x is not in the domain of A. Note that A(150) = 15000 and that A(200) = 0. Case 2: All of the wall is used. Let y be the length of fence added to one end of the wall, so that the side parallel to the wall has length 100 + y . Then 100 + 2y + 2x = 400, so y = 150 − x. We are to maximize the area A = x(100 + y ) = x(250 − x), 0 x 150. In this case A (x) = 0 when x = 125. And in this case A(150) = 15000, A(0) = 0, and A(125) = 15625. Answer: The maximum area is 15625 ft2 ; to attain it, use all the existing wall and build a square corral. C03S0M.087: First, R (x) = kM − 2kx; because k = 0, R (x) = 0 when x = M/2. Moreover, because R(0) = 0 = R(M ) and R(M/2) > 0, the latter is the maximum value of R(x). Therefore the incidence of the disease is the highest when half the susceptible individuals are infected. C03S0M.088: The trapezoid is shown next. It has altitude h = L cos θ and the length of its longer base is L + 2L sin θ, so its area is A(θ) = L2 (1 + sin θ) cos θ, − π 6 θ π . 2 Now dA/dθ = 0 when 1 − sin θ − 2 sin2 θ = 0; (2 sin θ − 1)(sin θ + 1) = 0; the only solution√ θ = π /6 because sin θ cannot equal −1 in the range of A. Finally, A(π /2) = 0, is √ A(−π /6) = 1 L2 3, and A(π /6) = 3 L2 3. The latter maximizes A(θ), and the fourth side of the 4 4 trapezoid then has length 2L. h L θ L L C03S0M.089: Let x be the width of the base of the box, so that the base has length 2x; let y be the height of the box. Then the volume of the box is V = 2x2 y , and for its total surface area to be 54 ft2 , we require 2x2 + 6xy = 54. Therefore the volume of the box is given by 12 V = V (x) = 2x2 27 − x2 3x = 2 (27x − x3 ), 3 √ 3 3. 0<x √ Now V (x) = 0 when x2 = 9, so that x = 3. Also V (x) → 0 as x → 0+ and V 3 3 = 0, so V (3) = 36 (ft3 ) is the maximum possible volume of the box. C03S0M.090: Suppose that the small cone has radius x and height y . Similar triangles that appear in a x R H vertical cross section of the cones (draw it! ) show that = . Hence y = H − x, and we seek to H −y H R maximize the volume V = 1 π x2 y . Now 3 V = V (x) = πH (Rx2 − x3 ), 3R 0 x R. πH x(2R − 3x). V (x) = 0 when x = 0 and when x = 2 R (in this case, y = H/3). But 3 3R 4π2 2 V (0) = 0 and V (R) = 0, so x = 3 R maximizes V . Finally, it is easy to find that Vmax = · R H , so 27 3 the largest fraction of the large cone that the small cone can occupy is 4/27. So V (x) = C03S0M.091: Let (x, y ) be the coordinates of the vertex of the trapezoid lying properly in the first quadrant and let θ be the angle that the radius of the circle to (x, y ) makes with the x-axis. The bases of the trapezoid have lengths 4 and 4 cos θ and its altitude is 2 sin θ, so its area is A(θ) = 1 (4 + 4 cos θ)(2 sin θ) = 4(1 + cos θ) sin θ, 2 0 θ π . 2 Now A (θ) = 4(cos θ + cos2 θ − sin2 θ) = 4(2 cos2 θ + cos θ − 1) = 4(2 cos θ − 1)(cos θ + 1). The only zero of A in its domain occurs at θ = π /3. At the endpoints, we have A(0) = 0 and A(π /2) = 4. √ But A(π /3) = 3 3 ≈ 5.196, so the latter is the maximum possible area of such a trapezoid. C03S0M.092: The square of the length of P Q is a function of x, G(x) = (x − x0 )2 + (y − y0 )2 , which we are to maximize given the constraint C (x) = y − f (x) = 0. Now dG dy = 2(x − x0 ) + 2(y − y0 ) dx dx When both vanish, f (x) = and dC dy = − f (x). dx dx dy x − x0 . The line containing P and Q has slope =− dx y − y0 y − y0 1 =− , x − x0 f (x) and therefore this line is normal to the graph at Q. C03S0M.093: If Ax + By + C = 0 is an equation of a straight line L, then not both A and B can be zero. Case 1: A = 0 and B = 0. Then L has equation y = −C/B and thus is a horizontal line. So the shortest segment from P (x0 , y0 ) to Q on L is a vertical segment that therefore meets L in the point Q(x0 , −C/B ). Therefore, because A = 0, the distance from P to Q is 13 y0 + C | By0 + C | |Ax0 + By0 + C | √ = . = B |B | A2 + B 2 Case 2: A = 0 and B = 0. Then L has equation x = −C/A and thus is a vertical line. So the shortest segment from P (x0 , y0 ) to Q on L is a horizontal segment that therefore meets L in the point Q(−C/A, y0 ). Therefore, because B = 0, the distance from P to Q is x0 + C | Ax0 + C | |Ax0 + By0 + C | √ = . = A |A| A2 + B 2 Case 3: A = 0 and B = 0. Then L is neither horizontal nor vertical, and the segment joining P (x0 , y0 ) to the nearest point Q(u, v ) on L is also neither horizontal nor vertical. The equation of L may be written in the form y=− A C −, B B so L has slope − A/B . Thus the slope of P Q is B/A (by the result in Problem 70), and therefore P Q lies on the line K with equation y − y0 = B (x − x0 ). A Consequently A(v − y0 ) = B (u − x0 ). But Q(u, v ) also lies on L, and so Au + Bv = −C . Thus we have the simultaneous equations Au + Bv = −C ; Bu − Av = Bx0 − Ay0 . These equations may be solved for u= −AC + B 2 x0 − ABy0 A2 + B 2 and v= A(−C − Ax0 − By0 ) A2 + B 2 and v − y0 = −BC − ABx0 + A2 y0 , A2 + B 2 and it follows that u − x0 = B (−C − Ax0 − By0 ) . A2 + B 2 Therefore (u − x0 )2 + (v − y0 )2 = = A2 (−C − Ax0 − By0 )2 B 2 (−C − Ax0 − By0 )2 + (A2 + B 2 )2 (A2 + B 2 )2 (A2 + B 2 )(−C − Ax0 − By0 )2 (Ax0 + By0 + C )2 = . 2 + B 2 )2 (A A2 + B 2 The square root of this expression then gives the distance from P to Q as |Ax0 + By0 + C | √ , A2 + B 2 and the proof is complete. 14 C03S0M.094: Let r be the radius of each semicircle and x the length of the straightaway. We wish to maximize A = 2rx given C = 2π r + 2x − 4 = 0. We use the method of auxiliary variables (as in the solution of Problem 83): dA dr dC dr = 2r + 2x and = 2π + 2. dx dx dx dx When both derivatives are zero, −r/x = dr/dx = −1/π , and so x = π r. Also 2π r + 2x = 4, and it follows 1 1 that r = and that x = 1. Answer: Design the straightaway 1 km long with semicircles of radius at π π each end. C03S0M.095: As the following diagram suggests, we are to minimize the sum of the lengths of the two diagonals. Fermat’s principle of least time may be used here, so we know that the angles at which the roads x 6−x meet the shore are equal, and thus so are the tangents of those angles: = . It follows that the pier 1 2 should be built two miles from the point on the shore nearest the first town. To be sure that we have found a minimum, consider the function that gives the total length of the two diagonals: f (x) = x2 + 1 + (6 − x)2 + 4, 0 x 6. √ (The domain certainly contains the global minimum value of f .) Moreover, f (0) = 1 + 40 ≈ 7.32, √ √ √ f (6) = 2 + 37 ≈ 8.08, and f (2) = 5 + 20 ≈ 6.71. This establishes that x = 2 yields the global minimum of f (x). Pier Lake x 6–x 1 2 Town 1 Town 2 2 4 cos θ C03S0M.096: The length of each angled path is . The length of the roadway path is 10 − . So sin θ sin θ the total time of the trip will be 5 32 − 12 cos θ + . 4 24 sin θ √ 5 ◦ θ◦ 29 29, so 21.80 T = T (θ) = Note that cos θ varies in the range 0 cos θ T (θ) = 90◦ . After simplifications, 12 − 32 cos θ ; 24 sin2 θ T (θ) = 0 when cos θ = 3 , so θ◦ ≈ 67.98◦ . With this value of θ, we find that the time of the trip is 8 √ 2 55 + 15 T= ≈ 2.486 (hours). 12 15 Because T ≈ 3.590 with θ◦ ≈ 21.80◦ and T ≈ 2.583 when θ◦ ≈ 90◦ , the value θ◦ ≈ 69.98◦ minimizes T , and the time saved is about 50.8 minutes. C03S0M.097: Denote the initial velocity of the arrow by v . First, we have dy 32x = m − 2 (m2 + 1); dx v mv 2 . Substitution of this value of x in the formula 32(m2 + 1) given for y in the problem yields the maximum height dy/dx = 0 when mv 2 = 32x(m2 + 1), so that x = ymax = m2 v 2 . 64(m2 + 1) For part (b), we set y = 0 and solve for x to obtain the range R= mv 2 . 16(m2 + 1) Now R is a continuous function of the slope m of the arrow’s path at time t = 0, with domain 0 m < + ∞. Because R(m) = 0 and R(m) → 0 as m → + ∞, the function R has a global maximum; because R is differentiable, this maximum occurs at a point where R (m) = 0. But dR v 2 (m2 + 1) − 2m2 , = · dm 16 (m2 + 1)2 so dR/dm = 0 when m = 1 and only then. So the maximum range occurs when tan α = 1; that is, when α = 1 π. 4 C03S0M.098: Here we have √ v2 2 R = R(θ) = (cos θ sin θ − cos2 θ) 16 for 1 4π θ 1 2 π. Now √ v2 2 R (θ) = (cos2 θ − sin2 θ + 2 sin θ cos θ); 16 R (θ) = 0 when cos 2θ + sin 2θ = 0, so that tan 2θ = −1. It follows that θ = 3π /8 (67.5◦ ). This yields the maximum range because R(π /4) = 0 = R(π /2). C03S0M.099: With initial guess x0 = 2.5 (the midpoint of the given interval [2, 3]), the iteration xn+1 = xn − f (xn ) (xn )2 − 7 = xn − f (xn ) 2xn of Newton’s method yields x1 = 2.65, x2 = 2.645754717, and x3 = 2.645751311. Answer: 2.6458. C03S0M.100: We get x0 = 1.5, x1 = 1.444444444, x2 = 1.442252904, and x3 = 1.442249570. Answer: 1.4422. C03S0M.101: With x0 = 2.5, we obtain x1 = 2.384, x2 = 2.371572245, x3 = 2.371440624, and x4 = 2.371440610. Answer: 2.3714. 16 C03S0M.102: With x0 = 5.5 we get x1 = 5.623872512, x2 = 5.623413258, and x3 = 5.623413252. Answer: 5.6234. If your calculator won’t raise numbers to fractional powers, you could solve instead the equation x4 − 1000 = 0. With x0 = 5.5 the results should be x1 = 5.627629602, x2 = 5.623417988, and x3 = 5.623413252. C03S0M.103: With x0 = −0.5 we obtain x1 = −0.333333333, x2 = −0.347222222, x3 = −0.347296353, and x4 = −0.347296355. Answer: −0.3473. C03S0M.104: With x0 = −0.5 we obtain x1 = −0.230769231, x2 = −0.254000237, x3 = −0.254101686, and x4 = −0.254101688. Answer: −0.2541. C03S0M.105: With f (x) = e−x − sin x and initial guess x0 = 0.6, five iterations of the formula of Newton’s method yields the approximate solution 0.588532744 of the equation f (x) = 0. C03S0M.106: With f (x) = cos x − ln x and initial guess x0 = 1.3, four iterations of the formula of Newton’s method yield the approximate solution 1.302964001 of the equation f (x) = 0. C03S0M.107: With x0 = −1.0 we obtain x1 = −0.750363868, x2 = −0.739112891, x3 = −0.739085133, and x4 = −0.739085133. Answer: −0.7391. C03S0M.108: With x0 = −0.75, we obtain x1 = −0.905065774, x2 = −0.877662556, x3 = −0.876727303, x4 = −0.876726215, and x5 = −0.876726215. Answer: −0.8767. C03S0M.109: With x0 = −1.5, we obtain x1 = −1.244861806, x2 = −1.236139793, x3 = −1.236129989, and x4 = −1.236129989. Answer: −1.2361. C03S0M.110: With x0 = −0.5 we obtain x1 = −0.858896298, x2 = −0.871209876, x3 = −0.871221514, and x4 = −0.871221514. Answer: −0.8712. C03S0M.111: The volume of a spherical segment of height h is V= 12 π h (3r − h) 3 if the sphere has radius r. If ρ is the density of water and the ball sinks to the depth h, then the weight of the water that the ball displaces is equal to the total weight of the ball, so 1 4 πρh2 (3r − h) = 2 πρr3 . 3 3 Because r = 2, this leads to the equation p(h) = 3h3 − 18h2 + 32 = 0. This equation has at most three [real] solutions because p(h) is a polynomial of degree 3, and it turns out to have exactly three solutions because p(−2) = −64, p(−1) = 11, p(2) = −16, and p(6) = 32. Newton’s method yields the three approximate solutions h = −1.215825766, h = 1.547852572, and h = 5.667973193. Only one is plausible, so the answer is that the ball sinks to a depth of approximately 1.54785 ft, about 39% of the way up a diameter. C03S0M.112: The iteration is x ←− x − x2 + 1 x2 − 1 = . 2x 2x With x0 = 2.0, the sequence obtained by iteration of Newton’s method is 0.75, −0.2917, 1.5685, 0.4654, −0.8415, 0.1734, −2.7970, −1.2197, −0.1999, 2.4009, 0.9922, −0.0078, 63.7100, . . . . 17 C03S0M.113: Let f (x) = x5 − 3x3 + x2 − 23x + 19. Then f (−3) = −65, f (0) = 19, f (1) = −5, and f (3) = 121. So there are at least three, and at most five, real solutions. Newton’s method produces three real solutions, specifically r1 = −2.722493355, r2 = 0.8012614801, and r3 = 2.309976541. If one divides the polynomial f (x) by (x − r1 )(x − r2 )(x − r3 ), one obtains the quotient polynomial x2 + (0.38874466)x + 3.770552031, which has no real roots—the quadratic formula yields the two complex roots −0.194372333 ± (1.932038153)i. Consequently we have found all three real solutions. C03S0M.114: Let f (x) = tan x − 1 . We iterate x 1 x x ←− x − . 1 sec2 x + 2 x tan x − The results are shown in the following table. The instability in the last one or two digits is caused by machine rounding and is common. Answers: To three places, α1 = 0.860 and α2 = 3.426. f[x ]:=Tan[x]−1/x g[x ]:=N[x−f[x]/f [x], 20] list={1.0,4.0}; g[list] 0.8740469203219249386, 3.622221245370322529 0.8604001629909660496, 3.440232462677783381 0.8603335904117901655, 3.425673797668214504 0.8603335890193797612, 3.425618460245614115 0.8603335890193797636, 3.425618459481728148 0.8603335890193797608, 3.425618459481728146 0.8603335890193797634, 3.425618459481728148 C03S0M.115: The number of summands on the right is variable, and we have no formula for finding its derivative. One thing is certain: Its derivative is not 2x2 . C03S0M.116: We factor: z 3/2 − x3/2 = (z 1/2 )3 − (x1/2 )3 = (z 1/2 − x1/2 )(z + z 1/2 x1/2 + x) and z − x = (z 1/2 )2 − (x1/2 )2 = (z 1/2 − x1/2 )(z 1/2 + x1/2 ). Therefore z 3/2 − x3/2 3x 3 z + z 1/2 x1/2 + x → 1/2 = x1/2 = z−x 2 z 1/2 + x1/2 2x as z → x. C03S0M.117: We factor: z 2/3 − x2/3 = (z 1/3 )2 − (x1/3 )2 = (z 1/3 − x1/3 )(z 1/3 + x1/3 ) and z − x = (z 1/3 )3 − (x1/3 )3 = (z 1/3 − x1/3 )(z 2/3 + z 1/3 x1/3 + x2/3 ). Therefore 18 z 2/3 − x2/3 z 1/3 + x1/3 2x1/3 2 = 2 /3 → = x−1/3 1/3 x1/3 + z 2/3 z−x 3 z +z 3x2/3 as x → x. C03S0M.118: The volume of the block is V = x2 y , and V is constant while x and y are functions of time t (in minutes). So 0= dV dy dx = 2xy + x2 . dt dt dt (1) We are given dy/dt = −2, x = 30, and y = 20, so by Eq. (1) dx/dt = 3 . Answer: At the time in question 2 the edge of the base is increasing at 1.5 cm/min. C03S0M.119: The balloon has volume V = 4 π r3 and surface area A = 4π r2 where r is its radius and V , 3 A, and r are all functions of time t. We are given dV /dt = +10, and we are to find dA/dt when r = 5. dV dr = 4π r2 , dt dt so 10 = 4π · 25 · Thus dr 10 1 = = . dt 100π 10π Also dA dr = 8π r , dt dt dr . dt dA dt r =5 = 8π · 5 · and therefore 1 = 4. 10π Answer: At 4 in. /s. 2 C03S0M.120: Let the nonnegative x-axis represent the ground and the nonnegative y -axis the wall. Let x be the distance from the base of the wall to the foot of the ladder; let y be the height of the top of the ladder above the ground. From the Pythagorean theorem we obtain x2 + y 2 = 100, so x Thus dx dy +y = 0. dt dt dy x dx dx 5280 22 =− · . We are given = = ft/s, and at the time when y = 1, we have dt y dt dt 3600 15 √ x = 100 − (0.01)2 = 99.9999 . At that time, dy dt y =0.01 =− √ 99.9999 22 · ≈ −1466.666 (ft/s), 0.01 15 almost exactly 1000 mi/h. This shows that in reality, the top of the ladder cannot remain in contact with the wall. If it is forced to do so by some latching mechanism, then a downward force much greater than that caused by gravity will be needed to keep the bottom of the latter moving at the constant rate of 1 mi/h. C03S0M.121: Let Q be the amount of water in the cone at time t, r the radius of its upper surface, and h its height. From similar triangles we find that h = 2r, so 19 Q= 12 2 1 πr h = πr3 = π h3 . 3 3 12 Now −50 = Therefore dQ 1 dh dh 200 = π h2 , so =− . dt 4 dt dt 36π dh 50 =− ≈ −1.7684 (ft/min). dt 9π C03S0M.122: Let x denote the distance from plane A to the airport, y the distance from plane B to the airport, and z the distance between the two aircraft. Then z 2 = x2 + y 2 + (3 − 2)2 = x2 + y 2 + 1 and dx/dt = −500. Now 2z dz dx dy = 2x + 2y , dt dt dt and when x = 2, y = 2. Therefore z = 3 at that time. Therefore, 3 · (−600) = 2 · (−500) + 2 · and thus dy dt x=2 C03S0M.123: is a clock! dy dt , x=2 = −400. Answer: Its speed is 400 mi/h. dV dy √ dy √ =3 y = −3 y , so = −1. Answer: At 1 in./min—a constant rate. The tank dt dt dt C03S0M.124: As in the solution of Problem 121, we find that when the height of water in the tank is y , 1 its volume is V = 12 π y 3 . For part (a), we have dV dy 1 √ +50 − 10 y = = πy2 . dt 4 dt So when y = 5, √ 25 dy 50 − 10 5 = π 4 dt , y =5 √ 1 (40 − 8 5) ≈ 1.40766 (ft/min). In part (b), 5π y =5 dV dy 1 √ = 25 − 10 y = π y 2 ; (1) dt 4 dt √ dy/dt = 0 when 25 = 10 y , so that y = 6.25 (ft) would seem to be the maximum height ever attained by the water. What actually happens is that the water level rises more and more slowly as time passes, approaching the limiting height of 6.25 ft as a right-hand limit, but never reaching it. This is not obvious; you must solve the differential equation in (1) (use the substitution y = u2 ) and analyze the solution to establish this conclusion. and therefore dy dt = 20 a2 − y0 = 2a, a consequence a − x0 of the two-point formula for slope and the fact that the line is tangent to the parabola at Q. Hence a2 − 2ax0 + y0 = 0. Think of this as a quadratic equation in the unknown a. It has two real solutions when the discriminant is positive: (x0 )2 − y0 > 0, and this establishes the conclusion in part (b). There are no real solutions when (x0 )2 − y0 < 0, and this establishes the conclusion in part (c). What if (x0 )2 − y0 = 0? C03S0M.125: The straight line through P (x0 , y0 ) and Q(a, a2 ) has slope 21 Section 4.2 C04S02.001: y = y (x) = 3x2 − 4x−2 : dy = 6x + 8x−3 , so dy = dx C04S02.002: y = y (x) = 2x1/2 − 3x−1/3 : 6x + 8 x3 dx. dy = x−1/2 + x−4/3 , so dx dy = x−1/2 + x−4/3 dx = 1 + x5/6 dx. x4/3 dy 1 = 1 − (4 − x3 )−1/2 · (−3x2 ), so dx 2 √ 3x2 3x2 + 2 4 − x3 √ √ 1+ dx = dx. 2 4 − x3 2 4 − x3 C04S02.003: y = y (x) = x − (4 − x3 )1/2 : dy = √ 1 − 1 x−1/2 1 1−2 x 2 √ : dy = − C04S02.004: y = y (x) = √ 2 dx = √ √ 2 dx. x− x (x − x ) 2 x (x − x ) C04S02.005: y = y (x) = 3x2 (x − 3)3/2 , so √ 3 9 dy = 6x(x − 3)3/2 + x2 (x − 3)1/2 dx = (7x2 − 12x) x − 3 dx. 2 2 C04S02.006: y = y (x) = x x2 + 4 (x2 − 4) − 2x2 dx = − 2 dx. , so dy = x2 − 4 (x2 − 4)2 (x − 4)2 C04S02.007: y = y (x) = x(x2 + 25)1/4 , so 1 3x2 + 50 dy = (x2 + 25)1/4 + x(x2 + 25)−3/4 · 2x dx = dx. 4 2(x2 + 25)3/4 C04S02.008: y = y (x) = (x2 − 1)−4/3 , so dy = − C04S02.009: y = y (x) = cos √ 8x dx. 3(x2 − 1)7/3 √ sin x dx. x, so dy = − √ 2x C04S02.010: y = y (x) = x2 sin x, so dy = (x2 cos x + 2x sin x) dx. C04S02.011: y = y (x) = sin 2x cos 2x, so dy = (2 cos2 2x − 2 sin2 2x) dx. C04S02.012: y = y (x) = (cos 3x)3 , so dy = −9 cos2 3x sin 3x dx. C04S02.013: y = y (x) = sin 2x 2x cos 2x − sin 2x dx. , so dy = 3x 3x2 C04S02.014: dy = (3x2 − 2x3 )e−2x dx. C04S02.015: y = y (x) = 1 x cos x + sin x dx. , so dy = 1 − x sin x (1 − x sin x)2 1 C04S02.016: dy = 1 − ln x dx. x2 C04S02.017: f (x) = 1 , so f (0) = 1. Therefore (1 − x)2 f (x) = C04S02.018: f (x) = − 1 ≈ f (0) + f (0)(x − 0) = 1 + 1 · x = 1 + x. 1−x 1 1 , so f (0) = − . Therefore 2 2(1 + x)3/2 f (x) = √ 1 1 ≈ f (0) + f (0)(x − 0) = 1 − x. 2 1+x C04S02.019: f (x) = 2(1 + x), so f (0) = 2. Therefore f (x) = (1 + x)2 ≈ f (0) + f (0)(x − 0) = 1 + 2x. C04S02.020: f (x) = −3(1 − x)2 , so f (0) = −3. Therefore f (x) = (1 − x)3 ≈ f (0)+ f (0)(x − 0) = 1 − 3x. √ C04S02.021: f (x) = −3 1 − 2x, so f (0) = −3; f (x) = (1 − 2x)3/2 ≈ f (0) + f (0)(x − 0) = 1 − 3x. C04S02.022: f (0) = −1, so L(x) = 1 − x. C04S02.023: If f (x) = sin x, then f (x) = cos x, so that f (0) = 1. Therefore f (x) = sin x ≈ f (0) + f (0)(x − 0) = 0 + 1 · x = x. C04S02.024: f (0) = 1, so L(x) = x. 1 1 C04S02.025: Choose f (x) = x1/3 and a = 27. Then f (x) = , so that f (a) = . So the linear 2 /3 27 3x 1 approximation to f (x) near a = 27 is L(x) = 2 + x. Hence 27 √ 79 3 25 = f (25) ≈ L(25) = ≈ 2.9259. 27 A calculator reports that f (25) is actually closer to 2.9240, but the linear approximation is fairly accurate, with an error of only about −0.0019. 1 1 x and a = 100. Then f (x) = √ , so that f (a) = . So the linear 20 2x 1 approximation to f (x) near a = 100 is L(x) = 5 + x. Hence 20 √ 101 102 = f (102) ≈ L(102) = = 10.1000. 10 C04S02.026: Choose f (x) = √ A calculator reports that f (25) is actually closer to 10.0995, but the linear approximation is quite accurate, with an error of only about −0.0005. 1 1 C04S02.027: Choose f (x) = x1/4 and a = 16. Then f (x) = , so that f (a) = . So the linear 32 4x3/4 3 1 approximation to f (x) near a = 16 is L(x) = + x. Hence 2 32 2 √ 4 63 = 1.96875. 32 15 = f (15) ≈ L(15) = A calculator reports that f (15) is actually closer to 1.96799. C04S02.028: Choose f (x) = √ x and a = 81. Then f (x) = 1 1 √ , so that f (a) = . So the linear 18 2x 9 1 + x. Hence 2 18 √ 161 80 = f (80) ≈ L(80) = ≈ 8.9444. 18 approximation to f (x) near a = 81 is L(x) = A calculator reports that f (15) is actually closer to 8.9443. 2 1 C04S02.029: Choose f (x) = x−2/3 and a = 64. Then f (x) = − 5/3 , so that f (a) = − . So the 1536 3x 5 1 linear approximation to f (x) near a = 64 is L(x) = − x. Hence 48 1536 95 65−2/3 = f (65) ≈ L(65) = ≈ 0.06185. 1536 A calculator reports that f (65) is actually closer to 0.06186. 3 1 Then f (x) = , so that f (a) = . So the linear 1 /4 4 4x 1 x. Hence 4 107 = f (80) ≈ L(80) = = 26.7500. 4 C04S02.030: Choose f (x) = x3/4 and a = 81. 27 approximation to f (x) near a = 81 is L(x) = + 4 803/4 A calculator reports that f (80) is actually closer to 26.7496. 45 1 π = π. 180 4 1√ the linear approximation to f (x) near a is L(x) = 2 2 C04S02.031: Choose f (x) = cos x and a = cos 43◦ = f 43 π 180 ≈L 1√ Then f (x) = − sin x, so that f (a) = − 2. So 2 √ 1 1 π + 1 − x 2. Hence 4 2 43 π 180 = π + 90 √ ≈ 0.7318. 90 2 A calculator reports that cos 43◦ is actually closer to 0.7314. C04S02.032: Choose f (x) = sin x and a = linear approximation to f (x) near a is 30 1 1√ π = π . Then f (x) = cos x, so that f (a) = 3. So the 180 6 2 √ 6−π 3 1√ L(x) = + x 3. 12 2 Hence 32 sin 32 = f π 180 ◦ 32 ≈L π 180 √ 90 + π 3 = ≈ 0.5302. 180 C04S02.033: Choose f (x) = ex and a = 0. Then f (x) = ex , so that f (a) = 1. So the linear approximation to f (x) near a is L(x) = x + 1. Thus e0.1 = f (0.1) ≈ L(0.1) = 1.1. A calculator reports that e0.1 ≈ 1.105171. 3 C04S02.034: Let f (x) = ln x and choose a = 1. Then the linear approximation to f near a is L(x) = x − 1, and L(1.1) = 0.1. (The actual value of f (1.1) is approximately 0.095310.) C04S02.035: Given x2 + y 2 = 1, we compute the differential of both sides and obtain 2x dx + 2y dy = 0; y dy = −x dx; dy x =− . dx y C04S02.036: ey dx + xey dy = 0, but ey is never zero, so x dy = − dx. Therefore dy 1 =− . dx x C04S02.037: Given x3 + y 3 = 3xy , we compute the differential of each side and obtain 3x2 dx + 3y 2 dy = 3y dx + 3x dy ; (y 2 − x) dy = (y − x2 ) dx; dy y − x2 =2 . dx y −x C04S02.038: Given: x ln y = 1. Then x dy + (ln y ) dx = 0, so y dy y ln y y exp(1/x) =− = −y (ln y )2 = − 2 = − . dx x x x2 C04S02.039: If f (x) = (1 + x)k , then f (x) = k (1 + x)k−1 , and so f (0) = k . approximation to f (x) near zero is L(x) = 1 + kx. Hence the linear C04S02.040: If C is the circumference of the circle and r its radius, then C = 2π r. Thus dC = 2π dr, and so ∆C ≈ 2π ∆r. With r = 10 and ∆r = 0.5, we obtain ∆C ≈ 2π (0.5) = π ≈ 3.1416. This happens to be the exact value as well (because C is a linear function of r). C04S02.041: If the square has edge length x and area A, then A = x2 . Therefore dA = 2x dx, and so ∆A ≈ 2x ∆x. With x = 10 and ∆x = −0.2, we obtain ∆A ≈ 2 · 10 · (−0.2) = −4. So the area of the square decreases by 4 in.2 . C04S02.042: The relationship between the surface area A and the radius r of the sphere is A = 4π r2 , and hence dA = 8π r dr. Thus ∆A ≈ 8π r ∆r. With r = 5 and ∆r = 0.2 we obtain ∆A ≈ 8π (5)(0.2) = 8π ≈ 25.1327 square inches. The true value is approximately 25.6354 square inches. C04S02.043: A [right circular] cylinder of base radius r and height h has volume V = π r2 h, and hence dV = π r2 dh + 2π rh dr. Therefore V ≈ π r2 ∆h + 2π rh ∆r. With r = h = 15 and ∆r = ∆h = −0.3 we find that ∆V ≈ (225π )(−0.3) + (450π )(−0.3) = 405 π , so the volume of the cylinder decreases by approximately 2 636.17 cm3 . C04S02.044: With volume V , height h, and radius r, we have V = 1 π r2 h. Because r = 14 is constant, 3 we may think of V as a function of r alone, so that 4 dV = 12 π r dh. 3 With r = 14, h = 7, and dh = 0.1, we find that dV = 1 π (196)(0.1) ≈ 20.5251. 3 The true value is exactly the same because V is a linear function of h. C04S02.045: Because θ = 45◦ , the range R of the shell is a function of its initial velocity v alone, and 1 R = 16 v 2 . Hence dR = 1 v dv . With v = 80 and dv = 1, we find that dR = 1 · 80 = 10, so the range is 8 8 increased by approximately 10 ft. C04S02.046: Because v is constant, R is a function of the angle of inclination θ alone, and hence dR = 12 v (cos 2θ) dθ. 8 With θ = π /4, dθ = π /180 (1◦ ), and v = 80, we obtain ∆R ≈ 1 π (6400)(0) = 0. 8 180 The true value of ∆R is approximately −0.2437 (ft). C04S02.047: Technically, if W = RI 2 , then dW = I 2 dR + 2IR dI . But in this problem, R remains constant, so that dR = 0 and hence dW = 2IR dI . We take R = 10, I = 3, and dI = 0.1, and find that dW = 6. So the wattage increases by approximately 6 watts. C04S02.048: With circumference C and radius r, we have C = 2π r. Therefore, given ∆r = +10, ∆C = 2π ∆r ≈ 20π (feet). Thus the wire should be lengthened by approximately 63 feet. C04S02.049: Let V be the volume of the ball and let r be its radius, so that V = 4 π r3 . Then the calculated 3 1 value of the volume is Vcalc = 4 (1000π ) ≈ 4188.7902 in.3 , whereas ∆V ≈ 4π (10)2 16 = 25π ≈ 78.5398 in.3 3 (the true value of ∆V is approximately 79.0317). C04S02.050: With volume V and radius r, we have V = 4 π r3 , and thus dV = 4π r2 dr. For |∆V | 3 require that 4π r2 |∆r| 1, so |∆r| 1, we 1 ≈ 0.0008 4π (10)2 inches. Thus the radius must be measured with error not exceeding 0.0008 inches. C04S02.051: With surface area S and radius r, we have S = 2π r2 (half the surface area of a sphere of radius r), so that dS = 4π r dr ≈ 4π (100)(0.01) = 4π . That is, ∆S ≈ 12.57 square meters. C04S02.052: With the notation of the preceding solution, we now require that |dS | S 0.0001; thus, at least approximately, 5 |4π r dr| |2π r2 | 0.0001. Hence 2 which implies that |dr| r dr r 0.0001, 0.00005. Answer: With percentage error not exceeding 0.005%. C04S02.053: We plotted f (x) = x2 and its linear approximation L(x) = 1 + 2(x − 1) on the interval [0.5, 1.5], and it was clear that the interval I = (0.58, 1.42) would be an adequate answer to this problem. We then used Newton’s nethod to find a “better” interval, which turns out to be I = (0.5528, 1.4472). √ C04S02.054: We plotted f (x) = x and its linear approximation L(x) = 1 + 1 (x − 1) on the interval 2 [0.3, 2.15], and it was clear that the interval I = (0.32, 1.98) would be adequate. We then used Newton’s method to “improve” the answer to I = (0.3056, 2.0944). C04S02.055: We plotted f (x) = 1/x and its linear approximation L(x) = 1 + 1 (2 − x) on the interval 2 4 [1.73, 2.32], and it was clear that the interval was a little too large to be a correct answer. We used Newton’s method to find more accurate endpoints, and came up with the answer I = (1.7365, 2.3035). Of course, any open subinterval of this interval that contains a = 2 is also a correct answer. 1 C04S02.056: We plotted f (x) = x1/3 and its linear approximation L(x) = 12 (x + 16) on the interval [6.3, 9.9] and it was thereby clear that the interval (6.5, 9.5) would suffice. We then used Newton’s method to find “better” endpoints and found (6.4023, 9.7976). C04S02.057: We plotted f (x) = sin x and its linear approximation L(x) = x on the interval [ −2.8, 2.8] and it was clear that the interval (−0.5, 0.5) would suffice. We then used Newton’s method to find “better” endpoints and found that (−0.6746, 0.6745) would suffice. C04S02.058: Graphically we find that |f (x) − (x + 1)| < 0.05 provided that −0.3339 < x < 0.3004, so I = (−0.3339, 0.3004). C04S02.059: We plotted f (x) = sin x and its linear approximation √ 2 π L(x) = 1− +x 2 4 on the interval [0.5, 1.1], and it was clear that the interval (0.6, 0.95) would be adequate. We then used Newton’s method to “improve” the endpoints and found that the interval (0.5364, 1.0151) would suffice. C04S02.060: We plotted f (x) = tan x and its linear approximation L(x) = 1 (2 − π + 4x) on the interval 2 [0.7, 0.9], from which it was clear that the interval (0.7, 0.85) would suffice. We then used Newton’s method to find “better” endpoints, with the result that the interval (0.6785, 0.8789) was nearly the best possible result. 6 Section 4.3 C04S03.001: f (x) = −2x; f is increasing on (−∞, 0) and decreasing on (0, +∞). Matching graph: (c). C04S03.002: f (x) = 2x − 2; f is increasing on (1, + ∞), decreasing on (−∞, 1). Matching graph: (b). C04S03.003: f (x) = 2x + 4; f is increasing on (−2, +∞), decreasing on (−∞, −2). Matching graph: (f). C04S03.004: f (x) = 3 x2 − 3; f (x) = 0 when x = ±2; f is increasing on (−∞, −2) and on (2, ∞), 4 decreasing on (−2, +2). Matching graph: (a). C04S03.005: f (x) = x2 − x − 2 = (x + 1)(x − 2); f (x) = 0 when x = −1 and when x = 2; f is increasing on (−∞, −1) and on (2, +∞), decreasing on (−1, 2). Matching graph: (d). C04S03.006: f (x) = 2 − 1 x − 1 x2 = − 1 (x2 + x − 6) = − 1 (x + 3)(x − 2); f (x) = 0 when x = −3 and 3 3 3 3 when x = 2. f is increasing on (−3, 2) and decreasing on (−∞, −3) and on (2, +∞). Matching graph: (e). C04S03.007: f (x) = 2x2 + C ; 5 = f (0) = C : f (x) = 2x2 + 5. C04S03.008: f (x) = 2x3/2 + C ; 4 = f (0) = C : f (x) = 2x3/2 + 4. C04S03.009: f (x) = − 1 1 + C ; 1 = f (1) = C − 1: f (x) = − + 2. x x C04S03.010: f (x) = −2e−3x + C ; the condition f (0) = 3 yields C = 5. C04S03.011: f (x) ≡ 3 > 0 for all x, so f is increasing for all x. C04S03.012: f (x) ≡ −5 < 0 for all x, so f is decreasing for all x. C04S03.013: f (x) = −4x, so f is increasing on (−∞, 0) and decreasing on (0, +∞). The graph of y = f (x) is shown next. 7.5 5 2.5 -3 -2 -1 1 2 3 -2.5 -5 -7.5 -10 C04S03.014: f (x) = 8x + 8 = 8(x + 1). Therefore f is increasing for x > −1 and decreasing for x < −1. 1 C04S03.015: f (x) = 6 − 4x. Therefore f is increasing for x < y = f (x) is shown next. 3 2 and decreasing for x > 3 . The graph of 2 4 2 1 2 3 -2 C04S03.016: f (x) = 3x2 − 12 = 3(x2 − 4) = 3(x + 2)(x − 2). Hence f is increasing for x > 2 and for x < −2, decreasing for x in the interval (−2, 2). C04S03.017: f (x) = 4x3 − 4x = 4x(x + 1)(x − 1). The intervals on which f (x) cannot change sign are x < −1, −1 < x < 0, 0 < x < 1, and 1 < x. Because f (−2) = −24, f (−0.5) = 1.5, f (0.5) = −1.5, and f (2) = 24, we may conclude that f is increasing if −1 < x < 0 or if x > 1, decreasing for x < −1 and for 0 < x < 1. The graph of y = f (x) is shown next. 4 3 2 1 -3 C04S03.018: Because f (x) = x > −1 and for x < −1. -2 -1 1 2 3 1 , f (x) > 0 for all x other than x = −1. Hence f is increasing for (x + 1)2 C04S03.019: f (x) = 12x3 + 12x2 − 24x = 12x(x + 2)(x − 1), so the only points where f (x) can change sign are −2, 0, and 1. If 0 < x < 1 then 12x > 0, x + 2 > 0, and x − 1 < 0, and therefore f (x) < 0 if 0 < x < 1. Therefore f is decreasing on the interval (0, 1). A similar analysis shows that f is also decreasing 2 on (−∞, −2) and is increasing on (−2, 0) and on (1, +∞). The graph of y = f (x) is shown next. 40 20 -4 -3 -2 -1 1 2 3 -20 C04S03.020: Note that f is continuous for all x, as is f (x) = f is increasing for all x. 2x2 + 1 ; also, f (x) > 0 for all x. Hence (x2 + 1)1/2 C04S03.021: After simplifications, f (x) = 2−x . 2ex/2 Hence f (x) can change sign only at x = 2. Therefore f is increasing if x < 2 and is decreasing if x > 2. The graph of y = f (x) is next. 0.8 0.6 0.4 0.2 1 2 3 4 5 -0.2 -0.4 -0.6 C04S03.022: Because f (x) = 2x(1 − x) , e2x f (x) can change sign only at x = 0 and at x = 1. Because f (−1) < 0, f (0.5) > 0, and f (2) < 0, f is 3 decreasing on (−∞, 0), increasing on (0, 1), and decreasing on (1, + ∞). The graph of y = f (x) is next. 0.3 0.25 0.2 0.15 0.1 0.05 -1 1 2 3 4 5 C04S03.023: Because f (x) = − (x − 1)(x − 3) , ex f (x) can change sign only at x = 1 and at x = 3. Also f (0) < 0, f (2) < 0, and f (4) < 0, so f is decreasing on (−∞, 1), increasing on (1, 3), and decreasing on (3, + ∞). The graph of y = f (x) is next. 0.5 0.4 0.3 0.2 0.1 1 2 3 4 C04S03.024: After simplifications, f (x) = 1 − ln(2x) . x2 Note that f (x) (and f (x)) are defined only for x > 0. Next, f (x) = 0 when ln 2x = 1; 2x = e; x= e . 2 Hence f (x) can change sign only when x = e/2 (x > 0). Because f (1) > 0 and f (3) < 0, f is increasing 4 on (0, e/2) and decreasing on (e/2, + ∞). The graph of y = f (x) is next. 0.6 0.4 0.2 2 4 6 8 -0.2 -0.4 -0.6 C04S03.025: f (0) = 0, f (2) = 0, f is continuous for 0 x 2, and f (x) = 2x − 2 exists for 0 < x < 2. To find the numbers c satisfying the conclusion of Rolle’s theorem, we solve f (c) = 0 to find that c = 1 is the only such number. C04S03.026: f (−3) = 81 − 81 = 0 = f (3), f is continuous everywhere, and f (x) = 18x − 4x3 exists for all x, including all x in the interval (−3, 3). Thus f satisfies the hypotheses of Rolle’s theorem. To find what √ value or values c might assume, we solve the equation f (x) = 0 to obtain the three values c = 0, c = 3 2, 2 √ and c = − 3 2. All three of these numbers lie in the interval (−3, 3), so these are the three possible values 2 for the number c whose existence is guaranteed by Rolle’s theorem. C04S03.027: Given: f (x) = 2 sin x cos x on the interval I = [0, π ]. Then f (x) = 2 cos2 x − 2 sin2 x = 2 cos 2x, and it is clear that all the hypotheses of Rolle’s theorem are satisfied by f on the interval I . Also f (x) = 0 when 2x = π /2 and when 2x = 3π /2, so the numbers whose existence is guaranteed by Rolle’s theorem are c = π /4 and c = 3π /4. C04S03.028: Here, f (x) = 10 −1/3 5 2/3 10 − 5x −x = ; x 3 3 3x1/3 f (x) exists for all x in (0, 5) and f is continuous on the interval 0 x 5 (the only point that might cause trouble is x = 0, but the limit of f and its value there are the same). Because f (0) = 0 = f (5), there is a solution c of f (x) = 0 in (0, 5), and clearly c = 2. C04S03.029: On the interval (−1, 0), f (x) = 1; on the interval (0, 1), we have f (x) = −1. Because f is not differentiable at x = 0, it does not satisfy the hypotheses of Rolle’s theorem, so there is no guarantee that the equation f (x) = 0 has a solution—and, indeed, it has no solution in (−1, 1). C04S03.030: Because f (x) = 2 3(2 − x)1/3 for x = 2, f (x) can never be zero, so the conclusion of Rolle’s theorem does not hold here. The reason is that f is not differentiable at every point of the interval 1 x 3; specifically, f (x) is undefined at x = 2. C04S03.031: If f (x) = xex , then f (1) = e = 0, so f does not satisfy the hypotheses of Rolle’s theorem on [0, 1]. Because f (x) = (x + 1)ex is zero only when x = −1, f also does not satisfy the conclusion of Rolle’s theorem there. 5 C04S03.032: It is clear that f satisfies the hypotheses of the mean value theorem, because every polynomial is continuous and differentiable everywhere. To find c, we solve f (1) − f (−1) 1+1 ; that is, 3c2 = . 1 − (−1) 1+1 √ √ Thus 3c2 = 1, with the two solutions c = 1 3 and c = − 1 3. Both these numbers lie in the interval 3 3 −1 x 1, so each is an answer to this problem. f (c) = C04S03.033: Here, f (x) = 6x + 6; f (−2) = −5 and f (1) = 4. So we are to solve the equation 6(c + 1) = 4 − (−5) = 3. 1 − (−2) 1 It follows that c = − . 2 C04S03.034: Here we note that f (x) = 1 2(x − 1)1/2 exists for all x > 1, so f satisfies the hypotheses of the mean value theorem for 2 solve x 5. To find c, we 1 (4)1/2 − (1)1/2 = ; 1 /2 5−2 2(c − 1) thus 2(c − 1)1/2 = 3, and so c = 13 4 = 3.25. Note that 2 < c < 5. C04S03.035: First, f (x) = 2 (x − 1)−1/3 is defined on (1, 2); moreover, f is continuous for 1 x 2 3 (the only “problem point” is x = 1, but the limit of f there is equal to its value there). To find c, we solve f (c) = 2 f (2) − f (1) 1−0 = = = 1. 2−1 1 3(c − 1)1/3 This leads to the equation (c − 1)1/3 = 2 , and thereby c = 3 35 27 . Note that c does lie in the interval (1, 2). C04S03.036: Because f (x) can be expressed as a rational function with denominator never zero if x = 0, it is both continuous and differentiable on (2, 3). Next, 3+ 1 − 2+ 1 1 5 3 2 = = 2 c 3−2 6 √ √ yields the information that c2 = 6, and thus that c = + 6 (not − 6; it’s not in the interval (2, 3)). f (c) = 1 − C04S03.037: First, f (x) = |x − 2| is not differentiable at x = 2, so does not satisfy the hypotheses of the mean value theorem on the given interval 1 x 4. Wherever f (x) is defined, its value is 1 or −1, but f (4) − f (1) 2−1 1 = = 4−1 3 3 is never a value of f (x). So f satisfies neither the hypotheses nor the conclusion of the mean value theorem on the interval 1 x 4. C04S03.038: Because f is not differentiable at x = 1, the hypotheses of the mean value theorem do not hold. The only values of f (x) are 1 (for x > 1) and −1 (for x < 1). Neither of these is equal to the everage slope of f on the interval 0 x 3: 6 f (3) − f (0) 3−2 1 = =, 3−0 3 3 so the conclusion of the mean value theorem also fails to hold. C04S03.039: The greatest integer function is continuous at x if and only if x is not an integer. Consequently all the hypotheses of the mean value theorem fail here: f is discontinuous at −1, 0, and 1, and also f (0) does not exist because (for one reason) f is not continuous at x = 0. Finally, the average slope of the graph of f is 1, but f (x) = 0 wherever it is defined. Thus the conclusion of the mean value theorem also fails to hold. C04S03.040: The function f (x) = 3x2/3 is continuous everywhere, but its derivative f (x) = 2x−1/3 does not exist at x = 0. Because f (x) does not exist for all x in (−1, 1), an essential hypothesis of the mean value theorem is not satisfied. Moreover, f (x) is never zero (the average slope of the graph of f on the interval −1 x 1), so the conclusion of the mean value theorem also fails to hold. C04S03.041: Let f (x) = x5 + 2x − 3. Then f (x) = 5x4 + 2, so f (x) > 0 for all x. This implies that f is an increasing continuous function, and therefore f (x) can have at most one zero in any interval. To show that f has at least one zero in the interval 0 x 1, it is sufficient to notice that f (1) = 0. Therefore the equation f (x) = 0 has exactly one solution in [0, 1]. C04S03.042: Let f (x) = e−x − x + 1. Then f (x) = −e−x − 1, so f (x) is decreasing and continuous on the interval [1, 2]. Therefore the equation f (x) = 0 has at most one solution in that interval. But f (1) = e−1 > 0 and f (2) = e−2 − 1 < 0. Because f is continuous, it has the intermediate value property, so f (x) has at least one zero in [1, 2]. Therefore the equation f (x) = 0 has exactly one solution there. That is, e−x = x − 1 for exactly one number x in [1, 2]. C04S03.043: Let f (x) = −3+ x ln x. Note that 1 < ln 2 < 1 because ln 2 ≈ 0.693147. Hence 1 < 2 ln 2 < 2, 2 so f (2) < 0. Also ln 4 = 2 ln 2, so 1 < ln 4 < 2, and hence 4 < 4 ln 4 < 8. Therefore f (4) > 0. Because f is differentiable on [2, 4], it is continuous there. Hence, by the intermediate value property of continuous functions, f (c) = 0 for some number c in [2, 4]. Thus c ln c = 3, and therefore the equation x ln x = 3 as at least one solution in [2, 4]. Next, f (x) = 1 + ln x is positive on [2, 4] (use inequalities similar to those in the previous paragraph if this is not perfectly clear), and therefore f is increasing on that interval. Hence the equation f (x) = 0 has at most one solution in [2, 4]. In conclusion, the equation x ln x = 3 has exactly one solution in [2, 4]. (By Newton’s method, the solution is approximately 2.85739078.) C04S03.044: Let f (x) = sin x − 3x + 1. Because f (x) = −3 + cos x is always negative, the graph of f is decreasing on every interval of real numbers, and in particular is decreasing on [ −1, 1]. Hence the equation f (x) = 0 can have at most one solution in that interval. Moreover, f (−1) ≈ 3.16 > 0 and f (1) ≈ −1.16 < 0. Because f is continuous on [ −1, 1], the intermediate value property of continuous functions guarantees that the equation f (x) = 0 has at least one solution in [ −1, 1]. So the equation f (x) = 0 has exactly one solution there. C04S03.045: The car traveled 35 miles in 18 minutes, which is an average speed of 250 ≈ 83.33 miles per 3 hour. By the mean value theorem, the car must have been traveling over 83 miles per hour at some time between 3:00 p.m. and 3:18 p.m. C04S03.046: A change of 15 miles per hour in 10 minutes is an average change of 1.5 miles per hour per minute, which is an average change of 90 miles per hour per hour. By the mean value theorem, the 7 instantaneous rate of change of velocity must have been exactly 90 miles per hour per hour at some time in the given 10-minute interval. C04S03.047: Let f (t) be the distance that the first car has traveled from point A on its way to point B at time t, with t measured in hours and with t = 0 corresponding to 9:00 a.m. (so that t = 1 corresponds to 10:00 a.m.). Let g (t) be the corresponding function for the second car. Let h(t) = f (t) − g (t). We make the very plausible assumption that the functions f and g are differentiable on (0, 1) and continuous on [0, 1], so h has the same properties. In addition, h(0) = f (0) − g (0) = 0 and h(1) = 0 as well. By Rolle’s theorem, h (c) = 0 for some c in (0, 1). But this implies that f (c) = g (c). That is, the velocity of the first car is exactly the same as that of the second car at time t = c. C04S03.048: Because f (0) does not exist, the function f (x) = x2/3 does not satisfy the hypotheses of the mean value theorem on the given interval. But consider the equation f (c) = f (27) − f (−1) ; 27 − (−1) (1) that is, 2 9−1 2 = =, 28 7 3c1/3 which leads to c1/3 = 7/3, and thus to c = satisfying Eq. (1). 343 27 ≈ 12.7. Because −1 < c < 27, there is indeed a number c C04S03.049: Because f (x) = 3 3 3√ 1+x −1 , (1 + x)1/2 − = 2 2 2 it is clear that f (x) > 0 for x > 0. Also f (0) = 0; it follows that f (x) > 0 for all x > 0. That is, 3 (1 + x)3/2 > 1 + x for x > 0. 2 C04S03.050: Proof: Suppose that f (x) is the constant K on the interval a x b. Let g (x) = Kx + f (a) − Ka. Then the graph of g is a straight line, and g (x) = K for all x. Consequently f and g differ by a constant on the interval a x b. But g (a) = Ka + f (a) − Ka = f (a), so g (x) = f (x) for all x in that interval. Therefore the graph of f is a straight line. C04S03.051: Proof: Suppose that f (x) is a polynomial of degree n − 1 on the interval I = [ a, b ]. Then f (x) has the form f (x) = an−1 xn−1 + an−2 xn−2 + . . . + a2 x2 + a1 x + a0 where an−1 = 0. Note that f (x) is the derivative of the function g (x) = 1 1 1 1 an−1 xn + an−2 xn−1 + . . . + a2 x3 + a1 x2 + a0 x. n n−1 3 2 8 By Corollary 2, f (x) and g (x) can differ only by a constant, and this is sufficient to establish that f (x) must also be a polynomial, and one of degree n because the coefficient of xn in f (x) is the same as the coefficient of xn in g (x), and that coefficient is nonzero. C04S03.052: Suppose that f (x) = 0 for x = x1 , x2 , . . . , xk in the interval [ a, b ]. By Rolle’s theorem, f (x) = 0 for some c1 in (x1 , x2 ), some c2 in (x2 , x3 ), . . . , and some ck−1 in (xk−1 , xk ). The numbers c1 , c2 , . . . , ck−1 are distinct because they lie in nonoverlapping intervals, and this proves the desired result. C04S03.053: First note that f (x) = 1 x−1/2 , and that the hypotheses of the mean value theorem are all 2 satisfied for the given function f on the given interval [100, 101]. Thus there does exist a number c between 100 and 101 such that √ 1 f (101) − f (100) √ = = 101 − 100. 1 /2 101 − 100 2c √ √ √ √ Therefore 1/ (2 c ) = 101 − 10, and thus we have shown that 101 = 10 + 1/ (2 c ) for some number c in (100, 101). √ √ Proof for part (b): If 0 c 10, then 0 c 100; because c > 100, we see that 0 c 10 is √ impossible. If 10.5 c then 110.25 c, which is also impossible because c < 110. So we may conclude √ that 10 < c < 10.5. Finally, 10 < √ c < 10.5 implies that √ 20 < 2 c < 21. Consequently 1 1 1 < 1 /2 < , 21 20 2c so 10 + √ 1 1 < 101 < 10 + . 21 20 The decimal expansion of 1/21 begins 0.047619047619 . . . , and therefore 10.0476 < √ 101 < 10.05. C04S03.054: Let f (x) = x7 + x5 + x3 + 1. Then f (−1) = −2, f (1) = 4, and f (x) = 7x6 + 5x4 + 3x2 . Now f (x) > 0 for all x except that f (0) = 0, so f is increasing on the set of all real numbers. This information together with the fact that f (continuous) has the intermediate value property establishes that the equation f (x) = 0 has exactly one [real] solution (approximately −0.79130272). C04S03.055: Let f (x) = (tan x)2 and let g (x) = (sec x)2 . Then f (x) = 2(tan x)(sec2 x) and g (x) = 2(sec x)(sec x tan x) = f (x) on (−π /2, π /2). Therefore there exists a constant C such that f (x) = g (x) + C for all x in (−π /2, π /2). Finally, f (0) = 0 and g (0) = 1, so C = f (0) − g (0) = −1. C04S03.056: The mean value theorem does not apply here because f (0) does not exist. C04S03.057: The average slope of the graph of f on the given interval [ −1, 2] is f (2) − f (−1) 5 − (−1) = =2 2 − (−1) 3 and f satisfies the hypotheses of the mean value theorem there. Therefore f (c) = 2 for some number c, −1 < c < 2. This implies that the tangent line to the graph of f at the point (c, f (c)) has slope 2 and is therefore parallel to the line with equation y = 2x because the latter line also has slope 2. 9 C04S03.058: To show that the graph of f (x) = x4 − x3 + 7x2 + 3x − 11 has a horizontal tangent line, we must show that its derivative f (x) = 4x3 − 3x2 + 14x + 3 has the value zero at some number c. Now f (x) is a polynomial, thus is continuous everywhere, and so has the intermediate value property; moreover, f (−1) = −18 and f (0) = 3, so f (c) = 0 for some number c in (−1, 0). (The value of c is approximately −0.203058.) C04S03.059: Use the definition of the derivative: g (0) = lim h→0 g (0 + h) − g (0) h = lim 1 1 + h2 sin 2h = lim 1 + h sin 2 h→0 h→0 = 1 +0 2 = 1 h 1 h 1 > 0. 2 (by the squeeze law) If x = 0 then g (x) = 1 + 2x sin 2 1 x − cos 1 x . Because cos(1/x) oscillates between +1 and −1 near x = 0 and 2x sin(1/x) is near zero for x close to zero, it follows that every interval about x = 0 contains subintervals on which g (x) > 0 and subintervals on which g (x) < 0. They are not clearly visible near x = 0 in the graph of y = g (x) (shown next) because they are very short intervals. 0.075 0.05 0.025 -0.2 -0.1 0.1 0.2 -0.025 -0.05 -0.075 C04S03.060: To prove a mathematical result it is frequently very helpful to restate exactly what it is that you must prove. In this case, to show that f is increasing on the unbounded open interval (2, + ∞), we need to show that if 2 < x1 < x2 , then f (x1 ) < f (x2 ). Suppose that 2 < x1 < x2 . Let a= 2 + x1 2 and b = 1 + x2 . Then [ a, b ] is a closed interval, and by hypothesis f is increasing there. Moreover, a < x1 (because a is the midpoint of the interval (2, x1 )), x1 < x2 , and x2 < b. So x1 and x2 are two numbers in [ a, b ] for which 10 x1 < x2 . Therefore f (x1 ) < f (x2 ). This is what we have agreed that it means for f to be increasing on the unbounded interval (2, + ∞). This concludes the proof. C04S03.061: Let h(x) = 1 − 1 x2 − cos x. Then h (x) = −x + sin x. By Example 8, sin x < x for all x > 0, 2 h(x) − 0 so h (x) < 0 for all x > 0. If x > 0, then = h (c) for some c > 0, so h(x) < 0 for all x > 0; that is, x−0 12 cos x > 1 − 2 x for all x > 0. C04S03.062: (a): Let j (x) = sin x − x + 1 x3 . Then 6 1 j (x) = cos x − 1 + x2 . 2 By the result in Problem 61, j (x) > 0 for all x > 0. Also, if x > 0, then Hence j (x) > 0 for all x > 0; that is, sin x > x − 1 x3 for all x > 0. 6 j (x) − 0 = j (c) for some c > 0. x−0 (b) By part (a) and Example 8 of the text, 1 x − x3 < sin x < x 6 for all x > 0. So π 1 − 36 6 π 36 3 < sin π π < ; 36 36 0.0871557 < sin 5◦ < 0.0872665; sin 5◦ ≈ 0.087. C04S03.063: (a): Let K (x) = 1 − 1 x2 + 2 14 24 x − cos x. Then 1 1 K (x) = −x + x3 + sin x = sin x − (x − x3 ). 6 6 By Problem 62, part (a), K (x) > 0 for all x > 0. So if x > 0, K (x) > 0 for all x > 0. That is, 1 1 cos x < 1 − x2 + x4 2 24 K (x) − 0 = K (c) for some c > 0. Therefore x−0 for all x > 0. (b): By Problem 61 and part (a), 1 1 1 1 − x2 < cos x < 1 − x2 + x4 2 2 24 for all x > 0. In particular, 1− 1 2 π 18 2 < cos π 1 <1− 18 2 π 18 2 + 1 24 π 18 4 ; hence 0.984769 < cos 10◦ < 0.984808. So cos 10◦ ≈ 0.985. C04S03.064: (a) Let h(x) = e−x − (1 − x) = e−x + x − 1 for x 0. Then h (x) = −e−x + 1 > 0 if x > 0 because e−x < 1 if x > 0. Hence h(x) is increasing for x > 0. But h(0) = 0 and h is continuous for all x 0, so h(x) > 0 if x > 0. That is, e−x > p1 (x) = 1 − x if x > 0. 11 (b) Let j (x) = 1 − x + 1 x2 − e−x for x 2 0. Then j (x) = −1 + x + e−x = e−x − (1 − x), which was shown positive for x > 0 in part (a). Because j (0) = 0 and j is continuous for x follows that j (x) > 0 if x > 0. That is, e−x < p2 (x) = 1 − x + (c) Let k (x) = e−x − 1 − x + 1 x2 − 1 x3 for x 2 6 12 x 2 if x > 0. 0. Then k (x) = −e−x + 1 − x + 12 x = p2 (x) − e−x . 2 By the result in part (b), k (x) > 0 if x > 0. But k is continuous for x Therefore k (x) > 0 for all x > 0. That is, e−x > p3 (x) = 1 − x + 0, it now 12 13 x− x 2 6 if 0 and k is increasing for x > 0. x > 0. (d) Continue in like fashion, or use induction to show that p2n−1 (x) < e−x < p2n (x) for every positive integer n and all x > 0. Substitution of x = 1 and n = 4 yields p7 (1) < 1 < p8 (1), e and thus (because all quantities involved are positive) 1 1 <e< . p8 (1) p7 (1) With the aid of a calculator (round up on the left, down on the right) we find thereby that 2.718263331 < e < 2.718446603. Therefore e ≈ 2.718 to three places. With the aid of a powerful computer and a symbolic algebra program you could use this method to show that e ≈ 2.71828182845904523536. 12 Section 4.4 C04S04.001: f (x) = 2x − 4; x = 2 is the only critical point. Because f (x) > 0 for x > 2 and f (x) < 0 for x < 2, it follows that f (2) = 1 is the global minimum value of f (x). The graph of y = f (x) is shown next. 10 8 6 4 2 -1 1 2 3 4 5 C04S04.002: f (x) = 6 − 2x, so x = 3 is the only critical point. If x < 3 then f is increasing, whereas f is decreasing for x > 3, so f (3) = 9 is the global maximum value of f (x). C04S04.003: f (x) = 3x2 − 6x = 3x(x − 2), so x = 0 and x = 2 are the only critical points. If x < 0 or if x > 2 then f (x) is positive, but f (x) < 0 for 0 < x < 2. So f (0) = 5 is a local maximum and f (2) = 1 is a local minimum. The graph of y = f (x) is shown next. 10 7.5 5 2.5 -2 -1 1 2 3 4 -2.5 -5 C04S04.004: f (x) = 3x2 − 3 = 3(x + 1)(x − 1), so x = 1 and x = −1 are the only critical points. If x < −1 or if x > 1, then f (x) > 0, whereas f (x) < 0 on (−1, 1). So f (−1) = 7 is a local maximum value and f (1) = 3 is a local minimum value. C04S04.005: f (x) = 3x2 − 6x + 3 = 3(x − 1)2 , so x = 1 is the only critical point of f . f (x) > 0 if x = 1, so the graph of f is increasing for all x; so f has no extrema of any sort. The graph of y = f (x) is shown 1 next. 6.6 6.4 6.2 -2 -1 1 2 3 4 5.8 5.6 5.4 C04S04.006: f (x) = 6x2 + 6x − 36 = 6(x + 3)(x − 2), so x = −3 and x = 2 are the only critical points. If x < −3 or if x > 2 then f (x) > 0, but f (x) < 0 on the interval (−3, 2). So f (−3) = 98 is a local maximum value of f (x) and f (2) = −27 is a local minimum value. C04S04.007: f (x) = −6(x − 5)(x + 2); f (x) < 0 if x < −2 and if x > 5, but f (x) > 0 for −2 < x < 5. Hence f (−2) = −58 is a local minimum value of f and f (5) = 285 is a local maximum value. The graph of y = f (x) is shown next. 400 300 200 100 -6 -4 -2 2 4 6 8 -100 C04S04.008: f (x) = −3x2 , so x = 0 is the only critical point. But f (x) < 0 if x = 0, so f is decreasing everywhere. Therefore there are no extrema. C04S04.009: f (x) = 4x(x − 1)(x + 1); f (x) < 0 for x < −1 and on the interval (0, 1), whereas f (x) > 0 for x > 1 and on the interval (−1, 0). Consequently, f (−1) = −1 = f (1) is the global minimum value of f (x) and f (0) = 0 is a local maximum value. Note that the [unique] global minimum value occurs at two 2 different points on the graph of f , which is shown next. 1.5 1 0.5 -2 -1 1 2 -0.5 -1 C04S04.010: f (x) = 15x2 (x + 1)(x − 1), so f (x) > 0 if x < −1 and if x > 1, but f (x) < 0 on (−1, 0) and on (0, 1). Therefore f (0) = 0 is not an extremum of f (x), but f (−1) = 2 is a local maximum value and f (1) = −2 is a local minimum value. The graph of y = f (x) is shown next. 2 1 -1 -0.5 0.5 1 -1 -2 C04S04.011: f (x) = 1 − 9x−2 , so the critical points occur where x = −3 and x = 3 (horizontal tangents); note that f is not defined at x = 0. If x2 > 9 then f (x) > 0, so f is increasing if x > 3 and if x < −3. If x2 < 9 then f (x) < 0, so f is decreasing on (−3, 0) and on (0, 3). Therefore f (−3) = −6 is a local maximum value and f (3) = 6 is a local minimum value for f (x). The graph of y = f (x) is next. 20 10 -4 -2 2 -10 -20 C04S04.012: Here, 3 4 f (x) = 2x − 2 2(x3 − 1) 2(x − 1)(x2 + x + 1) = = . x2 x2 x2 Because x2 + x + 1 > 0 for all x, the only critical point is x = 1; note that f is not defined at x = 0. Also f (x) has the sign of x − 1, so f (x) > 0 for x > 1 and f (x) < 0 for 0 < x < 1 and for x < 0. Consequently f (1) = 3 is a local minimum value of f (x). It is not a global minimum because f (x) → −∞ as x → 0− . The graph of y = f (x) is shown next. 40 20 -4 -2 2 4 -20 -40 C04S04.013: If f (x) = xe−2x , then f (x) = e−2x − 2xe−2x = 1 − 2x . e2x The only critical point occurs when x = 1 . The graph of f is increasing to the left of this point and 2 decreasing to the right, so f has a global maximum there. The graph of y = f (x) is next. 0.4 0.2 -0.5 0.5 1 1.5 2 -0.2 -0.4 -0.6 -0.8 -1 C04S04.014: If f (x) = x2 e−x/3 , then f (x) = 2xe−x/3 − 1 2 −x/3 x(6 − x) = . xe 3 3ex/3 Hence there are critical points at x = 0 and at x = 6. Because f (−2) ≈ −10.3879 < 0, f (3) ≈ 1.10364 > 0, and f (8) ≈ −0.37058 < 0, the graph of f is decreasing for x < 0 and for x > 6, increasing on (0, 6). Because f (x) 0 for all x and f (x) = 0 only if x = 0, there is a global minimum at (0, 0). Because f (x) → + ∞ as x → −∞, the maximum where x = 6 is local but not global. C04S04.015: If f (x) = (x + 4)2 e−x/5 , then 4 f (x) = 1 (6 − x)(4 + x) (24 + 2x − x2 )e−x/5 = . 5 5ex/5 Therefore the only critical points occur where x = −4 and x = 6. Because f (−5) ≈ −5.98022 < 0, f (0) = 4.8 > 0, and f (7) ≈ −0.54251 < 0, the graph of f is decreasing if x < −4 and if x > 6, increasing on (−4, 6). Because f (x) 0 for all x and f (x) = 0 only if x = −4, there is a global minimum at (−4, 0). Because f (x) → + ∞ as x → −∞, the maximum where x = 6 is local but not global. The graph of y = f (x) is next. 30 25 20 15 10 5 -5 -2.5 2.5 5 7.5 10 C04S04.016: If f (x) = 1 − ln x , x then f (x) = −2 + ln x . x2 So the only critical point occurs where x = e2 ≈ 7.38906. Because f (3) ≈ −0.10015 < 0 and f (12) ≈ 0.00337 > 0, the graph of f is decreasing if 0 < x < e2 and increasing if x > e2 . Hence there is a global extremum where x = e2 . C04S04.017: f (x) = 2 sin x cos x; f (x) = 0 when x is any integral multiple of π /2. In (0, 3), f (x) = 0 when x = π /2. Because f (x) > 0 if 0 < x < π /2 and f (x) < 0 if π /2 < x < 3, f (x) has the global maximum value f (π /2) = 1. C04S04.018: f (x) = −2 sin x cos x; f (x) = 0 when x = 0 and when x = π /2. f is increasing on (−1, 0) and on (π /2, 3), whereas f is decreasing on (0, π /2). So f has a global maximum at (0, 1) and a global minimum at (π /2, 0). C04S04.019: f (x) = 3 sin2 x cos x; f (x) = 0 when x = −π /2, 0, or π /2. f is decreasing on (−3, −π /2) and on (π /2, 3), but increasing on (−π /2, π /2). So f has a global minimum at (−π /2, −1) and a global maximum at (π /2, 1); there is no extremum at the critical point (0, 0). C04S04.020: f (x) = −4 cos3 x sin x vanishes at π /2 and at π . f is decreasing on (0, π /2) and on (π , 4), but increasing on (π /2, π ). Hence there is a global minimum at (π /2 , 0) and a global maximum at (π , 1). C04S04.021: f (x) = x sin x; f (x) = 0 at −π , 0, and π . f (x) > 0 on (−π , π ), f (x) < 0 on (−5, −π ) and on (π , 5). So f has a global maximum at (π , π ) and a global minimum at (−π , −π ). Note that the critical point (0, 0) is not an extremum. C04S04.022: Given: f (x) = cos x + x sin x on I = (−5, 5). First, f (x) = x cos x, so f (x) = 0 (for x in I ) when 5 x = 0, π x=± , 2 x=± 3π . 2 Hence f (x) < 0 on (−5, −3π /2), (−π /2, 0), and (π /2, 3π /2); f (x) > 0 on (−3π /2, −π /2), (0, π /2), and on (3π /2, 5). Therefore there are global minima at (−3π /2, −3π /2) and (3π /2, −3π /2), global maxima at (−π /2, π /2) and (π /2, π /2), and a local minimum at (0, 1). The global minima are global rather than local because f (±5) ≈ −4.510959. C04S04.023: If f (x) = ln x , x2 then f (x) = 1 − 2 ln x . x3 So f (x) = 0 when ln x = 1 ; that is, x = e1/2 . Because f (x) < 0 if 0 < x < e1/2 and f (x) > 0 when 2 x > e1/2 , the critical point at x = e1/2 is a global maximum. The graph of y = f (x) is next. 0.4 0.2 1 2 3 4 -0.2 -0.4 -0.6 -0.8 -1 C04S04.024: If f (x) = ln(1 + x) , 1+x then f (x) = 1 − ln(1 + x) . (1 + x)2 The only critical point, which occurs where f (x) = 0, is (e − 1, 1/e). Because f (1) ≈ 0.07671 > 0 and f (2) ≈ −0.01096 < 0, there is a global maximum at this critical point and there are no other extrema. C04S04.025: If f (x) = ex sin x, then f (x) = ex (cos x + sin x). Hence f (x) = 0 in (−3, 3) when tan x = −1, thus at x = a = −π /4 and at x = b = 3π /4. Because f (x) < 0 if −3 < x < a and if b < x < 3, the graph of f is decreasing on these intervals; it is increasing on (a, b). Inspection of the values of f (x) near −3 and near 3 shows that f (x) has a global minimum value at x = a and a global maximum value at x = b. C04S04.026: If f (x) = x3 exp(−x − x2 ), then f (x) = x2 (1 − x)(3+2x) exp(−x − x2 ). Hence f (x) = 0 when x = − 3 , when x = 0, and when x = 1. The corresponding values of f (x) are − 27 e−3/4 , 0, and e−2 . We 2 8 note that f (−2) ≈ −1.62402 < 0, f (−1) = 2 > 0, f (0.5) ≈ 0.23618 > 0, and that f (1.5) ≈ −15875 < 0. Hence the graph of f is decreasing on (−3, −1.5), increasing on (−1.5, 1), and decreasing on (1, 3). Because f (x) is near zero near the endpoints of its domain, there is a global minimum where x = −1.5 and a global maximum where x = 1. The point (0, 0) is not an extremum. C04S04.027: Let x be the smaller of the two numbers; then the other is x + 20 and their product is f (x) = x2 + 20x. Consequently f (x) = 2x + 20, so x = −10 is the only critical point of f . The graph of f is decreasing for x < −10 and increasing for x > −10. Therefore (−10, −100) is the lowest point on the graph of f . Answer: The two numbers are −10 and 10. 6 C04S04.028: We assume that the length turned upward is the same on each side—call it y . If the width of the gutter is x, then we have the constraint xy = 18, and we are to minimize the width x + 2y of the strip. Its width is given by the function f (x) = x + 36 , x x > 0, for which f (x) = 1 − 36 . x2 The only critical point in the domain of f is x = 6, and if 0 < x < 6 then f (x) < 0, whereas f (x) > 0 for x > 6. Thus x = 6 yields the global minimum value f (6) = 12 of the function f . Answer: The minimum possible width of the strip is 12 inches. C04S04.029: Let us minimize g (x) = (x − 3)2 + (3 − 2x − 2)2 = (x − 3)2 + (1 − 2x)2 , the square of the distance from (x, y ) on the line 2x + y = 3 to the point (3, 2). We have g (x) = 2(x − 3) − 4(1 − 2x) = 10x − 10, so x = 1 is the only critical point of g (x). If x > 1 then g (x) > 0, but g (x) < 0 for x < 1. Thus x = 1 minimizes g (x), and so the point on the line 2x + y = 3 closest to the point (3, 2) is (1, 1). As an independent check, note that the slope of the line segment joining (3, 2) and (1, 1) is 1 , whereas the slope of the line 2 2x + y = 3 is −2, so the segment and the line are perpendicular; see Miscellaneous Problem 70 of Chapter 3. C04S04.030: Base of box: x wide, 2x long. Height: y . Then the box has volume 2x2 y = 576, so y = 288x−2 . Its total surface area is A = 4x2 + 6xy , so we minimize A = A(x) = 4x2 + 1728 , x x > 0. Now A (x) = 8x − 1728 , x2 so the only critical point of A(x) occurs when 8x3 = 1728; that is, when x = 6. It is easy to verify that A (x) < 0 for 0 < x < 6 and A (x) > 0 for x > 6. Therefore A(6) is the global minimum value of A(x). Also, when x = 6 we have y = 8. Answer: The dimensions of the box of minimal surface area are 6 inches wide by 12 inches long by 8 inches high. C04S04.031: Base of box: x wide, 2x long. Height: y . Then the box has volume 2x2 y = 972, so y = 486x−2 . Its total surface area is A = 2x2 + 6xy , so we minimize A = A(x) = 2x2 + 2916 , x Now A (x) = 4x − 7 2916 , x2 x > 0. so the only critical point of A(x) occurs when 4x3 = 2916; that is, when x = 9. It is easy to verify that A (x) < 0 for 0 < x < 9 and that A (x) > 0 for x > 9. Therefore A(9) is the global minimum value of A(x). Answer: The dimensions of the box are 9 inches wide, 18 inches long, and 6 inches high. C04S04.032: If the radius of the base of the pot is r and its height is h (inches), then we are to minimize the total surface area A given the constraint π r2 h = 125. Thus h = 125/(π r2 ), and so A = π r2 + 2π rh = A(r) = π r2 + 250 , r r > 0. Now A (r) = 2π r − 250 ; r2 √ A (r) = 0 when r3 = 125/π , so that r = 5/ 3 π . The latter point is the only value of r at which A (r) can change sign (for r > 0), and it is easy to see that A (r) is positive when r is large positive, whereas A (r) is negative when r is near zero. Therefore we have located the global minimum of A(r), and it occurs when √ √ the pot has radius r = 5/ 3 π inches and height h = 5/ 3 π inches. Thus the pot will have its radius equal to its height, each approximately 3.414 inches. C04S04.033: Let r denote the radius of the pot and h its height. We are given the constraint π r2 h = 250, so h = 250/(π r2 ). Now the bottom of the pot has area π r2 , and thus costs 4π r2 cents. The curved side of the pot has area 2π rh, and thus costs 4π rh cents. So the total cost of the pot is C = 4π r2 + 4π rh; thus C = C (r) = 4π r2 + 1000 , r r > 0. Now C (r) = 8π r − 1000 ; r2 √ C (r) = 0 when 8π r3 = 1000, so that r = 5/ 3 π . It is clear that this is the only (positive) value of r at which C (r) can change sign, and that C (r) < 0 for r positive and near zero, but C (r) > 0 for r large positive. √ Therefore we have found the value of r that minimizes C (r). The corresponding value of h is 10/ 3 π , so the pot of minimal cost has height equal to its diameter, each approximately 6.828 centimeters. C04S04.034: If (x, y ) = (x, 4 − x2 ) is a point on the parabola y = 4 − x2 , then the square of its distance from the point (3, 4) is h(x) = (x − 3)2 + (4 − x2 − 4)2 = (x − 3)2 + x4 . We minimize the distance by minimizing its square: h (x) = 2(x − 3) + 4x3 ; h (x) = 0 when 2x3 + x − 3 = 0. It is clear that h (1) = 0, so x − 1 is a factor of h (x); h (x) = 0 is equivalent to (x − 1)(2x2 + 2x + 3) = 0. The quadratic factor in the last equation is always positive, so x = 1 is the only critical point of h(x). Also h (x) < 0 if x < 1, whereas h (x) > 0 for x > 1, so x = 1 yields the global minimum value h(1) = 5 for h(x). When x = 1 we have y = 3, so the point on the parabola y = 4 − x2 √ closest to (3, 4) is (1, 3), at distance 5 from it. 100 C04S04.035: If the sides of the rectangle are x and y , then xy = 100, so that y = . Therefore the x perimeter of the rectangle is 8 P = P (x) = 2x + 200 , x x > 0. Then P (x) = 2 − 200 ; x2 P (x) = 0 when x = 10 (−10 is not in the domain of P (x)). Now P (x) < 0 on (0, 10) and P (x) > 0 for x > 10, and so x = 10 minimizes P (x). A little thought about the behavior of P (x) for x near zero and for x large makes it clear that we have found the global minimum value for P : P (10) = 40. When x = 10, also y = 10, so the rectangle of minimal perimeter is indeed a square. C04S04.036: Let x denote the length of each side of the square base of the solid and let y denote its height. Then its total volume is x2 y = 1000. We are to minimize its total surface area A = 2x2 + 4xy . Now 1000 y = 2 , so x 4000 A = A(x) = 2x2 + , x > 0. x Therefore dA 4000 = 4x − 2 . dx x The derivative is zero when 4x3 = 4000; that is, when x = 10. Also A(x) is decreasing on (0, 10) and increasing for x > 10. So x = 10 yields the global minimum value of A(x). In this case, y = 10 as well, so the solid is indeed a cube. C04S04.037: Let the square base of the box have edge length x and let its height be y , so that its total volume is x2 y = 62.5 and the surface area of this box-without-top will be A = x2 + 4xy . So A = A(x) = x2 + 250 , x x > 0. Now A (x) = 2x − 250 , x2 so A (x) = 0 when x3 = 125: x = 5. In this case, y = 2.5. Also A (x) < 0 if 0 < x < 5 and A (x) > 0 if x > 5, so we have found the global minimum for A(x). Answer: Square base of edge length 5 inches, height 2.5 inches. C04S04.038: Let r denote the radius of the can and h its height (in centimeters). We are to minimize its total surface area A = 2π r2 + 2π rh given the constraint π r2 h = V = 16π . First we note that h = V /(π r2 ), so we minimize A = A(r) = 2π r2 + 2V , r Now A (r) = 4π r − 9 2V ; r2 r > 0. A (r) = 0 when 4π r3 = 2V = 32π —that is, when r = 2. Now A(r) is decreasing on (0, 2) and increasing for r > 2, so the global minimum of A(r) occurs when r = 2, for which h = 4. C04S04.039: Let x denote the radius and y the height of the cylinder (in inches). Then its cost (in cents) is C = 8π x2 + 4π xy , and we also have the constraint π x2 y = 100. So C = C (x) = 8π x2 + 400 , x x > 0. Now dC/dx = 16π x − 400/(x2 ); dC/dx = 0 when x = (25/π )1/3 (about 1.9965 inches) and consequently, when y = (1600/π )1/3 (about 7.9859 inches). Because C (x) < 0 if x3 < 25/π and C (x) > 0 if x3 > 25/π , we have indeed found the dimensions that minimize the total cost of the can. For simplicity, note that y = 4x at minimum: The height of the can is twice its diameter. C04S04.040: If the print width is x and its height is y (in inches), then the page area is A = (x + 2)(y + 4). We are to minimize A given xy = 30. Because y = 30/x, A = A(x) = 4x + 38 + 60 , x x > 0. Now A (x) = 4 − 60 ; x2 √ √ √ A (x√ = 0 when x = 15. But A (x) > 0 for x > 15 whereas √ (x) < 0 for 0 < x < 15. Therefore ) A x = 15 yields the global minimum value of A(x), which is 38 + 8 15, approximately 68.98 square inches. C04S04.041: Let (x, y ) = (x, x2 ) denote an arbitrary point on the curve. The square of its distance from (0, 2) is then f (x) = x2 + (x2 − 2)2 . Now f (x) = 2x(2x2 − 3), and therefore f (x) = 0 when x = 0, when x = − 3/2, and when x = + 3/2. Now f (x) < 0 if x < − 3/2 and if 0 < x < 3/2; f (x) > 0 if − 3/2 < x < 0 and if x > 3/2. Therefore x = 0 yields a local maximum for f ; the other two zeros of f (x) yield its global minimum. Answer: There are exactly two points on the curve that are nearest (0, 2); they are (+ 3/2, 3/2) and (− 3/2, 3/2). C04S04.042: Let (a, 0) and (0, b) denote the endpoints of the segment and denote its point of tangency by (c, 1/c). The segment then has slope −1/c2 , and therefore b−0 (1/c) − 0 1 =− 2 = . 0−a c c−a It follows that b = a/c2 and that c = a − c, so a = 2c and b = 2/c. The square of the length of the segment is then f (c) = 4c2 + 4 , c2 c > 0. Now f (c) = 8c − 10 8 ; c3 f (c) = 0 when c = −1 and when c = 1. We reject the negative solution. On the interval (0, 1), f is decreasing; f√ increasing for c > 1. Therefore c = 1 gives the segment of minimal length, which is is L = f (1) = 2 2. C04S04.043: If the dimensions of the rectangle are x by y , and the line segment bisects the side of length x, then the square of the length of the segment is f (x) = x 2 2 4096 x2 + 2, 4 x + y2 = x > 0, because y = 64/x. Now x 8192 − 3. 2 x √ √ When f (x) = 0, we must have x = +8 2, so that y √ 4 2. We have found the minimum of f because if = √ √ 0 < x < 8 2 then f (x) < 0, and f (x) > 0 if x > 8 2. The minimum length satisfies L2 = f 8 2 , so that L = 8 centimeters. f (x) = C04S04.044: Let y be the height of the cylindrical part and x the length of the radii of both the cylinder and the hemisphere. The total surface area is A = π x2 + 2π xy + 2π x2 = 3π x2 + 2π xy. But the can must have volume V = π x2 y + 2 π x3 , so 3 y= 1000 − 2 π x3 3 . π x2 Therefore A = A(x) = 5 2 2000 πx + , 3 x x > 0. Thus dA 10 2000 = πx − 2 . dx 3 x Now dA/dx = 0 when x = (600/π )1/3 ≈ 5.7588. Because dA/dx < 0 for smaller values of x and dA/dx > 0 for larger values, we have found the point at which A(x) attains its global minimum value. After a little arithmetic, we find that y = x, so the radius of the hemisphere and the radius and height of the cylinder should all be equal to (600/π )1/3 to attain minimal surface area. This argument contains the implicit assumption that y > 0. If y = 0, then x = (1500/π )1/3 ≈ 7.8159, for which A = (150)(18π )1/3 ≈ 575.747 cubic inches. But with x = y = (600/π )1/3 , we have A = (100)(45π )1/3 ≈ 520.940 cubic inches. So the solution in the first paragraph indeed yields the dimensions of the can requiring the least amount of material. 11 C04S04.045: If the end of the rod projects the distance y into the narrower hall, then we have the proportion y/2 = 4/x by similar triangles. So y = 8/x. The square of the length of the rod is then f (x) = (x + 2)2 + 4 + 8 x 2 x > 0. , It follows that f (x) = 4 + 2x − 64 128 − 3, x2 x √ and that f (x) = 0 when (x + 2)(x3 − 32) = 0. The only admissible solution is x = 3 32, which indeed minimizes f (x) by the usual argument (f (x) is very large positive if x is either large positive or positive and very close to zero). The minimum length is √ √ 3 3 L = 20 + 12 4 + 12 16 1 /2 ≈ 8.323876 (meters). C04S04.046: By similar triangles, y/1 = 8/x, and L1 + L2 = L = (x + 1)2 + (y + 8)2 1 /2 . We minimize L by minimizing f (x) = L2 = (x + 1)2 + 8 + f (x) = 2 + 2x − 8 x 2 , x > 0. 128 128 − 2; x3 x f (x) = 0 when 2x3 + 2x4 − 128 − 128x = 0, which leads to the equation (x + 1)(x − 4)(x2 + 4x + 16) = 0. The only relevant solution is x = 4. Because f (x) < 0 for x in the interval (−1, 4) and f (x) > 0 if x > 4, we have indeed found the global minimum of f . The corresponding value of y is 2, and the length of the √ shortest ladder is L = 5 5 feet, approximately 11 ft 2 in. C04S04.047: If the pyramid has base edge length x and altitude y , then its volume is V = 1 x2 y . From 3 Fig. 4.4.30 we see also that 2y = tan θ x and a = cos θ y−a where θ is the angle that each side of the pyramid makes with its base. It follows, successively, that 12 a y−a 2 = cos2 θ; sin2 θ = 1 − cos2 θ = = y 2 − 2ay y (y − 2a) = . (y − a)2 (y − a)2 y (y − 2a) sin θ = y−a y= x sin θ = 2 cos θ 1 /2 . y (y − 2a) y−a x 2 2y = x a x2 = Therefore (y − a)2 − a2 (y − a)2 1/2 y−a ; a 4a2 y 2 . y (y − 2a) V= y (y − 2a) ; 12 4a2 y 2 x y = V (y ) = , 3 3(y − 2a) y > 2a. Now dV 24a2 y (y − 2a) − 12a2 y 2 . = dy 9(y − 2a)2 The condition dV /dy = 0 then implies that 2(y − 2a) = y , and thus that y = 4a. Consequently the minimum volume of the pyramid is V (4a) = (4a2 )(16a2 ) 32 3 = a. (3)(2a) 3 The ratio of the volume of the smallest pyramid to that of the sphere is then 32/3 32 8 = =. 4π /3 4π π C04S04.048: Let x denote the distance from the noisier of the two discos. Let K be the “noise proportionality” constant. The noise level at x is then N (x) = 4K K + . x2 (1000 − x)2 N (x) = − 8K 2K + . 3 x (1000 − x)3 Now N (x) = 0 when 4(1000 − x)3 = x3 ; if so, it follows that x= (1000)41/3 ≈ 613.512. 1 + 4 1 /3 13 Because the noise level is very high when x is near zero and when x is near 1000, the last value of x minimizes the noise level—the quietest point is about 613.5 feet from the noisier of the two discos. C04S04.049: Let z be the length of the segment from the top of the tent to the midpoint of one side of its base. Then x2 + y 2 = z 2 . The total surface area of the tent is A = 4x2 + (4)( 1 )(2x)(z ) = 4x2 + 4xz = 4x2 + 4x(x2 + y 2 )1/2 . 2 Because the [fixed] volume V of the tent is given by V= 1 4 (4x2 )(y ) = x2 y, 3 3 we have y = 3V /(4x2 ), so A = A(x) = 4x2 + 1 (16x6 + 9V 2 )1/2 . x After simplifications, the condition dA/dx = 0 takes the form 8x 16x6 + 9V 2 1/2 − 1 16x6 + 9V 2 + 48x4 = 0, x2 √ which has solution x = 2−7/6 3 3V . Because this is the only positive solution of the equation, and because it is clear that neither large values of x nor values of x near zero will yield small values of the surface area, this is the desired value of x. C04S04.050: By similar triangles in Fig. 4.4.28, we have y/a = b/x, and thus y = ab/x. If L denotes the length of the ladder, then we minimize L2 = f (x) = (x + a)2 + (y + b)2 = (x + a)2 + b2 1 + a x 2 x > 0. , Now f (x) = 2(x + a) + 2b2 1 + a x − a ; x2 f (x) = 0 when x3 = ab2 , so that x = a1/3 b2/3 and y = a2/3 b1/3 . It’s clear that f is differentiable on its domain and that f (x) → +∞ as x → 0+ and as x → +∞. Therefore we have minimized L. With these values of x and y , we find that L = a2 + y 2 1/2 + x2 + b2 = a2/3 a2/3 + b2/3 = a2/3 + b2/3 3 /2 1 /2 1/2 = a2 + a4/3 b2/3 + b2/3 a2/3 + b2/3 1 /2 1/2 + a2/3 b4/3 + b2 = a2/3 + b2/3 1 /2 a2/3 + b2/3 1 /2 . Note that the answer is dimensionally correct. C04S04.051: Let x denote the length of each edge of the square base of the box and let y denote the height of the box. Then x2 y = V where V is the fixed volume of the box. The surface total area of this closed box is A = 2x2 + 4xy, and hence A(x) = 2x2 + 14 4V , x 0 < x < + ∞. Then 4V 4(x3 − V ) = , x2 x2 A (x) = 4x − so A (x) = 0 when x = V 1/3 . This is the only critical point of A, and A (x) < 0 if x is near zero while A (x) > 0 if x is large positive. Thus the global minimum value of the surface area occurs at this critical point. And if x = V 1/3 , then the height of the box is y= V V = 2/3 = V 1/3 = x, x2 V and hence the closed box with square base, fixed volume, and minimal surface area is a cube. C04S04.052: Let x denote the length of each edge of the square base of the box and let y denote its height. Then the box has fixed volume V = x2 y . The total surface area of the open box is A = x2 + 4xy , and hence A(x) = x2 + 4V , x 0 < x < + ∞. Next, A (x) = 2x − 4V 2(x3 − 2V ) = . x2 x2 Thus A (x) = 0 when x = (2V )1/3 . This critical point yields the global minimum value of A(x) because A (x) < 0 when x is small positive and A (x) > 0 when x is large positive. And at this critical point, we have y= V 2V 1 1 = = (2V )1/3 = x. x2 2 2 2 · (2V )2/3 Therefore the box with square base, no top, and fixed volume has minimal surface area when its height is half the edge length of its base. C04S04.053: Let r denote the radius of the base (and top) of the closed cylindrical can and let h denote its height. Then its fixed volume is V = π r2 h and its total surface area is A = 2π r2 + 2π rh. Hence A(r) = 2π r2 + 2π r · V 2V = 2π r 2 + , 2 πr r 0 < r < + ∞. Then A (r) = 4π r − 2V 4π r3 − 2V = ; 2 r r2 A (r) = 0 when r = (V /2π )1/3 . This critical point minimizes total surface area A(r) because A (r) < 0 when r is small positive and A (r) > 0 when r is large positive. And at this critical point we have h= V V 2 · (V /2π ) = = = 2(V /2π )1/3 = 2r. 2 2/3 πr π (V /2π ) (V /2π )2/3 Therefore the closed cylindrical can with fixed volume and minimal total surface area has height equal to the diameter of its base. 15 C04S04.054: Let r denote the radius of the circular base of the cylindrical can and let h denote its height. Then its fixed volume is V = π r2 h and the total surface area of the open cylindrical can is A = π r2 + 2π rh. Therefore A(r) = π r2 + 2π r · V 2V = πr2 + , πr2 r 0 < r < + ∞. Now A (r) = 2π r − 2V 2π r3 − 2V = , r2 r2 so A (r) = 0 when r = (V /π )1/3 . This critical point yields the global minimum value of A(r) because A (r) < 0 when r is small positive and A (r) > 0 when r is large positive. Moreover, at this critical point we have h= V V V /π = = = (V /π )1/3 = r. 2 2/3 πr π (V /π ) (V /π )2/3 Therefore the open cylindrical can with fixed volume and minimal total surface area has height equal to the radius of its base. C04S04.055: Finding the exact solution of this problem is quite challenging. In the spirit of mathematical modeling we accept the very good approximation that—if the thickness of the material of the can is small in comparison with its other dimensions—the total volume of material used to make the can may be approximated sufficiently accurately by multiplying the area of the bottom by its thickness, the area of the curved side by its thickness, the area of the top by its thickness, then adding these three products. Thus let r denote the radius of the inside of the cylindrical can, let h denote the height of the inside, and let t denote the thickness of its bottom and curved side; 3t will be the thickness of its top. The total (inner) volume of the can is the fixed number V = π r2 h. The amount of material to make the can will (approximately, but accurately) M = π r2 t + 2π rht + 3π r2 t = 4π r2 t + 2π rht, (1) so that M (r) = 4π r2 t + 2π rt · V = πr2 4π r2 + 2V r · t, 0 < r < L, where L is some rather large positive number that we don’t actually need to evaluate. (You can find L from Eq. (1) by setting h = 0 there and solving for r in terms of M and t.) Next, M (r) = 8π r − 2V r2 ·t = 2t(4π r3 − V ) ; r2 M (r) = 0 when r = (V /4π )1/3 . This critical point yields the global minimum value of M (r) because M (r) < 0 when r is small positive and M (r) > 0 when (V /2π )1/3 < r < L. (You need to verify that, under reasonable assumptions about the relative sizes of the linear measurements, L > (V /4π )1/3 .) And at this critical point, we have h= 4 · (V /4π ) V = = 4 · (V /4π )1/3 = 4r. πr2 (V /4π )2/3 16 Therefore the pop-top soft drink can of fixed internal volume V , with thickness as described in the problem, and using the minimal total volume of material for its top, bottom, and curved side, will have height (approximately) twice the diameter of its base. In support of this conclusion, the smallest commonly available pop-top can of a popular blend of eight vegetable juices has height about 9 cm and base diameter about 5 cm. Most of the 12-oz pop-top soft drink cans we measured had height about 12.5 cm and base diameter about 6.5 cm. C04S04.056: Let each edge of the square base of the box have length x and let y denote the height of the box. Then the fixed volume of the box is V = x2 y . The cost of its six faces is then (2x2 + 4xy ) · a and the cost to glue the edges together is (8x + 4y ) · b. Hence the total cost of material and construction will be C = (2x2 + 4xy ) · a + (8x + 4y ) · b. Because y = V /x2 , we have C (x) = 2x2 + 4x · V V · a + 8x + 4 · 2 · b = 2 x x 2x2 + 4V x · a + 8x + 4V x2 · b, 0 < x < + ∞. Next, C (x) = 4x − 4V x2 ·a+ 8− 8V x3 ·b = 4ax(x3 − V ) + 8b(x3 − V ) 4(ax + 2b)(x3 − V ) = . 3 x x3 Because x > 0, the only significant critical point of C (x) occurs when x = V 1/3 . Clearly C (x) < 0 when x is small positive and C (x) > 0 if x > V 1/3 . Therefore we have found the value of x that yields the global minimum value of the cost of the box. The corresponding value of the height of the box is y= V V = 2/3 = V 1/3 = x. x2 V Therefore the box of Problem 56 of minimal cost is a cube. It is remarkable and quite unexpected that the (positive) values of a and b do not affect the answer. 17 Section 4.5 C04S05.001: f (x) → +∞ as x → +∞, f (x) → −∞ as x → −∞. Matching graph: 4.5.13(c). C04S05.002: f (x) → +∞ as x → +∞, f (x) → +∞ as x → −∞. Matching graph: 4.5.13(a). C04S05.003: f (x) → −∞ as x → +∞, f (x) → +∞ as x → −∞. Matching graph: 4.5.13(d). C04S05.004: f (x) → −∞ as x → +∞, f (x) → −∞ as x → −∞. Matching graph: 4.5.13(b). C04S05.005: y (x) = 4x − 10, so the only critical point is because y (x) < 0 if x < 5 and y (x) > 0 if x > 5 . 2 2 5 2, − 39 , the lowest point on the graph of y 2 C04S05.006: The only critical point occurs where x = 3 . The graph is increasing if x < 3 , decreasing if 2 2 x > 3. 2 C04S05.007: y (x) = 12x2 − 6x − 90 is zero when x = − 5 and when x = 3. The graph of y is increasing 2 if x < − 5 , decreasing if − 5 < x < 3, and increasing if x > 3. Consequently there is a local maximum at 2 2 (−2.5, 166.75) and a local minimum at (3, −166). C04S05.008: The only critical points occur where x = − 7 and where x = 2 between them and decreasing otherwise. 5 3. The graph is increasing C04S05.009: y (x) = 12x3 + 12x2 − 72x = 12(x − 2)x(x + 3), so there are critical points at P (−3, −149), at Q(0, 40), and R(2, −24). The graph is decreasing to the left of P and between Q and R; it is increasing otherwise. C04S05.010: The critical points occur where x = − 8 , where x = 0, and where x = 5 . The graph is 3 2 increasing on − ∞, − 8 and on 0, 5 . The graph is decreasing on − 8 , 0 and on 5 , + ∞ . 3 2 3 2 C04S05.011: y (x) = 15x4 − 300x2 + 960 = 15(x + 4)(x + 2)(x − 2)(x − 4), so there are critical points at P (−4, −512), Q(−2, −1216), R(2, 1216), and S (4, 512). The graph is increasing to the left of P , between Q and R, and to the right of S ; it is decreasing otherwise. C04S05.012: The critical points occur at x = −5, x = −2, x = 0, x = 2, and x = 5. The graph of y is decreasing on (−∞, −5), on (−2, 0), and on (2, 5). The graph is increasing on (−5, −2), on (0, 2), and on (5, + ∞). C04S05.013: y (x) = 21x6 − 420x4 + 1344x2 = 21(x + 4)(x + 2)x2 (x − 2)(x − 4), so there are critical points at P (−4, 8192), Q(−2, −1280), R(0, 0), S (2, 1280), and T (4, −8192). The graph is increasing to the left of P , between Q and S (there is no extremum at R), and to the right of T ; it is decreasing otherwise. C04S05.014: The critical points occur where x = −3, x = −2, x = 0, x = 2, and x = 3. The graph of y is decreasing on (−∞, −3), on (−2, 0), and on (2, 3). It is increasing on (−3, −2), on (0, 2), and on (3, + ∞). 1 C04S05.015: f (x) = 6x − 6, so there is a critical point at (1, 2). Because f (x) < 0 if x < 1 and f (x) > 0 if x > 1, there is a global minimum at the critical point. The graph of y = f (x) is shown next. 20 15 10 5 -1 1 2 3 C04S05.016: f (x) = −8 − 4x is positive for x < −2, negative for x > −2. The graph is a parabola opening downward, with vertical axis, and vertex (and global maximum) at (−2, 13). C04S05.017: f (x) = 3(x2 − 4). There is a local maximum at (−2, 16) and a local minimum at (2, −16); neither is global. The graph of y = f (x) is shown next. 30 20 10 -4 -2 2 4 -10 -20 -30 C04S05.018: The function f is increasing on the set of all real numbers because f (x) = 3x2 + 3 is positive for all x. Thus f has no extrema of any kind. C04S05.19: f (x) = 3x2 − 12x + 9 = 3(x − 1)(x − 3), so f is increasing for x < 1 and for x > 3, decreasing for 1 < x < 3. It has a local maximum at (1, 4) and a local minimum at (3, 0). Its graph is shown next. 10 7.5 5 2.5 -2 2 4 6 -2.5 -5 C04S05.020: f (x) = 3x2 + 12x + 9 = 3(x + 1)(x + 3) is positive for x > −1 and for x < −3, negative for −3 < x < −1. So there is a local maximum at (−3, 0) and a local minimum at (−1, −4). There are intercepts at (−3, 0) and (0, 0). C04S05.021: f (x) = 3(x2 + 2x + 3) is positive for all x, so the graph of f is increasing on the set of all 2 real numbers; there are no extrema. The graph of f is shown next. 40 20 -4 -3 -2 -1 1 2 -20 -40 C04S05.022: f (x) = 3(x + 3)(x − 3): Local maximum at (−3, 54), local minimum at (3, −54). C04S05.023: f (x) = 2(x − 1)(x + 2)(2x + 1); there are global minima at (−2, 0) and (1, 0) and a local maximum at − 1 , 81 . The minimum value 0 is global because [clearly] f (x) = (x − 1)2 (x + 2)2 0 for all 2 16 x. The maximum value is local because f (x) → + ∞ as x → ± ∞. The graph of y = f (x) is next. 12 10 8 6 4 2 -4 -3 -2 -1 1 2 3 C04S05.024: f (x) = 2(x − 2)(2x + 3)(4x − 1): Global minimum value 0 at x = −1.5 and at x = 2, local maximum at (0.25, 37.515625). 3(1 − x) √ , so f (x) > 0 if 0 < x < 1 and f (x) < 0 if x > 1. Therefore there is a 2x local minimum at (0, 0) and a global maximum at (1, 2). The minimum is only local because f (x) → −∞ as x → + ∞. The graph of y = f (x) is shown next. C04S05.025: f (x) = 2 1 1 2 3 4 5 -1 -2 -3 -4 C04S05.026: Given: f (x) = x2/3 (5 − x): f (x) = 2 −1/3 2(5 − x) 10 − 2x − 3x 5(2 − x) (5 − x) − x2/3 = − x2/3 = = . x 1 /3 1 /3 3 3x 3x 3x1/3 Hence f (x) can change sign only at x = 2 and at x = 0. It’s clear that f is increasing for 0 < x < 2, decreasing for x < 0 and for x > 2. Thus there is a local maximum at (2, f (2)) and a local minimum at 3 (0, 0). Note that f (0) does not exist, but that f is continuous at x = 0. Neither extremum is global because f (x) → + ∞ as x → −∞ and f (x) → −∞ as x → + ∞. C04S05.027: f (x) = 15x2 (x − 1)(x + 1), so f is increasing for x < −1 and for x > 1, decreasing for −1 < x < 1. Hence there is a local maximum at (−1, 2) and a local minimum at (1, −2). The critical point at (0, 0) is not an extremum. The graph of y = f (x) is shown next. 6 4 2 -1.5 -1 -0.5 0.5 1 1.5 -2 -4 -6 C04S05.028: f (x) = 4x2 (x + 3) is positive for x > −3 and negative for x < −3; there is a horizontal tangent but no extremum at x = 0. There is a minimum at (−3, −27); it is global because f (x) = x4 + 4x3 = x4 1 + 4 x → +∞ as x → ± ∞. C04S05.029: f (x) = 4x(x − 2)(x + 2), so the graph of f is decreasing for x < −2 and for 0 < x < 2; it is increasing if −2 < x < 0 and if x > 2. Therefore the global minimum value −9 of f (x) occurs at x = ± 2 and the extremum at (0, 7) is a maximum, but not global because f (x) → +∞ as x → ± ∞. The graph of y = f (x) is shown next. 15 10 5 -3 -2 -1 1 2 3 -5 C04S05.030: Given f (x) = 1 = x−1 , x we see that f (x) = −x−2 = − 1 . x2 Therefore f (x) is negative for all x = 0, so f (x) is decreasing for all x = 0; there is an infinite discontinuity at x = 0. There are no extrema and no intercepts. Note that as x increases without bound, f (x) approaches zero. In sketching the graph of f it is very helpful to note that lim x→0− 1 = −∞ x and that lim+ x→0 1 = + ∞. x C04S05.031: Because f (x) = 4x − 3, there is a critical point at 3 , − 81 = (0.75, −10.125). The graph 4 8 of f is decreasing to the left of this point and increasing to its right, so there is a global minimum at this 4 critical point and no other extrema. The graph of y = f (x) is shown next. 20 15 10 5 -2 2 4 -5 -10 5 C04S05.032: The graph is a parabola, opening downward, vertical axis, vertex at − 12 , thus the highest point on the graph. 169 24 , which is C04S05.033: f (x) = 6(x − 1)(x + 2), so the graph of f is increasing for x < 2 and for x > 1, decreasing if −2 < x < 1. Hence there is a local maximum at (−2, 20) and a local minimum at (1, −7). The first is not a global maximum because f (10) = 2180 > 20; the second is not a global minimum because f (−10) = −1580 < −7. The graph of y = f (x) is shown next. 40 20 -4 -3 -2 -1 1 2 3 -20 C04S05.034: f (x) = 3x2 + 4 is positive for all x, so f (x) is increasing for all x; there are no extrema and (0, 0) is the only intercept. C04S05.035: f (x) = 6(5x − 3)(5x − 4), so there are critical points at (0.6, 16.2) and (0.8, 16). The graph of f is increasing for x < 0.6 and for x > 0.8; it is decreasing between these two points. Hence there is a local maximum at the first and a local minimum at the second. Neither is global because lim f (x) = lim x3 50 − x→∞ x→∞ 105 72 +2 x x = +∞ and, similarly, f (x) → −∞ as x → −∞. A graph of y = f (x) is shown next. 17 16.5 16 15.5 -0.5 0.5 1 5 1.5 C04S05.036: f (x) = 3(x − 1)2 is positive except at x = 1, so the graph is increasing for all x; there are no extrema, and the intercepts are at (0, −1) and (1, 0). C04S05.037: f (x) = 12x(x − 2)(x + 1), so f is decreasing for x < −1 and for 0 < x < 2, increasing for −1 < x < 0 and for x > 2. There is a local minimum at (−1, 3), a local maximum at (0, 8), and a global minimum at (2, −24). The latter is global rather than local because f (x) → + ∞ as x → ± ∞. The graph of f is shown next. 40 30 20 10 -2 -1 1 2 3 -10 -20 C04S05.038: f (x) = (x2 − 1)2 ; f (x) = 4x(x + 1)(x − 1). So f (x) is increasing for −1 < x < 0 and for x > 1, decreasing if x < −1 or if 0 < x < 1. The global minimum value is 0 = f (−1) = f (1) and there is a local maximum at (0, 1). C04S05.039: f (x) = 15x2 (x − 2)(x + 2), so f is increasing if | x | > 2 and decreasing if | x | < 2.There is a local maximum at (−2, 64) and a local minimum at (2, −64); the critical point at (0, 0) is not an extremum. The graph of y = f (x) appears next. 75 50 25 -2 -1 1 2 -25 -50 -75 C04S05.040: f (x) = 15(x + 1)(x − 1)(x + 2)(x − 2), so f is increasing if x < −2, if −1 < x < 1, and if x > 1, decreasing if −2 < x < −1 and if 1 < x < 2. So there are local maxima at (−2, −16) and (1, 38) and local minima at (−1, −38) and (2, 16). The only intercept is (0, 0). None of the extrema is global because lim f (x) = lim (3x5 − 25x3 + 60x) = lim x5 3 − x→∞ x→∞ x→∞ 25 60 +4 x2 x = +∞ and, similarly, f (x) → −∞ as x → −∞. C04S05.041: f (x) = 6(x2 + x + 1) is positive for all x, so the graph of f is increasing everywhere, with 6 no critical points and thus no extrema. The only intercept is (0, 0). The graph of f is shown next. 40 20 -3 -2 -1 1 2 -20 -40 C04S05.042: f (x) = 4x2 (x − 3), so f is increasing for x > 3 and decreasing for x < 3. Therefore there is a local (and global) minimum at (3, −27); (0, 0) and (4, 0) are the only intercepts. There is a horizontal tangent at (0, 0) but no extremum there. √ C04S05.043: √ (x) = 32x3 − 8x7 = √8x3 (x2 + 2)(x2 − 2), so the graph of f is increasing if x < − 2 f − √ and if 0√ x < 2 √ decreasing if − 2 < x < 0 and if x > 2. The global maximum value of f (x) is < but 16 = f 2 = f − 2 and there is a local minimum at (0, 0). The graph of f is shown next. 10 -2 -1 1 2 -10 -20 C04S05.044: Here we have f (x) = − 1 , 3x2/3 which is negative for all x = 0. Though f (0) is not defined, f is continuous at x = 0; careful examination of the behavior of f and f near zero shows that the graph has a vertical tangent at (0, 1); there are no extrema. C04S05.045: Given f (x) = x1/3 (4 − x), we find that f (x) = 1 −2/3 4(1 − x) (4 − x) − x1/3 = . x 3 3x2/3 Therefore f is increasing if x < 0 and if 0 < x < 1, decreasing if x > 1. Because f is continuous everywhere, including the point x = 0, it is also correct to say that f is increasing on (−∞, 1). The point (1, 3) is the highest point on the graph and there are no other extrema. Careful examination of the behavior of f (x) and f (x) for x near zero shows that there is a vertical tangent at the critical point (0, 0). The point (4, 0) is an 7 x-intercept. The graph of y = f (x) appears next. 2 -2 2 4 6 -2 -4 -6 C04S05.046: In this case f (x) = 8(x + 2)(x − 2) , 3x1/3 which is positive for x > 2 and for −2 < x < 0, negative for x < −2 and for 0 < x < 2. Note that f is continuous at x = 0 even though f (0) does not exist. Moreover, for x near zero, we have f (x) ≈ −16x2/3 and f (x) ≈ − 32 . 3x1/3 Consequently f (x) → −∞ as x → 0+ , whereas f (x) → +∞ as x → 0− . This is consistent with the observation that f (x) < 0 for all x near (but not equal to) zero. The origin is a local maximum and there are global minima where | x | = 2. C04S05.047: Given f (x) = x(x − 1)2/3 , we find that 2 5x − 3 f (x) = (x − 1)2/3 + x(x − 1)−1/3 = . 3 3(x − 1)1/3 Thus f (x) = 0 when x = 3 and f (x) does not exist when x = 1. So f is increasing for x < 3 and for x > 1, 5 5 decreasing if 3 < x < 1. Thus there is a local maximum at 3 , 0.3257 (y -coordinate approximate) and a 5 5 local minimum at (1, 0). Examination of f (x) and f (x) for x near 1 shows that there is a vertical tangent at (1, 0). There are no global extrema because f (x) → + ∞ as x → + ∞ and f (x) → −∞ as x → −∞. The graph of f is shown next. 0.8 0.6 0.4 0.2 -1 1 2 -0.2 -0.4 C04S05.048: After simplifications, we find that f (x) = 2 − 3x . 3x2/3 (2 − x)1/3 So f (x) = 0 when x = 2 and f (x) does not exist at x = 0 and at x = 2. Nevertheless, f is continuous 3 everywhere. Its graph is increasing for x < 2 and for x > 2, decreasing for 2 < x < 2. There is a 3 3 8 vertical tangent at (0, 0), which is not an extremum. There’s a horizontal tangent at 2 , 1.058 (ordinate 3 approximate), which is a local maximum. There is a cusp at (2, 0), which is also a local minimum. Note that f (x) ≈ x for | x | large; this aids in constructing the global sketch of the graph. C04S05.049: The graph of f (x) = 2x3 + 3x2 − 36x − 3 is shown next. 100 50 -6 -4 -2 2 4 -50 -100 C04S05.050: The graph of f (x) = 2x3 − 6x2 − 48x + 50 is shown next. 100 50 -4 -2 2 4 6 -50 -100 C04S05.051: The graph of f (x) = −2x3 − 3x2 + 36x + 15 is shown next. 100 50 -6 -4 -2 2 4 -50 -100 C04S05.052: The graph of f (x) = 3x4 + 8x3 − 18x2 + 5 appears next. 50 -4 -3 -2 -1 1 -50 -100 9 2 C04S05.053: The graph of f (x) = 3x4 − 8x3 − 30x2 + 72x + 45 is next. 200 100 -2 2 4 -100 C04S05.054: The graph of f (x) = 12x5 − 45x4 − 200x3 + 720x2 + 17 is shown next. 5000 4000 3000 2000 1000 -4 -2 2 4 -1000 -2000 C04S05.055: Let f (x) = x3 − 3x + 3. For part (a), we find that if we let x = −2.1038034027, then f (x) ≈ 7.58 × 10−9 . For part (b), we find that f (x) ≈ (x + 2.1038034027)(x2 − (2.1038034027)x + 1.4259887573). And in part (c), we find by the quadratic formula that the complex conjugate roots of f (x) = 0 are approximately 1.0519017014 + 0.5652358517i and 1.0519017014 − 0.5652358517i. C04S05.056: Let f (x) = x3 − 3x + q . Then f (x) = 3(x2 − 1), so the graph of y = f (x) will always have a local maximum at (−1, f (−1)) = (−1, q + 2) and a local minimum at (1, f (1)) = (1, q − 2). If the ordinates of these points have the same sign then the equation f (x) = 0 will have only one [real] solution—see Figs. 4.5.9 through 4.5.11. And this situation is equivalent to q + 2 < 0 or q − 2 > 0; that is, q < −2 or q > 2. If the ordinates have opposite signs, then the equation f (x) = 0 will have three real solutions, and this will occur if q − 2 < 0 < q + 2; that is, if −2 < q < 2. If q = ±2, then there will be exactly two real solutions because x3 − 3x + 2 = (x − 1)(x2 + x − 2) = (x − 1)2 (x + 2) and x3 − 3x − 2 = (x + 1)(x2 − x − 2) = (x + 1)2 (x − 2). 2 C04S05.057: If f (x) = [ x(x − 1)(2x − 1)] , then f (x) = 2x(x − 1)(2x − 1)(6x2 − 6x + 1), so the critical points of the√ graph of f will be (0, 0), 1 , 0 , (1, 0), 1 3 − 2 6 approximate), and 1 3 + 3 , 0.009259259 . The graph of f will be 6 10 √ 3 , 0.009259259 (ordinate decreasing for x < 0, increasing for 0<x< 1 6 √ decreasing for 1 6 3− increasing for 1 2 <x< decreasing for 1 6 3+ increasing for 3 1 6 √ 3 3− √ 3, < x < 1, 2 3+ √ 3, < x < 1, and 1 < x. There will be global minima at x = 0, x = 1 , and x = 1 and [equal] local maxima at x = 2 √ x= 1 3+ 3 . 6 C04S05.058: A Mathematica solution: poly = x∧3 − 3∗x + 1; r = (−1 + Sqrt[3])/2; x1 = r∧(−1/3) + r∧(1/3); x2 = r∧(4/3) + r∧(5/3); x3 = r∧(2/3) + r∧(7/3); (poly /. (poly /. (poly /. x → x1) // Expand // Simplify x → x2) // Expand // Simplify x → x3) // Expand // Simplify 0 0 0 N[x1] // chop N[x2] // chop N[x3] // chop 1.53209 −1.87939 0.347296 Thus we see three distinct real roots. C04S05.059: Given f (x) = 1 6 x(9x 4 − 5)(x − 1) , the Mathematica command Plot[ f[x], { x, −1, 2 }, PlotPoints → 97, PlotRange → {{ −0.8, 1.8 }, { −0.1, 1.0 }} ]; 11 1 6 3− √ 3 and generated the graph shown next. As predicted, the graph seems to have a “flat spot” on the interval [0, 1]. 1 0.8 0.6 0.4 0.2 -0.5 0.5 1 1.5 Then we modified the Plot command to restrict the range of y -values to the interval [ −0.00001, 0.00008]: Plot[ f[x], { x, −1, 2 }, PlotPoints → 197, PlotRange → {{ −0.18, 1.18 }, { −0.00001, 0.00008 }} ]; the graph generated by this command is next. 0.00008 0.00006 0.00004 0.00002 0.2 0.4 0.6 0.8 1 Then we used Mathematica to identify the extrema: soln = Solve[ f [x] == 0, x ] {{ x → 0 }, { x → 0 }, { x → 0 }, { x → 5 5 }, { x → }, 9 9 5 }, { x → 1 }, { x → 1 }, { x → 1 }, 9 14 − Sqrt[61] 14 + Sqrt[61] {x→ }, { x → }} 27 27 {x→ { x1 = soln[[1,1,2]], x2 = soln[[4,1,2]], x3 = soln[[7,1,2]], x4 = N[soln[[10,1,2]], 20], x5 = N[soln[[11,1,2]], 20] } { 0, 5 , 1, 0.22925001200345724466, 0.8077870250335797924 } 9 { y1 = f[x1], y2 = f[x2], y3 = f[x3], y4 = f[x4], y5 = f[x5] } { 0, 0, 0, 0.0000559441164359303138, 0.0000119091402810978625 } 12 The second graph makes it clear that (x1 , 0), (x2 , 0), and (x3 , 0) are local (indeed, global) minima, while (x4 , y4 ) and (x5 , y5 ) are local (not global) maxima. C04S05.060: After constructing the functions f (x) = x4 − 55x3 + 505x2 + 11000x − 110000 and g (x) = f (x) + ex2 where e = 1, we used Mathematica to find the zeros of these polynomials: NSolve[ f[x] == 0, x ] {{ x → −13.4468 }, { x → 9.38408 }, { x → 28.6527 }, { x → 30.4143 }} NSolve[ g[x] == 0, x ] {{ x → −13.4468 }, { x → 9.38459 }, { x → 29.5361 − 0.480808 I }, { x → 29.5371 + 0.480808 I }} In part (b), by changing the value of e, working up from e = 0 and down from e = 1, we bracketed the transition point. e = 0.7703; h[x ] := f[x] + e∗x∧2; NSolve[ h[x] == 0, x ] {{ x → −13.4478 }, { x → 9.37677 }, { x → 29.528 }, { x → 29.543 } e = 0.7704; h[x ] := f[x] + e∗x∧2; NSolve[ h[x] == 0, x ] {{ x → −13.4478 }, { x → 9.37676 }, { x → 29.5355 − 0.00665912 I }, { x → 29.5355 + 0.00665912 I }} 13 Section 4.6 C04S06.001: f (x) = 8x3 − 9x2 + 6, f (x) = 24x2 − 18x, f (x) = 48x − 18. C04S06.002: f (x) = 10x4 + 3 x1/2 + 1 x−2 , f (x) = 40x3 + 3 x−1/2 − x−3 , f (x) = 120x2 − 3 x−3/2 +3x−4 . 2 2 4 8 C04S06.003: f (x) = −8(2x − 1)−3 , f (x) = 48(2x − 1)−4 , f (x) = −384(2x − 1)−5 . C04S06.004: g (t) = 2t + 1 (t + 1)−1/2 , g (t) = 2 − 1 (t + 1)−3/2 , g (t) = 3 (t + 1)−5/2 . 2 4 8 C04S06.005: g (t) = 4(3t − 2)1/3 , g (t) = 4(3t − 2)−2/3 , g (t) = −8(3t − 2)−5/3 . C04S06.006: f (x) = (x + 1)1/2 + 1 x(x + 1)−1/2 , f (x) = (x + 1)−1/2 − 1 x(x + 1)−3/2 , 2 4 f (x) = − 3 (x + 1)−3/2 + 3 x(x + 1)−5/2 . 4 8 C04S06.007: h (y ) = (y + 1)−2 , h (y ) = −2(y + 1)−3 , h (y ) = 6(y + 1)−4 . C04S06.008: f (x) = 3 x−1/2 + 3 + 3 x1/2 , f (x) = − 3 x−3/2 + 3 x−1/2 , f (x) = 9 x−5/2 − 3 x−3/2 . 2 2 4 4 8 8 C04S06.009: g (t) = t(1 + 2 ln t), g (t) = 3 + 2 ln t, g (t) = C04S06.010: h (z ) = 2 . t (2z − 1)ez (4z 2 − 4z + 3)ez (8z 3 − 12z 2 + 18z − 15)ez , h (z ) = , h (z ) = . 2z 3/2 4z 5 / 2 8z 7 / 2 C04S06.011: f (x) = 3 cos 3x, f (x) = −9 sin 3x, f (x) = −27 cos 3x. C04S06.012: f (x) = −4 sin 2x cos 2x, f (x) = 8 sin2 2x − 8 cos2 2x, f (x) = 64 sin 2x cos 2x. C04S06.013: f (x) = cos2 x − sin2 x, f (x) = −4 sin x cos x, f (x) = 4 sin2 x − 4 cos2 x. C04S06.014: f (x) = 2x cos x − x2 sin x, f (x) = 2 cos x − 4x sin x − x2 cos x, f (x) = −2 sin x − 4 sin x − 6x cos x + x2 sin x. C04S06.015: f (x) = f (x) = x cos x − sin x (2 − x2 ) sin x − 2x cos x , f (x) = , x2 x3 (6 − x2 )x cos x + (3x2 − 6) sin x . x4 C04S06.016: Given: x2 + y 2 = 4. 2x + 2yy (x) = 0, y (x) = − x y (x) = − . y so y − xy (x) =− y2 y+ x2 y y2 C04S06.017: Given: x2 + xy + y 2 = 3. 1 =− y 2 + x2 4 = − 3. y3 y 2x + y . x + 2y 2x + y + x dy dy + 2y = 0, dx dx y (x) = − 2(1 + y (x) + [y (x)]2 ) 6(x2 + xy + y 2 ) 18 . =− =− x + 2y x + 2y (x + 2y )3 y (x) = − so C04S06.018: Given: x1/3 + y 1/3 = 1. 1 −2/3 1 −2/3 dy +y x = 0, 3 3 dx y (x) = − 2 3 y x −1/3 · so y (x) = − xy (x) − y 2y = x2 3 x5 1 /3 y x 2 /3 . . C04S06.019: Given: y 3 + x2 + x = 5. 3y 2 dy + 2x + 1 = 0, dx y (x) = − C04S06.020: Given: y (x) = − so 2x + 1 . 3y 2 2(1 + 3y (x)[ y (x)]2 ) 2[(2x + 1)2 + 3y 3 ] =− . 3[y (x)]2 9y 5 1 1 + = 1. xy − 1 1 dy = 0, −2 x2 y dx y (x) = −2 so y (x) = − y x 2 . y 2y 2 (x + y ) y xy (x) − y (x) · = =2 2 4 x x x x 3 . The last step is a consequence of the fact that, by the original equation, x + y = xy . C04S06.021: Given: sin y = xy . (cos y ) dy dy =y+x , dx dx y (x) = − so y (x) = − y . x − cos y [ x − cos y (x)]y (x) − y (x)[1 + y (x) sin y ] (y sin y + 2 cos y − 2x)y =− . (x − cos y )2 (x − cos y )3 C04S06.022: sin2 x + cos2 y = 1: 2 sin x cos x − 2y (x) sin y cos y = 0; y (x) = d2 y can be simplified (with the aid of the original equation) to dx2 d2 y cos2 x sin2 y − sin2 x cos2 y = ≡0 dx2 sin3 y cos3 y sin x cos x . sin y cos y if y is not an integral multiple of π /2. C04S06.023: f (x) = 3x2 − 6x − 45 = 3(x + 3)(x − 5), so there are critical points at (−3, 81) and (5, −175). f (x) = 6(x − 1), so the inflection point is located at (1, −47). C04S06.024: Critical points: (−3, 389) and (6, −340); inflection point: (1.5, 24.5). 2 C04S06.025: f (x) = 12x2 − 12x − 189 = 3(2x + 7)(2x − 9), so there are critical points at (−3.5, 553.5) and (4.5, −470.5). f (x) = 24x − 12, so the inflection point is located at (0.5, 41.5). C04S06.026: Critical points: (−6.25, −8701.56) and (3.4, 9271.08); inflection point: (−1.425, 284.76). √ C04S06.027: f (x) = 4x3 − 108x = 4x(x2 − 27), so there are critical points at (0, 237), − 3 3, −492 , √ and 3 3, −492 . Next, f (x) = 12x2 − 108 = 12(x − 3)(x + 3), so the inflection points are at (−3, −168) and (3, −168). C04S06.028: Critical point: (7.5, −1304.69); inflection points: (0, −250) and (5, −875). C04S06.029: f (x) = 15x4 − 80x3 = 5x3 (3x − 16), so there are critical points at (0, 1000) and 136 , − 181144 81 (approximately (5.333333, −2236.345679)). f (x) = 60x3 − 240x2 = 60x2 (x − 4), so the inflection points are at (0, 1000) and (4, −1048). √ √ √ √ C04S06.030: Critical points: −4 2, 8192 2 and 4 2, −8192 2 ; inflection points: (0, 0), (−4, 7168), and (4, −7168). C04S06.031: f (x) = 2x − 4 and f (x) ≡ 2, so there is a critical point at (2, −1) and no inflection points. Because f (2) = 2 > 0, there is a local minimum at the critical point. The first derivative test shows that it is in fact a global minimum. C04S06.032: f (x) = −6 − 2x; f (x) = −2. The only critical point is at (−3, 14), and it is a local maximum point because f (−3) = −2 < 0. There are no inflection points because f (x) never changes sign. The critical point is actually a global maximum by the first derivative test. C04S06.033: f (x) = 3(x + 1)(x − 1) and f (x) = 6x. f (−1) = 3 and f (−1) = −6, so the critical point at (−1, 3) is a local maximum. Similarly, the critical point at (1, −1) is a local minimum. The point (0, 1) is an inflection point because f (x) < 0 for x < 0 and f (x) > 0 for x > 0. The extrema are not global because f (x) → + ∞ as x → + ∞ and f (x) → −∞ as x → −∞. C04S06.034: f (x) = 3x(x − 2); f (x) = 6(x − 1). There is a critical point at (0, 0) and one at (2, −4). Now f (0) = −6 < 0, so there is a local maximum at (0, 0); f (2) = 6 > 0, so there is a local minimum at (2, −4). The only possible inflection point is (1, −2), and it is indeed an inflection point because f changes sign there. The extrema are not global. C04S06.035: If f (x) = xe−x , then f (x) = (1 − x)e−x and f (x) = (x − 2)e−x . Hence f is increasing for x < 1 and decreasing for x > 1; its graph is concave downward for x < 2 and concave upward for x > 2. Hence there is a global maximum at (1, e−1 ) and an inflection point at (2, 2e−2 ). The graph of y = f (x) is next. 0.2 -1 1 2 3 -0.2 -0.4 C04S06.036: If 3 4 5 f (x) = ln x , x then f (x) = 1 − ln x x2 and f (x) = −3 + 2 ln x . x3 Then f (x) = 0 when x = e and f (x) = 0 when x = e3/2 . Hence the graph of f is increasing for 0 < x < e, decreasing for x > e, concave downward for 0 < x < e3/2 and concave upward for x > e3/2 . Therefore there is a global maximum at (e, e−1 ) and an inflection point at e3/2 , 3 e−3/2 . The graph of f is shown next. 2 0.4 0.2 2 4 6 8 -0.2 -0.4 C04S06.037: f (x) = 5x4 + 2 and f (x) = 20x3 , so there are no critical points (f (x) > 0 for all x) and (0, 0) is the only possible inflection point. And it is an inflection point because f (x) changes sign at x = 0. C04S06.038: f (x) = 4x(x + 2)(x − 2); f (x) = 12x2 − 16. The critical points are located at (−2, −16), (0, 0), and (2, −16). Now f (−2) > 0 and f (2) > 0, so (−2, −16) and (2, −16) are local minimum points. But f √0) < 0, so (0, 0) is a local maximum point. The only possible inflection points are where 3x2 − 4 = 0; ( √ x = 2 3 and x = − 2 3. Because f (x) = 4(3x2 − 4), it is clear that f (x) changes sign at each of these 3 3 two points, so the corresponding points on the graph are inflection points. The local minima are actually global by a careful application of the first derivative test. C04S06.039: f (x) = 2x(x − 1)(2x − 1) and f (x) = 2(6x2 − 6x + 1). So the critical points are (0, 0), 1 1 2 , 16 , and (1, 0). The second derivative test indicates that the first and third are local minima and that the √ √ 1 1 second is a local maximum. The only possible inflection points are 1 3 − 3 , 36 and 1 3 + 3 , 36 . 6 6 2 Why are they both inflection points? The graph of f (x) = 2(6x − 6x + 1) is a parabola opening upward and with its vertex below the x-axis (because f (x) = 0 has two real solutions). The possible inflection points are located at the zeros of f (x), so it should now be clear that f (x) changes sign at each of the two possible inflection points. Because f (x) is never negative, the two local minima are actually global, but the local maximum is not (because f (x) → + ∞ as x → ± ∞). C04S06.040: f (x) = x2 (x + 2)(5x + 6) and f (x) = 4x(5x2 + 12x + 6). So the critical points occur where x = 0, x = −2, and x = − 6 . Now f (0) = 0, so the second derivative test fails here, but f (x) > 0 for x 5 near zero but x = 0, so (0, 0) is not an extremum. Next, f (−2) = −16 < 0, so (−2, 0) is a local maximum point; f (−1.2) = 5.76 > 0, so − 6 , 3456 is a local minimum point. The possible inflection points occur at 5 3125 x = 0, x= 1 5 −6 + √ 6, and x= 1 5 −6 − √ 6. In decimal form these are x = 0, x ≈ −0.710, and x ≈ −1.690. Because f (−2) = −16 < 0, f (−1) = 4 > 0, f (−0.5) = −2.5 < 0, and f (1) = 92 > 0, each of the three numbers displayed above is the abscissa of an inflection point of the graph of f . None of the extrema is global because f (x) → + ∞ as x → + ∞ and f (x) → −∞ as x → −∞. C04S06.041: f (x) = cos x and f (x) = − sin x. f (x) = 0 when x = π /2 and when x = 3π /2; f (π /2) = −1 < 0 and f (3π /2) = 1 > 0, so the first of these critical points is a local maximum and 4 the second is a local minimum. Because | sin x | 1 for all x, these extrema are in fact global. f (x) = 0 when x = π and clearly changes sign there, so there is an inflection point at (π , 0). C04S06.042: Local (indeed, global) maximum point: (0, 1); no inflection points. C04S06.043: f (x) = sec2 x 1 for −π /2 < x < π /2, so there are no extrema. f (x) = 2 sec2 x tan x, so there is a possible inflection point at (0, 0). Because tan x changes sign at x = 0 (and 2 sec2 x does not), (0, 0) is indeed an inflection point. C04S06.044: f (x) = sec x tan x and f (x) = sec3 x + sec x tan2 x = (sec x)(sec2 x + tan2 x). So (0, 1) is the only critical point and there are no possible inflection points. f (0) = 1, so the second derivative test shows that (0, 1) is a local minimum point. The first derivative test identifies (0, 1) as a global minimum point. C04S06.045: f (x) = −2 sin x cos x and f (x) = 2 sin2 x − 2 cos2 x. Hence (0, 1), (π /2, 0), and (π , 1) are critical points. By the second derivative test the first and third are local maxima and the second is a local minimum. (Because 0 cos2 x 1 for all x, these extrema are all global.) There are possible inflection points at (−π /4, 1/2), (π /4, 1/2), (3π /4, 1/2), and (5π /4, 1/2). A close examination of f (x) reveals that it changes sign at all four of these points, so each is an inflection point. C04S06.046: f (x) = sin3 x, −π < x < π : f (x) = 3 sin2 x cos x, f (x) = 6 sin x cos2 x − 3 sin3 x = 3(2 cos2 x − sin2 x) sin x. Global minimum at (−π /2, −1); global maximum at (π /2, 1); inflection points at x = 0 and at the four solutions of tan2 x = 2 in (−π , π ): approximately (−2.186276, −0.544331), (−0.955317, −0.544331), (0, 0), (0.955317, 0.544331), and (2.186276, 0.544331). C04S06.047: If f (x) = 10(x − 1)e−2x , then f (x) = 10(3 − 2x)e−2x and f (x) = 40(x − 2)e−2x . Thus the graph of f is increasing for 0 < x < 3 and decreasing for x > 3 . It is concave downward for 2 2 0 < x < 2 and concave upward if x > 2. Hence there is a global maximum at (1.5, 5e−3 ) ≈ (1.5, 0.248935) and an inflection point at (2, 10e−4 ) ≈ (2, 0.183156). The graph of f is next. 0.4 0.2 -1 1 2 3 -0.2 -0.4 -0.6 -0.8 -1 C04S06.048: If f (x) = (x2 − x)e−x , then f (x) = −(x2 − 3x + 1)e−x and f (x) = (x − 1)(x − 4)e−x . √ √ So the graph of f is decreasing for x < 1 3 − 5 and for x > 1 3 + 5 , increasing otherwise; it is 2 2 concave upward for x < 1 and for x > 4 and concave downward otherwise. Therefore there is a global 5 √ √ minimum at 1 3 − 5 , −0.161121 , a local maximum at 1 3 + 5 , 0.309005 , and inflection points 2 2 at (1, 0) and (4, 12e−4 ) ≈ (4, 0.219788). The graph of y = f (x) is next. 1 0.8 0.6 0.4 0.2 2 4 6 8 C04S06.049: If f (x) = (x2 − 2x − 1)e−x , then f (x) = −(x2 − 4x + 1)e−x and f (x) = (x − 1)(x − 5)e−x . √ √ Thus the graph of f is decreasing for x < 2 − 3 and for x > 2 + 3, increasing otherwise; it is concave upward for x < 1 and for x > 5 and concave downward otherwise. Thus there is a global minimum near √ √ 2 − 3, −1.119960 , a local maximum near 2 + 3, 0.130831 , and inflection points at (1, −2e−1 ) and (5, 14e−5 ). The graph of f is next. 2 1.5 1 0.5 2 4 6 -0.5 -1 C04S06.050: If f (x) = x exp(−x2 ), then f (x) = (1 − 2x2 ) exp(−x2 ) and f (x) = 2x(2x2 − 3) exp(−x2 ). √ √ 1 Thus the graph of f is decreasing for x < −√ 2 and for x > 1 2, increasing otherwise; it is concave 2 2 √ downward for x < − 1 6 and for 0 < x < 1 6, concave upward otherwise. Therefore there is a global 2 √2 √ minimum at − 1 2, −(2e)−1/2 , a global maximum at 1 2, (2e)−1/2 , and inflection points at (0, 0) and 2 2 near (−1.224745, −0.273278) and (1.224745, 0.273278). The graph of f is next. 0.4 0.2 -3 -2 -1 1 2 3 -0.2 -0.4 C04S06.051: We are to minimize the product of two numbers whose difference is 20; thus if x is the smaller, we are to minimize f (x) = x(x + 20). Now f (x) = 20 + 2x, so f (x) = 0 when x = −10. But 6 f (x) ≡ 2 is positive when x = −10, so there is a local minimum at (−10, −100). Because the graph of f is a parabola opening upward, this local minimum is in fact the global minimum. Answer: The two numbers are −10 and 10. C04S06.052: We assume that the length turned upward is the same on each side—call it y . If the width of the gutter is x, then we have the constraint xy = 18, and we are to minimize the width x + 2y of the strip. Its width is given by the function f (x) = x + 36 , x x > 0, for which f (x) = 1 − 36 x2 and f (x) = 72 . x3 The only critical point in the domain of f is x = 6, and f (x) > 0 on the entire domain of f . Consequently the graph of f is concave upward for all x > 0. Because f is continuous for such x, f (6) = 12 is the global minimum of f . C04S06.053: Let us minimize g (x) = (x − 3)2 + (3 − 2x − 2)2 = (x − 3)2 + (1 − 2x)2 , the square of the distance from the point (x, y ) on the line 2x + y = 3 to the point (3, 2). We have g (x) = 2(x − 3) − 4(1 − 2x) = 10x − 10; g (x) ≡ 10. So x = 1 is the only critical point of g . Because g (x) is always positive, the graph of g is concave upward on the set I of all real numbers, and therefore R (1, g (1)) = (1, 1) yields the global minimum for g . So the point on the given line closest to (3, 2) is (1, 1). C04S06.054: Base of box: x wide, 2x long. Height: y . Then the box has volume 2x2 y = 576, so y = 288x−2 . Its total surface area is A = 4x2 + 6xy , so we minimize A = A(x) = 4x2 + 1728 , z x > 0. Now A (x) = 8x − 1728 x2 and A (x) = 8 + 3456 . x3 The only critical point of A(x) occurs when 8x3 = 1728; that is, when x = 6. But A (x) > 0 for all x > 0, so the graph of y = A(x) is concave upward for all x > 0. Therefore A(6) is the global minimum value of A(x). Also, when x = 6 we have y = 8. Answer: The dimensions of the box of minimal surface area are 6 inches wide by 12 inches long by 8 inches high. C04S06.055: Base of box: x wide, 2x long. Height: y . Then the box has volume 2x2 y = 972, so y = 486x−2 . Its total surface area is A = 2x2 + 6xy , so we minimize A = A(x) = 2x2 + 2916 , x x > 0. Now A (x) = 4x − 2916 x2 and 7 A (x) = 4 + 5832 . x3 The only critical point of A occurs when x = 9, and A (x) is always positive. So the graph of y = A(x) is concave upward for all x > 0; consequently, (9, A(9)) is the lowest point on the graph of A. Answer: The dimensions of the box are 9 inches wide, 18 inches long, and 6 inches high. C04S06.056: If the radius of the base of the pot is r and its height is h (inches), then we are to minimize the total surface area A given the constraint π r2 h = 125. Thus h = 125/(π r2 ), and so A = π r2 + 2π rh = A(r) = π r2 + 250 , r r > 0. Hence A (r) = 2π r − 250 r2 and A (r) = 2π + 500 . r3 √ Now A (r) = 0 when r3 = 125/π , so that r = 5/ 3 π . This is the only critical point of A, and A (r) > 0 for all r, so the graph of y = A(r) is concave upward for all r in the domain of A. Consequently we have located √ √ the global minimum, and it occurs when the pot has radius r = 5/ 3 π inches and height h = 5/ 3 π inches. Thus the pot will have its radius equal to its height, each approximately 3.414 inches. C04S06.057: Let r denote the radius of the pot and h its height. We are given the constraint π r2 h = 250, so h = 250/(π r2 ). Now the bottom of the pot has area π r2 , and thus costs 4π r2 cents. The curved side of the pot has area 2π rh, and thus costs 4π rh cents. So the total cost of the pot is C = 4π r2 + 4π rh = C (r) = 4π r2 + 1000 , r r > 0. Now 1000 2000 and C (r) = 8π + 3 . r2 r √ C (r) = 0 when 8π r3 = 1000, so that r = 5/ 3 π . Because C (r) > 0 for all r > 0, the graph of y = C (r) is concave upward on the domain of C . Therefore we have found the value of r that minimizes C (r). √ The corresponding value of h is 10/ 3 π , so the pot of minimal cost has height equal to its diameter, each approximately 6.828 centimeters. C (r) = 8π r − C04S06.058: Let x denote the length of each side of the square base of the solid and let y denote its height. Then its total volume is x2 y = 1000. We are to minimize its total surface area A = 2x2 + 4xy . Now y = 1000/(x2 ), so A = A(x) = 2x2 + 4000 , x x > 0. Therefore A (x) = 4x − 4000 x2 and A (x) = 4 + 8000 . x3 The only critical point occurs when x = 10, and A (x) > 0 for all x in the domain of A, so x = 10 yields the global minimum value of A(x). In this case, y = 10 as well, so the solid is indeed a cube. C04S06.059: Let the square base of the box have edge length x and let its height be y , so that its total volume is x2 y = 62.5 and the surface area of this box-without-top will be A = x2 + 4xy . So A = A(x) = x2 + 8 250 , x x > 0. Now A (x) = 2x − 250 x2 and A (x) = 2 + 500 . x3 The only critical point occurs when x = 5, and A (x) > 0 for all x in the domain of A, so x = 5 yields the global minimum for A. Answer: Square base of edge length x = 5 inches, height y = 2.5 inches. C04S06.060: Let r denote the radius of the can and h its height (in centimeters). We are to minimize its total surface area A = 2π r2 + 2π rh given the constraint π r2 h = V = 16π . First we note that h = V /(π r2 ), so we minimize A = A(r) = 2π r2 + 2V , r r > 0. Now A (r) = 4π r − 2V r2 and A (r) = 4π + 4V . r3 The only critical point of A occurs when 4π r3 = 2V = 32π —that is, when r = 2. Now A (r) > 0 for all r > 0, so the graph of y = A(r) is concave upward for all r > 0. Thus the global minimum occurs when r = 2 centimeters, for which h = 4 centimeters. C04S06.061: Let x denote the radius and y the height of the cylinder (in inches). Then its cost (in cents) is C = 8π x2 + 4π xy , and we also have the constraint π x2 y = 100. So C = C (x) = 8π x2 + 400 , x x > 0. Now C (x) = 16π x − 400 x2 and C (x) = 16π + 800 . x3 The only critical point in the domain of C is x = 3 25/π (about 1.9965 inches) and, consequently, when y = 3 1600/π (about 7.9859 inches). Because C (x) > 0 for all x in the domain of C , we have indeed found the dimensions that minimize the cost of the can. For simplicity, note that y = 4x at the minimum: The height of the can is twice its diameter. C04S06.062: Let x denote the width of the print. Then 30/x is the height of the print, x + 2 is the width of the page, and (30/x) + 4 is the height of the page. We minimize the area A of the page, where A = A(x) = (x + 2) 30 +4 x = 4x + 38 + 60 , x 0 < x < ∞. Now 120 . x3 √ √ √ √ A (x) = 0 when x = 15 and A 15 = 120/ 15 15 > 0, so x √ 15 yields a local minimum of A(x). = √ √ In fact, A (x) < 0 if 0 < x < √ and A (x) >√ if 15 > x, so x = 15 yields the global minimum value of 15 0 √ A(x); this minimum value is 4 15 + 38 + 60/ 15 = 8 15 + 38 ≈ 68.983867 in.2 A (x) = 4 − 60 x2 and A (x) = C04S06.063: Given: f (x) = 2x3 − 3x2 − 12x + 3. We have 9 f (x) = 6(x − 2)(x + 1) and f (x) = 12x − 6. Hence (−1, 10) and (2, −17) are critical points and (0.5, −3.5) is a possible inflection point. Because f (x) > 0 if x > 0.5 and f (x) < 0 if x < 0.5, the possible inflection point is an actual inflection point, there is a local maximum at (−1, 10), and a local minimum at (2, −17). The extrema are not global because f (x) → + ∞ as x → + ∞ and f (x) → −∞ as x → −∞. The graph of f is next. 20 -3 -2 -1 1 2 3 4 -20 -40 C04S06.064: Given: f (x) = 3x4 − 4x3 − 5. Then f (x) = 12x3 − 12x2 = 12x2 (x − 1) and f (x) = 36x2 − 24x = 12x(3x − 2). So the graph of f is increasing for x > 1 and decreasing for x < 1 (even though there’s a horizontal tangent at x = 0), concave upward for x < 0 and x > 2 , concave downward on 0, 2 . There is a global minimum at 3 3 (1, −6), inflection points at (0, −5) and at 2 , − 151 . The x-intercepts are approximately −0.906212 and 3 27 1.682971. The graph of y = f (x) is next. -3.5 -4 -4.5 -5 -5.5 -1 -0.5 0.5 1 1.5 2 2 C04S06.065: If f (x) = 6 + 8x2 − x4 , then f (x) = −4x(x + 2)(x − 2) and f (x) = 16 − 12x√ So √ is . f 2 2 increasing for x < −2 and for 0 < x < 2, decreasing otherwise; its graph is concave upward on − 3 3, 3 3 and concave downward otherwise. Therefore the global maximum value of f is f (−2) = f (2) = 22 and there √ √ is a local minimum at f (0) = 6. There are inflection points at − 2 3, 134 and at 2 3, 134 . The graph 3 9 3 9 of f is next. 20 10 -3 -2 -1 1 -10 C04S06.066: Given: f (x) = 3x5 − 5x3 . Then 10 2 3 f (x) = 15x4 − 15x2 = 15x2 (x + 1)(x − 1) f (x) = 60x3 − 30x = 60x(x + r)(x − r) and where r= 1√ 2. 2 The graph is increasing for x < −1 and for x > 1, decreasing for −1 < x < 1 (although there is a horizontal tangent at the origin). It is concave upward on (−r, 0) and on (r, + ∞), concave downward on (−∞, −r) and on (0, r). Thus there is a local maximum at (−1, 2), a local minimum at (1, −2), and inflection points at (−r, 7r/4), (0, 0), and (r, −7r/4) (the last ordinate is approximately −1.237437). Finally, the x-intercepts are 0, − 5/3 , and 5/3 ≈ 1.29099. The graph of y = f (x) is shown next. 2 1 -1 -0.5 0.5 1 -1 -2 C04S06.067: If f (x) = 3x4 − 4x3 − 12x2 − 1, then f (x) = 12x3 − 12x2 − 24x = 12x(x − 2)(x + 1) and f (x) = 36x2 − 24x − 24 = 12(3x2 − 2x − 2). So the graph of f is decreasing for x < √1 and for 0 < x < 2 and increasing otherwise; it is concave upward − √ for x < 1 1 − 7 and for x > 1 1 + 7 and concave downward otherwise. So there is a local minimum 3 3 at (−1, √6), a local maximum at (0, −1), and a global minimum at (2√−33). There are inflection points at − , √ √ 1 1 1 1 7 , 27 −311 + 80 7 and at 3 1 + 7 , 27 −311 − 80 7 . The graph of y = f (x) is next. 3 1− 40 20 -3 -2 -1 1 2 3 -20 C04S06.068: Given: f (x) = 3x5 − 25x3 + 60x. Then f (x) = 15x4 − 75x2 + 60 = 15(x2 − 4)(x2 − 1) and f (x) = 60x3 − 150x = 30x(2x2 − 5). There are local maxima where x = −2 and x = 1, local minima where x = −1 and x = 2. Inflection points 11 occur where x = 0, x = − 5/2 , and x = 5/2 . The graph is next. 60 40 20 -2 -1 1 2 -20 -40 -60 C04S06.069: If f (x) = x3 (x − 1)4 , then f (x) = 3x2 (x − 1)4 + 4x3 (x − 1)3 = x2 (x − 1)3 (7x − 3) and f (x) = 6x(x − 1)4 + 24x2 (x − 1)3 + 12x3 (x − 1)2 = 6x(x − 1)2 (7x2 − 6x + 1). Hence the graph √ f is increasing for x < 3 and for x > 1, decreasing otherwise; concave upward for of √7 6912 0 < x < 1 3 − 2 and for x > 1 3 + 2 . So there is a local maximum at 3 , 823543 and a local 7 7 7 minimum at (1, 0). Also there are inflection points at (0, 0) and at the two points with x-coordinates √ 1 2 . The graph of y = f (x) is shown next. Note the scale on the y -axis. 7 3± 0.02 0.015 0.01 0.005 -0.5 0.5 1 1.5 -0.005 -0.01 C04S06.070: Given: f (x) = (x − 1)2 (x + 2)3 . Then f (x) = (x − 1)(x + 2)2 (5x + 1) and f (x) = 2(x + 2)(10x2 + 4x − 5). The zeros of f (x) are x = −2, x ≈ 0.535, and x ≈ −0.935. It follows that (1, 0) is a local minimum (from the second derivative test), that (−0.2, 8.39808) is a local maximum, and that (−2, 0) is not an extremum. Also, the second derivative changes sign at each of its zeros, so each of these three zeros is the abscissa of an inflection point on the graph. The graph is next. 20 15 10 5 -3 -2 -1 1 -5 -10 -15 C04S06.071: If f (x) = 1 + x1/3 then 12 2 f (x) = 1 −2/3 1 x = 2 /3 3 3x 2 2 f (x) = − x−5/3 = − 5/3 . 9 9x and Therefore f (x) > 0 for all x = 0; because f is continuous even at x = 0, the graph of f is increasing for all x, but (0, 1) is a critical point. Because f (x) has the sign of −x, the graph of f is concave upward for x < 0 and concave downward for x > 0. Thus there is an inflection point at (0, 1). Careful examination of the first derivative shows also that there is a vertical tangent at (0, 1). The graph is next. 2.5 2 1.5 1 0.5 -4 -2 2 4 -0.5 C04S06.072: Given: f (x) = 2 − (x − 3)1/3 . Then f (x) = − 1 3(x − 3)2/3 and 2 . 9(x − 3)5/3 f (x) = There is a vertical tangent at (3, 2) but there are no other critical points. The graph is decreasing for all x, concave down for x < 3, and concave up for x > 3. Because f is continuous for all x, there is an inflection point at (3, 2). The y -intercept is at (0, 3.44225) (ordinate approximate) and the x-intercept is at (11, 0). 3.5 3 2.5 2 1.5 1 0.5 -2 2 4 6 8 √ C04S06.073: Given f (x) = (x + 3) x, f (x) = 3(x + 1) √ 2x and f (x) = 3(x − 1) √. 4x x Note that f (x) has domain x 0. Hence f (x) > 0 for all x > 0; in fact, because f is continuous (from the right) at x = 0, f is increasing on [0, + ∞). It now follows that (1, 4) is an inflection point, but it’s not shown on the following figure for two reasons: First, it’s not detectable; second, the behavior of the graph near x = 0 is of more interest, and that behavior is not clearly visible when f is graphed on a larger interval. The point (0, 0) is, of course, the location of the global minimum of f and is of particular interest because 13 f (x) → + ∞ as x → 0+ . The graph of y = f (x) is next. 2.5 2 1.5 1 0.5 0.1 0.2 0.3 0.4 0.5 0.6 C04S06.074: Given: f (x) = x2/3 (5 − 2x). Then f (x) = 10 − 10x 3x1/3 and f (x) = − 20x + 10 . 9x4/3 If |x| is large, then f (x) ≈ −2x5/3 , which (because the exponent 5/3 has odd numerator and odd denominator) acts rather like −2x3 for |x| large (at least qualitatively). This aids in determining the behavior of f (x) for |x| large. The graph is decreasing for x < 0 and for x > 1, increasing on the interval (0, 1). It is concave upward for x < −0.5, concave downward for x > 0 and on the interval (−0.5, 0). There is a vertical tangent and a local minimum at the origin, a local maximum at (1, 3), an inflection point where x = −0.5, a dual intercept at (0, 0), and an x-intercept at x = 2.5. The graph is next. 6 4 2 -3 -2 -1 1 2 3 -2 C04S06.075: Given f (x) = (4 − x)x1/3 , we have 4(1 − x) 4(x + 2) and f (x) = − . 3x2/3 9x5/3 There is a global maximum at (1, 3), a vertical tangent, dual intercept, and inflection point at (0, 0), an √ x-intercept at (4, 0), and an inflection point at −2, −6 3 2 . The graph of f is next. f (x) = 2 -2 2 4 -2 -4 -6 14 6 C04S06.076: Given: f (x) = x1/3 (6 − x)2/3 . Then f (x) = 2−x − x)1/3 x2/3 (6 and f (x) = − x5/3 (6 8 . − x)4/3 If |x| is large, then (6 − x)2/3 ≈ x2/3 , so f (x) ≈ x for such x. This aids in sketching the graph, which has a local maximum where x = 2, a local minimum at (6, 0), vertical tangents at (6, 0) and at the origin. It is increasing for x < 2 and for x > 6, decreasing on the interval (2, 6), concave upward for x < 0, and concave downward on (0, 6) and for x > 6. All the intercepts have been mentioned, too. The figure on the left shows a “close-up” of the graph and the figure on the right gives a more distant view. 15 4 10 2 -2 5 2 4 6 -10 8 -5 5 10 15 20 -5 -2 -10 -4 C04S06.077: Figure 4.6.34 shows a graph that is concave downward, then concave upward, so its second derivative is negative, then zero, then positive. This matches the graph in (c). C04S06.078: Figure 4.6.35 shows a graph that is concave upward, then downward, so the second derivative will be positive, then zero, then negative. This matches the graph in (e). C04S06.079: Figure 4.6.36 shows a graph that is concave upward, then downward, then upward again, so the second derivative will be positive, then negative, then positive again. This matches the graph in (b). C04S06.080: Figure 4.6.37 shows a graph that is concave downward, then upward, then downward, so the second derivative is negative, then positive, then negative. This matches the graph in (f). C04S06.081: Figure 4.6.38 shows a graph that is concave upward, then almost straight, then strongly concave downward, so the second derivative must be positive, then close to zero, then large negative. This matches the graph in (d). C04S06.082: Figure 4.6.39 shows a graph that is concave upward, then downward, then upward, then downward. So the second derivative must be positive, then negative, then positive, and then negative. This matches the graph in (a). C04S06.083: (a): Proof: The result holds when n = 1. Suppose that it holds for n = k where k Then f (k) (x) = k ! if f (x) = xk . Now if g (x) = xk+1 , then g (x) = xf (x). So by the product rule, g (x) = xf (x) + f (x) = x (kxk−1 ) + xk = (k + 1)xk . Thus g (k+1) (x) = (k + 1)Dxk (xk ) = (k + 1)f (k) (x) = (k + 1)(k !) = (k + 1)!. 15 1. That is, whenever the result holds for n = k , it follows for n = k + 1. Therefore, by induction, it holds for all integers n 1. (b): Because the n th derivative of xn is constant, any higher order derivative of xn is zero. The result now follows immediately. C04S06.084: f (x) = cos x, f (x) = − sin x, f (3) = − cos x, and f (4) = sin x = f (x). It is now clear that f (n+4) (x) = f (n) (x) for all n 0 (we interpret f (0) (x) to mean f (x)). C04S06.085: dz dz dy = · . dx dy dx So d2 z dz d2 y dy d2 z dy = · 2+ · · . dx2 dy dx dx dy 2 dx C04S06.086: If f (x) = Ax2 + Bx + C , then A = 0. So f (x) = 2A = 0. Because f (x) never changes sign, the graph of f (x) can have no inflection points. C04S06.087: If f (x) = ax3 + bx2 + cx + d with a = 0, then both f (x) and f (x) exist for all x and f (x) = 6ax + 2b. The latter is zero when and only when x = −b/(3a), and this is the abscissa of an inflection point because f (x) changes sign at x = −b/(3a). Therefore the graph of a cubic polynomial has exactly one inflection point. C04S06.088: If f (x) = Ax4 + Bx3 + Cx2 + Dx + E , then both f (x) and f (x) are continuous for all x, and f (x) = 12Ax2 + 6Bx + 2C . In order for f (x) to change sign, we must have f (x) = 0. If so, then (because f (x) is a quadratic polynomial) either the graph of f (x) crosses the x-axis in two places or is tangent to it at a single point. In the first case, f (x) changes sign twice, so there are two points of inflection on the graph of f . In the second case, f (x) does not change sign, so f has no inflection points. Therefore the graph of a polynomial of degree four has either exactly two inflection points or else none at all. RT a − , so V −b V2 2a RT 2RT 6a p (V ) = 3 − and p (V ) = − 4. V (V − b)2 (V − b)3 V C04S06.089: First, p = p(V ) = From now on, use the constant values p = 72.8, V = 128.1, and T = 304; we already have n = 1. Then p= RT a − 2, V −b V 2a RT = , 3 V (V − b)2 3a RT = . 4 V (V − b)3 and The last two equations yield RT V 3 3(V − b) =2= , a(V − b)2 V and thus b = 1 V and V − b = 2 V . Next, 3 3 a= Finally, RT a = p + 2 , so V −b V R= V 3 RT V 3 RT 9V 3 RT 9 = = = VRT. 2 2 2 2(V − b) 2(2V /3) 8V 8 2V 3T p+ a V2 = 2V 3T p+ 16 9RT 8V = 2V p 3R + . 3T 4 8V p . We substitute this into the earlier formula for a, in order to determine that a = 3T 9 2 8 VRT = 3V p. In summary, and using the values given in the problem, we find that Therefore R = b= 1 V = 42.7, 3 a = 3V 2 p ≈ 3,583,859, and R = 8V p ≈ 81.8. 3T C04S06.090: (a): If f (c) > 0 and h > 0, then f (c + h) >0 h if h is close to zero. Thus f (c + h) > 0 for such h. Similarly, if f (c) > 0 and h < 0, then again f (c + h) >0 h if h is near zero, so f (c + h) < 0 for such h. So f (x) > 0 for x > c (but close to c) and f (x) < 0 for x < c (but close to c). By the first derivative test, f (c) is a local minimum for f . The proof in part (b) is very similar. C04S06.091: If f (x) = 1000x3 − 3051x2 + 3102x + 1050, then f (x) = 3000x2 − 6102x + 3102 and f (x) = 6000x − 6102. So the graph of f has horizontal tangents at the two points (1, 2101) and (1.034, 2100.980348) (coordinates exact) and there is a possible inflection point at (1.017, 2100.990174) (coordinates exact). Indeed, the usual tests show that the first of these is a local maximum, the second is a local minimum, and the third is an inflection point. The Mathematica command Plot[ f[x], { x, 0.96, 1.07 } ]; produces a graph that shows all three points clearly; it’s next. 2101.02 0.96 0.98 1.02 1.04 1.06 2100.98 2100.96 C04S06.092: If f (x) = [ x(1 − x)(9x − 7)(4x − 1)]4 , then f (x) = 4x3 (4x − 1)3 (x − 1)3 (9x − 7)3 (144x3 − 219x2 + 88x − 7) and 17 f (x) = 4x2 (4x − 1)2 (x − 1)2 (9x − 7)2 (77760x6 − 236520x5 + 274065x4 − 150400x3 + 39368x2 − 4312x + 147). Next, the graph of y = f (x) has horizontal tangents at the points (0, 0), (0.1052, 0.0119), 1 4, 0, (0.5109, 0.1539), 7 9, 0, (0.9048, 0.0044), and (1, 0) (all four-place decimals shown here are rounded approximations). The usual tests show that the first, third, fifth, and seventh of these are global minima and the other three are local maxima. Moreover, there are possible inflection points at (0, 0), (0.0609, 0.0061), (0.1499, 0.0069), (0.5971, 0.0870), 7 9, 1 4, 0 , (0.4246, 0.0865), 0 , (0.8646, 0.0026), (0.9446, 0.0023), and (1, 0) (again, all four-place decimals are rounded approximations). The usual tests show that the six of these points not extrema are indeed inflection points. The Mathematica command Plot[ f[x], { x, −0.05, 0.05 } ]; clearly shows the global minimum at (0, 0). The command Plot[ f[x], { x, 0.02, 0.1 } ]; clearly shows the inflection point near (0.0609, 0.0061). These graphs are shown next. 0.012 0.01 0.0015 0.008 0.001 0.006 0.004 0.0005 0.002 -0.04 -0.02 0.02 0.04 0.02 0.04 0.06 0.08 To see the other extrema and inflection points, plot y = f (x) on the intervals [0.08, 0.14], [0.12, 0.2], [0.23, 0.27], [0.3, 0.5], [0.48, 0.56], [0.54, 0.68], [0.76, 0.8], [0.8, 0.9], [0.87, 0.94], [0.92, 0.98], and (finally) [0.98, 1.04]. 18 Section 4.7 C04S07.001: It is almost always a good idea, when evaluating the limit of a rational function of x as x → ± ∞, to divide each term in numerator and denominator by the highest power of x that appears there. This technique occasionally succeeds with more complicated functions. Here we obtain x x 1 1 x lim = lim = lim = = 1. 1 1 x→∞ x + 1 x→∞ x x→∞ 1+0 1+ + xx x C04S07.002: x2 + 1 1 + (1/x2 ) 1+0 = → = 1 as x → −∞. x2 − 1 1 − (1/x2 ) 1−0 C04S07.003: x2 + x − 2 (x + 2)(x − 1) x2 + x − 2 = = x + 2 for x = 1, so → 1 + 2 = 3 as x → 1. x−1 x−1 x−1 C04S07.004: The numerator approaches −2 as x → −1, whereas the denominator approaches zero. Therefore this limit does not exist. C04S07.005: 2x2 − 1 2 − (1/x2 ) 2−0 = → = 2 as x → +∞. 2 − 3x x 1 − (3/x) 1−0 C04S07.006: Divide each term in numerator and denominator by e2x to obtain lim x→∞ 1 + ex e−2x + e−x 0+0 = lim = = 0. x→∞ 2e−2x + 1 2 + 32 x 0+1 C04S07.007: The numerator is equal to the denominator for all x = −1, so the limit is 1. If you’d prefer to write the solution more symbolically, try this: lim x→−1 C04S07.008: x2 + 2x + 1 x2 + 2x + 1 = lim 2 = lim 1 = 1. x→−1 x + 2x + 1 x→−1 (x + 1)2 5x3 − 2x + 1 5 − (2/x2 ) + (1/x3 ) 5−0+0 5 = → = as x → +∞. 3 + 4x2 − 2 7x 7 + (4/x) − (2/x3 ) 7+0−0 7 √ √ √ C04S07.009: Factor the numerator: x − 4 = ( x + 2)( x − 2). Thus the fraction is equal to x + 2 if √ x = 4. Therefore as x → 4, the fraction approaches 4 + 2 = 4. Alternatively, for a more symbolic solution, write √ √ √ √ x−4 ( x + 2)( x − 2) √ lim √ x + 2 = 4 + 2 = 2 + 2 = 4. = lim = lim x→4 x→4 x→4 x−2 x−2 For another approach, use the conjugate (as discussed in Chapter 2): √ √ √ √ x−4 (x − 4) ( x + 2) (x − 4) ( x + 2) √ lim √ x + 2 = 4 + 2 = 4. = lim √ = lim = lim x→4 x→4 x−4 x − 2 x→4 ( x − 2)( x + 2) x→4 C04S07.010: Divide each term in numerator and denominator by x3/2 , the highest power of x that appears in any term. (This works well for evaluating limits of rational functions as x → ± ∞, and sometimes works for 1 more complicated functions.) The numerator then becomes 2x−1/2 + x−3/2 , which approaches 0 as x → +∞; the denominator becomes x−1/2 − 1, which approaches −1 as x → +∞. Therefore the limit is 0. C04S07.011: Divide each term in numerator and denominator by x to obtain 8 x 1 √ 3 x2 − 2 x +1 → 0−0 =0 0+1 as x → −∞. C04S07.012: Divide each term in numerator and denominator by e6x . Then lim x→∞ 4e6x + 5 sin 6x 4 + 5e−6x sin 6x 4+0 1 = lim = =. 2x )3 −2x + 2)3 3 x→∞ (1 + 2e (e (0 + 2) 2 Note: | sin 6x | 1 for all x, so −e−6x by the squeeze law. e−6x sin 6x e−6x for all x 0. Hence e−6x sin 6x → 0 as x → + ∞ C04S07.013: Ignoring the radical for a moment, we divide each term in numerator and denominator by x2 , the higest power of x that appears in either. (Remember that this technique is effective with limits as x → ± ∞ but is not likely to be productive in other cases.) Result: lim x→+∞ 4x2 − x = lim x→+∞ x2 + 9 4 − (1/x) = 1 + (9/x2 ) √ 4−0 = 4 = 2. 1+0 C04S07.014: Divide each term in numerator and denominator by x, because in effect the “term of largest degree” in the numerator is x. Of course in the numerator we must divide each term under the radical by x3 ; the result is that 3 8 1 +3 2 1 x x → 4 3 3− x 1− as x → −∞. C04S07.015: As x → −∞, x2 + 2x = x(x + 2) is the product of two very large negative numbers, so x2 + 2x → +∞. Therefore lim x→−∞ x2 + 2x = + ∞ √ as well. If the large positive number −x is added to x2 + 2x, then the resulting sum also approaches + ∞. But arguments such as this are sometimes misleading (this subject will be taken up in detail in Sections 4.8 and 4.9), so it is more reliable to reason analytically, as follows: lim x→−∞ = lim x→−∞ = lim x→−∞ x2 + 2x − x = lim x→−∞ √ √ x2 + 2x − x x2 + 2x + x x2 + 2x − x2 √ = lim √ x→−∞ x2 + 2x + x x2 + 2x + x 2x 2x x √ = lim √ = lim x→−∞ x x2 + 2x + x x→−∞ x2 + 2x + x − x x 2 = +∞ 2 1− 1+ x 2 2 x2 + 2x x2 (see Note 1) 2 is slightly smaller than 1, so that x because, if x is large negative, then 1 + 1− 1+ 2 x is a very small positive number, approaching zero through positive values as x → −∞. √ Note 1: The minus sign is necessary because x < 0, and therefore x2 = −x, not x. It’s important not to miss this detail because the other sign will give the incorrect limit 1. Note 2: There are so many dangers associated with minus signs and negative numbers in this problem that it would probably be better to let u = −x and recast the problem in the form lim x→−∞ x2 + 2x − x = lim u2 − 2u + u , u→+∞ then multiply numerator and denominator by the conjugate of the numerator as in the previous calculation. C04S07.016: The following computation is correct: lim x→−∞ 2x − 4x2 − 5x = lim x→−∞ 4x2 − (4x2 − 5x) 5x √ √ = lim x→−∞ 2x + 2x + 4x2 − 5x 4x2 − 5x = lim x→−∞ 2+ √ 5 4x2 − 5x √ − x2 5 = lim x→−∞ 2− 5 4− x = −∞. See the Notes for the solution of Problem 15. Following the second note, we let u = −x and proceed as follows: lim x→−∞ 2x − 4x2 − 5x = lim u→+∞ −2u − 4u2 + 5u =− lim u→+∞ 2u + 4u2 + 5u = −∞ with little difficulty with negative numbers or minus signs. C04S07.017: Matches Fig. 4.7.20(g), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , and f (x) → 0 as x → ± ∞. C04S07.018: Matches Fig. 4.7.20(i), because f (x) → −∞ as x → 1+ , f (x) → + ∞ as x → 1− , and f (x) → 0 as x → ± ∞. C04S07.019: Matches Fig. 4.7.20(a), because f (x) → + ∞ as x → 1 and f (x) → 0 as x → ± ∞. C04S07.020: Matches Fig. 4.7.20(d), because f (x) → −∞ as x → 1 and f (x) → 0 as x → ± ∞. C04S07.021: Matches Fig. 4.7.20(f), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , f (x) → + ∞ as x → −1− , f (x) → −∞ as x → −1+ , and f (x) → 0 as x → ± ∞. C04S07.022: Matches Fig. 4.7.20(c), because f (x) → −∞ as x → 1+ , f (x) → + ∞ as x → 1− , f (x) → −∞ as x → −1− , f (x) → + ∞ as x → −1+ , and f (x) → 0 as x → ± ∞. C04S07.023: Matches Fig. 4.7.20(j), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , f (x) → −∞ as x → −1− , f (x) → + ∞ as x → −1+ , and f (x) → 0 as x → ± ∞. 3 C04S07.024: Matches Fig. 4.7.20(h), because f (x) → −∞ as x → 1+ , f (x) → + ∞ as x → 1− , f (x) → + ∞ as x → −1− , f (x) → −∞ as x → −1+ , and f (x) → 0 as x → ± ∞. C04S07.025: Matches Fig. 4.7.20(l), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , and f (x) → 1 as x → ± ∞. C04S07.026: Matches Fig. 4.7.20(b), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , f (x) → + ∞ as x → −1− , f (x) → −∞ as x → −1+ , and f (x) → 1 as x → ± ∞. C04S07.027: Matches Fig. 4.7.20(k), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , and because x2 1 =x+1+ x−1 x−1 f (x) = if x = 1, the graph of f has the slant asymptote with equation y = x + 1. C04S07.028: Matches Fig. 4.7.20(e), because f (x) → + ∞ as x → 1+ , f (x) → −∞ as x → 1− , f (x) → −∞ as x → −1− , f (x) → + ∞ as x → −1+ , and because f (x) = x3 x =x+ 2 , 2−1 x x −1 the graph of f has the slant asymptote y = x. C04S07.029: Given f (x) = 2 , we find that x−3 f (x) = − 2 (x − 3)2 and f (x) = 4 . (x − 3)3 So there are no extrema or inflection points, the only intercept is 0, − 2 , and f (x) → + ∞ as x → 3+ 3 whereas f (x) → −∞ as x → 3− . So the line x = 3 is a vertical asymptote. Also f (x) → 0 as x → ± ∞, so the line y = 0 is a [two-way] horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 10 7.5 5 2.5 2 4 6 -2.5 -5 -7.5 -10 C04S07.030: Given f (x) = 4 , we find that 5−x f (x) = 4 (5 − x)2 and 4 f (x) = 8 . (5 − x)3 So there are no extrema or inflection points, the only intercept is 0, 4 , x = 5 is a vertical asymptote, and 5 the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 10 7.5 5 2.5 2 4 6 8 10 -2.5 -5 -7.5 -10 3 , we find that (x + 2)2 6 f (x) = − and (x + 2)3 C04S07.031: Given f (x) = f (x) = 18 . (x + 2)4 So there are no extrema or inflection points, the only intercept is 0, 3 , the line x = −2 is a vertical 4 asymptote, and the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is shown next. 30 25 20 15 10 5 -5 -4 -3 -2 -1 4 , we find that (3 − x)2 8 f (x) = − and (3 − x)3 1 C04S07.032: Given f (x) = − f (x) = − 24 . (3 − x)4 So there are no extrema or inflection points, the only intercept is 0, − 4 , the line x = 3 is a vertical 9 asymptote, and the line y = 0 is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 2 4 6 -5 -10 -15 -20 C04S07.033: If f (x) = 1 , it follows that (2x − 3)3 6 f (x) = − and (2x − 3)4 5 f (x) = 48 . (2x − 3)5 1 So there are no extrema or inflection points, the only intercept is 0, − 27 , the line x = 3 is a vertical 2 asymptote, and the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 3 2 1 -1 1 2 3 4 -1 -2 -3 x+1 . Then x−1 2 f (x) = − (x − 1)2 C04S07.034: Given: f (x) = and f (x) = 4 . (x − 1)3 Thus there are no extrema or inflection points, the only intercepts are (−1, 0) and (0, −1), the line x = 1 is a vertical asymptote, and the line y = 1 is a [two-way] horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 10 7.5 5 2.5 -2 2 4 -2.5 -5 -7.5 -10 C04S07.035: Given: f (x) = x2 . Then x2 + 1 f (x) = (x2 2x + 1)2 and f (x) = 2(1 − 3x2 ) . (x2 + 1)3 √ Thus the only intercept is (0, 0), where there is a global minimum, there are inflection points at ± 1 3, 1 , 3 4 and the line y = 1 is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is shown next (without the asymptote). 0.8 0.6 0.4 0.2 -4 C04S07.036: Given: f (x) = -2 2 2x , we find that x2 + 1 6 4 f (x) = 2(1 − x2 ) (x2 + 1)2 and f (x) = 4x(x2 − 3) . (x2 + 1)3 √ √ There is a global minimum at (−1, −1), a global maximum at (1, 1), and the three points − 3, − 1 3 , 2 √ 1√ (0, 0), and 3, 2 3 are inflection points. Because f (x) → 0 as x → ± ∞, the y -axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is shown next. 1 0.5 -7.5 -5 -2.5 2.5 5 7.5 -0.5 -1 C04S07.037: If f (x) = 1 , then x2 − 9 f (x) = − 2x (x2 − 9)2 and f (x) = 6(x2 + 3) . (x2 − 9)3 So there is a local maximum at 0, − 1 , which is also the only intercept; there are vertical asymptotes at 9 x = −3 and at x = 3, and the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is shown next 1.5 1 0.5 -4 -2 2 4 -0.5 -1 -1.5 C04S07.038: Given: f (x) = x . Then 4 − x2 f (x) = x2 + 4 (x2 − 4)2 and f (x) = − 2x(x2 + 12) . (x2 − 4)3 Thus there are no extrema, the origin is the only intercept and the only inflection point, there are vertical asymptotes at x = ± 2, and the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 6 4 2 -4 -2 2 -2 -4 -6 7 4 C04S07.039: Given f (x) = 1 1 = , we find that x2 + x − 6 (x − 2)(x + 3) f (x) = − (x2 2x + 1 + x − 6)2 and 2(x2 + 3x + 7) . (x2 + x − 6)3 f (x) = 4 Thus there is a local maximum at − 1 , − 25 and the only intercept is at 0, − 1 . The lines x = −3 and 2 6 x = 2 are vertical asymptotes and the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is shown next. 2 1.5 1 0.5 -4 -2 2 4 -0.5 -1 -1.5 -2 C04S07.040: If f (x) = 2x2 + 1 2x2 + 1 = , then x2 − 2x x(x − 2) f (x) = − 2(2x2 + x − 1) x2 (x − 2)2 and f (x) = 2(4x3 + 3x2 − 6x + 4) . x3 (x − 2)3 So there is a local minimum at (−1, 1), a local maximum at 1 , −2 , and an inflection point close to 2 (−1.851708, 1.101708). There are no intercepts, the lines x = 0 and x = 2 are vertical asymptotes, and the line y = 2 is a horizontal asymptote (not shown in the figure). A Mathematica-generated graph of y = f (x) is next. 20 15 10 5 -2 -1 1 2 3 4 -5 -10 -15 -20 C04S07.041: Given: f (x) = x + 1 x2 + 1 = , we find that x x f (x) = x2 − 1 x2 and that f (x) = 2 . x3 Hence there is a local maximum at (−1, −2), a local minimum at (1, 2), and no inflection points or intercepts. The y -axis is a vertical asymptote and the line y = x is a slant asymptote (not shown in the figure). A 8 Mathematica-generated graph of y = f (x) is next. 10 5 -3 -2 -1 1 2 3 -5 -10 C04S07.042: If f (x) = 2x + e−x , then f (x) = 2ex − 1 ex and f (x) = e−x . Therefore the graph of f is decreasing for x < − ln 2 and increasing otherwise; it is concave upward for all x. Thus there is a global minimum at (− ln 2, 2 − 2 ln 2), no inflection points, and the only intercept is (0, 1). The line y = 2x is a slant asymptote to the right; there is no asymptote to the left. The graph of f is next. 7 6 5 4 3 2 1 -2 C04S07.043: If f (x) = -1 1 2 3 x2 1 =x+1+ , then x−1 x−1 f (x) = x(x − 2) (x − 1)2 and f (x) = 2 . (x − 1)3 So (0, 0) is a local maximum and the only intercept, there is a local minimum at (2, 4), the line x = 1 is a vertical asymptote, and the line y = x + 1 is a slant asymptote (not shown in the figure). A Mathematicagenerated graph of y = f (x) is shown next. 15 10 5 -2 -1 1 2 3 4 5 -5 -10 C04S07.044: Given: f (x) = 2x3 − 5x2 + 4x 1 x(2x2 − 5x + 4) = 2x − 1 + , = 2 − 2x + 1 x (x − 1)2 (x − 1)2 9 we first compute f (x) = 2(x − 2)(x2 − x + 1) (x − 1)3 and f (x) = 6 . (x − 1)4 So there is a local minimum at (2, 4), (0, 0) is the only intercept, there are no inflection points, the line x = 1 is a vertical asymptote, and the line y = 2x − 1 is a slant asymptote (not shown in the figure). A Mathematica-generated graph of y = x(f ) is next. 30 25 20 15 10 5 -1 1 2 3 -5 -10 C04S07.045: If f (x) = 1 , then (x − 1)2 f (x) = − 2 (x − 1)3 and f (x) = 6 . (x − 1)4 Hence there are no extrema or inflection points, the only intercept is (0, 1), the line x = 1 is a vertical asymptote, and the x-axis is a horizontal asymptote. A Mathematica-generated graph of y = f (x) is next. 20 15 10 5 -1 C04S07.046: Given f (x) = 1 2 3 1 , we first find that (1 + ex )2 f (x) = − 2ex (1 + ex )3 and f (x) = 2ex (2ex − 1) . (1 + ex )4 Thus f (x) < 0 for all x, so the graph of f is decreasing everywhere. Next, f (x) = 0 exactly when ex = 1 ; 2 that is, when x = − ln 2. It is no trouble to verify that − ln 2, 4 9 ≈ (−0.693147, 0.444444) is an inflection point of the graph of y = f (x). There can be no x-intercept, but 0, There are two horizontal asymptotes, y = 1 and y = 1, because lim f (x) = 1 x→−∞ and 10 lim f (x) = 0, x→∞ 1 4 is the y -intercept. and no other asymptotes. A graph of y = f (x), generated by Mathematica,shown next. 1 0.8 0.6 0.4 0.2 -4 -2 2 4 C04S07.047: If f (x) = ex , 1 + ex then f (x) = ex (1 + ex )2 and f (x) = ex (1 − ex ) . (1 + ex )3 Therefore the graph of f is increasing everywhere, concave upward for x < 0, and concave downward for x > 0. So there are no extrema, an inflection point at 0, 1 , and two horizontal asymptotes: the line y = 1 2 to the right, the line y = 0 to the left. The graph of f is next. 1 0.8 0.6 0.4 0.2 -4 -2 2 4 C04S07.048: If 1 , + e−x ex (e2x − 2ex − 1)(e2x + 2ex − 1) . (1 + e2x )3 √ It follows that f is increasing for x < 0, decreasing for x > 0, concave downward for ln 2 − 1 < x < √ ln 2 + 1 , and concave upward otherwise. Thus there is a global maximum at 0, 1 and inflection points 2 where the concave structure changes (the y -coordinates of these points are about 0.353553). The x-axis is a horizontal asymptote and the graph is shown next. f (x) = ex then f (x) = ex (1 − ex )(1 + ex ) (1 + e2x )2 and f (x) = 0.5 0.4 0.3 0.2 0.1 -4 -2 2 11 4 C04S07.049: If f (x) = 1 1 = , then x2 − x − 2 (x − 2)(x + 1) f (x) = 1 − 2x (x2 − x − 2)2 and f (x) = 6(x2 − x + 1) . (x2 − x − 2)3 Therefore 1 , − 4 is a local maximum and the only extremum, 0, − 1 is the only intercept, the lines x = −1 2 9 2 and x = 2 are vertical asymptotes, and the x-axis is a horizonal asymptote. A Mathematica-generated graph of y = f (x) is next. 6 4 2 -2 -1 1 2 3 -2 -4 -6 C04S07.050: If f (x) = 1 , then (x − 1)(x + 1)2 f (x) = 1 − 3x (x − 1)2 (x + 1)3 and f (x) = 4(3x2 − 2x + 1) . (x − 1)3 (x + 1)4 Therefore (0, −1) is the only intercept, 1 , − 27 is a local maximum and the only extremum, the lines 3 32 x = −1 and x = 1 are vertical asymptotes, and the x-axis is a horizontal asymptote. A Mathematicagenerated graph of y = f (x) is shown next. 20 15 10 5 -2 -1 1 2 -5 -10 -15 -20 C04S07.051: Given: f (x) = x2 − 4 4 = x − , we find that x x f (x) = x2 + 4 x2 and that f (x) = − 8 . x3 Therefore (−2, 0) and (2, 0) are the only intercepts, there are no inflection points or extrema, the y -axis is a vertical asymptote, and the line y = x is a slant asymptote (not shown in the figure). A Mathematica12 generated graph of y = f (x) is next. 20 15 10 5 -3 -2 -1 1 2 3 -5 -10 -15 -20 C04S07.052: If f (x) = ex − e−x , then ex + e−x f (x) = 4e2x (1 + e2x )2 and f (x) = 8e2x (1 − ex )(1 + ex ) . (1 + e2x )3 Therefore the graph of f is increasing for all x, concave upward if x < 0, and concave downward if x > 0. Thus there are no extrema and (0, 0) is an inflection point. The line y = 1 is a horizontal asymptote to the right and the line y = −1 is a horizontal asymptote to the left. The graph is next. 1 0.5 -3 -2 -1 1 2 3 -0.5 -1 C04S07.053: If f (x) = x3 − 4 4 = x − 2 , then 2 x x f (x) = x3 + 8 x3 and f (x) = − 24 . x4 √ Thus 3 4, 0 is the only intercept, there is a local maximum at (−2, −3) and no other extrema, and there are no inflection points. The y -axis is a vertical asymptote and the line y = x is a slant asymptote (not shown in the figure). A Mathematica-generated graph of y = f (x) is next. -4 -2 2 4 -5 -10 -15 -20 C04S07.054: If f (x) = x2 + 1 5 =x+2+ , then x−2 x−2 f (x) = x2 − 4x − 1 (x − 2)2 and 13 f (x) = 10 . (x − 2)3 Therefore the √ only intercept is 0, − 1 and there√ no inflection points. There is a local maximum at are 2 √ √ 2 − 5, 4 − 2 5 and a local minimum at 2 + 5, 4 + 2 5 . The line x = 2 is a vertical asymptote and the line y = x + 2 is a slant asymptote (not shown in the figure). A Mathematica-generated graph of y = f (x) is next. 30 20 10 -4 -2 2 4 6 8 10 -10 C04S07.055: The x-axis is a horizontal asymptote, and there are vertical asymptotes at x = 0 and x = 2. There are local minima at (−1.9095, −0.3132) and (1.3907, 3.2649) and a local maximum at (4.5188, 0.1630) (all coordinates approximate, of course), and inflection points at (−2.8119, −0.2768) and (6.0623, 0.1449). A Mathematica-generated graph of y = f (x) is next. 30 20 10 -2 -1 1 2 3 4 -10 A “close-up” of the graph for 3 8 is next, on the left, and another for −7 x x −1 is on the right. 0.16 0.2 0.14 0.1 0.12 4 5 6 7 8 -7 0.08 -6 -5 -4 -3 -2 -1 -0.1 0.06 -0.2 0.04 -0.3 C04S07.056: The line y = 1 is a horizontal asymptote (not shown in the figures) and the lines x = 0 and x = 4 are vertical asymptotes. There is a local minimum at (−1, 0) and inflection points at (−1.5300, 0.0983) and (2.1540, 0.9826) (numbers with decimal points are approximations). A Mathematica-generated graph 14 of y = f (x) is next, on the left; a close-up of the graph for −7.5 x −0.5 is on the right. 15 10 0.6 5 0.4 -2 2 4 6 -5 0.2 -10 -15 -7 -6 -5 -4 -3 -2 -1 C04S07.057: The x-axis is a horizontal asymptote and there are vertical asymptotes at x = 0 and x = 2. There are local minima at (−2.8173, −0.1783) and (1.4695, 5.5444) and local maxima at (−1, 0) and (4.3478, 0.1998). There are inflection points at the three points (−4.3611, −0.1576), (−1.2569, −0.0434), and (5.7008, 0.1769). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is next. 25 20 15 10 5 -2 2 4 -5 -10 -15 A “close-up” of the graph for −12 right. -12 -10 -8 -6 x -4 −0.5 is shown next, on the left; the graph for 3 x 10 is on the 0.2 -2 -0.1 -0.2 -0.3 0.15 0.1 0.05 -0.4 4 5 6 7 8 9 10 C04S07.058: The horizontal line y = 1 is an asymptote, as are the vertical lines x = 0 and x = 2. There are local maxima at (−5.6056, 1.1726) and (1.6056, −8.0861), local minima at (−1, 0), and (3, 0). (Numbers with decimal points are approximations.) There are inflection points at (−8.54627, 1.15324), (−1.29941, 0.228917), and (3.67765, 0.120408). A Mathematica-generated graph of y = f (x) is shown next, 15 on the left; the graph for −10 x −0.5 is on the right along with the horizontal asymptote. 20 2.5 15 10 2 5 1.5 -2 2 4 1 -5 -10 0.5 -15 -20 The graph for 2.5 -10 -8 10 is next, on the left; the graph for 10 x -6 x -4 -2 60 is on the right. 0.7 0.95 0.6 0.5 0.9 0.4 0.85 0.3 0.2 20 0.1 4 5 6 7 8 9 30 40 50 60 0.75 10 C04S07.059: The horizontal line y = 0 is an asymptote, as are the vertical lines x = 0 and x = 2. There are local minima at (−2.6643, −0.2160), (1.2471, 14.1117), and (3, 0); there are local maxima at (−1, 0) and (5.4172, 0.1296). There are inflection points at (−4.0562, −0.1900), (−1.2469, −0.0538), (3.3264, 0.0308), and (7.4969, 0.1147). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is shown next. 30 20 10 -2 2 -10 -20 -30 16 4 The graph for −10 -10 −0.5 is next, on the left; the graph for 2.5 x -8 -6 -4 10 is on the right. x 0.3 -2 -0.1 0.25 0.2 -0.2 0.15 -0.3 0.1 -0.4 0.05 -0.5 4 5 6 7 8 9 10 C04S07.060: The horizontal line y = 0 is an asymptote, as are the vertical lines x = 0 and x = 2. There are local minima at (−1.5125, −1.4172) and (3, 0), local maxima at (1.2904, −25.4845) and (9.2221, 0.0519) and inflection points at (−2.0145, −1.2127), (4.2422, 0.0145), and (14.2106, 0.0460). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is next, on the left; the graph for −10 x −0.75 is on the right. 1.5 40 1 20 0.5 -2 2 4 -10 -8 -6 -4 -2 -0.5 -20 -1 -40 The graph for 2.5 10 is next, on the left; the graph for 5 x 0.1 x 25 is on the right. 0.05 0.08 0.045 0.06 0.04 0.04 0.035 0.02 4 5 6 7 8 9 10 10 15 20 25 C04S07.061: The x-axis is a horizontal asymptote; there are vertical asymptotes at x = −0.5321, x = √ 0.6527, and x = 2.8794. There is a local minimum at (0, 0) and a local maximum at ( 3 2, −0.9008). There are no inflection points (Numbers with decimal points are approximations.) A Mathematics -generated graph 17 of y = f (x) is next. 4 2 -2 -1 1 2 3 4 5 -2 -4 C04S07.062: The x-axis is a horizontal asymptote; there is a vertical asymptote at x = −1.1038. There is a local minimum at (0, 0) and a local maximum at (2.1544, 4.3168). There are inflection points at (1.8107, 2.9787) and (2.4759, 3.4299). (Numbers with decimal points are approximations.) A Mathematicagenerated graph of y = f (x) is next. 6 4 2 -2 2 4 -2 -4 C04S07.063: The line y = x + 3 is a slant asymptote in both the positive and negative directions; thus there is no horizontal asymptote. There is a vertical asymptote at x = −1.1038. There are local maxima at (−2.3562, −1.8292) and (2.3761, 18.5247), local minima at (0.8212, 0.6146) and (5.0827, 11.0886). There are inflection points at (1.9433, 11.3790) and (2.7040, 16.8013). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is next, on the left; on the right the graph is shown on a wider scale, together with its slant asymptote. 25 25 20 20 15 15 10 10 5 5 -2 2 4 -10 -5 5 -5 -10 15 -10 -15 10 -5 -15 C04S07.064: The line 2y = x is a slant asymptote in both the positive and negative directions; thus there is no horizontal asymptote. There is a vertical asymptote at x = −1.4757. There is a local maximum at (−2.9821, −2.1859) and a local minimum at (0.7868, −2.8741). There are inflection points at 18 (−0.2971, 0.7736), (0.5713, 0.5566), (1, −2), and (9.1960, 4.6515). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is shown next, on the left; the graph is also shown on the right, on a wider scale, along with its slant asymptote. 15 10 10 5 5 -3 -2 -1 1 2 3 -6 -4 -2 2 4 6 -5 -5 -10 -10 -15 C04S07.065: Given f (x) = f (x) = x5 − 4x2 + 1 , we first find that 2x4 − 3x + 2 2x8 + 4x5 + 10x4 − 8x3 + 12x2 − 16x + 3 (2x4 − 3x + 2)2 f (x) = − and 2(30x8 + 24x7 − 40x6 + 90x5 − 102x4 − 28x3 + 24x2 + 7) . (2x4 − 3x + 2)3 The line 2y = x is a slant asymptote in both the positive and negative directions; thus there is no horizontal asymptote. There also are no vertical asymptotes. There is a local maximum at (0.2200976580, 0.6000775882), a local minimum at (0.8221567934, −2.9690453671), and inflection points at (−2.2416918017, −1.2782199626), (−0.5946286318, −0.1211409770), (0.6700908810, −1.6820255735), and (0.96490314661, −2.2501145861). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is next. 2 1 -4 -2 2 4 -1 -2 -3 C04S07.066: The line 2y = x is a slant asymptote in both the positive and negative directions; thus there is no horizontal asymptote. There also are no vertical asymptotes (the denominator in f (x) is never zero). There are x-intercepts where x = −2.05667157818, x = 0.847885655376, 19 and x = 1.929095045219 and 0, 2 is the y -intercept. (Numbers with decimal points are approximations throughout.) There are 5 local maximum at (−1.137867740647, 0.426255896993) and (0.472659729564, 0.585125167363) and local minima at (−0.394835802615, 0.318479939692) and (1.203561740743, −0.770172527937). There are inflection points at (−2.381297805169, −0.253127890429), (−0.775152255017, 0.373332211612), and (1.553505928225, −0.444592872637). (0.189601709800, 0.486606037763), (0.890253166310, −0.145751465654), A Mathematica-generated graph of y = f (x) is shown next, on the left; a wider view is on the right, along with the slant asymptote. 0.6 2 0.2 -2 3 0.4 1 -1 1 2 -6 -4 -2 2 -0.2 -0.4 6 -2 -0.6 4 -1 -3 -0.8 C04S07.067: The line 2y = x is a slant asymptote in both the positive and negative directions; thus there is no horizontal asymptote. There is a vertical asymptote at x = −1.7277. There are local maxima at (−3.1594, −2.3665) and (1.3381, 1.7792), local minima at (−0.5379, −0.3591) and (1.8786, 1.4388). There are inflection points at (0, 0), (0.5324, 0.4805), (1.1607, 1.4294), and (1.4627, 1.6727). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is next, on the left; the figure on the right shows the graph for −1.5 x 5. 2.5 4 2 2 1.5 -4 -2 2 4 1 -2 0.5 -4 -1 1 2 3 4 5 C04S07.068: The line 6x + 10 = 9y is a slant asymptote in both the positive and negative directions; thus there is no horizontal asymptote. There is a vertical asymptote at x = −0.8529. There are local maxima at (−1.3637, −0.0573) and (0.7710, 1.5254), local minima at 0, 3 and (1.1703, 1.4578). There are inflection 2 points at (0.5460, 1.5154), (1.0725, 1.4793), (1.3880, 1.9432), and (1.8247 , 2.6353). (Numbers with decimal points are approximations.) A Mathematica-generated graph of y = f (x) is shown next, on the left (with 20 the vertical asymptote but without the slant asymptote). The figure on the right shows the graph of f for −0.5 x 1.5. 4 1.56 3 2 1 -3 -2 1.54 1.52 -1 1 2 3 -0.5 -1 0.5 1 1.5 1.48 -2 1.46 C04S07.069: Sketch the parabola y = x2 , but modify it by changing its behavior near x = 0: Let y → +∞ as x → 0+ and let y → −∞ as x → 0− . Using calculus, we compute f (x) = 2(x3 − 1) x2 and f (x) = 2(x3 + 2) . x3 It follows that the √ graph of f is decreasing for 0 < x < 1 and for x < 0, increasing for x > 1. It is concave √ upward for x < − 3 2 and also for x > 0, concave downward for − 3 2 < x < 0. The only intercept is at √ − 3 2, 0 ; this is also the only inflection point. There is a local minimum at (1, 3). The y -axis is a vertical asymptote. A Mathematica-generated graph of y = f (x) is shown next. 10 7.5 5 2.5 -2 -1 1 2 3 -2.5 -5 -7.5 -10 C04S07.070: Because f (x) ≈ x3 when |x| is large, we obtain the graph of f by making “modifications” in the graph of y = x3 at and near the discontinuity of f (x) at x = 1. We are aided in sketching the graph of f by finding its x-intercepts—these are approximately −1.654 and 2.172—as well as its y -intercept 12 and its inflection points (2.22, 1.04) and (−0.75, 6.44) (also approximations). A Mathematica-generated graph of f 21 appears next. 80 60 40 20 -4 -2 2 -20 -40 -60 -80 22 4 Section 4.8 C04S08.001: You don’t need l’Hˆpital’s rule to evaluate this limit, but you may use it: o x−1 1 1 = lim =. x2 − 1 x→1 2x 2 lim x→1 C00S08.002: You don’t need l’Hˆpital’s rule to evaluate this limit, but you may use it: o lim x→∞ 3x − 4 3 3 = lim =. 2x − 5 x→∞ 2 2 C04S08.003: You don’t need l’Hˆpital’s rule to evaluate this limit, but you may use it (twice): o 2x2 − 1 4x 4 2 = lim = lim =. 2 + 3x x→∞ 10x + 3 x→∞ 10 5x 5 lim x→∞ C04S08.004: You don’t need l’Hˆpital’s rule to evaluate this limit (apply the definition of the derivative o to the evaluation of f (0) where f (x) = e3x ), but you may use it: lim x→0 e3x − 1 3e3x = lim = 3. x→0 x 1 C04S08.005: Without l’Hˆpital’s rule: o lim x→0 sin x2 sin x2 = 0 · 1 = 0. = lim x · x→0 x x2 (We used Theorem 1 of Section 2.3, lim x→0 With l’Hˆpital’s rule: o lim x→0 sin x = 1, and the product law for limits.) x sin x2 2x cos x2 = lim = 2 · 0 · 1 = 0. x→0 x 1 C04S08.006: You don’t need l’Hˆpital’s rule to evaluate this limit (see the solution to Problem 3 of Section o 2.3), but you may use it: 1 − cos lim + x x→0 √ x = lim+ sin x1/2 sin x1/2 1 = lim+ = 1/2 1 2 x→0 2x 1 −1/2 2x x→0 by Theorem 1 of Section 2.3. If you prefer a “pure” l’Hˆpital’s rule solution, you should substitute x = u2 o to obtain √ 1 − cos x 1 − cos u sin u cos u 1 lim = lim = lim = lim =. + u→0 u→0 2u u→0 x u2 2 2 x→0 Note that it was necessary to apply l’Hˆpital’s rule twice in the second solution. o C04S08.007: You may not use l’Hˆpital’s rule! The numerator is approaching zero but the denominator o is not. Hence use the quotient law for limits (Section 2.2): lim x→1 lim (x − 1) x−1 0 = x→1 = = 0. sin x lim sin x sin 1 x→1 1 Note that illegal use of l’Hˆpital’s rule in this problem will result in the incorrect value sec 1 ≈ 1.8508157177 o for the limit. C04S08.008: The numerator and denominator of the fraction are both approaching zero, so you may try using l’Hˆpital’s rule (twice): o lim x→0 1 − cos x sin x cos x = lim = lim . 3 2 x→0 3x x→0 6x x But the latter limit does not exist (the left-hand limit is −∞ and the right-hand limit is + ∞). So the hypotheses of Theorem 1 of Section 4.8 are not satisfied; this is a case in which l’Hˆpital’s rule (as stated in o Theorem 1) has failed. Other measures are needed. By l’Hˆpital’s rule (or by Problem 3 of Section 2.3), we o have lim x→0 1 − cos x sin x cos x 1 = lim = lim =. x→0 2x x→0 x2 2 2 Therefore 1 − cos x 1 ≈ 3 x 2x if x is close to zero. Consequently lim x→0 1 − cos x does not exist. x3 C04S08.009: Without l’Hˆpital’s rule we might need to resort to the Taylor series methods of Chapter 9 o to evaluate this limit. But l’Hˆpital’s rule may be applied (twice): o lim x→0 ex − x − 1 ex − 1 ex 1 = lim = lim =. x→0 x→0 2 x2 2x 2 C04S08.010: You may not apply l’Hˆpital’s rule: The numerator is approaching zero but the denominator o is not. But this means that the quotient law of limits (Section 2.2) may be applied instead: lim z →π /2 1 + cos 2z = 1 − sin 2z lim (1 + cos 2z ) z →π /2 lim (1 − sin 2z ) z →π /2 = 1−1 = 0. 1−0 This problem would be more interesting if the denominator were 1 − sin z . C04S08.011: applied: The numerator and denominator are both approaching zero, so l’Hˆpital’s rule may be o lim u →0 u tan u tan u + u sec2 u = lim u→0 1 − cos u sin u = lim u→0 sec u + u · sec2 u sin u = 1 + 1 · 12 = 2 . Note that we used the sum law for limits (Section 2.2) and the fact that lim u→0 u = 1, sin u a consequence of Theorem 1 of Section 2.3 and the quotient law for limits. 2 C04S08.012: Numerator and denominator are both approaching zero, so l’Hˆpital’s rule may be applied o (twice): lim x→0 x − tan x 1 − sec2 x = lim 3 x→0 x 3x2 = lim x→0 C04S08.013: lim x→∞ − 2 sec2 x tan x 6x ln x = lim x→∞ x · x1/10 1 1 −9/10 10 x =− 1 3 = lim x→∞ lim x→0 sec2 x sin x · cos x x =− 1 12 1 · ·1 = − . 31 3 10 = 0. x1/10 C04S08.014: Several applications of l’Hˆpital’s rule yield o lim r →∞ er er er er er = lim = lim = lim = = + ∞. r →∞ 4(r + 1)3 r →∞ 12(r + 1)2 r →∞ 24(r + 1) (r + 1)4 24 Even though the limit does not exist, the hypotheses of Theorem 1 of Section 4.8 are all satisfied, so the answer is correct. C04S08.015: C04S08.016: lim ln(x − 9) 1 = lim = 1. x→10 1 · (x − 9) x − 10 lim t2 + 1 2t 2 = lim = lim −1 = lim (2t) = + ∞. t→∞ 1 + ln t t→∞ t t→∞ t ln t x→10 t→∞ C04S08.017: Always verify that the hypotheses of l’Hˆpital’s rule are satisfied. o lim x→0 ex + e−x − 2 ex − e−x ex + e−x 1+1 = lim = lim = = 1. x→0 x cos x + sin x x→0 2 cos x − x sin x x sin x 2−0 C04S08.018: As x → (π /2)− , tan x → + ∞ and cos x → 0+ , so that ln(cos x) → −∞. Hence l’Hˆpital’s o rule may be tried: lim x→(π /2)− tan x sec2 x cos x − sec x = lim = lim = −∞ ln(cos x) x→(π/2)− − sin x x→(π /2)− sin x because sin x → 1 and sec x → + ∞ as x → (π /2)− . C04S08.019: Methods of Section 2.3 may be used, or l’Hˆpital’s rule yields o lim x→0 C04S08.020: lim x→0 sin 3x 3 cos 3x 3·1 3 = lim = =. tan 5x x→0 5 sec2 5x 5·1 5 ex − e−x ex + e−x = lim = 2. x→0 x 1 C04S08.021: The factoring techniques of Section 2.2 work well here, or l’Hˆpital’s rule yields o lim x→1 C04S08.022: lim x→2 x3 − 1 3x2 3 = lim =. 2−1 x→1 2x x 2 x3 − 8 3x2 12 3 = = lim =. x4 − 16 x→2 4x3 32 8 3 C04S08.023: Both numerator and denominator approach + ∞ as x does, so we may attempt to find the limit with l’Hˆpital’s rule: o lim x→∞ x + sin x 1 + cos x = lim , 3x + cos x x→∞ 3 − sin x but the latter limit does not exist (and so the equals mark in the previous equation is invalid). The reason: As x → + ∞, x runs infinitely many times through numbers of the form nπ where n is a positive even integer. At such real numbers the value of 1 + cos x 3 − sin x f (x) = is 2 . But x also runs infinitely often through numbers of the form nπ where n is a positive odd integer. At 3 these real numbers the value of f (x) is 0. Because f (x) takes on these two distinct values infinitely often as x → + ∞, f (x) has no limit as x → + ∞. This does not imply that the limit given in Problem 22 does not exist. (Read Theorem 1 carefully.) In fact, the limit does exist, and we have here the rare phenomenon of failure of l’Hˆpital’s rule. Other o techniques must be used to solve this problem. Perhaps the simplest is this: sin x 1+ x + sin x 1 1+0 x lim = lim cos x = 3 − 0 = 3 . x→∞ 3x + cos x x→∞ 3− x C04S08.024: First we try l’Hˆpital’s rule: o lim x→∞ (x2 + 4)1/2 = lim x→∞ x = lim 1 2 2 (x x→∞ 1 (x2 2 + 4)−1/2 · 2x x = lim x→∞ (x2 + 4)1/2 1 1 (x2 + 4)1/2 = lim . −1/2 · 2x x→∞ x + 4) Another failure of l’Hˆpital’s rule! Here are two ways to find the limit. o lim x→∞ (x2 + 4)1/2 x 2 = lim x→∞ 2 (x2 + 4)1/2 x = lim x→∞ x2 + 4 2x 1 = lim = lim = 1. 2 x→∞ 2x x→∞ 1 x Therefore the original limit is also 1. Second method: lim x→∞ C04S08.025: lim x→0 (x2 + 4)1/2 = lim x→∞ x x2 + 4 x2 1 /2 = lim x→∞ 1+ 4 x2 1 /2 = √ 1 = 1. 2x − 1 2x ln 2 ln 2 = lim x = ≈ 0.6309297536. x−1 x→0 3 ln 3 3 ln 3 C04S08.026: You may apply l’Hˆpital’s rule, but watch what happens: o lim x→∞ 2x 2x ln 2 2x (ln 2)2 2x (ln 2)3 = lim x = lim x = ··· . = lim x x→∞ 3 ln 3 x→∞ 3 (ln 3)2 x→∞ 3 (ln 3)3 3x If we knew that the original limit existed and was finite, we could conclude that it must be zero, but we don’t even know that it exists. But 4 3 2 3x = 27 8 x > 3x > ex , so lim x→∞ 3 2 x = + ∞, and therefore lim x→∞ 2x = lim x→∞ 3x 2 3 x = 0. C04S08.027: You can work this problem without l’Hˆpital’s rule, but if you intend to use it you should o probably proceed as follows: √ x2 − 1 lim √ = lim x→∞ x→∞ 4x2 − x = lim 1/2 x2 − 1 4x2 − x x→∞ 2x 8x − 1 = lim x→∞ 1 /2 x2 − 1 4x2 − x 1 /2 lim 2 8 = = lim x→∞ = 1 /2 1 4 1 /2 = 1 . 2 C04S08.028: As in the previous solution, √ x3 + x lim √ = x→∞ 2x3 − 4 = C04S08.029: lim x→0 lim x3 + x 2x3 − 4 lim 6x 12x x→∞ x→∞ 1/2 1 /2 = 1 2 x→∞ 1 /2 3x2 + 1 6x2 1 /2 1 =√ . 2 ln(1 + x) 1 = lim = 1. x→0 1 · (1 + x) x C04S08.030: It would be easier to establish first that lim x→∞ ln(ln x) = 0, ln x and it would follow immediately that the given limit is zero as well. But let’s see how well a direct approach succeeds. 1 ln(ln x) 1 x ln x = lim lim = lim =0 x→∞ x ln x x→∞ 1 + ln x x→∞ (1 + ln x)(x ln x) because 1 1 < (1 + ln x)(x ln x) x if x > 1.8554 and 1/x → 0 as x → ∞. C04S08.031: Three applications of l’Hˆpital’s rule yield o lim x→0 2ex − x2 − 2x − 2 2ex − 2x − 2 2ex − 2 2ex 1 = lim = lim = lim =. x→0 x→0 x→0 6 x3 3x2 6x 3 C04S08.032: Three applications of l’Hˆpital’s rule yield o 5 lim x→0 sin x − tan x cos x − sec2 x − sin x − 2 sec2 x tan x = lim = lim x→0 x→0 x3 3x2 6x = lim x→0 − cos x − 4 sec2 x tan2 x − 2 sec4 x −1 − 0 − 2 1 = =− . 6 6 2 C04S08.033: lim 2 − ex − e−x e−x − ex −e−x − ex 1 = lim = lim =− . 2 x→0 x→0 2x 4x 4 2 C04S08.034: lim e3x − e−3x 3e3x + 3e−3x = lim = 3. x→0 2x 2 x→0 x→0 C04S08.035: lim x→π /2 2x − π 2 1 = lim = = 1. 2 2x tan 2x sec2 π x→π /2 2 sec C04S08.036: The “direct approach” yields lim x→π /2 sec x sec x tan x tan x sec2 x sec x = lim = lim = lim = lim = ··· . 2x tan x x→π/2 sec x→π /2 sec x x→π /2 sec x tan x x→π /2 tan x Proceed instead as follows (without l’Hˆpital’s rule): o lim x→π /2 C04S08.037: lim x→2 C04S08.038: lim sec x 1 cos x 1 1 = lim · = lim = = 1. tan x x→π/2 cos x sin x 1 x→π /2 sin x x − 2 cos π x 1 + 2π sin π x 1+0 1 = lim = =. x→2 x2 − 4 2x 4 4 x→1/2 2x − sin π x 2 − π cos π x 2−π·0 1 = lim = =. 4x2 − 1 8x 4 2 x→1/2 C04S08.039: We first simplify (using laws of logarithms), then apply l’Hˆpital’s rule: o lim x→0+ ln(2x) = lim+ x→0 ln(3x)1/3 1 /2 1 2 (ln 2 1 3 (ln 3 + ln x) 3 = lim+ 2 x→0 + ln x) 1 x = 3. 1 2 x C04S08.040: One application of l’Hˆpital’s rule yields o lim x→0 ? = ln(1 + x) ln(1 − x2 ) lim x→0 x−1 . 2x The limit on the right-hand side does not exist, so we use left-hand and right-hand limits: lim x→0+ ln(1 + x) x−1 = lim+ = −∞ ln(1 − x2 ) 2x x→0 and Therefore the original limit does not exist. C04S08.041: Two applications of l’Hˆpital’s rule yield o 6 lim x→0− ln(1 + x) x−1 = lim− = + ∞. ln(1 − x2 ) 2x x→0 lim x→0 exp(x3 ) − 1 3x2 exp(x3 ) (6x + 9x4 ) exp(x3 ) = lim = lim x→0 x→0 x − sin x 1 − cos x sin x = lim x→0 (6 + 9x3 ) exp(x3 ) (6 + 0) · 1 = 6. = sin x 1 x Alternatively, three applications of l’Hˆpital’s rule yield o lim x→0 exp(x3 ) − 1 3x2 exp(x3 ) (6x + 9x4 ) exp(x3 ) = lim = lim x→0 x→0 x − sin x 1 − cos x sin x = lim x→0 (6 + 54x3 + 27x6 ) exp(x3 ) (6 + 54 · 0 + 27 · 0) · 1 = = 6. cos x 1 C04S08.042: The technique of multiplying numerator and denominator by the conjugate of the numerator succeeds just as it did in Sections 2.2 and 2.3. Use of l’Hˆpital’s rule yields o lim x→0 C04S08.043: lim x→0 (1 + 3x)1/2 − 1 = lim x→0 x (1 + 4x)1/3 − 1 = lim x→0 x 1 2 (1 1 3 (1 + 3x)−1/2 · 3 3 3 = lim =. x→0 2(1 + 3x)1/2 1 2 + 4x)−2/3 · 4 4 4 = lim =. x→0 3(1 + 4x)2/3 1 3 C04S08.044: Multiplication of numerator and denominator by the conjugate of the numerator is one way to find this limit; l’Hˆpital’s rule yields o lim x→0 (3 + 2x)1/2 − (3 + x)1/2 = lim x→0 x = lim x→0 1 2 (3 + 2x)−1/2 · 2 − 1 (3 + x)−1/2 2 1 1 1 − (3 + 2x)1/2 2(3 + x)1/2 1 1 1 =√ − √ = √ . 3 23 23 C04S08.045: If you want to use the conjugate technique to find this limit, you need to know that the conjugate of a1/3 − b1/3 is a2/3 + a1/3 b1/3 + b2/3 , and the algebra becomes rather long. Here l’Hˆpital’s rule o is probably the easy way: lim x→0 (1 + x)1/3 − (1 − x)1/3 = lim x→0 x = lim x→0 C04S08.046: lim x→π /4 C04S08.047: lim x→0 1 3 (1 + x)−2/3 + 1 (1 − x)−2/3 3 1 1 1 + 2/3 3(1 + x) 3(1 − x)2/3 √ −2 1 − tan x − sec2 x = lim = 4x − π 4 4 x→π /4 2 = 11 2 +=. 33 3 1 =− . 2 ln(1 + x2 ) 2x 0 = lim = = 0. ex − cos x x→0 (1 + x2 )(ex + sin x) (1 + 0)(1 + 0) C04S08.048: Because the numerator and denominator are both approaching zero as x → 2, it should be possible to factor x − 2 out of each, cancel, and proceed without l’Hˆpital’s rule. But if we use l’Hˆpital’s o o rule, the result is 7 lim x→2 x5 − 5x2 − 12 5x4 − 10x 80 − 20 60 1 = lim = = = . 10 − 500x − 24 9 − 500 x→2 10x x 5120 − 500 4620 77 The factor-and-cancel technique yields lim x→2 x5 − 5x2 − 12 (x − 2)(x4 + 2x3 + 4x2 + 3x + 6) = lim 10 − 500x − 24 9 + 2x8 + 4x7 + 8x6 + 16x5 + 32x4 + 64x3 + 128x2 + 256x + 12) x→2 (x − 2)(x x = lim x→2 = x4 + 2x3 + 4x2 + 3x + 6 x9 + 2x8 + 4x7 + 8x6 + 16x5 + 32x4 + 64x3 + 128x2 + 256x + 12 16 + 16 + 16 + 6 + 6 60 1 = = . 512 + 512 + 512 + 512 + 512 + 512 + 512 + 512 + 512 + 12 4620 77 In this problem l’Hˆpital’s rule seems the better choice. o C04S08.049: If f (x) = sin2 x , then x 2 sin x cos x = 2 · 0 · 1 = 0. 1 lim f (x) = lim x→0 x→0 The graph of y = f (x) is next. 0.6 0.4 0.2 -6 -4 -2 2 4 6 -0.2 -0.4 -0.6 C04S08.050: We don’t need l’Hˆpital’s rule—we could use Theorem 1 in Section 2.3—but we’ll use the o rule anyway: lim x→0 sin2 x 2 sin x cos x = lim = x→0 x2 2x lim x→0 sin x x lim cos x = x→0 The graph is next. 1 0.8 0.6 0.4 0.2 -6 -4 -2 2 C04S08.051: Here we have 8 4 6 lim x→0 cos x · 1 = 1 · 1 = 1. 1 lim x→π sin x cos x = lim = cos π = −1. x→π x−π 1 The graph is next. 0.2 -5 -2.5 2.5 5 7.5 10 12.5 -0.2 -0.4 -0.6 -0.8 -1 C04S08.052: One application of l’Hˆpital’s rule yields o cos x − sin x 1 = lim =− . 2x − π x→π/2 2 2 lim x→π /2 The graph is next. 0.1 -5 -2.5 2.5 5 7.5 -0.1 -0.2 -0.3 -0.4 -0.5 C04S08.053: Two applications of l’Hˆpital’s rule yield o lim x→0 1 − cos x sin x cos x 1 = lim = lim =. 2 x→0 2x x→0 x 2 2 The graph is next. 0.5 0.4 0.3 0.2 0.1 -10 -5 5 10 C04S08.054: Three applications of l’Hˆpital’s rule yield o lim x→0 x − sin x 1 − cos x sin x cos x 1 = lim = lim = lim =. 3 2 x→0 x→0 6x x→0 x 3x 6 6 9 The graph is next. 0.15 0.125 0.1 0.075 0.05 0.025 -15 -10 -5 5 10 15 C04S08.055: As x → −∞, f (x) = xe−x → −∞, but lim f (x) = lim x→∞ x→∞ x 1 = lim x = 0. x→∞ e ex Also f (x) = (1 − x)e−x and f (x) = (x − 2)e−x . It follows that the graph of f is increasing for x < 1, decreasing for x > 1, concave downward for x < 2, and concave upward for x > 2. The positive x-axis is a horizontal asymptote and the only intercept is (0, 0). The graph of y = f (x) is shown next. 0.2 -1 1 2 3 4 5 -0.2 -0.4 C04S08.056: Given f (x) = x1/2 e−x , we first use l’Hˆpital’s rule: o lim f (x) = lim x→∞ x→∞ x1/2 1 = lim = 0. x→∞ 2x1/2 ex ex Next, f (x) = 1 − 2x 2x1/2 ex and f (x) = 4x2 − 4x − 1 . 4x3/2 ex So the graph of f is increasing for 0 < x < 1 and decreasing for x > 1 . There is an inflection point 2 2 √ where x = 1 1 + 2 ; the y -coordinate is approximately 0.3285738758. The positive x-axis is a horizontal 2 asymptote and the only intercept is (0, 0). The graph of y = f (x) is shown next. 0.4 0.3 0.2 0.1 1 2 3 10 4 C04S08.057: If f (x) = x exp −x1/2 , then lim f (x) = lim x→∞ x→∞ x 2x1/2 2x1/2 2 = lim = lim 1/2 = lim = 0. x→∞ exp x1/2 x→∞ x x→∞ exp x1/2 exp x1/2 exp x1/2 Thus the positive x-axis is a horizontal asymptote. Next, f (x) = 2 − x1/2 2 exp x1/2 and f (x) = x1/2 − 3 . 4x1/2 exp x1/2 Hence the graph of f is increasing for 0 < x < 4 and decreasing for x > 4; it is concave downward if 0 < x < 9 and concave upward if x > 9. The only intercept is (0, 0). The graph of y = f (x) is shown next. 0.6 0.5 0.4 0.3 0.2 0.1 2.5 5 7.5 10 12.5 15 17.5 20 C04S08.058: Given: f (x) = x2 e−2x . By l’Hˆpital’s rule, o lim f (x) = lim x→∞ x→∞ x2 2x 2 = lim = lim = 0. 2x x→∞ 2e2x x→∞ 4e2x e So the positive x-axis is a horizontal asymptote. Also note that f (x) → + ∞ as x → −∞. Next, f (x) = 2x(1 − x)e−2x and f (x) = 2(2x2 − 4x + 1)e−2x , and it follows that the graph of f is decreasing if x < 0 and if x > 1, increasing if 0 < x < 1; it is concave √ √ upward for x < a = 1 2 − 2 and for x > b = 1 2 + 2 . It is concave downward for a < x < b, so there 2 2 are inflection points where x = a and where x = b. The graph of y = f (x) is next. 0.3 0.25 0.2 0.15 0.1 0.05 -1 C04S08.059: Given: f (x) = 1 2 3 4 ln x . So x lim f (x) = lim x→0+ x→0+ ln x = −∞ x because the numerator is approaching −∞ and the denominator is approaching 0 through positive values. Next, using l’Hˆpital’s rule, o 11 ln x 1 = lim = 0, x→∞ 1 · x x lim f (x) = lim x→∞ x→∞ so the positive y -axis is a horizontal asymptote and the negative y -axis is a vertical asymptote. Moreover, f (x) = 1 − ln x x2 and f (x) = −3 + 2 ln x , x3 and thus the graph of f is increasing if 0 < x < e, decreasing if x > e, concave downward for 0 < x < e3/2 , and concave upward if x > e3/2 . The inflection point where x = e3/2 is not visible because the curvature of the graph is very small for x > 3. The graph of y = f (x) is next. 0.4 0.2 2 4 6 8 -0.2 -0.4 -0.6 -0.8 -1 C04S08.060: Given: f (x) = ln x . + x1/3 x1/2 We used Mathematica to find f (x) and solve f (x) = 0, and thus discovered that there is a global maximum near (10.094566, 0.433088). Similarly, we found an inflection point near (20.379823, 0.416035). Clearly f (x) → −∞ as x → 0+ , so the negative y -axis is a vertical asymptote. Also lim f (x) = lim x→∞ x→∞ ln x = lim x→∞ x x1/2 + x1/3 1 −1/2 2x 1 = lim x→∞ + 1 x−2/3 3 1 1 /2 2x 1 = 0, + 1 x1/3 3 so the positive x-axis is a horizontal asymptote. The graph of y = f (x) is next. 0.4 0.3 0.2 0.1 10 20 30 40 -0.1 C04S08.061: The computation in the solution of Problem 55 establishes that lim x→∞ xn =0 ex (1) in the case n = 1. Suppose that Eq. (1) holds for n = k , a positive integer. Then (by l’Hˆpital’s rule) o lim x→∞ xk+1 (k + 1)xk = lim = (k + 1) x x→∞ e ex 12 lim x→∞ xk ex = (k + 1) · 0 = 0. Therefore, by induction, Eq. (1) holds for every positive integer n. Now suppose that k > 0 (a positive number not necessarily an integer). Let n be an integer larger than k . Then, for x large positive, we have 0< Therefore, by the squeeze law for limits, lim x→∞ xk xn < x. ex e xk = 0 for every positive number k . ex C03S08.062: Suppose that k is a positive real number. Then (by l’Hˆpital’s rule) o lim x→∞ ln x 1 1 = lim = lim = 0. x→∞ kxk−1 · x x→∞ kxk xk C04S08.063: Given: f (x) = xn e−x where n is a positive integer larger than 1. Then lim f (x) = lim x→∞ x→∞ xn =0 ex by the result in Problem 61. So the positive x-axis is a horizontal asymptote. Next, f (x) = (n − x)xn−1 ex and f (x) = (x2 − 2nx + n2 − n)xn−2 . ex Therefore f (x) = 0 at the two points (0, 0) and (n, nn e−n ). We consider only the part of the graph for which x > 0, and the graph of f is increasing for 0 < x < n and decreasing if x > n, so there is a local √ √ maximum at x = n. Next, f (x) = 0 when x = a = n − n and when x = b = n + n. It is easy to establish that f (x) > 0 if 0 < x < a and if x > b, but that f (x) < 0 if a < x < b. (Use the fact that the graph of g (x) = x2 − 2nx + n2 − n is a parabola opening upward.) Therefore the graph of f has two inflection points for x > 0. C04S08.064: Given: f (x) = x−k ln x where k is a positive constant. Then f (x) = 1 − k ln x , xk+1 and the sign of f (x) is the same as the sign of 1 − k ln x, which is positive if 0 < x < e1/k but negative if x > e1/k . Hence the graph of f will have a single local maximum where x = e1/k . Next, f (x) = k 2 ln x + k ln x − 2k − 1 , xk+2 so f (x) = 0 when x = exp 2k + 1 k2 + k , so the graph of f has at most one inflection point. Moreover, if x is near zero then f (x) < 0, whereas f (x) > 0 if x is large positive. Therefore the graph of f has exactly one inflection point. Finally, the result in Problem 62 shows that the positive x-axis is a horizontal asymptote. C04S08.065: The substitution y = 1 yields x lim xk ln x = lim x→0+ y →∞ − ln y =− yk 13 lim y →∞ 1 ky k = 0. C04S08.066: First suppose that n = −k < 0 where k is a positive integer. Then lim x→∞ (ln x)n 1 = lim = 0. x→∞ x(ln x)k x Next suppose that n = 0. Then lim x→∞ (ln x)n 1 = lim = 0. x→∞ x x Moreover, if n = 1, then by l’Hˆpital’s rule, o lim x→∞ (ln x)n ln x 1 = lim = lim = 0. x→∞ x x→∞ x x Assume that lim x→∞ (ln x)k =0 x for some positive integer k . Then, by l’Hˆpital’s rule, o lim x→∞ (ln x)k+1 (k + 1)(ln x)k (ln x)k = lim = (k + 1) lim = (k + 1) · 0 = 0. x→∞ x→∞ x x x Therefore, by induction for positive n, lim x→∞ (ln x)n = 0 for every integer n. x C04S08.067: In the following computations we take derivatives with respect to h in the first step. By l’Hˆpital’s rule, o lim h→0 f (x + h) − f (x − h) f (x + h) + f (x − h) 2f (x) = lim = = f (x). h→0 2h 2 2 The continuity of f (x) is needed for two reasons: It implies that f is also continuous, so the first numerator approaches zero as h → 0; moreover, continuity of f (x) is needed to ensure that f (x + h) and f (x − h) both approach f (x) as h → 0. C04S08.068: In the following computations we take derivatives with respect to h in the first two steps. By l’Hˆpital’s rule, o lim h→0 f (x + h) − 2f (x) + f (x − h) f (x + h) − f (x − h) = lim 2 h→0 h 2h = lim h→0 f (x + h) + f (x − h) 2f (x) = = f (x). 2 2 The continuity of f (x) is needed for the following reasons: We needed to know that f (x + h) and f (x − h) both approach f (x) as h → 0. We also needed to know that f was continuous so that the second numerator approaches zero as h → 0. There is a third reason, which you will see when you discover the reason for the presence of the term −2f (x) in the first numerator. C04S08.069: If f (x) = (2x − x4 )1/2 − x1/3 , 1 − x4/3 14 then both the numerator n(x) = (2x − x4 )1/2 − x1/3 and the denominator d(x) = 1 − x4/3 approach zero as x → 1, and both are differentiable, so l’Hˆpital’s rule may be applied. After simplifications we find that o n (x) = 3x2/3 − 6x11/3 − (2x − x4 )1/2 3x2/3 (2x − x4 )1/2 and d (x) = − 4x1/3 . 3 Therefore lim f (x) = lim x→1 x→1 n (x) −3x2/3 + 6x11/3 + (2x − x4 )1/2 −3 + 6 + 1 4 = = lim = = 1. 4 )1 2 d (x) x→1 4x(2x − x 4·1 4 C04S08.070: We are to show that if f (x) and g (x) both approach zero as x → + ∞, both f (x) and g (x) exist for arbitrarily large values of x, and the second limit in the next line exists, then lim x→∞ f (x) f (x) = lim . x→∞ g (x) g (x) Following the Suggestion, we let F (t) = f (1/t) and G(t) = g (1/t). Then, with t = 1/x, we have lim x→∞ f (x) F (t) F (t) = lim = lim + G(t) + G (t) g (x) t→0 t→0 provided that the last limit exists. Note that F and G are differentiable if t > 0 and t is close to zero. Hence lim x→∞ C04S08.071: lim x→∞ x e x lim x→∞ e2 e f (x) F (t) f (x) = lim = lim . g (x) t→0+ G (t) x→∞ g (x) x = lim ex = + ∞. x→∞ C04S08.072: The graph of C (t) for the case A = 1, k = 1, and x = 2 is next. 0.25 0.2 0.15 0.1 0.05 2 4 6 8 10 -0.05 -0.1 Next, dC A =√ dt k πt 1 x2 − 4kt2 2t exp − x2 4kt . Now dC/dt = 0 when 2x2 t = 4kt2 , so t = 0 or t = x2 /(2k ). The general shape of the graph shown here makes it clear that the former yields the minimum of C (t) (define C (0) = 0 and C will be continuous on [0, + ∞ ]) 15 and the latter yields the maximum, and the maximum pollutant concentration is the corresponding value of C (t); that is, Cmax = A x 2 . πe C04S08.073: If f (x) = xn e−x (with n a fixed positive integer), then f (x) = (n − x)xn−1 e−x . Because f (x) 0 for x 0, f (0) = 0, and f (x) → 0 as x → + ∞, f (x) must have a maximum value, and the critical point where x = n is the sole candidate. Hence the global maximum value of f (x) is f (n) = nn e−n . Next, f (n − 1) = (n − 1)n e−(n−1) < nn e−n , so n−1 n n en−1 1 =. en e < Therefore e< Also f (n + 1) = (n + 1)n e−(n+1) < n n−1 n = n−1 n −n = 1− 1 n −n . nn . Therefore, by similar computations, en 1+ 1 n n < e. When we substitute n = 106 (using a computer algebra program, of course) we find that 2.7182804690 < e < 2.7182831877 (round down on the left, up on the right). Thus, to five places, e = 2.71828. C04S08.074: We used Mathematica 3.0 to plot the graphs of y = ln x and y = x1/10 on the interval [1, 10]. The result is shown next. 2 1.5 1 0.5 2 4 6 8 10 The graph makes it clear that a solution of ln x = x1/10 is close to x0 = 3.1. With this initial estimate, a few iterations of Newton’s method yields the approximate solution x1 ≈ 3.05972667962080885461. 16 Next we let f (x) = (ln x) − x1/10 and followed the suggestion in Problem 74. The graph of f finally crossed the x-axis when viewed on the interval [1015 , 1016 ]. A few magnifications yielded the graph shown next. 0.008 0.006 0.004 0.002 -0.002 -0.004 -0.006 The scale on the x-axis ranges from 3.42 × 1015 to 3.44 × 1015 . Thus we have x2 ≈ 3.43 × 1015 . A few iterations of Newton’s method soon yielded the more accurate approximation x2 ≈ 3.43063112140780120278 × 1015 . 17 Section 4.9 C00S09.001: We use Theorem 1 of Section 2.3 and the quotient and product laws for limits in Section 2.2: lim x cot x = lim x→0 C04s09.002: lim x→0 1 cos x − x sin x x→0 x · cos x = 1 · 1 = 1. sin x sin x − x cos x x sin x x cos x + sin x = lim = lim = 0. x→0 x cos x + sin x x→0 2 cos x − x sin x x sin x = lim x→0 7x + 8 and the denominator x approach zero as x → 0, so l’Hˆpital’s o 4x + 8 rule may be applied. The trick is to use a law of logarithms to make the “numerator” easier to differentiate. C04S09.003: Both the “numerator” ln lim x→0 1 7x + 8 1 · ln = lim · [ln(7x + 8) − ln(4x + 8)] x 4x + 8 x→0 x = lim x→0 C04S09.004: x sin x x→0 lim e−x ln x = lim x→∞ x→∞ x→0 24 24 3 = =. (7x + 8)(4x + 8) 64 8 = 12 = 1. ln x 1 = lim = 0. x→∞ xex ex lim x e1/x − 1 = lim x→∞ 2 = lim ln sin x cos x = lim+ = lim+ (− sin x) = 0. csc x x→0 − csc x cot x sin x x→0 x→0 x→0 C04S09.007: 7 4 − 7x + 8 4x + 8 lim (sin x)(ln sin x) = lim + x→0+ C04S09.005: lim x2 csc2 x = lim C04S09.006: 1 1 x→∞ e1/x − 1 −(x−2 e1/x ) = lim = lim e1/x = 1. x→∞ x→∞ x−1 −(x−2 ) C04S09.008: Combine into a single fraction, then apply l’Hˆpital’s rule twice: o lim x→2 1 1 − x − 2 ln(x − 1) = lim x→2 ln(x − 1) − (x − 2) (x − 2) ln(x − 1) = lim x→2 C04S09.009: C04S09.010: lim x ln x = lim x→0+ x→0+ lim (tan x)(cos 3x) = lim x→π /2 C04S09.011: lim (x − π ) csc x = lim C04S09.012: x−2 + ln(x − 1) x−1 = lim x→2 1 1 (x − 1)2 =− . x 2 (x − 1)2 − x→π cos 3x −3 sin 3x 3 = lim = = −3. cot x −1 x→π /2 − csc2 x x−π 1 = lim = −1. x→π cos x sin x lim (x − sin x) exp(−x2 ) = lim x→∞ x−2 x−1 ln x 1 = lim = lim (−x) = 0. −1 + −x · x−2 x x→0 x→0+ x→π /2 x→π − x→∞ x − sin x 1 − cos x = lim = 0. x→∞ 2x exp(x2 ) exp(x2 ) 1 The last equality results from the observation that 0 as x → + ∞. 1 − cos x 2 for all x, whereas 2x exp(x2 ) → + ∞ C04S09.013: First combine terms to form a single fraction, apply l’Hˆpital’s rule once, then multiply each o term in numerator and denominator by 2x1/2 . Result: x→0+ = lim (cos x) − 1 x−1/2 2 x1/2 cos x + 1 x−1/2 sin x 2 = lim 1 1 √− x sin x lim (2x1/2 cos x) − 1 = −∞. 2x cos x + sin x x→0+ x→0+ The last limit follows because the numerator is approaching −1 as x → 0+ , while the denominator is approaching zero through positive values. C04S09.014: First combine terms to form a single fraction, then apply l’Hˆpital’s rule twice: o lim x→0 1 1 − x ex − 1 = lim x→0 ex − 1 − x ex − 1 ex 1 = lim =. = lim x x − 1) x x + xex x→0 e − 1 + xe x→0 2e x(e 2 C04S09.015: First combine terms to form a single fraction, then apply l’Hˆpital’s rule: o lim x→1+ x2 x 1 − +x−2 x−1 = lim+ x→1 2 = −∞. (1 − x)(2 + x) Note that l’Hˆpital’s rule is not actually required. o C04S09.016: First multiply numerator and denominator (the denominator is 1) by the conjugate of the numerator; l’Hˆpital’s rule is not required. o lim x→∞ √ x+1− √ x = lim √ x→∞ = lim √ x→∞ x+1− 1 √ x √ √ x+1+ x ·√ √ x+1+ x x+1−x 1 √ = lim √ √ = 0. x→∞ x+1+ x x+1+ x C04S09.017: In this solution we first combine the two terms into a single fraction, apply l’Hˆpital’s rule o a first time, make algebraic simplifications, then apply the rule a second time. lim x→0 1 1 − x ln(1 + x) = lim x→0 ln(1 + x) − x x ln(1 + x) 1 −1 1 − (1 + x) 1+x = lim = lim x x→0 x→0 x + (1 + x) ln(1 + x) + ln(1 + x) 1+x = lim x→0 −x −1 1 = lim =− x + (1 + x) ln(1 + x) x→0 1 + 1 + ln(1 + x) 2 C04S09.018: First multiply the “numerator” and the denominator (which is 1) by the conjugate of the numerator. 2 lim x→∞ x2 + x − = lim x2 − x x→∞ √ x2 + x − 1 √ x2 − x √ √ x2 + x + x2 − x √ ·√ x2 + x + x2 − x (x2 + x) − (x2 − x) 2x √ √ = lim √ = lim √ 2+x+ 2−x 2+x+ x→∞ x→∞ x x x x2 − x = lim x→∞ 2 1 1+ + x 1 1− x = 2 = 1. 1+1 The transition from the second line to the third did not involve l’Hˆpital’s rule. Instead we divided each o term in numerator and denominator by x, which becomes x2 when moved under the radical because x > 0. C04S09.019: The conjugate of a1/3 − b1/3 is a2/3 + a1/3 b1/3 + b2/3 because (a1/3 − b1/3 )(a2/3 + a1/3 b1/3 + b2/3 ) = a − b. Therefore we multiply “numerator” and denominator (which is 1) by the conjugate of the numerator. The result: lim x→∞ (x3 + 2x + 5)1/3 − x = lim x→∞ = lim x→∞ (x3 + 2x + 5)1/3 − x · (x3 + 2x + 5)2/3 + x(x3 + 2x + 5)1/3 + x2 (x3 + 2x + 5)2/3 + x(x3 + 2x + 5)1/3 + x2 x3 + 2x + 5 − x3 (x3 + 2x + 5)2/3 + x(x3 + 2x + 5)1/3 + x2 = lim x→∞ (x3 + 2x + 2x + 5 + x(x3 + 2x + 5)1/3 + x2 2 5 + x x2 = lim x→∞ 5)2/3 1+ 2 5 +2 xx 2 /3 + 1+ 2 5 +2 xx = 1/3 +1 0 = 0. 1+1+1 There was no need—certainly, no temptation—to use l’Hˆpital’s rule. o C04S09.020: Apply the natural logarithm function: ln lim xx x→0+ = lim+ ln(xx ) = lim+ x ln x x→0 x→0 1 ln x x = lim (−x) = 0. = lim = lim 1 x→0+ 1 x→0+ x→0+ −2 x x Therefore lim xx = e0 = 1. + x→0 C04S09.021: Apply the natural logarithm function: 3 ln = lim ln xsin x = lim (sin x)(ln x) + + lim xsin x x→0+ x→0 = lim z →0+ x→0 ln x 1 tan x 0 = lim = lim = 0. = csc x x→0+ −x csc x cot x x→0+ − x −1 sin x Therefore lim xsin x = e0 = 1. x→0+ C04S09.022: Apply the natural logarithm function: ln x→∞ x→∞ = e−1 = 2x − 1 2x + 1 x ln x→∞ x 2x − 1 2x + 1 x→∞ = lim −2x2 2x2 + 2x − 1 2x + 1 = lim Therefore lim x 2x − 1 2x + 1 lim = lim x→∞ = lim x→∞ ln(2x − 1) − ln(2x + 1) 1 x −4x3 − 2x2 + 4x3 − 2x2 −4x2 = lim = −1. x→∞ 4x2 − 1 4x2 − 1 1 . e C04S09.023: Apply the natural logarithm function: ln lim (ln x)1/x = lim ln(ln x)1/x = lim x→∞ x→∞ x→∞ ln(ln x) 1 = lim = 0. x→∞ x ln x x Therefore lim (ln x)1/x = e0 = 1. x→∞ C04S09.024: Apply the natural logarithm function: ln lim x→∞ 1− 1 x2 x = lim ln 1 − x→∞ ln = lim x→∞ = lim Therefore lim x→∞ 1− 1 x2 x x→∞ 1 x2 x2 − 1 x2 1 x x = lim x ln 1 − x→∞ 1 x2 2x 2 − ln(x2 − 1) − 2 ln x x2 − 1 x = lim = lim 1 1 x→∞ x→∞ −2 x x 2x 2 = lim = 0. x→∞ −2x 1 − x2 = e0 = 1. C04S09.025: Apply the natural logarithm function: ln lim x→0 sin x x 1/x2 = lim ln x→0 sin x x 1/x2 = lim x→0 1 sin x ln x2 x cos x 1 − ln(sin x) − ln x x = lim x cos x − sin x = lim = lim sin x 2 x→0 x→0 x→0 x 2x 2x2 sin x −x sin x = lim = lim x→0 2x2 cos x + 4x sin x x→0 4 sin x 1 1 x =− . =− 4 sin x 2·1+4·1 6 2 cos x + x − sin x x Therefore lim x→0 1/x2 = e−1/6 ≈ 0.8464817249. C04S09.026: Apply the natural logarithm function: ln lim (1 + 2x)1/(3x) x→0+ = lim ln(1 + 2x)1/(3x) x→0+ 2 ln(1 + 2x) 2 = lim = lim 1 + 2x = . + + 3x 3 3 x→0 x→0 Therefore lim (1 + 2x)1/(3x) = e2/3 ≈ 1.9477340411. x→0+ C04S09.027: Apply the natural logarithm function: ln lim x→∞ 1 cos 2 x (x4 ) 1 = lim ln cos 2 x→∞ x ln cos = lim x−4 x→∞ = lim − x→∞ Therefore lim x→∞ 1 cos 2 x 1 x2 (x4 ) = lim x4 ln cos x→∞ 1 x2 1 1 tan 2 2 x x = lim = lim x→∞ −4x−5 · x3 x→∞ −2x−2 2 tan 2 1 1 sec2 2 − sec2 2 3 x x = lim x = − 1. −3 x→∞ 4x 2 2 (x4 ) = e−1/2 ≈ 0.6065306597. C04S09.028: As x → 0+ , sin x → 0 through positive values and sec x → 1. So this is not an indeterminate form, and lim (sin x)sec x = 0. x→0+ C04S09.029: Apply the natural logarithm function: ln lim (x + sin x)x x→0+ = lim ln(x + sin x)x = lim x ln(x + sin x) x→0+ x→0+ 1 + cos x ln(x + sin x) −x2 (1 + cos x) = lim = lim x + sin x = lim −1 −2 ) x x + sin x x→0+ x→0+ −(x x→0+ = lim x→0+ x2 sin x − 2x(1 + cos x) 0·0−2·0·2 = = 0. 1 + cos x 1+1 Therefore lim+ (x + sin x)x = e0 = 1. x→0 C04S09.030: lim (tan x − sec x) = lim x→π /2 x→π /2 (sin x) − 1 cos x 0 = lim = = 0. cos x −1 x→π /2 − sin x C04S09.031: Apply the natural logarithm function: ln lim x1/(1−x) = lim ln x1/(1−x) = lim x→1 x→1 x→1 Therefore lim x1/(1−x) = e−1 ≈ 0.3678795512. x→1 5 ln x 1 = lim = −1. 1 − x x→1 −x C04S09.032: Apply the natural logarithm function: ln lim (x − 1)ln x = lim ln(x − 1)ln x = lim (ln x) ln(x − 1) + + x→1+ x→1 x→1 ln(x − 1) = lim = lim x→1+ (ln x)−1 x→1+ = lim x→1+ 1 x(ln x)2 x−1 = lim 1 x→1+ 1 − x − (ln x)−2 x (ln x)2 + 2 ln x 02 + 2 · 0 = = 0. −1 −1 Therefore lim (x − 1)ln x = e0 = 1. x→1+ C04S09.033: First combine the two terms to form a single fraction, then apply l’Hˆpital’s rule, and finally o simplify: x→2+ 1 1 − x−2 (x2 − 4)1/2 = lim x − 2 − (x2 − 4)1/2 1 − x(x2 − 4)−1/2 = lim x→2+ x(x2 − 4)−1/2 (x − 2) + (x2 − 4)1/2 (x2 − 4)1/2 (x − 2) = lim lim (x2 − 4)1/2 − x = −∞ x(x − 2) + (x2 − 4) x→2+ x→2+ because, in the last limit, the numerator is approaching −2 while the denominator is approaching zero through positive values. C04S09.034: Let Q = x5 − 3x4 + 17. We plan to multiply by the conjugate of Q1/5 − x. lim x→∞ (x5 − 4x4 + 17)1/5 − x = lim (Q1/5 − x) x→∞ = lim x→∞ Q4/5 = lim x→∞ + Q3/5 x Q − x5 + Q2/5 x2 + Q1/5 x3 + x4 17 − 3x4 . Q4/5 + Q3/5 x + Q2/5 x2 + Q1/5 x3 + x4 Now carefully divide each term in numerator and denominator by x4 . The numerator becomes 17 − 3, x4 which approaches −3 as x → + ∞. The first term in the denominator becomes (x5 − 3x4 + 17)4/5 = x4 x5 − 3x4 + 17 x5 4 /5 = 1− 3 17 + x x5 4 /5 , which approaches 1 as x → + ∞. The second term in the denominator becomes x · (x5 − 3x4 + 17)3/5 = x4 x5 − 3x4 + 17 x5 3 /5 = 1− 3 17 + x x5 3/5 , which also approaches 1 as x → + ∞, as do the third, fourth, and fifth terms in the demoninator. Therefore 6 lim x→∞ 3 (x5 − 4x4 + 17)1/5 − x = − . 5 C04S09.035: Given: f (x) = x1/x for x > 0. We plotted the graph of y = f (x) on the interval 10−6 x 1 and obtained strong evidence that f (x) approaches zero as x → 0+ . We also plotted y = f (x) on the interval 100 x 1000 and obtained some evidence that f (x) → 1 as x → + ∞. Then we verified these limits with l’Hˆpital’s rule as follows: o ln 1 1 ln x = lim = 0, x→∞ x x lim x1/x = lim x→∞ x→∞ so that lim x1/x = e0 = 1. x→∞ But lim x1/x is not indeterminate, because the exponent is approaching + ∞; this limit is clearly zero. x→0+ The graph that follows this solution indicates that the global maximum value of f (x) occurs close to 2.71828 (surely no coincidence). We found that f (x) = x1/x (1 − ln x) , x2 and it follows that the maximum value of f (x) is f (e) = e1/e ≈ 1.4446678610. 1.44467 1.44467 1.44467 1.44467 1.44467 2.71825 2.7183 2.71835 2.7184 2 C04S09.036: Given: f (x) = x1/(x ) for x > 0. We plotted f (x) on the interval 0.01 x 1 and obtained strong evidence that f (x) → 0 as x → 0+ . We plotted f (x) for 10 x 100 and obtained some evidence that f (x) → 1 as x → + ∞. The first conjecture is correct because, as x → 0+ , the exponent in f (x) is approaching + ∞ and therefore this limit is not an indeterminate form; it is clearly zero. Then ln lim f (x) = lim x→∞ x→∞ ln x 1 = lim = 0, x→∞ 2x · x x2 and therefore lim f (x) = e0 = 1. x→∞ Next we plotted y = f (x) for 1 x 4 and saw a clear maximum near where x = 1.65. We found that 2 (1 − 2 ln x)x1/(x ) f (x) = , x3 √ and it follows that the maximum is f ( e ) = e1/(2e) ≈ 1.2019433685. 7 C04S09.037: Given: f (x) = (x2 )1/x for x > 0. We plotted y = f (x) for 0.01 x 1 and obtained strong evidence that f (x) → 0 as x → 0+ . We plotted f (x) for 100 x 1000 and obtained weak evidence that f (x) → 1 as x → + ∞. (These two graphs follow this solution.) The first limit is clear because x2 is approaching zero while the exponent 1/x is approaching + ∞. For the second limit, we found ln lim f (x) = lim x→∞ x→∞ 2 ln x 2 = lim = 0, x→∞ x x so that lim f (x) = e0 = 1. x→∞ Next we plotted f (x) for 2.71827 x 2.71829 (by the “method of successive zooms”) and saw a clear maximum near where x = 2.71828. (The graph follows this solution.) We found that f (x) = (2 − 2 ln x)(x2 )1/x , x2 and it follows that the maximum value of f (x) is f (e) = e2/e ≈ 2.0870652286. 0.000025 1.08 0.00002 0.000015 1.06 0.00001 1.04 0.1 0.2 0.3 0.4 0.5 200 400 600 800 1000 2.08707 2.08707 2.08707 2.08707 2.08707 2.08707 2.08707 2.71828 2.71828 2.71829 2.71829 C04S09.038: Given: f (x) = x−x , x > 0. We plotted y = f (x) for 0.001 x 1; the graph is strong evidence that f (x) → 1 as x → 0+ . Then we plotted y = f (x) for 10 x 20; the graph is very strong evidence that f (x) → 0 as x → + ∞. To be sure, we computed ln lim f (x) x→0+ = lim (−x ln x) = lim x→0+ x→0+ − ln x = lim x = 0. x−1 x→0+ 1 = 0; we are not dealing with an indeterminate form here. xx The first graph showed a global maximum near the point where x = 0.4. We found that It is clear that lim x→∞ f (x) = − 1 + ln x , xx and therefore the global maximum value of f (x) is f (e−1 ) = e1/e ≈ 1.44466678610. 8 C04S09.039: We graphed f (x) = (1 + x2 )1/x for 0.001 x 1; the graph is extremely strong evidence that f (x) → 1 as x → 0+ . Then we graph y = f (x) for 100 x 1000; the graph is weak evidence that f (x) → 1 as x → + ∞. These two graphs are shown next. 2 1.8 1.08 1.6 1.06 1.4 1.04 1.2 0.2 0.4 0.6 0.8 1 200 400 600 800 1000 To be sure about these limits, we computed ln lim f (x) x→0+ = lim x→0+ ln(1 + x2 ) 2x = lim = 0, + 1 + x2 x x→0 and therefore f (x) → e0 = 1 as x → 0+ . Next, ln lim f (x) = lim x→∞ x→∞ ln(1 + x2 ) 2x = lim = 0, x→∞ 1 + x2 x and therefore f (x) → e0 = 1 as x → + ∞. Then we used the “method of successive zooms” and thereby found that the graph of y = f (x) for 1.9802 x 1.9804 shows a maximum near where x = 1.9803. (The graph follows this solution.) Then we found that f (x) = (1 + x2 )1/x 2x2 − (1 + x2 ) ln(1 + x2 ) x2 (1 + x2 ) but could not solve the transcendental equation 2x2 = (1 + x2 ) ln(1 + x2 ) exactly. So we used Newton’s method to solve f (x) = 0, and our conclusion is that the global maximum value of f (x) is approximately 2.2361202715 ≈ f (1.9802913004). 2.23612 2.23612 2.23612 2.23612 2.23612 2.23612 1.98025 1.9803 1.98035 1.9804 C04S09.040: Given: f (x) = 1+ 1 x2 x for 9 x > 0. We plotted y = f (x) for 0.001 x 0.2; the graph is very strong evidence that f (x) → 1 as x → 0+ . Then we plotted y = f (x) for 10 x 1000; the graph is good evidence that f (x) → 1 as x → + ∞. Both these limits are indeterminate, so we used l’Hˆpital’s rule: o ln 1 1+ 2 x lim x→0+ x x2 + 1 = lim x ln + x2 x→0 = lim x2 · x→0+ 2x 2 − ln(x2 + 1) − 2 ln x x2 + 1 x = lim = lim x−1 −(x−2 ) x→0+ x→0+ 2 2x − x x2 + 1 = lim x→0+ x2 (2x2 + 2 − 2x2 ) 2x = lim 2 = 0. x(x2 + 1) x→0+ x + 1 Repeat these computations with x → 0+ replaced with x → + ∞ to discover the same limit, zero. Thus f (x) → e0 = 1 as x → 0+ and as x → + ∞. Next we used the method of successive zooms to find that f (x) has a global maximum just a little to the right of the point where x = 0.5. Solving f (x) = 0 exactly seemed hopeless; we used Newton’s method to solve f (x) = 0 to find that the global maximum value of f (x) is approximately f (0.5049762122) ≈ 2.2361202715. C04S09.041: Given: f (x) = (x + sin x)1/x . The graph of y = f (x) for 0.01 x 1 provides strong evidence that f (x) → 0 as x → 0+ . The graph of y = f (x) for 10 x 1000 provides fairly good evidence that f (x) → 1 as x → + ∞. These graphs are shown next. 1.25 0.0175 0.015 1.2 0.0125 1.15 0.01 0.0075 1.1 0.005 1.05 0.0025 0.1 0.2 0.3 0.4 200 400 600 800 1000 Next, we verified these limits as follows: ln lim (x + sin x)1/x = lim x→∞ x→∞ ln(x + sin x) 1 + cos x = lim = 0, x→∞ x + sin x x and therefore lim (x + sin x)1/x = e0 = 1. x→∞ But (x + sin x)1/x is not indeterminate as x → 0+ because if x is very small and positive, then x + sin x is positive and near zero while 1/x is very large positive. Therefore lim+ (x + sin x)1/x = 0. x→0 Then a plot of y = f (x) for 0.5 x 2 revealed a global maximum near where x = 1.2. A plot of f for 1.2095 x 1.2097 (by the “method of repeated zooms”) showed the maximum near the midpoint of that 10 interval. That graph is shown next. The equation f (x) = 0 appeared to be impossible to solve exactly, so we used Newton’s method to find that the maximum of f (x) is very close to (1.2095994645, 1.8793598343). 1.87936 1.87936 1.87936 1.87936 1.87936 1.87936 1.20955 1.2096 1.20965 1.2097 (cos x−1) C04S09.042: Given: f (x) = exp(1/x2 ) . A plot of y = f (x) for 0.01 x 1 indicated that as x → 0+ , f (x) approaches a number between 0.60 and 0.61. A plot of y = f (x) for 10 x 1000 indicated oscillations between 0.9999 and 1.0, but dying out very rapidly while maintaining an upper bound of 1.0. Thus there is evidence that f (x) → 1 as x → + ∞. Analytically, we compute these limits as follows: ln lim x→0+ exp(1/x2 ) (cos x−1) = lim+ (cos x − 1) ln exp(1/x2 ) x→0 = lim x→0+ cos x − 1 − sin x 1 =− . = lim+ x2 2x 2 x→0 Therefore f (x) → e−1/2 ≈ 0.6065306597 as x → 0+ . As x → + ∞, cos x − 1 varies between 0 and −2 while exp(1/x2 ) → 1 from above (and quite rapidly). Hence f (x) → 1 as x → + ∞; note that f (x) 1 for all x > 0. Moreover, when cos x − 1 = 0 (which happens at each integral multiple of 2π ), f (x) = 1, and this is the maximum value of f (x). Examination of f (x) = − 2f (x)(cos x − 1) f (x) sin x − x3 x2 makes it clear that f (2nπ ) = 0 for every positive integer n, although this does not establish that there are no other locations of maxima. What is clear is that 1 is the maximum value of f (x) for x > 0. C04S09.043: Note that in using l’Hˆpital’s rule we are computing derivatives with respect to h. o ln lim (1 + hx)1/h h→0 = lim h→0 ln(1 + hx) x = lim = x, h→0 1 + hx h and therefore lim (1 + hx)1/h = ex . h→0 C04S09.044: The implication (through the notation) is that n → + ∞ while assuming only positive integral values. We need to differentiate with respect to n, thus we let n run through positive real values. Then if n is later restricted to positive integral values, the limit will be the same. 11 ln x 1+ n lim n→∞ n ln 1 + = lim 1 n n→∞ x n = lim ln n→∞ n+x n 1 n 1 1 − ln(n + x) − ln n n+x n = lim = lim 1 1 n→∞ n→∞ −2 n n nx x = lim = lim x = x. n→∞ n + x n→∞ 1+ n Therefore lim n→∞ 1+ x n n = ex . C04S09.045: We let f (x) = xtan x and applied Newton’s method to the equation f (x) = 0 with initial guess x0 = 0.45. Results: x1 = 0.4088273642, x2 = 0.4099763617, x3 = x4 = 0.4099776300; f (x4 ) = 0.6787405265. C04S09.046: Suppose that n 1/n [ p(x)] 1/n 2. Let Q = [ p(x)] . Then −x=Q−x = Qn − xn Qn−1 + Qn−2 x + Qn−3 x2 + · · · + Qxn−2 + xn−1 a2 an a3 + 2 · · · + n−1 x x x = n−1 . Q Qn−2 + n−2 + · · · + 1 xn−1 x a1 + Note that there are n terms in the last denominator and, apart from the last, each has the form (n−k)/n Qn−k [ p(x)] = xn−k xn−k where k is an integer and 1 = p(x) xn (n−k)/n n − 1. It now follows that, for each such k , k Qn−k a1 an a2 = 1+ + 2 + ··· + n n−k x x x x (n−k)/n →1 as x → + ∞. Therefore lim x→∞ 1/n [ p(x)] − x = lim a1 + x→∞ a2 an a3 + 2 + · · · + n−1 a1 x x x =. n n If n = 1, then lim x→∞ 1/n [ p(x)] − x = lim (x + a1 − x) = a1 = x→∞ a1 a1 =. 1 n This concludes the proof. C04S09.047: Replace b with x to remind us that it’s the only variable in this problem; note also that 0 < x < a. The surface area of the ellipsoid is then 12 A(x) = 2π ax x (a2 − a2 )1/2 a +2 arcsin 2 )1/2 a (a − x a = 2π x2 + 2π a2 · (a2 − x2 )1/2 a . (a2 − x2 )1/2 x arcsin Therefore (a2 − x2 )1/2 x arcsin a . lim A(x) = 2π a2 + 2π a2 lim x→a− a · x→a− (a2 − x2 )1/2 a Let u = (a2 − x2 )1/2 . Then u → 0+ as x → a− . Hence a lim A(x) = 2π a2 + 2π a2 x→a− = 2π a2 + 2π a2 lim x→a− x a lim √ u→0+ · lim u→0+ 1 1 − u2 arcsin u u = 4π a2 . C04S09.048: Part (a): We plan to show that dA/dn > 0, so it will follow that A is an increasing function of n. We assume throughout that A0 > 0, that 0 < r 1, and that n 1. For the purpose of computing dA/dn, we let n take on all real values in its range, not merely positive integral values. Now ln A = ln A0 + nt ln 1 + r , n so 1 dA n+r · = t ln A dn n + nt 1 1 − n+r n = t ln n+r n − rt . n+r Because A and t are positive, it suffices to show that the function defined by f (n) = is positive-valued for r > 0 and n takes the form g (x) = 1 n+r −r + (n + r) ln n+r n (1) 1. We substitute n = rx to simplify f ; the right-hand side in Eq. (1) 1 r + rx −r + (r + rx) ln r + rx rx =− 1 1 + ln 1 + 1+x x . It remains to show that g (x) is positive-valued. If n is a positive integer and r 1, then x > 1, so g (x) “starts” with the positive value g (1) ≈ 0.193147. Moreover, g (x) thenceforth decreases because its derivative, g (x) = − 1 , x(1 + x)2 is negative for all x > 0. Finally, it is obvious from the definition of g that g (x) → 0 as x → + ∞. Therefore g (x) remains positive, because if it once took on the value zero, it would thereafter attain a negative value 13 z , and then (continuing to decrease) would remain forever less than z —in which case it could not approach zero as x → + ∞. Part (b): L’Hˆpital’s rule yields o ln lim r n nt n+r n = lim nt ln = lim n→∞ 1+ −tn2 tn2 + n+r n n→∞ n→∞ = lim n→∞ = lim t · n→∞ = lim n→∞ ln(n + r) − ln n 1 n −tn3 + tn3 + tn2 r n(n + r) tnr rt = lim r = rt. n→∞ n+r 1+ n Therefore lim n→∞ 1+ r n nt = ert , and the result in Part (b) follows immediately. C04S09.049: Given: f (x) = | ln x |1/x for x > 0. The graph of y = f (x) for 0.2 x 0.3 shows f (x) taking on values in excess of 1028 , so it seems quite likely that f (x) → + ∞ as x → 0+ . Indeed, this is the case, because as x → 0+ , we see that | ln x | → +∞ and also the exponent 1/x is increasing without bound. Next we show the graph of y = f (x) for 0.5 x 1 (on the left) and for 0.9 x 1.1 (on the right). The first indicates an inflection point near where x = 0.8 and the second shows a clear global minimum at (1, 0). To find the inflection point, we redefined f (x) = (− ln x)1/x and computed f (x) = f (x) x4 (ln x)2 1 − x − 3x ln x − 2(ln x) ln(− ln x) + 2x(ln x)2 ln(− ln x) + (ln x)2 (ln(− ln x))2 (assisted by Mathematica, of course). We let g (x) = 1 − x − 3x ln x − 2(ln x) ln(− ln x) + 2x(ln x)2 ln(− ln x) + (ln x)2 (ln(− ln x))2 and applied Newton’s method to solve the equation g (x) = 0. With initial guess x0 = 0.8, six iterations yielded over 20 digits of accuracy, and the inflection point shown in the figure is located close to (0.8358706352, 0.1279267691). 0.12 0.1 0.4 0.08 0.3 0.06 0.2 0.04 0.1 0.5 0.6 0.7 0.8 0.02 0.9 0.9 0.95 1.05 1.1 It is clear that f (x) > 0 if 0 < x < 1 and if x > 1, so the graph of f has a global minimum at (1, 0). Still using f (x) = (− ln x)1/x , we computed 14 f (x) = f (x)(1 − (ln x)(ln(− ln x))) , x2 ln x and thereby found that lim f (x) = −1. x→1− Then we redefined f (x) = (ln x)1/x and computed f (x) = f (x) (1 − (ln x) ln(ln x)) . x2 ln x Then we found that lim f (x) = 1. x→1+ Thus the shape of the graph of f near (1, 0) is quite similar to the shape of y = | x | near (0, 0). Next we plotted the graph of y = f (x) for 1 x 2 (shown next, on the left) and for 4 x 10 (next, on the right). The first of these indicates an inflection point near where x = 1.2 and the second shows a clear local maximum near where x = 5.8. To locate the inflection point more accurately, we computed f (x) = f (x) x4 (ln x)2 1 − x − 3x ln x − 2(ln x) ln(ln x) + 2x(ln x)2 ln(ln x) + (ln x)2 (ln(ln x))2 and applied Newton’s method to the solution of g (x) = 0, where g (x) = 1 − x − 3x ln x − 2(ln x) ln(ln x) + 2x(ln x)2 ln(ln x) + (ln x)2 (ln(ln x))2 . 1.1025 0.8 5 0.6 6 7 8 9 10 1.0975 1.095 0.4 1.0925 1.09 0.2 1.0875 1.2 1.4 1.6 1.8 1.085 2 Beginning with the initial guess x0 = 1.2, seven iterations yielded more than 20 digits of accuracy; the inflection point is located close to (1.1163905964, 0.1385765415). To find the local maximum, we applied Newton’s method to the equation g (x) = 0, where g (x) = 1 − (ln x) ln(ln x). Beginning with the initial guess x0 = 5.8, six iterations yielded more than 20 digits of accuracy; the local maximum is very close to (5.8312001357, 1.1021470392). Finally, we plotted y = f (x) for 10 x 1000 (shown after this solution). The change in concavity indicates that there must be yet another inflection point near where x = 9. Newton’s method again yielded its approximate coordinates as (8.9280076968, 1.0917274397). The last graph also suggests that f (x) → 1 as x → + ∞. This is indeed the case; 15 ln lim f (x) = lim x→∞ x→∞ ln(ln x) 1 = lim = 0, x→∞ x ln x x and therefore f (x) → e0 = 1 as x → + ∞. 1.08 1.06 1.04 1.02 200 400 600 800 1000 C04S09.050: We applied the techniques of the previous solution to f (x) = | ln x |1/| ln x | , x > 0. The graph of y = f (x) for 0.0001 x 0.4 shows a maximum near where x = 0.06. We redefined f (x) = (− ln x)−1/ ln x and found that f (x) = f (x)(−1 + ln(− ln x)) , x(ln x)2 and it is easy to solve the equation f (x) = 0 for x = e−e . Hence there is a local maximum at (e−e , e1/e ) ≈ (0.0659880358, 1.4446678610). The limit of f (x) as x → 0+ is not clear from the graph, but ln lim f (x) x→0+ = lim x→0+ −x = 0, x ln x and therefore f (x) → e0 = 1 as x → 0+ . The graph of y = f (x) also indicates an inflection point near where x = 0.5, and application of Newton’s method to the equation f (x) = 0 with initial guess x0 = 0.5 reveals that its coordinates are very close to (0.5070215891, 0.5657817947). The graph of y = f (x) for 0.5 x 1.5 suggests a horizontal tangent and global minimum at (1, 0), but f (1) is not defined. Nevertheless, as x → 1, the base | ln x | approaches zero through positive values while the exponent 1/| ln x | approaches + ∞, so f (x) → 0 as x → 1. We collected graphical and numerical evidence that f (x) → 0 as x → 1 but have no formal proof. The graph of y = f (x) for 1.2 x 2 indicates an inflection point near where x = 1.7. We rewrote f in the form f (x) = (ln x)1/ ln x and applied Newton’s method to the equation f (x) = 0 to find that the inflection point is close to (1.6903045007, 0.2929095074). The graph also indicates a local maximum near where x = 15. Because f (x) = f (x)(1 − ln(ln x)) , x(ln x)2 it is easy to solve f (x) = 0 for x = ee , so that extremum is located at (ee , e1/e ) ≈ (15.154262, 1.444668). Hence both local maxima are global maxima. The graph also shows an inflection point near where x = 26, and Newton’s method reveals its approximate coordinates to be (26.5384454497, 1.4364458579). The graph of y = f (x) for 400 x 4000 did not indicate any particular limit as x → + ∞, but ln lim f (x) = lim x→∞ x→∞ and therefore f (x) → 1 as x → + ∞. 16 −x = 0, x ln x C04S09.051: Given: f (x) = | ln x || ln x | for x > 0. The graph of y = f (x) for 0.01 x 0.1 indicates that f (x) → + ∞ as x → 0+ , and this is clear. (That graph is next, on the left.) The graph also indicates a local minimum near where x = 0.7 (that graph is next, on the right.) 0.5 1000 0.6 0.7 0.8 0.9 0.95 800 0.9 600 0.85 400 0.8 200 0.75 0.02 0.04 0.06 0.08 0.7 0.1 We rewrote f (x) in the form f (x) = (− ln x)− ln x for 0 < x < 1 and found that f (x) = − (− ln x)− ln x (1 + ln(− ln x)) . x (1) Then it is easy to solve f (x) = 0 for x = e−1/e . So the graph of y = f (x) has a local minimum at (e−1/e , e−1/e ). Both the abscissa and the ordinate are approximately 0.6922006276. Next, the graph shows a cusp at the point (1, 1); there is a local maximum at that point, and | f (x) | → +∞ as x → 1. To see why, rewrite f (x) = (ln x)ln x for x > 1. Then f (x) = (ln x)ln x (1 + ln(ln x)) , x (2) and it is clear that f (x) → −∞ as x → 1+ . You can also use Eq. (1) to show that f (x) → + ∞ as x → 1− . This is not apparent from the graph of y = f (x) for 0.9 x 1.1, shown next (on the left). Next we plotted y = f (x) for 1 x 2 and found another local minimum near where x = 1.4. (The graph is next, on the right.) It is easy to solve f (x) = 0 (use Eq. (2)), and you’ll find that the coordinates of this point are (e1/e , e−1/e ), so the two local minima are actually global minima. Finally, it’s clear that f (x) → + ∞ as x → + ∞. 0.9 0.95 1.05 0.875 1.1 0.85 0.95 0.825 0.8 0.9 0.775 0.75 0.85 0.725 0.8 1.2 1.4 1.6 1.8 C04S09.052: If α is a fixed real number, then 1 ln lim exp − 2 x→0 x αx2 = lim αx2 ln exp − x→0 1 x2 Therefore lim exp − x→0 1 x2 17 αx2 = e−α . −αx2 = − α. x→0 x2 = lim 2 This means that the indeterminate form 00 may take on any positive real number value because that is the range of values of e−α . The only way that g (x) lim [ f (x)] x→a could be negative would be if f (x) were negative for all x near a, but then g (x) would take on irrational g (x) values and therefore the expression [ f (x)] would be undefined. Hence the limit of a 00 indeterminate form cannot be negative. The 00 form lim x1/| ln x| 1/2 (1) x→0+ has the value zero because ln lim x1/| ln x| 1/2 x→0+ ln x − ln x = lim − = lim −(− ln x)1/2 = −∞. x→0+ x→0+ | ln x |1/2 (− ln x)1/2 = lim x→0+ The 00 form lim x−1/(− ln x) + 1/3 (2) x→0 has the value + ∞ because ln lim x−1/(− ln x) 1/3 x→0+ = lim x→0+ − ln x = lim (− ln x)2/3 = + ∞. x→0+ (− ln x)1/3 Our thanks to Ted Shifrin for the examples in (1) and (2). C04S09.053: The figure on the left shows the graph of y = f (x) on the interval [ −1, 1]. The figure on the right shows the graph of y = f (x) on the interval [ −0.00001, 0.00001]. It is clear from the second figure that e ≈ 2.71828 to five places. When the removable discontinuity at x = 0 is removed in such a way to make f continuous there, the y -intercept will be e because ln lim x→0 1+ 1 x x = lim x ln x→0 = lim x→0 x+1 x = lim x→0 ln(x + 1) − ln x = lim 1 x→0 x x2 −x2 + x+1 x −x3 + x2 + x2 x2 x = lim = lim = 1, x→0 x(x + 1) x→0 x + 1 x(x + 1) 18 and therefore lim x→0 1+ 1 x x = e1 = e. -0.00001 2.71829 7 2.71829 6 2.71828 5 2.71828 4 2.71827 3 2.71827 -0.75 -0.5 -0.25 0.25 C04S09.054: Part (c): If v (t) = 0.5 0.75 mg 1 − e−kt/m , then k kt −kt/m g 1 − e−kt/m g − m2 e lim v (t) = lim · · = lim 1 1 m→∞ m→∞ k m→∞ k −2 m m g = lim · kte−kt/m = lim gte−kt/m = gt · 1 = gt m→∞ k m→∞ because k > 0 and t > 0. 19 0.00001 Chapter 4 Miscellaneous Problems C04S0M.001: dy = 3 (4x − x2 )1/2 (4 − 2x) dx. 2 C04S0M.002: dy = [24x2 (x2 + 9)1/2 + 4x3 (x2 + 9)−1/2 (2x) ] dx. C04S0M.003: dy = − 2 dx. (x − 1)2 C04S0M.004: dy = 2x cos(x2 ) dx = 2x cos x2 dx. C04S0M.005: dy = 2x cos C04S0M.006: dy = √ √ x − 1 x3/2 sin x dx. 2 sin 2x − 2x cos 2x dx. sin2 2x C04S0M.007: Let f (x) = x1/2 ; f (x) = 1 x−1/2 . Then 2 √ 6401 = f (6400 + 1) ≈ f (6400) + 1 · f (6400) 1 12801 = 80 + = = 80.00625. 160 160 √ (A calculator reports that 6401 ≈ 80.00624976.) C04S0M.008: Choose f (x) = 1 1 ; f (x) = − 2 . Choose x = 1 and ∆x = 0.000007. Then x x 1 = f (x + ∆x) ≈ f (x) + f (x) ∆x 1.000007 = 1 − 0.000007 = 0.999993. C04S0M.009: Let f (x) = x10 ; then f (x) = 10x9 . Choose x = 2 and ∆x = 0.0003. Then (2.0003)10 = f (x + ∆x) ≈ f (x) + f (x) ∆x = 210 + 10 · 29 · (0.0003) = 1024 + (5120)(0.0003) = 1024 + 1.536 = 1025.536. A calculator reports that (2.0003)10 ≈ 1025.537. C04S0M.010: Let f (x) = x1/3 ; then f (x) = 1 x−2/3 . Choose x = 1000 and ∆x = −1. Then 3 √ 3 999 = f (x + ∆x) ≈ f (x) + f (x) ∆x = (1000)1/3 + A calculator reports that √ 3 1 1 2999 · (−1) = 10 − = ≈ 9.996667. 2/3 3 · 100 300 3 · (1000) 999 ≈ 9.996665. C04S0M.011: Let f (x) = x1/3 ; then f (x) = 1 x−2/3 . Choose x = 1000 and ∆x = 5. Then 3 √ 3 1005 = f (x + ∆x) ≈ f (x) + f (x) ∆x = (1000)1/3 + 1 5 601 · 5 = 10 + = ≈ 10.0167. 3 · 100 60 3(1000)2/3 1 A calculator reports that √ 3 1005 ≈ 10.0166. C04S0M.012: Take f (x) = x1/3 ; then f (x) = 1 x−2/3 . Take x = 64 and ∆x = −2. Thus 3 (62)1/3 = f (x + ∆x) ≈ f (x) + f (x) ∆x = (64)1/3 + (−2) 1 3 (64)−2/3 = 95 ≈ 3.958. 24 C04S0M.013: Let f (x) = x3/2 ; then f (x) = 3 x1/2 . Let x = 25 and let ∆x = 1. Then 2 263/2 = f (x + ∆x) ≈ f (x) + f (x) ∆x 3 · (25)1/2 · 1 = 125 + 7.5 = 132.5. 2 = (25)3/2 + C04S0M.014: Take f (x) = x1/5 ; then f (x) = 1 x−4/5 . Take x = 32 and ∆x = −2. Then 5 (30)1/5 = f (x + ∆x) ≈ f (x) + f (x) ∆x = (32)1/5 + (−2) 1 5 (32)−4/5 = 2 − 1 79 = = 1.975. 40 40 C04S0M.015: Let f (x) = x1/4 ; then f (x) = 1 x−3/4 . Let x = 16 and ∆x = 1. Then 4 √ 4 17 = (17)1/4 = f (x + ∆x) ≈ f (x) + f (x) ∆x = (16)1/4 + C04S0M.016: With f (x) = x1/10 , f (x) = 1 1 65 ·1=2+ = = 2.03125. 4·8 32 4 · (16)3/4 1 −9/10 , 10 x (1000)1/10 = f (x + ∆x) ≈ f (x) + f (x) ∆x x = 1024, and ∆x = −24, we obtain 1 10 = (1024)1/10 + (−24) (1024)−9/10 = 2 − 3 1277 = ≈ 1.9953. 640 640 C04S0M.017: The volume V of a cube of edge s is given by V (s) = s3 . So dV = 3s2 ds, and thus with s = 5 and ∆s = 0.1 we obtain ∆V ≈ 3(5)2 (0.1) = 7.5 (cubic inches). C04S0M.018: With radius r and area A = π r2 , we have dA = 2π r dr. We take r = 10 and ∆r = −0.2 to obtain ∆A ≈ (2π )(10)(−0.2) = −4π (cm2 ). C04S0M.019: The volume V of a sphere of radius r is given by V (r) = 4 π r3 . Hence dV = 4π r2 dr, so 3 1 with r = 5 and Deltar = 10 we obtain ∆V ≈ 4π · 25 · C04S0M.020: Given V = we obtain 1 = 10π 10 (cm3 ). 1000 1000 , it follows that dV = − 2 dp. Therefore, with p = 100 and ∆p = −1, p p 2 ∆V ≈ − 1000 · (−1) = 0.1 (100)2 (cubic inches). C04S0M.021: If T = 2π L = 2π 32 Therefore if L = 2 and ∆L = L 32 1 12 , ∆T ≈ dT = 1/2 then , dT = π L 32 −1/2 · 1 π dL = 32 32 32 L 1 /2 dL. we obtain π 32 32 2 1/2 · 1 π4 π = · = ≈ 0.0327 12 32 12 96 (seconds). C04S0M.022: Here, dL = (−13)(1030 )E −14 dE . We take E = 110 and ∆E = +1 and obtain ∆L ≈ (−13)(1030 ) 110−14 (+1) ≈ −342 (hours). The actual decrease is L(110) − L(111) ≈ 2896.6 − 2575.1 ≈ 321.5 (hours). 1 C04S0M.023: First, f (x) = 1 + 2 , so f (x) exists for 1 < x < 3 and f is continuous for 1 x we are to solve that is, x 3. So f (3) − f (1) = f (c); 3−1 3− 1 3 −1+1 1 = 1 + 2. 2 c √ After simplifications we find that c2 = 3. Therefore, because 1 < c < 3, c = + 3. C04S0M.024: Every polynomial is continuous and differentiable everywhere, so all hypotheses are met. f (3) − f (−2) 26 − (−14) = = 8 = f (c) = 3c2 + 1, 3 − (−2) 5 so c2 = 7/3. Both roots lie in (−2, 3), so both + 7/3 and − 7/3 are solutions. C04S0M.025: Every polynomial is continuous and differentiable everywhere, so all hypotheses of the mean value theorem are satisfied. Then f (2) − f (−1) 8+1 = = 3 = f (c) = 3c2 , 2 − (−1) 3 so c2 = 1. But −1 does not lie in the interval (−1, 2), so the number whose existence is guaranteed by the mean value theorem is c = 1. C04S0M.026: Every polynomial is continuous and differentiable everywhere, so all hypotheses of the mean value theorem are satisfied. Then f (1) − f (−2) 1+8 = = 3 = f (c) = 3c2 , 1 − (−2) 3 3 so c2 = 1. But 1 does not lie in the interval (−2, 1), so the number whose existence is guaranteed by the mean value theorem is c = −1. C04S0M.027: Given: f (x) = 11 x5 on the interval [ −1, 2]. Because f (x) is a polynomial, it is continuous 5 and differentiable everywhere, so the hypotheses of the mean value theorem are satisfied on the interval [ −1, 2]. Moreover, f (2) − f (−1) = 2 − (−1) It follows that c4 = 11 5, so that c = ± 11 5 · 32 + 3 11 1/4 . 5 11 5 = 11 · 33 121 = = f (c) = 11c4 . 5·3 5 Only the positive root lies in the interval [ −1, 2], so the number whose existence is guaranteed by the mean value theorem is c = 11 1/4 . 5 √ C04S0M.028: Because f (x) = x is differentiable on (0, 4) and continuous on [0, 4], the hypotheses of the mean value theorem are satisfied. Moreover, f (4) − f (0) 2 1 1 = = = f (c) = 1/2 , 4−0 4 2 2c and it follows that c = 1. C04S0M.029: f (x) = 2x − 6, so f (x) = 0 when x = 3; f (x) ≡ 2 is always positive, so there are no inflection points and there is a global minimum at (3, −5). The y -intercept is (0, 4) and the x-intercepts are √ 3 ± 5 , 0 . The graph of f is shown next. 20 15 10 5 -2 2 4 6 -5 C04S0M.030: f (x) = 6(x − 3)(x +2), so there are critical points at (−2, 44) and (3, −81). f (x) = 12x − 6, so there is an inflection point at 1 , − 37 . The origin (0, 0) is a dual intercept and there are two other 2 2 x-intercepts. The graph of y = f (x) is next. 100 50 -4 -2 2 4 6 -50 -100 -150 C04S0M.031: f (x) = (3x4 − 5x2 + 60)x and f (x) = 15(x4 − x2 + 4), so f (x) > √ for all x and hence 0 √ (0, 0) is the only intercept. f (x) = 30x(2x2 − 1), so there are inflection points at − 1 2, − 233 2 , (0, 0), 2 8 √ √ and 1 2, 233 2 . The graph is actually concave upward between the first and second of these inflection 2 8 4 points and concave downward between the second and third, but so slightly that this is not visible on the graph of f that is shown next. 750 500 250 -3 -2 -1 1 2 3 -250 -500 -750 √ C04S0M.032: Given f (x) = (3 − x) x, we have f (x) = 3(1 − x) √ 2x and f (x) = − 3(x + 1) √. 4x x So there is a critical point at (1, 2), intercepts at (0, 0) and (3, 0), and the graph of f is concave down on the entire domain [0, + ∞) of f . Thus the critical point is a global maximum, there are no inflection points, and there is a local minimum at (0, 0). The graph of f is shown next. 2 1 1 2 3 4 5 -1 -2 -3 -4 C04S0M.033: Given f (x) = (1 x)x1/3 , we find that 2(2x + 1) . 9x5/3 √ So there are √ intercepts at (0, 0) and (1, 0) and a critical point at 1 , 3 3 2 . There is an inflection point 48 at − 1 , − 3 3 4 ; the graph of f is actually concave upward between this point and (0, 0) and concave 2 4 downward to its left, but the latter is not visible in the scale of the accompanying figure. The origin is also an inflection point and there is a vertical tangent there too. f (x) = 1 − 4x 3x2/3 and f (x) = − 0.5 -1 -0.5 0.5 1 1.5 2 2.5 -0.5 -1 -1.5 -2 C04S0M.034: Let g (x) = x5 + x − 5. Then g (2) = 29 > 0 while g (1) = −3 < 0. Because g (x) is a polynomial, it has the intermediate value property. Therefore the equation g (x) = 0 has at least one solution 5 in the interval 1 x 2. Moreover, g (x) = 5x4 + 1, so g (x) > 0 for all x. Consequently g is increasing on the set of all real numbers, and so takes on each value—including zero—at most once. We may conclude that the equation g (x) = 0 has exactly one solution, and hence that the equation x5 + x = 5 has exactly one solution. (The solution is approximately 1.299152792.) C04S0M.035: f (x) = 3x2 − 2, f (x) = 6x, and f (x) ≡ 6. C04S0M.036: f (x) = 100(x + 1)99 , f (x) = 9900(x + 1)98 , and f (x) = 970200(x + 1)97 . C04S0M.037: Given g (t) = g (t) = 1 1 − , t 2t + 1 2 1 − 2, (2t + 1)2 t g (t) = 2 8 − , t3 (2t + 1)3 and g (t) = C04S0M.038: h (y ) = 3 (3y − 1)−1/2 , h (y ) = − 9 (3y − 1)−3/2 , and h (y ) = 2 4 48 6 − 4. (2t + 1)4 t 81 8 (3y − 1)−5/2 . C04S0M.039: f (t) = 3t1/2 − 4t1/3 , f (t) = 3 t−1/2 − 4 t−2/3 , and f (t) = 8 t−5/3 − 3 t−3/2 . 2 3 9 4 C04S0M.040: g (x) = − 2x 6x2 − 18 216x − 24x3 , g (x) = 2 , and g (x) = . (x2 + 9)2 (x + 9)3 (x2 + 9)4 C04S0M.041: h (t) = − 4 8 24 , h (t) = , and h (t) = − . 2 3 (t − 2) (t − 2) (t − 2)4 C04S0M.042: f (z ) = 1 z −2/3 − 3 z −6/5 , f (z ) = − 2 z −5/3 + 18 z −11/5 , and f (z ) = 3 5 9 25 C04S0M.043: g (x) = − 10 −8/3 − 198 z −16/5 . 27 z 125 4 32 640 , g (x) = − , and g (x) = − . 3(5 − 4x)2/3 9(5 − 4x)5/3 27(5 − 4x)8/3 C04S0M.044: g (t) = 12(3 − t)−5/2 , g (t) = 30(3 − t)−7/2 , and g (t) = 105(3 − t)−9/2 . C04S0M.045: x−2/3 + y −2/3 dy dy = 0, so = −(y/x)2/3 . dx dx d2 y 2 x(dy/dx) − y 2 x(dy/dx) − y = − (y/x)−1/3 · =− · dx2 3 x2 3 x5/3 y 1/3 2 y + x1/3 y 2/3 2 x1/3 + y 1/3 2 =· = y 2 /3 · = (y/x5 )1/3 . 3 3 3 x5/3 y 1/3 x5/3 y 1/3 C04S0M.046: Given 2x2 − 3xy + 5y 2 = 25, we differentiate both sides with respect to x and obtain 4x − 3y − 3x dy dy + 10y = 0, dx dx so that dy 3y − 4x = . dx 10y − 3x We differentiate both sides of the second of these equations, again with respect to x, and find that d2 y (10y − 3x)(3y (x) − 4) − (3y − 4x)(10y (x) − 3) = . 2 dx (10y − 3x)2 Then replacement of y (x) with 3y − 4x yields 10y − 3x 6 d2 y 1550 =− 2 dx (10y − 3x)3 after simplifications that use the original equation. C04S0M.047: Given y 5 − 4y + 1 = x1/2 , we differentiate both sides of this equation (actually, an identity ) with respect to x and obtain 5y 4 dy dy 1 −4 = 1 /2 , dx dx 2x so that dy 1 √. = y (x) = dx 2(5y 4 − 4) x (1) Another differentiation yields d2 y 1 + 80x3/2 y 3 [ y (x)]3 =− , dx2 4(5y 4 − 4)x3/2 then substitution of y (x) from Eq. (1) yields y (x) = 40y 4 − 25y 8 − 20x1/3 y 3 − 16 . 4x3/2 (5y 4 − 4)3 C04S0M.048: Given: sin(xy ) = xy . The only solution of sin z = z is z = 0. Therefore xy = 0. Thus x = 0 or y = 0. This means that the graph of the equation sin(xy ) = xy consists of the coordinate axes. The y -axis is not the graph of a function, so the derivative is defined only for x = 0, and dy/dx = 0 for x = 0. Therefore also d2 y/dx2 = 0 for x = 0. C04S0M.049: Given x2 + y 2 = 5xy + 5, we differentiate both sides with respect to x to obtain 2x + 2y dy dy = 5y + 5x , dx dx so that Another differentiation yields y (x) = 2x − 5y dy = y (x) = . dx 5x − 2y 2[(y (x))2 − 5y (x) + 1] , 5x − 2y then substitution of y (x) from Eq. (1) yields y (x) = − 42(x2 − 5xy + y 2 ) 210 =− . (5x − 2y )3 (5x − 2y )3 In the last step we used the original equation in which y is defined implicitly as a function of x. C04S0M.050: x5 + xy 4 = 1: 5x4 + y 4 + 4xy 3 dy dy 5x4 + y 4 . = 0, so =− dx dx 4xy 3 dy dy 4xy 3 (20x3 + 4y 3 ) − (5x4 + y 4 )(4y 3 + 12xy 2 ) d2 y dx dx =− dx2 16x2 y 6 dy dy 20x4 y 3 + 4y 7 + (60x5 y 2 + 12xy 6 ) − 80x4 y 3 − 16xy 6 dx dx = 16x2 y 6 dy 4y 7 − 60x4 y 3 + (60x5 y 2 − 4xy 6 ) dx = 16x2 y 6 7 (1) = = 4y 5 − 60x4 y + (60x5 − 4xy 4 ) 16x2 y 4 dy dx 4y 5 − 60x4 y + (4xy 4 − 60x5 ) · 5x4 + y 4 4xy 3 16x2 y 4 16xy − 240x y + 20x5 y 4 + 4xy 8 − 300x9 − 60x5 y 4 = 64x3 y 7 8 54 20xy − 280x y − 300x9 5y 8 − 70x4 y 4 − 75x8 = = 64x3 y 7 16x2 y 7 8 44 8 4 4 5(y − 14x y − 15x ) 5(y + x )(y 4 − 15x4 ) = = . 16x2 y 7 16x2 y 7 8 But x4 + y 4 = 54 1 5(y 4 − 15x4 ) d2 y = . , so 2 x dx 16x3 y 7 C04S0M.051: Given: y 3 − y = x2 y : dy 2xy . = y (x) = − 2 dx x + 1 − 3y 2 (1) Then y (x) = − 2[ y + 2xy (x) − 3y (y (x))2 ] . x2 + 1 − 3y 2 (2) Substitution of y (x) from Eq. (1) in the right-hand side of Eq. (2) then yields y (x) = 2y [3x4 − 9y 4 + 6(x2 + 1)y 2 + 2x2 − 1] . (x2 + 1 − 3y 2 )3 C04S0M.052: (x2 − y 2 )2 = 4xy : dy dy ) = 4x + 4y ; dx dx dy dy (x2 − y 2 ) · x − (x2 − y 2 ) · y =x + y; dx dx dy (x + x2 y − y 3 ) = x3 − xy 2 − y ; dx dy x(x2 − y 2 ) − y x3 − xy 2 − y ; = = dx x + y (x2 − y 2 ) x + x2 y − y 3 2(x2 − y 2 )(2x − 2y d2 y = dx2 x + x2 y − y 3 3x2 − 2xy dy dy − y2 − − x3 − xy 2 − y dx dx 2 (x + x2 y − y 3 ) which upon simplification and substitution for dy/dx becomes d2 y 3xy (2 − xy ) = . dx2 (x + x2 y − y 3 )3 8 1 + x2 dy dy + 2xy − 3y 2 dx dx , C04S0M.053: f (x) = 4x3 − 32, so there is a critical point at (2, −48). f (x) = 12x2 , but there is no inflection point at (0, 0) because the graph of f is concave upward for all x. But (0, 0) is a dual intercept, √ and there is an x-intercept at 3 32, 0 . The graph of y = f (x) is shown next. 125 100 75 50 25 -2 -1 1 2 3 4 -25 -50 C04S0M.054: f (x) = 18x2 − x4 = x2 (18 − x2 ); f (x) = 36x − 4x3 = 4x(3 + x)(3 − x); f (x) = 12(3 − x2 ). There are global maxima√ (−3, 81) and (3, 81) and a local minimum at (0, 0). √ other two x-intercepts at The √ √ are at −3 2, 0 and 3 2, 0 . There are inflection points at − 3, 45 and 3, 45 . The graph of f is next. 50 -4 -2 2 4 -50 -100 -150 √ C04S0M.055: f (x) = 2x3 (3x2 − 4) and f (x) = 6x2 (5x2 − 4). There are global minima at − 2 3, − 32 3 27 √ and 2 √ , − 32 and a local maximum at (0, 0), which is also a dual intercept. There √ inflection√ are points 33 27 √ 95 95 at − 2 5, − 125 and 2 5, − 125 but not at (0, 0). There are x-intercepts at − 2, 0 and 2, 0 . 5 5 The graph of y = f (x) is shown next. 1.5 1 0.5 -2 -1 1 2 -0.5 -1 √ C04S0M.056: If f (x) = x x − 3, then 3(x − 2) f (x) = √ 2 x−3 and f (x) = 3(x − 4) . 4(x − 3)3/2 Therefore the graph of f is increasing for all x (the domain of f is [3, + ∞)), concave downward for x < 4, and concave upward for x > 4; there is an inflection point at (4, 4). The only intercept is (3, 0), which is 9 also a global minimum and the location of a vertical tangent. The graph of f is next. 4 3 2 1 3.2 3.4 3.6 3.8 4 C04S0M.057: If f (x) = x(4 − x)1/3 , then f (x) = 4(3 − x) 3(4 − x)2/3 and f (x) = 4(x − 6) . 9(4 − x)5/3 So there is a global maximum at (3, 3) intercepts at (0, 0) and (4, 0), a vertical tangent and inflection point √ at the latter, and an inflection point at 6, −6 3 2 . The graph is next. 2 -2 2 4 6 -2 -4 -6 C04S0M.058: If (x) = x−1 3 =1− , then x+2 x+2 f (x) = 3 (x + 2)2 and f (x) = − 6 . (x + 2)3 There are no critical points and no inflection points. The graph is increasing except at the discontinuity at x = −2. It is concave upward for x < −2 and concave downward if x > −2. The vertical line x = −2 and the horizontal line y = 1 are asymptotes, and the intercepts are 0, − 1 and (1, 0). The graph of y = f (x) 2 is next. 20 15 10 5 -4 -3 -2 -1 -5 -10 -15 -20 C04S0M.059: If f (x) = x2 + 1 5 =1+ 2 , then 2−4 x x −4 f (x) = − 10x − 4)2 (x2 and 10 f (x) = 10(3x2 + 4) . (x2 − 4)3 Thus there is a local maximum at 0, − 1 , no other extrema, no inflection points, and no intercepts. The 4 lines x = −2 and x = 2 are vertical asymptotes and y = 1 is a horizontal asymptote; the graph is next. 10 7.5 5 2.5 -4 -2 2 4 -2.5 -5 -7.5 -10 C04S0M.060: If f (x) = x , then (x − 2)(x + 1) f (x) = − (x2 x2 + 2 − x − 2)2 and f (x) = 2(x3 + 6x − 2) . (x + 1)3 (x − 2)3 There are no critical points; there is an inflection point with the approximate coordinates (0.3275, −0.1475). The graph is decreasing for all x other than −1 and 2, is concave upward on the intervals (−1, 0.3275) and (2, + ∞), and is concave downward on the intervals (0.3275, 2) and (−∞, −1). The asymptotes are y = 0, x = 2, and x = −1 and (0, 0) is the only intercept. The graph of y = f (x) is shown next. 10 7.5 5 2.5 -2 -1 1 2 3 -2.5 -5 -7.5 -10 C04S0M.061: If f (x) = 2x2 , then (x − 2)(x + 1) f (x) = − 2x(x + 4) (x − 2)2 (x + 1)2 and f (x) = 4(x3 + 6x2 + 4) . (x2 − x − 2)3 Thus (0, 0) is a local maximum and the only intercept; there is a local minimum at −4, 16 . There is an 9 inflection point with the approximate coordinates (−6.107243, 1.801610). The lines x = −1 and x = 2 are vertical asymptotes and the line y = 2 is a horizontal asymptote. The graph of y = f (x) for −12 < x < −2 is next, on the left; the graph for −2 < x < 4 is on the right. 10 2.5 7.5 2 5 2.5 1.5 -2 1 -1 1 -2.5 -5 0.5 -7.5 -12 -10 -6 -4 -10 -2 11 2 3 4 C04S0M.062: Given: f (x) = x3 1 1 =x+ + . First, −1 2(x + 1) 2(x − 1) x2 f (x) = x2 (x2 − 3) (x2 − 1)2 and f (x) = 2x(x2 + 3) . (x2 − 1)3 √ Inflection point and sole intercept: (0, 0). There is a local minimum where x = 3 and a local maximum √ where x = − 3. The graph is concave downward on the intervals (0, √ and (−∞, −1), concave upward 1) √ on the intervals (−1, 0) and (1, + ∞). The graph is increasing if x < − 3 and if x > 3 and is decreasing otherwise. The lines x = −1, x = 1, and y = x are asymptotes. The graph of y = f (x) is next. 6 4 2 -3 -2 -1 1 2 3 -2 -4 -6 C04S0M.063: Here we have f (x) = x3 (3x − 4), f (x) = 12x2 (x − 1), and f (x) = 12x(3x − 2). Hence there are intercepts at (0, 0) and 4 , 0 ; the graph is increasing for x > 1 and decreasing for x < 1; it is 3 concave upward for x > 2 and for x < 0, concave downward on the interval 0, 2 . Consequently there is a 3 3 global minimum at (1, −1) and inflection points at (0, 0) and 2 , − 16 . There are no asymptotes, no other 3 27 extrema, and f (x) → + ∞ as x → + ∞ and as x → −∞. The graph of f is shown next. 4 3 2 1 -2 -1 1 2 -1 C04S0M.064: Here we √ have f (x) = x2 (x2 − 2), f (x) = 4x(x + 1)(x − 1), and f (x) = 4(3x2 − 1). So √ there are intercepts at − 2, 0 , (0, 0), and 2, 0 . The graph is increasing on the intervals (1, + ∞) and (−1, 0), decreasing on the intervals (−∞, −1) and (0, 1). It is concave upward where x2 > 1 and concave 3 downward where x2 < 1 . There are global minima at (−1, −1) and (1, −1) and a local maximum at the 3 12 origin. There are inflection points at the two points where x2 = 1 . The graph is next. 3 1.5 1 0.5 -2 -1 1 2 -0.5 -1 C04S0M.065: If f (x) = x2 , then −1 x2 f (x) = − 2x 2 − 1)2 (x and f (x) = 2(3x2 + 1) . (x2 − 1)3 Thus (0, 0) is the only intercept and is a local maximum; there are no inflection points, the lines x = −1 and x = 1 are vertical asymptotes, and the line y = 1 is a horizontal asymptote. The graph of y = f (x) is next. 10 7.5 5 2.5 -2 -1 1 2 -2.5 -5 -7.5 -10 C04S0M.066: First, f (x) √ x(x2 − 12), f (x) = 3(x + 2)(x − 2), and f (x) = 6x. So there are intercepts = √ at 2 3, 0 , (0, 0), and −2 3, 0 . The graph is increasing for x > 2 and for x < −2; it is decreasing on the interval (−2, 2). It is concave upward for x > 0 and concave downward for x < 0. There is a local maximum at (−2, 16), a local minimum at (2, −16), and an inflection point at the origin. The graph of y = f (x) is shown next. 30 20 10 -4 -2 2 4 -10 -20 -30 C04S0M.067: If f (x) = −10 + 6x2 − x3 , then f (x) = 3x(4 − x) and f (x) = 6(2 − x). It follows that there is a local minimum at (0, −10), a local maximum at (4, 22), and an inflection point at (2, 6). The intercepts are approximately (−1.180140, 0), (1.488872, 0), (5.691268, 0), and [exactly] (0, −10). The graph 13 of f is next. 40 30 20 10 -2 2 4 6 -10 -20 -30 C04S0M.068: If f (x) = x , then 1 + x2 f (x) = 1 − x2 (x2 + 1)2 and f (x) = 2x(x2 − 3) . (x2 + 1)3 There is a global maximum at 1, 1 and a global minimum at √ 1, − 1 . The origin is the only intercept 2 2 √ 1− √ √ and an inflection point; there are also inflection points at 3, 4 3 and − 3, − 1 3 . The x-axis is the 4 only asymptote. The graph of y = f (x) is shown next. 0.4 0.2 -6 -4 -2 2 4 6 -0.2 -0.4 C04S0M.069: If f (x) = x3 − 3x, then f (x) = 3(x + 1)(x − 1) and f (x) = 6x. So there is a local maximum at (−1, 2), a local minimum at (1, −2), and an inflection point at (0, 0). There are also intercepts √ √ at − 3, 0 and 3, 0 but no asymptotes. The graph is next. 7.5 5 2.5 -3 -2 -1 1 2 3 -2.5 -5 -7.5 C04S0M.070: If f (x) = x4 − 12x2 √ x2 (x2 − 12), √ = then f (x) = 4x(x2 − 6) and f (x) = 12(x2 − 2). Hence there are global minima at − 6, −36 and 6, −36 and a local maximum at (0, √ There are 0). √ √ √ also intercepts at −2 3, 0 and 2 3, 0 . There are inflection points at − 2, −20 and 2, −20 and 14 no asymptotes. The graph of y = f (x) is shown next. 60 40 20 -4 -2 2 4 -20 C04S0M.071: If f (x) = x3 + x2 − 5x +3 = (x − 1)2 (x +3), then f (x) = (3x +5)(x − 1) and f (x) = 6x +2. So there is a local maximum at − 5 , 256 and a local minimum at (1, 0). There is an inflection point at 3 27 − 1 , 128 . A second x-intercept is (−3, 0) and there are no asymptotes. The graph of y = f (x) is next. 3 27 20 10 -4 -3 -2 -1 1 2 3 -10 x+1 x+2 2(x + 3) , f (x) = − 3 , f (x) = . The graph is decreasing for x > 0 and x2 x x4 for x < −2, increasing on the interval (−2, 0). It is concave downward for x < −3, concave upward for x > 0 and on the interval (−3, 0). The only intercept is (−1, 0) and there is a discontinuity where x = 0. There is a global minimum at (−2, −0.25) and an inflection point at −3, − 2 . As x → 0, f (x) → + ∞, so the 9 y -axis is a vertical asymptote. As |x| → +∞, f (x) → 0, so the x-axis is a horizontal asymptote. The graph of y = f (x) for −2 < x < 2 is next, on the left; the graph for −15 < x < −1 is on the right. C04S0M.072: f (x) = 20 -14 15 -12 -10 5 -1 -6 -4 -2 -0.05 10 -2 -8 -0.1 1 2 -0.15 -5 -10 -0.2 -15 -0.25 -20 C04S0M.073: The given function f (x) is expressed as a fraction with constant numerator, so we maximize f (x) by minimizing its denominator (x + 1)2 + 1. It is clear that x = −1 does the trick, so the maximum value of f (x) is f (−1) = 1. C04S0M.074: Let k be the proportionality constant for cost; if the pot has radius r and height h, we are to minimize total cost C = k (5)(π r2 ) + (1)(2π rh) 15 subject to the constraint π r2 h = 1. Then h = 1/ π r2 , so 2 r , C (r) = k 10π r − 2 r2 C = C (r) = k 5π r2 + r > 0. Now ; C (r) = 0 when r= 1 5π 1/3 and h= 25 π 1 /3 . It’s easy to establish in the usual way that these values minimize C , and it’s worth noting that when C is minimized, we also have h = 5r. C04S0M.075: Let x represent the width of the base of the box. Then the length of the base is 2x and, because the volume of the box is 4500, the height of the box is 4500/ 2x2 . We minimize the surface area of the box, which is given by 4500 4500 13500 + 2x · = 2x2 + , 0 < x < ∞. 2 2 2x 2x x √ Now f (x) = 4x − (13500/x2 ), so f (x) = 0 when x = 3 3375 = 15. Note that f (15) > 0, so surface area is minimized when x = 15. The box of minimal surface area is 15 cm wide, 30 cm long, and 10 cm high. f (x) = 2x2 + 4x · C04S0M.076: Let x represent the edge length of the square base of the box. Because the volume of the box is 324, the box has height 324/x2 . We minimize the cost C of materials to make the box, where C = C (x) = 3x2 + 4 · x · 324 1296 = 3x2 + , 2 x x 0 < x < ∞. Now C (x) = 6x − (1296/x2 ), so C (x) = 0 when x = 3 1296/6 = 6. Because C (6) > 0, the cost C is minimized when x = 6. The box we seek has a square base 6 in. on a side and height 9 in. C04S0M.077: Let x represent the width of the base of the box. Then the box has base of length 2x and height 200/x2 . We minimize the cost C of the box, where C = C (x) = 7 · 2x2 + 5 · 6x · 200 x2 + 5 · 2x2 = 24x2 + 6000 , x 0 < x < ∞. √ Now C (x) = 48x − (6000/x2 ), so C (x) = 0 when x = 3 125 = 5. Because C (5) > 0, the cost C is minimized when x = 5. The box of minimal cost is 5 in. wide, 10 in. long, and 8 in. high. C04S0M.078: If the zeros of f (x) are at a, b, and c (with a < b < c), apply Rolle’s theorem to f on the two intervals [ a , b ] and [ b, c ]. C04S0M.079: If the speed of the truck is v , then the trip time is T = 1000/v . So the resulting cost is C (v ) = 10000 + (1000) 1 + (0.0003)v 3/2 , v 16 so that C (v ) 10 = + 1 + (0.0003)v 3/2 . 1000 v Thus √ C (v ) 10 3 = − 2 + (0.0003) v. 1000 v 2 Then C (v ) = 0 when v = (200, 000/9)2/5 ≈ 54.79 mi/h. This clearly minimizes the cost, because C (v ) > 0 for all v > 0. C04S0M.080: The sum in question is S (x) = (x − a1 )2 + (x − a2 )2 + . . . + (x − an )2 . Now S (x) = 2(x − a1 ) + 2(x − a2 ) + . . . + 2(x − an ); S (x) = 0 when nx = a1 + a2 + . . . + an , so that x= 1 (a1 + a2 + . . . + an ) n (1) —the average of the n fixed numbers. It is clear that S is continuous and that S (x) → +∞ as |x| → +∞, so S (x) must have a global minimum value. Therefore the value of x in Eq. (1) minimizes the sum of the squares of the distances. C04S0M.081: First, given y 2 = x(x − 1)(x − 2), we differentiate implicitly and find that dy dy 3x2 − 6x + 2 = 3x2 − 6x + 2, so = . dx dx 2y √ The only zero of dy/dx in the domain is 1 − 1 3 , so there are two horizontal tangent lines (the y -coordinates 3 are approximately ±0.6204). Moreover, dx/dy = 0 when y = 0; that is, when x = 0, when x = 1, and when x = 2. So there are three vertical tangent lines. After lengthy simplifications, one can show that 2y d2 y 3x4 − 12x3 + 12x2 − 4 = . dx2 4y 3 The only zero of y (x) in the domain is about 2.4679, and there the graph has the two values y ≈ ± 1.3019. These are the two inflection points. The graph of the given equation is shown next. 1 0.5 0.5 1 1.5 2 2.5 -0.5 -1 C04S0M.082: Let x represent the length of the internal divider. Then the field measures x by 2400/x ft. We minimize the total length of fencing, given by: 17 f (x) = 3x + 4800 , x 0 < x < ∞. √ Now f (x) = 3 − 4800x−2 , which is zero only when x = 1600 = 40. Verification: f (x) > 0 if x > 40 and f (x) < 0 if x < 40, so f (x) is minimized when x = 40. The minimum length of fencing required for this field is f (40) = 240 feet. C04S0M.083: Let x represent the length of each of the dividers. Then the field measures x by 1800/x ft. We minimize the total length of fencing, given by: f (x) = 4x + 3600 , x 0 < x < ∞. Now f (x) = 4 − 3600x−2 , which is zero only when x = 30. Verification: f (x) > 0 if x > 30 and f (x) < 0 if x < 30, so f (x) is minimized when x = 30. The minimum length of fencing required for this field is f (30) = 240 ft. C04S0M.084: Let x represent the length of each of the dividers. Then the field measures x by 2250/x meters. We minimize the total length of fencing, given by: f (x) = 5x + 4500 , x 0 < x < ∞. Now f (x) = 5 − 4500x−2 , which is zero only when x = 30. Verification: f (x) > 0 if x > 30 and f (x) < 0 if x < 30, so f (x) is minimized when x = 30. The minimum length of fencing required for this field is f (30) = 300 meters. C04S0M.085: Let x represent the length of each of the dividers. Then the field measures x by A/x ft. We minimize the total length of fencing, given by: f (x) = (n + 2)x + 2A , x 0 < x < ∞. Now f (x) = n + 2 − 2Ax−2 , which is zero only when x = 2A/(n + 2). Verification: f (x) > 0 if x > 2A/(n + 2) and f (x) < 0 if x < 2A/(n + 2), so f (x) is minimized when x = 2A/(n + 2). The minimum length of fencing required for this field is f 2A n+2 √ 2A 2A n + 2 √ + n+2 2A = (n + 2) = 2A(n + 2) + 2A(n + 2) = 2 2A(n + 2) (ft). C04S0M.086: Let L be the line segment with endpoints at (0, c) and (b, 0) on the coordinate axes and suppose that L is tangent to the graph of y = 1/x2 at (x, 1/x2 ). We will minimize S , the square of the length of L, where S = b2 + c2 . We compute the slope of L in several ways: as the value of dy/dx at the point of tangency, as the slope of the line segment between (x, 1/x2 ) and (b, 0), and as the slope of the line segment between (x, 1/x2 ) and (0, c): 1 −0 2 2 − 3=x , x x−b 1 −c 2 2 − 3=x ; x x−0 so x = −2(x − b), −2x = x3 ( 1 − 3), x2 18 hence cx2 = 3, b= hence 3 x. 2 c= 3 . x2 Therefore 9 9x2 + ; 4 x 4 S (x) = S (x) = − 36 9 + x, x5 2 √ √ √ which is zero only when x = 2. Verification: S (x) < 0 if x < 2 and S (x) > 0 if x > 2. So S , hence √ √ the length of L, is minimized when x = 2. The length of this shortest line segment is 3 3. 2 C04S0M.087: Let L be the line segment with endpoints at (0, c) and (b, 0) on the coordinate axes and suppose that L is tangent to the graph of y = 1/x2 at (x, 1/x2 ). We compute the slope of L in several ways: as the value of dy/dx at the point of tangency, as the slope of the line segment between (x, 1/x2 ) and (b, 0), and as the slope of the line segment between (x, 1/x2 ) and (0, c): 1 −0 2 x2 − 3= , x x−b 1 −c 2 x2 − 3= ; x x−0 so x = −2(x − b), −2x = x3 ( 1 − 3), x2 hence cx2 = 3, b= hence 3 x. 2 c= 3 . x2 Now the length of the base of the right triangle is b and its height is c, so its area is given by A(x) = 1 3x 3 9 · · = , 2 2 x2 4x 0 < x < + ∞. Clearly A is a strictly decreasing function of x, so it has neither a maximum nor a minimum—not even any local extrema. C04S0M.088: Let L be the line segment in the first quadrant that is tangent to the graph of y = 1/x at (x, 1/x) and has endpoints (0, c) and (b, 0). Compute the slope of L in several ways: as the value of dy/dx at the point of tangency, as the slope of the line segment between (x, 1/x) and (b, 0), and as the slope of the line segment between (x, 1/x) and (0, c): 1 −0 1 x − 2= x x−b 1 −c 1 − 2=x : x x c− so b = 2x. 1 1 =, x x so c= 2 . x Therefore the area A of the triangle is A = A(x) = 1 1 · 2x · ≡ 1. 2 x Because A is a constant function, every such triangle has both maximal and minimal area. C04S0M.089: Let x be the length of the shorter sides of the base and let y be the height of the box. Then its volume is 3x2 y , so that y = 96/x2 . The total surface area of the box is 6x2 + 8xy , thus is given as a function of x by A(x) = 6x2 + 8x · 96 768 = 6x2 + , 2 x x 19 0 < x < + ∞. So A (x) = 12x − 768x−2 ; A (x) = 0 when x3 = 64, so that x = 4. Because A (x) > 0 for all x > 0, we have found the value of x that yields the global minimum value of A, which is A(4) = 288 (in.2 ). The fact that 288 is also the numerical value of the volume is a mere coincidence. C04S0M.090: Let x be the length of the shorter sides of the base and let y be the height of the box. Then its volume is 4x2 y , so that y = 200/x2 . The total surface area of the box is 8x2 + 10xy , thus is given as a function of x by A(x) = 8x2 + 10x · 200 2000 = 8x2 + , 2 x x 0 < x < + ∞. So A (x) = 16x − 2000x−2 ; A (x) = 0 when x3 = 125, so that x = 5. Because A (x) > 0 for all x > 0, we have found the value of x that yields the global minimum value of A, which is A(5) = 600 (in.2 ). C04S0M.091: Let x be the length of the shorter sides of the base and let y be the height of the box. Then its volume is 5x2 y , so that y = 45/x2 . The total surface area of the box is 10x2 + 12xy , thus is given as a function of x by A(x) = 10x2 + 12x · 45 540 = 10x2 + , 2 x x 0 < x < + ∞. So A (x) = 20x − 540x−2 ; A (x) = 0 when x3 = 27, so that x = 3. Because A (x) > 0 for all x > 0, we have found the value of x that yields the global minimum value of A, which is A(3) = 270 (cm2 ). C04S0M.092: Let x represent the width of the box. Then the length of the box is nx and its height is V /(nx2 ). We minimize the surface area A of the box, where A = A(x) = 2nx2 + V 2(n + 1)V · 2(n + 1)x = 2nx2 + , 2 nx nx 0 < x < ∞. Now A (x) = 4nx − 2(n + 1)V , nx2 Verification: A (x) > 0 for x > 3 and A (x) = 0 when (n + 1)V and A (x) < 0 for x < 2n2 minimizes A. The value of A(x), simplified, at this minimum point is 3 x= f (x) (1 − x)2/3 = lim x→±∞ x x→±∞ x2/3 /3 2 1 −1 = lim x = +1. x→±∞ 1 Then 20 (n + 1)V . 2n 2 (n + 1)V , so this critical point 2n 2 1 /3 2(n + 1)2 V 2 . n 3 C04S0M.093: First, m = lim 3 b = lim [f (x) − mx] = lim x→∞ x→∞ x1/3 (1 − x)2/3 − x x1/3 (1 − x)2/3 − x x2/3 (1 − x)4/3 + x4/3 (1 − x)2/3 + x2 x→∞ x2/3 (1 − x)4/3 + x4/3 (1 − x)2/3 + x2 x(1 − x)2 − x3 = lim 2/3 x→∞ x (1 − x)4/3 + x4/3 (1 − x)2/3 + x2 x − 2x2 = lim 2/3 4/3 + x4/3 (1 − x)2/3 + x2 x→∞ x (1 − x) 1 −2 x = lim 4 /3 2 /3 x→∞ 1−x 1−x + +1 x x 0−2 2 = =− . 1+1+1 3 = lim The limit is the same as x → −∞. So the graph of f (x) = x1/3 (1 − x)2/3 has the oblique asymptote y = x − 2 . 3 C04S0M.094: Let θ be the angle between your initial path and due north, so that 0 θ π /2, and if θ = π /2 then you plan to jog around a semicircle and not swim at all. Suppose that you can swim with speed v (in miles per hour). Then you will swim a length of 2 cos θ miles at speed v and jog a length of 2θ miles at speed 2v , for a total time of T (θ) = 2 cos θ 2θ 1 + = (θ + 2 cos θ), v 2v v 0 θ π /2. It’s easy to verify that this formula is correct even in the extreme case θ = π /2. It turns out that although T (θ) = 0 when θ = π /6, this value of θ actually maximizes T (θ); this function has an endpoint minimum not even at θ = 0, but at θ = π /2. Answer: Jog all the way. C04S0M.095: By l’Hˆpital’s rule, o lim x→2 Without l’Hˆpital’s rule, o lim x→2 x−2 1 1 = lim =. 2−4 x→2 2x x 4 x−2 x−2 1 1 = lim = lim =. x2 − 4 x→2 (x + 2)(x − 2) x→2 x + 2 4 C04S0M.096: By l’Hˆpital’s rule, lim o sin 2x 2 cos 2x 2·1 = lim = = 2. x→0 x 1 1 C04S0M.097: By l’Hˆpital’s rule, lim o 1 + cos x − sin x − cos x 1 = lim = lim =. x→π 2(x − π ) x→π (x − π )2 2 2 x→0 x→π C04S0M.098: By l’Hˆpital’s rule (applied three times), o lim x→0 x − sin x 1 − cos x sin x cos x 1 = lim = lim = lim =. 3 2 x→0 x→0 6x x→0 x 3x 6 6 C04S0M.099: By l’Hˆpital’s rule (applied three times), o 21 lim t→0 tan t − sin t sec2 t − cos t = lim t→0 t3 3t2 = lim t→0 2 sec2 t tan t + sin t 4 sec2 t tan2 t + 2 sec4 t + cos t 1 = lim =. t→0 6t 6 2 C04S0M.100: By l’Hˆpital’s rule, lim o x→∞ ln(ln x) 1 = lim = 0. x→∞ ln x ln x C04S0M.101: By l’Hˆpital’s rule, o lim (cot x) ln(1 + x) = lim x→0 x→0 ln(1 + x) 1 1 = lim = = 1. 2x x→0 (1 + x) sec tan x 1·1 C04S0M.102: By l’Hˆpital’s rule and the product rule for limits, o lim (e1/x − 1) tan x = lim + + x→0 x→0 e1/x − 1 e1/x sin x = lim+ 2 = lim+ cot x x x→0 x csc2 x x→0 2 e1/x = 1 · lim e1/x x→0+ = + ∞. Recall that l’Hˆpital’s rule is valid even if the resulting limit is + ∞ or −∞, although we are stretching the o hypotheses of the product rule a little here. We need a lemma. Lemma: If lim f (x) = p > 0 x→a+ and lim g (x) = + ∞, x→a+ then lim f (x) · g (x) = +∞. + x→a Proof: Given M > 0, choose δ > 0 such that, if x − a < δ then f (x) > p/2 and g (x) > 2M/p. Then, for such x, f (x) · g (x) > (p/2) · 2M/p = M . Because M may be arbitrarily large positive, this implies that f (x) · g (x) takes on arbitrarily large values if x > a and x is near a. That is, lim f (x) · g (x) = + ∞. x→a+ C04S0M.103: After combining the two fractions, we apply l’Hˆpital’s rule once, then use a little algebra: o lim x→0 1 1 − x2 1 − cos x sin x −2 1 − cos x − x2 (sin x) − 2x x = lim = lim . = lim 2 2 sin x x→0 x (1 − cos x) x→0 2x(1 − cos x) + x x→0 2(1 − cos x) + x sin x Now if x is close to (but not equal to) zero, (sin x)/x ≈ 1, so the numerator in the last limit is near −1. Moreover, for such x, cos x < 1 and x and sin x have the same sign, so the denominator in the last limit is close to zero and positive. Therefore the limit is −∞. C04S0M.104: We don’t need l’Hˆpital’s rule here, although it may be applied. Without it we obtain o lim x→∞ x2 x3 −2 x+2 x +3 3x2 − 2x3 = lim 3 = lim x→∞ x + 2x2 + 3x + 6 x→∞ Using l’Hˆpital’s rule (three times) we find that o 22 3 −2 0−2 x = = −2. 2 6 3 1+0+0+0 1+ + 2 + 3 xx x = lim x→∞ 3x2 − 2x3 x3 + 2x2 + 3x + 6 = lim x2 x3 −2 x+2 x +3 lim 6x − 6x2 6 − 12x −12 = lim = lim = −2. x→∞ 3x2 + 4x + 3 x→∞ 6x + 4 6 x→∞ x→∞ C04S0M.105: Here it is easier not to use l’Hˆpital’s rule: o lim x2 − x − 1 − x→∞ √ = lim √ x x→∞ x2 − x − 1 − x √ x2 − x − 1 + x x−2− x2 − 2x − 1 = lim √ √ = lim x→∞ x→∞ x2 − x − 1 + x C04S0M.106: ln lim x1/x = lim ln x1/x = lim x→∞ x→∞ x→∞ 1 x 1 1 1− − 2 − xx 1 x = + ∞. ln x 1 = lim = 0. Therefore lim x1/x = 1. x→∞ x x→∞ x C04S0M.107: First we need an auxiliary result: lim 2xe−2x = lim x→∞ x→∞ 2x 2 = lim = 0. 2x x→∞ 2e2x e Then ln lim (e2x − 2x)1/x = lim ln(e2x − 2x)1/x = lim x→∞ x→∞ = lim x→∞ x→∞ ln(e2x − 2x) x 2e2x − 2 2 − 2e−2x 2−0 = lim = = 2. 2 x − 2x x→∞ 1 − 2xe−2x e 1−0 Therefore lim (e2x − 2x)1/x = e2 . x→∞ C04S0M.108: Given: lim x→∞ ln lim x→∞ 1 − exp(−x2 ) 1 − exp(−x2 ) 1/x2 1/x2 = lim ln 1 − exp(−x2 ) x→∞ x→∞ 1 − exp(−x2 ) 1/x2 x→∞ = lim Therefore lim . 1/x2 = lim x→∞ ln 1 − exp(−x2 ) x2 2x exp(−x2 ) exp(−x2 ) 0 = lim = = 0. 2x [1 − exp(−x2 )] x→∞ 1 − exp(−x2 ) 1−0 = e0 = 1. C04S0M.109: This is one of the most challenging problems in the book. We deeply regret publication of this solution. First let u = 1/x. Then L = lim x · x→∞ 1+ 1 x x − e = lim+ u→0 Apply l’Hˆpital’s rule once: o 23 (1 + u)1/u − e . u u − (1 + u) ln(1 + u) u2 (1 + u) L = lim (1 + u)1/u u →0 + . Now apply the product rule for limits! L=e· lim u→0+ u − (1 + u) ln(1 + u) u2 (1 + u) . Finally apply l’Hˆpital’s rule twice: o L=e· lim u →0 + 1 − 1 − ln(1 + u) 2u + 3u2 =e· lim u→0+ −1 (1 + u)(2 + 6u) =e· −1 e =− . 1·2 2 Most computer algebra programs cannot evaluate the original limit. C04S0M.110: First replace b with x to remind us what the variable in this problem is. Thus x a A(x) = 2π ax ln +√ 2 − a2 a x x+ √ x2 − a2 a . Then lim A(x) = lim 2π x2 + + x→a x→a = 2π a2 + 2π a2 2π a2 x ln (x2 − a2 )1/2 lim+ x→a (x2 Then apply l’Hˆpital’s rule to the limit that remains: o lim x→a x ln(x + (x2 − a2 )1/2 ) − x ln a (x2 − a2 )1/2 = lim ln x + (x2 − a2 )1/2 + x+ (x2 x + (x2 − a2 )1/2 a x ln − a2 )1/2 x + (x2 − a2 )1/2 a x 1 · 1 + (x2 − a2 )−1/2 · 2x − ln a 2 )1/2 2 −a . 12 (x − a2 )−1/2 · 2x 2 x (x2 − a2 )1/2 ln x + (x2 − a2 )1/2 + · (x2 − a2 )1/2 + x − (x2 − a2 )1/2 ln a x + (x2 − a2 )1/2 = lim x x→a+ x→a = lim x→a+ (x2 − a2 )1/2 ln x + (x2 − a2 )1/2 + x − (x2 − a2 )1/2 ln a = 1. x Therefore A(b) → 4π a2 as b → a+ . 24 Section 5.2 C05S02.001: (3x2 + 2x + 1) dx = x3 + x2 + x + C . C05S02.002: (3t4 + 5t − 6) dt = 3 t5 + 5 t2 − 6t + C . 5 2 C05S02.003: (1 − 2x2 + 3x3 ) dx = 3 x4 − 2 x3 + x + C . 4 3 C05S02.004: C05S02.005: − 1 t2 dt = 1 + C. t (3x−3 + 2x3/2 − 1) dx = − 3 x−2 + 4 x5/2 − x + C . 2 5 √ 5 −x x4 C05S02.006: x5/2 − C05S02.007: 3 1 /2 2t C05S02.008: 2 3 − 2 /3 x3/4 x C05S02.009: C05S02.010: C05S02.011: C05S02.012: dx = (x5/2 − 5x−4 − x1/2 ) dx = 2 x7/2 + 5 x−3 − 2 x3/2 + C . 7 3 3 + 7 dt = t3/2 + 7t + C . dx = (2x−3/4 − 3x−2/3 ) dx = 8x1/4 − 9x1/3 + C . (x2/3 + 4x−5/4 ) dx = 3 x5/3 − 16x−1/4 + C . 5 √ 1 2x x − √ x dx = (2x3/2 − x−1/2 ) dx = 4 x5/2 − 2x1/2 + C . 5 (4x3 − 4x + 6) dx = x4 − 2x2 + 6x + C . 15 5 t−2 4 t dt = 15 4t C05S02.013: 1 dx = 7x 16 24 t + 5t−1 + C . 7ex/7 dx = 49ex/7 + C . C05S02.014: − 5t−2 dt = 11 1 · dx = ln | x | + C . Another correct answer is 7x 7 1 1 1 1 1 dx = ln | 7x | + C1 = ln | x | + ln 7 + C1 = ln | x | + C2 , 7x 7 7 7 7 the last step because the constant C2 is simply the constant C1 + C05S02.015: 1 7 ln 7. (x + 1)4 dx = 1 (x + 1)5 + C . Note that many computer algebra systems give the answer 5 C + x + 2x2 + 2x3 + x4 + 1 x5 . 5 10 C05S02.016: (t + 1) dt = 1 11 (t C05S02.017: 1 dx = (x − 10)7 C05S02.018: √ C05S02.019: √ C05S02.020: √ 3 z + 1 dz = + 1)11 + C . (x − 10)−7 dx = − 1 (x − 10)−6 + C = − 6 1 + C. 6(x − 10)6 (z + 1)1/2 dz = 2 (z + 1)3/2 + C . 3 x (1 − x)2 dx = (x1/2 − 2x3/2 + x5/2 ) dx = 2 x3/2 − 4 x5/2 + 2 x7/2 + C . 3 5 7 3 x (x + 1) dx = (x10/3 +3x7/3 +3x4/3 +x1/3 ) dx = C05S02.021: 2x4 − 3x3 + 5 dx = 7x2 C05S02.022: (3x + 4) √ dx = x 22 7x − 3 x + 5 x−2 dx = 7 7 3 13/3 9 + 10 x10/3 + 9 x7/3 + 3 x4/3 +C . 13 x 7 4 23 21 x − 32 14 x − 5 x−1 + C . 7 2 = 18 5/2 5x C05S02.023: x−1/2 9x2 + 24x + 16 dx = 9x3/2 + 24x1/2 + 16x−1/2 dx + 16x3/2 + 32x1/2 + C . (9t + 11)5 dt = C + 161051t + 1 C05S02.024: 7 (3z + 10) 1 54 (9t + 11)6 + C . Mathematica gives the answer 658845t2 441045t4 19683t6 + 359370t3 + + 72171t5 + . 2 2 2 dz = 1 (3z + 10)−7 dz = − 18 (3z + 10)−6 + C . 1 2x 1 −2x + C. e−e 2 2 C05S02.025: (e2x + e−2x ) dx = C05S02.026: (ex + e−x )2 dx = C05S02.027: (5 cos 10x − 10 sin 5x) dx = C05S02.028: (2 cos π x + 3 sin π x) dx = C05S02.029: (3 cos π t + cos 3π t) dt = C05S02.030: (4 sin 2π t − 2 sin 4π t) dt = − C05S02.031: Dx 1 2 (e2x + 2 + e−2x ) dx = 1 2 1 2x 1 e + 2x − e−2x + C . 2 2 sin 10x + 2 cos 5x + C . 2 3 sin π x − cos π x + C . π π 3 1 sin π t + sin 3π t + C . π 3π 2 1 cos 2π t + cos 4π t + C . π 2π sin2 x + C1 = sin x cos x = Dx − 1 cos2 x + C2 . Because 2 1 1 sin2 x + C1 = − cos2 x + C2 , 2 2 it follows that 2 C2 − C1 = 1 1 1 sin2 x + cos2 x = . 2 2 2 1 1−x+x 1 , F (x) = = . F1 (x) − F2 (x) = C1 for some constant (1 − x)2 2 (1 − x)2 (1 − x)2 C1 on (−∞, 1); F1 (x) − F2 (x) = C2 for some constant C2 on (1, +∞). On either interval, F1 (x) − F2 (x) = 1−x = 1. 1−x C05S02.032: F1 (x) = C05S02.033: sin2 x dx = cos2 x dx = 1 2 + 1 2 1 2 − 1 2 cos 2x dx = 1 x − 2 cos 2x dx = 1 x + 2 C05S02.034: (a): Dx tan x = sec2 x; 1 4 1 4 sin 2x + C and sin 2x + C . (b): tan2 x dx = sec2 x − 1 dx = (tan x) − x + C . C05S02.035: y (x) = x2 + x + C ; y (0) = 3, so y (x) = x2 + x + 3. C05S02.036: y (x) = 1 (x − 2)4 + C and y (2) = 1, so y (x) = 1 (x − 2)4 + 1. 4 4 C05S02.037: y (x) = 2 x3/2 + C and y (4) = 0, so y (x) = 2 x3/2 − 3 3 16 3. 1 1 + C and y (1) = 5, so y (x) = − + 6. x x √ √ C05S02.039: y (x) = 2 x + 2 + C and y (2) = −1, so y (x) = 2 x + 2 − 5. C05S02.038: y (x) = − C05S02.040: y = √ x + 9 dx = 2 (x + 9)3/2 + C ; 0 = y (−4) = 2 (−4 + 9)3/2 + C = 3 3 y (x) = 2 (x + 9)3/2 − 3 10 3 √ 2 3 √ · 5 5 + C; 5. C05S02.041: y (x) = 3 x4 − 2x1 + C ; y (1) = 1, so y (x) = 3 x4 − 2x−1 + 9 . 4 4 4 C05S02.042: y (x) = 15 32 3 1 3 3 9 x − x − 2 + C = x5 − x2 − 2 + . 5 2 2x 5 2 2x 5 C05S02.043: y (x) = 1 (x − 1)4 + C ; y (0) = 2 = 4 1 4 + C , so C = 7 . 4 C05S02.044: y (x) = 2 (x + 5)3/2 + C and y (4) = −3, so y (x) = 2 (x + 5)3/2 − 21. 3 3 C05S02.045: y (x) = 3e2x + C ; y (0) = 10, so C = 7. C05S02.046: y (x) = C + 3 ln x; y (1) = 7, so C = 7. C05S02.047: v (t) = 6t2 − 4t + C ; v (0) = −10, so v (t) = 6t2 − 4t − 10. Next, x(t) = 2t3 − 2t2 − 10t + K ; x(0) = 0, so K = 0. C05S02.048: v (t) = 10t − 15t2 + C ; v (0) = −5, so v (t) = 10t − 15t2 − 5. Next, x(t) = 5t2 − 5t3 − 5t + K ; x(0) = 5, so K = 5. C05S02.049: v (t) = 2 t3 + C ; v (0) = 3, so v (t) = 2 t3 + 3. Next, x(t) = 1 t4 + 3t + K ; x(0) = −7, so 3 3 6 K = −7. C05S02.050: v (t) = 10t3/2 + C ; v (0) = 7, so v (t) = 10t3/2 + 7. Next, x(t) = 4t5/2 + 7t + K ; x(0) = 5, so K = 5. 3 C05S02.051: v (t) = C − cos t; v (0) = 0, so v (t) = 1 − cos t. Next, x(t) = t − sin t + K ; x(0) = 0, so K = 0. C05S02.052: v (t) = 4 sin 2t + C ; v (0) = 4, so v (t) = 4 + 4 sin 2t. x(0) = −2, so K = 0. Next, x(t) = 4t − 2 cos 2t + K ; C05S02.053: Note that v (t) = 5 for 0 t 5 and that v (t) = 10 − t for 5 t 10. Hence x(t) = 5t + C1 for 0 t 5 and x(t) = 10t − 1 t2 + C2 for 5 t 10. Also C1 = 0 because x(0) = 0 and continuity of x(t) 2 requires that 5t + C1 and 10t − 1 t2 + C2 agree when t = 5. This implies that C2 = − 25 . The graph of x is 2 2 next. 40 35 30 25 20 15 10 5 2 4 6 8 10 C05S02.054: First note that v (t) = t for 0 t 5 and that v (t) = 5 for 5 t 10. Hence x(t) = 1 t2 + C1 2 for 0 t 5 and x(t) = 5t + C2 for 5 t 10. The condition x(0) = 0 implies that C1 = 0; continuity of x(t) implies that 1 t2 + C1 and 5t + C2 agree when t = 5, and this implies that C2 = − 25 . The graph of x(t) 2 2 is next. 40 35 30 25 20 15 10 5 2 4 6 8 10 C05S02.055: First, v (t) = t if 0 t 5 and v (t) = 10 − t if 5 t 10. Hence x(t) = 1 t2 + C1 if 0 t 5 2 and x(t) = 10t − 1 t2 + C2 if 5 t 10. Finally, C1 = 0 because x(0) = 0 and continuity of x(t) requires 2 that 1 t2 + C1 = 10t − 1 t2 + C2 when t = 5, so that C2 = −25. The graph of x(t) is next. 2 2 30 25 20 15 10 5 2 4 6 8 10 C05S02.056: The graph indicates that v (t) = 2t if 0 t 2.5, v (t) = 5 if 2.5 t 7.5, and v (t) = 20 − 2t if 7.5 t 10. Thus x(t) = t2 + C1 if 0 t 2.5, x(t) = 5t + C2 if 2.5 t 7.5, and x(t) = 20t − t2 + C3 4 if 7.5 t 10. Next, C1 = 0 because x(0) = 0, C2 = 6.25 because x(t) must be continuous when t = 2.5, and C3 = −62.5 because x(t) must be continuous at t = 7.5. The graph of x(t) is next. 40 35 30 25 20 15 10 5 2 4 6 8 10 In the solutions for Problems 57–78, unless otherwise indicated, we will take the upward direction to be the positive direction, s = s(t) for position (in feet) at time t (in seconds) with s = 0 corresponding to ground level, and v (t) velocity at time t in feet per second, a = a(t) acceleration at time t in feet per second per second. The initial position will be denoted by s0 and the initial velocity by v0 . C05S02.057: Here, a = −32, v (t) = −32t + 96, s(t) = −16t2 + 96t. The maximum height is reached when v = 0, thus when t = 3. The maximum height is therefore s(3) = 144. The ball remains aloft until s(t) = 0 for t > 0; t = 6. So it remains aloft for six seconds. C05S02.058: With initial velocity v0 , here we have a(t) = −32, v (t) = −32t + v0 , and s(t) = −16t2 + v0 t (because s0 = 0). The maximum altitude is attained when v = 0, which occurs when t = v0 /32. Therefore 400 = s(v0 /32) = (−16)(v0 /32)2 + (v0 )2 /32. It follows that 1 2 64 (v0 ) = 400, and therefore that v0 = 160 (ft/s). C05S02.059: Here it is more convenient to take the downward direction as the positive direction. Thus a(t) = +32, v (t) = +32t (because v0 = 0), and s(t) = 16t2 (because s0 = 0). The stone hits the bottom when t = 3, and at that time we have s = s(3) = 144. Answer: The well is 144 feet deep. C05S02.060: We have v (t) = −32t + v0 and s(t) = −16t2 + v0 t. Also 0 = s(4), so 4v0 = 256: v0 = 64. Thus v (t) = −32t + 64 and s(t) = −16t2 + 64t. The height of the tree is the maximum value of s(t), which occurs when v (t) = 0; that is, when t = 2. Therefore the height of the tree is s(2) = 64 (feet). C05S02.061: Here, v (t) = −32t + 48 and s(t) = −16t2 + 48t + 160. The ball strikes the ground at that value of t > 0 for which s(t) = 0: 0 = s(t) = −16(t − 5)(t + 2), 5 so t = 5. Therefore the ball remains aloft for 5 seconds. Its velocity at impact is v (5) = −112 (ft/s), so the ball strikes the ground at a speed of 112 ft/s. C05S02.062: First ball: v0 = 0, s0 = 576. So s(t) = −16t2 + 576. The first ball strikes the ground at that t > 0 for which s(t) = 0; t = 6. Second ball: The second ball must remain aloft from time t = 3 until time t = 6, thus for 3 seconds. Reset t = 0 as the time it is thrown downward. Then with initial velocity v0 , the second ball has velocity and position v (t) = −32t + v0 and s(t) = −16t2 + v0 t + 576 at time t. We require that s(3) = 0; that is, 0 = s(3) = −144 + 3v0 + 576, so that v0 = −144. Answer: The second ball should be thrown straight downward with an initial velocity of 144 ft/s. C05S02.063: One solution: Take s0 = 960, v0 = 0. Then v (t) = −32t and s(t) = −16t2 + 960. √ The ball hits the street for that value of t > 0 for which s(t) = 0—that is, when t = 2 √ 15. It therefore √ takes the ball approximately 7.746 seconds to reach the street. Its velocity then is v (2 15) = −64 15 (ft/s)—approximately 247.87 ft/s (downward), almost exactly 169 miles per hour. C05S02.064: Here we have v (t) = −32t + 320 and s(t) = −16t2 + 320t. After three seconds have elapsed the height of the arrow will be s(3) = 816 (ft). The height of the arrow will be 1200 feet when s(t) = 1200: 16t2 − 320t + 1200 = 0; 16(t − 5)(t − 15) = 0; t = 5 and t = 15 are both solutions. So the height of the arrow will be 1200 feet both when t = 5 (the arrow is still rising) and when t = 15 (the arrow is falling). The arrow strikes the ground at that value of t > 0 for which s(t) = 0: t = 20. So the arrow will strike the ground 20 seconds after it is released. C05S02.065: With v (t) = −32t + v0 and s(t) = −16t2 + v0 t, we have the maximum altitude s = 225 occurring when v (t) = 0; that is, when t = v0 /32. So 225 = s(v0 /32) = (−16)(v0 /32)2 + (v0 )2 /32 = (v0 )2 /64. It follows that v0 = +120. So the initial velocity of the ball was 120 ft/s. C05S02.066: Note that the units are meters and seconds. It is also more convenient to take the downward direction as the positive direction here. Thus v (t) = 9.8t and s(t) = 4.9t2 . √ √ The rock reaches the water when s(t) = 98: t = +2 √ So it takes the rock 2 5 seconds to reach the water. 5. Its velocity as it penetrates the water surface is v (2 5) ≈ 43.8 (m/s). C05S02.067: In this problem, v (t) = −32t + v0 = −32t and s(t) = −16t2 + s0 = −16t2 + 400. The ball reaches the ground when s = 0, thus when 16t2 = 400: t = +5. Therefore the impact velocity is v (5) = (−32)(5) = −160 (ft/s). C05S02.068: Here s0 will be the height of the building, so 6 v (t) = −32t − 25 and s(t) = −16t2 − 25t + s0 . The velocity at impact is −153 ft/s, so we can obtain the time of impact t by solving v (t) = −153: t = 4. At this time we also have s = 0: 0 = s(4) = (−16)(16) − (25)(4) + s0 , so that s0 = 356. Answer: The building is 356 feet high. C05S02.069: For this problem, we take s(t) = −16t2 + 160t and v (t) = −32t + 160. Because s = 0 when t = 0 and when t = 10, the time aloft is 10 seconds. The velocity is zero at maximum altitude, and that occurs when 32t = 160: t = 5. So the maximum altitude is s(5) = 400 (ft). C05S02.070: Let f (t) be the altitude of the sandbag at time t. Then f (t) = −16t2 + h, so the altitude of the sandbag will be h/2 when f (t) = h/2: t2 = h , 32 so t= 1√ 2h. 8 Let the ball have initial velocity v0 and altitude s(t) at time t. Then s(t) = −16t2 + v0 t. We require that s(t) = h/2 at the preceding value of t. That is, h = (−16) 2 √ Solution of this equation yields v0 = 4 2h. h 32 + (v0 ) 1√ 2h . 8 C05S02.071:√ Because v0 = −40, v (t) = −32t − 40. Thus s(t) = −16t2 − 40t + 555. Now s(t) = 0 when √ t = 1 −5 + 2 145 ≈ 4.77 (s). The speed at impact is | v (t) | for that value of t; that is, 16 145 ≈ 192.6655 4 (ft/s), over 131 miles per hour. C05S02.072: In this problem, v (t) = −gt and s(t) = − 1 gt2 + h. The rock strikes the ground when s(t) = 0, 2 √ so that t = 2h/g . The speed of the rock then is | − g 2h/g | = 2gh. C05S02.073: Bomb equations: a = −32, v = −32t, sB = s = −16t2 + 800. Here we have t = 0 at the time the bomb is released. Projectile equations: a = −32, v = −32(t − 2) + v0 , and sP = s = 16(t − 2)2 + v0 (t − 2), t 2. We require sB = sP = 400 at the same time. The equation sB = 400 leads to t = 5, and for sP (5) = 400, we must have v0 = 544/3 ≈ 181.33 (ft/s). C05S02.074: Let x(t) denote the distance the car has traveled t seconds after the brakes are applied; let v (t) denote its velocity and a(t) its acceleration at time t during the braking. Then we are given a = −40, so v (t) = −40t + 88 (because 60 mi/h is the same speed as 88 ft/s). The car comes to a stop when v (t) = 0; that is, when t = 2.2. The car travels the distance x(2.2) ≈ 96.8. Answer: 96.8 feet. C05S02.075: The deceleration a = k > 0 is unknown at first. But the velocity of the car is v (t) = −kt +88, and so the distance it travels after the brakes are applied at time t = 0 is 1 x(t) = − kt2 + 88t. 2 7 But x = 176 when v = 0, so the stopping time t1 is 88/k because that is the time at which v = 0. Therefore 1 88 176 = − k 2 k 2 + (88) 88 k = 3872 . k It follows that k = 22 (ft/s2 ), about 0.69g . C05S02.076: Let x(t) be the altitude (in miles) of the spacecraft at time t (hours), with t = 0 corresponding to the time at which the retrorockets are fired; let v (t) = x (t) be the velocity of the spacecraft at time t. Then v0 = −1000 and x0 is unknown. But the acceleration constant is a = +20000, so v (t) = (20000)t − 1000 and x(t) = (10000)t2 − 1000t + x0 . We want v = 0 exactly when x = 0—call the time then t1 . Then 0 = (20000)t1 − 1000, so t1 = 1/20. Also x(t1 ) = 0, so 0 = (10000) 1 400 − (1000) 1 20 + x0 . Therefore x0 = 50 − 25 = 25 miles. (Also t1 = 1/20 of an hour; that is, exactly three minutes.) C05S02.077: (a): With the usual coordinate system, the ball has velocity v (t) = −32t + v0 (ft/s) at time t (seconds) and altitude y (t) = −16t2 + v0 t (ft). We require y (T ) = 144 when v (T ) = 0: v0 = 32T , so 144 = −16T 2 + 32T 2 = 16T 2 , and thus T = 3 and v0 = 96. Answer: 96 ft/s. (b): Now v (t) = − 26 t + 96, y (t) = − 13 t2 + 96t. Maximum height occurs when v (t) = 0: t = 5 5 maximum height is y 240 13 =− 240 13 . The 240 13 2402 11520 + 96 · · = ≈ 886 ft. 5 132 13 13 Set up a coordinate system in which the Diana moves along the x-axis in the positive dv direction, with initial position x0 = 0 and initial velocity v0 = 0. Then = +0.032 feet per second per dt second, so C05S02.078: v (t) = (0.032)t + v0 = (0.032)t. Thus x(t) = (0.016)t2 + x0 = (0.016)t2 . The units are in feet, seconds, and feet per second. After one minute we take t = 60 to find that x(60) = 57.6 (feet). After one hour we take t = 3600 to find that x(3600) = 207360 (ft)—over 39 miles. After one day we take t = 86400; x(86400) = 119,439,360 (feet), approximately 22621 miles! At this point the speed of the Diana would be v (86400) = 2764.8 feet per second, approximately 1885 miles per hour! After 30 days, the Diana will have traveled well over 20 million miles and will be speeding along in excess of 56000 miles per hour. C05S02.079: Let a denote the deceleration constant of the car when braking. In the police experiment, we have the distance the car travels from x0 = 0 at time t to be 8 1 22 x(t) = − at2 + 25 · t 2 15 (the factor 22 converts 25 miles per hour to feet per second). When we solve simultaneously the equations 15 x(t) = 45 and x (t) = 0, we find that a = 1210 ≈ 14.938. When we use this value of a and substitute data 81 from the accident, we find the position function of the car to be x(t) = − 1 1210 2 · t + v0 t 2 81 where v0√ its initial velocity. Now when we solve simultaneously x(t) = 210 and x (t) = 0, we find that is v0 = 110 42 ≈ 79.209 feet per second, almost exactly 54 miles per hour. 9 9 Section 5.3 5 3i = 31 + 32 + 33 + 34 + 35 = 3 + 9 + 27 + 81 + 243. C05S03.001: i=1 6 √ C05S03.002: 2i = √ 2+ √ 4+ √ 6+ √ 8+ √ 10 + √ 12. i=1 5 1 11111 =++++. j+1 23456 C05S03.003: j =1 6 C05S03.004: j =1 (2j − 1) = 1 + 3 + 5 + 7 + 9 + 11. 6 C05S03.005: k=1 6 C05S03.006: k=1 1 11 1 1 1 =1+ + + + +. 2 k 4 9 16 25 36 (−1)k+1 111 1 1 1 =−+− + −. j2 1 4 9 16 25 36 5 xn = x + x2 + x3 + x4 + x5 . C05S03.007: n=1 5 C05S03.008: n=1 (−1)n+1 x2n−1 = x − x3 + x5 − x7 + x9 . 5 C05S03.009: 1 + 4 + 9 + 16 + 25 = n2 . n=1 6 C05S03.010: 1 − 2 + 3 − 4 + 5 − 6 = C05S03.011: 1 + C05S03.012: 1 + 1111 +++= 2345 n=1 5 k=1 11 1 1 ++ + = 4 9 16 25 n · (−1)n+1 . 1 . k 5 i=1 1 . i2 6 C05S03.013: 111 1 1 1 1 +++ + + = . 2 4 8 16 32 64 m=1 2m 5 11 (−1)n+1 1 1 1 C05S03.014: −+ − + = . 3 9 27 81 243 n=1 3n 1 5 C05S03.015: 24 8 16 32 ++ + + = 3 9 27 81 243 n=1 C05S03.016: 1 + √ 2+ √ 3+2+ √ 5+ √ 6+ n 2 3 √ . √ 7+2 2+3= 9 j. j =1 10 C05S03.017: x + x2 x3 x10 1n + + ··· + = x. 2 3 10 n n=1 C05S03.018: x − x3 (−1)n+1 2n−1 x5 x7 x19 . + − + ··· − = x 3 5 7 19 2n − 1 n=1 10 C05S03.019: Using Eqs. (3), (4), (6), and (7), we find that 10 i=1 10 (4i − 3) = 4 · 10 i=1 i−3· i=1 10 · 11 − 3 · 10 = 190. 2 1=4· C05S03.020: We use Eqs. (3), (4), (6), and (7) to obtain 8 8 j =1 (5 − 2j ) = 5 · 8 j =1 1−2· j =1 j =5·8−2· 8·9 = −32. 2 C05S03.021: We use Eqs. (3), (4), (6), and (8) to obtain 10 i=1 10 (3i2 + 1) = 3 · 10 i2 + i=1 i=1 1=3· 10 · 11 · 21 + 10 = 1165. 6 C05S03.022: We use Eqs. (3), (4), (7), and (8) and find thereby that 6 k=1 6 (2k − 3k 2 ) = 2 · k=1 6 k−3· k=1 k2 = 2 · 6·7 6 · 7 · 13 −3· = −231. 2 6 C05S03.023: First expand, then use Eqs. (3), (4), (6), (7), and (8): 8 r =1 8 (r − 1)(r + 2) = r =1 8 (r2 + r − 2) = 8 8 r2 + r =1 r =1 r−2· 1= r =1 8 · 9 · 17 8 · 9 + − 2 · 8 = 224. 6 2 C05S03.024: We use Eqs. (3), (4), (6), (7), and (9) to find that 5 5 i=1 (i3 − 3i + 2) = 6 C05S03.025: i=1 i3 − i2 = i=1 5 i3 − 3 · i=1 5 i+2· 1= i=1 25 · 36 5·6 −3· + 2 · 5 = 190. 4 2 62 · 72 6 · 7 · 13 − = 32 · 72 − 7 · 13 = 441 − 91 = 350. 4 6 2 10 C05S03.026: k=1 (4k 2 − 4k + 1) = 4 · 100 i2 = 100 · 101 · 201 = 50 · 101 · 67 = 338350. 6 i3 = 108 106 104 + + = 25502500. 4 2 4 C05S03.027: i=1 100 C05S03.028: i=1 C05S03.029: C05S03.030: 10 · 11 · 21 10 · 11 −4· + 10 = 1330. 6 2 lim 12 + 22 + · · · + n2 n(n + 1)(2n + 1) = lim = lim 3 n→∞ n→∞ n 6n 3 lim 13 + 23 + · · · + n3 n2 (n + 1)2 = lim = lim n→∞ n→∞ n4 4n4 n→∞ n→∞ n C05S03.031: i=1 (2i − 1) = 2 · n C05S03.032: 2 i=1 5 C05S03.033: A 5 = i=1 5 C05S03.034: A 5 = i=1 6 C05S03.036: A 6 = i=1 5 C05S03.037: A 5 = i=1 5 C05S03.038: A 5 = i=1 i−1 2 i=1 i=1 1 . 4 = · 2i + 3 5 5 · i=1 2 3i 5 9− i +1 4 · 2 6 3 33 = , 6 2 2· i 3 39 +3 · = . 2 6 2 13 − 3 · i−1 3 111 ·= . 2 6 4 A6 = i=1 6 A6 = i=1 5 2 i 5 A5 = 2 178 = , 5 25 9− 2i + 5 2 22 ·= . 5 5 5 A5 = 1 6 = , 5 25 2 i1 3 ·=. 55 5 i=1 +3 · 2 i−1 5 8 C05S03.040: A 8 = 5 i3 93 13 − 3 · ·= , 26 4 5 C05S03.039: A 5 = 1 . 3 n(n + 1)(2n + 1) n(n + 1) n(2n − 1)(2n + 1) −4· +n = . 6 2 3 A5 = 2i + 3 2 18 ·= , 5 5 5 2 i=1 i=1 (4i2 − 4i + 1) = 4 · i−1 1 2 ·=, 5 5 5 6 C05S03.035: A 6 = 1 1 1 + + 4 2n 4n2 = n(n + 1) − n = n2 . 2 n (2i − 1) = 1 1 1 + +2 3 2n 6n i=1 5 · 1 11 = . 5 25 2i + 5 5 A5 = i=1 5 3 378 = , 5 25 9− A5 = i=1 2 133 ·= , 8 16 3 2 · 3i − 3 5 8 A8 = i=1 2 258 = . 5 25 9− 2 · i−1 +1 4 3 513 = . 5 25 2 · 2 165 = . 8 16 10 i−1 10 C05S03.041: A 10 = i=1 10 3 · 10 1 81 = , 10 400 i=1 10 i−1 1 · ≈ 0.610509, 10 10 C05S03.042: A 10 = i=1 i 10 A 10 = 3 · 1 121 = . 10 400 i 1 · ≈ 0.710509. 10 10 A 10 = i=1 C05S03.043: When we add the two given equations, we obtain n 2· i=1 i = (n + 1) + (n + 1) + · · · + (n + 1) (n terms) n = n(n + 1), and therefore i= i=1 n(n + 1) . 2 C05S03.044: Following the directions in the problem, we get 23 − 13 = 3 · 12 + 3 · 1 + 1, 33 − 23 = 3 · 22 + 3 · 2 + 1, 43 − 33 = 3 · 32 + 3 · 3 + 1, 53 − 43 = 3 · 42 + 3 · 4 + 1, . . . n3 − (n − 1)3 = 3(n − 1)2 + 3(n − 1) + 1, (n + 1)3 − n3 = 3n2 + 3n + 1. When we add these equations, we get n (n + 1)3 − 1 = 3 · k=1 k2 + 3 · n(n + 1) + n, 2 so that n 3· k=1 3 3 3 1 k 2 = n3 + 3n2 + 3n − n2 − n − n = n3 + n2 + n. 2 2 2 2 Therefore n k2 = k=1 n C05S03.045: i=1 n C05S03.046: i=1 2n3 + 3n2 + n n(n + 1)(2n + 1) = . 6 6 i n(n + 1) 1 = → as n → ∞. 2 2 n 2n 2 2i n2 2 · 2 8 8n(n + 1)(2n + 1) → as n → ∞. = n 6n3 3 4 n C05S03.047: i=1 n C05S03.048: i=1 3i n i=1 5− n C05S03.050: i=1 · 3 81 81n2 (n + 1)2 → = as n → ∞. n 4n4 4 2i 2 +2 · n n n C05S03.049: 3 9− = 3i 1 · n n 3i n n = 5n · 2 · n f (xi ) ∆x = C05S03.051: i=1 2 4n(n + 1) + 2n · → 6 as n → ∞. 2 2n n i=1 7 1 3n(n + 1) → as n → ∞. − n 2n 2 2 3 3 27n(n + 1)(2n + 1) = 9n · − → 27 − 9 = 18 as n → ∞. n n 6n3 h bi b bh n(n + 1) 1 · · = 2· → bh as n → ∞. bnn n 2 2 C05S03.052: Let y be the length of the top half of the side of the polygon shown in Fig. 5.3.20 and let x be the distance from the center of the circle to the midpoint of that side. Then y π = sin r n x π = cos . r n and Hence the area of the large triangle in the figure is A= 1 π π (x)(2y ) = xy = r2 sin cos , 2 n n and therefore the total area of the regular n-sided polygon consisting of all n such triangles is π π cos . n n An = nr2 sin The length of the side of the polygon shown in the figure is 2y = 2r sin π , n and so the perimeter of the n-sided polygon is Cn = 2ny = 2nr sin π . n C05S03.053: Using the formulas derived in the solution of Problem 52, we have lim n→∞ An r π r = lim cos = . n→∞ 2 Cn n 2 Thus A = 1 rC . Hence if A = π r2 , then C = 2π r. 2 5 Section 5.4 n C05S04.001: lim n→∞ i=1 3 (2xi − 1) ∆x = 1 (2x − 1) dx. n C05S04.002: lim n→∞ i=1 n C05S04.003: lim n→∞ 2 (2 − 3xi−1 ) ∆x = −3 10 (x2 + 4) ∆x = i (2 − 3x) dx. (x2 + 4) dx. 0 i=1 n C05S04.004: lim n→∞ i=1 n C05S04.005: lim n→∞ √ 9 mi ∆ x = √ lim n→∞ i=1 lim n→∞ i=1 √ lim n→∞ i=1 1 ∆x = 1 + mi 8 √ 3 π /2 (cos 2xi−1 ) ∆x = lim n→∞ 1 /2 (sin 2π mi ) ∆x = lim n→∞ 5 i=1 i=1 n 5 f (xi ) ∆x = i=1 i=1 n 5 f (xi ) ∆x = C05S04.013: i=1 i=1 n 5 f (xi ) ∆x = C05S04.014: i=1 i=1 n i 5 2 i 5 3 · · √ i=1 1 5 = 15 11 · · 6 · 11 = . 125 6 25 1 5 = 11 9 · · 25 · 36 = . 625 4 25 i ≈ 8.382332347442. 2 1+ f (xi ) ∆x = i=1 e2x dx. 1 29 = . 1+i 20 6 C05S04.015: sin 2π x dx. 0 i=1 f (xi ) ∆x = C05S04.012: 1 exp(2xi ) ∆x = n C05S04.011: cos 2x dx. 0 i=1 n C05S04.010: 1 dx. 1+x 0 n C05S04.009: 25 − x2 dx. 0 n C05S04.008: (x3 − 3x2 + 1) dx. x dx. 5 25 − x2 ∆x = i n C05S04.007: 0 4 i=1 n C05S04.006: 3 (x3 − 3x2 + 1) ∆x = i i i 2 +1 · 1 1 39 = . 2 2 n 6 i=1 i 2 1+ 3i 5 i=1 n 5 f (xi ) ∆x = C05S04.017: 2 1+ f (xi ) ∆x = C05S04.016: i=1 i=1 n +2 1+ 3 f (xi ) ∆x = i=1 i=1 n 6 f (xi ) ∆x = C05S04.019: i 5 1+2 2+ i=1 cos i=1 1 /2 · · 331 1 = . 2 8 3i 5 −3 1+ 5 C05S04.018: i 2 · 294 3 = . 5 5 1 ≈ 4.220102178480. 5 iπ π π · =− . 6 6 6 C05S04.020: With a = 1, b = 6, n = 5, ∆x = 1, f (x) = ln x, xi = 1 + i · ∆x, and xi = xi , we obtain n 5 f (xi ) ∆x = i=1 i=1 n 5 f (xi ) ∆x = C05S04.021: i=1 i=1 n 5 f (xi ) ∆x = C05S04.022: i=1 i=1 n 5 f (xi ) ∆x = C05S04.023: i=1 i=1 n 5 f (xi ) ∆x = C05S04.024: i=1 i=1 n i=1 3 √ 4 . 25 i−1 2 +1 · 5 1+2 2+ i=1 n 6 f (xi ) ∆x = cos i=1 1 33 = . 2 2 2 3(i − 1) 5 i=1 i=1 i=1 = 1+ f (xi ) ∆x = C05S04.029: 1 5 · i−1 2 5 n 6 . 25 1+ f (xi ) ∆x = i=1 = i − 1 ≈ 6.146264369942. i=1 n 1 5 1 137 = . 1 + (i − 1) 60 6 i=1 C05S04.028: i−1 5 · 2 1+ f (xi ) ∆x = C05S04.027: 2 i=1 n C05S04.026: i−1 5 6 f (xi ) ∆x = C05S04.025: ln(1 + i) = ln 2 + ln 3 + ln 4 + ln 5 + ln 6 ≈ 6.579251212. +2 1+ i−1 2 3 −3 1+ i−1 5 1 /2 · · 1 247 = . 2 8 3(i − 1) 5 · 3 132 = . 5 5 1 ≈ 4.092967280401. 5 (i − 1)π π π ·=. 6 6 6 C05S04.030: With a = 1, b = 6, n = 5, ∆x = 1, f (x) = ln x, xi = 1 + i · ∆x, and xi = xi−1 , we obtain 2 n 5 f (xi ) ∆x = i=1 i=1 n 5 f (xi ) ∆x = C05S04.031: i=1 i=1 n 5 f (xi ) ∆x = C05S04.032: i=1 i=1 n 5 f (xi ) ∆x = C05S04.033: i=1 i=1 n 5 f (xi ) ∆x = C05S04.034: i=1 i=1 n 2i − 1 10 2 2i − 1 10 3 2i − 1 2 2 1+ i=1 i=1 n 6 i=1 ≈ 7.505139519609. 2i − 1 4 +1 · 1 = 18. 2 3 1+2 2+ 6 f (xi ) ∆x = i=1 49 . 200 6i − 3 10 i=1 n C05S04.039: = 1 /2 5 f (xi ) ∆x = i=1 1 5 · 1+ i=1 n C05S04.038: 33 . 100 2 5 f (xi ) ∆x = i=1 = 2i − 1 4 i=1 n C05S04.037: 1 5 1+ f (xi ) ∆x = C05S04.036: · 2 6086 = . 1 + 2i 3465 6 f (xi ) ∆x = C05S04.035: ln(i) = 0 + ln 2 + ln 3 + ln 4 + ln 5 ≈ 4.787491743. cos i=1 +2 1+ 2i − 1 4 · 1 575 = . 2 16 −3 1+ 6i − 3 10 · 3 1623 = . 5 40 2i − 1 10 1 /2 · 1 ≈ 4.157183161049. 5 (2i − 1)π π · = 0. 12 6 C05S04.040: With a = 1, b = 6, n = 5, ∆x = 1, f (x) = ln x, mi = 1 + i − obtain n 5 f (xi ) ∆x = i=1 ln i + i=1 n 5 f (xi ) ∆x = C05S04.041: i=1 i=1 n 5 f (xi ) ∆x = C05S04.042: i=1 n C05S04.043: i=1 i=1 2i n 2 · 1 2 = ln 1 2 · ∆x, and xi = mi , we 3 5 7 9 11 + ln + ln + ln + ln ≈ 5.783344297. 2 2 2 2 2 5 259775 = ≈ 1.837527940469. 5i + 2 141372 3i − 1 ≈ 7.815585306501. 3 2 8 8n(n + 1)(2n + 1) → = as n → ∞. n 6n3 3 3 n 4i n C05S04.044: i=1 n 2· C05S04.045: i=1 3 · 4 256n2 (n + 1)2 → 64 as n → ∞. = n 4n4 3i 3 18n(n + 1) 3 9 +1 · = + n · = 12 + → 12 as n → ∞. n n 2n2 n n n C05S04.046: i=1 4−3 1+ n C05S04.047: i=1 n i=1 +1 · 3 4i n C05S04.048: − 3 3 81n(n + 1)(2n + 1) + n · → 27 + 3 = 30 as n → ∞. = n 6n 3 n 4i 4 16n(n + 1) 256n2 (n + 1)2 − → 64 − 8 = 56 as n → ∞. ·= nn 4n 4 2n2 C05S04.049: Choose xi = xi = b 4 16 12 48n(n + 1) = ·n− ·n− → 4 − 24 = −20 as n → ∞. n n n 2n2 · 2 3i n 3· 4i n bi b and ∆x = . Then n n n bi n x2 dx = lim n→∞ 0 C05S04.050: Choose xi = xi = b i=1 · n(n + 1)(2n + 1) 3 b 1 b = b3 . = lim n→∞ n 6n 3 3 bi b and ∆x = . Then n n n n→∞ 3 bi n x3 dx = lim 0 2 i=1 · n2 (n + 1)2 4 b 1 · b = b4 . = lim 4 n→∞ n 4n 4 C05S04.051: If xi = 1 (xi−1 + xi ) for each i, then xi is the midpoint of each subinterval of the partition, 2 and hence {xi } is a selection for the partition. Moreover, ∆xi = xi − xi−1 for each i. So n n xi ∆xi = i=1 1 2 (xi i=1 + xi−1 )(xi − xi−1 ) n = x2 − x2−1 i i 1 2 i=1 = Therefore a x2 − x2 + x2 − x2 + x2 − x2 + · · · + x2 − x2 −1 1 0 2 1 3 2 n n = b 1 2 1 2 x2 − x2 = 1 (b2 − a2 ). n 0 2 m x dx = lim n→∞ i=1 xi ∆xi = 1 b2 − 1 a2 . 2 2 C05S04.052: Let P = {x0 , x1 , x2 , . . . , xn } be a partition of [ a, b ] and let {xi } be a selection for P . Then b a n kf (x) dx = lim n→∞ kf (xi ) ∆xi i=1 n =k lim n→∞ f (xi ) ∆xi i=1 4 b =k a f (x) dx. C05S04.053: Suppose that a < b. Let P = {x0 , x1 , x2 , . . . , xn } be a partition of [ a, b ] and let {xi } be a selection for P . Then b n c dx = lim n→∞ a c ∆xi i=1 = lim c · (x1 − x0 + x2 − x1 + x3 − x2 + · · · + xn − xn−1 ) n→∞ = lim c(xn − x0 ) = c(b − a). n→∞ The proof is similar in the case b < a. C05S04.054: Given any partition P = {x0 , x1 , x2 , . . . , xn } of [0, 1] and any positive integer M , there is a selection {xi } for P in which x1 = 1/M . The corresponding Riemann sum then satisfies the inequality n f (xi ) ∆xi i=1 f 1 M · (x1 − x0 ) = M (x1 − x0 ), which can be made artibrarily large by choosing M sufficiently large. Hence some Riemann sums remain arbitrariliy large as n → + ∞ and |P | → 0. Therefore the limit of Riemann sums for this function on this interval does not exist. There is no contradiction to Theorem 1 because f is not continuous on [0, 1]. C05S04.055: Whatever partition P is given, a selection {xi } with all xi irrational can be made because irrational numbers can be found in any interval of the form [ xi−1 , xi ] with xi−1 < xi . (An explanation of why this is possible is given after the solution of Problem C02S04.069 of this manual.) For such a selection, we have n n f (xi ) ∆xi = i=1 i=1 1 · ∆xi = 1 regardless of the choice of P or n. Similarly, by choosing xi rational for all i, we get n n f (xi ) ∆xi = i=1 i=1 0 · ∆xi = 0 regardless of the choice of P or n. Therefore the limit of Riemann sums for f on [0, 1] does not exist, and therefore 1 f (x) dx does not exist. 0 C05S04.056: With ∆x = 5/n and xi = i ∆x = 5i/n, we have 5 0 ex dx = lim exp(xi ) · n→∞ 5 n = lim 5 5/n e + e10/n + e15/n + · + e5n/n n = lim 5e5/n 1 + e5/n + e10/n + · · · + e5(n−1)/n n n→∞ n→∞ 5 5r (1 + r + r2 + · + rn−1 ) n = lim n→∞ where r = e5/n . By Eq. (17), 5 5r r n − 1 5e5/n e5 − 1 5(e5 − 1) · = lim · 5/n = lim . n→∞ n→∞ n(1 − e−5/n ) n r−1 n e −1 ex dx = lim n→∞ 0 As n → + ∞, the denominator in the last fraction has the indeterminate form 0 · ∞. By l’Hˆpital’s rule, o 5 − 2 e−5/n 1 − e−5/n n = lim = lim 5e−5/n = 5. = lim 1 1 n→∞ n→∞ n→∞ −2 n n lim n · 1 − e−5/n n→∞ Therefore 5 ex dx = 0 3i 3 and ∆x = . Then n n C05S04.057: Choose xi = xi = n n exp (xi ) ∆x = i=1 i=1 5 · (e5 − 1) = e5 − 1. 5 e−3i/n · 3 3 −3/n e + e−6/n + e−9/n + · · · + e−3n/n = n n = 3e−3/n 1 + e−3/n + e−6/n + · · · + e−3(n−1)/n n = 3r 1 + r + r2 + · · · + rn−1 n = 3r rn − 1 3(e−3 − 1) 3(1 − e−3 ) 3e−3/n e−3 − 1 = = . · = · −3/n n r−1 n e −1 n(1 − e3/n ) n(e3/n − 1) (where r = e−3/n ) Now by l’Hˆpital’s rule, o lim n · (e 3/n n→∞ 3 − 2 e3/n e3/n − 1 n − 1) = lim = lim 3e3/n = 3. = lim 1 1 n→∞ n→∞ n→∞ −2 n n Therefore 3 n e−x dx = lim n→∞ 0 exp (xi ) ∆x = lim C05S04.058: Let xi = xi = 2 + n n→∞ i=1 3i 3 and let ∆x = . Then n n n exp (xi ) ∆x = i=1 exp 2 + i=1 = 3(1 − e−3 ) 3(1 − e−3 ) = = 1 − e−3 . 3/n − 1) 3 n(e 3i n · 3 3 2+(3/n) = e + e2+(6/n) + e2+(9/n) + · · · + e2+(3n/n) n n 3 2 3/n + e2 · e6/n + e2 · e9/n + · · · + e2 · e3n/n e ·e n 6 = 3e2 e3/n 1 + e3/n + e6/n + · · · + e3(n−1)/n n = By l’Hˆpital’s rule, o 3 − 2 e−3/n 1 − e−3/n n = lim = lim = lim 3e−3/n = 3. 1 1 n→∞ n→∞ n→∞ −2 n n lim n · 1 − e −3/n n→∞ 3e2 e3/n e3 − 1 3e2 (e3 − 1) · 3/n = . n e −1 n(1 − e−3/n ) Therefore 5 n ex dx = lim n→∞ 2 n→∞ i=1 π kπ and ∆x = . Then n n C05S04.059: Let xi = xi = π 3e2 (e3 − 1) 3e2 (e3 − 1) = = e5 − e2 . 3 n(1 − e−3/n ) exp (xi ) ∆x = lim n sin x dx = lim n→∞ 0 sin (xi ) ∆x = lim n→∞ k=1 π n n sin k=1 kπ n = lim n→∞ π π cot , n 2n which is of the indeterminate form 0 · ∞. Hence (no need for l’Hˆpital’s rule) o π sin x dx = lim 2· n→∞ 0 C05S04.060: Let xi = xi = a + i · n π π π · cos 2n 2n = lim 2 cos π · 2n = 2 · 1 · 1 = 2. π π n→∞ 2n sin sin 2n 2n b−a b−a and ∆x = . Then n n n exp (xi ) ∆x = i=1 i=1 exp a + i · b−a n · b−a n = b−a · ea e(b−a)/n + ea e2(b−a)/n + ea e3(b−a)/n + · · · + ea en(b−a)/n n = b − a a (b−a)/n 1 + e(b−a)/n + e2(b−a)/n + · · · + e(n−1)(b−a)/n ·e e n = (b − a)ea e(b−a)/n eb−a − 1 b − a a (b−a)/n eb−a − 1 ·e e · (b−a)/n = n e −1 n e(b−a)/n − 1 = (b − a)ea eb−a − 1 n · 1 − e−(b−a)/n = (b − a) eb − ea . n · 1 − e−(b−a)/n But by l’Hˆpital’s rule, o lim n · 1 − e−(b−a)/n n→∞ = lim n→∞ = lim n→∞ 1 − e−(b−a)/n 1 n − b − a −(b−a)/n ·e n2 = lim (b − a)e−(b−a)/n = b − a. 1 n→∞ −2 n 7 Therefore b n e dx = lim exp (xi ) ∆x = lim x n→∞ a n→∞ i=1 C05S04.061: Let h = b − a, xi = xi = a + (b − a) eb − ea n · 1 − e−(b−a)/n (b − a) eb − ea b−a = eb − ea . ih h , and ∆x = . Then n n n sin (xi ) ∆x = i=1 = h n n sin a + i=1 ih n . According to Mathematica 3.0, n sin a + i=1 ih n = csc h h sin sin 2n 2 1 h 2a + h + 2 n = csc 1 b−a b+a+ 2 n b−a b−a sin sin 2n 2 . But then, b−a n n i=1 sin a + i · b−a n = b−a b−a b−a csc sin sin n 2n 2 2(b − a) b−a 2n = · sin b−a 2 sin 2n · sin 1 b−a b+a+ 2 n 1 b−a b+a+ 2 n → 2 · sin b−a 2 · sin b+a 2 as n → + ∞. But using one of the trigonometric identities that immediately precede Problems 59 through 62 in Section 7.4, we find that 2 · sin b−a 2 · sin b+a 2 b+a b−a − 2 2 = cos − cos b+a b−a + 2 2 = cos a − cos b. Therefore b a C05S04.062: Let xi = xi = a + i · b sin x dx = cos a − cos b. b−a b−a and let ∆x = . Then n n n cos x dx = lim n→∞ a cos (xi ) ∆x = lim n→∞ i=1 b−a n n i=1 cos a + i · b−a n . According to Mathematica 3.0, b−a n n i=1 cos a + i · b−a n = b−a a−b a−b b − a + n(b + a) csc sin cos n 2n 2 2n b−a 2n sin b − a cos b − a + n(b + a) → 2 · sin b − a = b−a 2 2n 2 sin 2n 2· 8 · cos b+a 2 as n → + ∞. Next, using one of the trigonometric identities that precede Problems 59 through 62 in Section 7.4, we find that 2 · sin b−a 2 · cos b+a 2 = sin b + sin(−a) = sin b − sin a. Therefore b a cos x dx = sin b − sin a. 9 Section 5.5 1 C05S05.001: √ √ 3x2 + 2 x + 3 3 x dx = x3 + 4 x3/2 + 9 x4/3 3 4 0 6 6 dx = − x2 x 3 C05S05.002: 1 1 C05S05.003: −2 1 C05S05.005: 0 14 4x 1 0 C05S05.007: (x4 − x3 ) dx = 3 (x + 1)3 dx = 1 4 C05S05.009: x4 + 1 dx = x2 √ x dx = 0 4 C05S05.010: 1 2 C05S05.011: 1 4 (x 4 + 1)4 = 0 √ 1 √ dx = 2 x x 0 1 1 0 −1 3 = 1 = 49 20 1 C05S05.013: −1 4 C05S05.014: 0 3 C05S05.015: 1 2 C05S05.016: 1 1 100 1 100 x x99 dx = 100 1 100 x 28 . 3 16 3. 0 = 1 −1 −1 −1 1 6 (x 2 (x2 + 1)3 dx = (2x + 1)3 dx = = 20 − (−4) = 24. 1 100 . = 0. − 1)6 3 1 = 4 0 = 192. 32 3. (x6 + 3x4 + 3x2 + 1) dx = 1 0 2 (7x5/2 − 5x3/2 ) dx = 2x7/2 − 2x5/2 1 C05S05.017: = 2.45. √ √ = 2 4 − 2 1 = 2. 4 x99 dx = (x − 1)5 dx = 49 60 . = 1. 4 (3x2 + 2x + 4) dx = x3 + x2 + 4x 0 = 0 1 = − 20 . 2 − 1 x4 4 13 1 x− 3 x 2 3 /2 3x 1 −1 C05S05.012: 55 12 . 7 . 24 = − 1 x4 4 15 5x 1 + 2 x5 + 1 x6 5 6 −2 15 5x −1 C05S05.008: −1 1 1 dx = − 3 4 x 3x (x4 − x3 ) dx = 2 C05S05.006: = = −2 + 6 = 4. 1 x3 (1 + x)2 dx = −1 0 3 0 C05S05.004: 1 0 3 17 35 7 x + 5 x +x +x 2 1 = (8x3 + 12x2 + 6x + 1) dx = 2x4 + 4x3 + 3x2 + x −1 1 1566 35 0 −1 ≈ 44.742857. = 0 − 0 = 0. 1 8 C05S05.019: x2/3 dx = 9 1+ √ x 8 3 5/3 5x 1 C05S05.020: 2 9 dx = 1 √ 4 3t dt = 2 3 /2 3t 0 √ 2 2 e3t dt = 0 du 1 =− 2 u u 2 1 dt = t 2 C05S05.025: 1 10 C05S05.026: 5 1 C05S05.027: 0 = 93 5. √ (1 + 2 x + x) dx = x + 4 x3/2 + 1 x2 3 2 √ 4 3 0 = 16 3 1 dx = x 3 = e+ −1 √ 2 = 0 1 10 = ln 10 − ln 5 = ln 5 1 0 cos 2x dx = 1 2 10 = ln 2. 5 (e2x − 2ex + 1) dx = π sin2 x cos x dx = 0 π C05S05.031: 0 2 C05S05.032: 0 π /2 C05S05.033: 0 1 3 1 sin π t π cos 3x dx = 1 3 = 0. 0 π /4 2 1 2 (sin x) sin3 x π sin 5x dx = − 1 cos x 5 cos π t dt = π /2 sin 2x sin x cos x dx = 0 C05S05.030: = 0. = ln 2 − 0 = ln 2 ≈ 0.6931471805599453094. ln x 0 π /4 248 3. 2 ln t e − 4e + 5 1 5 = e2 − 2e + ≈ 0.7579643925. 2 2 2 C05S05.029: 1 +e e = 11 1 =− + = . 32 6 2 (ex − 1)2 dx = π /2 − 1 23 2 2 e − = (e3 − 1) ≈ 12.723691282. 3 3 3 2 C05S05.028: 1 e 9 3. 2 3t/2 e 3 e3t/2 dt = 0 3 C05S05.024: 1 ex − e−x dx = ex + e−x −1 C05S05.023: 4 . 9 = 1 1 C05S05.022: = 1 1 C05S05.021: 3 10 5 dx = − (2x + 3)2 2x + 3 3 C05S05.018: 0 π 0 0 = 1. 4 = 0. = 2. 5 2 = 0. 0 sin 3x π /2 0 = −1. 3 2 1 2x e − 2ex + x 2 1 = 0 12 3 e − 2e + 1 + 2 2 5 C05S05.034: sin 0 2 C05S05.035: cos 0 π /8 C05S05.036: 5 πx 10 πx dx = − cos 10 π 10 πx dx = 4 4 πx sin π 4 sec2 2t dt = 1 2 0 tan 2t = 0 2 = 0 π /8 0 10 . π 4 . π = 1. 2 C05S05.037: Choose xi = i/n, ∆x = 1/n, x0 = 0, and xn = 1. Then the limit in question is the limit of a Riemann sum for the function f (x) = 2x − 1 on the interval 0 x 1, and its value is therefore 1 0 (2x − 1) dx = x2 − x 1 0 = 1 − 1 = 0. C05S05.038: Choose xi = i/n, ∆x = 1/n, x0 = 0, and xn = 1. Then the limit in question is the limit of a Riemann sum for the function f (x) = x2 on the interval 0 x 1, and therefore n lim n→∞ i=1 i2 = n3 1 x2 dx = 0 1 . 3 C05S05.039: Choose xi = i/n, ∆x = 1/n, x0 = 0, and xn = 1. Then the limit in question is the limit of a Riemann sum for the function f (x) = x on the interval 0 x 1, and therefore lim n→∞ 1 + 2 + 3 + ··· + n = lim n→∞ n2 n i=1 i = n2 1 x dx = 0 1 . 2 C05S05.040: Choose xi = i/n, ∆x = 1/n, x0 = 0, and xn = 1. Then the limit in question is the limit of a Riemann sum for the function f (x) = x3 on the interval 0 x 1, and therefore lim n→∞ 13 + 23 + 33 + · · · + n3 = lim n→∞ n4 n i=1 i3 = n4 1 x3 dx = 0 1 . 4 C05S05.041: Choose xi = i/n, ∆x = 1/n, x0 = 0, and xn = 1. Then the given limit is the limit of a √ Riemann sum for the function f (x) = x on the interval 0 x 1, and therefore lim n→∞ √ 1+ √ 2+ √ √ 3 + ··· + n √ = lim n→∞ nn n i=1 √ i √= nn 1 0 √ x dx = 2 . 3 C05S05.042: Choose xi = i/n, ∆x = 1/n, x0 = 0, and xn = 1. Then the given limit is the limit of a Riemann sum for f (x) = sin π x on the interval 0 x 1, and therefore n lim n→∞ i=1 1 πi sin = n n 3 1 0 sin π x dx = 2 . π C05S05.043: The graph is shown next. The region represented by the integral consists of two triangles above the x-axis, one with base 3 and height 3, the other with base 1 and height 1. So the value of the integral is the total area 9 + 1 = 5. 2 2 3 2.5 2 1.5 1 0.5 -2 -1 1 2 C05S05.044: The graph is next. The region represented by the integral consists of two triangles above the x-axis, one with base 11 and height 11, the other with base 7 and height 7, for a total area of 121 + 49 = 85 . 3 3 6 6 3 10 8 6 4 2 -3 -2 -1 1 2 3 C05S05.045: The graph is next. The region represented by the integral consists of two triangles, one above the x-axis with base 2 and height 2, the other below the x-axis with base 3 and height 3, so the total value of the integral is 2 − 9 = − 5 . 2 2 2 1 1 2 3 4 5 -1 -2 -3 C05S05.046: The graph is next. The region represented by the integral consists of two triangles, one above the x-axis with base 5 and height 5, the other below the x-axis with base 7 and height 7, so the total value 2 2 4 of the integral is 25 4 − 49 4 = −6. 4 2 1 2 3 4 5 6 -2 -4 -6 C05S05.047: If y = √ 25 − x2 for 0 5, then x x2 + y 2 = 25, 0 5, x and 0 5. y Therefore the region represented by the integral consists of the quarter of the circle x2 + y 2 = 25 that lies in the first quadrant. This circle has radius 5, and therefore the value of the integral is 25 π . The region is 4 shown next. 5 4 3 2 1 1 C05S05.048: If y = √ 6x − x2 for 0 x2 − 6x + y 2 = 0; x that is, 2 3 4 5 6, then (x − 3)2 + y 2 = 9 where 0 x 6, 0 y 3. So the region represented by the integral consists of the half of the circle (x − 3)2 + y 2 = 9 (radius 3, center (3, 0)) that lies on and above the x-axis. The circle has radius 3, and therefore the value of the integral is 9 2 π . The semicircle is shown next. 3 2.5 2 1.5 1 0.5 1 2 3 5 4 5 6 C05S05.049: 0 x if 0 x2 x 1. Hence 1 1 if 0 x 8 if 1 x3 x √ 1 + x2 2. Hence 1 + x √ x 1 + x3 1+x x and x √ x if 0 1. So 1 + x2 x 1+ x5 if x 1. So 1 + x2 1 + x5 if 2 5. Therefore x 1 1 + x2 2 1 for all t. Therefore sin √ 2 x dx 0 C05S05.054: If 0 x 1 4 π, then 2 2 cos2 x 1 for such x, and therefore x 1 4 π. cos x 1. 1 + cos2 x 3 2 1 2 if 0 1 dx = 2. 0 √ 1 2 x for such x. Therefore 5. The inequality in Problem 52 now follows from the comparison property. C05S05.053: sin t Thus √ 1 1 + x2 1 1 + x5 x 3 1. The inequality in Problem 51 now follows from the comparison property for definite integrals. C05S05.052: x2 if 2 9 for such x. Therefore 1 + x3 1 √ 1+ x x 1+x 2. The inequality in Problem 50 now follows from the comparison property. C05S05.051: x2 if 0 1 + x for such x. Therefore 1. Hence, by the comparison property, the inequality in Problem 49 follows. C05S05.050: x if 1 1 + x2 1 1 + cos2 x 2 if 0 x 2 3 So 1π · 24 π /4 0 1 dx 1 + cos2 x 2π ·, 34 and the inequality in Problem 54 follows immediately. C05S05.055: If 0 x 1, then 1 1 2 1+x 1 1+x 1 · (1 − 0) 2 2; 1; 1 0 1 dx 1+x 6 1 · (1 − 0). 1 4 π. Hence So the value of the integral lies between 0.5 and 1.0. C05S05.056: If 4 x 9, then 2 3 √ 1+ 1 4 3; x √ 4; x 1 √ 1+ x 1 ; 3 1 · (9 − 4) 4 9 4 1 √ dx 1+ x Hence the value of the given integral lies between C05S05.057: If 0 x 1 6 π, 5 4 1 · (9 − 4). 3 and 5 . 3 then √ 3 2 3 4 cos x 1; 1; cos2 x 3 π · −0 4 6 π /6 cos2 x dx 1· cos2 x dx π −0 . 6 π . 6 0 Therefore π /6 π 8 C05S05.058: If 0 x 1 4 π, 0 then √ 2 ; 2 0 sin x 0 sin2 x 0 2 sin2 x 16 1 ; 2 1; 16 + 2 sin2 x 4 π 4· 4 17; √ 16 + 2 sin2 x π /4 17; 16 + 2 sin x dx 2 0 √ π 17 . 4 Therefore π /4 3.14159 16 + 2 sin2 x dx 3.2384. 0 C05S05.059: Suppose that f is integrable on [ a, b ] and that c is a constant. Then 7 n b n cf (x) dx = lim ∆x→0 a cf (xi ) ∆x = lim c · ∆x→0 i=1 f (xi ) ∆x i=1 n =c· lim ∆x→0 f (xi ) ∆x b =c f (x) dx. a i=1 C05S05.060: Suppose that f and g are integrable on [ a, b ] and that f (x) by way of contradiction that b I1 = b f (x) dx > a g (x) for all x in [ a, b ]. Suppose g (x) dx = I2 . a Let = I1 − I2 . Choose n a positive integer so large that every Riemann sum for f based on a regular partition with n or more subintervals lies within /3 of I1 and every Riemann sum for g based on a regular partition with n or more subintervals lies within /3 of I2 . (This can be done by determining n1 for I1 , n2 for I2 , and letting n be the maximum of n1 and n2 .) Then every Riemann sum for f based on a regular partition with n or more subintervals exceeds every Riemann sum for g based on a regular partition with n or more subintervals. Let P be such a partition and let {xi } be a selection for P . Then n n f (xi ) ∆x > i=1 g (xi ) ∆x. i=1 But this is impossible because f (xi ) g (xi ) for all i, 1 i n. This contradiction shows that I1 establishes the first comparison property of the definite integral. C05S05.061: Suppose that f is integrable on [ a, b ] and that f (x) Then by the first comparison property, b b f (x) dx a g (x) dx = M x a b a I2 and M for all x in [ a, b ]. Let g (x) ≡ M . = M (b − a). The proof of the other inequality is similar. C05S05.062: Suppose that a < c < b and that f is integrable on [ a, b ]. Then f is integrable on [ a, c ] and on [ c, b ]. Let {Rn } and {Sn } be sequences of Riemann sums, the former for f on [ a, c ], the latter for f on [ c, b ], such that, for each positive integer n, Rn and Sn each have norm less than (b − a)/n. Then c lim Rn = n→∞ f (x) dx and b lim Sn = n→∞ a f (x) dx. c Then because Rn + Sn is a Riemann sum of norm less than (b − a)/n for f on [ a, b ], we have c b f (x) dx + a c 30 C05S05.063: 1000 + contains 0 f (x) dx = lim Rn n→∞ + b lim Sn = lim (Rn + Sn ) = n→∞ V (t) dt = 1000 + (0.4)t2 − 40t n→∞ 30 0 = 160 (gallons). Alternatively, the tank V (t) = (0.4)t2 − 40t + 1000 8 f (x) dx. a gallons at time t 0, so at time t = 30 it contains V (30) = 160 gallons. C05S05.064: In 1990 the population in thousands was 125 + 20 8 + (0.5)t + (0.03)t2 dt = 125 + 8t + (0.25)t2 + (0.01)t3 t=0 = 125 + 160 + 100 + 80 = 465 20 0 (thousands). C05S05.065: In Fig. 5.5.11 of the text we see that 12 − 4x 9 1 x 3−x . 2 Therefore 2 3 2 1 1 dx x 3 . 4 Another way to put it would be to write 2 1 1 dx = 0.708333 ± 0.041667. x For a really sophisticated answer, you could point out that, from the figure, it appears that the low estimate of the integral is about twice as accurate as (has half the error of) the high estimate. So 2 1 C05S05.066: Because L(x) f (x) 1 1 22 13 dx ≈ · + · ≈ 0.6944. x 33 34 L(x) + 0.07 if 0 1 L(x) dx 0 1 f (x) dx 0 1, it follows that x [L(x) + 0.07] dx. 0 That is, 3 4 1 0 1 dx 1 + x2 41 . 50 Another way to put it would be to write 1 0 1 dx = 0.785 ± 0.035. 1 + x2 Or, for the reasons given in the solution of Problem 65, 1 0 1 1 41 2 3 dx ≈ · + · ≈ 0.7733. 1 + x2 3 50 3 4 9 Section 5.6 C05S06.001: C05S06.002: C05S06.003: C05S06.004: C05S06.005: C05S06.006: C05S06.007: 1 2−0 C05S06.010: C05S06.011: C05S06.012: x4 dx = 0 1 4−1 4 √ 2 4 0 1 5−0 4 4 sin 2x dx = π 0 2 1 e2t dt = −1 1 dx = x 3 √ 1 1 −1 1 9x3 5 = 0 4 1 = 2 . 3 = 1 2t e 2 3 = 0 π /2 = 0 1 1 · − cos 2x π 2 1 2 1 2 1 −2 e− e 2 2 125 . 4 1 = −1 14 . 9 2 . π π = 0. 0 1 2 e4 − 1 ≈ 1.8134302039. 4e2 = 3 − (−1) = 4. −1 (y 5 − 1) dy = 4 −4 2 1 · − cos 2x π 2 sin 2x dx = 0 16 . 3 = 12 · (x + 1)3/2 33 x + 1 dx = 0 1 C05S06.016: √ 26 . 3 = 0. −4 1 · 2x1/2 3 x−1/2 dx = 0 −1 C05S06.015: 4 1 14 ·x 54 = = 16. 1 13 ·x 83 1 3 0 2 123 · (x + 1)3/2 23 x2 dx = x3 dx = 1 1 − (−1) C05S06.014: 1 4 −4 4 4 14 . 9 = 1 · 4x2 4 0 π /2 3 4 8x dx = −4 5 1 4−1 C05S06.013: 1 · 4x2 4 8x dx = 1 4 − (−4) 1 π−0 0 1 2 3 /2 ·x 33 x dx = 16 . 5 = 3x2 (x3 + 1)1/2 dx = 1 4 − (−4) 2 π 2 0 1 4−0 1 3−0 1 15 ·x 25 1 1 2−0 C05S06.0008: C05S06.009: 2 16 6y 4 dx = 1 (x3 + 2)2 dx = −y 2 1 = 19 . 2 1 −3/2 2 x dx = − x−1/2 3 3 1 (x6 + 4x3 + 4) dx = −1 1 4 1 =− 1 2 · −1 3 2 17 x + x4 + 4x 7 = 1 = −1 1 . 3 58 . 7 3t − 5 dt = t4 3 C05S06.017: 1 −1 C05S06.018: 339 ++ 852 − π C05S06.019: 1 1 1 C05S06.022: 2 1 0 0 √ 1 0 −2 x2 − 1 dx = π /3 C05S06.026: 0 8 C05S06.027: 4 11 C05S06.028: 6 0 C05S06.029: −1 0 = 1 √ 0 −1 −2 2 x dx = 0 1 1 − x4 dx + dt = −1 π /3 1 cos 3x 3 = = 2 1 12 π sin x dx + 4 x− 0 π /2 2 = 1 2 = −1 5 . 2 35 . 24 √ 1 (e − 1)2 − (1+1) = ≈ 1.086161. e e dx = x √ 7−4 2 ≈ 0.44771525. 3 6 2 1 (x2 − 1) dx = 444 + + = 4. 333 8 = ln 2 ≈ 0.6931471805599453094. 4 = ln 10 − ln 5 = ln 1 1 − x3 dx = x − x5 5 π e+ 0 11 ln(x − 1) 0 −1 1 x · |x| 2 | x | dx = 2 . 3 = ln 8 − ln 4 = ln 1 2 13 1 t − t − t−1 3 4 (1 − x2 ) dx + 8 4 5 . Alternatively, 2 1 and x2 − 1 < 0 if | x | < 1, we split the integral into three: (x2 − 1) dx + ln x ≈ −5.95428758. 1 e2 − 1 1 e − e−1 = ≈ 1.1752011936. 2 2 2e x − x dx + 0 if | x | 1 dx = x−1 π /2 C05S06.030: 0 −2 = 0. 1 sin 3x dx = − 1 dx = x = −1 0 (ex − e−x ) dx = ex + e−x x dx = C05S06.025: Because x2 − 1 2 1 1 2x−1 e 2 e2x − 1 dx = ex x− 3 8 /3 3 5 / 3 9 2 / 3 x −x +x 8 5 2 π 1 t2 − 1 + t−2 4 1 e2x−1 dx = 2 C05S06.024: 22 . 81 √ 3 2 dt = 0 C05S06.023: −2 (−x) dx + −1 1 t − t−1 2 2 C05S06.021: 0 | x | dx = −1 1 (x5/3 − x2/3 + 3x−1/3 ) dx = 1 sin2 x 2 sin x cos x dx = 2 =− √ 219 36 4 12 6 9 3 3 ++ · 22 / 3 = − = 73 − 96 4 8 52 40 5 40 0 C05S06.020: −1 3 5 3 −2 3t3 2t (3t−3 − 5t−4 ) dt = x2 − x + 3 √ dx = 3 x −2 = 3 10 = ln 2. 5 0 1 + x − x4 4 −1 1 π x − x2 dx = − π 2 cos x 4 12 13 13 13 1 1 1 3π + π π + π − π − π + π3 = π2 + π3 = . 4 2 3 8 24 4 12 12 2 2 3 π /2 + 0 1 0 =1− 1 31 1 +1− = . 5 4 20 1 2 13 πx − x 2 3 π π /2 0 C05S06.031: −3 0 C05S06.032: −3 x3 − 9x dx = 0 x3 − 2x2 − 15x dx − 0 1 4 2 3 15 2 x− x− x 4 3 2 = 3 x3 − 9x dx − −3 − 5 0 14 92 x− x 4 2 0 −3 − 3 14 92 x− x 4 2 = 0 81 81 81 + = . 4 4 2 x3 − 2x2 − 15x dx 1 4 2 3 15 2 x− x− x 4 3 2 5 = 0 117 1375 863 + = . 4 12 6 C05S06.033: Height: s(t) = 400 − 16t2 . Velocity: v (t) = −32t. Time T of impact occurs when T 2 = 25, so that T = 5. Average height: 1 5 5 s(t) dt = 0 1 16 3 400t − t 5 3 5 = 0 800 ≈ 266.666667 (ft). 3 Average velocity: 1 5 5 v (t) dt = 0 1 − 16t2 5 5 0 = −80 (ft/s). C05S06.034: The average value of P (t) over the time interval [0, 10] is 1 10 10 P (t) dt = 0 1 10 10 13 t + 5t2 + 100t 150 = 0 452 ≈ 150.666667. 3 As an independent check, P (0) + P (2) + P (4) + P (6) + P (8) + P (10) 100 + 120.08 + 140.32 + 160.72 + 181.28 + 202 ≈ ≈ 150.733. 6 6 C05S06.035: Clearly the tank empties itself in the time interval [0, 10]. So the average amount of water in the tank during that time interval is 1 10 10 V (t) dt = 0 1 10 50 3 t − 500t2 + 5000t 3 10 = 0 5000 ≈ 1666.666667 (L). 3 As an independent check, V (0) + V (2) + V (4) + V (6) + V (8) + V (10) 5000 + 3200 + 1800 + 800 + 200 + 0 = ≈ 1833.333. 6 6 C05S06.036: Noon corresponds to t = 12 and 6 p.m. corresponds to t = 18. So the average temperature over that time interval was 1 6 18 12 1 T (t) dt = 6 40 π π (t − 10) 2π (t − 10) − 3 cos 12 18 = 12 10 1 + √ π 3 + 8π ≈ 88.69638782. As an independent check, T (12) + T (14) + T (16) + T (18) 85 + 88.6603 + 90 + 88.6603 ≈ ≈ 88.08. 4 4 C05S06.037: The average temperature of the rod is 3 1 10 10 T (x) dx = 0 1 4 20x2 − x3 10 3 10 200 ≈ 66.666667. 3 = 0 As an independent check, T (0) + T (2) + T (4) + T (6) + T (8) + T (10) 0 + 64 + 96 + 96 + 64 + 0 = ≈ 53.33. 6 6 C05S06.038: Because r2 + x2 = 1, the radius of the circular cross section at x is r = area is A(x) = π (1 − x2 ), so the average area of such a cross section is 1 1 1 0 A(x) dx = π x − 13 x 3 1 = 0 √ 1 − x2 . Hence its 2 π ≈ 2.0943951024. 3 C05S06.039: Similar triangles yield y/r = 2/1, so r = y/2. So the area of the cross section at y is A(y ) = π (y/2)2 , and thus the average area of such a cross section is 1 2 2 1 π3 y 2 12 A(y ) dt = 0 2 0 π . 3 = C05S06.040: The velocity of the sports car at time t is v (t) = at, so its final velocity is v (T ) = aT and its average velocity is 1 T T v (t) dt = 0 1 T 12 at 2 T = 0 1 aT. 2 The position of the sports car at time t is x(t) = 1 at2 , so its final position is x(T ) = 1 aT 2 and its average 2 2 position is 1 T T x(t) dt = 0 C05S06.041: First, A(x) = 3(9 − x2 ) for −3 1 6 3 −3 x A(x) dx = 1 T 13 at 6 t = 0 12 aT . 6 3. Hence the average value of A on that interval is 1 27x − x3 6 3 −3 = 18. √ Next, A(x) = 18 has the two solutions x = ± 3 in the interval [ −3, 3], so there are two triangles having 4 the same area as the average. One is shown next. 8 6 4 2 -3 -2 -1 C05S06.042: First, A(x) = x(10 − x) for 0 1 10 10 1 x A(x) dx = 0 2 3 10. So the average value of A on that interval is 10 1 1 5x2 − x3 10 3 = 0 50 . 3 √ The equation A(x) = 50 has the two solutions 5 3 ± 3 , so there are exactly two rectangles with the same 3 3 area as the average rectangle. One of them is shown next. 10 8 6 4 2 2 4 6 8 √ C05S06.043: The area function is A(x) = 2x 16 − x2 for 0 1 4 The equation A(x) = 32 3 4 0 A(x) dx = x 1 2 − (16 − x2 )3/2 4 3 has the two solutions x = 2 5 2 3 3± √ 10 4, so the average value of A is 4 = 0 32 . 3 5 , so there are two rectangles having the same area as the average. One is shown next. 4 3 2 1 -4 -2 2 4 C05S06.044: The area is A(x) = 2x(16 − x2 ) and its average value on [0, 4] is 1 4 4 A(x) dx = 0 1 1 16x2 − x4 4 2 4 = 32. 0 The equation A(x) = 32 has three solutions, but only two lie in the domain 0 x 4 of A; they are approximately 1.07838 and 3.35026. So there are two rectangles having the same area as the average area. Both are shown next. 15 12.5 10 7.5 5 2.5 -4 -2 2 4 C05S06.045: f (x) = (x2 + 1)17 . C05S06.046: g (t) = √ t2 + 25 . C05S06.047: h (z ) = (z − 1)1/3 . C05S06.048: A (x) = 1 . x C05S06.049: Because f (x) = − x 10 (et − e−t ) dt, 6 part (1) of the fundamental theorem of calculus (Section 5.6) implies that f (x) = −(ex − e−x ) = e−x − ex . To use part (2) of the fundamental theorem of calculus, first compute 10 f (x) = x (et − e−t ) dt = −ex − e−x + e10 − e−10 , and differentiation now shows that f (x) = e−x − ex . x C05S06.050: G(x) = f (t) dt, so G (x) = f (x) = x . x2 + 1 f (t) dt, so G (x) = f (x) = √ 2 x C05S06.051: G(x) = x + 4. 0 x C05S06.052: G(x) = f (t) dt, so G (x) = f (x) = sin3 x. 0 x C05S06.053: G(x) = f (t) dt, so G (x) = f (x) = √ x3 + 1 . 1 C05S06.054: Let u(x) = x2 . Then x2 f (x) = u 1 + t3 dt = g (u) = 1 + t3 dt. 0 0 Therefore f (x) = Dx g (u) = g (u) · u (x) = 2x 1 + u3 = 2x 1 + x6 . C05S06.055: Let u(x) = 3x. Then 3x f (x) = u sin t2 dt = g (u) = 2 sin t2 dt. 2 Therefore f (x) = Dx g (u) = g (u) · u (x) = 3 sin u2 = 3 sin 9x2 . C05S06.056: Let u(x) = sin x. Then sin x f (x) = 0 1 − t2 dt = g (u) = u 0 1 − t2 dt. Therefore f (x) = Dx g (u) = g (u) · u (x) = (cos x) 1 − u2 = (cos x) 1 − sin2 x = (cos x) | cos x |. C05S06.057: Let u(x) = x2 . Then 7 x2 f (x) = u sin t dt = g (u) = 0 sin t dt. 0 Therefore f (x) = Dx g (u) = g (u) · u (x) = 2x sin u = 2x sin x2 . For an independent verification, note that f (x) = − cos t x2 0 = 1 − cos x2 , and therefore that f (x) = 2x sin x2 . C05S06.058: Let u(x) = sin x. Then sin x f (x) = u (t2 + 1)3 dt = g (u) = 1 (t2 + 1)3 dt. 1 Therefore f (x) = Dx g (u) = g (u) · u (x) = (cos x)(u2 + 1)3 = (cos x)(1 + sin2 x)3 . C05S06.059: Let u(x) = x2 + 1. Then x2 +1 f (x) = 1 1 dt = g (u) = t u 1 1 dt. t Therefore f (x) = Dx g (u) = g (u) · u (x) = 2x · 1 1 = 2x · 2 . u x +1 C05S06.060: Let u(x) = ex . Then ex f (x) = u ln(1 + t2 ) dt = g (u) = 1 ln(1 + t2 ) dt. 1 Therefore f (x) = Dx g (u) = g (u) · u (x) = ln(1 + u2 ) · ex = ex ln(1 + e2x ). x C05S06.061: y (x) = 1 C05S06.062: y (x) = 1 dt. t π + 4 1 dt. 1 + t2 x 1 x C05S06.063: y (x) = 10 + 1 + t2 dt. 5 x C05S06.064: y (x) = 2 + tan t dt. 1 8 C05S06.065: The fundamental theorem does not apply because the integrand is not continuous on [ −1, 1]. We will see how to handle integrals such as the one in this problem in Section 9.8. One thing is certain: Its value is not −2. C05S06.066: Suppose that f is differentiable on [ a, b ]. Then the average value of f (x) on [ a, b ] is 1 b−a b f (x) dx = a 1 · f (x) b−a b a = f (b) − f (a) , b−a the average rate of change of f on [ a, b ]. C05S06.067: If 0 x 2, then x g (x) = 0 Thus g (0) = 0 and g (2) = 4. If 2 x g (x) = g (2) + x 2 2 Therefore g (4) = 8 and g (6) = 4. If 6 x g (x) = g (6) + (8 − 2t) dt = 4 + 8t − t2 x x g (x) = g (8) + x x f (t) dt = 4 + 6 2 = 4 + 8x − x2 − 16 + 4 = 8x − x2 − 8. (−4) dt = 4 − 4x + 24 = 28 − 4x. 10, then (2t − 20) dt = −4 + t2 − 20t 8 x 8, then continuity of g at x = 6 implies that 6 Thus g (8) = −4. Finally, if 8 2t dt = x2 . 0 6, then continuity of g at x = 2 implies that x f (t) dt = 4 + x f (t) dt = x 8 = −4 + x2 − 20x − 64 + 160 = x2 − 20x + 92, and therefore g (10) = −8. Next, g (x) is increasing where f (x) > 0 and decreasing where f (x) < 0, so g is increasing on (0, 4) and decreasing on (4, 10). The global maximum of g will therefore occur at (4, 8) and its global minimum at (10, −8). The graph of y = g (x) is next. 7.5 5 2.5 2 4 6 8 10 -2.5 -5 -7.5 C05S06.068: If 0 x 2, then x g (x) = 0 x f (t) dt = (t + 1) dt = 0 12 t +t 2 x = 0 12 x + x. 2 Therefore g (0) = 0 and g (2) = 4. Continuity of g at x = 2 then requires that if 2 9 x 4, then x g (x) = g (2) + 3 dt = 4 + 3t 2 and therefore g (4) = 10. Then if 4 x g (x) = g (4) + 4 Thus g (6) = 10. If 6 x 2 = 3x − 2, 6, (15 − 3t) dt = 10 + 15t − x x 32 t 2 x 4 = 10 + 15x − 32 3 x − 60 + 24 = 15x − x2 − 26. 2 2 8, then x g (x) = 10 + 6 and so g (8) = 3. And if 8 x 1 12 − t dt = 10 − t 2 4 x 6 = 10 − 12 1 x + 9 = 19 − x2 , 4 4 10, then x g (x) = 3 + 8 3 t − 16 2 dt = 32 x − 16x + 83, 4 and it follows that g (10) = −2. Next, g is increasing where f (x) > 0 and decreasing where f (x) < 0, so g is increasing on (0, 5) and decreasing on (5, 10). The global maximum of g will therefore occur at (5, 11.5) and its global minimum is at (10, −2). The graph of y = f (x) is shown next. 10 8 6 4 2 2 4 6 8 10 -2 C05S06.069: The local extrema of g (x) occur only where g (x) = f (x) = 0, thus at π , 2π , and 3π , and at the endpoints 0 and 4π of the domain of g . Figure 5.6.19 shows us that g is increasing on (0, π ) and (2π , 3π ), decreasing on (π , 2π ) and (3π , 4π ), so there are local minima at (0, 0), (2π , −2π ), and (4π , −4π ); there are local maxima at (π , π ) and (3π , 3π ). The global maximum occurs at (3π , 3π ) and the global minimum occurs at (4π , −4π ). To find the inflection points, we used Newton’s method to solve the equation f (x) = 0 and found the x-coordinates of the inflection points—it’s clear from the graph that there are four of them—to be approximately 2.028758, 4.913180, 7.978666, and 11.085538. (The values of f (x) at these four points are approximately 1.819706, −4.814470, 7.916727, and −11.040708.) The graph of y = g (x) is 10 shown next. 5 2 4 6 8 10 12 -5 -10 C05S06.070: By reasoning similar to that in the solution of Problem 69, g has a global minimum at (0, 0) (not shown on the following graph), a global maximum at (π , 1.851937) (numbers given in decimal form are approximations), a local minimum at (2π , 1.418152), a local maximum at (3π , 1.674762), and a local minimum at (4π , 1.492161). Newton’s method gives the approximate coordinates of the points on the graph of f where g has inflection points to be (4.493409, 0.017435), (7.725252, 0.017400), and (10.904122, 0.017348). The graph of y = g (x) is next. 1.8 1.6 1.4 1.2 2 4 6 0.8 11 8 10 12 Section 5.7 C05S07.001: Let u = 3x − 5. Then du = 3 dx, so dx = 17 (3x − 5) 6 (4x + 7) dx = 1 4 du. Thus 1 17 1 18 1 18 u du = u +C = (3x − 5) + C. 3 54 54 dx = C05S07.002: Let u = 4x + 7. Then dx = 1 1 3 1 4 du. Thus u−6 du = − 1 −5 1 + C. u +C =− 20 20(4x + 7)5 C05S07.003: The given substitution yields 1 1 /2 1 12 u x +9 du = u3/2 + C = 2 3 3 C05S07.004: The given substitution yields 1 −1/3 1 1 u du = u2/3 + C = 2x3 − 1 6 4 4 C05S07.005: The given substitution yields 1 1 1 sin u du = − cos u + C = − cos 5x + C . 5 5 5 C05S07.006: The given substitution yields 1 1 1 cos u du = sin u + C = sin kx + C . k k k C05S07.007: The given substitution yields 1 1 1 sin u du = − cos u + C = − cos 2x2 + C . 4 4 4 3 /2 + C. 2 /3 + C. C05S07.008: Let u = x1/2 ; then x = u2 and dx = 2u du. So √ ex √ dx = x √ eu · 2u du = 2eu + C = 2 exp x + C. u C05S07.009: The given substitution yields u5 du = 16 1 6 u + C = (1 − cos x) + C . 6 6 C05S07.010: Let u = 5 + 2 sin 3x. Then du = 6 cos 3x dx, so that cos 3x dx = cos 3x dx = 5 + 2 sin 3x 1 6 u du = 1 6 du. Thus 1 1 ln u + C = ln(5 + 2 sin 3x) + C. 6 6 1 (x + 1)7 + C . 7 C05S07.011: If necessary, let u = x + 1. In any case, (x + 1)6 dx = C05S07.012: If necessary, let u = 2 − x. In any case, 1 (2 − x)5 dx = − (2 − x)6 + C . 6 C05S07.013: If necessary, let u = 4 − 3x. In any case, (4 − 3x)7 dx = − C05S07.014: If necessary, let u = 2x + 1. In any case, √ 1 2x + 1 dx = 1 (4 − 3x)8 + C . 24 1 (2x + 1)3/2 + C . 3 1 2 dx = (7x + 5)1/2 + C . 7 7x + 5 C05S07.015: If necessary, let u = 7x + 5. In any case, √ C05S07.016: If necessary, let u = 3 − 5x. In any case, dx 1 = + C. 2 (3 − 5x) 5(3 − 5x) C05S07.017: If necessary, let u = π x + 1. In any case, sin(π x + 1) dx == − C05S07.018: If necessary, let u = πt . In any case, 3 cos 1 cos(π x + 1) + C . π πt 3 πt dt = sin + C. 3 π 3 1 sec 2θ + C . 2 C05S07.019: If necessary, let u = 2θ. In any case, sec 2θ tan 2θ dθ = C05S07.020: If necessary, let u = 5x. In any case, 1 csc2 5x dx = − cot 5x + C . 5 C05S07.021: 1 e1−2x dx = − e1−2x + C . 2 C05S07.022: x exp x2 dx = C05S07.023: x2 exp 3x3 − 1 dx = 1 exp x2 + C . 2 1 exp 3x3 − 1 + C . 9 If you prefer integration by substitution, let u = 3x3 − 1. 1 exp 2x3/2 + C . 3 C05S07.024: x1/2 exp 2x3/2 dx = C05S07.025: 1 1 dx = ln | 2x − 1 | + C . 2x − 1 2 C05S07.026: 1 1 dx = ln | 3x + 5 | + C . 3x + 5 3 C05S07.027: 1 1 (ln x)2 dx = (ln x)3 + C . x 3 C05S07.028: 1 dx = ln | ln x | + C . x ln x C05S07.029: x + e2x 1 dx = ln x2 + e2x + C . For integration by substitution, let u = x2 + e2x . x2 + e2x 2 C05S07.030: ex + e−x 2 dx = e2x − e−2x + C. 2 e2x + 2 + e−2x dx = 2x + Mathematica version 3.0 returns the antiderivative in the form 2 C− (e−x + ex ) 2 2 (1 + e2x ) 2 + e4x (e−x + ex ) 2 2 (1 + e2x ) 2 + 2e2x (e−x + ex ) x 2 (1 + e2x ) . Then the successive commands Together and Expand simplify it to the form we first gave here. 2 C05S07.031: If necessary, let u = x2 − 1, so that du = 2x dx; that is, x dx = 1 du. Then 2 x x2 − 1 dx = 1 1 /2 1 1 du = u3/2 + C = (x2 − 1)3/2 + C. u 2 3 3 Editorial Comment (DEP): In my opinion a much better, faster, shorter, and more reliable method runs as follows. Given x x2 − 1 dx, it should be evident that the antiderivative involves (x2 − 1)3/2 . Because Dx (x2 − 1)3/2 = 32 (x − 1)1/2 · 2x = 3x x2 − 1, 2 the fact that if c is constant then Dx cf (x) = cf (x) now allows us to modify the initial guess (x2 − 1)3/2 for the antiderivative by multiplying it by 1 . Therefore 3 x x2 − 1 dx = 12 (x − 1)3/2 + C. 3 All that’s necessary is to remember the constant of integration. Note that this technique is self-checking, requires less time and space, and is immune to the dangers of various illegal substitutions. The computation of the “correction factor” (in this case, 1 ) can usually be done mentally. 3 10 3 1 − 2t2 44 11 C05S07.032: If necessary, let u = 1 − 2t2 . In any case, 3t 1 − 2t2 C05S07.033: If necessary, let u = 2 − 3x2 . In any case, 1 x(2 − 3x2 )1/2 dx = − (2 − 3x2 )3/2 + C . 9 C05S07.034: If necessary, let u = 2t2 + 1. In any case, √ C05S07.035: If necessary, let u = x4 + 1. In any case, x3 (x4 + 1)1/2 dx = C05S07.036: If necessary, let u = x3 + 1. In any case, x2 (x3 + 1)−1/3 dx = C05S07.037: If necessary, let u = 2x3 . In any case, C05S07.038: If necessary, let u = t2 . In any case, dt = − + C. t 1 dt = (2t2 + 1)1/2 + C . 2 2t 2 + 1 x2 cos(2x3 ) dx = t sec2 t2 dt = 14 (x + 1)3/2 + C . 6 13 (x + 1)2/3 + C . 2 1 sin(2x3 ) + C . 6 1 tan t2 + C . 2 C05S07.039: If necessary, let u = −x2 or let u = x2 . In the latter case du = 2x dx, so that x dx = In any case, x exp(−x2 ) dx = − du. 1 2 du. 1 exp(−x2 ) + C. 2 C05S07.040: If necessary, let u = x2 or let u = 1 + x2 . In the latter case, du = 2x dx, so that x dx = In either case, 3 1 2 x 1 dx = ln(1 + x2 ) + C. 1 + x2 2 C05S07.041: If necessary, let u = cos x. In any case, 1 cos3 x sin x dx = − cos4 x + C . 4 C05S07.042: There are two choices for the substitution. The substitution u = 3z results in 1 1 1 sin5 u cos u du = sin6 u + C = sin6 3z + C. 3 18 18 sin5 3z cos 3z dz = Alternatively, the substitution u = sin 3z yields du = 3 cos 3z dz , so that cos 3z dz = sin5 3z cos 3z dz = 1 3 du. Then 15 16 1 u du = u +C = sin6 3z + C. 3 18 18 C05S07.043: If necessary, let u = tan θ. In any case, tan3 θ sec2 θ dθ = 1 tan4 θ + C . 4 C05S07.044: If necessary, let u = sec θ. Then du = sec θ tan θ dθ, and hence sec3 θ tan θ dθ = (sec2 θ) · sec θ tan θ dθ = C05S07.045: If necessary, let u = √ x. In any case, C05S07.046: If necessary, let u = 1 + √ u2 du = 13 1 u + C = sec3 θ + C. 3 3 x−1/2 cos x1/2 dx = 2 sin x1/2 + C . x. In any case, √ dx x (1 + √ 2 x) = −2 √ + C. 1+ x C05S07.047: If necessary, let u = x2 + 2x + 1. This yields (x2 + 2x + 1)4 (x + 1) dx = 14 15 12 u du = u +C = (x + 2x + 1)5 + C. 2 10 10 Alternatively, (x2 + 2x + 1)4 (x + 1) dx = (x + 1)9 dx = C05S07.048: If necessary, let u = x2 + 4x + 3. In any case, 1 (x + 1)10 + C. 10 (x + 2) dx (x2 1 = − (x2 + 4x + 3)−2 + C . 4 + 4x + 3) 3 C05S07.049: If necessary, let u = x2 + 4x + 3, so that du = (2x + 4) dx. Thus (x + 2) dx = case, x+2 1 dx = ln(x2 + 4x + 3) + C, x2 + 4x + 3 2 although, because the quadratic can be negative, a better answer is 1 ln | x2 + 4x + 3 | + C . 2 C05S07.050: If necessary, let u = x2 + ex + 1, so that du = (2x + ex ) dx. In any case, 4 1 2 du. In any 2x + ex 1 dx = − 2 + C. (x2 + ex + 1)2 x + ex + 1 2 C05S07.051: 1 dt 1 =− (t + 1)3 2(t + 1)2 4 C05S07.052: √ 0 4 C05S07.053: dx = (2x + 1)1/2 2x + 1 x x2 + 9 dx = 0 4 C05S07.054: Given: I = 1 2 1 =− 4 0 1 1 −− 18 8 = 5 . 72 = 3 − 1 = 2. 4 12 (x + 9)3/2 3 = 0 125 98 −9= . 3 3 √4 √ (1 + x ) √ dx. Let u = 1 + x. Then x 1 dx du = x−1/2 dx, so 2 du = √ . 2 x Thus √4 (1 + x ) √ dx = x Therefore I = 2u4 du = √ 25 2 u +C = 1+ x 5 5 5 + C. 486 64 422 − = . 5 5 5 8 C05S07.055: Given: J = √ t t + 1 dt. Let u = t + 1. Then du = dt, and so 0 J= 8 t=0 = (u − 1)u1/2 du = 8 t=0 2 2 (t + 1)5/2 − (t + 1)3/2 5 3 u3/2 − u1/2 du = 8 = 0 396 4 −− 5 15 2 5 / 2 2 3 /2 u −u 5 3 = 8 t=0 1192 . 15 Alternatively, J= 9 u=1 (u − 1)u1/2 du = 9 1 u3/2 − u1/2 du = 2 5 /2 2 3 /2 u −u 5 3 9 = 1 4 396 −− 5 15 = 1192 . 15 Thus you have at least two options: 1. Make the substitution for u in place of t, find the antiderivative, then express the antiderivative in terms of t before substituting the original limits of integration, or 2. Make the substitution for u in place ot t, find the antiderivative, then substitute the new limits of integration in terms of u. Whichever option you use, be sure to use the notation correctly (as shown here). π /2 C05S07.056: 0 sin x cos x dx = 1 sin2 x 2 π /2 = 0 5 1 1 −0= . 2 2 π /6 C05S07.057: 0 √ C05S07.058: π 0 π /2 C05S07.059: sin 2x cos3 2x dx = − √ t2 t2 t sin dt = − cos 2 2 π /2 0 x x dx = 2 tan 2 2 π /2 C05S07.061: Given: K = 0 1 1 −− 128 8 = 15 . 128 = 0 − (−1) = 1. 0 2 (1 + 3 sin θ)5/2 15 (1 + 3 sin θ)3/2 cos θ dθ = sec2 =− π 0 C05S07.060: π /6 1 cos2 2x 8 π /2 = 0 64 2 62 − = . 15 15 15 π /2 0 = 2 − 0 = 2. exp(sin x) cos x dx. Let u = sin x. Then du = cos x dx. Hence 0 1 K= 1 eu du = eu 0 2 C05S07.062: Given: I = 1 1+ln 2 I= 1 2 1 C05S07.064: π 2 /4 1+ln 2 = 1 1 1 (ln 2)2 + 2 ln 2 (1 + ln 2)2 − = ≈ 0.9333736875. 2 2 2 1 1 . Then du = − 2 dx, and therefore x x e−1/x dx = − x2 π2 = e − 1 ≈ 1.71828182845904523546. 1 + ln x 1 dx. Let u = 1 + ln x. Then du = dx, and so x x 12 u 2 u du = C05S07.063: Let u = 0 1/2 1 /2 e−u du = e−u 1 = e−1/2 − e−1 = 1 √ √ sin x cos x √ dx = x √ sin x 2 √ e −1 ≈ 0.2386512185. e π2 π 2 /4 = 0 − 1 = −1. C05S07.065: sin2 x dx = 1 − cos 2x 1 1 1 1 dx = x − sin 2x + C = x − sin x cos x + C . 2 2 4 2 2 C05S07.066: cos2 x dx = 1 1 1 1 1 + cos 2x dx = x + sin 2x + C = x + sin x cos x + C . 2 2 4 2 2 π π C05S07.067: sin2 3t dt = 0 0 1 C05S07.068: cos2 π t dt = 0 1 − cos 6t t 1 dt = − sin 6t 2 2 12 t 1 + sin 2π t 2 4π 1 = 0 π = 0 1 1 −0= . 2 2 C05S07.069: tan2 x dx = sec2 x − 1 dx = −x + tan x + C . C05S07.070: tan2 3t dt = sec2 3t − 1 dt = −t + 6 1 tan 3t + C . 3 π π −0= . 2 2 sin3 x dx = C05S07.071: π /2 C05S07.072: sin x − cos2 x sin x dx = − cos x + π /2 cos3 x dx = 0 0 1 cos3 x + C . 3 cos x − sin2 x cos x dx = sin x − 1 sin3 x 3 π /2 = 0 2 2 −0= . 3 3 C05S07.073: If you solve the equation 1 1 sin2 θ + C1 = − cos2 θ + C2 2 2 for C2 − C1 , you will find its value to be 1 . That is, cos2 θ + sin2 θ = 1. Two functions with the same 2 derivative differ by a constant; in this case sin2 θ − (− cos2 θ) = 1. The graphs of f (x) = 1 sin2 x and 2 g (x) = − 1 cos2 x are shown next. 2 0.4 0.2 1 2 3 4 5 6 -0.2 -0.4 C05S07.074: If you solve the equation 1 tan2 θ + C1 = 2 be 1 . That is, sec2 θ = 1 + tan2 θ. The graphs of f (x) = 2 1 2 1 2 sec2 θ + C2 for C1 − C2 , you will find its value to tan2 x and g (x) = 1 sec2 x are shown next. 2 2 1.5 1 0.5 1 C05S07.075: First, if f (x) = 2 3 4 5 6 x , then 1−x f (x) = (1 − x) · 1 − x · (−1) 1 = . (1 − x)2 (1 − x)2 Next, if u = 1 − x then x = 1 − u and dx = − du, so dx = (1 − x)2 − du 1 1 = + C2 = + C2 . u2 u 1−x 7 If g (x) = 1 , then 1−x g (x) − f (x) = 1 x 1−x − = ≡ 1. 1−x 1−x 1−x As expected, because g (x) and f (x) have the same derivative on (1, + ∞), they differ by a constant there. (This also holds on the interval (−∞, 1).) The graphs of y = f (x) and y = g (x) are shown next. 2 1.5 1 0.5 -2 -1 1 2 3 4 -0.5 -1 -1.5 -2 C05S07.076: The substitution u = x2 involves du = 2x dx, so that x dx = 1 2 du. Thus 1 2 du 1 u x2 =· + C1 = + C1 . 2 (1 − u) 2 1−u 2(1 − x2 ) x dx = (1 − x2 )2 Next, if we let u = 1 − x2 , then du = −2x dx, so that x dx = − 1 du. This yields 2 x dx = (1 − x2 )2 − 1 du 1 1 2 = + C2 = + C2 . 2 u 2u 2(1 − x2 ) To reconcile these results, let f (x) = x2 2(1 − x2 ) and g (x) = 1 . 2(1 − x2 ) Then g (x) − f (x) = 1 x2 1 − x2 1 − = ≡. 2) 2) 2(1 − x 2(1 − x 2(1 − x2 ) 2 That is, because f (x) = g (x) on the interval (−1, 1), f (x) and g (x) differ by a constant there. (This result also holds on the interval (1, + ∞) and on the interval (−∞, −1).) The graphs of y = f (x) and y = g (x) 8 are shown next. 1 0.75 0.5 0.25 -2 -1 1 2 -0.25 -0.5 -0.75 -1 C05S07.077: Suppose that f is continuous and odd. The substitution u = −x requires dx = − du, so that 0 0 f (x) dx = −a a − f (−u) du = a 0 f (−u) du = − a a f (u) du = − 0 f (x) dx. 0 (The last equality follows because in the next-to-last integral, u is merely a “dummy” variable of integration, and can simply be replaced with x.) Therefore a 0 f (x) dx = −a a f (x) dx + −a 0 f (x) dx = − a a f (x) dx + 0 f (x) dx = 0. 0 C05S07.078: If f is even and continuous, then the substitution u = −x yields 0 0 f (x) dx = −a a −f (−u) du = a 0 f (−u) du = a a f (u) du = 0 f (x) dx. 0 (The last equality is merely replacement of the dummy variable of integration by a different dummy variable.) Therefore a −a f (x) dx = 0 −a a f (x) dx + a f (x) dx = 0 a f (x) dx + 0 0 a f (x) dx = 2 f (x) dx. 0 C05S07.079: Because the tangent function is continuous and odd on [ −1, 1], it follows from the result in Problem 77 that 1 tan x dx = 0. −1 Next, f (x) = x1/3 is continuous and odd on [ −1, 1] while g (x) = (1 + x2 )7 is continuous and even there, so (Exercise!) their quotient is odd. Thus 1 −1 x1/3 dx = 0. (1 + x2 )7 Finally, h(x) = x17 is continuous and odd on [ −1, 1] while j (x) = cos x is continuous and even there, so (Exercise!) their product is odd. Thus 9 1 x17 cos x dx = 0. −1 √ C05S07.080: Both f (x) = x10 sin x and g (x) = x5 1 + x4 are continuous and odd on [ −5, 5], so 5 −5 −x10 sin x + x5 dx = 0 1 + x4 by the result in Problem 77. Hence, using the result in Problem 68, 5 −5 3x2 − x10 sin x + x5 1 + x4 5 dx = 5 3x2 dx = 2 −5 5 3x2 dx = 2 x3 0 0 = 2 · 125 = 250. C05S07.081: Given b I= f (x + k ) dx, a let u = x + k . Then x = u − k , dx = du, u = a + k when x = a, and u = b + k when x = b. Hence b +k I= f (u) du = a+k b +k f (x) dx a+k (because it doesn’t matter whether the variable of integration is called u or x). C05S07.082: Given b J=k f (kx) dx, a let u = kx. Then du = k dx, u = ka when x = a, and u = kb when x = b. Hence kb J=k ka 1 f (u) du = k kb f (x) dx. ka C05S07.083: (a) Du [ eu (u − 1)] = eu (u − 1) + eu · 1 = ueu . (b) Let u = x1/2 . Then x = u2 , dx = 2u du, u = 0 when x = 0, and u = 1 when x = 1. Therefore 1 exp √ 1 dx = x 0 1 2ueu du = 0 2eu (u − 1) 0 = 0 − (−2e0 ) = 2. C05S07.084: (a) Du (sin u − u cos u) = cos u − 1 · cos u + u sin u = u sin u. x = u2 , dx = 2u du, u = 0 when x = 0, and u = π when x = π 2 . Therefore π2 0 π √ sin x dx = 0 (b) Let u = √ π 2u sin u du = 2 · sin u − u cos u 10 0 = 2 · (−π cos π ) − 2 · 0 = 2π . x . Then Section 5.8 C05S08.001: To find the limits of integration, we solve 25 − x2 = 9 for x = ±4. Hence the area is 4 1 (25 − x2 − 9) dx = 16x − x3 3 −4 4 128 128 −− 3 3 = −4 = 256 . 3 C05S08.002: To find the limits of integration, we solve 16 − x2 = −9 for x = ± 5. Thus the area is 5 1 (16 − x2 + 9) dx = 25x − x3 3 −5 5 250 250 −− 3 3 = −5 = 500 . 3 C05S08.003: To find the limits of integration, we solve x2 − 3x = 0 for x = 0, x = 3. Hence the area is 3 0 (3x − x2 ) dx = 3 32 13 x− x 2 3 = 0 9 9 −0= . 2 2 C05S08.004: To find the limits of integration, we solve x3 − 9x = 0 for x = −3, x = 0, and x = 3. Therefore the area of the region shown in Fig. 5.8.18 is 3 0 (9x − x3 ) dx = 3 92 14 x− x 2 4 = 0 81 81 −0= . 4 4 C05S08.005: To find the limits of integration, we solve 12 − 2x2 = x2 for x = ±2. Therefore the area is 2 −2 2 (12 − 3x2 ) dx = 12x − x3 −2 = 16 − (−16) = 32. C05S08.006: To find the limits of integration, we solve 2x − x2 = 2x2 − 4x for x = 0, x = 2. Therefore the area of the figure is 2 0 2 (6x − 3x2 ) dx = 3x2 − x3 0 = 4 − 0 = 4. C05S08.007: To find the limits of integration, we solve 4 − x2 = 3x2 − 12 for x = ±2. So the area is 4 (16 − 4x2 ) dx = 16x − x3 3 −2 2 2 = −2 64 64 −− 3 3 = 128 . 3 C05S08.008: To find the limits of integration, we solve 12 − 3x2 = 4 − x2 for x = ±2. So the area of the region is 2 (8 − 2x2 ) dx = 8x − x3 3 −2 2 2 = −2 32 32 −− 3 3 = 64 . 3 C05S08.009: √To find the limits of integration, we solve x2 − 3x = 6 for x = a = x = b = 1 3 + 33 . So the area is 2 1 1 2 3− √ 33 and b b 3 1 (6 − x + 3x) dx = 6x + x2 − x3 2 3 2 a a √ √ √ 45 + 11 33 45 − 11 33 11 33 = − = . 4 4 2 C05S08.010: To find the limits of integration, we solve x2 − 3x = x for x = 0, x = 4. So the area is 4 0 1 C05S08.011: 0 1 −1 dx = (x + 1)2 x+1 1 1 C05S08.013: 0 −1 2 C05S08.015: 0 x2 − (−1) dx = 1 dx = x+1 = 0 3 = 1 32 32 −0= . 3 3 = 0 1 . 4 1 . 4 14 15 x− x 4 5 (x3 − x4 ) dx = 2 C05S08.014: 1 12 14 x− x 2 4 (x − x3 ) dx = 3 C05S08.012: 4 1 (x − x2 + 3x) dx = 2x2 − x3 3 13 x +x 3 1 = 0 1 . 20 2 = −1 14 4 −− 3 3 = 6. 2 ln(x + 1) 0 = ln 3 − ln 1 = ln 3 ≈ 1.0986122887. C05S08.016: To find the limits of integration, we first solve 4x − x2 = 0 for x = 0 and x = 4. Hence the area of R is 4 0 1 (4x − x2 ) dx = 2x2 − x2 3 4 = 0 32 . 3 C05S08.017: To find the limits of integration, we first solve y 2 = 4 for y = ±2. So the area of the region R is 2 −2 (4 − y 2 ) dy = 4y − 13 y 3 2 = −2 16 16 −− 3 3 = 32 . 3 C05S08.018: To find the limits of integration, we first solve x4 − 4 = 3x2 for x = ±2. So the area of R is 1 (3x2 − x4 + 4) dx = 4x + x3 − x5 5 −2 2 2 = −2 48 48 −− 5 5 = 96 . 5 √ C05S08.019: To find the limits of integration, we first solve 8 − y 2 = y 2 − 8 for y = a = −2 2 and √ y = b = 2 2. So the area of R is b 2 (16 − 2y ) dy = 16y − y 3 3 b 2 a a √ 64 2 = − 3 2 √ 64 2 − 3 √ 128 2 = . 3 C05S08.020: To find the limits of integration, we solve x1/3 = x3 for x = −1, x = 0, and x = 1. Thus the region R comes in two parts, one in the first quadrant and one in the third. The area of the first is 1 0 3 4 /3 1 4 x −x 4 4 x1/3 − x3 dx = 1 = 0 1 . 2 The area of the second is 0 x3 − x1/3 dx = −1 0 1 4 3 4 /3 x− x 4 4 −1 =0− − 1 2 = 1 . 2 Therefore R has area 1. C05S08.021: The area of the region—shown next—is 2 A= 0 2x − x2 dx = x2 − 13 x 3 2 = 0 4 . 3 4 3 2 1 0.5 C05S08.022: 2 −2 8 − 2x2 dx = 8x − 1 23 x 3 1.5 2 = −2 2 32 32 −− 3 3 = 64 . 3 C05S08.023: We find the limits of integration by solving y 2 = 25 for y = ±5. The area of the region— shown next—is therefore 5 −5 25 − y 2 dy = 25y − 13 y 3 5 = −5 250 250 −− 3 3 = 500 . 3 4 2 5 10 15 20 25 30 -2 -4 C05S08.024: Area: 4 −4 32 − 2y 2 dy = 32y − 23 y 3 4 = −4 256 256 −− 3 3 = 512 . 3 C05S08.025: We find the limits of integration by solving x2 = 2x + 3 for x = −1, x = 3. Thus the area of the region—shown next—is 3 3 A= −1 2x + 3 − x2 dx = 3x + x2 − 13 x 3 3 =9− − −1 5 3 = 32 . 3 10 8 6 4 2 -1 1 2 3 C05S08.026: We find the limits of integration by solving x2 = 2x + 8 for x = −2, x = 4. The area is 4 2x + 8 − x2 dx = 8x + x2 − −2 4 13 x 3 = −2 80 28 −− 3 3 = 36. C05S08.027: We first find the limits of integration by solving y 2 = y + 6 for y = −2, y = 3. The area of the region—shown next—is therefore 3 A= y + 6 − y 2 dy = 6y + −2 12 13 y− y 2 3 3 = −2 27 22 −− 2 3 = 125 . 6 3 2 1 2 4 6 8 10 -1 -2 C05S08.028: Solving y 2 = 8 − 2y for y = −4, y = 2 yields the limits of integration, and the area of the region is A= 1 (8 − 2y − y 2 ) dy = 8y − y 2 − y 3 3 −4 2 2 = −4 28 80 −− 3 3 = 36. C05S08.029: The two graphs meet at the right-hand endpoint of the given interval, where x = π /4. Therefore the area of the region they bound—shown next—is π /4 A= 0 π /4 (cos x − sin x) dx = sin x + cos x = 0 4 √ 2 − 1 ≈ 0.414213562373. 1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 C05S08.030: The two graphs meet at the left-hand endpoint x = −3π /4 of the given interval. The area they bound over that interval is A= 0 0 −3π /4 (cos x − sin x) dx = sin x + cos x = 1+ −3π /4 √ 2 ≈ 2.414213562373. C05S08.031: Solution of 4y 2 + 12y + 5 = 0 yields the limits of integration y = a = − 5 and y = b = − 1 . 2 2 Hence the area of the region bounded by the two given curves—shown next—is 4 −12y − 5 − 4y 2 dy = −5y − 6y 2 − y 3 3 b A= a b = a 7 25 −− 6 6 = 16 . 3 0.5 -10 10 20 30 -0.5 -1 -1.5 -2 -2.5 -3 C05S08.032: Solution of x2 = 3 + 5x − x2 yields the limits of integration x = a = − 1 and x = b = 3. 2 Hence the area bounded by the gives curves is b A= a (3 + 5x − 2x2 ) dx = 3x + 52 23 x− x 2 3 b = a 27 19 −− 2 24 = 343 . 24 C05S08.033: Solution of 3y 2 = 12y − y 2 − 5 yields the limits of integration y = a = Hence the area of the region bounded by the given curves—shown next—is b A= a 4 (−5 + 12y − 4y 2 ) dy = −5y + 6y 2 − y 3 3 5 b = a 25 7 −− 6 6 = 1 2 16 . 3 and y = b = 5 2. 2.5 2 1.5 1 0.5 5 10 15 20 C05S08.034: We solve x2 = 4(x − 1)2 for x = a = 2 and x = b = 2 to find the limits of integration. The 3 area of the region bounded by the two given curves is b A= a b (8x − 4 − 3x2 ) dx = 4x2 − 4x − x3 a =0− − 32 27 = 32 . 27 C05S08.035: The only solution of x + 1 = 1/(x + 1) in the given interval I = [0, 1] is x = 0. The area between the two curves over I is thus 1 A= 0 x+1− 1 x+1 dx = x+ 12 x − ln(x + 1) 2 1 = 0 3 − ln 2 ≈ 0.80685281944005469. 2 The region bounded by the two curves over the interval I is shown next. 2 1.75 1.5 1.25 0.2 0.4 0.6 0.8 1 0.75 0.5 C05S08.036: It’s clear that the first two curves cross only at (0, 1). Hence the area of the region bounded by all three—shown below—is 1 A= 0 (x + 1 − e−x ) dx = 12 x + x + e−x 2 6 1 = 0 31 e+2 + −1= ≈ 0.8678794412. 2e 2e 2 1.5 1 0.5 -0.2 0.2 0.4 0.6 0.8 1 1.2 C05S08.037: It’s clear that the first two curves cross only at (0, 1). Thus the area of the region bounded by all three—shown next—is 1 A= 0 1 (ex − e−x ) dx = ex + e−x =e+ 0 1 (e − 1)2 −2= ≈ 1.0861612696. e e 3.5 3 2.5 2 1.5 1 0.5 -0.2 0.2 0.4 0.6 0.8 1 1.2 C05S08.038: It is clear that the first two curves cross only at (0, 1). Hence the area bounded by all three—shown next—is 10 A= 0 1 1 − x + 1 10x + 1 = ln 11 − dx = ln(x + 1) − 1 ln(10x + 1) 10 10 0 1 ln 101 ≈ 1.936383221. 10 1.75 1.5 1.25 1 0.75 0.5 0.25 2 4 7 6 8 10 C05S08.039: The following figure makes it clear that the area bounded by the three curves is 1 A= 0 x exp(−x2 ) dx = − 1 exp(−x2 ) 2 1 0 =− 1 1 e−1 += ≈ 0.3160602794. 2e 2 2e 0.4 0.3 0.2 0.1 -0.2 0.2 0.4 0.6 0.8 1 1.2 C05S08.040: Solving 8 =4−x x+2 shows that the two curves cross where x = 0 and where x = 2. Hence the area between them is 2 A= 0 4−x− 8 x+2 dx = 4x − 12 x − 8 ln(x + 2) 2 2 0 = 6 − 8 ln 2 ≈ 0.4548225555. 4.5 4 3.5 3 2.5 0.5 1 1.5 2 C05S08.041: The curves meet where x = −1 and where x = 1, so the area of the region they bound—shown next—is 1 1 1 (1 − x3 − x2 + x) dx = x + x2 − x3 − x4 2 3 4 −1 1 3 2 1 -1 -0.5 0.5 C05S08.042: First we solve 8 1 1 = −1 11 5 −− 12 12 = 4 . 3 x3 − x = 1 − x4 : x4 + x3 − x − 1 = 0; x3 (x + 1) − (x + 1) = 0; (x + 1)(x3 − 1) = 0; (x + 1)(x − 1)(x2 − x + 1) = 0; x = −1 or x = 1 (x2 − x + 1 = 0 has no real solutions). Hence the two curves meet at (−1, 0) and (1, 0). Therefore the area of the region they bound is A= 1 −1 (1 − x4 − x3 + x) dx = x + 12 14 15 x− x− x 2 4 5 1 = −1 21 11 −− 20 20 = 8 . 5 The region itself is shown next. 1 0.5 -1 -0.5 0.5 1 -0.5 -1 C05S08.043: We solve x2 = x3 − 2x to find x = −1, x = 0, and x = 2. So the two given curves meet at (−1, 1), (0, 0), and (2, 4), as shown in the following figure. The area of the region on the left is A1 = 0 −1 (x3 − 2x − x2 ) dx = 0 14 13 x − x − x2 4 3 −1 =0− − 5 12 The area of the region on the right is 2 A2 = 0 1 1 (2x + x2 − x3 ) dx = x2 + x3 − x4 3 4 Therefore the total area bounded by the two regions is A1 + A2 = 4 3 2 1 -1 -0.5 0.5 1 -1 9 1.5 2 37 . 12 2 = 0 8 . 3 = 5 . 12 C05S08.044: We solve x3 = 2x3 + x2 − 2x for x = −2, x = 0, and x = 1, indicating that the two given curves meet at the three point (−2, −8), (0, 0), and (1, 1), as shown in the following figure. The area of the region on the left is 0 A1 = −2 0 14 13 x + x − x2 4 3 (x3 + x2 − 2x) dx = −2 =0− − 8 3 = 8 . 3 The area of the region on the right is 1 A2 = 0 Therefore the total area of the two regions is A1 + A2 = -2 -1.5 -1 1 13 14 x− x 3 4 (2x − x2 − x3 ) dx = x2 − = 0 5 . 12 37 . 12 -0.5 0.5 1 -2 -4 -6 -8 C05S08.045: The first integral is I1 = 3 −3 3 4 4x(9 − x2 )1/2 dx = − (9 − x2 )3/2 3 = 0. −3 The second is I2 = 3 −3 5 9 − x2 dx = 5 3 −3 9 − x2 dx. √ Because the graph of y = 9 − x2 is a semicircle of radius 3, centered at the origin, and lying in the first and second quadrants, I2 is thus five times the area of such a semicircle, so that I2 = 5 · 1 45 · π · 32 = π, 2 2 and because I1 = 0, this is also the value of the integral given in Problem 45. C05S08.046: Let u = x2 , so that du = 2x dx and x dx = 3 I= 0 x(81 − x4 )1/2 dx = 1 2 9 0 1 2 du. Then (81 − u2 )1/2 du = 1 2 9 0 (81 − x2 )1/2 dx. (1) √ The graph of y = 81 − x2 is a quarter circle of radius 9 centered at (0, 0) and lying in the first quadrant, so the last expression in Eq. (1) is half the area of that quarter circle. Therefore I= 11 81 · · π · 92 = π. 24 8 10 C05S08.047: We solve the equation of the ellipse for y= b2 (a − x2 )1/2 , a and hence the area of the ellipse is given by a A=4 0 b2 4b (a − x2 )1/2 dx = a a a 0 (1) a2 − x2 dx. The last integral in Eq. (1) is the area of the quarter-circle in the first quadrant with center at the origin and radius a, and therefore A= 4b 1 2 · π a = π ab. a4 C05S08.048: The area bounded by the parabolic segment is 1 −1 13 x 3 (1 − x2 ) dx = x − 1 = −1 2 2 −− 3 3 = 4 . 3 The triangle has base 2 and height 1, so its area is 1, and the area of the parabolic segment is indeed the area of the triangle. 4 3 times C05S08.049: By solving the equations of the line and the parabola simultaneously, we find that A = (−1, 1) and B = (2, 4). The slope of the tangent line at C is the same as the slope 1 of the line through A and B , and it follows that C has x-coordinate 1 and thus y -coordinate 1 . It now follows that the distance from A 2 4 √ √ √ to B is AB = 3 2, the distance from B to C is BC = 3 29, and the distance from A to C is AC = 3 5. 4 4 Heron’s formula then allows us to find the area of triangle ABC ; it is the square root of the product of √ √ 3√ 4 2 + 5 + 29 , 8 − √ √ 3√ 3√ 5+ 4 2 + 5 + 29 , 4 8 The product can be simplified to is 729 64 , √ √ √ 3√ −3 2 + 4 2 + 5 + 29 , 8 and − √ √ 3√ 3√ 29 + 4 2 + 5 + 29 . 4 8 so the area of triangle ABC is 2 −1 (x + 2 − x2 ) dx = 27 8. The area of the parabolic segment 9 4 27 =·, 2 38 exactly as Archimedes proved in more general form over 2000 years ago. Mathematica did the arithmetic for us in this problem. If you prefer to do it by hand, show that the line through C perpendicular to the tangent line there has equation y = 3 − x. Show that this line meets the √ through AB at the point line 4 − 5 , 11 . Show that the perpendicular from C to that line has length h = 9 2. Show that AB has length 8 √8 8 3 2. Then triangle ABC has base AB and height h, so its area is 1 · 2 9√ 2 8 √ 27 ·3 2= . 8 C05S08.050: The area of the part of the region R lying over [1, b ] is 11 b 1 1 1 dx = − x2 x b 1 1 =1− . b When we evaluate the limit of this expression as b → + ∞, we find that the area of R is 1. We will return to this and related topics in Section 9.8. C05S08.051: The graph of the cubic y = 2x3 − 2x2 − 12x meets the x-axis at x = −2, x = 0, and x = 3. The graph is above the x-axis for −2 < x < 0 and below it for 0 < x < 3, so the graph of the cubic and the x-axis form two bounded plane regions. The area of the one on the left is 0 A1 = −2 (2x3 − 2x2 − 12x) dx = 14 23 x − x − 6x2 2 3 0 23 14 x− x 3 2 3 = −2 32 3 and the area of the one on the right is 3 A2 = 0 (12x + 2x2 − 2x3 ) dx = 6x2 + Therefore the total area required in Problem 51 is A1 + A2 = = 0 63 . 2 253 . 6 C05S08.052: On the one hand, the area in question is A= h 1312 px + qx + rx 3 2 (px2 + qx + r) dx = −h h = −h 23 ph + 2rh. 3 But 1 1 2 1 h [ f (−h) + 4f (0) + f (h) ] = h ph2 − qh + r + 4r + ph2 + qh + r = h(2ph2 + 6r) = ph3 + 2rh, 3 3 3 3 and this establishes the result sought in Problem 52. This problem figures significantly in the subsection on parabolic approximations in Section 5.9. C05S08.053: Given y 2 = x(5 − x)2 , the loop lies above and below the interval [0, 5], so by symmetry (around the x-axis) its area is 5 2 0 (5 − x)x1/2 dx = 2 2 (25x3/2 − 3x5/2 ) 15 5 0 =2· 20 √ 40 √ 5= 5. 3 3 C05S08.054: Given y 2 = x2 (x + 3), the loop lies above and below the interval [ −3, 0], so by symmetry (around the x-axis) its area is A=2 The minus sign is required because x < 0 and yields 3 A=2 0 √ 0 −3 √ −x x + 3 dx. x + 3 > 0 for −3 < x < 0. Next, the substitution u = x + 3 (3 − u)u1/2 du = 2 2u3/2 − 2 5 /2 u 5 12 3 0 √ 2√ =2· 2·3 3− ·9 3 5 = 24 √ 3. 5 C05S08.055: We applied Newton’s method to the equation x2 − cos x = 0 to find that the two curves y = x2 and y = cos x meet at the two points (−0.824132, 0.679194) and (0.824132, 0.679194) (numbers involving decimals are approximations). Let a = −0.824132 and b = 0.824132. The region between the two curves (shown next) has area b a (cos x − x2 ) dx = sin x − 13 x 3 b a ≈ 1.09475. 1 0.8 0.6 0.4 0.2 -1 -0.5 0.5 1 C05S08.056: We applied Newton’s method to f (x) = sin x − x2 + 2x to find the positive solution b = 2.316934 of f (x) = 0 (numbers involving decimals are approximations). Clearly a = 0 is the other solution; the points of intersection of the two graphs are (0, 0) and (2.316934, 0.734316). The area bounded by the two graphs is b a 1 (sin x − x2 + 2x) dx = x2 − x3 − cos x 3 b a ≈ 2.90108. The region bounded by the two curves is shown next. 3 2 1 -1 1 2 3 -1 C05S08.057: We used Newton’s method to solve f (x) = 0 where f (x) = 1 − x2 + 1, 1 + x2 and found the points at which the two curves intersect to be (−1.189207, 0.414214) and (1.189207, 0.414214) (numbers involving decimals are approximations). With b = 1.189207 and a = −b, the area bounded by the two curves (shown next) is b A= a 1 − x2 + 1 1 + x2 1 dx = arctan(x) − x3 + x 3 13 b a ≈ 3.00044 (the antiderivative was computed with the aid of Formula 17 of the endpapers). 1 0.5 -1.5 -1 -0.5 0.5 1 1.5 -0.5 -1 C05S08.058: We used Newton’s method with f (x) = 2x − x2 − x4 + 16 to find the negative solution x = a of f (x) = 0; it turns out that the corresponding point where the two curves meet is (−1.752172, −6.574449) (numbers involving decimals are approximations). The other solution is b = 2, so the curves also meet at (2, 0). The area between them is thus b a (2x − x2 − x4 + 16) dx = x2 − 13 15 x − x + 16x 3 5 b a ≈ 46.8018. The region bounded by the two curves is shown next. 5 -2 -1 1 2 -5 -10 -15 C05S08.059: The curves y = x2 and y = k − x2 meet at the two points where x = a = − (k/2)1/2 and x = b = (k/2)1/2 . So the area between the two curves is b a (k − 2x2 ) dx = k x − b 23 x 3 = a √ 2 2 3/2 k. 3 When we set the last expression equal to 72, we find that k = 18. C05S08.060: By symmetry, it is sufficient to work in the first quadrant. Then the curves y = k and √ y = 100 − x2 cross at the point x = b = 100 − k . When we solve the equation b 0 100 − x2 − k dx = kb + 10 b 100 − x2 dx for k , we find that k = 50 2 − 21/3 . C05S08.061: We are interested in the region (or regions) bounded by the graphs of the two functions f (x) = x and g (x) = x(x − 4)2 . First we plot the two curves, using Mathematica 3.0, to guide our future computations. 14 Plot[ { f[x], g[x] }, { x, −1, 6 }, PlotRange → {− 1, 11 } ]; 10 8 6 4 2 -1 1 2 3 4 5 6 It is easy to see that the curves cross where x = 0, x = 3, and x = 5. Note that g (x) f (x) if 0 x 3 and that f (x) g (x) if 3 x 5. Hence we compute the values of two integrals and add the results to get the total area of the (bounded) regions bounded by the two curves. a1 = Integrate[ g[x] - f[x], { x, 0, 3 } ] 63 4 a2 = Integrate[ g[x] - f[x], { x, 3, 5 } ] 16 3 a1 + a2 253 12 N[ %, 20 ] 21.083333333333333333 C05S08.062: With f (x) = x2 and g (x) = x(x − 4)2 , we can easily use Mathematica 3.0 to find where the graphs cross and to plot the graphs. Solve[ f[x] == g[x], x ] {{ x → 0 }, { x → 9− √ 2 17 }, { x → 9+ √ 2 15 17 }} Plot[ { f[x], g[x] }, {x, −2, 8 }, PlotRange → {− 4, 54 } ]; 50 40 30 20 10 -2 2 4 6 8 As in the solution of Problem 61, two integrals are required to find the total area bounded by the two curves. r1 = 0; r2 = (9 − Sqrt[17])/2; r3 = (9 + Sqrt[17])/2; a1 = Integrate[ g[x] - f[x], { x, r1, r2 } ] √ 51 17 − 107 8 a2 = Integrate[ f[x] - g[x], { x, r2, r3 } ] √ 51 17 4 a1 + a2 // Together √ 153 17 − 107 8 N[ %, 20 ] 65.479395089937758902 C05S08.063: We are given f (x) = (x − 2)2 and g (x) = x(x − 4)2 . We begin by plotting the graphs of both functions. Plot[ { f[x], g[x] }, { x, −2, 6 }, PlotRange → {− 4, 20 } ]; 20 15 10 5 -2 2 16 4 6 √ The Solve command in Mathematica returns exact solutions all of which include the number i = −1 , but their numerical values are numbers such as 5.48929 + 10−51 i, so all solutions are pure real, as indicated in the preceding graph. We entered the numerical approximations to the roots: r1 = 0.22154288174161126812; r2 = 3.28916854644830996908; r3 = 5.48928857181007876279; Then we computed two integrals and added the results to find the total area bounded by the two curves. a1 = Integrate[ g[x] − f[x], { x, r1, r2 } ] 17.96479813911499075801 a2 = Integrate[ f[x] − g[x], { x, r2, r3 } ] 7.39746346656350500268 a1 + a2 25.36226160567849576069 C05S08.064: Given f (x) = 5 ln x − 2x + 3, we first plotted the graph of f : Plot[ f[x], { x, 0.5, 7 } ]; 2 1 1 2 3 4 5 6 7 -1 7 Beginning with initial estimates x0 = 10 and x0 = 6, we then applied Newton’s method, initially carrying 40 significant figures, to approximate the solutions of f (x) = 0 very accurately. Five iterations produced the results r1 = 0.73696591726375532670; r2 = 5.96460858839295659664; (accurate to the number of digits shown) and a single integral yielded the (approximate) value of the area: Integrate[ f[x], { x, r1, r2 } ] 8.89522349387660371502 C05S08.065: Given f (x) = 10 ln x and g (x) = (x − 5)2 , we first plotted the graphs of both functions: 17 Plot[ { f[x], g[x] }, { x, 0.7, 10.5 } ]; 30 25 20 15 10 5 2 4 6 8 10 We then applied Newton’s method, carrying 50 digits initially, to approximate the solutions of f (x) = g (x). The initial guesses x0 = 2.1 and x0 = 9.8 produced (after seven iterations) the following solutions, accurate to the number of digits shown: r1 = 2.19561040829474667642305282648227; r2 = 9.77472510546165546062557839624217; and a single integral yields a very good approximation to the area bounded by the two curves: Integrate[ f[x] − g[x], { x, r1, r2 } ] 86.148905476732141854653048449271365936 C05S08.066: Let f (x) = ex and g (x) = 10(1 + 5x − x2 ). First we plotted the graphs of f and g : Plot[ { f[x], g[x] }, { x, −1, 4.5 } ]; 80 60 40 20 -1 1 2 3 4 -20 -40 Clearly the graphs cross near x = −0.2 and near x = 4.0. Using these numbers as initial estimates of the solutions of f (x) = g (x), seven iterations of Newton’s method—carrying 60 digits initially—yielded the two solutions, accurate to the number of digits shown here: r1 = −0.17697970125714609944601323479679; r2 = 3.94429748860047291454039853228598; Finally, a single integration yields a very accurate approximation to the area bounded by the two curves: 18 Integrate[ g[x] − f[x], { x, r1, r2 } ] 174.00159869810014633984111106677060 C05S08.067: Given f (x) = e−x/2 and g (x) = x4 − 6x2 − 2x + 4, we first plotted the graphs of f and g : Plot[ { f[x], g[x] }, { x, −2.5, 2.7 } ]; 10 5 -2 -1 1 2 -5 Clearly the graphs cross at four points, near where x = −2.2, −0.9, 0.6, and 2.5. Using these estimates as initial values, seven iterations of Newton’s method (carrying 60 digits initially) yields these four solutions of f (x) = g (x), accurate to the number of digits shown here: r1 = −2.21286965649560596689486746473117; r2 = −0.90416380610482135626736226590897; r3 = 0.60360685508638066445123518977576; r4 = 2.49090865808642591763321220641991; Then three integrations, followed by the sum of the results, gives a very accurate approximation to the total area bounded by the graphs of f and g : a1 = Integrate[ f[x] − g[x], { x, r1, r2 } ] 3.29201409095125227617309525829577 a2 = Integrate[ g[x] − f[x], { x, r2, r3 } ] 3.03877187423119505551620253633379 a3 = Integrate[ f[x] − g[x], { x, r3, r4 } ] 10.50223144411747460353190208812036 a1 + a2 + a3 16.83301740929992193522119988274992 19 Section 5.9 C05S09.001: With ∆x = 1, f (x) = x, n = 4, and xi = i · ∆x, we have Tn = ∆x · f (x0 ) + f (xn ) + 2 · 2 n−1 f (xi ) = 8, i=1 which is also the true value of the given integral. C05S09.002: With ∆x = 0.2, f (x) = x2 , n = 5, and xi = 1 + i · ∆x, we have Tn = ∆x · f (x0 ) + f (xn ) + 2 · 2 n−1 f (xi ) = i=1 117 = 2.34. 50 7 ≈ 2.333333. 3 √ C05S09.003: With ∆x = 0.5, f (x) = x, n = 5, and xi = i · ∆x, we have The exact value of the integral is Tn = ∆x · f (x0 ) + f (xn ) + 2 · 2 The exact value of the integral is ≈ 0.6497385976 ≈ 0.65. 1 , n = 4, and xi = 1 + i · ∆x, we have x2 ∆x · f (x0 ) + f (xn ) + 2 · 2 The exact value of the integral is f (xi ) i=1 2 ≈ 0.666667. 3 C05S09.004: With ∆x = 0.5, f (x) = Tn = n−1 n−1 f (xi ) = i=1 141 = 0.705000 ≈ 0.71. 200 2 ≈ 0.666667. 3 C05S09.005: With ∆x = π /6, f (x) = cos x, n = 3, and xi = i · ∆x, we have Tn = ∆x · f (x0 ) + f (xn ) + 2 · 2 n−1 f (xi ) = i=1 √ π 2+ 3 12 ≈ 0.9770486167 ≈ 0.98. The exact value of the integral is 1. C05S09.006: With ∆x = π /4, f (x) = sin x, n = 4, and xi = i · ∆x, we have Tn = ∆x · f (x0 ) + f (xn ) + 2 · 2 n−1 f (xi ) i=1 = √ π 1+ 2 4 ≈ 1.8961188979 ≈ 1.90. The exact value of the integral is 2. C05S09.007: With ∆x = 1, f (x) = x, n = 4, and mi = i − 1 2 · ∆x, we have n Mn = (∆x) · f (mi ) = 8. i=1 1 The exact value of the integral is also 8. C05S09.008: With ∆x = 0.2, f (x) = x2 , n = 5, and mi = 1 + i − n Mn = (∆x) · f (mi ) = i=1 · ∆x, we have 1 2 233 = 2.33. 100 7 ≈ 2.333333. 3 √ C05S09.009: With ∆x = 0.2, f (x) = x, n = 5, and mi = i − The exact value of the integral is · ∆x, we have 1 2 n Mn = (∆x) · The exact value of the integral is i=1 f (mi ) ≈ 0.6712800859 ≈ 0.67. 2 ≈ 0.666667. 3 C05S09.010: With ∆x = 0.5, f (x) = 1 , n = 4, and mi = 1 + i − x2 n Mn = (∆x) · The exact value of the integral is f (mi ) = i=1 · ∆x, we have 1 2 7781792 ≈ 0.65. 12006225 2 ≈ 0.666667. 3 C05S09.011: With ∆x = π /6, f (x) = cos x, n = 3, and mi = i − n Mn = (∆x) · f (mi ) = i=1 · ∆x, we have 1 2 √ π√ 2+ 6 12 ≈ 1.01. The exact value of the integral is 1. C05S09.012: With ∆x = π /4, f (x) = sin x, n = 4, and mi = i − n Mn = (∆x) · f (mi ) = i=1 π 4 sin 1 2 · ∆x, we have π 3π 5π 7π + sin + sin + sin 8 8 8 8 ≈ 2.05. The exact value of the integral is 2. C05S09.013: With ∆x = 0.5, n = 4, f (x) = x2 , and xi = 1 + i · ∆x, we have Tn = ∆x · 2 f (x0 ) + f (xn ) + 2 · n−1 i=1 f (xi ) = 35 = 8.75 4 and Sn = ∆x 26 · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + · · · + 4f (xn−1 ) + f (xn ) ] = ≈ 8.67. 3 3 The true value of the integral is also 26 (see Problem 29). 3 C05S09.014: With a = 1, b = 5, n = 4, ∆x = 1, xi = 1 + i · ∆x, and f (x) = 2 1 , we obtain x Tn = ∆x · f (x0 ) + f (xn ) + 2 · 2 n−1 f (xi ) = i=1 101 ≈ 1.6833333333 60 and Sn = ∆x 73 · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 ) ] = ≈ 1.6222222222. 3 45 5 The true value is 1 5 1 dx = x ln x 1 = ln 5 ≈ 1.6094379124. C05S09.015: With a = 0, b = 2, n = 4, ∆x = 0.5, f (x) = e−x , and xi = i · ∆x, we obtain Tn = n−1 ∆x · f (x0 ) + f (xn ) + 2 · 2 f (xi ) i=1 ≈ 0.8826039513 and Sn = ∆x · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 ) ] ≈ 0.8649562408. 3 2 The true value of the integral is 2 e−x dx = − e−x 0 C05S09.016: With ∆x = 0.25, n = 4, f (x) = Tn = ∆x · 2 √ 0 =1− 1 ≈ 0.8646647168. e2 1 + x, and xi = i · ∆x, we have f (x0 ) + f (xn ) + 2 · n−1 f (xi ) i=1 1.2181903242 ≈ 1.22 and Sn = ∆x · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + · · · + 4f (xn−1 ) + f (xn ) ] ≈ 1.2189451569 ≈ 1.22 3 The true value of the integral is C05S09.017: With ∆x = Tn = 2√ 2 2 − 1 ≈ 1.2189514165 ≈ 1.22. 3 √ 1 , n = 6, f (x) = 1 + x3 , and xi = i · ∆x, we have 3 ∆x · 2 f (x0 ) + f (xn ) + 2 · n−1 i=1 f (xi ) ≈ 3.2598849023 ≈ 3.26 and Sn = ∆x · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + · · · + 4f (xn−1 ) + f (xn ) ] ≈ 3.2410894400 ≈ 3.24. 3 The true value of the integral is approximately 3.24131. The antiderivative of f is known to be a nonelementary function. C05S09.018: With ∆x = 0.5, n = 6, f (x) = 1 , and xi = i · ∆x, we have 1 + x4 3 Tn = ∆x · 2 f (x0 ) + f (xn ) + 2 · n−1 f (xi ) 22392745 ≈ 1.0980035065 ≈ 1.10 20394056 = i=1 and Sn = ∆x 192249939 · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + · · · + 4f (xn−1 ) + f (xn ) ] = ≈ 1.1090309786 ≈ 1.11. 3 173349476 The true value of the integral is approximately 1.0984398680. The antiderivative of f is an elementary function but is not easy to find by hand; techniques of the later sections of Chapter 7 are required. C05S09.019: With a = 1, b = 5, n = 8, ∆x = 0.5, f (x) = (1 + ln x)1/3 , and xi = 1 + i · ∆x, we obtained Tn = ∆x · 2 f (x0 ) + f (xn ) + 2 · n−1 f (xi ) i=1 ≈ 5.0139700316 and Sn = ∆x · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + · · · + 4f (xn−1 ) + f (xn ) ] ≈ 5.0196759933. 3 The true value of the integral is approximately 5.02005. C05S09.020: With a = 0, b = 1, n = 10, ∆x = 0.1, f (x) = we obtained Tn = ∆x · 2 f (x0 ) + f (xn ) + 2 · n−1 i=1 ex − 1 if 0 < x x f (xi ) 1, f (0) = 1, and xi = i · ∆x, ≈ 1.3183187746 and Sn = ∆x · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + · · · + 4f (xn−1 ) + f (xn ) ] ≈ 1.3179023254. 3 The true value of the integral is approximately 1.31790. C05S09.021: With ∆x = 0.25 and n = 6, we have Tn = ∆x [3.43 + 2 · (2.17 + 0.38 + 1.87 + 2.65 + 2.31) + 1.97 ] = 3.02 2 and Sn = ∆x [3.43 + 4 · (2.17 + 1.87 + 2.31) + 2 · (0.38 + 2.65) + 1.97 ] ≈ 3.07167. 3 C05S09.022: With ∆x = 1 and n = 10, we have Tn = ∆x [23 + 2 · (8 − 4 + 12 + 35 + 47 + 53 + 50 + 39 + 29) + 5 ] = 283 2 and Sn = ∆x [23 + 4 · (8 + 12 + 47 + 50 + 29) + 2 · (−4 + 35 + 53 + 39) + 5 ] = 286. 3 4 C05S09.023: We read the following data from Fig. 5.9.16: 0 1 2 3 4 5 6 7 8 9 10 250 300 320 327 318 288 250 205 158 110 80 With ∆x = 1 and n = 10, we have Tn = ∆x [250 + 2 · (300 + 320 + 327 + 318 + 288 + 250 + 205 + 158 + 110) + 80 ] = 2441 2 and Sn = ∆x 7342 [250 + 4 · (300 + 327 + 288 + 205 + 110) + 2 · (320 + 318 + 250 + 158) + 80 ] = ≈ 2447.33. 3 3 C05S09.024: We read the following data from Fig. 5.9.17: 0 3 6 9 12 15 18 21 24 27 30 19 15 13 20 24 18 13 8 4 9 16 With ∆x = 3 and n = 10, we have Tn = ∆x 849 [19 + 2 · (15 + 13 + 20 + 24 + 18 + 13 + 8 + 4 + 9) + 16 ] = . 2 2 Divide by 30 to obtain an approximation to the average value of the temperature over the 30-day period; the result is 14.15. Next, ∆x [19 + 4 · (15 + 20 + 18 + 8 + 9) + 2 · (13 + 24 + 13 + 4) + 16 ] = 423; 3 Divide by 30 to get an average temperature of 14.10. Sn = C05S09.025: With ∆x = 50 and n = 12, we find Tn = ∆x [0 + 2 · (165 + 192 + 146 + 63 + 42 + 84 + 155 + 224 + 270 + 267 + 215) + 0 ] = 91150. 2 This result is in square feet. Divide by 9 to convert to square yards, then divide by 4840 to convert to acres; the result is approximately 2.093 acres. Next, Sn = ∆x 281600 [0 + 4 · (165 + 146 + 42 + 155 + 270 + 215) + 2 · (192 + 63 + 84 + 224 + 267) + 0 ] = . 3 3 As before, divide by 9 and by 4840 to obtain the estimate 2.155 acres. C05S09.026: With a = 1, b = 2.7, n = 100, ∆x = (b − a)/n, xi = 1 + i · ∆x, and f (x) = Sn = 1 , we find that x ∆x [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn ) ] ≈ 0.993252 < 1. 3 The same computation with b = 2.8 yields Sn ≈ 1.02962 > 1, and this is enough to show that 2.7 < e < 2.8. We can be sure of both inequalities because Simpson’s error estimate indicates that in each case the error is less than 24 · (2.8 − 1)5 ≈ 2.52 × 10−8 . 180 · 1004 5 2 C05S09.027: We have f (x) = 3 , so the constant in the error estimate is K2 = 2. With a = 1, b = 2, x and n subintervals, we require K2 (b − a)3 < 0.0005, 12n2 which implies that n2 > 333.33 and thus that n > 18.2; n = 19 will suffice. (In fact, with only n = 13 subintervals, the trapezoidal estimate of 0.6935167303 differs from the true value of ln 2 by less than 0.00037.) C05S09.028: We have f (4) (x) = 24x−5 , so the constant in Simpson’s error estimate is K4 = 24. With a = 1, b = 2, and n subintervals, we require K4 (b − a)5 < 0.000005, 180n4 which implies that n2 > 26666.7 and thus that n > 12.78. Because n must be even, the answer is that n should be 14. (With n = 14 subintervals, the difference between S14 ≈ 0.6931479839 and the true value of ln 2 is less than 8.1 × 10−7 .) C05S09.029: If p(x) = ax3 + bx2 + cx + d is a polynomial of degree 3 or smaller, then p(4) (x) ≡ 0, so the constant K4 in Simpson’s error estimate is zero, which implies that the error will also be zero no matter what the value of the even positive integer n may be. C05S09.030: Let f (x) = (4 − 2x) − (6x2 − 7x) = 4+5x − 6x2 . Choose a = − 1 , b = 4 , n = 2, ∆x = (b − a)/n, 2 3 and xi = i · ∆x. Then S2 = ∆x 1331 [ f (x0 ) + 4f (x1 ) + f (x2 ) ] = . 3 216 C05S09.031: With the usual meanings of the notation, we have Mn + Tn = (∆x) · [ f (m1 ) + f (m2 ) + f (m3 ) + · · · + f (mn ) ] + = ∆x · [ f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn ) ] 2 ∆x · [ f (x0 ) + 2f (m1 ) + 2f (x1 ) + 2f (m2 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + 2f (mn ) + f (xn ) ] = 2T2n . 2 The result in Problem 31 follows immediately. C05S09.032: With α = 10◦ (which we convert to radians) and f (x) = 1 1 − (k sin x)2 , we use n = 10, a = 0, b = π /2, ∆x = (b − a)/n, and xi = i · ∆x and obtain Sn = ∆x · [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn ) ] ≈ 1.5737921309. 3 We multiply by 4(L/g )1/2 to find the period T of the pendulum to be approximately 2.0109178213 (seconds). The same computation with α = 50◦ yields T ≈ 2.1070088018 (seconds). The limiting value of T as α → 0 is approximately 2.00709. 6 C05S09.033: Suppose first that f (x) > 0 and f (x) > 0 for a x b. Then the graph of f is concave upward on [ a, b ]. Now examine Fig. 5.9.11, where it is shown that the midpoint approximation is the same as the tangent approximation. The tangent line will lie under the graph of f because the graph is concave upward; the chord connecting (xi−1 , f (xi−1 )) and (xi , f (xi )) will lie over the graph. Hence every term in the midpoint sum will underestimate the area under the graph of f and every term in the trapezoidal sum will overestimate it. Thus b Mn < f (x) dx < Tn a no matter what the choice of n. If f (x) < 0 on [ a, b ], then Figs. 5.9.11 and 5.9.12 show that the inequalities will be reversed. C05S09.034: Simpson’s approximation to the given integral is √ 1+4 e+e 10 1/2 1 S2 = e + 4e + e = . 6 6 Therefore √ 1+4 e+e ≈ e − 1, 6 √ and this leads to the approximation 5e − 4 e − 7 ≈ 0. We solve for e: √ 5e − 7 ≈ 4 e; 25e2 − 70e + 49 ≈ 16e; 25e2 − 86e + 49 ≈ 0; √ 1 e ≈ 50 86 + 2496 = 1 25 √ 43 + 4 39 ; e ≈ 2.71919967974. Therefore e ≈ 2.7192. In Chapter 9 you will see ways to approximate e with much more precision and with very little additional work. C05S09.035: We estimate the integral 100000 90000 1 dx first with the midpoint rule: ln x 10000 ≈ 872.476; ln(95000) then with the trapezoidal rule: (5000) 1 1 + ln(90000) ln(100000) ≈ 872.600. The first is an underestimate and the second is an overestimate because the graph of y = 1/(ln x) is concave upward for x > 0. The true value of the integral, incidentally, is approximately 872.5174045. 7 Chapter 5 Miscellaneous Problems C05S0M.001: 5 1 (5x−3 − 2x−2 + x2 ) dx = − x−2 + 2x−1 + x3 + C . 2 3 C05S0M.002: (x1/2 + 3x + 3x3/2 + x2 ) dx = C05S0M.003: (1 − 3x)9 dx = − C05S0M.004: 7 7(2x + 3)−3 dx = − (2x + 3)−2 + C . 4 C05S0M.005: (9 + 4x)1/3 dx = C05S0M.006: 24(6x + 7)−1/2 dx = 8(6x + 7)1/2 + C . C05S0M.007: x3 (1 + x4 )5 dx = C05S0M.008: 3x2 (4 + x3 )1/2 dx = C05S0M.009: 3 x(1 − x2 )1/3 dx = − (1 − x2 )4/3 + C . 8 C05S0M.010: 3x(1 + 3x2 )−1/2 dx = (1 + 3x2 )1/2 + C . C05S0M.011: (7 cos 5x − 5 sin 7x) dx = C05S0M.012: 5 sin3 4x cos 4x dx = 2 3 / 2 3 2 6 5 /2 1 3 x + x + x + x + C. 3 2 5 3 1 (1 − 3x)10 + C . 30 3 (9 + 4x)4/3 + C . 16 1 (1 + x4 )6 + C . 24 2 (4 + x3 )3/2 + C . 3 1 (25 cos 7x + 49 sin 5x) + C . 35 5 sin4 4x + C . 16 C05S0M.013: If u = x4 , then du = 4x3 dx, so that x3 dx = x3 (1 + x4 )1/2 dx = 1 du. Hence 4 1 12 1 (1 + u)1/2 du = · (1 + u)3/2 + C = (1 + x4 )3/2 + C. 4 43 6 C05S0M.014: If u = sin x then du = cos x dx. Hence sin2 x cos x dx = C05S0M.015: If u = 1 + x1/2 then du = x−1/2 1+ 2 x1/2 dx = u2 du = 13 1 u + C = sin3 x + C. 3 3 1 −1/2 dx, so that x−1/2 dx = 2 du. Therefore x 2 2 du = u2 2u−2 du = −2u−1 + C = − 1 2 √ + C. 1+ x 1 −1/2 dx, so that x−1/2 dx = 2 du. Hence x 2 C05S0M.016: If u = x1/2 then du = x−1/2 1+ 2 x1/2 dx = 2 du = (1 + u)2 2(1 + u)−2 du = −2(1 + u)−1 + C = − 2 √ + C. 1+ x 1 du. Hence 12 1 1 1 cos u du = sin u + C = sin 4x3 + C. 12 12 12 C05S0M.017: If u = 4x3 then du = 12x2 dx, so that x2 dx = x2 cos 4x3 dx = C05S0M.018: If u = x + 1, then x = u − 1 and dx = du. Hence x(x + 1)14 dx = (u − 1)u14 du = (u15 − u14 ) du = 1 16 1 15 1 1 u − u +C = (x + 1)16 − (x + 1)15 + C. 16 15 16 15 1 du. Thus 2 1 15 12 du = u +C = (x + 1)15 + C. 30 30 C05S0M.019: If u = x2 + 1, then du = 2x dx, so that x dx = 1 14 u 2 x(x2 + 1)14 dx = An unpleasant alternative is to expand (x2 + 1)14 using the binomial formula, multiply by x, then integrate the resulting polynomial to obtain 1 30 1 28 7 26 91 24 91 22 1001 20 1001 18 x+x+x+ x+ x+ x+ x 30 2 2 6 2 10 6 + 429 14 1001 12 1001 10 91 8 91 6 7 4 1 2 x+ x+ x + x + x + x + x + C. 2 6 10 2 6 2 2 1 du. Then 4 1 1 1 cos u du = sin u + C = sin x4 + C. 4 4 4 C05S0M.020: If u = x4 , then du = 4x3 du, so that x3 dx = x3 cos x4 dx = C05S0M.021: If u = 4 − x then x = 4 − u and dx = − du. Hence x(4 − x)1/2 dx = = (u − 4)u1/2 du = (u3/2 − 4u1/2 ) du 2 5 / 2 8 3 /2 2 8 u − u + C = (4 − x)5/2 − (4 − x)3/2 + C. 5 3 5 3 C05S0M.022: If u = x4 + x2 , then du = (4x3 + 2x) dx, so that (x + 2x3 ) dx = (x + 2x3 )(x4 + x2 )−3 dx = 1 du. Thus 2 1 −3 1 1 + C. u du = − u−2 + C = − 4 + x2 )2 2 4 4(x C05S0M.023: If u = x4 , then du = 4x3 dx, so that 2x3 dx = 2 1 du. Therefore 2 1 (1 + u)−1/2 du = (1 + u)1/2 + C = 2 2x3 (1 + x4 )−1/2 dx = 1 + x4 + C. C05S0M.024: If u = x2 + x, then du = (2x + 1) dx, so that (x2 + x)−1/2 (2x + 1) dx = x C05S0M.025: y (x) = u−1/2 du = 2u1/2 + C = 2 x2 + x + C. x (3t2 + 2t) dt + 5 = t3 + t2 + 5 = x3 + x2 + 5. 0 0 x C05S0M.026: y (x) = x 3t1/2 dt + 20 = 2t3/2 4 C05S0M.027: If + 20 = 2x3/2 − 16 + 20 = 2x3/2 + 4. 4 dy 1 = (2x + 1)5 , then y (x) = (2x + 1)6 + C . Then dx 12 1 23 2 = y (0) = + C implies that C = , 12 12 and therefore y (x) = 1 23 (2x + 1)6 + . 12 12 x C05S0M.028: y (x) = x 2(t + 5)−1/2 dt + 3 = 4(t + 5)1/2 4 4 x C05S0M.029: y (x) = 1 x C05S0M.030: y (x) = 0 t−1/3 dt + 1 = 3 2 /3 t 2 x +1= 1 √ + 3 = 4(x + 5)1/2 − 12 + 3 = 4 x + 5 − 9. 3 2 /3 3 3x2/3 − 1 x − +1= . 2 2 2 x (1 − cos t) dt = t − sin t 0 = x − sin x. C05S0M.031: First convert 90 mi/h to 132 ft/s (just multiply by 22 ). Let x(t) denote the distance (in 15 feet) the automobile travels after its brakes are first applied at time t = 0 (s). Then x(t) = −11t2 + 132t, so the automobile first comes to a stop when v (t) = x (t) = −22t + 132 = 0; that is, when t = 6. Therefore the total distance it travels while braking will be x(6) = 396 (ft). C05S0M.032: If the stone is dropped at time t = 0 (s), then its altitude at time t will be y (t) = −11250t2 + 450 (ft). The stone reaches the ground when y (t) = 0, so that t = 1 . Its impact velocity will be x (t) = v (t) = −22500t 5 evaluated when t = 1 ; that is, v 1 = −4500 (ft/s). Thus the stone remains aloft for 0.2 s and its impact 5 5 speed will be 4500 ft/s. C05S0M.033: Let v0 denote the initial velocity of the automobile and let x(t) denote the distance it has skidded since its brakes were applied at time t = 0 (units are in feet and seconds). Then x(t) = −20t2 + v0 t. Let T denote the time at which the automobile first comes to a stop. Then x(T ) = 180 and That is, 3 x (T ) = 0. −20T 2 + v0 T = 180 and − 40T + v0 = 0. The second of these equations implies that v0 = 40T , and substitution of this datum in the first of these equations yields −20T 2 + 40T 2 = 180, so that T 2 = 9. Thus T = 3, and therefore v0 = 120. Hence the initial velocity of the automobile was 120 feet per second, slightly less than 82 miles per hour (slightly less than 132 kilometers per hour). C05S0M.034: Assume that the car begins its journey at time t = 0 and at position x = 0 (units are in feet and seconds). Then x(t) = 4t2 because both initial position and initial velocity are zero. It will reach a speed of 60 miles per hour—that is, 88 feet per second—when x (t) = 88; that is, when y = 11. At that point the car will have traveled a distance of x(11) = 484 feet. C05S0M.035: Let q denote the acceleration of gravity on the planet Zorg. First consider the ball dropped from a height of 20 feet. Then its altitude at time t will be y (t) = − 1 qt2 + 20 (because its initial velocity 2 is v0 = 0). Then the information that x(2) = 0 yields the information that q = 10. Now suppose that the ball is dropped from an initial height of 200 feet. Then its altitude√ time t will be y (t) = −5t2 + 200, so at the ball will reach then ground when 5t2 = 200; that is, when t = 2 10. Its velocity during its descent will √ √ be v (t) = y (t) = −10t, so its impact velocity will be v 2 10 = −20 10 feet per second. Thus its impact √ √ speed will be | v 2 10 | = 20 10 ≈ 63.25 feet per second. C05S0M.036: First we need to find the initial velocity v0 you can impart to the ball that you throw straight upward. Let y (t) denote the altitude of the ball at time t (units are in feet and seconds), with the throw occurring at time t = 0. Then y (t) = −16t2 + v0 t (we make the simplifying assumption that y0 = y (0) = 0). The ball reaches its maximum height when its velocity v (t) = y (t) = −32t + v0 is zero; let T denote the time at which that event occurs. Then x(T ) = 144 when v (T ) = 0; that is, we must solve simultaneously the equations −16T 2 + v0 T = 144 and − 32T + v0 = 0. The second of these equations yields v0 = 32T , and substitution in the first yields −16T 2 + 32T 2 = 144, so that T = 3, and thus v0 = 96. We now assume that you can impart the same initial velocity to the ball on both Zorg and Mesklin (the latter assumption is surely invalid, but it’s clearly the intent of the problem). On Zorg, the altitude function of the ball will be y (t) = −5t2 + 96t, so its velocity will be v (t) = −10t + 96. The ball will reach its maximum altitude when v (t) = 0, thus when t = 9.6. Thus the maximum altitude reached by the ball will be y (9.6) = 460.8 (feet). On Mesklin, the altitude function of the ball will be y (t) = −11250t2 + 96t, so its velocity at time t will be v (t) = −22500t + 96. The ball reaches its maximum altitude when v (t) = 0, 8 8 so that t = 1875 ≈ 0.004267 (seconds). Its maximum altitude will therefore be y 1875 = 128 = 0.2048 feet; 625 that is, only 2.4576 inches, a little less than 6.25 centimeters. C05S0M.037: First we need to find the deceleration constant a of the car. Let x(t) denote the distance (in feet) the car has skidded at time t (in seconds) if its brakes are applied at time t = 0. Then x(t) = − 1 at2 +44t. 2 (We converted 30 miles per hour to 44 feet per second.) Thus in the first skid, if the car first comes to a stop at time T , then both x(T ) = 44 and 4 x (T ) = 0. Because the velocity of the car is v (t) = x (t) = −at + 44, we must solve simultaneously the equations 1 − aT 2 + 44T = 44 2 The second of these equations yields T = and − aT + 44 = 0. 44 , then substitution in the first equation yields a − 1 442 442 · + = 44; 2a a − 22 44 + = 1; a a 22 = 1; a a = 22. Now suppose that the initial velocity of the car is 60 miles per hour; that is, 88 feet per second. With the same meaning of x(t) as before, we now have x(t) = −11t2 + 88t. Thus v (t) = x (t) = −22t + 88, so the car comes to a stop when t = 4. The distance it now skids will be x(t) = 176 feet. The point of the problem is that doubling the initial speed of the car quadruples the stopping distance. C05S0M.038: When the fuel is exhausted, the acceleration of the rocket is that solely due to gravity, so the velocity stops increasing and begins to decrease. In Fig. 5.MP.1 this appears to occur about at time t = 1.8. When the parachute opens, the velocity stops decreasing because of the upward force due to the parachute; this appears to occur close to time t = 3.2. The rocket reached its maximum altitude when its velocity was zero, close to time t = 2.8. The rocket landed at time t = 5. Its maximum height was about 240 feet and the pole atop which it landed was about 110 feet high. One of the more interesting aspects of this problem is that during the free fall of the rocket, its velocity changed from about 200 feet per second at time t = 1.8 to about −80 feet per second at time t = 3.2, implying that the acceleration of gravity on the planet where this all took place is about 200 feet per second per second. 100 100 17 = 17 · C05S0M.039: i=1 100 1 = 17 · 100 = 1700. 1 1 − k k+1 C05S0M.040: k=1 =1− i=1 = 1− 1 2 + 11 − 23 + 11 − 34 +···+ 1 1 − 99 100 + 1 1 − 100 101 1 100 = . 101 101 10 C05S0M.041: n=1 10 (3n − 2)2 = 9 · 16 sin C05S0M.042: n=1 n=1 10 n2 − 12 · n=1 10 n+4· n=1 1 = 9· 10 · 11 · 21 10 · 11 − 12 · + 4 · 10 = 2845. 6 2 nπ = 4 · (1 + 0 − 1 + 0) = 0. 2 n C05S0M.043: On [1, 2], we have lim n→∞ i=1 ∆x = xi 2 1 √ 1 √ dx = 2 x x n C05S0M.044: On [0, 3], we have lim n→∞ i=1 (xi )2 − 3xi ∆x = 5 3 0 2 1 √ = 2 2 − 2. (x2 − 3x) dx = 13 32 x− x 3 2 3 0 9 =− . 2 n C05S0M.045: On [0, 1], we have lim n→∞ = 2π (1 + x2 )3/2 3 1 = 0 2π xi 1 + (xi )2 ∆x = i=1 1 2π x 1 + x2 dx 0 √ 2π 2 2−1 . 3 C05S0M.046: First note that lim n→∞ 110 + 210 + 310 + 410 + · · · + n10 = lim n→∞ n11 n k=1 k n 10 · 1 n (1) is the limit of Riemann sums for f (x) = x10 on [0, 1] with xi = i/n and ∆x = 1/n. Hence the value of the limit in Eq. (1) is 1 x10 dx = 0 1 11 x 11 1 = 0 1 . 11 Alternatively, one can show that n k 10 = k=1 n(n + 1)(2x + 1)(n2 − n + 1)(3n6 + 9n5 + 2n4 − 11n3 + 3n2 + 10n − 5) . 66 Then division by n11 yields 5 1 1 1 5 1 1 +, − 8+ 6− 4+ 2+ 10 66n 2n n n 6n 2n 11 and the value of the limit as n → + ∞ is now clear. C05S0M.047: If f (x) ≡ c (a constant), then for every partition of [ a, b ] and every selection for each such partition, we have f (xi ) = c. Therefore n n f (xi ) ∆x = i=1 i=1 n c· b−a 1 1 = c(b − a) · = c(b − a) · n · = c(b − a). n n n i=1 Then, because every Riemann sum is equal to c(b − a), this is also the limit of those Riemann sums. Therefore, by definition, b a f (x) dx = c(b − a). C05S0M.048: Because f is continuous on [ a, b ], the definite integral of f exists there; that is, the appropriate Riemann sums have a limit as ∆x → 0. For every partition of [ a, b ] and every selection for every such partition, we have f (xi ) 0, so every Riemann sum is a sum of nonnegative numbers. Therefore their limit is nonnegative (if the limit were negative, then at least one Riemann sum would have to be negative). Therefore b f (x) dx 0. a C05S0M.049: Given: f continuous on [ a, b ] and f (x) > 0 there. Let m be the global minimum value of f on [ a, b ]; then m > 0. Hence, by the second comparison property (Section 5.5), 6 b f (x) dx m(b − a) > 0. a 1 C05S0M.050: 0 1 (1 − x2 )3 dx = C05S0M.051: (2x) C05S0M.052: 1 + x1/3 x1/2 1 /2 0 − (3x ) 1 C05S0M.054: 0 2 C05S0M.056: 0 1 /2 x √ √ 35 17 x− x 5 7 dx = 1 = 0 31 16 −= . 57 35 √ √ 2 2 3 /2 2 3 dx = x + 1/2 + C. 3 3x 3 −3/2 2− x 3 2 (x−1/2 + 2x−1/6 + x1/6 ) dx = 2x1/2 + 12 5/6 6 7/6 x + x + C. 5 7 1 1 2x−2 − x dx = − 2x−1 − x2 + C . 2 4 1 dt 1 = 2 (3 − 2t) 2(3 − 2t) x1/2 cos x3/2 dx = C05S0M.055: dx = 3 −1/2 4 − x3 dx = 2x2 C05S0M.053: (1 − 3x2 + 3x4 − x6 ) dx = x − x3 + = 0 11 1 −=. 26 3 2 sin x3/2 + C . 3 2 x2 (9 − x3 )1/2 dx = − (9 − x3 )3/2 9 2 =− 0 2 52 − (−6) = . 9 9 1 1 1 sin dt = cos + C . t2 t t C05S0M.057: 2 C05S0M.058: 1 2t + 1 dt = 2(t2 + t)1/2 (t2 + t)1/2 2 1 √ √ = 2 6 − 2 2. u 1 /3 3 du = − (1 + u4/3 )−2 + C . 4/3 )3 8 (1 + u C05S0M.059: π /4 C05S0M.060: π /4 (cos t)−1/2 sin t dt = 0 4 C05S0M.061: 1 (1 + t1/2 )2 dt = t1 / 2 1 u2 C05S0M.062: 1− C05S0M.063: Let u = I= 1 u 4 − 2(cos t)1/2 0 = −(23/4 ) − (−2) = 2 − 23/4 . (t1/2 + 2 + t−1/2 ) dt = 1 1 /3 du = 3 1 1− 4 u 2 3/2 t + 2t + 2t1/2 3 4 = 1 52 14 38 − = . 3 3 3 4 /3 + C. 1 1 1 , so that x = and dx = − 2 du. Then x u u (4x2 − 1)1/2 dx = x4 − u4 u2 4 −1 u2 1 /2 du = − u2 4 −1 u2 1 /2 du. Next move one copy of u from outside the square root to inside, where it becomes u2 , and we see 7 I=− u(4 − u2 )1/2 du = 1 1 1 (4 − u2 )3/2 + C = 4− 2 3 3 x 3 /2 + C. To make this answer more appealing, multiply numerator and denominator by x3 ; when the x3 in the numerator is moved into the 3/2-power, it becomes x2 and the final version of the answer is I= 1 C05S0M.064: The area is −1 1 C05S0M.065: The area is 0 (4x2 − 1)3/2 + C. 3x3 14 x 4 (1 − x3 ) dx = x − (x4 − x5 ) dx = 1 = −1‘ 3 5 −− 4 4 1 15 16 x− x 5 6 = 0 = 2. 11 1 −= . 56 30 √ √ C05S0M.066: Solve 3y 2 − 6 = y 2 to find that the two curves cross where y = a = − 3 and y = b = 3. Hence the area between them is b a 2 (6 − 2y 2 ) dy = 6y − y 3 3 b a √ √ = 4 3 − −4 3 √ = 8 3. C05S0M.067: Solve x4 = 2 − x2 to find that the two curves cross where x = ± 1. Hence the area between them is 1 1 (2 − x2 − x4 ) dx = 2x − x3 − x5 3 5 −1 1 1 22 22 −− 15 15 = −1 = 44 . 15 C05S0M.068: Solve x4 = 2x2 − 1 for x = ± 1 to find where the curves cross. The area between them is then 1 −1 (x4 − 2x2 + 1) dx = 15 23 x − x +x 5 3 1 8 8 −− 15 15 = −1 = 16 . 15 C05S0M.069: Solve (x − 2)2 = 10 − 5x to find that the two curves cross where x = −3 and where x = 2. The area between them is 2 1 1 (10 − 5x − (x − 2)2 ) dx = 6x − x2 − x3 2 3 −3 2 = −3 22 27 −− 3 2 = 125 . 6 C05S0M.070: Solve x2/3 = 2 − x2 to find that the two curves cross where x = ± 1. Thus the area between them is 1 3 (2 − x2 − x2/3 ) dx = 2x − x3 − x5/3 3 5 −1 1 1 = −1 16 16 −− 15 15 = 32 . 15 √ C05S0M.071: If y = 2x − x2 , then y 0 and x2 − 2x + y 2 = 0, so that (x − 1)2 + y 2 = 1. Therefore the graph of the integrand is the top half of a circle of radius 1 centered at (1, 0), and so the value of the integral is the area of that semicircle: 8 2 0 2x − x2 dx = 1 π · π · 12 = . 2 2 √ C05S0M.072: If y = 6x − 5 − x2 , then y 0 and x2 − 6x + 5 + y 2 = 0, so that (x − 3)2 + y 2 = 4. Hence the graph of the integrand is the top half of a circle of radius 2 centered at (3, 0), and so the value of the integral is the area of that semicircle: 5 6x − 5 − x2 dx = 1 1 · π · 22 = 2π . 2 C05S0M.073: If x x2 = 1 + 2 1 + [ f (t)] dt, 1 then differentiation of both sides of this identity with respect to x (using Part 1 of the fundamental theorem of calculus, Section 5.6) yields 2x = 2 1 + [ f (x)] , 2 so that 4x2 = 1 + [ f (x)] . Therefore if x > 1, one solution is f (x) = √ 4x2 − 1 . C05S0M.074: Let x F (x) = φ(t) dt. a Then G(x) = F (h(x)), so that G (x) = F (h(x)) · h (x). But F (x) = φ(x) by the fundamental theorem of calculus (Section 5.6). Therefore G (x) = φ(h(x)) · h (x). √ C05S0M.075: Because f (x) = 1 + x2 is increasing on [0, 1], the left-endpoint approximation will be an underestimate of the integral and the right-endpoint approximation will be an overestimate. With n = 5, ∆x = 1 , and xi = i · ∆x, we find the left-endpoint approximation to be 5 5 i=1 f (xi−1 ) ∆x ≈ 1.10873 and the right-endpoint approximation is 5 i=1 f (xi ) ∆x ≈ 1.19157. The average of the two approximations is 1.15015 and half their difference is 0.04142, and therefore 1 0 1 + x2 dx = 1.15015 ± 0.04143. The true value of this integral is approximately 1.1477935747. When we used n = 4 subintervals the error in the approximation was larger than 0.05. C05S0M.076: With n = 6, a = 0, b = π , ∆x = (b − a)/n, and xi = i · ∆x, the trapezoidal approximation is 9 √ 1 √ 1√ 1√ T6 = 2 + 2 1 + 6+ 2+ 12 2 2 √ 3 1+ π ≈ 2.8122538625. 2 3 1− + 2 Simpson’s approximation is S6 = √ 1 4+2 2+ 18 √ 6+4 1− √ 3 +4 2 sin2 1+ √ x dx 2 3 π ≈ 2.8285015468. 2 The exact value of the integral is π √ 0 1 − cos x dx = = √ π 2 0 √ π 2 0 √ 1 − cos x dx = 2 2 sin √ x dx = 2 2 π π 0 sin 0 √ x x dx = − 2 2 cos 2 2 π 0 √ = 2 2. C05S0M.077: M5 ≈ 0.2866736772 and T5 ≈ 0.2897075147. Because the graph of the integrand is concave upward on the interval [1, 2], 2 M5 < 1 1 dx < T5 x + x2 for the reasons given in the solution of Problem 33 of Section 5.9. C05S0M.078: First, 2 (xi−1 ) Therefore xi−1 xi 2 1 2 2 (xi−1 ) + xi−1 xi + (xi ) 3 2 (xi ) . xi . Then (xi ) (xi − xi−1 ) = 1 1 2 2 3 3 (xi−1 ) + xi−1 xi + (xi ) (xi − xi−1 ) = (xi ) − (xi−1 ) , 3 3 13 b − a3 . 3 Because all Riemann sums for f (x) = x2 on [ a, b ] have the same limit, this must be the same limit as the limit of the particular Riemann sums of this problem. This shows that and when such expressions are summed for i = 1, 2, 3, . . . , n, the result is b x2 dx = a 13 (b − a3 ). 3 C05S0M.079: Suppose that 0 < a < b, that n is a positive integer, that P = {x0 , x1 , x2 , . . . , xn } is √ a partition of [ a, b ], and that ∆xi = xi − xi−1 for 1 i n. Let xi = xi−1 xi for 1 i n. Then S = {x1 , x2 , . . . , xn } is a selection for P because xi−1 = for 1 i (xi−1 )2 < √ xi−1 xi < n. Next, 10 (xi )2 = xi n i=1 1 ∆xi = (xi )2 = = n xi − xi−1 = xi−1 xi i=1 1 1 − x0 x1 + n 1 1 − xi−1 xi i=1 1 1 − x1 x2 + 1 1 − x2 x3 1 1 − xn−2 xn−1 + ··· + + 1 1 − xn−1 xn 1 1 11 − = −. x0 xn ab Therefore n i=1 1 11 ∆xi → − 2 (xi ) ab as |P | → 0. But n i=1 1 ∆xi (xi )2 b is a Riemann sum for a 1 dx. x2 Because f is continuous on [ a, b ], all such Riemann sums converge to the same limit, which must therefore be the same as the particular limit just computed. Therefore 1 11 dx = −. x2 ab b a C05S0M.080: Suppose that 0 < a < b, that n is a positive integer, that P = {x0 , x1 , x2 , . . . , xn } is a partition of [ a, b ], and that ∆xi = xi − xi−1 for 1 i n. Let xi = for 1 i 2 3 3 /2 (xi ) 3 /2 − (xi−1 ) xi − xi−1 n. Then S = {x1 , x2 , . . . , xn } is a selection for P because xi = 2 3 √ √ 3 xi xi 2 − − √ √ xi−1 xi−1 3 2 √ √ √ xi − xi−1 xi 2 = ·√ · √ 3 xi − xi−1 2 √ √ + xi xi−1 + xi−1 √ √ xi + xi−1 2 √ 2 xi + xi xi−1 + xi−1 =· . √ √ 3 xi + xi−1 At this point we’d like to be able to claim that the last term in the last equation is less than √ 2 xi + xi + xi xi =√ = xi , · √ 3 2 xi xi because this would establish that xi < xi . But while we increase the numerator by replacing xi−1 with the larger xi , we are also increasing the denominator. So we need a technical lemma whose proof you may prefer to ignore. Lemma: If 0 < a < b, then 11 √ √ b + ab + a 3b 3b √ . <√= √ 2 b+ a 2b Proof: Suppose that 0 < a < b. Then √ a < b; √ a < ab; √ √ 2a < a + ab < b + ab; √ √ 2b + 2 ab + 2a < 3b + 3 ab; √ √√ √ 2(b + ab + a) < 3 b b+ a ; √ √ √ b + ab + a 3b √ < . √ 2 b+ a This concludes the proof of the lemma. Therefore, if 0 < xi−1 < xi , then √ xi + xi xi−1 + xi−1 3√ < xi . √ √ xi + xi−1 2 √ It now follows that xi < xi , and thus that xi < xi for 1 i n. In much the same way, one can establish that xi−1 < xi for 1 i n. Hence S is indeed a selection for P . Consequently, n n xi ∆xi = i=1 2 3 /2 3 /2 · (xi ) − (xi−1 ) 3 i=1 = 2 (x1 )3/2 − (x0 )3/2 + (x2 )3/2 − (x1 )3/2 + · · · + (xn )3/2 − (xn−1 )3/2 3 = 2 · b3/2 − a3/2 . 3 √ Finally, because f (x) = x is continuous on [ a, b ], all the Riemann sums for f there converge to the same limit. Some of these sums have been shown to converge to 2 b3/2 − a3/2 . Therefore they all converge to 3 that limit, and thus b a √ x dx = 2 3/2 b − a3/2 . 3 12 Section 6.1 n C06S01.001: With a = 0 and b = 1, lim n→∞ n ∆x n→∞ 2 (xi ) i=1 1 n C06S01.003: With a = 0 and b = 1, lim n→∞ 1 (sin π xi ) ∆x = 0 i=1 3 (xi ) − 1 ∆x = i=1 = 1 1 . 2 1 cos π x π 1 = 0 2 . π 3 3 2 n→∞ 2 sin π x dx = − n C06S01.004: With a = −1 and b = 3, lim = 1. 0 1 1 dx = − x2 x 2 = 2x dx = x2 0 i=1 C06S01.002: With a = 1 and b = 2, lim 1 1 2xi ∆x = −1 (3x2 − 1) dx = x3 − x = 24. −1 C06S01.005: With a = 0 and b = 4, n lim n→∞ xi 4 2 (xi ) + 9 ∆x = i=1 4 C06S01.006: The limit is 2 4 √ ln x 6 0 1 C06S01.010: The limit is 0 0 1 = e0 − e−1 = 1 − . e 1 (2x + 1)3/2 3 2x + 1 dx = x 1 dx = ln(x2 + 9) x2 + 9 2 4 =9− 0 6 1 26 = . 3 3 1 1 1 ln 45 − ln 9 = ln 5. 2 2 2 = 0 1 2x exp(−x2 ) dx = − exp(−x2 ) 0 1 = e0 − e−1 = 1 − . e n C06S01.011: With a = 1 and b = 4, Eq. (2) in Section 6.3. 0 125 98 −9= . 3 3 = ln 4 − ln 2 = 2 ln 2 − ln 2 = ln 2. 2 0 C06S01.009: The limit is = 4 1 dx = x e−x dx = −e−x 0 C06S01.008: The limit is 0 1 1 C06S01.007: The limit is 4 12 (x + 9)3/2 3 x2 + 9 dx = x lim n→∞ n→∞ 1 2 [ f (xi )] ∆x = 2π xf (x) dx. Compare this with 1 i=1 n C06S01.012: With a = −1 and b = 1, lim 4 2π xi f (xi ) ∆x = 2 [ f (x) ] dx. −1 i=1 n 2 C06S01.013: With a = 0 and b = 10, lim n→∞ 1 + [ f (xi ) ] 10 ∆x = i=1 2 1 + [ f (x) ] dx. 0 C06S01.014: With a = −2 and b = 3, we have n lim n→∞ 2π mi 2 1 + [ f (mi ) ] ∆x = 3 −2 i=1 1 2π x 2 1 + [ f (x) ] dx. 100 C06S01.015: M = 0 25 C06S01.016: M = 0 = 1000 − 0 = 1000 (grams). 0 25 0 10 C06S01.018: M = 100 (60 − 2x) dx = 60x − x2 10 C06S01.017: M = 1 12 x dx = x 5 10 x(10 − x) dx = 5x2 − = 875 − 0 = 875 (grams). 0 10 13 x 3 0 10 πx 100 πx dx = − cos 10 π 10 10 sin 0 500 500 −0= (grams). 3 3 = = 0 100 100 −− π π = 200 (grams). π C06S01.019: The net distance is 10 0 10 (−32) dt = − 32t 0 = −320 and the total distance is 320. C06S01.020: The net distance is 5 5 (2t + 10) dt = t2 + 10t 1 and because v (t) = 2t + 10 1 0 for 1 = 75 − 11 = 64 5, this is the total distance as well. t C06S01.021: The net distance is 10 0 Because v (t) = 4t − 25 10 (4t − 25) dt = 2t2 − 25t 0 for 0 6.25 − 0 10 v (t) dt + 0 C06S01.022: Because v (t) 5 6.25 and v (t) t 2 .5 | 2t − 5 | dt = 0 t = 200 − 250 = −50. 0 for 6.25 t 10, the total distance is 625 225 425 + = = 106.25. 8 8 4 v (t) dt = 6.25 0 for 0 0 5, the net and total distance are both equal to (5 − 2t) dt + 5 2 .5 (2t − 5) dt = 25 25 25 + = = 12.5. 4 4 2 C06S01.023: The net distance is 3 3 4t3 dt = t4 −2 Because v (t) 0 for −2 t −2 = 81 − 16 = 65. 0, the total distance is 0 − −2 3 4t3 dt + 4t3 dt = 16 + 81 = 97. 0 C06S01.024: The net distance is 2 1 t− 0 .1 Because v (t) 0 for 0.1 1 t2 1 12 1 t+ 2 t dt = = 0 .1 3 2001 1701 − =− = −8.505. 2 200 200 1, the total distance is 8.505. t C06S01.025: Because v (t) = sin 2t equal to π /2 0 0 for 0 π /2, the net distance and the total distance are both t 1 sin 2t dt = − cos 2t 2 π /2 1 1 −− 2 2 = 0 = 1. C06S01.026: The net distance is π /2 cos 2t dt = 0 Because v (t) = cos 2t 0 for π /4 0 = 0 − 0 = 0. 0 π /2, the total distance is t π /4 π /2 1 sin 2t 2 π /2 cos 2t dt − 11 + = 1. 22 cos 2t dt = π /4 C06S01.027: The net distance is 1 −1 Because v (t) = cos π t − 0 for −1 −0.5 1 1 sin π t π cos π t dt = −1 −0.5 and for 0.5 t 0 .5 cos π t dt + −1 −0.5 cos π t dt − 1, the total distance is t 1 = 0 − 0 = 0. cos π t dt = 0 .5 1 2 1 4 ++=. πππ π C06S01.028: The net distance is π π (sin t + cos t) dt = 0 But v (t) = sin t + cos t 0 for 3π /4 3 π /4 0 sin t − cos t 0 = 1 − (−1) = 2. π , so the total distance is t (sin t + cos t) dt − π (sin t + cos t) dt = 1 + √ 3 π /4 2+ √ √ 2 − 1 = 2 2. C06S01.029: The net distance is 10 0 (t2 − 9t + 14) dt = but because v (t) = t2 − 9t + 14 2 0 0 for 2 v (t) dt − 7 2 t 13 92 t − t + 14t 3 2 10 = 0 70 ≈ 23.333333, 3 7, the total distance is 10 v (t) dt + v (t) dt = 7 3 38 125 63 + + = 65. 3 6 2 C06S01.030: The net distance is 6 0 Because v (t) = t3 − 8t2 + 15t < 0 for 3 3 0 5 (t3 − 8t2 + 15t) dt − 3 t 6 1 4 8 3 15 2 t− t+ t 4 3 2 (t3 − 8t2 + 15t) dt = 0 = 18 − 0 = 18. 5, the total distance is (t3 − 8t2 + 15t) dt + 6 5 (t3 − 8t2 + 15t) dt = 63 16 91 86 + + = ≈ 28.666667. 4 3 12 3 C06S01.031: If v (t) = t3 − 7t + 4 for 0 t 3, then v (t) < 0 for α = 0.602705 < t < β = 2.29240 (numbers with decimal points are approximations), so the net distance is 3 0 14 72 t − t + 4t 4 2 (t3 − 7t + 4) dt = 3 = 0 3 4 but the total distance is approximately α 0 β (t3 − 7t + 4) dt − α (t3 − 7t + 4) dt + 3 β (t3 − 7t + 4) dt ≈ 1.17242 + 3.49165 + 3.06923 = 7.73330. C06S01.032: Because v (t) = t3 − 5t2 + 10 < 0 if α = 1.175564 < t < β = 4.50790 (numbers with decimal points are approximations), the net distance is 5 0 (t3 − 5t2 + 10) dt = 14 53 t − t + 10t 4 3 5 0 =− 25 ≈ −2.083333, 12 but the total distance is α 0 (t3 − 5t2 + 10) dt − β α (t3 − 5t2 + 10) dt + 5 β (t3 − 5t2 + 10) dt ≈ 10.9126 + 15.2724 + +2.27654 ≈ 28.4615. C06S01.033: Here, v (t) = t sin t − cos t is negative for 0 decimals are approximations), so the net distance is π 0 t < α = 0.860333589019 (numbers with π (t sin t − cos t) dt = −t cos t =π 0 but the total distance is α − 0 (t sin t − cos t) dt + π α (t sin t − cos t) dt ≈ 0.561096338191 + 3.702688991781 = 4.263785329972. C06S01.034: The velocity v (t) = sin t + t1/2 cos t is negative for α = 2.167455 < t < β = 5.128225 (numbers with decimals are approximations), so the net distance is 2π 0 (sin t + t1/2 cos t) dt = −0.430408 and the total distance is 4 α 0 (sin t + t1/2 cos t) dt − β (sin t + t1/2 cos t) dt + α 2π (sin t + t1/2 cos t) dt β ≈ 2.02380 + 4.05657 + 1.60236 = 7.68273. We used Mathematica’s NIntegrate command for all four definite integrals. The numbers α and β were found using Newton’s method. C06S01.035: If n is a large positive integer, ∆x = r/n, xi = i · ∆x, and xi is chosen in the interval [ xi−1 , xi ] for 1 i n, then the area of the annular ring between xi−1 and xi is approximately 2π xi ∆x and its average density is approximately ρ (xi ), so a good approximation to the total mass of the disk is n 2π xi ρ (xi ) ∆x. i=1 But this is a Riemann sum for r 2π xρ(x) dx, 0 and therefore such sums approach this integral as ∆x → 0 and n → + ∞ because 2π xρ(x) is (we presume) continuous for 0 x r. But such Riemann sums also approach the total mass M of the disk and this establishes the equation in Problem 35. 10 C06S01.036: The mass is M = 2π x2 dx = 0 5 C06S01.037: The mass is M = 0 23 πx 3 10 = 0 2000π ≈ 2094.395102. 3 1 2π x(25 − x2 ) dx = − π (x4 − 50x2 ) 2 5 = 0 625π ≈ 981.747704. 2 C06S01.038: The maximum height will be 5 0 5 (160 − 32t) dt = 160t − 16t2 = 400 (feet). 0 C06S01.039: The amount of water that flows into the tank from time t = 10 to time t = 20 is 20 10 (100 − 3t) dt = 100t − 20 C06S01.040: Answer: 32 t 2 20 10 = 1400 − 850 = 550 (gallons). 20 1000(16 + t) dt = 16000t + 500t2 0 = 520000. 0 C06S01.041: Answer: 20 375000 + 0 (1000(16 + t) − 1000 5 + 20 1 t 2 dt = 375000 + 11000t + 250t2 = 695000. 0 C06S01.042: Let n be a positive integer, let P = {t0 , t1 , t2 , . . . , tn } be a partition of the interval [0, 365], let ∆t = 365/n, and let ti be a point in [ ti−1 , ti ] for 1 i n. Then the approximate rainfall in Charleston in the time interval ti−1 t ti will be r (ti ) ∆t. Hence the total rainfall in a year will be approximately 5 n r (ti ) ∆t. i=1 But this is a Riemann sum for 365 R= r(t) dt, 0 and if r(t) is piecewise continuous (I’ve lived there; I can assure you that r(t) is not continuous, but merely piecewise continuous) then these Riemann sums approach R as a limit (as n → + ∞). Thus R gives the average total annual rainfall in Charleston. C06S01.043: We solved r(0) = 0.1 and r(182.5) = 0.3 for a = 0.2 and b = 0.1. Then we found that 365 r(t) dt = 0 73 · 4π 4π t 2π t − sin 365 365 365 = 73 0 inches per year, average annual rainfall. C06S01.044: Let n be a positive integer, let ∆t = (b − a)/n, let P = {t0 , t1 , t2 , . . . , tn } be a regular partition of [ a, b ], and let ti be a number in [ ti−1 , ti ] for 1 i n. Then the amount of water that flows into the tank in the interval [ ti−1 , ti ] will be approximately r (ti ) ∆t. Hence the total amount of water that flows into the tank between times t = a and t = b will be approximately n r (ti ) ∆t. i=1 The error in this approximation will approach zero as t → + ∞, and the sum itself is a Riemann sum for r(t) on [ a, b ], and therefore—if r is piecewise continuous (physical considerations dictate that it must be)—the amount of water that flows into the tank during that interval must be b Q= r(t) dt. a C06S01.045: If f (x) = x1/3 on [0, 1], n is a positive integer, ∆x = 1/n, and xi = xi = i · ∆x, then n lim n→∞ i=1 i1/3 = lim n→∞ n4 /3 n f (xi ) ∆x = i=1 1 0 x1/3 dx = 3 4 /3 x 4 1 = 0 3 . 4 C06S01.046: Let r denote the radius of the spherical ball and partition the interval [0, r ] into n subintervals all having the same length ∆x = r/n. Let xi = i · ∆x, so that P = {x0 , x1 , x2 , . . . , xn } is a regular partition of [0, r ]. Let xi be the midpoint of [ xi−1 , xi ], so that {xi } is a selection for P . Then the volume of the 2 spherical shell with inner radius xi−1 and outer radius xi will be approximately 4π (xi ) ∆x, so that the total volume of the spherical ball will be closely approximated (if n is large) by n 2 4π (xi ) ∆x. i=1 But this sum is a Riemann sum for f (x) = 4π x2 on the interval [0, r ]. The error in this approximation to the volume V of the spherical ball will approach zero as n → + ∞, but it also approaches the integral shown next, and therefore 6 r V= r 43 πx 3 4π x2 dx = 0 = 0 43 πr . 3 C06S01.047: Let n be a large positive integer, let ∆x = 1/n, and let xi = i · ∆x. Then P = {x0 , x1 , x2 , . . . , xn } is a regular partition of [0, 1]. Let xi be the midpoint of the interval [ xi−1 , xi ]. Then the weight of the spherical shell with inner radius xi−1 and outer radius xi will be approximately 2 100 (1 + xi ) · 4π (xi ) ∆x (the approximate volume of the shell multiplied by its approximate average density). Therefore the total weight of the ball will be approximately n 2 i=1 100 (1 + xi ) · 4π (xi ) ∆x. This is a Riemann sum for f (x) = 100(1 + x) · 4π x2 on [0, 1], and such sums approach the total weight W of the ball as n → + ∞. Therefore 1 W= 0 100(1 + x) · 4π x2 dx = C06S01.048: Given v (x) = k cos r F= 2π kx cos 0 = 4kr2 − 400 3 π x + 100π x4 3 1 = 0 700 π ≈ 733.038286 (pounds). 3 πx , the flow rate is 2r πx 2r dx = 4 πx πx · 2kr2 cos + π krx sin π 2r 2r r 0 8kr2 4k (π − 2)r2 = . π π C06S01.049: Because the pressure P is inversely proportional to the fourth power of the radius r, the values r = 1.00, 0.95, 0.90, 0.85, 0.80, and 0.75 yield the values r−4 = 1.00, 1.22774, 1.52416, 1.91569, 2.44141, and 3.16049. We subtract 1 from each of the latter and then multiply by 100 to obtain percentage increase in pressure (and multiply each value of r by 100 to convert to percentages). Result: 100 95 90 85 80 75 0.000 22.774 52.416 91.569 144.141 216.049 C06S01.050: The amount of dye injected is A = 4 (in milligrams) and the concentration function is c(t) = 40te−t , 0 t 10 (t in seconds). Then 10 c(t) dt = (−40 − 40t)e−t Thus the patient’s cardiac output is approximately 7 0 = 40 − 440e−10 . F= 60A 6e10 = 10 ≈ 6.003 40 − 440e−10 e − 11 liters per minute (the factor of 60 in the first fraction converts seconds to minutes). C06S01.051: Given: A = 4.5 (mg). Simpson’s rule applied to the concentration data c = 0, 2.32, 9.80, 10.80, 7.61, 4.38, 2.21, 1.06, 0.47, 0.18, 0.0 measured at the times t = 0, 1, 2, . . . , 10 (seconds) yields 10 0 c(t) dt ≈ S10 = 1 1 · 0 + 4 · (2.32) + 2 · (9.8) + 4 · (10.8) + 2 · (7.61) + 4 · (4.38) 3 1919 = 38.38. 50 Hence—multiplying by 60 to convert second to minutes—the cardiac output is approximately + 2 · (2.21) + 4 · (1.06) + 2 · (0.47) + 4 · (0.18) + 1 · 1 = F≈ 60A ≈ 7.0349 38.38 L/min (liters per minute). The Mathematica code is straightforward, although you must remember that an array’s initial subscript is 1 rather than zero (unless you decree it otherwise): c = {0, 232/100, 98/10, 108/10, 761/100, 438/100, 221/100, 106/100, 47/100, 18/100, 0}; c[[1]] + c[[11]] + 4∗Sum[c[[i]], {i, 2, 10, 2}] + 2∗Sum[c[[i]], {i, 3, 9, 2}] 5757 50 N[ 5757/(3∗50), 12 ] 38.3800000000 N[ (60*45/10)/(3838/100), 12 ] 7.03491401772 C06S01.052: Here we have A = 5.5, a = 200, b = 8, c = 7.1, d = 0.5, and f (t) = atb exp(−ctd ) yielding the graph of concentration (in mg/L) as a function of t (in seconds). Result: 12 c(t) 10 (mg/L) 8 6 4 2 0 2.5 5 7.5 10 12.5 15 17.5 20 t (seconds) 8 Next, 20 I= 0 f (t) dt ≈ 67.4709, so the cardiac output of the patient is F≈ 60A ≈ 4.891 I liters per minute. 9 Section 6.2 1 C06S02.001: The volume is V = 0 4 C06S02.002: The volume is V = 4 1 0 .1 π C06S02.005: The volume is V = 4 −3 = 8π − 0 = 8π . 0 = 8π . 0 1 π π dx = − 2 x x 0 .1 π π sin2 x dx = π 0 3 π . 5 4 0 C06S02.006: The volume is V = 0 12 πy 2 π y dy = 0 C06S02.004: The volume is V = = 12 πx 2 π x dx = 0 C06S02.003: The volume is V = 1 15 πx 5 π x4 dx = π (9 − x2 )2 dx = π = −π − (−10π ) = 9π . 1 − cos 2x 1 1 dx = π x − sin 2x 2 2 4 15 x − 6x3 + 81x 5 3 = −3 π = 0 12 π. 2 1296π ≈ 814.300816. 5 C06S02.007: Rotation of the given figure around the x-axis produces annular rings of inner radius y = x2 √ √ and outer radius y = x (because x2 x if 0 x 1). Hence the volume of the solid is 1 V= π √ x 2 0 12 15 x− x 2 5 − x4 dx = π 1 11 − 25 =π 0 = 3 π. 10 C06S02.008: Rotation of the given region around the vertical line x = 5 produces annular rings with inner √ radius 5 − y and outer radius 5 − 1 y . Hence the volume of the solid is 4 16 V= π 0 = 5− 1 y 4 2 − (5 − π 320y 3/2 − 84y 2 + y 3 48 5 C06S02.009: The volume is 1 √ 2 y) 16 dy = π 0 10y 1/2 − 7 12 y+ y dy 2 16 16 = 64π − 0 = 64π ≈ 201.062. 0 5 π dx = π ln x x 1 = π ln 5 ≈ 5.05619832. C06S02.010: Rotation of the region between the given curves around the y -axis produces annular rings with outer radius y + 6 and inner radius y 2 ; solution of the equation y + 6 = y 2 yields the limits of integration y = −2 and y = 3. Hence the volume of the solid in question is V= 3 −2 π (y + 6)2 − y 4 dy = π = π 36y + 6y 2 + C06S02.011: Volume: V = 13 15 y− y 3 5 3 −2 (36 + 12y + y 2 − y 4 ) dy 3 = −2 612π 664π −− 5 15 2 1 π (1 − x2 )2 dx = π x − x3 + x5 3 5 −1 1 1 = 500π ≈ 523.598776. 3 1 = −1 16π ≈ 3.351032. 15 1 C06S02.012: Volume: V = 0 13 25 17 x− x+ x 3 5 7 π (x − x3 )2 dx = π C06S02.013: A horizontal cross section “at” y has radius x = 1 0 1 C06S02.014: The volume is 12 y 2 π (1 − y ) dy = π y − 1 2x e 2 e2x dx = 0 1 √ 1 = 0 8π ≈ 0.2393594403. 105 1 − y , so the volume of the solid is 1 = 0 π . 2 12 (e − 1) ≈ 3.19452805. 2 = 0 C06S02.015: The region between the two curves extends from y = 2 to y = 6 and the radius of a horizontal √ cross section “at” y is x = 6 − y . Therefore the volume of the solid is 6 V= 2 12 y 2 π (6 − y ) dy = π 6y − 6 2 = 18π − 10π = 8π . C06S02.016: The region bounded by the two curves extends from y = 0 to y = 1. When it is rotated √ around the vertical line x = 2, it generates annular regions with outer radius 2 + 1 − y and inner radius √ 2 − 1 − y . Hence the volume of the solid is 1 V= π 2+ 0 =π − 1−y 16 (1 − y )3/2 3 2 − 2− 1−y 1 0 =0− − 16π 3 2 = 1 dy = π 0 8(1 − y )1/2 dy 16π ≈ 16.755161. 3 C06S02.017: When the region bounded by the given curves is rotated around the horizontal line y = −1, it generates annular regions with outer radius x − x3 + 1 and inner radius 1. Hence the volume of the solid thereby generated is 1 V= 0 1 π (x − x3 + 1)2 − 12 dx = π = π x2 + 13 14 25 17 x− x− x+ x 3 2 5 7 0 (2x + x2 − 2x3 − 2x4 + x6 ) dx 1 = 0 121π 121π −0= ≈ 1.8101557671. 210 210 C06S02.018: When the region bounded by the given curves is rotated around the x-axis, it generates annular regions with outer radius ex and inner radius e−x . Hence the volume of the solid thereby generated is 1 V= 0 (π e2x − π e−2x ) dx = π 1 2x 1 −2x e+e 2 2 1 = 0 1 1 1 π (e2 + e−2 − e0 − e0 ) = π e − 2 2 e 2 ≈ 8.67769369. C06S02.019: When the region bounded by the given curves is rotated around the y -axis, it generates √ circular regions; the one “at” y has radius x = y , so the volume generated is 4 V= π y dy = π 0 2 12 y 2 4 = 8π . 0 C06S02.020: When the region bounded by the given curves is rotated around the x-axis, it generates √ circular regions; the one “at” x has radius y = 16 − x, so the volume generated is 16 V= 0 12 x 2 π (16 − x) dx = π 16x − 16 = 128π . 0 C06S02.021: When the region bounded by the given curves is rotated around the horizontal line y = −2, √ it generates annular regions with outer radius x − (−2) and inner radius x2 − (−2). Hence the volume of the solid thereby generated is 1 V= √ π x+2 0 2 − (2 + x2 )2 dx = π 1 8 3/2 1 2 4 3 1 5 x + x− x− x 3 2 3 5 =π = 0 1 0 (4x1/2 + x − 4x2 − x4 ) dx 49π ≈ 5.1312680009. 30 C06S02.022: When the region bounded by the given curves is rotated around the horizontal line y = −1, it generates annular regions with outer radius 8 − x2 + 1 and inner radius x2 + 1, and therefore the volume of the solid thus generated is 2 V= −2 π (9 − x2 )2 − (x2 + 1)2 dx = 2π = 2π 80x − 20 3 x 3 2 = 0 2 0 (80 − 20x2 ) dx 640π ≈ 670.206433. 3 C06S02.023: When the region bounded by the given curves is rotated around the vertical line x = 3, it √ generates annular regions with outer radius 3 − y 2 and inner radius 3 − y . Hence the volume of the solid thereby generated is 1 V= 0 π (3 − y 2 )2 − (3 − = π 4y 3/2 − √ 2 1 dy = π y) 0 12 1 y − 2y 3 + y 5 2 5 1 = 0 (6y 1/2 − y − 6y 2 + y 4 ) dy 17π ≈ 5.340708. 10 C06S02.024: When the region bounded by the given curves is rotated around the horizontal line y = −1, it generates annular regions with outer radius 3 and inner radius 2e−x + 1 with locations varying from x = 0 to x = 1. Therefore the volume of the solid thereby generated is 1 V= 0 π · 32 − π · (2e−x + 1)2 dx = π 1 0 (9 − 4e−2x − 4e−x − 1) dx 1 = π 8x + 2e−2x + 4e−x 0 = π (8 + 2e−2 + 4e−1 − 2 − 4) = 2π 1 + 1 e 2 ≈ 11.7564313695185460. C06S02.025: The volume generated by rotation of R around the x-axis is π V= 0 π π sin2 x dx = π 0 1 − cos 2x 1 1 dx = π x − sin 2x 2 2 4 3 π = 0 π2 ≈ 4.9348022005. 2 C06S02.026: The volume is V= 1 πx dx = π 2 π cos2 −1 1 −1 1 + cos π x dx = π 2 1 1 x + sin π x 2 2 1 = −1 π π −− 2 2 = π. C06S02.027: The curves y = cos x and y = sin x cross at π /4 and the former is above the latter for 0 x < π /4. So the region between them, when rotated around the x-axis, generates annular regions with outer radius cos x and inner radius sin x. Therefore the volume of the solid generated will be π /4 V= 0 π /4 π (cos2 x − sin2 x) dx = π cos 2x dx = π 0 1 sin 2x 2 π /4 = 0 π . 2 C06S02.028: The curve y = cos x and the horizontal line y = 1 cross where x = − 1 π and where x = 1 π , 2 3 3 and the curve is above the line between those two points. So when the region they bound is rotated around the x-axis, it generates annular regions with outer radius cos x and inner radius 1 . So the volume of the 2 solid thereby generated will be V= π /3 π cos2 x − −π /3 = 1 4 π /3 dx = π √ −π /3 √ π π 3 3 + 2π − − 3 3 + 2π 24 24 1 + 2 cos 2x 1 dx = π (x + sin 2x) 4 4 = √ π /3 −π /3 π 3 3 + 2π ≈ 3.0052835900. 12 C06S02.029: The volume is π /4 V= π /4 π tan2 x dx = π 0 0 π /4 =π 0 sin2 x dx = π cos2 x π /4 0 π /4 (sec2 x − 1) dx = π (tan x) − x = 0 1 − cos2 x dx cos2 x π (4 − π ) ≈ 0.6741915533. 4 C06S02.030: When the region bounded by the given curves is rotated around the x-axis, it produces annular regions with outer radius 1 and inner radius tan x. Hence the volume of the solid generated is π /4 V= 0 π 1 − tan2 x dx = π π /4 = π 2x − tan x = 0 π /4 0 1 − sec2 x + 1 dx π (π − 2) ≈ 1.7932095470. 2 C06S02.031: The two curves cross near a = −0.532089, b = 0.652704, and c = 2.87939. When the region between the curves for a x b is rotated around the x-axis, the solid it generates has approximate volume b V1 = a π (x3 + 1)2 − (3x2 )2 dx ≈ 1.68838 − (−1.39004) = 2.99832. When the region between the curves for b approximate volume c V2 = b x c is rotated around the x-axis, the solid it generates has π (3x2 )2 − (x3 + 1)2 dx ≈ 265.753 − (−1.68838) ≈ 267.442. 4 The total volume generated is thus V1 + V2 ≈ 270.440. C06S02.032: The two curves cross near a = −1.28378 and b = 1.53375. When the region between the curves is rotated around the x-axis, the volume of the solid it generates is b V= a π (x + 4)2 − (x4 )2 dx = π 16x + 4x2 + b 13 19 x− x 3 9 a ≈ 94.0394 − (−42.7288) ≈ 136.768. C06S02.033: The two curves cross near a = −0.8244962453 and b = 0.8244962453. When the region between them is rotated around the x-axis, the volume of the solid it generates is b V≈ a π (cos2 x) − x4 dx = π 10x − 4x5 + 5 sin 2x 20 b a ≈ 1.83871 − (−1.83871) ≈ 3.67743. C06S02.034: The two curves cross near a = 0.3862368706 and b = 1.9615690350. When the region between them is rotated around the x-axis, the volume of the solid generated is approximately b V= a = π sin2 x − (x − 1)4 dx π 40x2 − 10x − 40x3 + 20x4 − 4x5 − 5 sin 2x 20 b a ≈ 2.48961 − (−0.51503) ≈ 3.00464. C06S02.035: The two curves cross at the two points where x = 6, and the one on the right meets the x-axis at (3, 0). When the region they bound is rotated around the x-axis, the volume of the solid it generates is 6 V= 0 6 π x dx − 3 2π (x − 3) dx = π 12 x 2 6 0 6 − π x2 − 6x 3 = (18π − 0) (0 + 9π ) = 9π . C06S02.036: The top half of the ellipse is the graph of the function f (x) = b a a2 − x2 , −a x a. Hence when the region bounded above by the graph of y = f (x) and below by the x-axis is rotated around the x-axis, the volume of the ellipsoid thereby swept out will be V= a 2 π [ f (x) ] dx = π −a b2 x(3a2 − x2 ) 3a2 a = −a 4 π ab2 . 3 C06S02.037: The right half of the ellipse is the graph of the function g (y ) = a b b2 − y 2 , −b y b. Hence when the region bounded on the right by the graph of x = g (y ) and on the left by the y -axis is rotated around the y -axis, the volume of the ellipsoid thereby swept out will be V= b −b 2 π [ g (y ) ] dy = π a2 y (3b2 − y 2 ) 3b 2 5 a = −a 42 π a b. 3 C06S02.038: Part (a): The volume of the part of the solid from x = 1 to x = b is b V (b) = 1 1 π e−2x dx = − π e−2x 2 b 1 π (e−2 − e−2b ). 2 = 1 Hence the volume of the unbounded solid is lim V (b) = b→∞ π ≈ 0.21258417. 2e2 Part (b): The volume of the part of the solid from x = 1 to x = b is b V (b) = 1 π dx = π ln x x b = π ln b. 1 Therefore the volume of the unbounded solid is lim V (b) = lim π ln b = + ∞. b→∞ b→∞ C06S02.039: Locate the base of the observatory in the xy -plane with the center at the origin and the diameter AB on the x-axis. Then the boundary of the base has equation x2 + y 2 = a2 . A typical vertical cross-section of the observatory has√ its base a chord of that circle perpendicular to the x-axis at [say] x, as so that the length of this chord is 2 a2 − x2 . The square of this length gives the area of that vertical cross section, and it follows that the volume of the observatory is a V= −a 4(a2 − x2 ) dx = 4 (3a2 x − x3 ) 3 a = −a 83 8 a − − a2 3 3 = 16 3 a. 3 C06S02.040: Locate the base of the solid in the xy -plane with the center at the origin and the diameter AB on the x-axis. Then the boundary of the base has equation x2 + y 2 = a2 . A typical vertical cross-section of the solid has as its base a chord of that circle perpendicular to the x-axis at [say] x, so that the length of √ this chord is 2 a2 − x2 . This chord is the diameter of that semicircular cross section, which therefore has √ radius a2 − x2 and thus area 1 π (a2 − x2 ). Hence the volume of the solid is 2 a −a x(3a2 − x2 ) 6 π2 (a − x2 ) dx = π 2 a = −a 23 πa , 3 exactly as expected, for after all the solid is a hemisphere of radius a. C06S02.041: Locate the base of the solid in the xy -plane with the center at the origin and the diameter AB on the x-axis. Then the boundary of the base has equation x2 + y 2 = a2 . A typical vertical cross-section of the base has as its base a chord of that circle perpendicular to the x-axis at [say] x, so that the length of √ this chord is 2 a2 − x2 . This chord is one of the three equal sides of the vertical cross section—which is an equilateral triangle—and it follows that this triangle has Base: b = 2 a2 − x2 So the area of this triangle is a −a 1 2 bh √ = √ and height: h= √ 3 b= 2 3a2 − 3x2 . 3 (a2 − x2 ), and therefore the volume of the solid is 3 (a2 − x2 ) dx = √ 3 (3a2 x − x3 ) 3 6 a = −a √ 433 a. 3 C06S02.042: A cross section of the solid perpendicular to the x-axis at x has base of length √ 2 its area is x − x2 = x − 2x5/2 + x4 . So the volume of the solid is 1 0 x − 2x5/2 + x4 dx = 1 2 4 7 /2 1 5 x− x + x 2 7 5 1 = 0 √ x − x2 , so 9 . 70 C06S02.043: The volume of the paraboloid is h Vp = h 2π px dx = π px2 0 = π ph2 . 0 If r is the radius of the cylinder, then the equation y 2 = 2px yields r2 = 2ph, so the volume of the cylinder is Vc = π r2 h = 2π ph2 = 2Vp . C06S02.044: Consider a cross section of the pyramid parallel to its base and at distance x from the vertex of the pyramid. Similar triangles show that the length y and width z of this cross section are proportional to x, so that its area is f (x) = kx2 . Moreover, f (h) = A, so that kh2 = A, and therefore k = A/(h2 ). Thus the total volume of the pyramid will be h V= h f (x) dx = 0 0 A · x2 dx = h2 h Ax3 3h2 = 0 Ah3 1 = Ah. 2 3h 3 C06S02.045: Consider a cross section of the pyramid parallel to its base and at distance x from the vertex of the pyramid. Similar triangles show that the lengths of the edges of this triangular cross section are proportional to x. If the edges have lengths p, q , and r, then Heron’s formula tells us that the area of this triangular cross section is 1 4 (p + q + r)(p + q − r)(p − q + r)(q + r − p) , and thus the area g (x) of this triangular cross section is proportional to x2 ; that is, g (x) = kx2 for some constant k . But g (h) = A, which tells us that kh2 = A and thus that k = A/(h2 ). So the total volume of the pyramid will be h V= 0 h g (x) dx = 0 A · x2 dx = h2 h Ax3 3h 2 = 0 Ah3 1 = Ah. 3h2 3 C06S02.046: Set up a coordinate system in which the origin is at the center of the sphere and the sphere-with-hole is symmetric around the y -axis. Then a horizontal cross section of the solid “at” location y is an annular region with inner radius 3 and outer radius x = 25 − y 2 . Therefore the volume of the sphere-with-hole is V= 5 −4 π (25 − y 2 − 9) dy = 2π 1 (48y − y 3 ) 3 4 = 0 256π ≈ 268.082573. 3 As an independent check, the volume of the sphere-without-hole is about 523.598776. C06S02.047: Set up a coordinate system in which one cylinder has axis the x-axis and the other has axis the y -axis. Introduce a z -axis perpendicular to the xy -plane and passing through (0, 0). We will find the volume of the eighth of the intersection that lies in the first octant, where x, y , and z are nonnegative, then multiply by 8 to find the answer. 7 A cross section of the eighth perpendicular to the z -axis (thus parallel to the xy -plane) is a square; if this cross section meets the z -axis at z , then one of its edges lies in the yz -plane and reaches from the z -axis (where y = 0) to the side of the cylinder symmetric around the x-axis. That cylinder has equation the same as the equation as the circle in which it meets the yz -plane: y 2 + z 2 = a2 . Hence the edge of the square √ under consideration has length y = a2 − z 2 . So the area of the square cross section “at” z is a2 − z 2 . So the volume of the eighth of the solid in the first octant is a 0 13 z 3 (a2 − z 2 ) dz = a2 z − a 0 = a3 − 13 2 a = a3 . 3 3 16 3 a. 3 As an independent check, it’s fairly easy to see that the sphere of radius a centered at the origin is enclosed in the intersection and occupies most of the volume of the intersection; the ratio of the volume of the intersection to the volume of that sphere is Therefore the total volume of the intersection of the two cylinders is 16 3 3a 4 3 3 πa = 4 ≈ 1.273240, π a very plausible result. C06S02.048: Set up a coordinate system in which the center of the sphere is located at the origin and the flat part of the spherical segment is perpendicular to the y -axis. Now consider a cross section of the spherical segment, lying between y = r − h and y = r on the y -axis, and “at” location y . This cross section is a circular disk of radius x = r2 − y 2 , so its has cross-sectional area π (r2 − y 2 ). Therefore the volume of the spherical segment will be V (h) = 1 π (r2 − y 2 ) dy = π r2 y − y 3 3 r −h r r =π r −h 23 1 1 r − (r − h)(3r2 − (r − h)2 ) = π h2 (3r − h). 3 3 3 Note that the answer is dimensionally correct (the product of three lengths; see page 156 of the text). It also meets the “test of extremes,” known in the past as “the exception that proves the rule,” which in modern translation would be “the exceptional case that tests the rule.” Specifically, V (0) = 0, V (r) = 2 π r3 , and 3 V (2r) = 4 π r3 , exactly as should be the case. 3 C06S02.049: The cross section of the torus perpendicular to the y -axis “at” y is an annular ring with outer radius x = b + a2 − y 2 and inner radius x = b − a2 − y 2 , a consequence of the fact that the circular disk that generates the torus has equation (x − b)2 + y 2 = a2 . So the cross section has area π b+ a2 − y 2 2 a2 − y 2 − b− 2 = 4π b a2 − y 2 . Therefore the volume of the torus is V = 4π b a −a a2 − y 2 dy = 4π b · 12 πa 2 because the integral is the area of a semicircle of radius a centered at the origin. Therefore V = 2π 2 a2 b. C06S02.050: Simpson’s approximation is S4 = 25π 640000π · 602 + 4 · 552 + 2 · 502 + 4 · 352 + 02 = ≈ 670206.433. 3 3 8 C06S02.051: First, r is the value of y = R − kx2 when x = 1 h, so r = R − 1 kh2 = R − δ where 4δ = kh2 . 2 4 Next, the volume of the barrel is h/2 V =2 0 π (R − kx2 )2 dx = 2π = 2π R2 x − 2 1 Rkx3 + k 2 x5 3 5 = π h R2 − 1 1 24 Rkh2 + kh 6 80 = π h R2 − 2 1 Rδ + δ 2 3 5 = = h/2 0 R2 − 2Rkx2 + k 2 x4 dx h/2 12 1 1 25 R h− Rkh3 + kh 2 12 160 = 2π 0 = π h R2 − 1 1 R · 4δ + · 16δ 2 6 80 πh 3 · 3R2 − 2Rδ + δ 2 3 5 πh 2 · 2R2 + R2 − 2Rδ + δ 2 − δ 2 3 5 = πh 2 · 2R2 + (R − δ )2 − δ 2 3 5 = πh 2 · 2R2 + r2 − δ 2 . 3 5 C06S02.052: Suppose that the depth of water in the clepsydra is h. A horizontal cross section of the water “at” y , 0 y h, is a circular disk of radius x = (y/k )1/4 , so the total volume of water in the clepsydra will be h V (h) = 0 π· y 1 /2 2π h3/2 dy = . k 1 /2 3k 1/2 Thus if the depth of water is y , we find that its volume is V (y ) = 2π y 3 / 2 . 3k 1 / 2 But even without this computation the fundamental theorem of calculus tells us that dV π = 1 /2 · y 1 /2 . dy k Therefore—using Torricelli’s law— −cy 1/2 = dV dy dV dy π = · = 1/2 · y 1/2 · , dt dy dt dt k and now it follows that √ dy ck =− , dt π which is a constant. That is, the water level falls at a constant rate. To use the clepsydra as a clock (as it was used in ancient Egypt, Greece, and Rome), put a ruler vertically in the clepsydra. The ancient Egyptians were so sophisticated with these clocks that they had different rulers for different water temperatures, as the less viscous warm water would run out at a slightly greater (but still constant) rate. According to the American Heritage Dictionary of the English Language (Boston: Houghton Mifflin, 1969, 1970), the name of the device is derived from the Greek word klepsudra, “water stealer” (from the “stealthy” flow of the water), from the words kleptein, “to steal” (an English relative is kleptomania) and hud or, “water” (an English relative is hydrosphere ). 9 C06S02.053: The factors 27 and 3.3 in the following computations convert cubic feet to cubic yards and cubic yards to dollars. The trapezoidal approximation gives T6 = (10)(3.3) · (1513 + 2 · 882 + 2 · 381 + 2 · 265 + 2 · 151 + 2 · 50 + 0) ≈ 3037.83 (2)(27) and Simpson’s approximation gives S6 = (10)(3.3) · (1513 + 4 · 882 + 2 · 381 + 4 · 265 + 2 · 151 + 4 · 50 + 0) ≈ 3000.56. (3)(27) To the nearest hundred dollars, each answer rounds to $3000. C06S02.054: Please don’t give this solution away, particularly to a differential equations student. Note first that if V (t) is the volume of water in the bowl at time t, then dV /dt = −k · A(t) where k is a positive constant. Let y (t) be the depth of water in the bowl at time t. Consider the time interval [ t, t + ∆t ]. Assume that the water level drops by the amount ∆y . Then the change in the volume of water in the bowl is ∆V ≈ A(t) ∆y where A(t) is the area of the water surface at time t. Thus ∆V ∆y ≈ A(t) · , ∆t ∆t and if we let ∆t → 0, the error in this approximation will approach zero; therefore A(t) · dy dV = ; dt dt A(t) · dy = −k A(t); dt dy = −k. dt That is, the water level in the bowl drops at a constant rate. C06S02.055: First “finish” the frustum; that is, complete the cone of which it is a frustum. We measure all distances from the vertex of the completed cone and perpendicular to the bases of the frustum. Let H be the height of the cone and suppose that H − h y H , so that a cross section of the cone at distance y from its vertex is a circular cross section of the frustum. The area A(y ) of such a cross section is proportional to the square of its radius, which is proportional to y 2 , so that A(y ) = ky 2 where k is a positive proportionality constant. By similar triangles, H −h H =, r R H= hR . R−r A(y ) = π R2 y 2 . H2 and so Also, A(H ) = kH 2 = π R2 , and it follows that k= π R2 , H2 so that Therefore the volume of the frustum is V= H H −h π R2 y 2 π R2 dy = · H 3 − (H − h)3 . 2 H 3H 2 10 Next note that A(H − h) = π r2 = π R2 (H − h)2 . Therefore H2 π R2 π R2 3H 2 h − 3Hh2 + h3 = 2 2 (R − r)2 2 3H 3h R 3h3 R2 3h3 R − + h3 2 (R − r) R−r = π 3h3 R2 − 3h3 R(R − r) + h3 (R − r)2 3h2 πh 3R2 − 3R(R − r) + (R − r)2 3 = πh πh 2 (3R2 − 3R2 + 3rR + R2 − 2Rr + r2 ) = (R + Rr + r2 ). 3 3 V= = C06S02.056: The solid is formed by rotating around the x-axis the plane region above the x-axis and common to the two circular disks bounded by the two circles with equations x + 1a 2 2 + y 2 = a2 and x − 1a 2 2 + y 2 = a2 . Let R denote the half of that region that lies in the first quadrant. Then the solid of Problem 56 has volume V double that obtained when R is revolved around the x-axis. The curve that forms the upper boundary of R has equation y = f (x) = 1 2 3a2 − 4ax − 4x2 , and therefore a/2 V =2 2 π [ f (x) ] dx = 2π 0 32 1 1 a x − ax2 − x3 4 2 3 a/2 = 0 5 π a3 . 12 Note that the answer is dimensionally correct. Moreover, the solid occupies 31.25% of the volume of either sphere, and thus the answer also passes the test of plausibility (see page 155 of the text). C06S02.057: The solid of intersection of the two spheres can be generated by rotating around the x-axis the region R common to the two circles shown in the following figure. y (a,0) Radius a x Radius b < a x=c The circle on the left, of radius a and centered at the origin, has equation x2 + y 2 = a2 . The circle on the right, of radius b < a and centered at (a, 0), has equation (x − a)2 + y 2 = b2 . The x-coordinate of their two points of intersection can be found by solving x2 − a2 = (x − a)2 − b2 for 11 x= 2a2 − b2 = c. 2a The part of R above the x-axis is comprised of two smaller regions, one to the left of the vertical line x = c and one to its right. The width of the region R1 on the left—measured along the x-axis—is h1 = c − (a − b) = 2a2 − b2 b(2a − b) −a+b= . 2a 2a The width of the region R2 on the right is h2 = a − c = b2 . 2a One formula will tell us the volume of the solid generated by rotation of either R1 or R2 around the x-axis. Suppose that C is the circle of radius r centered at the origin and that 0 < h < r. Let us find the volume generated by rotation of the region above the x-axis, within the circle, and to the right of the vertical line x = r − h around the x-axis. It is V (r, h) = r r −h π (r2 − x2 ) dx = π r2 x − 13 x 3 r r −h = ··· = π h2 (3r − h) 3 (we worked this problem by hand—it is not difficult, merely tedious—but checked our results with Mathematica 3.0). Hence the volume of intersection of the two original spheres is V (b, h1 ) + V (a, h2 ) = · · · = π b3 (8a − 3b) 12a (also by hand, tedious but not difficult, and checked with Mathematica). C06S02.058: As n → ∞, xi−1 → xi . Continuity of f then implies that f (xi−1 ) → f (xi ), so that n lim n→∞ i=1 π 2 2 [ f (xi−1 )] + f (xi−1 )f (xi ) + [ f (xi )] 3 n ∆x = lim n→∞ b = a 12 i=1 π 2 · 3 [ f (xi )] ∆x 3 2 π [ f (x)] dx. Section 6.3 C06S03.001: The region R, bounded by the graphs of y = x2 , y = 0, and x = 2, is shown next. If we rotate R around the y -axis, then the vertical strip in R “at” x will move around a circle of radius x and the height of the strip will be y = x2 , so the volume generated will be 2 2π x3 dx = 2π 0 2 14 x 4 0 = 8π ≈ 25.1327412287. 5 4 3 2 1 0.5 1 1.5 2 √ C06S03.002: The region R to be rotated around the y -axis is bounded above by the graph of y = x and √ √ below by the graph of y = − x. Hence the vertical strip “at” x has height 2 x and is rotated around a circle of radius x, so the volume swept out by R is 4 V= 4π x3/2 dx = 0 8 5 /2 πx 5 4 = 0 256π ≈ 160.8495438638. 5 C06S03.003: To obtain all the cylindrical shells, we need only let x range from 0 to 5, so the volume of the solid is 5 V= 0 2π x(25 − x2 ) dx = π 25x2 − 2 C06S03.004: The volume is 0 2 C06S03.005: The volume is 0 3 C06S03.006: The volume is 0 14 x 2 5 = 0 625π ≈ 981.7477042468. 2 14 x 2 2 2π x(8 − 2x2 ) dx = 2π 4x2 − 14 x 2 2 2π x(8 − 2x2 ) dx = 2π 4x2 − 2π y (9 − y 2 ) dy = π 9y 2 − 14 y 2 0 0 3 = 0 = 16π ≈ 50.2654824574. = 16π ≈ 50.2654824574. 81π ≈ 127.2345024704. 2 C06S03.007: A horizontal strip of the region R (shown next) “at” y stretches from x = y to x = 3 − 2y and thus has length 3 − 3y . Hence the volume swept out by rotation of R around the x-axis is 1 V= 0 1 2π y (3 − 3y ) dy = π 3y 2 − 2y 3 1 0 = π ≈ 3.1415926536. 1.2 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 2.5 3 -0.2 C06S03.008: The horizontal strip of the region R “at” y moves around a circle of radius 5 − y and has length y 1/2 − 1 y , so the volume swept out by R is 2 4 V= 0 2π (5 − y ) y 1/2 − 10 3/2 5 2 2 5/2 1 3 1 y dy = 2π y − y− y +y 2 3 4 5 6 80 64 32 − 20 − + 3 5 3 = 2π = 4 0 136π ≈ 28.4837733925. 15 C06S03.009: A horizontal strip of the region R “at” y (0 y 2) stretches from y 2 /4 to x = (y/2)1/2 and is rotated around a circle of radius y , so the volume swept out when R is rotated around the x-axis is 4 V= 2π y 0 = 2π y 2 1 /2 − y2 dy = 2π 4 √ 2 5 /2 14 − y y 5 16 3 C06S03.010: The volume is 0 2 = 0 4 0 √ 2 3/2 1 3 y −y 2 4 6π ≈ 3.7699111843. 5 2π x(3x − x2 ) dx = 2π x3 − 14 x 4 3 = 0 27π ≈ 42.4115008235. 2 C06S03.011: The graph of y = 4x − x3 crosses the x-axis at x = −2, x = x-axis for −2 < x < 0 and above it for 0 < x < 2. A vertical strip “at” x for and moves around a circle of radius x, whereas a vertical strip “at” x for −2 and moves around a circle of radius −x. Therefore the total volume swept rotated around the y -axis is 2 V= 0 = 2π 2π x(4x − x3 ) dx + 43 15 x− x 3 5 2 = −2 0 −2 dy (−2π x)(x3 − 4x) dx = 2 −2 0, and x = 2. It is below the 0 x 2 has height 4x − x3 x 0 has height −(4x − x3 ) out when the given region is 2π x(4x − x3 ) dx 256π ≈ 53.6165146213. 15 C06S03.012: A horizontal strip “at” y of the given region has length x = y 3 − y 4 and is rotated around a circle of radius y − (−2) = y + 2, so the volume swept out when this region is rotated around the horizontal line y = −2 is 2 1 V= 0 1 C06S03.013: The volume is 0 4 C06S03.014: The volume is 0 1 14 15 16 y− y− y 2 5 6 2π (y + 2)(y 3 − y 4 ) dy = 2π = = 4π ≈ 0.8377580410. 15 0 13 15 x− x 3 5 2π x(x − x3 ) dx = 2π 4π ≈ 0.8377580410. 15 1 14 y 4 4 2π y (16 − y 2 ) dy = 2π 8y 2 − 0 = 128π ≈ 402.1238596595. 0 C06S03.015: A vertical strip “at” x of the given region has height y = x − x3 and moves around a circle of radius 2 − x, so the volume the region sweeps out is 1 V= 0 2π (2 − x)(x − x3 ) dx = 2π x2 − 2 C06S03.016: The volume is 0 2π x · x3 dx = 2π 13 14 15 x− x+ x 3 2 5 15 x 5 2 1 11π ≈ 2.3038346126. 15 = 0 64π ≈ 40.2123859659. 5 = 0 C06S03.017: If 0 x 2, then a vertical strip of the given region “at” x has height x3 and moves around a circle of radius 3 − x, so the volume swept out is 2 V= 0 2π (3 − x)x3 dx = 2π 34 15 x− x 4 5 2 56π ≈ 35.1858377202. 5 = 0 C06S03.018: If 0 y 8, then a horizontal strip of the given region “at” y has length 2 − y 1/3 and moves around a circle of radius y , so the volume generated by rotation of that region around the x-axis is 8 V= 0 2π y (2 − y 1/3 ) dy = 2π y 2 − 3 7 /2 y 7 8 0 = 2π 64 − 384 7 = 128π ≈ 57.4462656656. 7 C06S03.019: If −1 x 1, then a vertical strip of the given region “at” x has height x2 and moves around a circle of radius 2 − x, so the volume generated by rotating the given region around the vertical line x = 2 is V= 1 −1 1 C06S03.020 The volume is 0 0 1 C06S03.022: The volume is 0 1 = 2π −1 2π x(x − x2 ) dx = 2π 1 C06S03.021: The volume is = 2π 23 14 x− x 3 4 2π (2 − x)x2 dx = 2π 13 14 x− x 3 4 2π y (y 1/2 − y ) dy = 2π 1 = 2π 0 4 21 −1− + 3 53 3 1 = 0 2 5 /2 1 3 y −y 5 3 2π (2 − y )(y 1/2 − y ) dy = 2π 4 3 /2 2 1 − y 2 − y 5/2 + y 3 y 3 5 3 5 11 + 12 12 1 0 = 8π ≈ 8.3775804096. 3 π ≈ 0.5235987756. 6 1 = 0 2π ≈ 0.4188790205. 15 (2y 1/2 − 2y − y 3/2 + y 2 ) dy = 8π ≈ 1.6755160819. 15 C06S03.023: If 0 x 1, then a vertical strip of the given region “at” x has height x − x2 and moves around a circle of radius x − (−1) = x + 1, so the volume swept out by rotating the region around the vertical line x = −1 is 1 V= 0 2π (x + 1)(x − x2 ) dx = 2π 1 0 12 14 x− x 2 4 (x − x3 ) dx = 2π 1 = 0 π ≈ 1.5707963268. 2 C06S03.024: If 1 x 2, then a vertical strip of the given region “at” x has height 1/x2 and moves around a circle of radius x, so the volume generated by rotating the region around the y -axis is 2 1 2π x · 1 dx = 2π x2 2 1 2 1 dx = 2π ln x x 1 = 2π ln 2 ≈ 4.3551721806. C06S03.025: The volume is 1 1 V= 2π x exp(−x2 ) dx = 0 2 C06S03.026: The volume is 0 − π exp(−x2 ) 0 =− π π (e − 1) − (− π ) = ≈ 1.9858653038. e e 2π x dx = π ln(1 + x2 ) 1 + x2 2 0 = π ln 5 ≈ 5.0561983221. √ C06S03.027: If 0 x π , then a vertical strip of the given region “at” x has height 2 sin x2 and is rotated around a circle of radius x. Hence the volume thereby generated is √ V= π √ 4π x sin x2 dx = 0 2 C06S03.028: The volume is 1 − 2π cos x2 2π (x + 1) · 0 π = 2π − (−2π ) = 4π ≈ 12.5663706144. 1 dx = 2π x2 2 1 1 1 +2 xx dx = 2π − 1 + ln x x 2 1 = −π + 2π ln 2 − (−2π ) = π (1 + 2 ln 2) ≈ 7.4967648342. C06S03.029: The curves cross to the right of the y -axis at the points x = a = 0.17248 and x = b = 1.89195 (numbers with decimals are approximations). The quadratic is above the cubic if a < x < b, so the volume of the region generated by rotation of R around the y -axis is b V= a 1 1 1 2π x(6x − x2 − x3 − 1) dx = 2π − x5 − x4 + 2x3 − x2 5 4 2 b a ≈ 23.2990983139. C06S03.030: The curves cross at x = a = 0.506586 and x = b = 1.95208 (numbers with decimals are approximations). The linear function is above the quartic for a < x < b, so the volume generated by rotation of r around the y -axis is b V= a 2π x(10x − 5 − x4 ) dx = 2π 10 3 5 2 1 6 x− x− x 3 2 6 b a ≈ 39.3184699459. C06S03.031: We used Newton’s method to find that the two curves cross at x = a = −0.8241323123 and x = b = −a. But because the region R is symmetric around the y -axis, the interval of integration must be [0, b ]. Also cos x x2 on this interval, so the volume generated by rotation of R around the y -axis is 4 b 0 2π x(cos x − x2 ) dx = 2π cos x + x sin x − b 14 x 4 0 ≈ 1.0602688478. C06S03.032: The two curves clearly cross where x = 0. We used Newton’s method to find that they also cross where x = b = 1.4055636328 (numbers with decimals are approximations). Because cos x (x − 1)2 on [0, b ], the volume generated by rotation of R around the y -axis is b 2π x[cos x − (x − 1)2 ] dx = 2π cos x + x sin x − 0 b 14 23 12 x+ x− x 4 3 2 0 ≈ 2.7556103644. C06S03.033: We used Newton’s method to find that the two curves cross where x = a = 0.1870725959 and x = b = 1.5758806791. Because cos x 3x2 − 6x + 2 on [ a, b ], the volume generated by rotation of R around the y -axis is b V= a 2π x(cos x − 3x2 + 6x − 2) dx = 2π cos x + x sin x − x2 + 2x3 − 34 x 4 b a ≈ 8.1334538068. C06S03.034: The region R is symmetric around the y -axis and the two curves cross to the right of the y -axis where x = b = 1.7878717268 (numbers with decimal points are approximations). Because 3 cos x − cos 4x for 0 < x < b, the volume generated when R is rotated around the y -axis is b V= 2π x(3 cos x + cos 4x) dx = 2π 3 cos x + 0 1 1 cos 4x + 3x sin x + x sin 4x 16 4 b 0 ≈ 12.0048972158. C06S03.035: The slant side of the cone is the graph of f (x) = h − hx r for 0 x r. Therefore the volume of a cone of radius r and height h is r V= 1 2 hx3 hx − 2 3r 2π xf (x) dx = 2π 0 r = 0 12 π r h. 3 C06S03.036: The volume of the paraboloid is √ V= 0 2ph y2 2π y h − 2p 1 2 y4 hy − dy = 2π 2 8p √ 2ph 0 1 = 2π ph2 − ph2 2 = π ph2 . C06S03.037: The top half of the ellipse is the graph of y = f (x) = b2 (a − x2 )1/2 , a −a x a. The height of a vertical cross section of the ellipse “at” the number x is therefore 2f (x), so the volume of the ellipsoid will be a V= 0 4π xf (x) dx = 4π b a a 0 x(a2 − x2 )1/2 dx = 5 4π b 1 − (a2 − x2 )3/2 a 3 a = 0 4π b 3 4 · a = π a2 b. 3a 3 Note the lower limit of integration: zero, not −a. C06S03.038: Place the sphere of radius r with its center at the origin in the vertical xy -plane, so that the sphere intersects the xy -plane in the circle with equation x2 + y 2 = r2 . We will find the volume of the spherical segment that occupies the region for which r − h y r where 0 h 2r. Note first that if the horizontal line y = r − h meets the circle at the point where x = a > 0, then a2 + (r − h)2 = r2 , and it √ follows that a = 2rh − h2 . So if x is between 0 and a, then the vertical strip above x from y = r − h to √ the top of the sphere (where y = r2 − x2 ) has length (r2 − x2 )1/2 − (r − h) and will be rotated around a circle of radius x. Hence the volume of the spherical segment is a V = 2π r2 − x2 − (r − h) x 0 dx = π (r2 − x2 ) 3(r − h) − 2 3 a r2 − x2 = π (r2 − 2rh + h2 ) 3(r − h) − 2 3 = π (r − h)2 3(r − h) − 2(r − h) − 3r2 (r − h) + 2r3 3 = 0 π3 π π h2 (r − 3r2 h + 3rh2 − h3 − r3 + 3r2 h) = (3rh2 − h3 ) = (3r − h). 3 3 3 r2 − 2rh + h2 − r2 3(r − h) − 2r = π (r − h)3 − 3r3 + 3r2 h + 2r3 3 C06S03.039: The torus is generated by rotating the circular disk D in the xy -plane around the y -axis; the boundary of D has equation (x − b)2 + y 2 = a2 where 0 < a b. Thus D has its center at (b, 0) on the positive x-axis. If b − a x b + a, then a vertical slice through D at x has height 2 a2 − (x − b)2 , and so the volume of the torus is b+a V= b−a 2π x · 2 a2 − (x − b)2 dx. The substitution u = x − b, with x = u + b and dx = du, transforms this integral into V= a −a 2π (u + b) · 2 a2 − u2 du = 4π 1 = 4π − (a2 − u2 )3/2 3 a + 4π b a a a2 − u2 du = 0 + 4π b · −a −a u(a2 − u2 )1/2 + b(a2 − u2 )1/2 du −a 12 π a = 2π 2 a2 b 2 because the last integral represents the area of a semicircle of radius a. C06S03.040: The two curves meet at (−1, 1) and (2, 4) and the graph of the linear equation is above that of the quadratic for −1 < x < 2. If a vertical strip of the region R they bound, above the point x on the x-axis, is rotated around the vertical line x = −2, then it has height x + 2 − x2 and moves around a circle of radius x + 1. Hence the volume of the solid generated in part (a) is V= 2 −1 2π (x + 2)(x + 2 − x2 ) dx = 2π = 2π 4x + 2x2 − 13 14 x− x 3 4 2 = 2π −1 6 2 −1 (4 + 4x − x2 − x3 ) dx 28 23 + 3 12 = 45π ≈ 70.6858347058. 2 But if that vertical strip is rotated around the vertical line x = 3, then the radius of its circular path is now 3 − x, and therefore the volume of the solid generated in part (b) is 2 V= −1 2 2π (3 − x)(x + 2 − x2 ) dx = 2π = 2π 6x + 12 43 14 x− x+ x 2 3 4 −1 2 = 2π −1 (6 + x − 4x2 + x3 ) dx 22 47 + 3 12 = 45π ≈ 70.6858347058. 2 The equality of the answers in parts (a) and (b) is merely a coincidence. √ C06S03.041: If −a x a, then the vertical cross section through the disk “at” x has length 2 a2 − x2 and moves through a circle of radius a − x, so the volume of the so-called pinched torus the disk generates is V= a −a = 4π a 2π (a − x) · 2(a2 − x2 )1/2 dx 1 (a2 − x2 )1/2 dx − 4π − (a2 − x2 )3/2 3 −a a The value of the last integral is 1 2 2 πa a −a 1 = 4π a · π a2 + 0 = 2π 2 a3 . 2 because it represents the area of a semicircle of radius a. C06S03.042: First, Dx (x − 1)ex = (x − 1)ex + ex = xex , and therefore xex dx = (x − 1)ex + C. Therefore the volume of the solid of part (b) is 1 V= 0 1 2π xex dx = 2π (x − 1)ex 0 = 0 − (−2π ) = 2π . C06S03.043: The next figure shows the central cross section of the sphere-with-hole; the radius of the hole is a, its height is h, and the radius of the sphere is b. b h/2 a The figure shows that 1 h = 2 2 yields √ b2 − a2 , and substitution in the volume formula V = 4 π (b2 − a2 )3/2 of Example 3 7 V= 4 π· 3 h 2 3 π3 h, 6 = quite independent of the values of a or b. C06S03.044: The method of cross sections yields volume 16 V= 0 π (25 − y − 9) dy = π 16y − 16 12 y 2 = 128π ≈ 402.1238596595. 0 The method of cylindrical shells yields volume 5 V= 3 5 25 2 1 4 x− x 2 4 2π x(25 − x2 ) dx = 2π = 3 625π 369π − = 128π . 2 2 C06S03.045: (a) The method of cross sections yields volume 5 V= 0 π x(5 − x)2 dx = 5 π 3x4 − 40x3 + 150x2 12 = 0 625π ≈ 163.6246173745. 12 (b) The method of cylindrical shells yields volume 5 V= 4π x 3 /2 0 (5 − x) dx = 2π 4x 5 /2 4 − x7/2 7 5 0 √ 400π 5 = ≈ 401.4179846309. 7 (c) The method of cylindrical shells yields volume 5 V= 0 4π x1/2 (5 − x)2 dx = 4π 50 3/2 2 x − 4x5/2 + x7/2 3 7 5 = 0 √ 1600π 5 ≈ 535.2239795079. 21 C06S03.046: (a) The method of cross sections yields volume 0 V= π x2 (x + 3) dx = π x3 + −3 14 x 4 0 = −3 27π ≈ 21.2057504117. 4 (b) In the method of parallel slabs, we must note that the radius of the vertical strip “at” x is −x and that √ x2 = −x because x 0. Thus the volume is V= 0 −3 (−2π x)(−2x)(x + 3)1/2 dx = 4π 0 x2 (x + 3)1/2 dx. −3 The substitution u = x + 3 converts the last integral into 3 V = 4π 0 (u − 3)2 u1/2 du = 4π = 4π 6u 3/2 12 5/2 2 7/2 − u +u 5 7 3 0 3 0 (9u1/2 − 6u3/2 + u5/2 ) du √ √ 144 3 576π 3 = 4π · = ≈ 89.5498657542. 35 35 (c) The method of cylindrical shells yields volume 8 V= 0 −3 2π (x + 3)(−2x)(x + 3)1/2 dx = −4π 0 x(x + 3)3/2 dx. −3 Then the substitution u = x + 3 transforms the last integral into V = −4π = −4π 3 0 (u − 3)u3/2 du = −4π 2 7 /2 6 5 /2 u −u 7 5 3 0 (u5/2 − 3u3/2 ) du √ 432π 3 ≈ 67.1623993156. 35 3 = 0 C06S03.047: Given f (x) = 1 + x2 x4 − 5 500 and g (x) = x4 , 10000 the curves cross where f (x) = g (x). The Mathematica command Solve[ f[x] == g[x], x ] returns two complex conjugate solutions and two real solutions, x = ±10. Hence the volume of the solid obtained by rotating the region between the two curves around the y -axis can be computed in this way (we include extra steps for the reader’s benefit): Integrate[ 2∗Pi∗x∗(f[x] - g[x]), x ] π x2 + (% /. π x4 7π x6 − 10 10000 x → 10) − (% /. 400π x → 0) N[%, 20] 1256.6370614359172954 To find the volume of water the birdbath will hold when full, we need to find the highest points on the graph of y = f (x). Solve[ D[ f[x], x] == 0, x ] √ √ {{ x → 0 }, { x → −5 2 }, { x → 5 2 }} f[ 5∗Sqrt[2] ] 6 Thus the amount of water the birdbath will hold can be found as follows: Integrate[ 2∗Pi∗x∗(6 - f[x]), x ] 5π x2 − π x4 π x6 + 10 1500 9 (% /. x → 5∗Sqrt[2]) - (% /. 250π 3 x → 0) N[%, 20] 261.79938779914943654 10 Section 6.4 Note: In problems 1–20, we will also provide the exact answer (when the antiderivative is elementary) and an approximation to the exact answer (in every case). This information is not required of students working these problems, but it provides an opportunity for extra practice for them in techniques of integration (after they complete Chapter 7) and in numerical integration (which most of them have already completed). The approximations are accurate to the number of decimal places given. 1 C06S04.001: The length is 1 + 4x2 dx = 0 3 C06S04.002: The length is 1 1 (4 + 25x3 )1/2 dx ≈ 14.7554. (The antiderivative is nonelementary.) 2 1 /2 2 C06S04.003: The length is 1 + 36x2 (x − 1)2 0 1 9 + 16x2/3 3 1 C06S04.004: The length is −1 100 C06S04.005: The length is 0 = √ 2 −1 1 0 1 + e2 − √ 5− √ 2 + ln 1 + √ C06S04.010: First, ds = the arc length is π /4 ≈ 3.1678409049. 1 + e2x dx √ 2+ 2 C06S04.009: The arc length is = 1+ 1 1 + e2 ≈ 2.0034971116. 1 x2 2 + ln 2 − ln 1 + √ dy (1 + 16y 6 )1/2 dy ≈ 18.2471. C06S04.008: The arc length is = tanh−1 2 − tanh−1 dx ≈ 2.85552. √ 1 (1 + 4x2 )1/2 dx = 50 40001 + sinh−1 200 ≈ 10001.6228669180. 4 √ 1 sinh−1 4 − sinh−1 2 − 2 5 2 C06S04.007: The length is 1 /2 1 + 4(y − 2)2 0 17 + dx ≈ 6.6617. 1/2 1 C06S04.006: The length is 1√ 2 5 + sinh−1 2 ≈ 1.4789428575. 4 1 /2 dx √ 5 ≈ 1.2220161771. 1 + tan2 x dx = sec x dx because sec x > 0 on the interval 0 π /4 sec x dx = 0 ln | sec x + tan x | 4 C06S04.011: The surface area is = ln 1 + 0 4 2 π /4. Hence ≈ 0.8813735870. 2π x2 (1 + 4x2 )1/2 dx = √ π 1032 65 − sinh−1 8 ≈ 816.5660537285. 32 2π x(1 + 4x2 )1/2 dx = π 653/2 − 1 ≈ 273.8666397863. 6 0 C06S04.012: The surface area is √ x 0 1 1 C06S04.013: The surface area is 0 2π (x − x2 )(4x2 − 4x + 2)1/2 dx π√ = 2 + 5 sinh−1 1 ≈ 1.1429666793. 16 1 C06S04.014: The surface area is 0 2π (4 − x2 )(1 + 4x2 )1/2 dx √ 5π = 22 5 + 13 sinh−1 2 ≈ 33.3601584259. 32 1 C06S04.015: The surface area is 0 = π√ 7 5 + 1 + 6 sinh−1 2 ≈ 13.2545305651. 6 1 C06S04.016: The surface area is 0 = 2π (2 − x)(1 + 4x2 )1/2 dx 2π (x − x3 )(9x4 − 6x2 + 2)1/2 dx √ π√ 5 2 + 5 + 3 sinh−1 1 + 3 sinh−1 2 ≈ 1.8945156885. 27 C06S04.017: Given f (x) = ln(x2 − 1), 2 1 + [ f (x)] = 1 + 4x2 (x2 + 1)2 =2 . (x2 − 1)2 (x − 1)2 Hence the surface area of revolution around the y -axis is 3 A= 2 2π x · x2 + 1 dx = π (5 + 6 ln 2 − 2 ln 3) ≈ 21.8706952193. x2 − 1 4 C06S04.018: The surface area is 2π x 1 + 1 = 1 4x 1 /2 4 dx = π (4x2 + x)1/2 dx 1 √ √ π 132 17 − 18 5 + sinh−1 2 − sinh−1 4 ≈ 49.4162355383. 32 C06S04.019: If f (x) = ln(x + 1), then 2 1 + [ f (x)] = 1 + 1 x2 + 2x + 2 = . (x + 1)2 (x + 1)2 Therefore the surface area of revolution around the line x = −1 is 1 A= 0 2π (x + 1) · √ √ √ x2 + 2x + 2 dx = π 2 5 − 2 + sinh−1 2 − sinh−1 1 ≈ 11.3731443412. x+1 4 C06S04.020: The surface area is 1 2 C06S04.021: The length is π (2 + x5/2 )(4 + 25x3 )1/2 dx ≈ 3615.28. (1 + 2x2 ) dx = x + 0 2 23 x 3 2 = 0 22 . 3 2 3 /2 dy = y 3 5 C06S04.022: The length is y 1 /2 1 C06S04.023: First let f (x) = 1 √ 10 5 − 2 = ≈ 6.7868932583. 3 13 1 x+ . Then 6 2x 1 2 1 −2 x− x 2 2 1 + [ f (x)]2 = 1 + 5 2 =1+ 1 4 1 1 −4 1 11 x − + x = x4 + + x−4 = 4 24 4 24 2 1 2 1 −2 x+ x 2 2 . Therefore the length is 1 2 3 (x2 + x−2 ) dx = 1 C06S04.024: First let g (y ) = 1 + [ g (y )]2 = 1 + 1 2 3 13 1 x− 3 x = 1 13 1 −− 3 3 = 14 . 3 14 1 y + 2 . Then 8 4y 1 3 1 −3 y− y 2 2 2 =1+ 1 6 1 1 −6 1 11 y − + y = y 6 + + y −6 = 4 24 4 24 2 1 3 1 −3 y+ y 2 2 . Therefore the length is 1 2 2 (y 3 + y −3 ) dy = 1 C06S04.025: First solve for y = f (x) = 1 + [ f (x)]2 = 1 + x3 − 1 −3 x 4 1 4 1 −2 y− y 4 2 1 2 2 = 1 31 1 −− 16 8 = 33 = 2.0625. 16 2x6 + 1 1 1 = x4 + x−2 . Then 8x2 4 8 2 = 1 + x6 − 1 1 1 −6 1 −6 + x = x6 + + x= 2 16 2 16 x3 + 1 −3 x 4 2 . Therefore the length is 2 x3 + 1 2 1 −3 1 4 1 −2 x dx = x− x 4 4 8 C06S04.026: First solve for x = g (y ) = 1 + [ g (y )]2 = 1 + y 2 − 1 −2 y 4 = 1 127 1 123 −= = 3.84375. 32 8 32 3 + 4y 4 1 1 = y −1 + y 3 . Then 12y 4 3 2 = 1 + y4 − 1 1 −4 1 1 −4 + y = y4 + + y= 2 16 2 16 y2 + Therefore the length is 2 1 y2 + 1 −2 1 3 1 −1 dy = y y− y 4 3 4 C06S04.027: Given f (x) = 1 2 2 = 1 61 1 59 − = ≈ 2.4583333333. 24 12 24 (ex + e−x ), 2 1 + [ f (x)] = 1 + 1x e − e−x 4 Therefore the length of the given arc is 3 2 = 1x e + e−x 2 2 . 1 −2 y 4 2 . 1 L= 0 1 1x 1x e + e−x dx = e − e−x 2 2 = 0 e2 − 1 ≈ 1.1752011936. 2e C06S04.028: If f (x) = 1 x2 − ln x, then 8 x1 − 4x 2 1 + [ f (x)] = 1 + 2 (x2 + 4)2 . 16x2 = Therefore the length of the given arc is 2 L= 1 x2 + 4 12 dx = x + ln x 4x 8 2 = 1 3 + ln 2 ≈ 1.0681471806. 8 C06S04.029: Because 2 1 −1/2 x 2 1 + [ f (x)]2 = 1 + = 4x + 1 , 4x the surface area is 1 0 1 1/2 πx 2 1 /2 4x + 1 x 1 dx = π (4x + 1) 1 /2 0 1 dx = π (4x + 1)3/2 6 1 0 √ 5 5 −1 = π ≈ 5.3304135003. 6 C06S04.030: Because 1 + [ f (x)]2 = 1 + 9x4 , the surface area is 2 2π x3 (1 + 9x4 )1/2 dx = 1 1 π (1 + 9x4 )3/2 27 2 = 1 √ √ 145 145 − 10 10 π ≈ 199.4804797017. 27 C06S04.031: First, 1 + [ f (x)]2 = 1 + x4 − 1 4x4 2 = x8 + 1 1 = + 2 16x8 x4 + 1 4x4 2 . Therefore the surface area of revolution is 2 2π x x4 + 1 1 4x4 2x5 + 1 =π C06S04.032: Let g (y ) = 2 dx = π 1 −3 x 2 1 6 1 −2 x− x 3 4 dx 2 = 1 339 π ≈ 66.5624943479. 16 1 4 1 −2 y + y . Then 8 4 13 16 2 1 + [ g (y )]2 = y + 2 + y −6 = y + y −3 . 4 4 Therefore the surface area is 2 1 π y 4 + y −2 dy = π 15 y − y −1 5 4 2 = 1 67π ≈ 21.0486707791. 10 C06S04.033: Let f (x) = (3x)1/3 . Then 1 + [ f (x)]2 = 1 /2 1 (3x)4/3 1+ . Therefore the surface area of revolution is 9 0 1 2π x 1 + (3x)4/3 1 /2 C06S04.034: Given f (x) = 1 2 √ 82 82 − 1 = π ≈ 258.8468426921. 9 3 /2 9 π dx = 9x2 + (3x)2/3 27x 0 (ex + e−x ), 2 1 + [ f (x)] = 1 + 1x e − e−x 4 2 1x e + e−x 2 = 2 2 = [ f (x)] . Therefore the surface area of revolution around the x-axis is 1 A= 2 2π [ f (x)] dx = 0 C06S04.035: Given f (x) = x2 − 1 8 π 2x e + 4x − e−2x 4 ln x, 1 1 π2 e + 4 − e−2 ≈ 8.8386516600. 4 = 0 2, x 2 1 + [ f (x)] = 1 + 2x − 1 8x 2 (16x2 + 1)2 . 64x2 = Therefore the surface area of revolution around the y -axis is 2 A= 1 2π x · 2 16x2 + 1 1 4 dx = π x + x3 8x 4 3 = 1 115π ≈ 30.1069296. 12 C06S04.036: The length of one arch of the sine curve is π L1 = 1 + cos2 x 1 /2 dx. 0 To find the arc length L2 of half the ellipse, we take y = 2 − 2x2 dy 2x =− , so 1 + 1 /2 dx (2 − 2x2 ) dy dx 1/2 , −1 2 = x 1. Then 1 + x2 . 1 − x2 Thus L2 = 1 −1 1 + x2 1 − x2 1 /2 dx. Let x = cos u. Then dx = − sin u du, and L2 = 0 π 1 + cos2 u sin2 u 1 /2 (− sin u) du = π 1 + cos2 u 1 /2 du = L1 . 0 This concludes the solution, but an additional comment is appropriate. The Mathematica command 5 Integrate[ Sqrt[ 1 + (Cos[u])∧2 ], {u, 0, Pi} ] elicits the response √ 2 2 EllipticE 1 2 . √ Because elliptic integrals are involved, this strongly suggests that the antiderivative of 1 + cos2 u is nonelementary. In this case none of the techniques of Chpater 7 will produce an antiderivative; y ou should therefore, if necessary, approximate the value of L1 by using Simpson’s approximation. Your result should be close to Mathematica’s approximation 3.820197789027712017904762. C06S04.037: We take f (x) = √ 1 + cos2 x, ∆x = π /6, xi = i · ∆x, and compute ∆x [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + 2f (x4 ) + 4f (x5 ) + f (x6 )] 3 √ √ √ π = 4 + 2 2 + 2 5 + 4 7 ≈ 3.8194031934. 18 S6 = C06S04.038: We take f (x) = S10 = = √ 1 + 4x2 , ∆x = 1/10, xi = i · ∆x, and compute ∆x [ f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 2f (x8 ) + 4f (x9 ) + f (x10 )] 3 √ √ √ √ √ √ √ 1 2√ 4√ 29 + 41 + 61 + 89 + 1+ 5+ 5 2 + 26 + 34 + 74 + 106 30 5 5 ≈ 1.4789423874. The exact value of the arc length is 1 0 √ 2 5 + sinh−1 2 1 + 4x2 dx = ≈ 1.4789428575. 4 C06S04.039: The line segment from (r1 , 0) to (r2 , h) is part of the line with equation y = f (x) = and 1 + [ f (x)]2 = h(x − r1 ) , r2 − r 1 1+ h2 . (r2 − r1 )2 Therefore the area of the conical frustum is r2 r1 2π x 1 + [ f (x)]2 dx = π x2 (r2 − r1 )2 + h2 (r2 − r1 )2 1/2 r2 r1 With r = (r1 + r2 )/2 the “average radius” of the frustum and L = yields the result in Eq. (6) of the text. = π (r1 + r2 ) h2 + (r2 − r1 )2 . h2 + (r2 − r1 )2 its slant height, this C06S04.040: The spherical surface S of radius r centered at the origin may be generated by rotating the √ graph of f (x) = r2 − x2 around the x-axis for −r x r. Because 6 1 + [ f (x)]2 = √ the surface area of S is r −r 2π (r2 − x2 )1/2 · r dx = 2 − x2 )1/2 (r r , r2 − x2 r r 2π r dx = 2π rx −r −r = 2π r2 − (−2π r2 ) = 4π r2 . A perceptive student may notice that we have glossed over a problem with an improper integral here (Section 7.8). Perhaps such a student could be informed that the first simplification is valid for −r < x < r, and that changing the value of an integrable function at two points cannot change the fact of its integrability nor can it change the value of the definite integral. C06S04.041: Let f (x) = (1 − x2/3 )3/2 for 0 lies in the first quadrant in Fig. 6.4.16. Next, 1. Then the graph of f is the part of the astroid that x 1 + [ f (x)]2 = 1 , x1/3 so it would appear that the total length of the astroid is 1 4 0 1 dx. x1/3 But this integral is not a Riemann integral—the integrand approaches + ∞ as x → 0+ . (It is an improper integral, the topic of Section 9.8.) But we can avoid the difficulty at x = 0 by integrating from the midpoint √ of the graph of f to x = 1. That midpoint occurs where y = x, so that 2x2/3 = 1, and thus x = a = 1 2. 4 Therefore the total length of the astroid is 1 1 L=8 a x1/3 3 2 /3 x 2 dx = 8 1 a 33 − 24 =8· = 6. C06S04.042: The surface area is 1 2 0 2π x dx = 2 x1/3 1 12 5/3 πx 5 2π x2/3 dx = 0 1 = 0 12π ≈ 7.5398223686. 5 The integral is improper, but see the remarks attached to the solution of Problem 40. C06S04.043: We will solve this problem by first rotating Fig. 6.4.18 through an angle of 90◦ . Let f (x) = (r2 − x2 )1/2 . Think of the sphere as generated by rotation of the graph of f around the x-axis for −r x r. Suppose that the spherical zone Z of height h is the part of the sphere between x = a and x = a + h, where −r a+h a r. Also, 2π f (x) 1 + [ f (x)]2 = 2π (r2 − x2 )1/2 · (r2 r = 2π r. − x2 )1/2 Therefore the area of the spherical zone Z is a+h A= a a+h 2π r dx = 2π rx = 2π r(a + h) − 2π ra = 2π rh. a 7 As noted in the statement of the problem, the area of Z depends only on the radius r of the sphere and the height (width) h of the zone and not on the location of the two planes (at a and a + h) that determine the zone. You can also “test” the answer by substituting the value 2r for h. C06S04.044: The top half of the loop is the graph of √ x 4 − x2 √ f (x) = 42 for 0 x 2. Next, 1 + [f (x)]2 √ (6 − x2 )2 2 6 − x2 = · . 8(4 − x2 ) 4 (4 − x2 )1/2 = Thus 2π f (x) 1 + [ f (x)]2 = π x(6 − x2 ) , 8 and therefore the surface area generated by rotating the loop of Fig. 6.4.17 around the x-axis is 2 A= 0 π x(6 − x2 ) π2 dx = x (12 − x2 ) 8 32 2 0 = π − 0 = π. C06S04.045: The right-hand endpoint of the cable is located at the point (S, H ), so that H = kS 2 . It follows that y (x) = H2 x, S2 dy 2H = 2 x. dx S so that Therefore the total length of the cable is L= S −S 1+ dy dx 2 S dx = 2 1+ 0 4H 2 2 x dx. S4 C06S04.046: The following Mathematica computations yield Simpson’s approximation to the integral in question. We used s in place of S and h in place of H , and first let s = 8000; h = 380; and then we set up the integrand f (x) = 2 1 + 4h2 x2 . s4 f[x ] := 2*Sqrt[1 + 4∗h∗h∗x∗x/(s∗s∗s∗s)] n = 20; delta = s/n; x[i ] := i∗delta (delta/3)∗(N[f[x[0]],20] + N[f[x[n]],20] + 4∗Sum[N[f[x[i]],20], { i, 1, n − 1, 2 } ] + 2∗Sum[N[f[x[i]],20], { i, 2, n − 2, 2 } ]) 16024.034190838711974 8 Next, n = 40 yielded the (probably) better approximation 16024.034190963156976. We repeated with n = 80, n = 160, n = 320, n = 640, and n = 1280. The last two agreed in the first eleven digits to the right of the decimal; with n = 1280 we obtained 16024.034190971453043. If you prefer exact results, f (x) dx = x s4 + 4h2 x2 s2 2hx + sinh−1 4 s 2h s2 + C, and (with the given values of s and h) s 0 f (x) dx = 1+ 4h 2 s2 2h + sinh−1 2 s 2h s √ 1600000 19 = 40 40361 + sinh−1 19 200 9 ≈ 16024.03419097145305045313786697. Section 6.5 b C06S05.001: The work is W = b 1 b a 0 b C06S05.005: The work is W = 4 F (x) dx = F (x) dx = 1 −1 a 5 32 x −x 2 (3x − 1) dx = 1 b = 30. −2 10 F (x) dx = a C06S05.004: The work is W = 5 F (x) dx = a C06S05.003: The work is W = 10 dx = 10x −2 a C06S05.002: The work is W = 1 1 F (x) dx = 10x−2 dx = − 10 x = 1 65 1 − = 32. 2 2 10 1 = −1 − (−10) = 9. 4 −3x1/2 dx = − 2x3/2 sin π x dx = − 0 1 cos π x π = −16 − 0 = −16. 1 = −1 1 1 − = 0. ππ C06S05.006: The spring constant is 10 N/m, so the force function for this spring is F (x) = 10x. Hence the work done is −2/5 W= 10x dx = 5x2 0 −2/5 0 = 4 = 0.8 5 (N·m). C06S05.007: The spring constant is 30 lb/ft, so the force function for this spring is F (x) = 30x. Hence the work done is 1 W= 1 30x dx = 15x2 = 15 0 0 (ft·lb). C06S05.008: The force function is F (x) ≡ 100, so the work done is 10 W= 10 100 dx = 100x 0 = 1000 0 (ft·lb). C06S05.009: With k = 16 × 109 , we find that the work done is W= 6000 5000 k k dx = − x2 x 6000 = 5000 k 16 = × 105 30000 3 (mi·lb). We multiply by 5280 to convert the answer to 2.816 × 109 ft·lb. C06S05.010: With water density ρ = 62.4 (lb/ft3 ) and cross-sectional area function A(y ) ≡ 25π , we find that the work done to fill the tank is 10 W= 0 10 ρyA(y ) dy = 780π y 2 0 = 78000π ≈ 245044.226980 (ft·lb). C06S05.011: By similar triangles, if the height of water in the tank is y and the radius of the circular water surface is r, then r = 1 (10 − y ). Hence the area of the water surface is A(y ) = 1 π (10 − y )2 . With 2 4 water density ρ = 62.4 (lb/ft3 ), the work done to fill the tank is 1 10 W= ρyA(y ) dy = 0 13 π (3y 4 − 80y 3 + 600y 2 ) 10 10 0 = 13000π ≈ 40840.704497 (ft·lb). C06S05.012: By similar triangles, if the height of water in the tank is y and the radius of the circular water surface is r, then r = 1 y . Hence the area of the water surface is A(y ) = 1 π y 2 . With water density 2 4 ρ = 62.4 (lb/ft3 ), the work done to fill the tank is 10 W= 39 4 πy 10 ρyA(y ) dy = 0 5 C06S05.013: The work is 0 10 0 = 39000π ≈ 122522.113490 250 3 y + 1250y 2 3 50(y +10) · 5π y dy = π 5 = 0 (ft·lb). 125000π ≈ 130899.694 3 (ft·lb). C06S05.014: Suppose that the bottom of the tank is located where y = a and the top where y = b. Let n be a positive integer and let P = {y0 , y1 , y2 , . . . , yn } be a partition of the interval [ a, b ]. Let S = {y1 , y2 , . . . , yn } be a selection for P . Consider the liquid in the tank between the levels yi−1 and yi where 1 i n. Let A(yi ) be the cross-sectional area of the liquid at level yi Then the weight of the liquid between yi−1 and yi will be approximately ρA(yi ) ∆yi (where ∆yi = yi − yi−1 ) and this liquid must be lifted the approximate distance h − yi in pumping all the liquid in the tank to the level y = h. Hence the total work done will be approximately n i=1 ρ(h − yi )A(yi ) ∆yi . But the error in these approximations will approach zero as n → + ∞ and the maximum of the ∆yi approaches zero. Because this sum is a Riemann sum, it has as its limit the following integral; therefore the work to pump all the liquid in the tank to the level y = h will be b W= a 10 C06S05.015: W = 0 ρ(h − y )A(y ) dy. 10 ρ(15 − y ) · 25π dy = π 23400y − 780y 2 0 = 156000π ≈ 490088.454 (ft·lb). C06S05.016: Set up the following coordinate system: The x-axis and y -axis cross at the center of one end of the tank, so that the equation of the circular (vertical) cross section of the tank is x2 + y 2 = 9. Then the gasoline must be lifted to the level y = 10. A horizontal cross section of the tank at level y is a rectangle of length 10 and width 2x where x and y satisfy the equation x2 + y 2 = 9 of the end of the tank, and thus 2x = 2(9 − y 2 )1/2 . Thus the amount of work required to pump all the gasoline in the tank into automobiles will be W= 3 2 −2 3 = 900 −3 = 9000 · · 10 · (10 − y ) · 45 dy 9 − y2 10 9 − y2 − y 9 − y2 1 1 · π · 9 + 900 (9 − y 2 )3/2 2 3 = 40500π ≈ 127234.5 2 (ft·lb). dy 3 −3 Now assume that the tank uses a 1.341 hp motor. If it were to operate at 100% efficiency, it would pump all the gasoline in 40500π ≈ 2.875161 33000 · 1.341 minutes, using 1 kW for 2.875161 minutes. This would amount to 0.047919356 kWh, costing about 0.345 cents. Assuming that the pump is only 30% efficient, the actual cost would be about 1.15 cents. C06S05.017: With water density ρ = 62.4 lb/ft3 , the work will be 10 −10 ρ(50 + y ) · π (100 − y 2 ) dy = 10 26 π 60000y + 600y 2 − 200y 3 − 3y 4 5 −10 = 4160000π ≈ 13069025 (ft·lb). C06S05.018: Place the origin at the center of the base of the hemisphere, at indicated in Fig. 6.6.16. Then the cross-sectional area at y is A(y ) = π (100 − y 2 ) for 0 y 10 and the oil that ends up at level y is lifted a total distance 60 + y , so the work is 10 0 50(y + 60)A(y ) dy = π 300000y + 2500y 2 − 1000y 3 − 25 4 y 2 10 0 = 2125000π ≈ 6675884 (ft·lb). C06S05.019: Let y = 0 at the surface of the water in the well, so the top of the well is at y = 100. The weight of water in the bucket is 100 − 1 y when the bucket is at level y , so the total work done in lifting the 4 water to the top of the well is 100 W= 0 100 − y 4 dy = 100y − 12 y 8 100 = 8750 0 (ft·lb). C06S05.020: Set up a coordinate system in which the bottom of the rope is initially located at y = 0 and the top of the building at y = 100. When the bottom of the rope has been lifted to position y (0 y 100), then the weight of the part of the rope still dangling from the top is (100 − y )/4 (pounds). The work required to lift the rope a short distance ∆y is approximately 1 (100 − y ) ∆y , and therefore the total work to lift the 4 rope to the top of the building will be 100 W= 0 1 1 (100 − y ) dy = 25y − y 2 4 8 100 = 1250 0 (ft·lb). (1) This solution in effect partitions the process of lifting the rope to the top of the building. You may prefer the alternative of partitioning the rope. Imagine a short section of the rope near position y and of length ∆y . This section of rope weighs 1 ∆y pounds and will be lifted the distance 100 − y to the top of 4 the building, so the work to lift this short section of rope will be 1 (100 − y ) ∆y . So the total work to lift all 4 such short sections of the rope will be exactly the same as that shown in Eq. (1). Finally, the answer may be checked in the following way: Stiffen the rope and turn it 90◦ around its center (where y = 50). This requires no net work as the rope will balance at its center. Then lift the 25-lb rope 50 feet to the top of the building, requiring 25 · 50 = 1250 ft·lb of work. C06S05.021: The weight of the rope and the water will be w(y ) = 100 + (100 − y )/4 when the bucket is y feet above the water surface. So the work to lift the rope and the water to the top of the well will be W= w(y ) dy = 125y − 3 12 y 8 100 = 11250 0 (ft·lb). x2 C06S05.022: The work is W = x1 A · p(Ax) dx. Let V = Ax; then dV = A dx. Therefore V2 W= pV dV. V1 C06S05.023: Given: pV 1.4 = c. When p1 = 200, V1 = 50. Hence c = 200 · 507/5 = 200 · 507/5 , so c p= V 7/5 = 200 · 50 V 7 /5 . Therefore the work done by the engine in each cycle is 500 W= 50 50 V 200 · 7 /5 500 5 dV = 200 · 507/5 · − V −2/5 2 = 2500(10 − 103/5 ) ≈ 15047.320736. 50 The answer is in “inch-pounds” (in.·lb); divide by 12 to convert the answer into W ≈ 1253.94339468 footpounds. C06S05.024: Set up a coordinate system in which the center of the hemisphere is at the origin, with a diameter lying on the x-axis and the y -axis perpendicular to the base, so that the highest point of the hemisphere has coordinates (0, 60). Now imagine a horizontal thin circular slice of its contents at position y and having radius x, so that x2 + y 2 = 3600. If the thickness of this slice is dy , then its volume is dV = π 3600 − y 2 dy , so its weight is 40π 3600 − y 2 dy . This is the force acting on the slice, which is to be lifted a distance 60 − y , so the work used in lifting this slice is 40π 3600 − y 2 (60 − y ) dy . Therefore the total work to pump all the liquid to the level of the top of the tank is 60 W= 0 1 C06S05.025: W = 0 40π 3600 − y 2 (60 − y ) dy = 216000000π ≈ 6.78584 × 108 (ft·lb). √ 60π (1 − y ) y dy = 16π ≈ 50.265482 (ft·lb). C06S05.026: Set up a coordinate system in which the center of the tank is at the origin and the ground surface coincides with the horizontal line y = −3. Imagine a horizontal cross section of the tank at position y , −3 y ≤ 3. The equation of the circle x2 + y 2 = 9 gives us the width 2x of this rectangular cross section: 1 /2 2x = 2 9 − y 2 . The length of the cross section is 20, so if we denote its thickness by dy then its volume is 40 9 − y 2 1 /2 dy . To fill this slab with gasoline weighing 40 pounds per cubic foot, which is to be lifted the distance y + 3 feet, requires dW = 1600(y + 3) 9 − y 2 the tank is W= 3 −3 1600(y + 3) 9 − y 2 = 1600 3 −3 y 9 − y2 1 /2 1 /2 1 /2 dy ft·lb of work. So the work required to fill dy dy + 4800 3 −3 9 − y2 3 /2 1 /2 dy. The first integral is zero because it involves the evaluation of 9 − y 2 at y = 3 and at y = −3. The second is the product of 4800 and the area of a semicircle of radius 3, so the answer is that the total work is 21600π ≈ 67858.401318 ft·lb. 4 C06S05.027: It is convenient to set up a coordinate system in which the center of the tank is at the origin, the x-axis horizontal, and the y -axis vertical. A horizontal cross section at y is circular with radius x satisfying x2 + y 2 = 144, so the work to fill the tank is W= 12 −12 (50π )(y + 12) 144 − y 2 dy = π 86400y + 3600y 2 − 200y 3 − = 950400π − (−432000π ) = 1382400π ≈ 4.342938 × 106 25 4 y 2 12 −12 (ft·lb). C06S05.028: Set up a coordinate system in which the y -axis is vertical, y = 0 corresponds to the bottom of the cage, and y = 40 to its top. The work done in lifting the monkey from y = 0 to y = 40 is 20 · 40 = 800 ft·lb. Now we compute the work done in lifting the chain. Suppose that the free end of the chain is lifted from y = 0 to y = 10 and is at position y . Then the length of chain lifted is y , so its weight is 1 y . Then the free end of the chain is lifted from y = 10 to y = 40. 2 When the free end is at position y , part of the chain is doubled; let z denote the length of each part that is doubled. The length of the part of the chain not doubled is 40 − y . So 40 − y + 2z = 50, and it follows that z = 5 + 1 y . So the weight of the chain lifted is 1 z = 5 + 1 y . Hence the work done to lift the chain is 2 2 2 4 10 0 1 y dy + 2 40 10 51 + y dy = 25 + 262.5 = 287.5 24 (ft·lb). Therefore the total amount of work done in lifting monkey and chain is 800 + 287.5 = 1087.5 ft·lb. C06S05.029: Let the string begin its journey stretched out straight along the x-axis from x = 0 to √ x = 500 2. Imagine it reaching its final position by simply pivoting at the origin up to a 45◦ angle while remaining straight. A small segment of the string initially at location x and of length dx is lifted from y = 0 √ to the final height y = x/ 2, so the total work done in lifting the string is √ 500 2 W= 0 x 1 √ dx = √ x2 16 2 32 2 √ 500 2 0 √ 15625 2 = 2 (ft·oz). Divide by 16 to convert the answer into ft·lb. Answer: Approximately 690.533966 ft·lb. To check the answer without using calculus, note that the string is lifted an average distance of 250 feet. Multiply this by the weight of the string in pounds to obtain the answer in ft·lb. C06S05.030: Set up a coordinate system in which the center of the sphere is located at the origin. The cross section of the liquid in the tank “at” position y (−R y R) is circular with radius x = R2 − y 2 . Hence the work to fill the tank is W= R −R (y + H ) ρπ R2 − y 2 dy 1 1 1 = ρπ R2 Hy + R2 y 2 − Hy 3 − y 4 2 3 4 2 = ρπ 2R3 H − R3 H 3 = R −R 4 πρR3 H. 3 This is the product of the volume 4 π R3 of the tank, the weight density ρ of the liquid, and the distance H 3 from the ground to the center of the tank, and the result in Problem 30 now follows. 5 C06S05.031: Set up a coordinate system with the y -axis vertical and the x-axis coinciding with the bottom of one end of the trough. A horizontal section of the trough at y is 2 − y feet below the water surface, so the total force on the end of the trough is given by 2 F= 0 2 (2)(2 − y )ρ dy = ρ 4y − y 2 = 4ρ = 249.6 (lb). 0 C06S05.032: Set up a coordinate system in which the origin is at the lowest point of the triangular end of √ the trough and the y -axis is vertical. A narrow horizontal strip at height y has width 2x = 2 y 3. Therefore 3 the total force on the end of the trough is given by 3 2 F= √ 3√ 3−y 2 3 ρ 0 √ 32 23 3 =ρ y− y 2 9 2√ y3 3 3√ 23 = 0 3 2 dy = ρ √ 2√ 3y − y 2 3 3 3 0 27 ρ = 210.6 8 dy (lb). C06S05.033: Set up a coordinate system in which one end of the trough lies in the [vertical ] xy -plane with its base on the x-axis and bisected by the y -axis. Thus the trapezoidal end of the trough has vertices at the points (1, 0), (−1, 0), (2, 3), and (−2, 3). Because the width of a horizontal section at height y is 2x = 2 (y + 3), the total force on the end of the trough is 3 3 F= 0 2 2 ρ (y + 3)(3 − y ) dy = ρ 6y − y 3 3 9 3 = 12ρ = 748.8 (lb). 0 C06S05.034: Describe the end of the tank by the inequality x2 + y 2 16, so that a horizontal section at 1 /2 level y has width 2x = 2 16 − y 2 . Note that ρ = 50 in this problem. Then the total force on the end of the tank is F= 4 −4 ρ(2)(4 − y ) 16 − y 2 1 /2 dy = 2ρ 4 −4 4 16 − y 2 1 /2 dy − 2ρ 4 −4 y 16 − y 2 1/2 dy. The last integral involves the evaluation of (16 − y 2 )3/2 at y = 4 and at y = −4, so its value is zero. The next-to-last is the product of 8ρ and the area of a semicircle of radius 4, so its value is (8ρ)(8π ) = 64ρπ = 3200π ≈ 10053.1 (lb). C06S05.035: Let ρ = 62.4 lb/ft3 and set up a coordinate system in which the y -axis is vertical and y = 0 is the location of the bottom of the square gate. Then the pressure at a horizontal cross section of the plate at location y will be (15 − y )ρ and the area of the cross section will be 5 dy , so the total force on the gate will be 5 F= 0 5ρ(15 − y ) dy = ρ 75y − 52 y 2 5 = 19500 (lb). 0 C06S05.036: Let ρ = 62.4, as usual. Put the origin at the center of the circle. Then the force on the gate is 6 3 F= −3 2ρ(13 − y ) 9 − y 2 1 /2 dy = 26ρ 3 −3 1 /2 9 − y2 dy − 2ρ 3 −3 y 9 − y2 1 /2 dy. The last integral involves the evaluation of (9 − y 2 )1/2 at y = 3 and at y = −3, so its value is zero. The next-to-last is the product of 26ρ and the area of a semicircle of radius 3, so its value is the product of 26ρ and 9 π : 117πρ ≈ 22936.139645 (lb). 2 C06S05.037: Let ρ denote the density of water, as usual. Place the origin at the low vertex of the triangular gate with the y -axis vertical. Then a horizontal cross section of the gate at y has width 8 y and depth 15 − y , 5 so the total force on the gate will be 5 F= 0 8 83 ρy (15 − y ) dy = ρ 12y 2 − y 5 15 5 = 14560 (lb). 0 C06S05.038: Place the origin at the center of the diameter of the gate with the y -axis vertical; let ρ = 62.4 lb/ft3 denote the density of water. The equation of the semicircle is x2 + y 2 = 16, y 0. So the force on the gate is F= 0 −4 2ρ(10 − y ) 16 − y 2 1 /2 dy = 20ρ 0 −4 16 − y 2 1 /2 dy − 2ρ 0 −4 y 16 − y 2 1 /2 dy. The next-to-last integral is the product of 20ρ and the area of a quarter-circle of radius 4, so its value is 80πρ. Therefore 1 F = 80πρ − 2ρ − (16 − y 2 )3/2 3 0 = 80πρ + −4 128 ρ ≈ 18345.230527 3 (lb). C06S05.039: Set up a coordinate system in which the bottom of the vertical face of the dam lies on the x-axis and the y -axis is vertical, with the origin at the center of the bottom of the vertical face of the dam. The vertical face occupies the interval 0 y 100; form a regular partition of this interval and suppose that [ yi−1 , yi ] is one of the subintervals in this partition. Horizontal lines perpendicular to the vertical face through the endpoints of this interval determine a strip on the slanted face of length 200 (units are in feet) and (by similar triangles) width √ √ 1002 + 302 1002 + 302 (yi − yi−1 ) = ∆y. 100 100 If yi lies in this interval, then the strip on the slanted face opposite [ yi−1 , yi ] has approximate depth 100 − yi . So the total force of water on this strip—acting normal to the slanted face—is √ 1002 + 302 200ρ(100 − yi ) · ∆y 100 where ρ = 62.4 lb/ft3 is the density of the water. Therefore the total force the water exerts on the slanted face of the dam—normal to that face—is 100 F= 0 200ρ(100 − y ) · √ 1002 + 302 dy 100 100 = ρ(109)1/2 2000y − 10y 2 0 √ = 100000ρ 109 ≈ 6.514721 × 107 7 (lb). √ The horizontal component of this force can be found by multiplying the total force by 10/ 109, and is thus 6.24 × 107 (lb). C06S05.040: Given f (x) = 1 + 12 14 x− x, 5 500 water filling the birdbath occupies the space region S generated by rotation around the y -axis of the plane region R bounded on the left by the y -axis, above by the line y = 6, and below and on the right by the graph √ of f (x) for 0 x α = 5 2 . With measurements in inches, fresh water has weight density ρ= 62.4 1728 pounds per cubic inch; we will convert to ft·lb at the end of the solution. Solution (a): By the method of nested cylindrical shells. Such a shell, with centerline the y -axis and meeting the positive x-axis at x, has radius x, height 6 − f (x) and the water comprising it has been lifted an average distance of 40 + f (x) + 6 − f (x) 2 inches. Hence the work to lift all the water from ground level to fill the space region S is α W= 0 = π· 2πρx [6 − f (x)] · 40 + f (x) + 6 − f (x) 2 dx 377x2 533x4 403x6 13x8 13x10 − + + − 48 3600 540000 3600000 900000000 α = 0 28925π . 216 We divide by 12 to convert the answer to 28925π ≈ 35.058089315233 2592 (ft·lb). Solution (b): By the method of parallel slabs. A horizontal slab “at” location y , 1 x= 50 − 10 y 6, has radius 30 − 5y . Hence the work to pump water originally at ground level to fill the region S will be 6 W= 1 = π· πρ(40 + y ) · 50 − 10 650y 65y 2 + + 9 72 30 − 5y dy 2756 1261y 13y 2 − − 45 135 90 6 30 − 5y = 1 28925π , 216 and this solution concludes in the same way, and with the same result, as the previous solution. 8 Section 6.6 C06S06.001: By symmetry, the centroid is located at (2, 3). C06S06.002: By symmetry, the centroid is at (2, 3). C06S06.003: By symmetry, the centroid is at (1, 1). 9 and 2 C06S06.004: The area of the triangle is 3 My = 0 x(3 − x) dx = 3 32 13 x− x 2 3 = 0 27 18 9 − =. 2 2 2 Hence x = 1, and by symmetry y = 1 as well. C06S06.005: The area of the triangular region is A = 4, and 4 My = 0 Hence x = x 2− 1 x 2 dx = x2 − 13 x 6 4 = 16 − 0 32 16 = . 3 3 4 . Next, 3 4 Mx = 0 and therefore y = 1 · 2 4−x 2 2 dx = 1 8 4 0 (16 − 8x + x2 ) dx = 1 1 16x − 4x2 + x3 8 3 4 = 0 8 , 3 2 . 3 C06S06.006: The area of the triangular region is A = 1. Moreover, x = 1 by symmetry. Using additivity of moments, 1 12 1 Mx = 2 x dx = . 2 3 0 Therefore y = 1 . 3 C06S06.007: The area of the region is 2 A= x2 dx = 0 13 x 3 2 = 0 8 . 3 Next, 2 My = 14 x 4 x3 dx = 0 Mx = Therefore (x, y ) = 1 2 2 x4 dx = 0 36 , . 25 C06S06.008: The area of the region is 1 2 =4 and 0 15 x 10 2 = 0 16 . 5 3 A=2 0 (9 − x2 ) dx = 2 9x − 3 13 x 3 0 = 2 · (27 − 9) = 36. Also x = 0 by symmetry. Finally, Mx = Therefore y = 1 2 3 (92 − x4 ) dx = −3 81x − 15 x 5 3 0 = 243 − 243 972 = . 5 5 972 27 = . 5 · 36 5 C06S06.009: The area of the region is 2 A=2 0 (4 − x2 ) dx = 2 4x − 13 x 3 2 0 = 2· 8− 8 3 = 32 . 3 Now x = 0 by symmetry, but Mx = − 1 2 =− Therefore y = − 2 −2 2 (4 − x2 )2 dx = − 15 83 x − x + 16x 5 3 0 2 =− 0 (x2 − 4)2 dx = − 32 64 − + 32 5 3 2 0 (x4 − 8x2 + 16) dx =− 256 . 15 256 3 8 · =− . 15 32 5 C06S06.010: By symmetry, x = 0. Next, 1 2 Mx = The area of the region is 2 (x2 + 1)2 dx = −2 206 . 15 28 103 , and hence y = . 3 70 C06S06.011: The area of the region is 2 A=2 0 2 Mx = −2 Therefore y = (4 − x2 ) dx = 32 , 3 and 1 256 (4 − x2 )2 dx = . 2 15 256 · 3 8 = ; x = 0 by symmetry. 15 · 32 5 C06S06.012: The area of the region is 3 A= 0 3 Mx = 0 (18 − 2x2 ) dx = 36 and 1 · 18 · (18 − 2x2 ) dx = 324. 2 Therefore (x, y ) = (0, 9). Alternatively, you can find the centroid by symmetry. 2 C06S06.013: The area of the region is A = 1. Moreover, 1 My = 3 4 3x3 dx = 0 Therefore (x, y ) = and 94 9 x dx = . 2 10 1 Mx = 0 39 , . 4 10 C06S06.014: The area of the region is 4 A= x dx = 16 , 3 x3/2 dx = 64 , 5 √ 0 4 My = 0 2 Mx = 0 Therefore (x, y ) = and y (4 − y 2 ) dy = 4. 12 3 . , 54 C06S06.015: The parabola and the line meet at the two points P (−3, −3) and Q(2, 2). Hence A= 2 (6 − x − x2 ) dx = −3 My = 2 (6x − x2 − x3 ) dx = − −3 Mx = 125 , 6 125 , 12 1 1 125 (6 − x2 )2 − x2 dx = . 2 2 3 2 −3 1 − ,2 . 2 Therefore the centroid is located at the point C06S06.016: First note that x = y by symmetry. Next, 1 A= √ 0 x − x2 dx = 1 My = x3/2 − x3 0 1 3 and 3 . 20 dx = 99 , . 20 20 Therefore the centroid is located at the point C06S06.017: The region has area A = and 1 . Next, 12 1 My = 0 Mx = 1 2 (x3 − x4 ) dx = 1 0 1 20 (x4 − x6 ) dx = 3 1 . 35 and Thus (x, y ) = 3 12 , . 5 35 π . Next, 2 C06S06.018: By symmetry, x = π A= sin x dx = 2 and 0 π Mx = 0 1 π sin2 x dx = . 2 4 π Therefore y = . 8 C06S06.019: First note that y = x by symmetry. Next, r My = x(r2 − x2 )1/2 dx = 0 and the area of the quarter circle is 1 − (r2 − x2 )3/2 3 r = 0 13 r 3 12 π r . Therefore the centroid is at 4 4r 4r (x, y ) = , . 3π 3π C06S06.020: By Pappus’s first theorem, 12 πr 4 · (2π x) = 23 πr . 3 13 r 4r Therefore x = 3 = . By symmetry, y = x. 12 3π πr 4 C06S06.021: By symmetry, y = x. Because y = (r2 − x2 )1/2 , 1+ dy dx 2 = r2 r2 . − x2 So r My = 0 The length of the quarter circle is rx dx = (r2 − x2 )1/2 r − r(r2 − x2 )1/2 1 2r π r, so x = . 2 π C06S06.022: By Pappus’s second theorem, 1 π r · (2π x) = 2π r2 . 2 Therefore—with the aid of symmetry—x = 2r = y. π C06S06.023: First, 4 = r2 . 0 r My = 0 x h− hx r dx = 12 hr 6 and A= 1 rh. 2 Therefore x = r/3. By interchanging the roles of x and y , we find that y = x. Next, the midpoint of the hypotenuse is (r/2, h/2) and its slope is −h/r. The line L from (0, 0) to the midpoint has equation y= h x. r If x = r/3, then y = h/3, so (r/3, h/3) lies on the line L. The distance from (0, 0) to (r/3, h/3) is D1 = 12 (r + h2 )1/2 ; 3 D2 = 12 (r + h2 )1/2 . 2 the distance from (0, 0) to (r/2, h/2) is Therefore D1 2 =, D2 3 and this concludes the proof. C06S06.024: V = 2π r · 3 1 rh 2 C06S06.025: A = 2· πr 2 r2 + h2 = π r(r2 + h2 )1/2 = π rL. = 12 π r h. 3 C06S06.026: (a) Part (1): the rectangle. Its area is A = r2 h, and thus r2 My = xh dx = 0 r2 Mx = 0 12 hr ; 22 12 1 h dx = h2 r2 . 2 2 1 Part (2): the triangle. Its area is A = h(r2 − r1 ). By the result in Problem 23, 2 1 1 x = r2 + (r1 − r2 ) = (r1 + 2r2 ), 3 3 and y = h/3. Therefore M y = Ax = h (r1 + 2r2 )(r1 − r2 ) 6 M x = Ay = h2 (r1 − r2 ). 6 and Part (3): the trapezoid. By additivity of moments, My = 12h h2 2 hr2 + (r1 + 2r2 )(r1 − r2 ) = (r1 + r1 r2 + r2 ); 2 6 6 5 Mx = 12 h2 h2 h r2 + (r1 − r2 ) = (r1 + 2r2 ). 2 6 6 Answer to (a): x= 2 1 r 2 + r 1 r 2 + r2 My = 1 , A 3(r1 + r2 ) y= 1 h(r1 + 2r2 ) Mx = . A 3(r1 + r2 ) Answer to (b): By Pappus’s first theorem, V = 2π xA = 2π My = 1 (r1 + r2 ), so the lateral area is 2 C06S06.027: The radius of revolution is A = 2π · πh 2 2 (r + r1 r2 + r2 ). 31 r 1 + r2 2 (r2 − r1 )2 + h2 1 /2 = π (r1 + r2 )L. C06S06.028: The lateral surface is generated by rotating around the axis of the cylinder a vertical line of length h. Its midpoint is at (r, h/2) and the radius of the circle around which the midpoint moves is r, so the lateral surface area is A = (2π r)h. Alternatively, from Problem 27, A = π (r1 + r2 )L = 2π rh. C06S06.029: The semicircular region has centroid (x, y ) where x = 0 (by symmetry) and y = b + earlier work). So, for the semicircular region, Mx = b+ 4a 3π · 4a (by 3π 12 πa . 2 For the rectangle, we have Mx = b · 2ab = ab2 . 2 The sum of these two moments is the moment of the entire region: 23 π2 a + a b + ab2 . 3 2 Mx = When we divide this moment by the area 2ab + y= 12 π a of the entire region, we find that 2 4a2 + 3π ab + 6b2 . 12b + 3π a Of course x = 0 by symmetry. For Part (b), we use the fact that the radius of the circle of rotation is y , so the volume generated by rotation around the x-axis is V = 2π yA = 2π y 2ab + 12 πa 2 6 = π y (4ab + π a2 ) = π ay (4b + π a) = 1 π ay (12b + 3π a); 3 that is, 1 π a(4a2 + 3π ab + 6b2 ). 3 V= C06S06.030: (a) First note that h A= 2 3 /2 h 3 2py dy = 0 2p . Now r2 = 2ph, so A= 2 h 3 2ph = 0 so x = ph /2 3ph 1 = . But ph = r2 , so x = 2rh/3 4r 2 2 But 1 1 (2py ) dy = ph2 , 2 2 h My = 2 rh. 3 3 8 r. 12 r , so 2 1 V = π r2 h. 2 Part (b): V = 2π xA = 2π My = π ph2 . But ph = C06S06.031: Because y = √ 1 − x2 , as in Example 6 of this section we have ds = √ 1 dx. 1 − x2 Hence the moment with respect to the y -axis is sin α Mx = 2 0 √ y dx = 1 − x2 sin α 2x = 2 sin α. 0 Clearly the arc length is s = 2α, and therefore the y -coordinate of the centroid is y= sin α . α C06S06.032: Because d = 1− sin α α and h = 1 − cos α, three applications of l’Hˆpital’s rule then produce the value of the limit: o lim α→0 d α − sin α 1 − cos α = lim = lim α→0 α − α cos α α→0 1 − cos α + α sin α h = lim α→0 sin α cos α 1 = lim =. α→0 3 cos α − α sin α α cos α + 2 sin α 3 7 C06S06.033: Let f (x) = x and g (x) = x2 . The area of the region bounded by the graphs of f and g is 1 A= 0 (x − x2 ) dx = 11 1 −=. 23 6 The moments with respect to the coordinate axes are 1 My = 0 1 Mx = 0 (x2 − x3 ) dx = 11 1 −= 34 12 12 1 (x − x4 ) dx = · 2 2 and 11 − 35 = 1 . 15 Therefore the centroid is located at the point (x, y ) = C 12 , 25 . The axis L of rotation is the line y = x; the line through the centroid perpendicular to L has equation y= and this perpendicular meets L at the point P d= 1 9 − 2 20 9 −x 10 99 . The distance from P to C is , 20 20 2 + 2 9 − 5 20 2 = √ 2 . 20 Because d is the radius of the circle through which C is rotated, the volume generated is (by the first theorem of Pappus) V = 2π dA = 2π · √ √ π2 21 ·= ≈ 0.07404804897. 20 6 60 C06S06.034: We let f (x) = xm and g (x) = xn . Then we used Mathematica 3.0: A = Integrate[ f[x] − g[x], { x, 0, 1 } ] 1 1 − 1+m 1+n Then we compute the moments: My = Integrate[ x∗(f[x] − g[x]), { x, 0, 1 } ]; Mx = Integrate[ (1/2)∗( (f[x])∧2 − (g[x])∧2 ), { x, 0, 1 } ]; Thus the centroid has coordinates { xc, yx } = { My/A, Mx/A } // Simplify { (1 + m)(1 + n) 1 + m + n + mn , } (2 + m)(2 + n) 1 + 2m + 2n + 4mn 8 For selected values of m and with n = m + 1 we check to see if it’s True that the centroid lies within the region: m = 1; n = m + 1; yc < xc∧m yc > xc∧n True True m = 2; n = m + 1; yc < xc∧m yc > xc∧n True True m = 3; n = m + 1; yc < xc∧m yc > xc∧n False True Therefore if m = 3 and n = 4, then the centroid does not lie within the region. 9 Section 6.7 C06S07.001: If f (x) = 10x , then f (x) = 10x ln 10 by Eq. (28). 2 C06S07.002: If f (x) = 21/x , then f (x) = − C06S07.003: If f (x) = 3x = 4x 2 2 · 21/x ln 2 by Eq. (29). 3 x x 3 4 3 4 , then f (x) = C06S07.004: If f (x) = log10 cos x, then f (x) = − x ln 3 by Eq. (28). 4 sin x by Eq. (40). (ln 10) cos x C06S07.005: If f (x) = 7cos x , then f (x) = − (7cos x ln 7) · sin x by Eq. (29). 2 C06S07.006: If f (x) = 2x · 3x , then 2 2 f (x) = (2x ln 2) · 3x + 3x ln 3 · 2x · 2x by the product rule and Eq. (29). √ C06S07.007: If f (x) = 2x x 3/2 = 2x , then f (x) = C06S07.008: If f (x) = log100 10x = x log100 10 = C06S07.009: If f (x) = 2ln x , then f (x) = x 3 1/2 x3/2 2 ln 2 . x 2 1 1 x, then f (x) ≡ . 2 2 1 ln x 2 ln 2 . x x x C06S07.010: If f (x) = 78 = 7(8 ) , then f (x) = 78 · 8x · (ln 7) · (ln 8). C06S07.011: If f (x) = 17x , then f (x) = 17x ln 17. √ C06S07.012: If f (x) = 2 x , then f (x) = 1 −1/2 √x 2 ln 2 . x 2 C06S07.013: If f (x) = 101/x , then f (x) = − √ C06S07.014: If f (x) = 3 1−x2 x 1 101/x ln 10 . x2 , then f (x) = − x(1 − x2 )−1/2 3 x x √ 1−x2 ln 3 . 2 C06S07.015: If f (x) = 22 = 2(2 ) , then f (x) = 22 · 2x · (ln 2) . C06S07.016: If f (x) = log2 x, then f (x) = C06S07.017: If f (x) = log3 √ x2 + 4 = f (x) = 1 by Eq. (39). x ln 2 1 log3 (x2 + 4), then by Eq. (40) we find that 2 1 1 x · · 2x = 2 . 2 (x2 + 4) ln 3 (x + 4) ln 3 C06S07.018: If f (x) = log10 (ex ) = x log10 e, then f (x) = log10 e = 1 1 (by Eq. (40)). ln 10 C06S07.019: If f (x) = log3 (2x ) = x log3 2, then f (x) = log3 2 = ln 2 (by Eq. (40)). ln 3 C06S07.020: If f (x) = log10 (log10 x), then by Eq. (40) f (x) = (log10 e) ln (log10 x) = (log10 e) ln [(log10 e) ln x ] = (log10 e) [ln (log10 e) + ln (ln x)] . Therefore 1 x ln x f (x) = (log10 e) = 1 . (x ln x) ln 10 C06S07.021: If f (x) = log2 (log3 x), then by Eq. (35) f (x) = (log2 e) ln (log3 x) = (log2 e) ln [(log3 e) ln x ] = (log2 e) [ln (log3 e) + ln (ln x)] . Therefore 1 x ln x f (x) = (log2 e) = 1 . (x ln x) ln 2 C06S07.022: If f (x) = π x + xπ + π π , then f (x) = π x ln π + π xπ−1 . C06S07.023: If f (x) = exp (log10 x), then f (x) = f (x) = exp (log10 x) . Mathematica reports that x ln 10 x−1+(1/ ln 10) , ln 10 but a moment’s work shows that the two answers are the same. 3 3 3 C06S07.024: If f (x) = π x = π (x ) , then f (x) = (3x2 )π x ln π by Eq. (29). 32 x + C by Eq. (30). 2 ln 3 C06S07.025: 32x dx = C06S07.026: x · 10−x dx = − 2 C06S07.027: 2 √ 10−x + C. 2 ln 10 √ 2x 2·2 x √ dx = + C. ln 2 x Comment: The easiest way to find an antiderivative such as this is to make an “educated guess” as to the form of the answer, differentiate it, and then modify the guess so it becomes correct. Here, for example, √ we guess 2 x for the antiderivative. Then Dx 2 √ x =2 √ x ln 2 · Dx x =2 1 /2 √ x √ 1 ln 2 2 x ln 2 · √ = ·√ . 2 2x x 2 to correct it. ln 2 Of course this technique will succeed only if the correction consists of multiplication by a constant. If something else is needed, make a better guess or try integration by substitution. √ Thus we should multiply 2 x by 2 C06S07.028: 101/x 101/x dx = − + C. 2 x ln 10 C06S07.029: Given: 3 x2 · 7x +1 dx. Let u = x3 + 1. Then du = 3x2 dx, so that x2 dx = 1 du. Thus 3 3 3 x2 · 7x +1 dx = 1u 7u 7x +1 7 du = +C = + C. 3 3 ln 7 3 ln 7 C06S07.030: First, x log10 x = x (log10 e) ln x = 1 dx = x log10 x C06S07.031: log2 x dx = x x ln x . Thus ln 10 ln 10 dx = (ln 10) x ln x 1 dx = (ln 10) · ln (ln x) + C. x ln x (log2 e) ln x 1 dx = x ln 2 ln x 1 2 dx = (ln x) + C . x 2 ln 2 C06S07.032: If necessary, use the substitution u = 2x . In any case, x (2x ) 3(2 x ) dx = 3(2 ) + C. (ln 3)(ln 2) C06S07.033: Taking logarithms transforms the equation R = kW m into ln R = ln k + m ln W , an equation linear in the two unknown coefficients ln k and m. We put the data given in Fig. 6.7.13 into the array datapoints = { {25, 131}, {67, 103}, {127, 88}, {175, 81}, {240, 75}, {975, 53} }, then entered the Mathematica command logdatapoints = N[Log[datapoints], 10] to obtain the logarithms of the values of W and R to ten significant figures. We then set up the graph pts = ListPlot[ logdatapoints ]; to see if the data points lay on a straight line. They very nearly did. Next we used Mathematica’s Fit command to find the coefficients of the equation of the straight line that best fit the data points (by minimizing the sum of the squares of the deviations of the data points from the straight line—see Miscellaneous Problem 51 of Chapter 13). The command Fit[ logdatapoints, {1, x}, x ] finds the best-fitting linear combination a · 1 + b · x to the given data. The result was 5.67299196 − 0.24730011 x which told us that ln k ≈ 5.67299, so that k ≈ 290.903, and that m ≈ − 0.2473. Thus we obtained the formula R = (290.903) · W −0.2473 . This formula predicts the following values for R: predicted W R experimental R 25 131.231 131 67 102.839 103 127 87.7966 3 88 175 81.1045 81 240 75.0105 75 975 53.0356 53 The agreement between the predicted and experimental results is quite good. The graph of the logarithms of the data points and the line y = 5.67299 − (0.2473)x is shown next. 4.8 4.6 4.4 4.2 4 5 6 7 C06S07.034: We solved this problem in the same way as Problem 33. The straight line that best fit the logarithms of the data points had equation y = 3.87045 − (1.39711)x, so that k = exp(3.87045) ≈ 47.9640 and m ≈ −1.39711. Thus we found the formula p = (47.9640)V −1.39711 . The agreement with the experimental data was even better than in Problem 33. The best-fitting straight line and the logarithms of the data points are shown next. 3.5 3 2.5 2 1.5 0.5 1 1.5 2 C06S07.035: If f (x) = x · 2−x , then f (x) = (1 − x ln 2) · 2−x , so f (x) = 0 when x = a = 1/ ln 2. Because f (x) > 0 if x < a and f (x) < 0 if x > a, we have found the highest point on the graph of f ; it is 1 1 , 1/(ln 2) ln 2 2 ln 2 ≈ (1.4426950409, 0.5307378454) . C06S07.036: Clearly the graphs of f (x) = 2−x and g (x) = (x − 1)2 cross at the point (0, 1). When the graphs of f and g are plotted with the same coordinate axes, it becomes evident that the other point of intersection has x-coordinate near x0 = 1.6. We applied Newton’s method to h(x) = f (x) − g (x) and found the sequence x1 = 1.5789150927, x2 = 1.5786206934, and x3 = x4 = a = 1.5786206361 of improving approximations. Because f (x) > g (x) if 0 < x < a, the area bounded by the graphs of f and g is 4 a A= 0 [ f (x) − g (x) ] dx = x2 − x − 1 3 2−x x− 3 ln 2 a 0 ≈ 0.5617703346. C06S07.037: Let f (x) = 2−x and g (x) = (x − 1)2 . We saw in the solution of Problem 36 that the graphs meet where x = 0 and where x = a ≈ 1.5786206361 (obtained by applying Newton’s method to the equation f (x) − g (x) = 0 with initial estimate x0 = 1.6). There we saw also that f (x) > g (x) if 0 < x < a. So the method of parallel cross sections gives the volume of revolution around the x-axis as a V= π2 −2x 0 − (x − 1) 4 1 2−2x dx = π 2x − x − 2x + x − x5 − 5 2 ln 2 2 3 a 4 0 ≈ 1.343088216395. C06S07.038: Let f (x) = 32−x and g (x) = (3x − 4)2 . We plotted the graphs of f and g for 0.5 x 2 and saw that the graphs cross near where x = 0.6 and near where x = 1.7. We applied Newton’s method to the equation f (x) − g (x) = 0 using these values as initial estimates and obtained the following results: n First xn Second xn 1 0.622814450454 1.722194918580 2 0.623229033865 1.721721015948 3 0.623229171888 1.721720799346 4 0.623229171888 1.721720799346 Let a be the first x4 and b the second. Note that f (x) > g (x) for a < x < b. So the area of the region bounded by the graphs of f and g is approximately b A= a [ f (x) − g (x)] dx = 12x2 − 16x − 3x3 − 32−x ln 3 b a ≈ 1.645167893979. C06S07.039: By definition of z , x, and y , respectively, we have az = c, ax = b, and by = c. Therefore axy = by = c = az . Because a > 0 and a = 1, it now follows that z = xy . C06S07.040: lim x→0+ 1 1 = lim = 0; 1/x k→∞ 1 + 2k 1+2 lim− x→0 1 1 = lim = 1. 1/x k→−∞ 1 + 2k 1+2 C06S07.041: Beginning with the equation xy = 2, we first write y · ln x = ln 2, then differentiate implicitly with respect to x to obtain y dy + ln x = 0. x dx Thus dy y ln 2 . =− =− dx x ln x x(ln x)2 5 Section 6.8 1 2 C06S08.001: arcsin arcsin − 1 2 C06S08.002: arccos arccos − π π because sin 6 6 = π =− , 6 1 2 1 2 = = arcsin arccos 1 π and − 2 2 1√ 2 2 π π because cos 3 3 2π , 3 = π 6 π . Similarly, 2 1√ 3 2 π , 4 and arcsin − 1 and 0 2 π 3 π . Similarly, π , 4 and arccos − = = 1√ 2 2 = π =− . 3 1√ 3 2 π π < 0 < . Similarly, 2 2 √ π arctan(−1) = − , and arctan 3 4 = C06S08.003: arctan(0) = 0 because tan(0) = 0 and − arctan(1) = π , 4 C06S08.004: arcsec(1) = 0 because sec(0) = 1 and 0 arcsec(−1) = π , π , 3 arcsec(2) = C06S08.005: If f (x) = sin−1 x100 , then f (x) = √ and √ arcsec − 2 100x99 . 1 − x200 1 C06S08.007: If f (x) = sec−1 (ln x), then f (x) = x | ln x | (ln x)2 − 1 C06S08.008: If f (x) = ln tan−1 x , then f (x) = 1 . (1 + x2 ) arctan x C06S08.009: If f (x) = arcsin(tan x), then f (x) = √ sec2 x 1 − tan2 x . x + arctan x. 1 + x2 C06S08.011: If f (x) = sin−1 ex , then f (x) = √ ex . 1 − e2x √ C06S08.012: If f (x) = arctan x, then f (x) = 1 √. 2(1 + x) x C06S08.013: If f (x) = cos−1 x + sec−1 1 x and 0 < x < 1, then 1 π . 3 π . Similarly, ex . 1 + e2x C06S08.006: If f (x) = arctan(ex ), then f (x) = C06S08.010: If f (x) = x arctan x, then f (x) = 0 = . = 3π . 4 5π . 6 f (x) = − √ = −√ But if −1 < x < 0, then f (x) = − √ = −√ 1 − 1 − x2 1 − 1 − x2 1 x2 · 1 · x 1 x 2 −1 1 2 1 −√ = −√ . 2 −2 − 1 1−x xx 1 − x2 1 1 x2 · · x x−2 − 1 = −√ 1 1 + 1 1 − x2 x2 · · x−2 − 1 x 1 1 2 √ − . = −√ 2 −2 − 1 1−x (−x) x 1 − x2 In the last line in the second derivation, we needed to replace x < 0 with −x > 0 in order to move it under the radical. C06S08.014: If f (x) = arccot x−2 , then f (x) = − 1 1 2x 2x · −2x−3 = · =4 . −4 −4 x4 1+x 1+x x +1 C06S08.015: If f (x) = csc−1 x2 , then f (x) = − 2x 2x 2 √ =− √ =− √ . |x2 | x4 − 1 x2 x4 − 1 x x4 − 1 C06S08.016: If f (x) = arccos x−1/2 , then f (x) = − √ C06S08.017: If f (x) = 1 1 · − x−3/2 −1 2 1−x = 1 1 =√ . √√ −1 2x x − 1 2x x 1 − x 1 1 . = (arctan x)−1 , then f (x) = − 2 )(arctan x)2 arctan x (1 + x 2 arcsin x C06S08.018: If f (x) = (arcsin x)2 , then f (x) = √ . 1 − x2 C68S08.019: If f (x) = tan−1 (ln x), then f (x) = C06S08.020: If f (x) = arcsec f (x) = √ √ 1 1 + (ln x) +1 · 1 1 = . 2 x x 1 + (ln x) x2 + 1, then 1 x2 2 (x2 + 1) − 1 · x x 12 √ =2 (x + 1)−1/2 · 2x = . 2 + 1) x2 2 (x + 1)|x| (x 2 C06S08.021: If f (x) = tan−1 ex + cot−1 e−x , then f (x) = C06S08.022: If f (x) = exp(arcsin x), then f (x) = ex −e−x 2ex − = . 2x −2x 1+e 1+e 1 + e2x exp(arcsin x) √ . 1 − x2 C06S08.023: If f (x) = sin(arctan x), then f (x) = cos(arctan x) . 1 + x2 But this problem has a twist. A reference right triangle with acute angle θ = arctan x, adjacent side 1, √ opposite side x, and hypotenuse 1 + x2 shows that sin(arctan x) = √ x . 1 + x2 Therefore, by the quotient rule, f (x) = (1 + x2 )1/2 − x2 (1 + x2 )−1/2 1 + x2 − x2 1 = = . 2 2 )3/2 1+x (1 + x (1 + x2 )3/2 The second version of the derivative is potentially more useful than the first version. C06S08.024: If f (x) = sec(sec−1 ex ), then f (x) = ex wherever it is defined (for x (if x > 0). C06S08.025: If f (x) = 0), so f (x) = ex arctan x , then (1 + x2 )2 f (x) = (1 + x2 )2 · 1 − 4x(1 + x2 ) arctan x 1 − 4x arctan x 1 + x2 = . 2 )4 (1 + x (1 + x2 )3 C06S08.026: If f (x) = (sin−1 2x2 )−2 , then f (x) = (−2)(sin−1 2x2 )−3 · √ C06S08.027: Given: tan−1 x + tan−1 y = 1 8x √ · 4x = − . −1 4 2 )3 1 − 4x4 1 − 4x (sin 2x π : 2 1 1 dy + · = 0, 1 + x2 1 + y 2 dx so dy 1 + y2 . =− dx 1 + x2 So the slope of the line tangent to the graph at P (1, 1) is −1, and therefore an equation of that line is y − 1 = −(x − 1); that is, y = 2 − x. C06S08.028: Given: sin−1 x + sin−1 y = √ 1 + 1 − x2 π : 2 1 1 − y2 · dy = 0, dx So the slope of the line tangent to the graph at P line is 11 2, 2 3 so √ 3 1 − y2 dy . =−√ dx 1 − x2 is − 1 3 √ 3, and therefore an equation of that √ √ 3 3 y− =− 2 3 1 x− 2 C06S08.029: Given: (sin−1 x)(sin−1 y ) = ; that is, √ 3 23 y=− x+ . 3 3 π2 : 16 sin−1 y sin−1 x dy √ + · = 0, 1 − x2 1 − y 2 dx So the slope of the line tangent to the graph at P is y− √ 1 2 so √ 1√ 1√ 2=− x− 2 2 2 C06S08.030: Given: (sin−1 x)2 + (sin−1 y )2 = 2, 1 2 ; √ dy (1 − y 2 )1/2 sin−1 y =− . dx (1 − x2 )1/2 sin−1 x 2 is −1, and therefore an equation of that line that is, y = −x + √ 2. 5π 2 : 36 2 sin−1 x 2 sin−1 y dy √ + · = 0, 1 − x2 1 − y 2 dx dy (1 − y 2 )1/2 sin−1 x =− . dx (1 − x2 )1/2 sin−1 y √ √ So the slope of the line tangent to the graph at P 1 , 1 3 is − 1 3. Therefore an equation of that line is 22 6 √ 1√ 1√ 1 3 y− 3=− 3 x− ; that is, y = (7 − 2x). 2 6 2 12 1 C06S08.031: 0 1 dx = 1 + x2 2 √ 2 0 π π −0= . 4 4 1 /2 arcsin x 1 √ dx = 2−1 xx arcsec x √ 0 C06S08.033: = 1 dx = 1 − x2 1 /2 C06S08.032: 1 arctan x so = π π −0= . 6 6 = ππ π −= ≈ 0.261799387799. 3 4 12 0 2 √ 2 In comparison, Mathematica 3.0 yields the result 2 √ C06S08.034: 2 √ −2/ 3 −2 1 √ dx = 2−1 xx − arctan 1 √ dx = x x2 − 1 arcsec | x | √ 1 2−1 x √ −2/ 3 −2 = 2 √ 2 =− ππ π += . 6 4 12 ππ π − =− . 6 3 6 The answer is negative because the integrand is negative for −2 x 2 −√ . 3 C06S08.035: Let x = 3u. Then dx = 3 du, and as x ranges from 0 to 3, u ranges from 0 to 1. Therefore 3 0 1 dx = 9 + x2 1 0 3 1 du = 9 + 9u 2 3 1 0 1 1 du = arctan u 1 + u2 3 4 1 = 0 π π −0= ≈ 0.261799387799. 12 12 Alternatively, the antiderivative can be expressed as a function of x before evaluation: 3 0 1 dx = 9 + x2 1 0 3 1 du = 9 + 9u 2 3 3 1 x arctan 3 3 = = 0 1 0 1 1 du = arctan u 1 + u2 3 1 0 π π −0= ≈ 0.261799387799. 12 12 Note that in the latter case the original x-limits of integration must be restored before evaluation of the antiderivative. C06S08.036: Let x = 4u. Then dx = 4 du, and as x ranges from 0 to √ 1 2 3. Thus √ 0 12 1 √ dx = 16 − x2 √ 3 /2 0 4 √ du = 16 − 16u2 √ 0 C06S08.037: Let u = 2x, so that 4x2 = u2 and dx = √ 1 1 dx = 2 2 1 − 4x √ 1 2 1 √ du = 1 − u2 √ 12 = 2 3, u ranges from 0 to √ 3 /2 arcsin u = 0 π π −0= . 3 3 du. Then 1 1 1 du = arcsin u + C = arcsin 2x + C. 2 2 2 1−u C06S08.038: Let u = 3 x. Then x = 2 u, dx = 2 3 1 2 dx = 9x2 + 4 3 3 /2 √ 2 3 du, and 9x2 + 4 = 4u2 + 4. Thus 1 1 1 3x du = arctan u + C = arctan + C. 4u2 + 4 6 6 2 C06S08.039: Let x = 5u. Then dx = 5 du and x2 − 25 = 25u2 − 25 = 25(u2 1). Thus 1 √ dx = 2 − 25 xx 5 1 1 |x| √ du = arcsec | u | + C = arcsec + C. 2−1 5 5 5 5 · 5u u Mathematica 3.0 reports that 1 1 √ dx = C − arctan 5 x x2 − 25 C06S08.040: Let u = 2 x, so that 2x = 3u and dx = 3 x(4x2 1 3 dx = 1 /2 2 − 9) = C06S08.041: ex dx = 1 + e2x 3 2 √ 5 x2 − 25 . du. Then 1 du − 9)1/2 3 2 2 u(9u 1 1 1 2 |x| du = arcsec | u | + C = arcsec + C. 3 3 3 3u(u2 − 1)1/2 ex dx = arctan (ex ) + C . 1 + (ex )2 If you prefer integration by substitution, use u = ex , so that du = ex dx. Then ex dx = 1 + e2x 1 du = arctan u + C = arctan (ex ) + C. 1 + u2 5 C06S08.042: Let u = 1 x3 . Then x3 = 5u, 3x2 dx = 5 du, and x6 + 25 = 25(u2 + 1). Thus 5 x6 x2 5 dx = + 25 3 1 1 1 du = arctan u + C = arctan 25(u2 + 1) 15 15 13 x 5 + C. C06S08.043: Let u = 1 x3 . Then 5u = x3 , 3x2 dx = 5 du, and x6 − 25 = 25(u2 − 1). So 5 1 √ dx = 6 − 25 xx = 3x2 dx = 3x3 (x6 − 25)1/2 1 15 5 1 /2 3 · 5u [25(u2 − 1)] du 1 1 1 | x3 | du = arcsec | u | + C = arcsec + C. 15 15 5 u(u2 − 1)1/2 Mathematica 3.0 gives the answer in the form C + 1 arctan 15 C06S08.044: Let u = x3/2 . Then du = 3 x1/2 dx, so x1/2 dx = 2 x1/2 2 dx = 3 1+x 3 1 5 2 3 x6 − 25 . du and 1 2 2 du = arctan u + C = arctan x3/2 2 1+u 3 3 + C. C06S08.045: The radicand is 1 x(1 − x) = x − x2 = −(x2 − x) = − (4x2 − 4x) 4 1 1 1 1 = − (4x2 − 4x + 1) + = 1 − (2x − 1)2 = (1 − u2 ) 4 4 4 4 if we let u = 2x − 1. If so, du = 2 dx, and then 1 x(1 − x) dx = 1 2 √ 2 du = arcsin u + C = arcsin(2x − 1) + C. 1 − u2 The more “obvious” substitution x = u2 , so that u = x1/2 and dx = 2u du, leads to 1 x(1 − x) dx = 2u u2 (1 − du = 2 u2 ) u √ du | u | 1 − u2 √ 1 du = 2 arcsin u + C = 2 arcsin x + C. 1 − u2 √ Replacement of | u | with u here is permitted because u = x > 0. Test your skill at trigonometry by showing √ that f (x) = arcsin(2x − 1) and g (x) = 2 arcsin x differ by a constant (if 0 x 1). =2 √ C06S08.046: Let u = sec x, so that du = sec x tan x dx. Then sec x tan x dx = 1 + sec2 x 1 du = arctan u + C = arctan(sec x) + C. 1 + u2 Mathematica 3.0 returns the amazing answer C − [arctan(cos x)] (3 + cos 2x) sec2 x . 2(1 + sec2 x) C06S09.047: Let u = x50 , so that du = 50x49 dx. Then 6 x49 1 dx = 100 1+x 50 1 1 1 du = arctan u + C = arctan x50 + C. 2 1+u 50 50 C06S08.048: Let u = x5 , so that du = 5x4 dx. Then √ x4 1 dx = 10 5 1−x √ 1 1 1 du = arcsin u + C = arcsin x5 + C. 2 5 5 1−u C06S08.049: 1 dx = arctan(ln x) + C . (Use u = ln x if necessary.) x [1 + (ln x)2 ] C06S08.050: arctan x 1 dx = (arctan x)2 + C . (Use u = arctan x if necessary.) 2 1+x 2 1 C06S08.051: 0 1 1 dx = arctan(2x − 1) 1 + (2x − 1)2 2 1 = 0 π π −− 8 8 = π ≈ 0.7853981634. 4 If integration by substitution is preferred, let u = 2x − 1, du = 2 dx, and do not forget to change the limits of integration to u = −1 and u = 1. 1 C06S08.052: 0 x3 1 dx = 1 + x4 4 4x3 1 dx = ln(1 + x4 ) 1 + x4 4 1 0 1 = 0 1 ln 2 (ln 2 − ln 1) = ≈ 0.1732867951. 4 4 If integration by substitution is preferred, let u = 1 + x4 , du = 4x3 dx, and do not forget to change the limits of integration to u = 1 and u = 2. C06S08.053: Let u = ln x (if necessary) to find that 1 2 C06S08.054: 1 e 1 e dx = x 1 − (ln x)2 arcsin(ln x) 1 2 1 √ dx = 2−1 xx = = arcsec x 1 π π −0= . 2 2 π π −0= . 3 3 C06S08.055: If u = x1/2 , then du = 1 x−1/2 dx. . Moreover, u = 1 when x = 1 and u = 2 Therefore 3 1 1 dx = 1/2 (1 + x) 2x 3 1 1 1 −1/2 2x + (x1/2 )2 √ = √ dx = 3 1 3 arctan u = 3 when x = 3. 1 du 1 + u2 √ arctan √ 3 = x 1 1 ππ π −= ≈ 0.2617993878. 3 4 12 C06S08.056: First note that cos−1 x = C − sin−1 x for some constant C for all x in the interval (0, 1). In particular, C = cos−1 1 2 + sin−1 7 1 2 = ππ π +=. 3 6 2 Therefore sin−1 x + cos−1 x = π /2 if 0 < x < 1. This formula also holds if x = 0 and if x = 1, and therefore it holds for all x in the closed interval 0 x 1. C06S08.057: Suppose that u < −1 and let x = −u. Then x > 0, so y = arcsec | u | = arcsec x, and then the chain rule yields dy −1 dy dx 1 1 1 √ · (−1) = . = · = = =√ du dx du | x | x2 − 1 u u2 − 1 x (−x)2 − 1 (−x) (−x)2 − 1 C06S08.058: If a > 0 and u = ax, then du = a dx, a2 − u2 = a2 − a2 x2 = a2 (1 − x2 ), and a2 − u2 = a 1 − x2 . Therefore √ a2 1 du = − u2 a u √ dx = arcsin x + C = arcsin 2 a a 1−x + C. C06S08.059: If a > 0 and u = ax, then du = a dx and a2 + u2 = a2 + a2 x2 = a2 (1 + x2 ). So a2 1 du = + u2 a 1 1 u dx = arctan x + C = arctan 2) +x a a a a2 (1 + C. C06S08.060: If a > 0 and u = ax, then du = a dx, u2 − a2 = a2 x2 − a2 = a2 (x2 − 1), and u2 − a2 = a x2 − 1 . Therefore u √ 1 du = − a2 √ u2 a2 x a 1 1 u dx = arcsec | x | + C = arcsec + C. 2−1 a a a x C06S08.061: If x > 1, then 1 f (x) = x2 1− 1 = 1 x2 2 x x2 − x x2 = 1 √ . | x | x2 − 1 If x < −1, then 1 f (x) = x2 1− 1 x2 1 = (−x)2 1− 1 x2 1 = (−x) (−x)2 − (−x) x2 2 1 √ . | x | x2 − 1 = C06S08.062: If x > 1 or if x < −1, then Dx cos−1 1 x =− 1 1 1− 2 x 8 ·− 1 x2 1 = x2 1 1− 2 x . Now apply the result in Problem 61. But at this point you can conclude only that sec−1 x = cos−1 1 x + Ci for some constant C1 if x > 1 and for some possibly different constant C2 if x < −1. Substitute x = 2 and then x = −2 in the last equation to verify that Ci = 0 in each case. C06S08.063: Let g denote the “alternative secant function,” so that y = g (x) if and only if sec y = x and either 0 y < π /2 or π y < 3π /2. We differentiate implicitly the identity sec y = x on both the intervals 0 < y < π /2 and π < y < 3π /2 and find that (sec y tan y ) dy = 1, dx so that g (x) = 1 1 1 =± . =± √ 2−1 2y−1 sec y tan y xx x sec Then Fig. 6.8.13 shows that g (x) < 0 if x < −1 and that g (x) > 0 if x > 1. Therefore the choice of the plus sign in the previous equation is correct in both cases: 1 g (x) = √ . x x2 − 1 C06S08.064: We begin with the identity tan(A + B ) = tan A + tan B . 1 − tan A tan B (1) Let x = tan A, y = tan B , and suppose that xy < 1. We will treat only the case in which x and y are both positive; the other three cases are similar. In this case, Eq. (1) implies that 0 < A + B < π /2, so we may apply the inverse tangent function to both sides of the identity in (1) to obtain A + B = arctan x+y , 1 − xy and therefore arctan x + arctan y = arctan Now we turn to part (b). We have 9 x+y . 1 − xy 11 + arctan 2 3 = arctan 1 1− 6 5 6 = arctan 1 = π . 5 4 6 11 + 3 3 + arctan 1 = arctan arctan 1 7 1− 9 2 3 + arctan 1 = arctan 3 + arctan 1 8 7 4 7 9 31 + 4 7 = arctan 25 = π . = arctan 3 25 4 1− 28 120 − 119 arctan 120 1+ 119 1 28561 239 = arctan 28441 = π . 1 28561 4 · 239 28441 1 2 arctan = arctan 5 2 5 1 1− 25 = arctan 1 4 arctan = arctan 5 4 arctan 5 10 = arctan ; 24 12 10 120 12 = arctan ; 25 119 1− 144 1 1 120 1 π − arctan = arctan − arctan = 5 239 119 239 4 (by (iii)). C06S08.065: See the following figure for the meanings of the variables. 12 θ 4 x We are required to maximize the angle θ, and from the figure and the data given in the problem we may express θ as a function of the distance x of the billboard from the motorist: θ = θ(x) = arctan 16 4 − arctan , x x 0 < x < + ∞. After simplifications we find that dθ 4 16 =2 − . dx x + 16 x2 + 256 10 The solution of θ (x) = 0 certainly maximizes θ because θ is near zero if x is near zero and if x is large positive. We solve θ (x) = 0: 4(x2 + 16) = x2 + 256; 3x2 = 192; x2 = 64; x = 8. Answer: The billboard should be placed so that it will be 8 meters (horizontal distance) from the eyes of passing motorists. Many alert students have pointed out that such a billboard wouldn’t be visible long enough to be effective. This illustrates that once you have used mathematics to solve a problem, you must interpret the results! C06S08.066: As in the figure that follows this solution, assume that the observer’s eyes are at the height L above the floor and at horizontal distance W from the painting. Let h be the height of the painting and let y be the distance from the floor to the bottom of the painting. We are to maximize the angle θ = θ1 + θ2 where θ1 = arctan y+h−L W and θ2 = arctan L−y . W With the aid of the arctangent addition formula (part (a) of Problem 64) we find—after simplifications—that θ(y ) = arctan Wh . W 2 + (y − L)(y − L + h) (1) Now θ is maximized when tan θ is maximized (because the tangent function is increasing on (0, π /2)), and this occurs when the denominator of the fraction in Eq. (1) is minimal. So we let f (y ) = W 2 +(y −L)(y −L+h) and apply calculus: f (y ) = y − L + y − L + h = 2y − 2L + h; f (y ) = 0 when y =L− h . 2 This value of y clearly minimizes f (y ), and the center of the painting is then at height y+ h hh =L− + =L 2 2 2 above the floor—exactly at the height of the observer’s eyes. θ1 h θ2 L y W 11 C06S08.067: If f (x) = (a2 − x2 )1/2 , then 1 + [ f (x)]2 = 1 + x2 a2 =2 , a2 − x2 a − x2 so the circumference of a circle of radius a is √ a/ 2 C=8 √ 0 √ a/ 2 a x dx = 8 · a arcsin a a2 − x2 0 =8 πa − 0 = 2π a. 4 The integration was carried out by using the result in Problem 58. C06S08.068: By the method of nested cylindrical shells, the volume is 1 V= 0 1 2π x dx = π arctan(x2 ) 1 + x4 = 0 π2 π2 −0= ≈ 2.4674011003. 4 4 C06S08.069: Let Aa denote the area under the graph of y (x) for 0 a Aa = 0 Then lim Aa = a→∞ x a. Therefore a 1 dx = 1 + x2 arctan x = arctan a. 0 π by Eq. (2) and Fig. 6.8.4. 2 C06S08.070: Let y be the height of the elevator (measured upward from ground level) and let θ be the angle that your line of sight to the elevator makes with the horizontal (θ > 0 if you are looking up, θ < 0 if dθ dy down). You’re to maximize given = −25. dt dt tan θ = y − 100 , 50 so θ = tan−1 y − 100 50 . Therefore dθ dθ dy 1/50 −25 · 50 = · = −25 · 2 = 2500 + (y − 100)2 . dt dy dt 1 + [(y − 100)/50] (1) To find the value of y that maximizes f (y ) = dθ/dt, we need only minimize the last denominator in Eq. (1): y = 100. Answer: The elevator has maximum apparent speed when it’s at eye level. C06S08.071: For x > 1: f (x) = arcsec x + A: 1 = f (2) = π + 1, 3 so A=1− π . 3 For x < −1: f (x) = −arcsec x + B ; 1 = f (−2) = − 2π + B, 3 12 so B =1+ 2π . 3 π 2π Therefore f (x) = arcsec x + 1 − if x > 1, f (x) = −arcsec x + 1 + if x < −1. The graph of y = f (x) 3 3 is next. 1.2 1 0.8 0.6 0.4 0.2 -3 -2 -1 1 2 3 C06S08.072: If |x| < 1, then Dx arctan √ x 1 − x2 Therefore 1 = 1+ x2 1 − x2 arctan √ · (1 − x2 )1/2 + x2 (1 − x2 )−1/2 1 =√ . 1 − x2 1 − x2 x 1 − x2 = C + arcsin x for some constant C if −1 < x < 1. Now substitute x = 0 to show that C = 0. C06S08.073: If f (x) = arctan(x2 − 1)1/2 for x > 1, then f (x) = 1 x 1 1 =√ · (x2 − 1)−1/2 · 2x = 2 2 . 2−1 1 + (x2 − 1) 2 x (x − 1)1/2 xx Therefore arcsec x = C + arctan(x2 − 1)1/2 for some constant C for all x > 1. Now substitute x = show that C = 0. √ 2 to If g (x) = π − arctan(x2 − 1)1/2 for x < −1, then (using the earlier result) Therefore 1 g (x) = − √ . x x2 − 1 arcsec x = C + π − arctan x2 − 1 √ for some constant C for all x < −1. Now substitute x = − 2 to show that C = 0. C06S08.074: The graph of f on [0.001, 5] indicates a global maximum near x = 1.4. We used Newton’s method to show that its location is close to (1.3917452003, 0.8033644570). 13 C06S08.075: The graph of f on [0, 8] (following this solution) indicates a global maximum near x = 2.7. We used Newton’s method to show that its location is close to (2.6892200292, 0.9283427321). 0.9 0.8 0.7 0.6 2 4 6 8 C06S08.076: The graph of f on [1, 30] indicates a global maximum near x = 8.3. We used Newton’s method to show that its location is close to (8.3332645728, 1.3345303526). 14 Section 6.9 C06S09.001: If f (x) = cosh(3x − 2), then f (x) = 3 sinh(3x − 2). √ √ cosh x √ C06S09.002: If f (x) = sinh x, then f (x) = cosh x1/2 · Dx x1/2 = . 2x C06S09.003: If f (x) = x2 tanh f (x) = 2x tanh 1 , then x 1 x + x2 sech2 1 x ·− 1 x2 = 2x tanh 1 x − sech2 1 x C06S09.004: If f (x) = sech e2x , then f (x) = −2e2x sech e2x tanh e2x . 3 C06S09.005: If f (x) = coth3 4x = (coth 4x) , then 2 f (x) = 4 · (3 coth 4x) −(csch 4x)2 = −12 coth2 4x csch2 4x. C06S09.006: If f (x) = ln sinh 3x, then f (x) = 1 3 cosh 3x (cosh 3x) · 3 = = 3 coth 3x. sinh 3x sinh 3x C06S09.007: If f (x) = ecsch x , then f (x) = ecsch x · Dx csch x = −ecsch x csch x coth x. C06S09.008: If f (x) = cosh ln x = cosh (ln x), then f (x) = (sinh ln x) · Dx (ln x) = sinh(ln x) x − x−1 x2 − 1 . = = x 2x 2x2 C06S09.009: If f (x) = sin (sinh x), then f (x) = [cos (sinh x)] · Dx (sinh x) = (cosh x) cos (sinh x) . C06S09.010: If f (x) = tan−1 (tanh x), then f (x) = 1 sech2 x 1 − tanh2 x sech2 x · Dx (tanh x) = = = . 1 + tanh2 x 1 + tanh2 x 1 + tanh2 x 2 − sech2 x C06S09.011: If f (x) = sinh x4 = sinh(x4 ), then f (x) = 4x3 cosh x4 . C06S09.012: If f (x) = sinh4 x = (sinh x)4 , then f (x) = 4 sinh3 x cosh x. C06S09.013: If f (x) = 1 , then (by the reciprocal rule) x + tanh x f (x) = − 1 + sech2 x (tanh2 x) − 2 = . 2 (x + tanh x) (x + tanh x)2 1 . C06S09.014: If f (x) = cosh2 x − sinh2 x, then f (x) ≡ 1 by Eq. (4), and therefore f (x) ≡ 0. Alternatively, f (x) = 2 cosh x sinh x − 2 sinh x cosh x ≡ 0. C06S09.015: If necessary, use the substitution u = x2 , du = 2x dx to show that x sinh x2 dx = C06S09.016: By Eq. (11), cosh2 3u = cosh2 3u dx = 1 2 1 cosh x2 + C. 2 (cosh 6u + 1). Therefore 1 1 1 1 sinh 6u + u + C = sinh 3u cosh 3u + u + C 12 2 6 2 (the last equality by Eq. (9)). C06S09.017: By Eq. (5) we have tanh2 3x dx = 1 − sech2 3x dx = x − 1 tanh 3x + C. 3 √ C06S09.018: Let u = x = x1/2 . Then x = u2 , so dx = 2u du. Therefore (with the aid of Eq. (17)) √ √ √ sech x tanh x sech u tanh u √ dx = · 2u du = −2 sech u + C = −2 sech x + C. u x C06S09.019: Let u = sinh 2x. Then du = 2 cosh 2x dx, so sinh2 2x cosh 2x dx = C06S09.020: (cosh x) sinh 3x 1 dx = ln (cosh 3x) + C . cosh 3x 3 tanh 3x dx = C06S09.021: −3 12 1 1 u du = u3 + C = sinh3 2x + C. 2 6 6 sinh x dx = −2 (cosh x) −2 +C =− 1 sech2 x + C . 2 C06S09.022: By Eqs. (11) and (12), sinh4 x = sinh2 x = 1 4 2 = 1 (cosh 2x − 1) 2 2 = 1 cosh2 2x − 2 cosh 2x + 1 4 1 31 1 (cosh 4x + 1) − 2 cosh 2x + 1 = − cosh 2x + cosh 4x. 2 82 8 Therefore sinh4 x dx = 1 (sinh 4x − 8 sinh 2x + 12x) + C. 32 C06S09.023: Let u = coth x. Then du = − csch2 x dx. So coth x csch2 x dx = − 1 1 1 u du = − u2 + C = − coth2 x + C = − csch2 x + C1 . 2 2 2 2 C06S09.024: This is not easy. Here’s one solution: sech x dx = 2 dx = x + e−x e 2ex dx = 2 arctan (ex ) + C. 1 + (ex )2 If multiplication of numerator and denominator by ex seems artificial, try a substitution: Let u = ex , so that du = ex dx, and thus dx = e−x du = u−1 du. Then sech x dx = ex 2 dx = + e−x 2u−1 du = u + u−1 2 du = 2 arctan u + C = 2 arctan (ex ) + C. u2 + 1 Mathematica 3.0 apparently uses a rationalizing substitution like that found in the discussion following Miscellaneous Problem 134 of Chapter 7 (but adapted to hyperbolic rather than trigonometric integrals); it obtains sech x dx = 2 arctan tanh x + C. 2 C68S09.025: If necessary, let u = 1 + cosh x or let u = cosh x. But simply “by inspection,” sinh x dx = ln (1 + cosh x) + C. 1 + cosh x C06S09.026: Let u = ln x. Then du = 1 dx. So x sinh(ln x) dx = x sinh u du = cosh u + C = cosh(ln x) + C. Alternatively, first simplify the integrand: sinh(ln x) 1 exp(ln x) − exp(− ln x) x − x−1 x2 − 1 = = = = x 2x 2x 2x2 2 Then sinh(ln x) 1 dx = x 2 1− 1 x2 dx = 1 2 x+ 1 x +C = 1− 1 x2 . x2 + 1 + C. 2x C06S09.027: One solution: 1 1 dx = (ex + e−x )2 4 2 ex + e−x 2 dx = 1 4 sech2 x dx = Another solution: Let u = ex , so that du = ex dx; thus dx = e−x du = (ex 1 dx = + e−x )2 = u−1 du = (u + u−1 )2 u dx = u4 + 2u2 + 1 u2 1 tanh x + C. 4 1 du. Then u u−1 dx + 2 + u−2 1 u(u2 + 1)−2 du = − (u2 + 1)−1 + C 2 1 1 2 sech x 1 1 e−x = − · 2x +C =− x · x +C =− + C. +C =− · x 2 e +1 2 e + e−x 4e e + e−x 4ex 3 The two answers appear quite different, but they are both correct (although they differ by the constant 1 ). 4 C06S09.028: Let u = ex − e−x if you wish, but by inspection we find ex + e−x dx = ln ex − e−x + C. ex − e−x You may also continue the previous calculations as follows: · · · = ln ex − e−x − ln 2 + C1 = ln ex − e−x + C1 = ln | sinh x | + C1 . 2 Mathematica 3.0 obtains ex + e−x dx = −x + ln | −1 + e2x | + C. ex − e−x C06S09.029: f (x) = √ C06S09.030: f (x) = C06S09.031: f (x) = 1 2 · Dx (2x) = √ . 2 1 + 4x 1 + 4x2 1 (x + 1)2 −1 · Dx x2 − 1 = √ = 2x 2x √ = . 2 + 2x |x | x2 + 2 1 1 1 /2 √. = √ 2 · Dx x 2(1 − x) x 1 − ( x) 1/2 C06S09.032: If f (x) = coth−1 x2 + 1 f (x) = x4 1 1− (x2 + , then · Dx x2 + 1 2 1)1/2 1/2 1 1 x 1 1 · (x2 + 1)−1/2 · 2x = − 2 · √ =− √ . 1 − x2 − 1 2 x x2 + 1 x x2 + 1 (Compare this result with Eq. (33).) 1 , then x C06S09.033: If f (x) = sech−1 f (x) = − =− 1 1 x 1− 1 x2 x 1− 1 x2 · Dx ·− 1 x2 1 x x = x2 1− 1 x2 = x √ . | x | x2 − 1 (You need to write x2 in the form | x |2 in order to move one copy of | x | underneath the radical, where it becomes x2 . Compare this result with Eq. (29).) C06S09.034: If f (x) = csch−1 ex , then f (x) = − | ex | √ 1 1 · ex = − √ . 2x 1+e 1 + e2x 4 3 /2 C06S09.035: If f (x) = sinh−1 x , then 1 /2 3 f (x) = sinh−1 x 2 1/2 −1 · Dx sinh 3 sinh−1 x x= . 2(1 + x2 )1/2 C06S09.036: If f (x) = sinh−1 (ln x), then 1 f (x) = · Dx (ln x) = 2 1 + (ln x) 1 2 . x 1 + (ln x) C06S09.037: If f (x) = ln tanh−1 x , then f (x) = C06S09.038: If f (x) = 1 1 · Dx tanh−1 x = . −1 2 ) tanh−1 x tanh x (1 − x 1 , then tanh−1 3x 1 f (x) = − tanh−1 3x 2 · Dx tanh−1 x = − 3 (1 − 9x2 ) tanh−1 3x 2. C06S09.039: Let x = 3u. Then dx = 3 du, so √ 1 dx = +9 √ x2 C06S09.040: Let y = 1 4y 2 − 9 dy = 3 9u2 du = +9 √ 1 x du = arcsinh u + C = arcsinh + C. 3 +1 u2 3 3 u. Then dy = du, so 2 2 3 2 √ 1 9u2 −9 du = 1 2 √ 1 1 1 2 du = arccosh u + C = arccosh y + C. 2 2 3 −1 u2 C06S09.041: Let x = 2u. Then dx = 2 du, so I= 1 1 /2 = 1 dx = 4 − x2 1 tanh−1 u 2 1 x=1/2 1 = x=1/2 2 1 du = 4 − 4u 2 2 1 x=1/2 1 1 x tanh−1 2 2 = 1 /2 1 du 1 − u2 1 tanh−1 2 1 2 − tanh−1 1 4 . Now use Eq. (36) to transform the answer into a more familiar form: I= 1 4 ln 1+ 1− 1 2 1 2 − ln 1+ 1− 1 4 1 4 = 1 4 ln 3 − ln 5 3 = 19 ln ≈ 0.1469466662. 45 C06S09.042: Use the substitution in the solution of Problem 41, but now we must use Eq. (42b) rather than Eq. (42a): 5 10 I= 5 = 1 4 ln C06S09.043: Let x = 10 1 x 1 dx = coth−1 4 − x2 2 2 5+1 − ln 5−1 +1 −1 5 2 5 2 = 1 4 = 5 ln 1 coth−1 (5) − coth−1 2 7 3 − ln 2 3 = 5 2 9 1 ln ≈ −0.1104581881. 4 14 √ √ 2 2 u. Then 4 − 9x2 = 4 − 4u2 and dx = du. Therefore 3 3 1 2 √ dx = 3 x 4 − 9x2 = 4 3u C06S09.044: Let x = 5u: dx = 5 du, = 1 du 1 − u2 3 1 1 √ du = − sech−1 | u | + C = − sech−1 x + C. 2 2 u 1 − u2 1 2 1 √ dx = 2 + 25 xx √ √ √ x2 + 25 = 5 u2 + 1. So 5 du 25u u2 + 1 √ 1 5 u √ 1 u2 1 1 x du = − csch−1 | u | + C = − csch−1 + C. 5 5 5 +1 C06S09.045: Let u = ex : du = ex dx. Hence √ ex dx = e2x + 1 √ 1 du = sinh−1 u + C = sinh−1 (ex ) + C. u2 + 1 C06S09.046: Let u = x2 : du = 2x dx and √ x 1 dx = 2 −1 √ x4 √ 1 dx = 1 − e2x √ u2 − 1. So 1 1 1 du = cosh−1 u + C = cosh−1 x2 + C. 2 2 −1 u2 C06S09.047: Let u = ex : x = ln u and ds = √ x4 − 1 = 1 du. Thus u 1 du = − sech−1 | u | + C = − sech−1 (ex ) + C. u 1 − u2 √ C06S09.048: Let u = sin x: du = cos x dx. Therefore cos x 1 + sin x 2 dx = √ 1 du = sinh−1 u + C = sinh−1 (sin x) + C. 1 + u2 C06S09.049: sinh x cosh y + cosh x sinh y − sinh(x + y ) ex − e−x ey + e−y ex + e−x ey − e−y ex+y − e−x−y · + · − 2 2 2 2 2 1 x+y = e + ex−y − e−x+y − e−x−y + ex+y − ex−y + e−x+y − e−x−y − 2ex+y + 2e−x−y = 0. 4 Therefore sinh(x + y ) = sinh x cosh y + cosh x sinh y . = 6 C06S09.050: Given cosh2 x − sinh2 x = 1 (for all x), we divide both sides by cosh2 x (which is never zero) to find that cosh2 x sinh2 x 1 − = ; 2 2 cosh x cosh x cosh2 x 1 − tanh2 x = sech2 x that is, (for all x). If we instead divide by sides by sinh2 x (which is zero only when x = 0), we find that cosh2 x sinh2 x 1 − = ; sinh2 x sinh2 x sinh2 x coth2 x − 1 = csch2 x that is, if x = 0. C06S09.051: Substitute y = x in the identity cosh(x + y ) = cosh x cosh y + sinh x sinh y to prove that cosh 2x = cosh2 x + sinh2 x. Then, with the aid of Eq. (4), we find that cosh 2x = 2 cosh2 x − 1, cosh2 x = so that 1 (1 + cosh 2x) . 2 C06S09.052: First, x (t) = kA sinh kt + kB cosh kt, and therefore x (t) = k 2 A cosh kt + k 2 B sinh kt = k 2 x(t). C06S09.053: The length is a L= a 1 + sinh2 x dx = 0 a cosh x dx = sinh x 0 = sinh a. 0 C06S09.054: The volume is π V= π sinh2 x dx = 0 = π 4 π 4 π 0 e2x − 2 + e−2x dx = 1 2π 1 11 e − 2π − e−2π − + 2 2 22 = π 4 1 2x 1 e − 2x − e−2x 2 2 π 0 π (−2π + sinh 2π ) ≈ 205.3515458383. 4 C06S09.055: Beginning with the equation A(θ) = 1 cosh θ sinh θ − 2 cosh θ 1 (x2 − 1)1/2 dx, we take the derivative of each side with respect to θ (using the fundamental theorem of calculus on the right-hand side). The result is A (θ) = = 1 1 cosh2 θ + sinh2 θ − (cosh2 θ − 1)1/2 sinh θ 2 2 1 1 1 1 cosh2 θ − sinh2 θ = . cosh2 θ + sinh2 θ − sinh2 θ = 2 2 2 2 7 Therefore A(θ) = 1 θ + C for some constant C . Evaluation of both sides of this equation when θ = 0 yields 2 the information that C = 0, and therefore A(θ) = 1 θ. 2 C06S09.056: By l’Hˆpital’s rule, o lim x→0 sinh x cosh x = lim = cosh 0 = 1. x→0 x 1 We do not require l’Hˆpital’s rule for the other two limits: o lim tanh x = lim x→∞ lim x→∞ x→∞ ex − e−x 1 − e−2x 1−0 = 1; = lim = x + e−x x→∞ 1 + e−2x e 1+0 cosh x ex + e−x 1 − e−2x 1−0 1 = lim = lim = =. x x x→∞ x→∞ e 2e 2 2 2 C06S09.057: Let y = sinh−1 1. Then 1 = sinh y = ey − e−y ; 2 ey − e−y = 2; e2y − 2ey − 1 = 0; √ √ 2± 4+4 u= = 1 ± 2. 2 √ y But u = e > 0, so u = 1 + 2. Hence u2 − 2u − 1 = 0 sinh−1 1 = y = ln u = ln 1 + √ 2 where u = ey ; ≈ 0.8813735870. C06S09.058: If x = 0, then by Eq. (34), sinh−1 1 = ln x 1 + x 1 +1 x2 = ln 1 + x √ x2 + 1 |x| = csch−1 x by Eq. (39). C06S09.059: Let y = sinh−1 x and remember that cosh y > 0 for all y . Hence sinh y = x; dy 1 = = dx cosh y C06S09.060: Let y = sech−1 x, 0 < x (cosh y ) 1 2 cosh y = dy = 1; dx 1 2 1 + sinh y =√ 1 . 1 + x2 1. Recall that sech y > 0 for all y . Hence: sech y = x; (− sech y tanh y ) 8 dy = 1; dx 1 1 dy 1 =− =± . =± √ 2 dx sech y tanh y x 1 − x2 x 1 − sech y Now x > 0 but dy < 0 (because y = sech−1 x is decreasing on (0, 1)). Therefore dx dy 1 =− √ , 0 < x < 1. dx x 1 − x2 C06S09.061: Let f (x) = tanh−1 x and g (x) = 1 ln 2 1+x 1−x for −1 < x < 1. Equation (36) states that f (x) = g (x). To prove this, note that f (x) = and g (x) = 1 1 ln(1 + x) − ln(1 − x), so that 2 2 1 1 1 g (x) = + 2 1+x 1−x 1 1 − x2 = 1−x+1+x 1 , = 2(1 − x2 ) 1 − x2 and therefore f (x) = g (x). Hence f (x) = g (x) + C for some constant C and for all x in (−1, 1). To evaluate C , note that 0 = f (x) = tanh−1 0 = g (0) + C = 1 ln 1 + C = 0 + C = C. 2 Therefore f (x) = g (x) if −1 < x < 1. C06S09.062: Given: x = sinh y = ey − e−y : 2 ey − e−y = 2x; u2 − 2xu − 1 = 0 where e2y − 2xey − 1 = 0; √ 2x ± 4x2 + 4 u= =x± 2 u = ey ; u=x+ x2 + 1 because ey > 0. √ Therefore y = ln u = ln x + x2 + 1 for all real x. C06S09.063: Let u = ey . Then x = coth y = Therefore u2 + 1 = xu2 − x; Therefore y = ln u = ey + e−y e2y + 1 u2 + 1 = 2y =2 . ey − e−y e −1 u −1 (x − 1)u2 = x + 1; 1 1 x+1 ln u2 = ln for all x such that | x | > 1. 2 2 x−1 9 u2 = x+1 . x−1 x2 + 1; and g (x) = 1 ln 2 x+1 x−1 1 f (x) = 1 − x2 C06S09.064: Let f (x) = coth−1 x and g (x) = 1 1 ln(x + 1) − ln(x − 1), so 2 2 1 1 1 g (x) = − 2 x+1 x−1 = for | x | > 1. Then (x − 1) − (x + 1) 1 1 . =− 2 = 2 − 1) 2(x x −1 1 − x2 Therefore there are constants C1 and C2 such that f (x) = g (x) + C1 −1 Let us now express y = coth if x > 1 and f (x) = g (x) + C2 if x < −1. 2 in a more manageable form. ey + e−y = 2; ey − e−y coth y = 2; ey + e−y = 2ey − 2e−y ; 3e−y = ey ; e2y = 3; 2y = ln 3; y= 1 ln 3; 2 coth−1 2 = 1 ln 3. 2 Thus C1 = f (2) − g (2) = 1 1 2+1 ln 3 − ln = 0. 2 2 2−1 Therefore Eq. (37) holds for all x > 1. Repeat this argument, unchanged except for a few minus signs, with y = coth−1 2 to show that C2 = 0 as well. This establishes Eq. (37). C06S09.065: Let −1 f (x) = csch x and g (x) = ln 1 + x √ 1 + x2 |x| for x = 0. Then f (x) = − If x > 0, then g (x) = ln 1+ g (x) = √ 1 + x2 x 1 √ . | x | 1 + x2 . Thus x · 1 (1 + x2 )−1/2 · 2x − 1 − (1 + x2 )1/2 x 2 · 2 )1/2 x2 1 + (1 + x = 1 x2 (1 + x2 )−1/2 − 1 − (1 + x2 )1/2 · x 1 + (1 + x2 )1/2 = 1 x2 − (1 + x2 )1/2 − 1 − x2 · 2 )1/2 1 + (1 + x x(1 + x2 )1/2 =− 1 1 √ =− . 2 )1/2 x(1 + x | x | 1 + x2 10 But if x < 0, then g (x) = ln g (x) = 1− √ 1 + x2 x , and hence −x · 1 (1 + x2 )−1/2 · 2x − 1 + (1 + x2 )1/2 x 2 · x2 1 − (1 + x2 )1/2 = 1 −x2 (1 + x2 )−1/2 − 1 + (1 + x2 )1/2 · x 1 − (1 + x2 )1/2 = 1 −x2 − (1 + x2 )1/2 + 1 + x2 · 1 − (1 + x2 )1/2 x(1 + x2 )1/2 = 1 1 √ =− . 2 )1/2 x(1 + x | x | 1 + x2 Therefore there exist constants C1 and C2 such that f (x) = g (x) + C1 if x > 0 and f (x) = g (x) + C2 if x < 0. Let us now express u = csch−1 1 in a more useful way. csch u = 1; eu 2 = 1; − e−u eu − e−u = 2; e2u − 1 = 2eu ; √ √ 2± 4+4 2u u u e − 2e − 1 = 0; e= = 1 ± 2. 2 √ √ Therefore eu = 1 + 2, and so f (1) = csch−1 1 = u = ln 1 + 2 . But g (1) = ln √ 1 2 + 1 1 = ln 1 + √ 2. Therefore C1 = 0, and so csch −1 x = ln 1 + x √ 1 + x2 |x| if x > 0. This argument may be repeated for u = csch−1 (−1) with few changes other than minus signs here and there, and it follows that C2 = 0 as well. This establishes Eq. (39). C06S09.066: A plot of f (x) = x +2 and g (x) = cosh x for −1 x 2.5 reveals intersections near x = −0.7 and x = 2.1. We applied Newton’s method to the equation f (x) − g (x) = 0 and found that the curves cross very close to the two points (−0.7252637249, 1.2747362751) and (2.0851860142, 4.0851860142). With a the abscissa of the first of these points and b the abscissa of the second, the area between the two curves is b A= a [ f (x) − g (x)] dx = 1 (4x + x2 − 2 sinh x) 2 11 b a ≈ 2.7804546672. C06S09.067: Given: f (x) = e−2x tanh x. First, ex − e−x 1 − e−2x = lim = 0. −x ) x→∞ e2x + 1 +e lim f (x) = lim x→∞ e2x (ex x→∞ Next, a plot of y = f (x) for 0 x 1 reveals a local maximum near where x = 0.45. We applied Newton’s method to solve f (x) = 0 numerically and after seven iterations found that the maximum is close to the point (0.4406867935, 0.1715728753). Indeed, f (x) = − e4x − 2e2x − 1 , 2e2x (e2x + 1)2 so f (x) = 0 when e4x − 2e2x − 1 = 0; that is, when √ √ 2± 4+4 e2x = = 1 ± 2, so that 2 x= √ 1 ln 1 + 2 2 ≈ 0.4406867935. C06S09.068: If f (x) = e−x sinh−1 x, then by l’Hˆpital’s rule, o lim f (x) = lim x→∞ x→∞ sinh−1 x 1 √ = lim = 0. x∞ ex 1 + x2 ex Next, a plot of y = f (x) for 0 x 2 reveals a local maximum near where x = 0.85. The equation f (x) = 0 is transcendental and we were unable to solve it exactly, but Newton’s method revealed that the high point on the graph of f is close to (0.8418432341, 0.3296546569). Note: To solve f (x) = 0 you must solve 1 1 ln(x2 + 1) = x + √ . 2+1 2 x The graph of y = f (x) for 0 is a global maximum. 10 is next. It provides convincing evidence that the extremum we found x 0.3 0.25 0.2 0.15 0.1 0.05 2 C06S09.069: Given: y (x) = y0 + 4 6 8 10 1 (−1 + cosh kx). Then k dy = sinh kx dx and d2 y = k cosh kx. dx2 So k Therefore d2 y =k dx2 1+ 2 1 + [ y (x)] = k dy dx 1 + sinh2 kx = k 2 . Moreover, 12 cosh2 kx = k cosh kx. y (0) = y0 + 1 0 (−1 + cosh 0) = y0 + = y0 k k and y (0) = sinh(k · 0) = sinh(0) = 0. C06S09.070: Using the coordinate system in Fig. 6.9.4 with units in feet, the cable has the shape of the graph of y (x) = 30 + 1 (−1 + cosh kx). k We also know that 50 = y (100) = 30 + 1 (−1 + cosh 100k ), k and it follows that g (k ) = 0 where g (k ) = cosh(100k ) − 20k − 1. A plot of y = g (x) reveals a solution near x = 0.004, and Newton’s method reveals the more accurate approximation k ≈ 0.003948435453. We then found that the approximate length of the high-voltage line is 100 L=2 0 100 2 1 + [ y (x)] dx = 2 0 cosh kx dx = 2 · 13 1 sinh kx k 100 0 ≈ 205.2373736258 (ft). Chapter 6 Miscellaneous Problems C06S0M.001: The net distance is 3 v (t) dt = 0 13 12 t − t − 2t 3 2 3 0 =− 3 3 −0=− . 2 2 Because v (t) < 0 for 0 < t < 2, the total distance is 2 − 3 v (t) dt + 0 10 3 10 −0 + − + 3 2 3 v (t) dt = 2 31 ≈ 5.166667. 6 = C06S0M.002: Because t2 − 4 < 0 for 1 < t < 2 but t2 − 4 > 0 for 2 < t < 4, whereas v (t) the net and total distance are both 2 − 1 (t2 − 4) dt + 4 2 13 t − 4t 3 (t2 − 4) dt = − C06S0M.003: Because v (t) < 0 for 0 < t < 3 /2 0 1 2 2 1 but v (t) > 0 for v (t) dt = − cos 4 13 t − 4t 3 + 1 π (2t − 1) 2 1 2 = 2 0 for 1 t 4, 5 32 37 + = ≈ 12.333333. 3 3 3 < t < 3 , the net distance is 2 3 /2 0 =1−0=1 and the total distance is 1 /2 − 3 /2 v (t) dt + 0 1 /2 1 C06S0M.004: The volume is x3 dx = 0 4 C06S0M.005: The volume is x1/2 dx = 1 2 C06S0M.006: The volume is x3 dx = 1 1 C06S0M.007: The volume is 0 1 C06S0M.008: The volume is −1 v (t) dt = (1 − 0) + (1 − (−1)) = 3. 1 14 x 4 0 2 3 /2 x 3 14 x 4 4 = 1 2 1 =4− π (x2 − x4 ) dx = π x100 dx = 1 1 −0= . 4 4 = 16 2 14 −= ≈ 4.666667. 3 3 3 1 15 = = 3.75. 4 4 13 15 x− x 3 5 1 101 x 101 1 = −1 1 = 0 2π ≈ 0.4188790205. 15 2 ≈ 0.0198019802. 101 C06S0M.009: Between time t = 0 and time t = 12, the rainfall in inches is 12 0 1 12 1 (t + 6) dt = t+ t 12 2 24 12 0 = 12 − 0 = 12. C06S0M.010: The curves meet at (0, 0) and at (1, 1), and the quadratic is higher than the cubic between those points. A cross section of the solid perpendicular to the x-axis is a square of base length 2x − x2 − x3 and thus the cross section has area A(x) = (2x − x2 − x3 )2 . Hence the volume of the solid is 1 1 V= 1 43 3 1 1 x − x4 − x5 + x6 + x7 3 5 3 7 A(x) dx = 0 = 0 22 ≈ 0.2095238095. 105 C06S0M.011: The region R of Problem 10 is bounded above by the graph of f (x) = 2x − x2 and below by the graph of g (x) = x3 for 0 x 1. A cross section of the solid S of Problem 11 perpendicular to the x-axis at x is an annular region with outer radius f (x) and inner radius g (x), thus of cross-sectional area 2 2 A(x) = π [ f (x) ] − π [ g (x) ] . Therefore the volume of S is 1 V= A(x) dx = π 0 43 1 1 x − x4 + x5 − x7 3 5 7 1 = 0 41 π − 0 ≈ 1.2267171314. 105 C06S0M.012: The region R of this problem is bounded above by the graph of f (x) = x2 + 1 and below by the graph of g (x) = 2x4 for −1 x 1 and (important) is symmetric around the y -axis. If R is rotated around the x-axis, the solid S that it generates has as cross sections perpendicular to the x-axis at x annular 2 2 regions with outer radius f (x) and inner radius g (x), thus of cross-sectional area A(x) = π [ f (x) ] − π [ g (x) ] . Therefore the volume of S is 1 V1 = A(x) dx = π x + −1 1 23 15 49 x+ x− x 3 5 9 = −1 64π 64π −− 45 45 = 128π ≈ 8.9360857702. 45 If R is rotated around the y -axis to form the solid T , then (using the symmetry of R around the y -axis) the method of cylindrical shells yields the volume of T as 1 V2 = 0 1 2 2π x [ f (x) − g (x) ] dx = π x2 + x4 − x6 2 3 1 = 0 5π ≈ 2.6179938780. 6 C06S0M.013: Each cross section perpendicular to the x-axis has area A(x) = the helix is 20 m= 0 17π x 32 (8.5) · A(x) dx = 20 = 0 85π ≈ 33.3794219444 8 1 16 π , so the total mass of (grams). C06S0M.014: Most of the natural ways to solve this problem involve algebraic difficulties. For example, the side of the frustum should not be part of the graph of y = mx, even though this would seem to yield the simplest choice. In each case, both the method of cross sections and the method of cylindrical shells lead to difficulties. Here’s the simplest solution we’ve found. Write r for r1 and s for r2 . Sketch a trapezoid in the first quadrant with vertices at (0, 0), (h, 0), (h, s), and (0, r). Then an equation of the top edge of the trapezoid is s−r x. h The frustum is produced by rotating the trapezoidal region around the x-axis, and its volume is y=r+ h V= π r+ 0 = π r2 x + s−r x h 2 h dx = π r(s − r) 2 1 x+ h 3 r2 + 0 s−r h 2r(s − r) x+ h h 2 x3 0 s−r h 2 x2 dx 1 = π r2 h + r(s − r)h + (s − r)2 h 3 1 πh 2 πh 2 2 = π h(3r2 + 3rs − 3r2 + s2 − 2rs + r2 ) = r + rs + s2 = r1 + r 1 r 2 + r 2 . 3 3 3 2 C06S0M.015: Let z denote the distance from P to the origin. A horizontal cross-section of the elliptical cone “at” z (thus at distance z from P ) is an ellipse with major axis and minor axis each proportional to z . So the area A(z ) of this cross section is proportional to z 2 : A(z ) = kz 2 where k is a positive constant. But A(h) = kh2 = π ab by the result of Problem 47 of Section 5.8, and hence k = π ab/h2 . Therefore A(z ) = π abz 2 /h2 . So the volume of the elliptical cone is h V= 0 h π ab 2 π ab 3 z dz = z h2 3h 2 = 0 1 π abh, 3 one-third the product of the area of the base and the height of the elliptical cone. C06S0M.016: Because (a − h, r) lies on the ellipse, And so V= a π y 2 dx = a−h But r2 = a a−h π b2 1 − b2 b2 r2 h(2a − h), so 2 h = . Therefore a2 a 2a − h V = 1 πr2 h 3 2 a−h a x2 a2 + r b dx = π 2 = 1. Therefore r2 = 2ah − h2 2 b. a2 b2 h2 (3a − h) . 3a2 3a − h . 2a − h C06S0M.017: Because (a + h, r) lies on the hyperbola, (a + h)2 r2 − 2 = 1. a2 b It follows that r2 = b2 (2ah + h2 ) . a2 (1) Moreover, the equation of the hyperbola may be written in the form y2 = b2 2 (x − a2 ). a2 Therefore the “segment of the hyperboloid” has volume a+h V= π y 2 dx = a π b2 a2 a+h a a+h = π b2 3 x − 3a2 x 3a2 = (x2 − a2 ) dx = π b2 a2 13 x − a2 x 3 a+h a 1b 2 π h (3a + h). 3 a2 But by Eq. (1), b2 = = a π b2 3 a + 3a2 h + 3ah2 + h3 − 3a3 − 3a2 h − a3 + 3a3 3a2 2 a2 r2 1 h2 a2 r2 3a + h 1 . So V = π 2 (3a + h) = πr2 h . 2 2ah + h 3a h(2a + h) 3 2a + h 3 t C06S0M.018: V (t) = 1 t C06S0M.019: V = 2 π (f (x)) dx = π 1 − π (1 + 3t)2 − 16 . Thus 6 π 2 π (f (x)) = [(2)(1 + 3x)(3)] = π (1 + 3x). 6 2 π (f (x)) dx = 1 Therefore f (x) = √ 1 + 3x . t C06S0M.020: V (t) = 1 2π xf (x) dx = 2 π 9 √ 3 /2 1 + 3 t2 V (t) = 2π tf (t) = 2 π 9 Therefore f (x) = 1 π 1 2 , so V (t) = π (f (t)) = 2 . Therefore f (x) = . t t x 3 2 − 8 , so 1 + 3t3 (6t) = 2π t 1 + 3 t2 . 1 + 3x2 . C06S0M.021: The graphs of f (x) = sin 1 π x and g (x) = x cross at (0, 0) and (1, 1), and g (x) < f (x) if 2 0 < x < 1. When the region they bound is rotated around the y -axis, the method of cylindrical shells yields the volume of the solid thus generated to be 1 V= 0 Now let u = 2π x [ f (x) − g (x) ] dx = 2π 1 πx − x2 2 x sin 0 dx. πx 2u , so that x = . This substitution yields 2 π 2 4 u sin u − 2 u2 π π π /2 V = 2π 0 = · 8 16 3 (sin u − u cos u) − u π 3π 2 2 du = π π /2 = 0 π /2 0 8 16 u sin u − 2 u2 π π du 8 16 π 3 8 2π − 2· =− ≈ 0.4520839871. π 3π 8 π 3 C06S0M.022: If −1 x 2, then a thin vertical strip of the region above x is rotated in a circle of radius x + 2. Therefore the volume generated is 2 V= −1 = C06S0M.023: 2π (x + 2)(x + 2 − x2 ) dx = π 8x + 4x2 − 56π 23π −− 3 6 = 23 14 x− x 3 2 2 −1 45π ≈ 70.6858347058. 2 dy = 1 x1/2 − 1 x−1/2 , so 2 2 dx 1+ dy dx 2 = 1 1 /2 2x + 1 x−1/2 2 2 (1) . So the length of the curve is 4 L= 1 1 1/2 1 −1/2 1 dx = x1/2 + x3/2 x +x 2 2 3 4 4 = 1 14 4 10 −= . 3 3 3 C06S0M.024: We use the result in Eq. (1) in the solution of Problem 23: ds = 1 x−1/2 + 1 x1/2 dx. The 2 2 graph of f (x) = 1 x3/2 − x1/2 lies below the x-axis for 1 x 3 and above it for 3 x 4. Hence the 3 radius of the cirle of rotation is −f (x) in the former case and f (x) in the latter case. So the area of the left-hand part of the surface is AL = − 3 3 2π f (x) ds = 2π 1 = 2π 1 1 13 1 x + x2 − x 2 6 18 11 1 + x − x2 23 6 3 dx 33311 1 +−−−+ 2 2 2 2 6 18 = 2π 1 = 16π . 9 The area of the right-hand part of the surface is 3 AR = 2π f (x) ds = 2π 1 1 3 12 1 x− x− x 18 6 2 32 8 333 − −2− + + 9 3 222 = 2π = 4 3 7π . 9 Therefore the total area of the surface of revolution around the x-axis is AL + AR = 16π 7π 23π + = ≈ 8.0285145592. 9 9 9 There is no such difficulty in part (b), in which the graph of f is rotated around the y -axis. The area of the surface thereby generated is 4 A= 1 1 /2 1 3 / 2 x +x 2 2 4 2π x ds = 2π 1 1 = 2π 1 3 /2 1 5 /2 x +x 3 5 4 = 2π 1 dx 8 32 1 1 + −− 3 5 35 = 256π ≈ 53.6165146213. 15 C06S0M.025: Let x = f (y ) = 3 (y 4/3 − 2y 2/3 ). Then 8 2 1 + [ f (y ) ] = 1 + 9 64 4 1/3 4 −1/3 −y y 3 3 2 = 1 2/3 1 1 −2/3 (1 + y 2/3 )2 ++y = , y 4 24 4y 2/3 and therefore ds = 1 (y 1/3 + y −1/3 ) dy . Hence the length of the graph of g from y = 1 to y = 8 is 2 8 L= 8 1 ds = 1 1 3 4 /3 3 2 / 3 1 1 /3 + y −1/3 ) dy = +y (y y 2 8 4 8 1 =9− 9 63 = = 7.875. 8 8 C06S0M.026: Let x = g (y ) = 3 (y 4/3 − 2y 2/3 ), 1 y 8. As in the solution of Problem 25, we find that 8 ds = 1 (y 1/3 + y −1/3 ) dy . So the surface area generated by revolving the graph of g around the x-axis will be 2 8 A= 1 8 2π y ds = π 1 (y 4/3 + y 2/3 ) dy = π 3 7 /3 3 5 / 3 +y y 7 5 8 = 1 2592π 36π 2556π − = ≈ 229.4260235022. 35 35 35 √ But the graph of x = g (y ) crosses the y -axis where y = a = 2 2, so two integrals are required to find the surface area generated by rotating the graph around the y -axis. They are 5 a A1 = − 1 2π g (y ) ds = − 3π 8 a 1 (y 5/3 − y − 2y 1/3 ) dy = −π 9 8 /3 32 9 4 /3 − y y− y 64 16 16 a = 1 57π 64 and 8 A2 = 2π g (y ) ds = a 3π 8 8 a (y 5/3 − y − 2y 1/3 ) dy = π 9 8 /3 32 9 4/3 y − y− y 64 16 16 Therefore the answer in part (b) is A1 + A2 = x = a 33π . 2 1113π ≈ 54.63425974. 64 C06S0M.027: Let f (x) = 1 x3/2 − x1/2 , 1 3 8 2 1 + [ f (x)] = 1 + 4. Then 2 1 1/2 1 −1/2 x −x 2 2 = 1 11 x + + x−1 = 4 24 1 1/2 1 −1/2 x +x 2 2 2 . Therefore ds = 1 x1/2 + x−1/2 dx. Therefore the area of the surface generated when the graph of f is 2 rotated around the vertical line x = 1 is 4 A= 1 = C06S0M.028: 44π 8π −− 5 5 = b a 2π r2 − x2 √ 1 2 5/2 x − 2x1/2 5 (x3/2 − x−1/2 ) dx = π 4 1 52π ≈ 32.6725635973. 5 dy x , so 1 + = −√ 2 − x2 dx r A= 4 2π (x − 1) ds = π dy dx r dx = 2 − x2 r 2 = b r2 . Therefore r2 − x2 b 2π r dx = 2π rx a a = 2π r(b − a) = 2π rh. C06S0M.029: This is merely a matter of substituting 2r for h in the area formula A = 2π rh derived in Problem 28. Thus the area of a sphere of radius r is A = 2π r · 2r = 4π r2 . √ C06S0M.030: Let f (x) = 2x3 and g (x) = 2 x. The region R bounded by the graphs of f and g lies in the first quadrant and the two curves cross at the origin and at (1, 2). The graph of f is also the graph of x = h(y ) = (y/2)1/3 and the graph of g is also the graph of x = j (y ) = (y/2)2 . The graph of g is above the graph of f on the interval 0 x 1 and the graph of j is to the left of the graph of h on the interval 0 y 2. Part (a): R is rotated around the x-axis, generating a solid of volume V1 . To find V1 by the method of cross sections, we first simplify 2 2 [ g (x)] − [ f (x)] = 4x − 4x6 , and therefore 1 V1 = π 0 (4x − 4x6 ) dx = π 2x2 − 47 x 7 To find V1 by the method of cylindrical shells, we evaluate 6 1 = 0 10π ≈ 4.4879895051. 7 2 0 2 2π y [ h(y ) − j (y )] dy = π 22 / 3 y 4 / 3 − 0 13 3 · 22 / 3 7 / 3 1 4 y dy = π y −y 2 7 8 2 = 0 10π . 7 Part (b): R is rotated around the y -axis, generating a solid of volume V2 . To find V2 by the method of cylindrical shells, we evaluate 2 V2 = 0 2π x [ g (x) − f (x)] dx = π 1 0 (4x3/2 − 4x4 ) dx = π 8 5 /2 4 5 x −x 5 5 1 4π ≈ 2.5132741229. 5 = 0 To find V2 by the method of cross sections, we first simplify 2 2 [ h(y )] − [ j (y )] = y 2/3 y4 1 − = 8 · 21 / 3 y 2 / 3 − y 4 . 16 16 22/3 Then 2 V2 = 0 π· 2 1 π 8 · 21/3 y 2/3 − y 4 dy = 24 · 21/3 y 5/3 − y 5 16 80 = 0 4π . 5 Part (c): R is rotated around the horizontal line y = −1, generating a solid of volume V3 . To find V3 by the method of cross sections, we simplify 2 2 [ g (x) + 1] − [ f (x) + 1] = 4x1/2 + 4x − 4x3 − 4x6 . Then 1 V3 = 0 π (4x1/2 + 4x − 4x3 − 4x6 ) dx = π 56x3/2 + 42x2 − 21x4 − 12x7 21 1 0 = 65π ≈ 9.7239772611. 21 To find V3 by the method of cylindrical shells, we first simplify the integrand: 2π (y + 1) [ h(y ) − j (y )] = 1 1 1 π (y + 1)(2 · 22/3 y 1/3 − y 2 ) = π 22/3 y 1/3 + 22/3 y 4/3 − y 2 − y 3 . 2 2 2 Then 2 V3 = π 0 =π 22 / 3 y 1 / 3 + 22 / 3 y 4 / 3 − 12 13 y− y 2 2 dy 3 · 22/3 4/3 3 · 22/3 7/3 1 3 1 4 + − y− y y y 4 7 6 8 2 = 0 65π . 21 Part (d): Finally, R is rotated around the vertical line x = 2, thereby generating a solid of volume V4 . To evaluate V4 by the method of cross sections, we first simplify 2 2 [2 − j (y )] − [2 − h(y )] = 25/3 y 1/3 − Then 7 y 2 /3 y4 − y2 + . 16 22 / 3 2 V4 = π 25/3 y 1/3 − 0 y 2 /3 y4 − y2 + 2 /3 16 2 dy 3 · 22/3 4/3 3 · 21/3 5/3 1 3 15 =π y y y − − y+ 2 10 3 80 2 = 0 38π ≈ 7.9587013891. 15 To evaluate V4 by the method of cylindrical shells, we first simplify (2 − x) [ g (x) − f (x)] = 4x1/2 − 2x3/2 − 4x3 + 2x4 . Then 1 V4 = 2π 0 8 3 /2 4 5 /2 2 x − x − x4 + x5 3 5 5 4x1/2 − 2x3/2 − 4x3 + 2x4 dx = 2π 1 = 0 38π . 15 C06S0M.031: Denote the spring constant by K . The information given in the problem yields 5 2 5 2 3 K (x − L) dx = 5 (x − L) dx = 5 1 (x − L)2 2 5 =5 2 2 3 2 K (x − L) dx; (x − L) dx; 1 (x − L)2 2 3 ; 2 (5 − L)2 − (2 − L)2 = 5(3 − L)2 − 5(2 − L)2 ; 25 − 10L + L2 − 4 + 4L − L2 = 45 − 30L + 5L2 − 20 + 20L − 5L2 ; 4L = 4. Therefore the natural length of the spring is L = 1 (ft). C06S0M.032: Set up a coordinate system in which y = 50 is the position of the windlass and the lowest point P of the cable is initially at y = 0. When P is at location y (0 y 50), the length of the cable is 50 − y , so the total weight on the windlass is 1000 + 5 · (50 − y ) (lb). Therefore the work to wind in 25 feet of the cable is 25 W= 0 (1000 + 250 − 5y ) dy = 1250y − 52 y 2 25 = 29687.5 0 (ft·lb). C06S0M.033: Set up a coordinate system in which the center of the tank is at the origin and the y axis is vertical. A horizontal cross section of the oil at positive y (−R y R) is circular with radius x = R2 − y 2 , so its area is π (R2 − y 2 ). Hence the work to pump the oil to its final position y = 3R is W= R −R = πρ (3R − y )πρ(R2 − y 2 ) dy = πρ 14 1 y − Ry 3 − R2 y 2 + 3R3 y 4 2 8 R −R (y 3 − 3Ry 2 − R2 y + 3R3 ) dy R = πρ −R 74 94 R+ R 4 4 = 4πρR4 . C06S0M.034: Set up a coordinate system with the axis of the cone lying on the y -axis and with a diameter of the base of the cone lying on the x-axis. Now a horizontal slice of the cone at height y has radius given by x = 1 (1 − y ); the units here are in feet. Therefore the work done in building the anthill is 2 1 W= 0 1 π (150y )π (1 − y )2 dy = 4 4 75 4 y − 100y 3 + 75y 2 2 1 = 0 25 π ≈ 9.82 8 (ft·lb). C06S0M.035: Set up a coordinate system in which the center of the earth is at the origin and the hole extends upward along the vertical y -axis, with its top where y = R, the radius of the earth in feet. A 1-pound weight at position y (0 y R) weighs y/R pounds, so the total work to lift the weight from y = 0 to y = R is R W= 0 y dy = R y2 2R R = 0 R 3960 · 5280 = = 10454400 2 2 (ft·lb). The assumption of constant density of the earth is required to draw the conclusion that the gravitational force is proportional to the distance from the center of the earth. C06S0M.036: Set up a coordinate system in which the center of the earth is at the origin and the hole extends along the nonnegative y -axis from y = 0 to y = R = 3960 · 5280, the radius of the earth in feet. Imagine a thin cylindrical horizontal slab of dirt (or basalt, or whatever) in the hole at distance y from the center of the earth. As it moves from its initial position y to its final position R, its weight varies: If it is at position u, y u R, then its weight will be u R (350π ) · du where du denotes its thickness. The total work required to lift this slab from its initial position (u = y ) to the surface (u = R) is then R y 350π u u2 du = 350π R 2R R = y 350π R2 − y 2 . 2R Therefore the total work required to lift all the dirt (or basalt, or whatever) from the hole to the surface of the earth is R W= 0 350π 350π 1 R2 − y 2 dy = R2 y − y 3 2R 2R 3 R = 0 350π R2 ≈ 1.6023407560 × 1017 3 (ft·lb). It is intriguing to note that the answer may be written in the form R W= 0 R 350π y u du dy. R C06S0M.037: If the coordinate system is chosen with the origin at the midpoint of the bottom of the dam and with the x-axis horizontal, then the equation of the slanted edge of the dam is y = 2x − 200 (with units in feet). Therefore the width of the dam at level y is 2x = y + 200. Let ρ = 62.4 be the density of water in pounds per cubic foot. Then the total force on the dam is 100 F= 0 ρ(100 − y )(y + 200) dy = ρ 20000y − 50y 2 − 9 13 y 3 100 = 0 3500000ρ = 72800000 3 (lb). C06S0M.038: The answer may be obtained from the answer to Problem 37 by multiplying the latter by √ sec 30◦ : The force is 2/ 3 times as great, or approximately 8.4062199194 × 107 pounds. The analytical approach here is to introduce the additional factor sec(π /6) into the integral in the solution of Problem 37, but because this factor is a constant, one may as well simply multiply the answer by the same factor. C06S0M.039: The volume of the solid is c V= 2π y + 0 1 c c 2√ 4π 2 5/2 2 y dy = ·y + y 3 /2 c c 5 3c = 8π 0 1 3/2 1 −1/2 c +c . 5 3 It is clear that there is no maximum volume, because V → +∞ as c → 0+ . But V → +∞ as c → +∞ as √ well, so there is a minimum volume; V (c) = 0 when c = 1 5, so this value of c minimizes V . 3 C06S0M.040: Here we have 1+ 2 dy dx = x4 + 1 4x4 2 . Therefore 2 L= x4 + 1 4x4 dx = 3011 , 480 x5 + 1 4x3 dx = 339 , 32 1 2 My = 1 2 Mx = 1 = 15 1 x+ 5 12x3 · x4 + x10 x2 x2 1 + + − 50 40 24 288x6 and 1 4x4 2 dx = 1 2 x9 x x 1 + + + 5 20 12 48x7 dx 1057967 = 20.66341796875. 51200 = 1 Therefore x= My 5085 = ≈ 1.68880770508 L 3011 and y= Mx 3173901 = ≈ 3.294 − 6862388. L 963520 C06S0M.041: Here, 2 13 33 (y + y −3 ) dy = , 2 16 2 1 · 2 2 L= 67 14 (y + y −2 ) dy = . 2 20 1 My = 1 Mx = 1 1 4 1 −2 y+ y 8 4 · (y 3 + y −3 ) dy = 1179 , 512 Therefore x= 393 ≈ 1.116477 352 and C06S0M.042: First, 10 y= 268 ≈ 1.624242. 165 and dy dx 1+ 2 = 2 1 1/2 1 −1/2 x +x 2 2 . Therefore 4 1 1 /2 10 (x + x−1/2 ) dx = , 2 3 4 1 3 /2 128 (x + x1/2 ) dx = , 2 15 4 L= 1 · 2 1 My = 1 Mx = 1 1 3 /2 x − x1/2 3 and 1 · (x1/2 + x−1/2 ) dx = − . 2 Therefore x= 128 3 64 · = 15 10 25 and y=− 13 3 · =− . 2 10 20 C06S0M.043: To begin with, dx dy 1+ it follows that the length of the curve is L = 2 = 1 1/3 1 −1/3 +y y 2 2 2 ; 63 . Next, 8 8 31 999 · · (y 4/3 − 2y 2/3 ) · (y 1/3 + y −1/3 ) dy = = 7.8046875 82 128 8 My = 1278 1 4 /3 + y 2/3 ) dy = (y ≈ 36.51428571. 2 35 1 Mx = 1 Therefore x= 111 ≈ 0.991071427 112 and y= 1136 ≈ 4.636734694. 245 C06S0M.044: The two curves meet at (0, 0) and at (1, 1). So 1 A= 0 1 My = 0 (2x − x2 − x3 ) dx = (2x2 − x3 − x4 ) dx = 1 Mx = 0 5 , 12 13 , 60 and 1 41 (2x − x2 − x3 )2 dx = . 2 210 Therefore x= 13 = 0.52 25 and y= 11 82 ≈ 0.4685714286. 175 and C06S0M.045: The curves meet at (2, 1), at (0, 0), and at (2, −1). It follows that y = 0 by symmetry and that we may compute x by using only the upper half of the figure. In that case we have 1 A= 0 1 My = 0 Therefore x = (y 2 + 1 − 2y 4 ) dy = 14 15 and 1 32 (y 2 + 1)2 − (2y 4 )2 dy = . 2 45 16 ≈ 0.7619047619. 21 C06S0M.046: Given a triangle in the plane, set up a coordinate system in such a way that the lowest vertex of the triangle is at the origin, there is a vertex in the first quadrant at (a, b), and a vertex in the second quadrant at (−a, c). Thus the y -axis passes through the midpoint of the side opposite the vertex at the origin, and hence a median of the triangle lies on the y -axis. We will show that the y -coordinate of the centroid also lies on the y -axis. Then, by rotating the triangle to plane the other two vertices at the origin in a similar way, we may conclude that y lies on all three medians. Then interchange the roles of x and y to conclude that x lies on the intersection of the medians as well. The left side of the triangle has equation y = h(x) = −cx/a, the right side has equation y = g (x) = bx/a, and the top side has equation y = f (x) = b + b−c (x − a). 2a Hence the moment of the triangle with respect to the y -axis is Mx = ML + MR where ML denotes its moment to the left of the y -axis and MR its moment to the right. Now a MR = bx + 0 = = b−c 2 b−c b x− x − x2 2a 2 a 1 2 b−c 3 b−c 2 b3 bx + x− x− x 2 6a 4 3a dx a = 0 1 2 b−c 2 b−c 2 b 2 ba + a− a− a 2 6 4 3 6b + 2b − 2c − 3b + 3c − 4b 2 b+c 2 a= a. 12 12 Moreover, ML = − 0 −a bx + b−c 2 b−c c x− x + x2 2a 2 a dx = · · · = b+c 2 a 12 by extremely similar computations. Thus the triangle balances on the y -axis, and therefore y = 0. In light of the opening remarks, this completes the proof. C06S0M.047: 2π y · π ab 4 4b = π ab2 , and it follows that y = . 2 3 3π C06S0M.048: Note that x = y . The area of the quarter ring is A= 1 π (π b2 − π a2 ) = (b2 − a2 ), 4 4 and the volume generated by rotating it around the x-axis is V= 2323 πb − πa . 3 3 12 2 π π (b3 − a3 ) = (2π y ) · · (b2 − a2 ). 3 4 Part (a): Consequently Therefore V = y= Part (b): lim x = b →a 2 3 3 3 π (b − a ) 122 2 2 π (b − a ) = 4(b2 + ab + a2 ) = x. 3π (b + a) 12a2 2a = = lim y . b →a (3π )(2a) π C06S0M.049: (a) The area A of the triangle T can be computed in several ways; we chose the most direct which, elementary, is easy to do by hand. Let O denote the vertex of the triangle at (0, 0), C = C (c, 0), A = A(a, 0), B = B (a, b), and D = D(c, d). Then A is the area of triangle OCD plus the area of trapezoid CABD minus the area of triangle OAB : A= = cd (a − c)(b + d) ab + − 2 2 2 cd ab ad bc cd ab ad − bc + + − − − = . 2 2 2 2 2 2 2 (b) In Problem 46 we saw that the centroid of a triangle lies on the intersection of its medians. From plane geometry we also know that the point of intersection is two-thirds of the way from any vertex to the midpoint of the opposite side. The midpoint of L has y -coordinate (b + d)/2, and hence y= 2 b+d b+d · = . 3 2 3 b + d ad − bc 1 · = π (b + d)(ad − bc). 3 2 3 1 ad − bc ad − bc (d) pw = A = , so p = . 2 2 w b+d (e) S = 2π · · w = π w(b + d). 2 b+d 1 b+d 1 (f) V = 2π yA = 2π · · pw = π pw · = pS . 3 2 3 3 (c) V = 2π yA = 2π · C06S0M.050: Let n = 2k . Inscribe the 2k -gon with opposite vertices on the x-axis. Let T be one of the triangles formed by a side of the polygon and two radii of the circle. The perpendicular from the origin to the midpoint of the side of the polygon has length (in the notation of Problem 49) p = r cos π . k By part (f) of Problem 49, V= 1 π r cos S. 3 k Now let k → +∞ and replace S with 4π r2 to obtain Archimedes’ result V= 43 πr . 3 C06S0M.051: A Mathematica solution: First let f (x) = xm and g (x) = xn where m and n are positive integers and n > m. 13 a = Integrate[ f[x] - g[x], { x, 0, 1 } ] 1 1 − m+1 n+1 area = a /. m → 1 1 1 − 2 n+1 Formulas (11) and (12) in the text give the moments my = Integrate[ x∗(f[x] - g[x]), { x, 0, 1 } ]; mx = Integrate[ (1/2)*((f[x])∧2 - (g[x])∧2), { x, 0, 1 } ]; My = my /. m→1 n−1 3(n + 2) Mx = mx /. m → 1 1 1 − 3 2n + 1 2 Hence the centroid has coordinates { xc, yc } = { My/area, Mx/area } // Simplify { (n + 1) 2(n + 1) , } 3(n + 2) 3(2n + 1) Limit[ { xc, yc }, n → Infinity ] { 21 ,} 33 Obviously this is the centroid of the triangle with vertices (0, 0), (1, 0), and (1, 1)—which the area of the region bounded by the graphs of f and g “exhausts” as n → + ∞. C06S0M.052: By Example 1 in Section 6.6, the semicircular disk has area and centroid a1 = 9∗Pi/2; c1 = { 0, 4/Pi }; Hence its moment with respect to the x-axis is mx1 = a1∗c1[[2]] 18 (Recall that the Mathematica command list[[n]] extracts the n th entry from the k -dimensional array list = { a1, a2, a3, . . . , ak }.) The square are area and centroid a2 = 4; 14 c2 = { 0, −1 }; so its moment with respect to the x-axis is mx2 = a2∗c2[[2]] −4 Therefore the region R has area and x-moment a = a1 + a2 9π 4+ 2 mx = mx1 + mx2 14 So the y -coordinate of its centroid is yc - mx/a // Simplify 28 8 + 9π A numerical approximation to this result is N[yc] 0.771896 The radius of revolution around the line y = −4 is r = 4 + yc // Together 12(5 + 3π ) 8 + 9π So the volume of revolution is v = 2∗Pi∗r∗a // Simplify 12π (3π + 5) C06S0M.053: Let u = 1 − 2x. Then dx = − 1 du, and so 2 dx 1 =− 1 − 2x 2 1 1 1 du = − ln | u | + C = − ln | 1 − 2x | + C. u 2 2 C06S0M.054: Let u = 1 + x3/2 . Then du = 3 x1/2 dx, so that x1/2 dx = 2 x1/2 2 dx = 3 1 + x3/2 2 3 du. Hence 1 2 2 du = ln u + C = ln 1 + x3/2 + C. u 3 3 C06S0M.055: Let u = 1 + 6x − x2 . Then du = (6 − 2x) dx, so that (3 − x) dx = 15 1 2 du. Thus 3−x 1 dx = 1 + 6x − x2 2 1 1 1 du = ln | u | + C = ln | 1 + 6x − x2 | + C. u 2 2 C06S0M.056: Let u = ex + e−x . Then du = (ex − e−x ) dx, so that 1 du = (ln u) + C = ln ex + e−x + C. u ex − e−x dx = ex + e−x C06S0M.057: Let u = 2 + cos x. Then du = − sin x dx, and thus sin x dx = − 2 + cos x 1 du = −(ln u) + C = − ln(2 + cos x) + C. u 2 C06S0M.058: 2 e−1/x 1 dx = e−1/x + C . (Optional substitution: u = −1/x2 .) x3 2 √ C06S0M.059: Let u = 10 x . Then √ du = (10 x √ 1 10 x ln 10 ln 10) · x−1/2 dx = √ · dx. 2 2 x Therefore √ 10 √ x x dx = √ 2 2u 2 · 10 x du = +C = + C. ln 10 ln 10 ln 10 1 dx, so that x 1 1 1 1 dx = du = − + C = − + C. x(ln x)2 u2 u ln x C06S0M.060: Let u = ln x. Then du = C06S0M.061: Let u = 1 + ex . Then du = ex dx, and thus ex (1 + ex )1/2 dx = u1/2 du = 2 3 /2 2 u + C = (1 + ex )3/2 + C. 3 3 1 dx, and therefore x 2 2 u1/2 du = u3/2 + C = (1 + ln x)3/2 + C. 3 3 C06S0M.062: Let u = 1 + ln x. Then du = 1 (1 + ln x)1/2 dx = x C06S0M.063: 2x · 3x dx = 6x dx = 6x + C. ln 6 C06S0M.064: Let u = 1 + x2/3 . Then du = dx 3 = 2 x1/3 (1 + x2/3 ) 2 −1/3 x dx. Hence 3 1 3 3 du = (ln u) + C = ln(1 + x2/3 ) + C. u 2 2 16 C06S0M.065: The revenue realized upon selling after t months will be f (t) = B · 2 + t 12 · 2−t/12 , for which f (t) = B · Thus f (t) = 0 when 12 − 24 ln 2 − t ln 2 . 144 · 2t/12 12 − 24 ln 2 ≈ −6.68765951. ln 2 But this value of t is negative, and in addition f (t) < 0 for all larger values of t. Thus the revenue is a decreasing function of t for all t 0. Therefore the grain should be sold immediately. t= C06S0M.066: The profit will be f (t) = 800 exp 1 2 √ t − 1000 exp 1 10 t . Now 100 2 exp √ 1 t − t1/2 exp 10 t , 1 /2 t We used Newton’s method to solve f (t) = 0 and found the solution to be approximately 11.7519277504. To make sure that this value of t maximizes the profit, we graphed f (t) for 0 t 21 (the graph is shown following this solution). The profit upon cutting and selling after about 11.75 years will be approximately $1202.37. f (t) = 1 2 1200 1000 800 600 400 200 5 10 15 20 -200 C06S0M.067: If lots composed of x pooled samples are tested, there will be 1000/x lots, so there will be 1000/x tests. In addition, if a lot tests positive, there will be x additional tests. The probability of a lot testing positive is 1 − (0.99)x , so the expected number of lots that require additional tests will be the product of the number of lots and the probability 1 − (0.99)x that a lot tests positive. Hence the total number of tests to be expected will be f (x) = if x 1000 1000 1000 + [1 − (0.99)x ] · x = + 1000 − 1000(0.99)x x x x 2. Next, 1000 − 1000(0.99)x ln(0.99); x2 1 f (x) = 0 when = (0.99)x ln(100/99); x2 f (x) = − x2 = x= (0.99)−x ; ln(100/99) (0.99)−x/2 ln(100/99) . 17 The form of the last equation is exactly what we need to implement the method of repeated substitution (see Problems 23 through 25 of Section 3.9 of the regular edition, Section 3.10 of the early transcendentals version, of the text). We substitute our first “guess” x0 = 10 into the right-hand side of the last equation, thus obtaining a “better” (we hope) value x1 and continue this process until the digits in these successive approximations stabilize. Results: x1 = 10.488992, x2 = 10.514798, x3 = 10.516161, x4 = 10.516233, and x5 = 10.516237 = x6 . The method is not as fast as Newton’s method but the formula is simpler. (The graph of y = f (x) follows this solution to convince you that we have actually found the minimum value of f .) We must use an integral number of samples, so we find that f (10) ≈ 195.618 and f (11) = 195.571, so there should be 90 lots of 11 samples each and one lot of 10 for the most economical results. Alternatively, it might be simpler to use 10 samples in every lot; the extra cost would be only about 24 cents. The total cost of the batch method will be about $978, significantly less than the $5000 cost of testing each sample individually. 500 450 400 350 300 250 10 C06S0M.068: If f (x) = 1 x2 − 2 1 4 20 30 40 ln x, then 2 1 + [ f (x) ] = 1 + x − 1 4x 2 = x2 + 1 1 = + 2 16x2 x+ 1 4x 2 . Therefore the arc length is e 1 x+ 1 4x dx = 12 1 x + ln x 2 4 e = 1 C06S0M.069: If f (x) = sin−1 3x, then f (x) = C06S0M.070: If f (x) = tan−1 7x, then f (x) = 1 12 1 2e2 − 1 + e− = ≈ 3.4445280495. 42 2 4 1 1− (3x)2 · Dx (3x) = √ 3 . 1 − 9x2 7 . 1 + 49x2 C06S0M.071: If g (t) = sec−1 t2 , then g (t) = 1 (t2 )2 | t2 | −1 C06S0M.072: If g (t) = tan−1 et , then g (t) = · Dt t2 = t2 √ 2t 2 =√ . 4−1 4−1 t tt et . 1 + e2t C06S0M.073: If f (x) = sin−1 (cos x), then f (x) = 1 2 1 − (cos x) · Dx (cos x) = − √ 18 sin x sin x =− . | sin x | 1 − cos2 x C06S0M.074: If f (x) = sinh−1 2x, then f (x) = √ 1 C06S0M.075: If g (t) = cosh−1 10t, then g (t) = 1 1 u · Du 2 1 u 1− (10t)2 −1 · Dt (10t) = √ 10 , t> 100t2 − 1 1 10 . 1 , then u C06S0M.076: If h(u) = tanh−1 h (u) = 2 . 1 + 4x2 =− 1 u2 1 1 =− 2 · = − 1 u2 u −1 1 − u2 u2 provided that | u | > 1. 1 , then x2 1 · Dx 2 1 1− x2 C06S0M.077: If f (x) = sin−1 f (x) = 1 x2 = 1 x = 1 1− 1 x4 · −2 2 =− √ . 4−1 x3 xx 1 , then x C06S0M.078: If f (x) = tan−1 1 f (x) = 1+ 1 x2 · Dx x2 −1 1 · =− 2 x2 + 1 x2 x +1 provided that x = 0. √ C06S0M.079: If f (x) = arcsin x, then f (x) = 1 1 1 1 · Dx x1/2 = √ · x−1/2 = √ √. √2 2 1−x 2 1−x x 1 − ( x) C06S0M.080: If f (x) = x sec−1 x2 , then f (x) = x (x2 )2 | x2 | −1 · Dx x2 + sec−1 x2 = √ 2 + sec−1 x2 −1 x4 C06S0M.081: If f (x) = tan−1 (x2 + 1), then f (x) = C06S0M.082: If f (x) = sin−1 f (x) = 1 1 − (1 − x2 ) √ 1 2x · Dx (x2 + 1) = 4 . 1 + (x2 + 1)2 x + 2x2 + 2 1 − x2 , then · Dx (1 − x2 )1/2 = √ 1 x2 · 1 x √ . (1 − x2 )−1/2 · (−2x) = − 2 | x | 1 − x2 C06S0M.083: If f (x) = ex sinh ex , then f (x) = e2x cosh ex + ex sinh ex . 19 C06S0M.084: If f (x) = ln cosh x, then f (x) = sinh x = tanh x. cosh x C06S0M.085: If f (x) = tanh2 3x + sech2 3x, then f (x) ≡ 1 by Eq. (5) in Section 6.9. Therefore f (x) ≡ 0. Alternatively, f (x) = 6 tanh 3x sech2 3x − 6 sech 3x sech 3x tanh 3x ≡ 0. C06S0M.086: If f (x) = sinh−1 f (x) = 1 √ 1+ x2 − 1 x2 − 1, then · Dx (x2 − 1)1/2 = √ 2 C06S0M.087: If f (x) = cosh−1 f (x) = √ √ x2 + 1, then 1 √ x2 + 1 2 1 x 1 √ · (x2 − 1)−1/2 · 2x = . 2−1 2 1+x | x | x2 − 1 −1 · Dx (x2 + 1)1/2 = √ 1 x2 · 12 x √ (x + 1)−1/2 · 2x = . 2 | x | x2 + 1 C06S0M.088: If f (x) = tanh−1 (1 − x2 ), then f (x) = 1 −2x 2 · (−2x) = 2 =3 . 2 )2 4 1 − (1 − x 2x − x x − 2x C06S0M.089: Let u = 2x. Then du = 2 dx, so √ 1 1 dx = 2 2 1 − 4x √ 1 1 1 du = sin−1 u + C = sin−1 2x + C. 2 2 2 1−u C06S0M.090: Let u = 2x. Then du = 2 dx, so 1 1 dx = 1 + 4x2 2 1 1 1 du = arctan u + C = arctan 2x + C. 1 + u2 2 2 C06S0M.091: Let x = 2u. Then dx = 2 du and √ 1 dx = 4 − x2 √ 4 − x2 = √ √ 4 − 4u2 = 2 1 − u2 . Therefore 2 x √ + C. du = arcsin u + C = arcsin 2 2 1 − u2 C06S0M.092: Let x = 2u. Then dx = 2 du and 4 + x2 = 4(1 + u2 ). Thus 1 dx = 4 + x2 2 1 du = 4(1 + u2 ) 2 1 1 1 x du = arctan u + C = arctan + C. 1 + u2 2 2 2 C06S0M.093: Let u = ex . Then du = ex dx and √ ex dx = 1 − e2x √ √ 1 − e2x = √ 1 − u2 . Hence 1 du = arcsin u + C = arcsin ex + C. 1 − u2 C06S0M.094: Let u = x2 . Then du = 2x dx and 1 + x4 = 1 + u2 . Therefore x 1 dx = 1 + x4 2 1 1 1 du = arctan u + C = arctan x2 + C. 1 + u2 2 2 20 Mathematica 3.0 returns the answer − 1 arctan 2 1 x2 + C, demonstrating that your intelligence and knowledge are still required to interpret what the computer tells you. C06S0M.095: Let u = 2 x, so that x = 3 u. Then dx = 3 2 √ 1 31 dx = · 2 23 9 − 4x √ 3 2 √ 9 − 4x2 = √ √ 9 − 9u2 = 3 1 − u2 . So 1 1 1 2x du = arcsin u + C = arcsin 2 2 2 3 1−u C06S0M.096: Let u = 2 x, so that x = 3 u. Then dx = 3 2 1 31 dx = · 2 9 + 4x 29 du and 3 2 + C. du and 9 + 4x2 = 9 + 9u2 = 9(1 + u2 ). Thus 1 1 1 2x du = arctan u + C = arctan 2 1+u 6 6 3 + C. C06S0M.097: The integrand resembles the derivative of the inverse tangent of something, so we let u = x3 . Then du = 3x2 dx, so that x2 dx = 1 du. Then 3 x2 1 dx = 1 + x6 3 1 1 1 du = arctan u + C = arctan x3 + C. 1 + u2 3 3 C06S0M.098: If necessary, use the substitution u = sin x, but by inspection, cos x dx = arctan(sin x) + C. 1 + sin2 x Note that while expressions such as sin(arctan x) can be simplified to algebraic functions, expressions such as arctan(sin x) normally cannot be further simplified. C06S0M.099: Let u = 2x. Then du = 2 dx and 1 1 √ dx = 2 x 4x2 − 1 1 2u √ 1 du = u2 − 1 √ 4x2 − 1 = √ u2 − 1. Thus 1 √ du = arcsec | u | + C = arcsec | 2x | + C. u u2 − 1 Mathematica 3.0 returns the equivalent alternative answer 1 1 √ dx = − arctan √ x 4x2 − 1 4x2 − 1 which in turn is equal to arctan + C, 4x2 − 1 + C . C06S0M.100: The integrand resembles the derivative of the inverse secant function, so we try the substi√ √ tution u = x2 . Then du = 2x dx and x4 − 1 = u2 − 1, and thus 1 1 √ dx = 2 x x4 − 1 2x 1 √ dx = 2 x2 x4 − 1 1 1 1 √ du = arcsec | u | + C = arcsec x2 + C. 2 2 u u2 − 1 Mathematica 3.0 returns the equivalent answer 21 which itself is equal to 1 arctan 2 1 1 1 √ dx = − arctan √ 4−1 4−1 2 xx x + C, x4 − 1 + C . C06S0M.101: The integrand is slightly evocative of the derivative of the inverse secant function, so we √ √ try the substitution u = ex , so that e2x − 1 = u2 − 1 . Then du = ex dx, so √ 1 e2x − 1 dx = ex √ ex dx = e2x − 1 1 √ du = arcsec | u | + C = arcsec (ex ) + C. u u2 − 1 Mathematica 3.0 returns the equivalent answer arctan e2x − 1 + C. C06S0M.102: Use the substitution u = x3 if necessary, but by inspection x2 cosh x3 dx = C06S0M.103: Let u = constant k . Because √ 1 sinh x3 + C. 3 x if necessary, but by inspection an antiderivative is f (x) = k cosh x1/2 for some f (x) = k sinh x1/2 · Dx x1/2 = k sinh x1/2 , 2x1/2 it follows that k = 2 and therefore that √ √ sinh x √ dx = 2 cosh x + C. x C06S0M.104: Let u = 3x − 2 if necessary, but by inspection an antiderivative is f (x) = k tanh(3x − 2) for some constant k . Because f (x) = 3k sech2 (3x − 2), it follows that k = 1 , and thus that 3 sech2 (3x − 2) dx = 1 tanh(3x − 2) + C. 3 Mathematica 3.0 returns the equivalent answer 1 sech2 (3x − 2) dx = − tanh(2 − 3x) + C. 3 Figure 6.9.3 shows why the two answers are really the same. C06S0M.105: Let u = arctan x if necessary, but evidently arctan x 1 dx = (arctan x)2 + C. 2 1+x 2 C06S0M.106: Let u = 2x, so that √ 1 1 dx = 2 4x2 − 1 √ 4x2 − 1 = √ √ u2 − 1 and du = 2 dx. Then 1 1 1 du = cosh−1 u + C = cosh−1 2x + C. 2 2 u2 − 1 22 C06S0M.107: Let u = Therefore √ 2 3 x, so that x = 1 31 dx = · 23 4x2 + 9 √ 3 2 u. Then dx = x 1 dx = 2 +1 √ x4 du and √ 4x2 + 9 = √ √ 9u2 + 9 = 3 u2 + 1. 1 1 2x 1 du = sinh−1 u + C = sinh−1 2 2 3 u2 + 1 C06S0M.108: We expect to see an integrand containing thus √ 3 2 √ + C. u2 + 1, so let u = x2 . Then du = 2x dx, and 1 1 1 du = sinh−1 u + C = sinh−1 x2 + C. 2 2 +1 u2 C06S0M.109: The volume is √ 1/ 2 V= √ 0 2π x dx. 1 − x4 Let u = x2 . Then du = 2x dx, and hence √ 2π x dx = π 1 − x4 √ 1/ 2 Therefore V = π arcsin x2 = 0 √ 1 du = π arcsin u + C = π arcsin x2 + C. 1 − u2 π2 ≈ 1.6449340668. 6 C06S0M.110: The volume is 1 V= 0 √ 2π x dx. x4 + 1 Let u = x2 . Then du = 2x dx, and therefore √ 2π x dx = x4 + 1 √ π du = π sinh−1 u + C = π sinh−1 x2 + C. u2 + 1 1 Thus V = π sinh−1 x2 = π sinh−1 1 = π ln 1 + √ 2 0 ≈ 2.7689167860. C06S0M.111: By Eq. (36) of Section 6.9, −1 tanh 1 x 1 1 1+ x 1 x+1 = ln = coth−1 x = ln 1 2 2 x−1 1− x by Eq. (37) (provided that |x| > 1). By Eq. (35), if 0 < x cosh −1 1 x = ln by Eq. (38). Note that √ 1 + x 1 −1 x2 = ln 1 + x √ 1 then 1 − x2 √ x2 x2 = x because x > 0. C06S0M.112: If k = 0 and x(t) = A cosh kt + B sinh kt, then 23 = ln 1+ √ 1 − x2 x = sech−1 x x (t) = kA sinh kt + kB cosh kt and x (t) = k 2 A cosh kt + k 2 B sinh kt = k 2 x(t). If x(0) = 1 and x (0) = 0, then A = 1 and kB = 0, so A = 1 and B = 0. If x(0) = 0 and x (0) = 1, then A = 0 and kB = 1, so A = 0 and B = 1/k . C06S0M.113: The graphs of y = cos x and y = sech x are shown following this solution, graphed for 0 x 6. One wonders if there is a solution of cos x = sech x in the interval (0, π /2). We graphed y = f (x) = sech x − cos x for 0 x 1 and it was clear that f (x) > 0 if x > 0.25. We graphed y = f (x) for 0 x 0.25 and it was clear that f (x) > 0 if x > 0.06. We repeated this process several times and could see that f (x) > 0 if x > 0.0002. Instability in the hardware or software made further progress along these lines impossible. Methods of Section 7.8 can be used to show the desired inequality, but Ted Shifrin provided the following elegant argument. First, sech x 1 sec x if x is in I = [0, π /2). Moreover, sech x = sec x only for x = 0 in that interval; otherwise, sech x < sec x for x in J = (0, π /2). So sech2 x < sec2 x if x is in J . But tanh x = tan x if x = 0. Therefore tanh x < tan x for x in J because Dx tanh x < Dx tan x for such x. That is, sinh x sin x < cosh x cos x if x is in J . All the expressions involved here are positive, so sinh x cos x < cosh x sin x if x is in J . That is, − cosh x sin x + sinh x cos x < 0 for x in J . But cosh x cos x = 1 if x = 0, and we have now shown that Dx (cosh x cos x) < 0 if x is in J . Therefore cosh x cos x < 1 for x in J . In other words, sech x cos x for x in I and sech x > cos x for x in J . Because cos x 0 < sech x for π /2 < x < 3π /2 and because cos x is increasing, while sech x is decreasing, for 3π /2 x < 2π , it follows that the least positive solution of f (x) = 0 is slightly larger than 3π /2, about 4.7, exactly as the figure suggests. We then applied Newton’s method to the solution of f (x) = 0 with x0 = 4.7, with the following results: x1 ≈ 4.7300338216, x2 ≈ 4.7300407449, x3 ≈ 4.7300407449. Thus x3 is a good approximation to the least positive solution of cos x cosh x = 1. 1 0.5 1 2 3 -0.5 -1 24 4 5 6 C06S0M.114: If f (x) = sinh−1 (tan x), then f (x) = √ sec2 x sec2 x sec2 x =√ = = | sec x |. | sec x | 1 + tan x sec2 x 2 Therefore sec x dx = sinh−1 (tan x) + C if sec x > 0; that is, if k is an odd integer and k π/ 2 < x < (k + 2)π /2. On the intervals where sec x < 0, we find that sec x dx = − sinh−1 (tan x) + C. Next, if g (x) = tan−1 (sinh x), then g (x) = cosh x cosh x = = sech x, 2 1 + sinh x cosh2 x and therefore sech x dx = tan−1 (sinh x) + C for all x. C06S0M.115: Given f (x) = x1/2 , let F (x) = f (x) − ln x. Then F (x) = 1 1 x1/2 − 2 −= , 1 /2 x 2x 2x so F (x) = 0 when x = 4. Clearly F is decreasing on (0, 4) and increasing on (0, + ∞), so the global minimum value of F (x) is F (4) = 2 − ln 4 = 2 − 2 ln 2 > 2 − 2 · 1 because ln 2 < 1. Therefore f (x) > ln x for all x > 0. For part (b), we need to solve x1/3 − ln x = 0. The iteration of Newton’s method takes the form x ←− x − x1/3 − ln x . 1 −2/3 1 − x 3 x Beginning with x0 = 100, we get x5 ≈ 93.354461. q 1/p For part (c), suppose that j (x) = x1/p is tangent to the graph of g (x) = ln x at the point (q, ln q ). Then = ln q and j (q ) = g (q ). Hence 1 (1/p)−1 1 =; q p q q 1/p = p; p = ln q = ln pp = p ln p; ln p = 1, 25 so p = e. Section 7.2 C07S02.001: Let u = 2 − 3x. Then du = −3 dx, and so (2 − 3x)4 dx = − 1 3 (2 − 3x)4 (−3) dx = − 1 3 u4 du = − C07S02.002: Let u = 1 + 2x. Then x = 1 (u − 1) and dx = 2 1 1 dx = 2 (1 + 2x) 2 u−2 du = − 1 2 115 1 · u + C = − (2 − 3x)5 + C. 35 15 du, so that 1 1 +C =− + C. 2u 2(1 + 2x) C07S02.003: Let u = 2x3 − 4. Then du = 6x2 dx, so that x2 (2x3 − 4)1/2 dx = 1 6 1 6 (2x3 − 4)1/2 · 6x2