2ndEdAPCalc SOLN - Section 1.1 Chapter 1 Prerequisites for...

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Unformatted text preview: Section 1.1 Chapter 1 Prerequisites for Calculus Section 1.1 Exercises 1. 1. y 2 4(3 3) 2. 3 3 2x x 3 3 2(x 2x 2 4(0) 1) 2 2 0 1 2 x y 1 2 ( 3) 2 2 4 3. Quick Review 1.1 1 1 2. s Section 1.1 Lines (pp. 1–9) x y x y 2 4. 2 1 x y 8 1 1 0 ( 3) 0 0 2 2 3 5 0 4 6 5. (a, c) y 5 3. m 2 5 3 4 1 1 4. m 2 3 ( 3) ( 1) 1 B 5. (a) 3(2) A 5 Yes 4( 1) 13 5 5 (b) m No 9 (x2 7. d 2( 2) No x1)2 (0 1 2 ( 2) 1 3 1 6. (a, c) 2( 1) 5 2 5 Yes (b) 1 1 x 5 1 6 6. (a) 7 7 5 1 4 4 (b) 3(3) 5 4 3 y 5 5 (y2 1)2 y1)2 5 0)2 (1 x A B 2 (x2 x1)2 (1 8. d = 2)2 (y2 1 3 y1)2 2 1 (b) m 2 ( 1) 1 ( 2) 7. (a, c) y B 16 9 1 1 3 5 42 3 ( 1)2 1 3 A 5 25 9 x 5 3 9. 4x 3y 7 3y 4x 4 x 3 y 10. 2x (b) m 5y 7 7 3 33 12 0 3 8. (a, c) 0 y 5 3 A 5y 2x 3 y 2 x 5 3 5 5 B x 1 2 Section 1.1 8. continued 26. The line contains (0, 0) and (5, 2). 32 11 (b) m 5 (undefined) 0 m This line has no slope. 9. (a) x (b) y y 2 3 27. 3x 10. (a) x 1 (b) y 0 2 0 y 1) 1[x 1(x 3 3 4 1 ( 1)] 1 1) 1 2(x 16. y y y 12 (c) 1(x 14. y y 17. m 3x (b) y-intercept: 3 (b) y 15. y 12 (a) Slope: 12. (a) x 13. y 2 5 3 x 4 y (b) y 0 0 4y 4y 4 3 11. (a) x 2 5 2 x 5 0) 2[x 2(x 3 [ 10, 10] by [ 10, 10] ( 4)] 0 4) 0 28. x y 30 3 20 2 3 (x 0) 0 2 3 x 2 2y 2y 2 x 2 (a) Slope: 1 (b) y-intercept: 2 (c) 3x 3x y 18. m 0 1 2 1 1 y 0(x y [ 10, 10] by [ 10, 10] 0 1 0 29. 1) 1 2 19. m 1 2 (undefined) 0 0 ( 2) 2 Vertical line: x 20. m y 2 2 3 [x 4 4y 3(x 4y 3x 3x 21. y 2) 1 x 3 4 x 3 1 4 4 3 (b) y-intercept: 4 3 4 (c) 1 4 2 4y 3x 3 4 ( 2)] y y 4 (a) Slope: 2 1 ( 2) x 3 y 4 [ 10, 10] by [ 10, 10] 30. y 2 2x 4 (a) Slope: 2 2 22. y 1x 2 or y 23. y 1 x 2 1 x 3 2 (b) y-intercept: 4 3 24. y x (c) 1 [ 10, 10] by [ 10, 10] 25. The line contains (0, 0) and (10, 25). m y 25 10 5 x 2 0 0 25 10 5 2 Section 1.1 31. (a) The desired line has slope (0, 0): y 1(x 0) 0 or y 1 and passes through x. 1 1 (b) The desired line has slope (0, 0): y 1(x 0) 0 or y x. 1 and passes through 39. (a) y (c) 1 2 1 and passes through 2 1 (x 2 2) 1 x 2 2 or y (b) We seek a horizontal line through ( 2, 4): y line through 1, 1 :y 2 1 . 2 1 1, : x 2 (b) We seek a vertical line through 2 1 f (x) 1) 7 x 2 Since f (x) 4 4 3 , 2 ( 1) 2 3 (x 2 f (x) 2 7 (5) 2 Check: f (5) 36. m 7 3 x 2 2 3 16, as expected. 2 7 we have m and b 2 3 2 2) 3 x 2 Since f (x) 4 2 (6) 3 y y 3 1 y [1975, 1995] by [20,000, 35,000] (d) When x 2000, y 1,060.4233(2000) 2,077,548.669 43,298. In 2000, the construction workers’ average annual compensation will be about $43,298. 42. (a) When y (b) When y 2 When x 3 and b 2 2. 2(x x 2 x 8) 4 8 2 x 1, so x c. 1, so y d. 2, so x 2c. 2, so y 2d. and y x 2 and k so k ( 2) ( 8) 44. (a) m 3 k 1, respectively, so the slopes are 1. The lines are parallel when 2 k 1, 2. 68 69.5 0.4 0 (b) m 10 68 4 0.4 (c) m 6 2 x k 43. (a) The given equations are equivalent to y (b) The lines are perpendicular when k 2. 2 38. x c y 0, we have d x 0, we have c y 0, we have d 0, we have The x-intercept is 2c and the y-intercept is 2d. 3 4 3 . 2 7, as expected. 2, we have m 3 ( 2) 37. y 2 2,077,548.669 (c) When x 3 x 2 29.413. 41. y 1 (x 3) 4 yx34 yx1 This is the same as the equation obtained in Example 5. 3 2 ( 1) 3 (6) 2 Check: f (6) 2 3 1. 7 2 7 (x 2 1,060.4233x 9.013 4. 34. (a) The given line is horizontal, so we seek a horizontal 9 3 (d) When x 30, y 0.680(30) She weighs about 29 pounds. (b) The slope is 1,060.4233. It represents the approximate rate of increase in earnings in dollars per year. 3. 33. (a) The given line is vertical, so we seek a vertical line through ( 2, 4): x 2. 35. m [15, 45] by [15, 45] 40. (a) y ( 2, 2): y 9.013 (b) The slope is 0.68. It represents the approximate average weight gain in pounds per month. 32. (a) The given equation is equivalent to y 2x 4. The desired line has slope 2 and passes through ( 2, 2): y 2(x 2) 2 or y 2x 2. (b) The desired line has slope 0.680x 3 5 10 4.7 4 1.5 0.4 58 3.6 5 0.7 2 k 1 , so 1 3.75 degrees/inch 16.1 degrees/inch 7.1 degrees/inch (d) Best insulator: Fiberglass insulation Poorest insulator: Gypsum wallboard The best insulator will have the largest temperature change per inch, because that will allow larger temperature differences on opposite sides of thinner layers. 4 Section 1.1 p d 45. Slope: k 10.94 1 100 0 9.94 100 y 6 0.0994 atmospheres per meter (–1, 4) At 50 meters, the pressure is p 0.0994(50) 46. (a) d(t) 1 (2, 3) 5.97 atmospheres. 45t (–1, 1) (b) x 6 (2, 0) y [0, 6] by [ 50, 300] 6 (c) The slope is 45, which is the speed in miles per hour. (d) Suppose the car has been traveling 45 mph for several hours when it is first observed at point P at time t 0. (e) The car starts at time t 47. (a) y 5632x (2, 3) 0 at a point 30 miles past P. (–1, 1) 11,080,280 2732x 5,362,360 50. (d) The median price is increasing at a rate of about $5632 per year in the Northeast, and about $2732 per year in the Midwest. It is increasing more rapidly in the Northeast. 48. (a) Suppose x F is the same as x C. 9 x 32 5 9 x 32 5 x 1 4 x 5 y (c , d ) W (a , b ) x X Z (g , h ) 32 x (–1, –2) (b) The rate at which the median price is increasing in dollars per year (c) y 6 (2, 0) (e , f ) Y Suppose that the vertices of the given quadrilateral are x 40 Yes, 40 F is the same as (a, b), (c, d), (e, f ), and (g, h). Then the midpoints of the 40 C. (b) cb d c ed f , ,X , , 2 2 2 2 e gf h g ah b Y , , and Z , . When these four 2 2 2 2 consecutive sides are W a points are connected, the slopes of the sides of the resulting figure are: [ 90, 90] by [ 60, 60] d It is related because all three lines pass through the point ( 40, 40) where the Fahrenheit and Celsius temperatures are the same. 49. The coordinates of the three missing vertices are (5, 2), ( 1, 4) and ( 1, 2), as shown below. y 6 (2, 3) (5, 2) (–1, 1) (2, 0) 6 x WX: c f d 2 e 2 h 2 XY: e g 2 fh 2 ZY: e g 2 hb 2 WZ: g a 2 f b 2 a c f e b a 2 d f 2 c e h g d c f e b a 2 h b 2 g a 2 bd 2 ac 2 h g d c Opposite sides have the same slope and are parallel. Section 1.1 4 3 51. The radius through (3, 4) has slope 0 0 4 . 3 The tangent line is tangent to this radius, so its slope is 3 (x 4 3 x 4 3 x 4 y y y 3) 1 4/3 3 . We seek the line of slope 4 3 that passes through (3, 4). 4 4 9 4 25 4 4 52. (a) The equation for line L can be written as A x B y C , so its slope is B B (x A y a) A . The perpendicular line has slope B B A B (x A a) 2 2 Ax B (x a) b for y in the equation for line L gives: b C a) ABb AC (A2 B B and passes through (a, b), so its equation is A b. (b) Substituting (x Ax 1 A/B B2)x B2 a B2 a x AC ABb AC ABb A2 B 2 Substituting the expression for x in the equation for line L gives: B2 a A By By By y AC ABb By C A2 B 2 A(B2a AC ABb) C(A2 B2) 2 2 A B A2 B 2 2 2 2 2 AB a A C A Bb A C B2C A2 B2 A2Bb A2 b B2C AB2a A2 B2 BC ABa A2 B2 B2a The coordinates of Q are (c) Distance A a)2 (x B2a (y B 2 A Bb A 2 a 2 B2 A AC BC 2 A(C Aa) 2 Bb A2 (A2 B2)(C Aa (A2 B2)2 Aa (C A C Bb)2 2 2 Aa A2 Aa B Bb B2 Bb A 2 C B 2 BC ABa A2 B 2 B2) 2 A2b ABa B2b 2 B2 A 2 B(C B2 A2(C Aa Bb)2 (A2 B2)2 A2 b a ABb a(A2 A2 B2 AC BC ABa . A2 B2 , 2 b)2 AC ABb A2 B 2 B2a ABb A2b AC 2 Bb) 2 Aa A2 B2 B2(C Aa Bb)2 (A2 B2)2 Bb)2 BC 2 b ABa b(A2 A2 B2 B2 ) 2 5 6 Section 1.2 s Section 1.2 Functions and Graphs 5. x2 16 Solutions to x2 16: x 4, x 4 Test x 6 ( 6)2 36 16 x2 16 is false when x 4 Test x 0: 02 0 16 x2 16 is true when 4 x Test x 6: 62 36 16 x2 16 is false when x 4. Solution set: ( 4, 4) (pp. 9–19) Exploration 1 1. y3 g f, y4 Composing Functions fg 2. Domain of y3: [ 2, 2] Range of y3: [0, 2] y1: 6. 9 x2 0 Solutions to 9 x2 0: x 3, x 3 Test x 4: 9 ( 4)2 9 16 70 9 x2 0 is false when x 3. Test x 0: 9 02 9 0 9 x2 0 is true when 3 x 3. Test x 4: 9 42 9 16 70 9 x2 0 is false when x 3. Solution set: [ 3, 3] [ 4.7, 4.7] by [ 2, 4.2] y2: 7. Translate the graph of f 2 units left and 3 units downward. [ 4.7, 4.7] by [ 2, 4.2] 8. Translate the graph of f 5 units right and 2 units upward. y3: 9. (a) x2 x2 3)(x (x [ 4.7, 4.7] by [ 2, 4.2] 3. Domain of y4: [0, ); Range of y4: ( y4: (b) , 4] f (x) 5 9 3) x y1(x) 4 x2 2 4 (y2(x)) 4 ( x)2 1 x 4 x, x 1 5 0 1 x 1. 3x 1 5x 3 2x 4 x 2 Solution: [ 2, ) 3. x 4 4x34 1x7 Solution set: [ 1, 7] 0 No solution 11. (a) 0. x 2. (b) 2. 3 4. x 2 5 x2 5 or x 2 5 x 3 or x 7 Solution set: ( , 3] [7, ) 6 5 x 3 (b) f (x) = 0 Quick Review 1.2 2. x(x 2) 0 Solutions to x(x 2) 0: x 0, x 2 Test x 1: 1( 1 2) 3 0 x(x 2) 0 is true when x Test x 1: 1(1 2) 10 x(x 2) 0 is false when 0 Test x 3: 3(3 2) 3 0 x(x 2) 0 is true when x Solution set: ( , 0) (2, ) 3 or x 5 [ 2, 6] by [ 2, 6] y2 ( y1(x)) y1( y2(x)) 4 4 0 0 f (x) 6 x2 5 x2 1 No real solution 10. (a) f (x) 4. y3 y4 4 12. (a) (b) f (x) = 4 x7 4 x7 16 x 9 Check: 9 7 f (x) x7 x7 x Check: 3 3 f (x) x1 x1 x x x f (x) 1 1 x 16 4; it checks. 1 1 1 6 6 7 2 2 8 7 3 3 27 28 1; it checks. Section 1.2 Section 1.2 Exercises d2 r2 1. Since A 2 8. (a) Since we require 2 , the formula is A d , where A 4 (b) ( x 0, the domain is ( 7 , 0]. , 0] (c) represents area and d represents diameter. 2. Let h represent height and let s represent side length. s2 2 h2 s2 [ 10, 3] by [ 4, 2] (d) None 12 s 4 h2 s2 h2 32 s 4 x 0, the domain is ( 2 , 3]. 0, the domain is (b) [0, ) 3 h 9. (a) Since we require 3 2 (c) s Since side length and height must be positive, the formula 3 is h 2 s. [ 4.7, 4.7] by [ 6, 6] (d) None s s 2 10. (a) Since we require x ( , 2) (2, ). s h (b) Since s 2 ( s 3. S 6e2, where S represents surface area and e represents edge length. 1 x , 0) 2 can assume any value except 0, the range is (0, ). (c) 4 r3, where V represents volume and r represents 4. V 3 radius. 5. (a) ( (b) ( , ) or all real numbers [ 4.7, 4.7] by [ 6, 6] , 4] (d) None (c) 11. (a) ( , ) or all real numbers (b) ( , ) or all real numbers (c) [ 5, 5] by [ 10, 10] (d) Symmetric about y-axis (even) 6. (a) ( , ) or all real numbers [ 6, 6] by [ 3, 3] (b) [ 9, ) (d) None (c) 12. (a) ( , ) or all real numbers (b) The maximum function value is attained at the point (0, 1), so the range is ( , 1]. (c) [ 5, 5] by [ 10, 10] (d) Symmetric about the y-axis (even) 7. (a) Since we require x 1 0, the domain is [1, ). (b) [2, ) [ 6, 6] by [ 3, 3] (c) (d) Symmetric about the y-axis (even) [ 3, 10] by [ 3, 10] (d) None 8 Section 1.2 13. (a) Since we require x 0, the domain is ( , 0]. 18. (a) This function is equivalent to y [0, ). (b) [0, ) x3, so its domain is (b) [0, ) (c) (c) [ 10, 3] by [ 1, 2] [ 2, 5] by [ 2, 8] (d) None 14. (a) Since we require x ( , 0) (0, ). (b) Note that (d) None 0, the domain is 19. Even, since the function is an even power of x. 1 can assume any value except 0, so 1 x 1 x 20. Neither, since the function is a sum of even and odd powers of x. 21. Neither, since the function is a sum of even and odd powers of x (x1 2x0). can assume any value except 1. The range is ( , 1) (1, ). 22. Even, since the function is a sum of even powers of x (x2 3x0). (c) 23. Even, since the function involves only even powers of x. 24. Odd, since the function is a sum of odd powers of x. 25. Odd, since the function is a quotient of an odd function (x3) and an even function (x2 1). [ 4, 4] by [ 4, 4] (d) None 15. (a) Since we require 4 x2 26. Neither, since, (for example), y( 2) 0, the domain is [ 2, 2]. (b) Since 4 x2 will be between 0 and 4, inclusive (for x in the domain), its square root is between 0 and 2, inclusive. The range is [0, 2]. (c) 41/3 and y(2) 0. 27. Neither, since, (for example), y( 1) is defined and y(1) is undefined. 28. Even, since the function involves only even powers of x. 29. (a) [ 9.4, 9.4] by [ 6.2, 6.2] [ 4.7, 4.7] by [ 3.1, 3.1] Note that f (x) x 3 2, so its graph is the graph of the absolute value function reflected across the x-axis and then shifted 3 units right and 2 units upward. (d) Symmetric about the y-axis (even) 16. (a) This function is equivalent to y all real numbers. 3 x2, so its domain is (c) (b) ( ,) (c) ( (b) [0, ) , 2] 30. (a) The graph of f (x) is the graph of the absolute value function stretched vertically by a factor of 2 and then shifted 4 units to the left and 3 units downward. [ 2, 2] by [ 1, 2] (d) Symmetric about the y-axis (even) 17. (a) Since we require x2 ( , 0) (0, ) (b) Since 1 x2 0, the domain is [ 10, 5] by [ 5, 10] (b) ( 0 for all x, the range is (1, ). , ) or all real numbers (c) [ 3, ) 31. (a) (c) [ 4, 4] by [ 1, 5] (d) Symmetric about the y-axis (even) [ 4.7, 4.7] by [ 1, 6] (b) ( , ) or all real numbers (c) [2, ) Section 1.2 32. (a) 44. Line through ( 1, 0) and (0, m 3 0 3 1 0 ( 1) 3): 3, so y Line through (0, 3) and (2, m [ 4, 4] by [ 2, 3] (b) ( 13 20 , ) or all real numbers 3 1): 4 2 2, so y 3, 3, 3x 2x f (x) (c) [0, ) 3x 1 0 2x x x 3 0 2 33. (a) 45. Line through ( 1, 1) and (0, 0): y x Line through (0, 1) and (1, 1): y 1 Line through (1, 1) and (3, 0): [ 3.7, 5.7] by [ 4, 9] (b) ( , ) or all real numbers m (c) ( , ) or all real numbers so y 34. (a) 0 3 1 1 1 (x 2 1 2 1) [ 2.35, 2.35] by [ 1, 3] 35. Because if the vertical line test holds, then for each x-coordinate, there is at most one y-coordinate giving a point on the curve. This y-coordinate would correspond to the value assigned to the x-coordinate. Since there is only one y-coordinate, the assignment would be unique. 36. If the curve is not y 0, there must be a point (x, y) on the curve where y 0. That would mean that (x, y) and (x, y) are two different points on the curve and it is not the graph of a function, since it fails the vertical line test. Line through (1, f (x) x 3 1) and (3, m 40. No x, x 2, 0 1 x x 0 1 2 3 x x x x 0 2 x T T 2 1, x 1 x 3 2, 5 , 3 x x T x 3T 2 T 3T 2 2 2 x 5 5 3 2 1 2 1 3 1 , 3 1 T 2 T 2 (b) g( f (x)) 0 5 2 x T 0 T x 49. (a) f (g(x)) x x T 2 T 2 0 A, 0 x A, 2 1 2 3 4 1 3 T , 0 and (T, 1): 2 10 2 2 , so y x T (T/2) T T 48. f (x) Line through (2, 1) and (5, 0): m 2) x A, x 1 2 1 x 3 1 x 43. Line through (0, 2) and (2, 0): y 1 (x 3 x 1 0 1 A, 2, 0, 2, 0, x 0 2, 0, 41. Line through (0, 0) and (1, 1): y Line through (1, 1) and (2, 0): y 1): y 2 47. Line through f (x) 39. Yes 2x 2 2x 1 x 2 1) and (0, 0): y 1, 38. Yes so y 1 1 x, 2 37. No 42. f (x) x Line through (0, 2) and (1, 0): y (c) [0, ) f (x) 0 1 3 , 2 x 0 1, 46. Line through ( 2, , ) or all real numbers 3 2 1 1 x 2 (b) ( 1 x 2 1 x, f (x) f (x) 1 , 2 (x 2 x 2T 3) 2 2 3 (x 5) (x 2 10x 25) x 2 10x 22 (c) f (g(0)) 02 2 (d) g( f (0)) 02 10 0 (e) g(g( 2)) (f) f ( f (x)) x2 5 [( 2) (x 5) 3 2 2 22 3] 5 2 x 22 3 10 12 3 2 9 10 Section 1.2 50. (a) f (g(x)) (x 1) 1 (b) g( f (x)) (x 1) 1 x (c) f (g(x)) 0 (d) g( f (0)) 0 1) 1 1 x g f: x (e) g(g( 2)) (f) f ( f (x)) (2 (x 1) 3 1 4 [ 4.7, 4.7] by [ 2, 4] 2 51. (a) Enter y1 f (x) x 7, y2 g(x) y3 ( f g)(x) y1(y2(x)), and y4 (g f )(x) y2(y1(x)) f g: Domain: ( , Range: [0, ) x, (b) ( f g)(x) g f: (g f )(x) [ 3, 20] by [ 4, 4] Domain: [0, ) Range: [ 7, ) (b) ( f g)(x) x Domain: [7, ) Range: [0, ) 7 (g f )(x) 52. (a) Enter y1 f (x) 1 x 2, y2 g(x) y3 ( f g)(x) y1(y2(x)), and y4 (g f )(x) y2(y1(x)) f g: x [1, ) ( x 2)2 3 (x 2) 3, x x 1, x 2 (x2 3) 2 54. (a) Enter y1(x) [ 10, 70] by [ 10, 3] 1] 2x x f (x) 2 x2 1 ,y 32 1 3x 2 1 , x y3 ( f g)(x) y1(y2(x)), and y4 (g f )(x) y2(y1(x)). Use a “decimal window” such as the one shown. f g: 7 x, [ 9.4, 9.4] by [ 6.2, 6.2] Domain: ( , 2) (2, ) Range: ( , 2) (2, ) g f: [ 6, 6] by [ 4, 4] Domain: [0, ) Range: ( , 1] g f: [ 9.4, 9.4] by [ 6.2, 6.2] Domain: ( , 3) ( 3, ) Range: ( , 3) ( 3, ) 2 [ 2.35, 2.35] by [ 1, 2.1] Domain: [ 1, 1] Range: [0, 1] (b) ( f g)(x) 1 ( x)2 (g f )(x) 1 x2 (b) ( f g)(x) 1 x, x 53. (a) Enter y1 f (x) x 2 3, y2 g(x) y3 ( f g)(x) y1(y2(x)), and y4 (g f )(x) y2(y1(x)). f g: 3x 1 2x 3x 1 2x 1 3 2(3x 1) (2 (3x 1) 3(2 0 x 7x ,x 7 2, x, x 3 (g f )(x) 2x x 2 2 1 3 2x x 1 1 3 7x ,x 7 Domain: [ 2, ) Range: [ 3, ) x, x 2 2 3(2x 1) (x 2(x 3) (2x [ 10, 10] by [ 10, 10] x) ,x x) 3 3 3) ,x 1) 3 Section 1.2 55. 11 58. [ 5, 5] by [ 2, 5] We require x 2 and x 2 is ( 4 , [ 2.35, 2.35] by [ 1.55, 1.55] 4 We require x 2 0 (so that the square root is defined) ( 0 (to avoid division by zero), so the domain (2, ). For values of x in the domain, x2 2) x2 and hence 1 4 and 4 can attain any positive x2 , 1) x2 (1, ). For values of x in the 1 can attain any value in [ 1, 0) (0, ), so 1 can also attain any value in [ 1, 0) 4 value, so the range is (0, ). (Note that grapher failure may 0, so the domain is ( 1, 1) domain, x 2 3 1 Therefore, 1 3 x (0, ). can attain any value in 1 cause the range to appear as a finite interval on a ( grapher. (Note that grapher failure can cause the intervals in the 56. , 1] (0, ). The range is ( , 1] (0, ). range to appear as finite intervals on a grapher.) 59. (a) y 1.5 [ 5, 5] by [ 2, 5] x2 We require 9 x2 and 9 0 (so that the fourth root is defined) any value in (0, 9]. Therefore, value in (0, , 0 2 x 0 (to avoid division by zero), so the domain is ( 3, 3). For values of x in the domain, 9 2 –2 2 3], and 4 x2 9 2 . The range is 3 , 4 x 2 can attain any 9 –1.5 x 2 can attain (b) y can attain any value in 1.5 or approximately [1.15, ). 3 –2 (Note that grapher failure may cause the range to appear as 0 a finite interval on a grapher.) 2 x –1.5 57. 60. (a) y 2 [ 4.7, 4.7] by [ 3.1, 3.1] x2 We require 9 ( , 3) domain, 9 ( 3, 3) , 0) (0, 9], so ( , 0) (0, ( , 0) –2 (3, ). For values of x in the x 2 can attain any value in ( value in ( 0, so the domain is 3 , 0) 2 3 , 3 9 2 3 , 2 3 9 x2 can attain any . The range is 9 or approximately ( 1 –1 –2 x 2 can attain any value in 9]. Therefore, –1 , 0) [0.96, ). 9 (Note that grapher failure can cause the intervals in the range to appear as finite intervals on a grapher.) 2 x 12 Section 1.2 60. continued (b) (d) Since ( f g)(x) f ( x) x , f (x) x 2. The completed table is shown. Note that the absolute value sign in part (d) is optional. y 2 g(x) 1 –2 –1 1 2 x f (x) x2 x 1 x x2 5 1 x 1 1 ( f g)(x) 5 x, x 1 –2 61. (a) 1 x y 1.3 –2 1 x –1 1 2 0 x2 x x, x x, x 0 64. (a) Note that the data in the table begins at x 20. (We do not include the initial investment in the data.) The power regression equation is y 27.1094x 2.651044. x (b) –1.3 (b) y [0, 30] by [ 20,000, 180,000] 1.3 –2 –1 1 2 (c) When x 30, y 223,374. According to the power regression equation, the investment will grow to approximately $223,374. x (d) The linear regression equation is y 12,577.97x 177,275.52. When x 30, y 200,064. According to the linear regression equation, the investment will grow to approximately $200,064. –1.3 62. (a) y 65. (a) Because the circumference of the original circle was 8 and a piece of length x was removed. 1.3 (b) r –2 0 2 8 x 2 x 2 4 x y 1.3 4 16 4x (b) r2 16 –1.3 16 16 (c) h x2 42 x2 2 0 2 16 x 42 x x2 42 x2 42 16 x 2 –2 4x x2 –1.3 63. (a) Since ( f g)(x) g(x) 5 x2 5, g(x) 1 x, we know that g(x) 1 1 x 1, so g(x) . g(x) x1 1 1 (c) Since ( f g)(x) f x, f (x) . x x (b) Since ( f g)(x) 1 x 2. (d) V 12 rh 3 18 x2 16 x 3 2 2 (8 x)2 16 x x 2 24 2 x2 Section 1.3 66. (a) Note that 2 mi 10,560 ft, so there are 8002 feet of river cable at $180 per foot and (10,560 feet of land cable at $100 per foot. The cost is C(x) 180 8002 x2 100(10,560 x) (b) x2 x) C(0) $1,200,000 C(500) $1,175,812 C(1000) $1,186,512 C(1500) $1,212,000 C(2000) $1,243,732 C(2500) $1,278,479 C(3000) $1,314,870 Values beyond this are all larger. It would appear that the least expensive location is less than 2000 ft from point P. 5. [ 5, 5] by [ 2, 5] [ 3, 3] by [ 1, 3] (c) The functions y1 y2, y2 y1, and y1 y2 all have domain [0, 1], the same as the domain of y1 y2 found in part (b). Domain of Domain of y2 y2 y1 3 3 x 2. Using a calculator, 3 1.5 3 5 x for x 0; 0.192. 17 17 2.5713 24 24 1.8882 5 1.4567 10 1.4567 1.0383 7. 500(1.0475)5 $630.58 $1201.16 (x 3y2)2 (x4y3)3 x 6y4 x12y9 6 12 4 9 x 18 y y 5 1 x18y5 a3b c4 22 a4c b3 1 a6b c8 a6 b4c8 ( f g)(x), 4 a6 2 b 43 1 6 a 2b 4 b3 a4c 2 32 bc a4 c c 82 a2 bc6 Section 1.3 Exercises 1. The graph of y 2x is increasing from left to right and has the negative x-axis as an asymptote. (a) 2. The graph of y 1. x 3 6. x10 x x 10. s Section 1.3 Exponential Functions 3 4.729. 3. Using a calculator, 3 x (b) The product will be even, since ( f g)( x) f ( x) g( x) ( f (x)) ( g(x)) f (x) g(x) ( f g)(x). x 0; 2 0. 2.924. The domain of a quotient of two functions is the intersection of their domains with any zeros of the denominator removed. Exponential Functions for x for x 2 (d) The domain of a sum, difference, or product of two functions is the intersection of their domains. Exploration 1 x 1. Using a calculator, 52/3 9. : (0, 1] (pp. 20–26) x 5 5 8. 1000(1.063)3 : [0, 1) 68. (a) Yes. Since ( f g)( x) f ( x) g( x) f (x) g(x) the function ( f g)(x) will also be even. x x Quick Review 1.3 5. x5 x x (b) Domain of y1: [0, ) Domain of y2: ( , 1] Domain of y3: [0, 1] y1 x 6. 2 2 4. x x x 67. (a) 13 3 x or, equivalently, y 1x , is 3 decreasing from left to right and has the positive x-axis as an asymptote. (d) 3. The graph of y 3 x is the reflection about the x-axis of the graph in Exercise 2. (e) [ 5, 5] by [ 2, 5] 2. x 0 3. x 0 4. x 0 2x, is the 4. The graph of y 0.5 x or, equivalently, y reflection about the x-axis of the graph in Exercise 1. (c) 5. The graph of y 2 and has the line y x 2 is decreasing from left to right 2 as an asymptote. (b) 6. The graph of y 1.5x 2 is increasing from left to right and has the line y 2 as an asymptote. (f) 14 Section 1.3 7. 16. [ 4, 4] by [ 8, 6] [ 6, 6] by [ 2, 6] Domain: ( , ) Range: ( , 3) x-intercept: 1.585 y-intercept: 2 x 1.3863 17. 8. [ 6, 6] by [ 3, 5] x 0.6309 18. [ 4, 4] by [ 2, 10] Domain: ( , ) Range: (3, ) x-intercept: None y-intercept: 4 [ 6, 6] by [ 3, 5] 9. x 19. 1.5850 x y y 1 1 2 1 3 3 4 5 2 [ 4, 4] by [ 4, 8] Domain: ( , ) Range: ( 2, ) x-intercept: 0.405 y-intercept: 1 2 2 10. 20. x y y 1 1 [ 4, 4] by [ 8, 4] 3 Domain: ( , ) Range: ( , 1) x-intercept: None y-intercept: 2 2x 2 2x 11. 9 (3 ) 12. 163x 3 3 1 2x 8 (2 3 2x 14. 1x 27 (3 3x ) 2 6x 21. x 1 ) 3 15. 8 4 212x 13. 5 3 4x 3 (24)3x 2 2 y y 1 3x 3 2 4 5 3 9 7 4 [ 6, 6] by [ 2, 6] x 2.3219 16 Section 1.3 22. x ratio y 1 8.155 31. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve 2.718 2 A1 22.167 3 60.257 2.718 2.718 4 163.794 23. Let t be the number of years. Solving 500,000(1.0375)t 1,000,000 graphically, we find that t 18.828. The population will reach 1 million in about 19 years. 24. (a) The population is given by P(t) 6250(1.0275)t, where t is the number of years after 1890. Population in 1915: P(25) 12,315 Population in 1940: P(50) 24,265 (b) Solving P(t) 50,000 graphically, we find that t 76.651. The population reached 50,000 about 77 years after 1890, in 1967. 25. (a) A(t) 6.6 1 t/14 2 (b) Solving A(t) 1 graphically, we find that t 38.1145. There will be 1 gram remaining after about 38.1145 days. 26. Let t be the number of years. Solving 2300(1.06)t 4150 graphically, we find that t 10.129. It will take about 10.129 years. (If the interest is not credited to the account until the end of each year, it will take 11 years.) 27. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A(1.0625)t 2A, which is equivalent to 1.0625t 2. Solving graphically, we find that t 11.433. It will take about 11.433 years. (If the interest is credited at the end of each year, it will take 12 years.) 28. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A1 1 15 0.0625 12t 2A, which is equivalent to 12 12t 0.0625 2. Solving graphically, we find that 12 t 11.119. It will take about 11.119 years. (If the interest is credited at the end of each month, it will take 11 years 2 months.) 29. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve Ae 0.0625t 2A, which is equivalent to e 0.0625t 2. Solving graphically, we find that t 11.090. It will take about 11.090 years. 30. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A(1.0575) t 3A, which is equivalent to 1.0575t 3. Solving graphically, we find that t 19.650. It will take about 19.650 years. (If the interest is credited at the end of each year, it will take 20 years.) 1 t 0.0575 365t 3A, which is equivalent to 365 0.0575 365t 3. Solving graphically, we find that 365 19.108. It will take about 19.108 years. 32. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve Ae 0.0575t 3A, which is equivalent to e 0.0575t 3. Solving graphically, we find that t 19.106. It will take about 19.106 years. 33. After t hours, the population is P(t) 2 t/0.5 or, equivalently, P(t ) 22t. After 24 hours, the population is P(24) 2 48 2.815 1014 bacteria. 34. (a) Each year, the number of cases is 100% 20% 80% of the previous year’s number of cases. After t years, the number of cases will be C(t ) 10,000(0.8)t. Solving C(t ) 1000 graphically, we find that t 10.319. It will take about 10.319 years. (b) Solving C(t ) 1 graphically, we find that t It will take about 41.275 years. 41.275. 35. Since x 1, the corresponding value of y is equal to the slope of the line. If the changes in x are constant for a linear function, then the corresponding changes in y are constant as well. 36. (a) When t 0, B 100e0 100. There were 100 bacteria present initially. (b) When t 6, B 100e 0.693(6) 6394.351. After 6 hours, there are about 6394 bacteria. (c) Solving 100e 0.693t 200 graphically, we find that t 1.000. The population will be 200 after about 1 hour. Since the population doubles (from 100 to 200) in about 1 hour, the doubling time is about 1 hour. 37. (a) Let x 0 represent 1900, x 1 represent 1901, and so on. The regression equation is P(x) 6.033(1.030)x. [0, 100] by [ 10, 90] (b) The regression equation gives an estimate of P(0) 6.03 million, which is not very close to the actual population. (c) Since the equation is of the form P(x) the annual rate of growth is about 3%. 38. (a) The regression equation is P(x) P(0) 1.030x, 4.831(1.019)x. [0, 100] by [ 5, 30] (b) P(90) 26.3 million (c) Since the equation is of the form P(x) P(0) 1.019x, the annual rate of growth is approximately 1.9%. 16 Section 1.4 39. 5422(1.018)19 s Section 1.4 Parametric Equations (pp. 26–31) 7609.7 million 40. (a) Exploration 1 Parametrizing Circles 1. Each is a circle with radius a . As a increases, the radius of the circle increases. [ 5, 5] by [ 2, 10] In this window, it appears they cross twice, although a third crossing off-screen appears likely. (b) x change in Y1 change in Y2 [ 4.7, 4.7] by [ 3.1, 3.1] 1 3 5 0 8 t : t : t 3 : 2 4 7 2. 0 2 2 2 3 [ 4.7, 4.7] by [ 3.1, 3.1] 4 It happens by the time x (c) Solving graphically, x 4. 0.7667, x 2, x 4. (d) The solution set is approximately ( 0.7667, 2) (4, ). 41. Since f (1) 4.5 we have ka we have ka 1 0.5. Dividing, we have a 0.5 0 4.5 0.5 ka ka 1 a2 4.5, and since f ( 1) [ 4.7, 4.7] by [ 3.1, 3.1] [ 4.7, 4.7] by [ 3.1, 3.1] 9 2 t 4: 3 Since f (x) k a x is an exponential function, we require a 0, so a 3. Then ka 4.5 gives 3k 4.5, so k 1.5. The values are a 3 and k 1.5. 42. Since f (1) 1.5 we have ka 1.5, and since f ( 1) we have ka 1 6. Dividing, we have ka ka 1 6 [ 4.7, 4.7] by [ 3.1, 3.1] 0 t 4: 1.5 6 a2 0.25 a 0.5 Since f (x) k a x is an exponential function, we require a 0, so a 0.5. Then ka 1.5 gives 0.5k 1.5, so k 3. The values are a 0.5 and k 3. [ 4.7, 4.7] by [ 3.1, 3.1] Let d be the length of the parametric interval. If d you get d of a complete circle. If d 2 complete circle. If d 2, 2 , you get the 2 , you get the complete circle but portions of the circle will be traced out more than once. For example, if d 4 the entire circle is traced twice. Section 1.4 Exploration 2 3. 1. a 2 2, b 3 2 t Parametrizing Ellipses 3: [ 12, 12] by [ 8, 8] initial point: (0, 3) terminal point: (0, 3) a 2, b 4: [ 12, 12] by [ 8, 8] a 2, b 5: t2 initial point: ( 3, 0) terminal point: (3, 0) [ 12, 12] by [ 8, 8] a 3 2 t 2, b 6: 3 initial point: (0, 3) terminal point: ( 3, 0) [ 12, 12] by [ 8, 8] 2. a 3, b 4: [ 9, 9] by [ 6, 6] t5 initial point: ( 3, 0) terminal point: ( 3, 0) 4. For 0 t 2 the complete circle is traced once clockwise beginning and ending at (2, 0). For t 3 the complete circle is traced once clockwise beginning and ending at ( 2, 0). For 2 t starting at (0, 3 the half circle below is traced clockwise 2 a 5, b 4: [ 9, 9] by [ 6, 6] a 6, b 4: 2) and ending at (0, 2). [ 9, 9] by [ 6, 6] a 7, b 4: [ 9, 9] by [ 6, 6] 3. If a b , then the major axis is on the x-axis and the minor on the y-axis. If a b , then the major axis is on the y-axis and the minor on the x-axis. 17 18 Section 1.4 4. 0 t 2 Exploration 3 : [ 6, 6] by [ 4, 4] 0 t Graphing the Witch of Agnesi 1. We used the parameter interval [0, ] because our graphing calculator ignored the fact that the curve is not defined when t 0 or . The curve is traced from right to left across the screen. x ranges from to . 2. : t 2 2 : [ 5, 5] by [ 2, 4] [ 6, 6] by [ 4, 4] 0 t 0 3 : 2 t 2 : [ 5, 5] by [ 2, 4] [ 6, 6] by [ 4, 4] 0 t 4: 2 t : [ 6, 6] by [ 4, 4] [ 5, 5] by [ 2, 4] Let d be the length of the parametric interval. If d you get d of a complete ellipse. If d 2 2, 2 , you get the complete ellipse. If d 2 , you get the complete ellipse but portions of the ellipse will be traced out more than once. For example, if d 4 the entire ellipse is traced twice. 5. 0 t For t 2 2 , the entire graph described in part 1 is drawn. The left branch is drawn from right to left across the screen starting at the point (0, 2). Then the right branch is drawn from right to left across the screen stopping at the 2: point (0, 2). If you leave out 2 and 2 , then the point (0, 2) is not drawn. [ 6, 6] by [ 4, 4] initial point: (5, 0) terminal point: (5, 0) t For 0 2 , the right branch is drawn from right to left across the screen stopping at the point (0, 2). If you leave out 3: t For 2 , then the point (0, 2) is not drawn. t 2 , the left branch is drawn from right to left across the screen starting at the point (0, 2). If you leave [ 6, 6] by [ 4, 4] initial point: ( 5, 0) terminal point: ( 5, 0) 2 t 3 : 2 [ 6, 6] by [ 4, 4] initial point: (0, 2) terminal point: (0, 2) Each curve is traced clockwise from the initial point to the terminal point. out 2 , then the point (0, 2) is not drawn. 3. If you replace x 2 cot t by x 2 cot t, the same graph is drawn except it is traced from left to right across the screen. If you replace x 2 cot t by x 2 cot ( t), the same graph is drawn except it is traced from left to right across the screen. Quick Review 1.4 1. m y y 3 4 8 1 5 (x 3 5 x 3 5 3 1) 29 3 5 3 8 19 Section 1.4 2. y Section 1.4 Exercises 4 3. x 1. Graph (c). Window: [ 4, 4] by [ 3, 3], 0 2 4. When y 0, we have and 3. When x x 9 1, so the x-intercepts are y2 16 0, we have 3 0, we have and 4. When x x2 16 1, so the x-intercepts are y 9 4 1, which has no real solution, so there are no y-intercepts. 6. When y 0, we have 0 1, so the x-intercept is 0, we have 2y2 When x and x 1, so the y-intercepts are 2 t 2 4. Graph (b). Window: [ 15, 15] by [ 15, 15], 0 t 2 5. (a) The resulting graph appears to be the right half of a hyperbola in the first and fourth quadrants. The parameter a determines the x-intercept. The parameter b determines the shape of the hyperbola. If b is smaller, the graph has less steep slopes and appears “sharper.” If b is larger, the slopes are steeper and the graph appears more “blunt.” The graphs for a 2 and b 1, 2, and 3 are shown. 2 0, we have 2 t 3. Graph (d). Window: [ 10, 10] by [ 10, 10], 0 1, so the y-intercepts are 4 and 4. 5. When y t 2. Graph (a). Window: [ 2, 2] by [ 2, 2], 0 2 1. 1 2 1 . 2 [ 10, 10] by [ 10, 10] 7. (a) 2(1)2(1) + 12 3 3 3 Yes (b) 2( 1)2( 1) ( 2) 2 8. (a) 9(1)2 4 3 3 [ 10, 10] by [ 10, 10] This appears to be the left half of the same hyberbola. (c) Yes 27 27 27 Yes 18(1) 4( 3) 9 18 36 27 9( 1)2 18( 1) 4(3)2 9 18 36 63 [ 10, 10] by [ 10, 10] 27 27 27 Yes 27 27 27 No 5 2x 5 t 2x 3 5 2t 2t 2t t tan2 t 1 by a standard trigonometric identity. Substituting tan t gives 1 3y 1 3y 1 3y One must be careful because both sec t and tan t are discontinuous at these points. This might cause the grapher to include extraneous lines (the asymptotes of the hyperbola) in its graph. The extraneous lines can be avoided by using the grapher’s dot mode instead of connected mode. (d) Note that sec2 t 3t 3t (b) 3y No 2 (b) 9(1) 9. (a) 2x 3 18(1) 4(3)2 9 18 36 27 2 (c) ( 2)2 3 2 12 3 3 3 1 (c) ( 1)2 21 1 (b) x2 a y2 b x y for sec t and for a b 1. (e) This changes the orientation of the hyperbola. In this case, b determines the y-intercept of the hyperbola, and 1 a determines the shape. The parameter interval 2 10. (a) The equation is true for a 0. , 2 a or (b) The equation is equivalent to “ a a2 a.” Since a2 a is true for a 0 and a2 a is true for a 0, at least one of the two equations is true for all real values of a. Therefore, the given equation a2 a is true for all real values of a. (c) The equation is true for all real values of a. 22 gives the upper half of the hyperbola. The parameter interval 2 , 3 2 gives the lower half. The same values of t cause discontinuities and may add extraneous lines to the graph. Substituting x for tan t in the identity sec2 t a y2 x2 1. b a and y for sec t b tan2 t 1 gives 20 Section 1.4 6. (a) 10. (a) [ 6, 6] by [ 4, 4] [ 4.7, 4.7] by [ 3.1, 3.1] The graph is a circle of radius 2 centered at (h, 0). As h changes, the graph shifts horizontally. Initial and terminal point: (4, 0) (b) (b) x2 4 y2 2 cos2 t sin2 t 1 The parametrized curve traces all of the ellipse defined by x2 4 y2 2 1. 11. (a) [ 6, 6] by [ 4, 4] The graph is a circle of radius 2 centered at (0, k). At k changes, the graph shifts vertically. (c) Since the circle is to be centered at (2, 3), we use h 2 and k 3. Since a radius of 5 is desired, we need to change the coefficients of cos t and sin t to 5. x 5 cos t 2, y 5 sin t 3, 0 t 2 (d) x 5 cos t 3, y 2 sin t 4, 0 t [ 4.7, 4.7] by [ 3.1, 3.1] Initial point: (0, 2) Terminal point: (0, 2 7. (a) (b) x2 4 y2 2 2) sin2 t cos2 t 1 The parametrized curve traces the right half of the ellipse defined by defined by x [ 3, 3] by [ 2, 2] Initial point: (1, 0) Terminal point: ( 1, 0) x2 4 4 2 y2 2 1 (or all of the curve y2). 12. (a) (b) x 2 y 2 cos2 t sin 2 t 1 The parametrized curve traces the upper half of the circle defined by x 2 y 2 1 (or all of the semicircle defined by y 1 x 2). [ 9, 9] by [ 6, 6] 8. (a) Initial and terminal point: (0, 5) (b) x2 4 y2 5 sin2 t cos2 t 1 The parametrized curve traces all of the ellipse defined [ 3, 3] by [ 2, 2] by Initial and terminal point: (0, 1) (b) x 2 y 2 sin2 (2 t) cos2 (2 t) 1 The parametrized curve traces all of the circle defined by x 2 y 2 1. x2 4 y2 5 1. 13. (a) 9. (a) [ 3, 3] by [ 1, 3] No initial or terminal point. (b) y 9t 2 (3t)2 x 2 The parametrized curve traces all of the parabola defined by y x 2. [ 3, 3] by [ 2, 2] Initial point: ( 1, 0) Terminal point: (0, 1) 2 2 2 14. (a) 2 y cos ( t) sin ( t) 1 (b) x The parametrized curve traces the upper half of the circle defined by x 2 y 2 1 (or all of the semicircle defined by y 1 x2). [ 3, 3] by [ 1, 3] Initial point: (0, 0) Terminal point: None Section 1.4 (b) y t ( t)2 x 2 The parametrized curve traces the left half of the parabola defined by y x 2 (or all of the curve defined by x y). 21 19. (a) 15. (a) [ 9, 9] by [ 6, 6] No initial or terminal point. (b) y 4t 7 2(2t 5) 3 2x 3 The parametrized curve traces all of the line defined by y 2x 3. [ 1, 5] by [ 1, 3] Initial point: (0, 0) Terminal point: None 20. (a) (b) y t x The parametrized curve traces all of the curve defined by y x (or the upper half of the parabola defined by x y 2). [ 6, 6] by [ 4, 4] 16. (a) No initial or terminal point. (b) y 1 t 2 (1 t) 2 x x2 The parametrized curve traces all of the line defined by y x 2. 21. (a) [ 3, 9] by [ 4, 4] No initial or terminal point. (b) x sec2 t 1 tan2 t y 2 The parametrized curve traces all of the parabola defined by x y 2. [ 3, 3] by [ 2, 2] 17. (a) Initial point: (0, 1) Terminal point: (1, 0) (b) y 1 t 1 x x1 The Cartesian equation is y x 1. The portion traced by the parametrized curve is the segment from (0, 1) to (1, 0). [ 3, 3] by [ 2, 2] No initial or terminal point. Note that it may be necessary to use a t-interval such as [ 1.57, 1.57] or use dot mode in order to avoid “asymptotes” showing on the calculator screen. 22. (a) (b) x 2 y 2 sec2 t tan2 t 1 The parametrized curve traces the left branch of the hyperbola defined by x 2 y 2 1 (or all of the curve defined by x y2 1). 18. (a) [ 2, 4] by [ 1, 3] Initial point: (3, 0) Terminal point: (0, 2) (b) y 2t (2t 2) 2 The Cartesian equation is y 2 (3 3 3t) 2 x 3 2 2 x 3 2 2. The portion traced by the curve is the segment from (3, 0) to (0, 2). [ 6, 6] by [ 5, 1] 23. (a) No initial or terminal point. Note that it may be necessary to use a t-interval such as [ 1.57, 1.57] or use dot mode in order to avoid “asymptotes” showing on the calculator screen. (b) y2 2 x2 sec2 t tan2 t [ 6, 6] by [ 2, 6] 1 The parametrized curve traces the lower branch of the hyperbola defined by defined by y y2 2 2 x2 x2 1). 1 (or all of the curve Initial point: (4, 0) Terminal point: None t 4 (4 t) 4 x x4 (b) y The parametrized curve traces the portion of the line defined by y x 4 to the left of (4, 0), that is, for x 4. 22 Section 1.4 24. (a) 30. The vertex of the parabola is at ( 1, 1), so the left half of the parabola is given by y x 2 2x for x 1. Substituting t for x, we obtain one possible parametrization: x t, y t 2 2t, t 1. [ 1, 5] by [ 1, 3] Initial point: (0, 2) Terminal point: (4, 0) 4 t2 4x (b) y The parametrized curve traces the right portion of the curve defined by y 4 x, that is, for x 0. 25. (a) 31. For simplicity, we assume that x and y are linear functions of t and that the point (x, y) starts at (2, 3) for t 0 and passes through ( 1, 1) at t 1. Then x f (t), where f (0) 2 and f (1) 1. x t Since slope x f (t) g(0) 3t y No initial or terminal point, since the t-interval has no beginning or end. The curve is traced and retraced in both directions. (b) y cos 2t cos2 t sin2 t 1 2 sin2 t 1 2x 2 2x 2 1 The parametrized curve traces the portion of the parabola defined by y 2x 2 1 corresponding to 1 x 1. g(t) 4t x (b) x t 2 3 y 2 3 The parametrized curve traces the lower half of the parabola defined by x y 2 3 (or all of the curve defined by y x 3). 27. Using ( 1, 3) we create the parametric equations x 1 at and y 3 bt, representing a line which goes through ( 1, 3) at t 0. We determine a and b so that the line goes through (4, 1) when t 1. Since 4 1 a, a 5. Since 1 3 b, b 4. Therefore, one possible parametrization is x 1 5t, y 3 4t, 0 t 1. 28. Using ( 1, 3) we create the parametric equations x 1 at and y 3 bt, representing a line which goes through ( 1, 3) at t 0. We determine a and b so that the line goes through (3, 2) at t 1. Since 3 1 a, a 4. Since 2 3 b, b 5. Therefore, one possible parametrization is x 1 4t, y 3 5t, 0 t 1. 29. The lower half of the parabola is given by x y 2 1 for y 0. Substituting t for y, we obtain one possible parametrization: x t 2 1, y t, t 0. g(t), where 13 10 3 3 4, 4t. 2 3t, y 3 4t, t 0. 32. For simplicity, we assume that x and y are linear functions of t and that the point (x, y) starts at ( 1, 2) for t 0 and passes through (0, 0) at t 1. Then x f (t), where f (0) 1 and f (1) 0. Since slope x t x ( 1) y Initial point: None Terminal point: ( 3, 0) 3t. Also, y 1. y t f (t) 1t g(t) 0 1 ( 1) 0 1 g(t), where g(0) y t Since slope [ 4, 5] by [ 4, 2] 2 3, One possible parametrization is: Also, y 26. (a) 2 3 and g(1) Since slope [ 3, 3] by [ 2, 2] 12 10 2t 0 1 2 t. 2 and g(1) 2 0 2 1, 0. 2, 2t. One possible parametrization is: x 1 t, y 2 2t, t 0. 33. The graph is in Quadrant I when 0 y 2, which corresponds to 1 t 3. To confirm, note that x(1) and x(3) 0. 2 34. The graph is in Quadrant II when 2 y 4, which corresponds to 3 t 5. To confirm, note that x(3) and x(5) 2. 0 35. The graph is in Quadrant III when 6 y 4, which corresponds to 5 t 3. To confirm, note that x( 5) 2 and x( 3) 0. 36. The graph is in Quadrant IV when 4 y 0, which corresponds to 3 t 1. To confirm, note that x( 3) 0 and x(1) 2. 37. The graph of y x 2 2x 2 lies in Quadrant I for all x 0. Substituting t for x, we obtain one possible parametrization: x t, y t 2 2t 2, t 0. x 3 lies in Quadrant I for all x 0. 38. The graph of y Substituting t for x, we obtain one possible parametrization: x t, y t 3, t 0. Section 1.5 s Section 1.5 Functions and Logarithms (pp. 32–40) 39. Possible answers: (a) x a cos t, y (b) x a cos t, y (c) x a cos t, y (d) x a sin t, 0 a cos t, y a sin t, 0 t t a sin t, 0 a sin t, 0 2 2 t t 23 Exploration 1 4 Testing for Inverses Graphically 1. It appears that ( f g)(x) (g f )(x) and g may be inverses of each other. 4 40. Possible answers: x, suggesting that f (a) f and g: (a) x a cos t, y (b) x a cos t, y (c) x a cos t, y (d) x b sin t, 0 a cos t, y t b sin t, 0 b sin t, 0 t t b sin t, 0 41. Note that m OAQ 2 2 4 t 4 [ 4.7, 4.7] by [ 3.1, 3.1] t, since alternate interior angles (b) f g: formed by a transversal of parallel lines are congruent. OQ AQ Therefore, tan t 2 , so x x 2 tan t 2 cot t. [ 4.7, 4.7] by [ 3.1, 3.1] Now, by equation (iii), we know that 2 (c) g f : (AQ) AO AQ (AQ) AO AB (cos t)(x) [ 4.7, 4.7] by [ 3.1, 3.1] (cos t)(2 cot t) 2. It appears that f g identity function. 2 cos2 t . sin t 2 AB sin t 2 g, suggesting that f may be the (a) f and g: Then equation (ii) gives y gf 2 cos2 t sin t sin t 2 2 cos2 t 2 sin2 t. [ 4.7, 4.7] by [ 3.1, 3.1] The parametric equations are: x 2 cot t, y 2 sin2 t, 0 (b) f g: t Note: Equation (iii) may not be immediately obvious, but it may be justified as follows. Sketch segment QB. Then [ 4.7, 4.7] by [ 3.1, 3.1] OBQ is a right angle, so AB AQ ABQ AQO, which gives (c) g f : AQ . AO 42. (a) If x2 x1 then the line is a vertical line and the first parametric equation gives x x1, while the second will give all real values for y since it cannot be the case that y2 y1 as well. Otherwise, solving the first equation for t gives t (x x1)/(x2 x1). Substituting that into the second equation gives y y1 [(y2 y1)/(x2 x1)](x x1) which is the point-slope form of the equation for the line through (x1, y1) and (x2, y2). Note that the first equation will cause x to take on all real values, because (x2 x1) is not zero. Therefore, all of the points on the line will be traced out. [ 4.7, 4.7] by [ 3.1, 3.1] 3. It appears that ( f g)(x) (g f )(x) and g may be inverses of each other. (a) f and g: [ 4.7, 4.7] by [ 3.1, 3.1] (b) f g: (b) Use the equations for x and y given in part (a), with 0 t 1. [ 4.7, 4.7] by [ 3.1, 3.1] x, suggesting that f 24 Section 1.5 5. Substituting t for x, one possible answer is: 3. continued (c) g f : x t, y 1 t 1 ,t 2. 6. Substituting t for x, one possible answer is: x t, y t, t 3. 7. [ 4.7, 4.7] by [ 3.1, 3.1] 4. It appears that ( f g)(x) (g f )(x) x, suggesting that f and g may be inverse of each other. (Notice that the domain of f g is (0, ) and the domain of g f is ( , ).) (a) f and g: [ 10, 10] by [ 10, 10] (4, 5) 8. [ 4.7, 4.7] by [ 3.1, 3.1] (b) f g: [ 10, 10] by [ 10, 10] 8 , 3 3 (2.67, 3) 9. (a) [ 4.7, 4.7] by [ 3.1, 3.1] (c) g f : [ 10, 10] by [ 10, 10] (1.58, 3) [ 4.7, 4.7] by [ 3.1, 3.1] Exploration 2 Supporting the Product Rule 1. They appear to be vertical translations of each other. (b) No points of intersection, since 2x values of x. 0 for all 10. (a) [ 10, 10] by [ 10, 10] [ 1, 8] by [ 2, 4] ( 1.39, 4) 2. This graph suggests that each difference (y3 constant. y1 y2) is a (b) No points of intersection, since e of x. x 0 for all values Section 1.5 Exercises 1. No, since (for example) the horizontal line y the graph twice. 2. Yes, since each horizontal line intersects the graph only once. [ 1, 8] by [ 2, 4] 3. y3 y1 y2 ln (ax) Thus, the difference y3 ln x ln a ln x ln x ln a y1 y2 is the constant ln a. f (g(1)) 2. (g f )( 7) 3. ( f g)(x) 3 x 2 4. (g f )(x) f (2) g( f ( 7)) f (g(x)) x 1 g( 2) f (x 2 1) 2/3 3 g( f (x)) 3 g( x (x 1) (x 2/3 1) 2 1 1 3. Yes, since each horizontal line intersects the graph at most once. 4. No, since (for example) the horizontal line y intersects the graph twice. Quick Review 1.5 1. ( f g)(1) 2 intersects 1) 5 3 (x 2 1) 1 0.5 5. Yes, since each horizontal line intersects the graph only once. 6. No, since (for example) the horizontal line y the graph at more than one point. 2 intersects Section 1.5 7. 13. y 2x 3 y 3 2x y 3 x 2 Interchange x and y. [ 10, 10] by [ 10, 10] Yes, the function is one-to-one since each horizontal line intersects the graph at most once, so it has an inverse function. x 3 y 2 f 1 x (x) . Verify. 1 )(x) (f f 8. 3 2 x f 3 2 x 2 3 3 2 (x 3) 3 (2x 3) x [ 10, 10] by [ 10, 10] 1 (f f )(x) 1 f No, the function is not one-to-one since (for example) the horizontal line y 0 intersects the graph twice, so it does not have an inverse function. (2x 3) 2 2x 2 3 x 9. 14. y 5 4x 4x 5 y 5 x y 4 [ 10, 10] by [ 10, 10] Interchange x and y. No, the function is not one-to-one since (for example) the horizontal line y 5 intersects the graph more than once, so it does not have an inverse function. y 10. f 5 1 (x) x 45 x 4 Verify. 1 )(x) (f f 5 f x 5 4 5 (5 4 5 x 4 x) x 1 f )(x) x 3 15. y x y 1 1 (y 12. 4x) (5 5 11. No, the function is not one-to-one since each horizontal line intersects the graph twice, so it does not have an inverse function. 4x) f Yes, the function is one-to-one since each horizontal line intersects the graph only once, so it has an inverse function. [ 10, 10] by [ 10, 10] 1 (5 4x 4 4 (f [ 5, 5] by [ 20, 20] x3 1) 1/3 x Interchange x and y. 1)1/3 (x f 1 (x) y (x 1)1/3 or 1 )(x) 3 f( x ( (f 1 f )(x) 3 1 f 1) 1)3 x 1 3 (x 3 Yes, the function is one-to-one since each horizontal line intersects the graph at most once, so it has an inverse function. x Verify. (f f [ 9, 9] by [ 2, 10] 3 (x3 1 (x 1) 1 1) 1) 1 3 x3 x x 25 26 Section 1.5 x2 y 1, x 1 16. y 2 x ,x y 1 For x 0 (f 0 2 (the domain of f ), 1 f )(x) x 1 1 20. y x )(x) f( x 1) f )(x) 1 f (x (x x 2 x ,x 17. y ), (x 2)2 1 1 2 f 1 1 1 or x 1/2 x For x 1 x 1 )(x) (f f x f( x 1)2 2 x Interchange x and y. For x 1 (f 1/2 x or x f )(x) f 1 )(x) (f f For x f( f )(x) 1 f ), x)2 ( (x 2) (x 2/3)3/2, x y 3/2 x2 x x 0 1 f x 3/2 1 0 (the domain of f 1 )(x) for x f (x 3/2 ) f )(x) 1 (x 3/2 2/3 f 2/3 (x ) 2)2, x 2)2 y, x ) x 2 (x ) x 1 x 0 y or 1 x1/2 )(x) f ), 1 (1/ x)2 x 0 (the domain of f ), f )(x) 1 x3 1 y 31 y f 1 11 x2 x2 1/x 2 1 3 y 1 y 2 1 1 Interchange x and y. ( x)1/2 x or 2 f Verify. 3 1 x (x) 1 3 or x 1 x1/3 Verify. For x (f f (f x x (x) x x3 y 2 1 1 0 (the domain of f 1 22. y Interchange x and y. f 1) 1 1 1 For x 2/3 3/2 y 2 y 1 x 2 x 1 0 1 (x) (f f 2 x 1 For x ), 0, (the domain of f ), (x (x 2x Verify. For x 19. y 1 1) 1)2 1 x Verify. (f 2 x (x) 1 x 1 y y (f f 2 Interchange x and y. Interchange x and y. f 1 ,x x2 1 ,x y 1 y 21. y x x x 2x 2 x 0 y 3/2 3/2 (x 2 (x x x2 x 2/3, x 18. y x) 1 (x 0, (the domain of f ), 1 (f 1 0 (the domain of f x 1 (the domain of f ), 1 x Verify. For x 1] ), 2( 2 x)2 x 1 1) x [( 0 (x) 1) 1 0 (the domain of f x) f x 1 y (x) ( 1 x x ( y 2) 1 y x (x 1 1 2 Verify. 1) 2 1) , x x 2 1), x 2 (x x x 1) 2 x Interchange x and y. 0, (the domain of f ), 1 (f 1) 2x y 2 (x 2 y 1) x ( (x y 1 1 (the domain of f 1 For x 1)1/2 1 or (x Verify. For x (f f 2)2) 2 y (x) ( (x 2 x Interchange x and y. f 1 f 1 0 (the domain of f 1 )(x) f (2 (f f x) x) [(2 ( ) 2 x) 2]2 x (f x 1 1 1 1 3 (1/ x)3 1 )(x) f f )(x) x 1 f13 x 3 3 1/x3 x x x x Section 1.5 2x x 23. y 1 3 x x 24. y xy 3y 2x 2x 1 2)x 1 3y x 1 y 2y x 3 xy 3y (y xy 1 xy x 2y 3 x(y 1 x f 1 1 x 2x x y f 3x 2 1 )(x) f 2 1 x 1 3x x2 1 3x x2 5x 5 1 f )(x) f 3 1 3 2x x 1 3 1 5x 5 2) 2) (f 1 f )(x) x 3 1 3 1 1 3 x x x x Graph of f 3(x 2(x 1) 1) x 3 2 3 2 3 2 3 1 3) 3) 3(x 2) (x 2) x et t : x2 Graph of y 3) 3) t, y1 25. Graph of f: x1 1 2 x 2(x (x 5x 5 2 3 1x f 2 1) 3) 3 1 (2x (2x 5x 5 3 (x 3) 3(2x (2x 1) 2(x 2x x f 1 3 x 2x x )(x) 2x x 2x x x 1 2x 1 1 3x 2 2(1 3x) (x (1 3x) 3(x (f 2x x Verify. (f f 1 3 3 1 (x) Verify. (f f 2y 3 1 Interchange x and y. 3x 2 (x) 1) 2y y x 3y 2 Interchange x and y. y 3 2 e , y2 t t, y3 x: x3 t [ 6, 6] by [ 4, 4] 26. Graph of f: x1 Graph of f Graph of y 1 t, y1 : x2 x: x3 [ 6, 6] by [ 4, 4] 3t 3t, y2 t, y3 t t 27 28 Section 1.5 27. Graph of f: x1 Graph of f 1 t, y1 : x2 Graph of y t 2 33. t 2 , y2 t t, y3 t x: x3 [ 10, 5] by [ 7, 3] Domain: ( , 3) Range: ( , ) 34. [ 4.5, 4.5] by [ 3, 3] 28. Graph of f: x1 Graph of f 1 t, y1 3 t, y2 : x2 Graph of y t 3 t t, y3 t x: x3 [ 5, 10] by [ 5, 5] Domain: ( 2, ) Range: ( , ) 35. [ 4.5, 4.5] by [ 3, 3] 29. Graph of f: x1 Graph of f 1 t, y1 : x2 Graph of y ln t ln t, y2 t t, y3 t x: x3 [ 3, 6] by [ 2, 4] Domain: ( 1, ) Range: ( , ) 36. [ 4.5, 4.5] by [ 3, 3] 30. Graph of f: x1 Graph of f 1 t, y1 : x2 Graph of y log t [ 2, 10] by [ 2, 4] log t, y2 t, y3 x: x3 t Domain: (4, ) Range: ( , ) t 37. (1.045)t 2 t ln(1.045) ln 2 t ln 1.045 ln 2 t [ 4.5, 4.5] by [ 3, 3] 31. Graph of f: x1 Graph of f 1 Graph of y sin sin 1 1 t t, y2 t, y3 x: x3 15.75 Graphical support: t, y1 : x2 ln 2 ln 1.045 t t [ 2, 18] by [ 1, 3] 38. e 0.05t ln e 0.05t [ 3, 3] by [ 2, 2] 32. Graph of f: x1 Graph of f Graph of y 1 0.05t t, y1 : x2 x: x3 tan 1 1 t, y2 t, y3 t t ln 3 0.05 3 ln 3 ln 3 20 ln 3 t tan t Graphical support: [ 5, 35] by [ 1, 4] [ 6, 6] by [ 4, 4] 21.97 Section 1.5 39. e x e x e x 3 3 x e e x(e x 3 (e x)2 3e x x ex 0 x e e x(0) ) 1 3 1 1 2 0 ( 3)2 2(1) 5 3 e 43. y 100 2 2 3 ln 100 y x 100 y x log2 x 5 x 0.96 or 0.96 2 1 100 1 y 100 log2 1 y 100 log2 1 y 100 y log2 y y log2 100 y log2(2 x) 4(1)(1) 2 x x Graphical support: Interchange x and y. y [ 4, 4] by [ 4, 8] 40. 2x 2 2x x 5 f x 2x(2x 5 (2x)2 5(2x) 1 (x) x 100 x log2 x 100 x Verify. 5 2 log2 (f f 0 2 x) 1 )(x) f log2 x 100 x 2x(0) 100 2x 2x 5 x 0 2 5 1 4(1)(1) ( 5) 2(1) 21 2 1 2log2 100 2 5 log2 x 100 x log2 1 21 2.26 or 2.26 2 100 x x Graphical support: 100 100 x x 1 [ 4, 4] by [ 4, 8] 41. ln y 2t 4 e ln y y e 2t (f 4 42. ln(y 1) ln 2 ln(y 1) x e ln(y 100x 100 x) x 4 e 2t 100x (100 x 1) y 1 y 2xe x ex x f )(x) f 1 1 100 2 ln 2 x ln x ln x 1 ln x ln 2 log2 e x(x)(2) 1 100 100 2 1 x 100 2 x 1 log2 100(1 100 2 x) log2 1 2x log2(2 x) 100 x 29 30 Section 1.5 44. y 50 1.1 1 1 1.1 1.1 x 50 y (b) f ( f (x)) 50 log1.1 y log1.1(1.1 ) 50 y 50 log1.1 y log1.1 f log1.1 1 (x) 46. (a) Amount 1 1 log1.1 50 y y log1.1 y 50 y (b) 8 x 50 x log1.1 x 50 1 )(x) x f log1.1 t x 50 x 0 1 1/x x for all x 0 1 t/12 2 1 1 8 13 2 36 47. 500(1.0475)t log1.1 1.1 8 There will be 1 gram remaining after 36 hours. 50 1 1 t/12 2 1 t/12 2 1 t/12 2 t 3 12 Verify. (f f x, since x 1 x f 1 Interchange x and y: y x x 2) 1 x x (1 x2 50 y ( f (x))2 1 x x x 1 45. (a) f ( f (x)) 1000 x 50 x 1.0475t 2 t ln(1.0475 ) ln 2 t ln 1.0475 ln 2 50 1.1log1.1 1 50 x x t 1 14.936 It will take about 14.936 years. (If the interest is paid at the end of each year, it will take 15 years.) 50 50 ln 2 ln 1.0475 x x 48. 375,000(1.0225)t 50x (50 x 50x 50 x) 1.0225t x 8 3 ln(1.0225t) t ln 1.0225 (f 1 f )(x) f 1 1 50 1.1 t x 1,000,000 8 3 8 ln 3 ln ln(8/3) ln 1.0225 44.081 It will take about 44.081 years. 50 1 1.1 x log1.1 50 50 1 1.1 log1.1 50 1.1 x) 50(1 49. (a) y 2539.852 636.896 ln x (b) When x 75, y 209.94. About 209.94 million metric tons were produced. x (c) 636.896 ln x 400 636.896 ln x 50 2539.852 2939.852 ln x log1.1 1 1.1 x log1.1(1.1x) x x 2939.852 636.896 2939.852 e 636.896 101.08 According to the regression equation, Saudi Arabian oil production will reach 400 million metric tons when x 101.08, in about 2001. 31 Section 1.6 50. (a) y 590.969 (b) The expression a logb(x c) d is defined when x c 0, so the domain is (c, ). Since a logb(x c) d attains every real value for some value of x, the range is ( , ). 152.817 ln x (b) When x 85, y 87.94. About 87.94 million metric tons were produced. (c) 590.969 152.817 ln x 120 152.817 ln x ln x 56. (a) Suppose f (x1) 710.969 710.969 152.817 ax1 cx1 104.84 51. (a) Suppose that f (x1) f (x2). Then mx1 b so mx1 mx2. Since m 0, this gives x1 mx y b y b mx2 x2. b f 1 (x) x b 1 . The graphs of the m d) bd adx2 bcx1 bd bcx1 (ad bc)x2 bc 0, this means that x1 ax x2. b b d a)x dy m, respectively. Since each of 1 and m dx b cx a dx b f 1(x) cx a slopes. a (c c ,f 1 (x) tal asymptote is y 52. (a) y2 is a vertical shift (upward) of y1, although it’s difficult to see that near the vertical asymptote at x One might use “trace” or “table” to verify this. 0. (b) Each graph of y3 is a horizontal line. (c) The graphs of y4 and y ln(e y2 y1) a, ln a, y1 y2 a are the same. ln a, ln a 0). Since f (x) is undefined at dx b → cx a d (c c 1 f (x1) d . c d , so the horizonc 0). Since f 1 (x) is a , the vertical asymptote is x c undefined at x a . c The horizontal asymptote of f becomes the vertical asymptote of f 1 and vice versa due to the reflection of the graph about the line y ln x x. ln a 53. If the graph of f (x) passes the horizontal line test, so will the graph of g(x) f (x) since it’s the same graph reflected about the x-axis. Alternate answer: If g(x1) g(x2) then f (x1) f (x2), f (x1) f (x2), and x1 x2 since f is one-to-one. g(x2). Then b a → , so the horizontal d c d , the vertical asymptote is x c (d) As x → inverses will be perpendicular lines with nonzero ax cx , f (x) asymptote is y x m is the negative reciprocal of the other, the graphs of the 54. Suppose that g(x1) b dy b cy a (c) As x → 1 , m respectively, then the inverse functions will have slopes f (x1) bcx2 b)(cx1 y (d) If the original functions have slopes m and y1 (ax2 Interchange x and y: m inverses will be parallel lines with nonzero slope. y2 dy x inverse functions will have slope (d) ax cx (cy (c) If the original functions both have slope m, each of the e y2 y1 d) adx2 bc)x1 cxy The slopes are reciprocals. 1 and m bcx2 adx1 (b) y mx y m d adx1 Since ad Interchange x and y. x cx2 b)(cx2 b x m b acx1x2 (ad b f (x2). Then: ax2 acx1x2 According to the regression equation, oil production will reach 120 million metric tons when x 104.84, in about 2005. (b) y d (ax1 710.969 e 152.817 x b 1 , f (x2) f (x2), and x1 and x2 since f is one-to-one. 55. (a) The expression a(b c x) d is defined for all values of x, so the domain is ( , ). Since b c x attains all positive values, the range is (d, ) if a 0 and the range is ( , d) if a 0. s Section 1.6 Trigonometric Functions (pp. 41–51) Exploration 1 Functions Unwrapping Trigonometric 1. (x1, y1) is the circle of radius 1 centered at the origin (unit circle). (x2, y2) is one period of the graph of the sine function. 32 Section 1.6 2. The y-values are the same in the interval 0 t 2. 3. The y-values are the same in the interval 0 t 3. 4. 4. The x1-values and the y2-values are the same in each interval. 5. y2 tan t: [ 3, 3] by [ 2, 2] Using trace, cos t and sin t are being computed for 0, 15, 30, … , 360 degrees. Quick Review 1.6 y2 csc t: 1. 180 2. y2 180 40 4. 45 450 180 2.5 3. sec t: 60 3 2 9 143.24 4 180 5. y2 cot t: [0, 2 ] by [ 1.5, 1.5] For each value of t, the value of y2 ratio y1 x1 tan t is equal to the x 0.6435, x 2.4981 6. . For each value of t, the value of y2 csc t is equal to the 1 ratio . y1 [0, 2 ] by [ 1.5, 1.5] x For each value of t, the value of y2 sec t is equal to the 1.9823, x 4.3009 7. 1 ratio . x1 For each value of t, the value of y2 ratio x1 y1 cot t is equal to the . Exploration 2 2 Finding Sines and Cosines 1. The decimal viewing window [ 4.7, 4.7] by [ 3.1, 3.1] is square on the TI-82/83 and many other calculators. There are many other possibilities. x , 3 2 by [ 2, 2] 0.7854 or 4 ,x 3.9270 or 5 4 8. f ( x) 2( x)2 3 2x 2 3 f (x) The graph is symmetric about the y-axis because if a point (a, b) is on the graph, then so is the point ( a, b). ( x)3 3( x) x 3 3x (x 3 3x) f (x) The graph is symmetric about the origin because if a point (a, b) is on the graph, then so is the point ( a, b). 9. f ( x) [ 4.7, 4.7] by [ 3.1, 3.1] 2. Using the Ask table setting for the independent variable on the TI-83 we obtain 10. x 0 Section 1.6 Exercises 1. Arc length 5 (2) 8 10 2. Radius 175 180 5 4 72 7 3.274 Section 1.6 3. Angle 4. Angle 16. (a) Period 1 radian or about 28.65 2 7 14 3 /2 6 4 2 33 2 (b) Amplitude 1 (c) [ 4, 4] by [ 2, 2] radian or 45 5. (a) The period of y sec x is 2 , so the window should have length 4 . One possible answer: [0, 4 ] by [ 3, 3] 17. (a) Period (b) Domain: Since csc (3x (b) The period of y csc x is 2 , so the window should have length 4 . One possible answer: [0, 4 ] by [ 3, 3] (c) The period of y cot x is , so the window should have length 2 . One possible answer: [0, 2 ] by [ 3, 3] 6. (a) The period of y sin x is 2 , so the window should have length 4 . One possible answer: [0, 4 ] by [ 2, 2] 2 3 3x (k k , or x 1) 3 k 3 equivalent to x ) 1 sin (3x ) , we require . This requirement is for integers k. (c) Since csc (3x ) 1, the range excludes numbers between 3 2 5 and 3 2 1. The range is ( , 5] [1, ). (d) (b) The period of y cos x is 2 , so the window should have length 4 . One possible answer: [0, 4 ] by [ 2, 2] (c) The period of y tan x is , so the window should have length 2 . One possible answer: [0, 2 ] by [ 3, 3] 7. Since sin 6 8. Since sin 6 is in the range , 22 0.5, sin 1(0.5) 4 is the range 2 2 4 180 , sin radian or 6 , 22 2 2 3 22 , 33 ,) (d) 1 x and radian or 1.3734 radians or 0.7954 radian or 45.5730 . 19. (a) Period 3 2 k 2 (b) Domain: We require 3x 1.5 (b) Amplitude (k 2) 6 for odd integers k. for odd integers k. This requirement is equivalent to x k for odd integers k. 6 (c) Since the tangent function attains all real values, the range is ( , ). (d) by [ 4, 4] 2 2 (b) Amplitude 3 (c) [ 2 , 2 ] by [ 4, 4] 2 1/2 (b) Amplitude 4 5 2 (b) Amplitude , by [ 8, 8] 22 2 20. (a) Period 2 (b) Domain: ( ,) (c) Range: Since sin 2x (c) [ 4 , 4 ] by [ 10, 10] 15. (a) Period 30 . 2 (c) Since sin (4x ) 1, the range extends from 2 3 1 to 2 3 5. The range is [1, 5]. Therefore, x (c) [ 2 , 2 ] by [ 2, 2] 14. (a) Period 180 2 4 (b) Domain: ( 2 2 (b) Amplitude 13. (a) Period x and by [ 8, 8] , by [ 8, 8] 22 10. Using a calculator, cos 1(0.7) (c) 4 1 6 sin 2 1 9. Using a calculator, tan 1( 5) 78.6901 . 12. (a) Period of y sin 18. (a) Period 45 . 4 11. (a) Period of y 22 , 33 (d) 6 /3 4 (c) [ 3, 3] by [ 5, 5] [ , ] by [ 3, 3] 3 1, the range is [ 2, 2]. 34 Section 1.6 82 21. Note that 152 8 17 Since sin cos and 1 tan 22. Note that 15 , 8 52 , 2 15 , 17 cos sin cot csc 1 cos sec 122 y r x y sin cot ( 3) 4 , 5 4 x r 22 2 1 y r 22 5 , 3 tan y x 2 2 1, cot sec r x 2 2 2 2 x 1.190 and x 7 6 11 6 30. The equation is equivalent to tan x 1 1 x 1 tan ( 1) 4 , 3 5 4 32. Let 2 1 22 , 17 22 2 2 2 x 2 3 4 6 2 1 cos2 9 13 . Then sin2 1 19 tan 13 2 1 1 , 2 2 are x 6 is y so the solutions in 5 . 6 and x 92 13 88 . 13 9/13 9 88/13 sin cos so Therefore, 88 0.959. 33. (a) Using a graphing calculator with the sinusoidal regression feature, the equation is y 1.543 sin (2468.635x 0.494) 0.438. (b) The frequency is 2468.635 radians per second, which is equivalent to 2468.635 2 392.9 cycles per second (Hz). The note is a “G.” 2 12 34. (a) b 6 cos 1 1 , so 3 1 3 are x 1.911 and x 80 (c) k 30 2 80 30 2 the solution 1.911. 2 (when the temperature is 30 F) and its maximum at t 8 (when the temperature is 80 F). The value of h is 2 1.911. 25. 55 (d) The function should have its minimum at t Since the cosine function is even, the solutions in the x 9 , 13 and sin 2 [0, 0.01] by [ 2.5, 2.5] 28. This equation is equivalent to cos x interval 72 11 1 4.332. The solutions are 27. This equation is equivalent to sin x x so 0.771. 11 1 7 , 11 and cos (b) It’s half of the difference, so a in the interval 0 k , where k is any . Another solution in 4.332. x tan x is , k , where k is any 4 . Then 0 sin 11 cos tan sin 1, 11 sin 2 2 2 17 cos 72 11 26. The angle cos 1( 0.7) 2.346 is the solution to this equation in the interval 0 x . Since the cosine function is even, the value cos 1( 0.7) 2.346 is also a solution, so any value of the form cos 1( 0.7) 2k is a solution, where k is an integer. In 2 x 4 the solutions are x cos 1( 0.7) 2 8.629 and x cos 1( 0.7) 4 10.220. the interval 0 is integer. r y r y 2 1, so the . Since the period of y integer. This is equivalent to x y x x y 4 x 2 all solutions are of the form x sin cos csc 2k , where k 1.190 is the solution to this 2 is tan 1(2.5) x 2k or x 31. Let x r cos equation in the interval 0 , we 5 , 12 13 , 12 2, csc 25. The angle tan 1(2.5) 2 2. Then: 2 sin 2 In summary: tan r x ( 2)2 24. Note that r 17 8 5. Then: 3 , 5 sec and 7 and 6 2 are x sin x has period 2 , the solutions are solution in the interval 2 cos 3 , 4 1 sin csc x is any integer. 8 , 15 13. 2 11 . Since y 6 all of the form x sin cos 17 , 15 sin 5/13 5 12/13 12 cos 12 5 sin and cos . 13 13 12 5 , cos , tan 13 13 1 1 12 , sec 5 tan cos 1 13 5 sin 23. Note that r 15 . 17 tan Since tan have x 2 8 17 1 8 , 17 Therefore: sin 29. The solutions in the interval 0 2 sin2 1 cot 17. 8 2 5. Equation: y 25 sin 6 (x 5) 55 Section 1.6 (e) (b) Amplitude Period 35. (a) Amplitude 2 (c) Horizontal shift (d) Vertical shift (b) Average 365 (c) sin x (sin x) cos 4 101 37 37 62 ( 12) 2 (cos x) 4 1 2 1 62 F 12 F (cos x) sin 4 1 (sin x) 25 36. (a) Highest: 25 Lowest: 25 4 7 4 Vertical shift: 0 2 (2 /365) (b) Period 0.785 that is, or 5.498 that is, 37 2) 1.414 (that is, Horizontal shift [ 1, 13] by [ 10, 100] 35 2 (sin x cos x) 2 Therefore, sin x 43. (a) 25 F This average is the same as the vertical shift because 2 sin ax cos x 2 sin x 4 . 4 (b) See part (a). (c) It works. the average of the highest and lowest values of y (d) sin ax sin x is 0. 37. (a) cot ( x) cos ( x) sin ( x) (sin ax) cos (x) sin (x) cot (x) f(x) g(x) f ( x) g( x) f (x) g(x) so f g csc (x) Then f ( x) so 39. Assume that f is even and g is odd. Then f ( x)g( x) ( f (x))( g(x)) odd. cos (ax) 2 sin ax 4 tan . 1b a (c) It works. (d) sin x 1 f (x) cos ax) (b) See part (a). 1 sin (x) 1 f (x) 4 2 (sin ax 44. (a) One possible answer: y a 2 b 2 sin x (b) Assume that f is odd. 1 (cos ax) sin 4 1 is odd. The situation g f 38. (a) csc ( x) (cos ax) 2 1 So, sin (ax) is similar for . 1 sin ( x) 1 2 (b) Assume that f is even and g is odd. Then (sin ax) cos 4 1 f is odd. tan 1b a 1b sin (x) cos tan a sin (x) a2 1 f (x)g(x) so ( fg) is a2 b2 cos (x) sin tan a cos (x) b2 (a sin x 1b a b a2 b2 b cos x) 40. If (a, b) is the point on the unit circle corresponding to the angle , then ( a, b) is the point on the unit circle corresponding to the angle ( ) since it is exactly half way around the circle. This means that both tan ( ) and b tan ( ) have the same value, . and multiplying through by the square root gives the 41. (a) Using a graphing calculator with the sinusoidal regression feature, the equation is y 3.0014 sin (0.9996x 2.0012) 2.9999. sin tan a (b) y 3 sin ( x 2) 3 42. (a) desired result. Note that the substitutions cos tan 1b a a 1b a a2 b b2 a2 b2 and depend on the requirement that a is positive. If a is negative, the formula does not work. 45. Since sin x has period 2 , sin3 (x 2 ) sin3 (x). This function has period 2 . A graph shows that no smaller number works for the period. [ 2 , 2 ] by [ 2, 2] The graph is a sine/cosine type graph, but it is shifted and has an amplitude greater than 1. [ 2 , 2 ] by [ 1.5, 1.5] 36 Chapter 1 Review 46. Since tan x has period , tan (x ) tan x . This function has period . A graph shows that no smaller number works for the period. 9. Since 4x 3y 4 . 3 4 (x 3 4 x 3 y y [ 2 , 2 ] by [ 1, 5] 4) 12 20 3 10. Since 3x 30 4, the slope of the given line (and hence the slope of the desired line) is 2 47. The period is 60 4 x 3 12 is equivalent to y 5y 1 is equivalent to y . One possible graph: the given line is 3 5 3 x 5 1 , 5 the slope of and the slope of the perpendicular line is 5 . 3 5 (x 3 5 x 3 19 3 1 2 1 y 3 y y , 60 60 by [ 2, 2] 2 48. The period is 60 1 . 30 2) 11. Since x One possible graph: 3 1 is equivalent to y 3 2 of the given line is 3 x 2 and the slope of the perpendicular 2 3 line is . 2 (x 3 2 x 3 y y 11 , by [ 2, 2] 60 60 3(x 3x 1 (x 2 1 x 2 2. y y 3. x 4. m 1) 9 m 2 2 1 6 ( 3) y y 7. y 3) 8 4 2 2) 6 m y 2 y 53 2 23 5 2 (x 3) 3 5 2 21 5 5 3x 2 5 4 3 1 (4) 2 3 0 ( 2) 34 2 (x 4) 7 2 6 x 7 7 2 7 2 7 2 15. [ 3, 3] by [ 2, 2] 3 8. Since 2x y 2 is equivalent to y 2x 2, the slope of the given line (and hence the slope of the desired line) is 2. y 2(x 3) 1 y 2x 5 1 2 14. The line passes through (4, 2x y 2 4 4 ( 2) 1 (x 2 1 x 2 f (x) 5) and (3, 0) 5 3 5 2 2 f (x) 3 2 ( 5) 30 Check: f (4) 2(x 6. m 5 x 3 13. m 0 y 5. y 0 y ( 6) 1) 2 8 3 12. The line passes through (0, s Chapter 1 Review Exercises (pp. 52–53) 1. y y 1) 3, the slope Symmetric about the origin. 16. [ 3, 3] by [ 2, 2] Symmetric about the y-axis. 1, as expected. 2) and ( 3, 0). Chapter 1 Review 17. 29. (a) Since the square root requires 16 is [ 4, 4]. x2 37 0, the domain (b) For values of x in the domain, 0 16 x 2 0 16 x 2 4. The range is [0, 4]. 16, so (c) [ 6, 6] by [ 4, 4] Neither 18. [ 9.4, 9.4] by [ 6.2, 6.2] 30. (a) The function is defined for all values of x, so the domain is ( , ). [ 1.5, 1.5] by [ 0.5, 1.5] (b) Since 32 (1, ). Symmetric about the y-axis. 19. y( x) Even ( x)2 1 x2 20. y( x) Odd ( x)5 ( x)3 21. y( x) Even 1 22. y( x) sec ( x) tan ( x) cos( x) 1 y(x) 1 cos x attains all positive values, the range is (c) x5 ( x) x x3 x y(x) y(x) [ 6, 6] by [ 4, 20] sin ( x) cos2 ( x) 31. (a) The function is defined for all values of x, so the domain is ( , ). sin x cos2 x sec x tan x (b) Since 2e ( 3, ). y(x) x attains all positive values, the range is (c) Odd 23. y( x) Odd x4 x3 ( x)4 1 ( x)3 2( x) 24. y( x) 1 sin ( x) Neither even nor odd 1 26. y( x) Even ( x) 1 1 2x y(x) [ 4, 4] by [ 5, 15] sin x 25. y( x) x cos ( x) Neither even nor odd 4 x4 x3 1 2x x 32. (a) The function is equivalent to y cos x 2x x 4 1 k 4 x 27. (a) The function is defined for all values of x, so the domain is ( , ). (b) Since x attains all nonnegative values, the range is [ 2, ). k 2 tan 2x, so we require for odd integers k. The domain is given by for odd integers k. (b) Since the tangent function attains all values, the range is ( , ). (c) (c) , 22 [ 10, 10] by [ 10, 10] 28. (a) Since the square root requires 1 ( , 1]. x 0, the domain is by [ 8, 8] 33. (a) The function is defined for all values of x, so the domain is ( , ). (b) Since 1 x attains all nonnegative values, the range is [ 2, ). (b) The sine function attains values from 1 to 1, so 2 2 sin (3x ) 2, and hence 3 2 sin (3x ) 1 1. The range is [ 3, 1]. (c) (c) [ 9.4, 9.4] by [ 3, 3] [ , ] by [ 5, 5] 38 Chapter 1 Review 34. (a) The function is defined for all values of x, so the domain is ( , ). 5 x 2, which attains (b) The function is equivalent to y all nonnegative values. The range is [0, ). (c) 39. First piece: Line through (0, 1) and (1, 0) m 0 1 y 1 0 x 1 1 1 1 or 1 x Second piece: Line through (1, 1) and (2, 0) m 1 1 y [ 8, 8] by [ 3, 3] 35. (a) The logarithm requires x (3, ). 0 2 3 0, so the domain is (b) The logarithm attains all real values, so the range is ( , ). (c) (x y x 1 1 1) 1 2 or 2 1 2 f (x) 1 x, x, x 0 1 x x 1 2 40. First piece: Line through (0, 0) and (2, 5) m y 5 2 5 x 2 0 0 5 2 Second piece: Line through (2, 5) and (4, 0) [ 3, 10] by [ 4, 4] 36. (a) The function is defined for all values of x, so the domain is ( , ). (b) The cube root attains all real values, so the range is ( , ). m y y (c) f (x) 0 4 5 2 5 2 5 (x 2 5 x 2 5x , 2 2) 5 5x 2 10 or 10 0 x 2 2 5x , 2 10 (Note: x 5 2 x 4 2 can be included on either piece.) [ 10, 10] by [ 4, 4] 37. (a) The function is defined for is [ 4, 4]. 4 x 4, so the domain x , 4 x 4, (b) The function is equivalent to y which attains values from 0 to 2 for x in the domain. The range is [0, 2]. 41. (a) ( f g)( 1) f (g( 1)) (b) (g f )(2) g( f (2)) f ( f (x)) f (d) (g g)(x) g(g(x)) [ 6, 6] by [ 3, 3] x 2, so the domain x 42. (a) ( f g)( 1) (b) See the graph in part (c). The range is [ 1, 1]. 3 f (0) 0 1 1/ x 2 2 2 2 1 1) 2 (b) (g f )(2) g( f (2)) (c) ( f f )(x) [ 3, 3] by [ 2, 2] x, x f ( g( 1)) f( (c) 2 5 or 2 2 1 2.5 1 x x 1 2 1 1/x g 4 1 1/2 + 2 1 x (c) ( f f )(x) 2 1 1 2 g 1 1 f (1) 1 (c) 38. (a) The function is defined for is [ 2, 2]. 1 f f ( f (x)) f (2 (d) (g g)(x) g(g(x)) g( 0 g(2 2 2) x) 3 x 2 1) 3 g(0) 3 3 0 1 x) (2 x x 1 1 1 Chapter 1 Review 43. (a) ( f g)(x) f (g(x)) f( x 2) (x 2 2) x, x (g f )(x) cos2 t 2 x 4 2 x 2) y 4 2 y 4 in the identity 1 gives the Cartesian equation 1, or x 2 y2 16. The left half of the 47. (a) 2 (2 2 sin2 t and sin t circle is traced by the parametrized curve. g( f (x)) g(2 x 4 (b) Substituting cos t 39 x) 2 4 x 2 (b) Domain of f g: [ 2, ) Domain of g f: [ 2, 2] [ 8, 8] by [ 10, 20] (c) Range of f g: ( , 2] Range of g f: [0, 2] 44. (a) ( f g)(x) Initial point: (4, 15) Terminal point: ( 2, 3) The line segment is traced from right to left starting at (4, 15) and ending at ( 2, 3). f (g(x)) f( 1 x) 1 4 (g f )(x) 1 (b) Substituting t 2 x into y 11 2t gives the Cartesian equation y 11 2(2 x), or y 2x 7. The part of the line from (4, 15) to ( 2, 3) is traced by the parametrized curve. x x g( f (x)) g( x) 1 x 48. (a) (b) Domain of f g: ( , 1] Domain of g f: [0, 1] (c) Range of f g: [0, ) Range of g f: [0, 1] [ 8, 8] by [ 4, 6] Initial point: None Terminal point: (3, 0) The curve is traced from left to right ending at the point (3, 0). 45. (a) [ 6, 6] by [ 4, 4] Initial point: (5, 0) Terminal point: (5, 0) The ellipse is traced exactly once in a counterclockwise direction starting and ending at the point (5, 0). x 5 (b) Substituting cos t cos2 t x 5 2 sin2 t y 2 and sin t y 2 in the identity 1 gives the Cartesian equation 2 1. (b) Substituting t x 1 into y 4 2t gives the Cartesian equation y 4 2(x 1), or y 6 2x. The entire curve is traced by the parametrized curve. 49. (a) For simplicity, we assume that x and y are linear functions of t, and that the point (x, y) starts at ( 2, 5) for t 0 and ends at (4, 3) for t 1. Then x f (t), where f (0) 2 and f (1) 4. Since slope x f (t) x t 4 1 6t ( 2) 0 2 6, 2 6t. 46. (a) g(t), where g(0) slope The entire ellipse is traced by the curve. Also, y y t y g(t) 3 1 5 0 2t 5 5 and g(1) 2, 5 2t. One possible parametrization is: x 2 6t, y 5 2t, 0 t [ 9, 9] by [ 6, 6] Initial point: (0, 4) 3 Terminal point: None (since the endpoint is not 2 included in the t-interval) The semicircle is traced in a counterclockwise direction starting at (0, 4) and extending to, but not including, (0, 4). 1 3. Since 40 Chapter 1 Review 50. For simplicity, we assume that x and y are linear functions of t and that the point (x, y) passes through ( 3, 2) for t 0 and (4, 1) for t 1. Then x f (t), where f (0) 3 and f (1) 4. Since x t slope = x f (t) 4 ( 3) 0 1 7t 7, 3 3 7t. Also, y Since g(t), where g(0) slope y t y g(t) 1 1 t 2 and g(1) ( 2) 0 2 1. 1 2 t. One possible parametrization is: x 3 7t, y 2 t, t . 51. For simplicity, we assume that x and y are linear functions of t and that the point (x, y) starts at (2, 5) for t 0 and passes through ( 1, 0) for t 1. Then x f (t), where f (0) 2 and f (1) 1. Since x t slope 12 10 3, x Also, y g(t), where g(0) slope y t 0 1 5 0 f (t) g(t) 3x 3t. 5 5 [ 6, 12] by [ 4, 8] 5t. 55. Using a calculator, sin 1(0.6) 36.8699 . sin 2 40 , 7 sin x 2 (x) x csc 1 )(x) 1 f(f f 2 x 3 3 2 1 f )(x) 2 x (2 f 1 3 x) 1 f ( f (x)) (2 2 (2 3 3x 3 3 40 7 3x) 3x) x Interchange x and y. 40 , 3 7 , 3 0.2 x y y = f (–x) x (3, 0) [ 6, 6] by [ 4, 4] x , sec sin cos 1 cos 60. (a) The given graph is reflected about the y-axis. (–1, 0) 2)2, x x2 y2 tan 59. e 0.2x 4 ln e 0.2x ln 4 0.2x ln 4 ln 4 x 5 ln 4 –3 y 40 . 7 (b) Since the period of sin x is 2 , the solutions are x 3.3430 2k and x 6.0818 2k , k any integer. (b) (x 40 49 40 3 (0, 2) 54. (a) y 32 7 1 58. (a) Note that sin 1( 0.2) 0.2014. In [0, 2 ), the solutions are x sin 1( 0.2) 3.3430 and x sin 1( 0.2) 2 6.0818. (x)) 2 (f , 3 , 7 cos 1 tan 1 sin cot 3 Verify. (f f 1.1607 radians or Therefore, 3 1 and 0 cos2 1 Interchange x and y. f 0.6435 radians or y 3 y 3 7 57. Since cos y 2 x 2 56. Using a calculator, tan 1( 2.3) 66.5014 . 3x 2 2 (b) 0. Since 5t 52. One possible parametrization is: x t, y t(t 4), t 2. 2 3t 5 and g(1) 5, y One possible parametrization is: x 2 3t, y 5 5t, t 0. 53. (a) y x2 y f 1(x) x2 Verify. For x 0 (the domain of f 1) ( f f 1)(x) f ( f 1(x)) f ( x 2) [( x 2) 2]2 ( x)2 x For x 2 (the domain of f ), ( f 1 f )(x) f 1( f (x)) f 1((x 2)2) (x 2)2 2 x2 2 (x 2) 2 x 2 –3 Chapter 1 Review (b) The given graph is reflected about the x-axis. y 62. (a) V (b) 100,000 10,000x, 0 x 41 10 V 55,000 10,000x 55,000 10,000x 45,000 x 4.5 The value is $55,000 after 4.5 years. 100,000 3 (1, 0) (–3, 0) –3 3 x 63. (a) f (0) (b) f (2) (c) y = – f (x ) (0, –2) –3 (c) The given graph is shifted left 1 unit, stretched vertically by a factor of 2, reflected about the x-axis, and then shifted upward 1 unit. 90 units 90 52 ln 3 32.8722 units [0, 4] by [ 20, 100] y y = –2f (x + 1) + 1 (–4, 1) 64. 1500(1.08)t 3 1.08t (0, 1) –4 5000 1500 10 3 ln (1.08)t ln t ln 1.08 x 2 ln t (–1, –3) (d) The given graph is shifted right 2 units, stretched vertically by a factor of 3, and then shifted downward 2 units. 4 y y = 3f (x – 2) – 2 (2, 4) 10 3 10 3 ln (10/3) ln 1.08 t 15.6439 It will take about 15.6439 years. (If the bank only pays interest at the end of the year, it will take 16 years.) 4 2t 65. (a) N(t) (b) 4 days: 4 24 1 week: 4 27 (c) N(t) 4 2t 2t ln 2t t ln 2 x –2 4 (–1, –2) (3, –2) 61. (a) 5000 y (3, 2) –3 3 ln 500 ln 2 2000 2000 500 ln 500 ln 500 8.9658 There will be 2000 guppies after 8.9658 days, or after nearly 9 days. (d) Because it suggests the number of guppies will continue to double indefinitely and become arbitrarily large, which is impossible due to the finite size of the tank and the oxygen supply in the water. 3 (–3, 2) t 64 guppies 512 guppies x 66. (a) y 20.627x 338.622 (–1, –1) (1, –1) –3 (b) y [0, 30] by [ 100, 1000] 3 (3, 2) (–1, 1) –3 3 (1, –1) (–3, –2) –3 x (b) When x 30, y 957.445. According to the regression equation, about 957 degrees will be earned. (c) The slope is 20.627. It represents the approximate annual increase in the number of doctorates earned by Hispanic Americans per year. 67. (a) y 14.60175 1.00232x (b) Solving y 25 graphically, we obtain x 232. According to the regression equation, the population will reach 25 million in the year 2132. (c) 0.232% 42 Section 2.1 Chapter 2 Limits and Continuity 4. (a) lim p(s) (b) lim p(s) (pp. 55–65) 3 5(2) 2 2. f (2) 5 4(2) 23 3. f (2) sin 4 2 2 sin 0 (b) lim F(x) 2 (c) lim F(x) does not exist, because the left- and right-hand limits are not equal. 4 2 3 1 x x 8. x d2 d2 x c 3x x 4 6. (a) lim G(x) 1 (b) lim G(x) 3 2 5 1 (c) lim G(x) 1 x→2 3 x→2 d2 c x2 (d) F(0) c2 x 7. x –3 x→0 c2 c 4 x→0 x→0 x 6. x d2 d2 c x 18 (x 2x x 2x2 x 1 c 3)(x 6) x3 3 2 10. 5. (a) lim F(x) 1 3 1 3 0 4 4 9. 4 11 12 1 22 (d) p( 2) 2 2(2 ) 5. x 3 s→−2 s→ 2 Quick Review 2.1 4. f (2) 3 (c) lim p(s) s Section 2.1 Rates of Change and Limits 1. f (2) 3 s→ 2 x(2x 1) (2x 1)(x 1) x→2 x 6, x x x 1 ,x 3 1 2 Section 2.1 Exercises 1. (a) lim f (x) (d) G(2) 3 7. lim 3x 2(2x x→ 1/2 1) 3 12 2 2 1 2 1 3 1 4 ( 2) 3 2 Graphical support: 3 x→3 (b) lim f (x) –2 x→3 (c) lim f (x) does not exist, because the left- and [ 3, 3] by [ 2, 2] x→3 right-hand limits are not equal. (d) f (3) = 1 8. lim (x x→ 4 3)1998 (4 3)1998 ( 1)1998 1 Graphical support: 2. (a) lim g(t) 5 (b) lim g(t) 2 t→ 4 t→ 4 (c) lim g(t) does not exist, because the left- and [ 4.001, 3.999] by [0, 5] t→ 4 right-hand limits are not equal. (d) g( 4) 9. lim (x 3 x→1 3x 2 2x 3. (a) lim f (h) 4 (b) lim f (h) Graphical support: 4 h→0 h→0 –4 h→0 (d) f (0) 4 (1)3 1 2 (c) lim f (h) 17) [ 3, 3] by [ 25, 25] 3(1)2 3 2 2(1) 17 17 15 43 Section 2.1 10. lim y2 5y y y→2 22 6 2 2 5(2) 2 6 20 4 5 16. lim ln (sin x) Graphical support: 11. lim y 4y [ 2 ( 3) 4( 3) ( 3)2 3 3 y2 y→ 3 3 ln 1 2 0 Graphical support: [ 3, 3] by [ 5, 10] 2 ln sin x→ /2 3 0 6 , ] by [ 3, 1] 17. You cannot use substitution because the expression x is not defined at x 2. Since the expression is not defined at points near x 2, the limit does not exist. 0 Graphical support: 18. You cannot use substitution because the expression not defined at x 0. Since 1 x2 2 1 is x2 becomes arbitrarily large as x approaches 0 from either side, there is no (finite) limit. (As [ 5, 5] by [ 5, 5] 12. lim int x int x→1/2 1 2 we shall see in Section 2.2, we may write lim 1 .) 2 x→0 x 0 Note that substitution cannot always be used to find limits of the int function. Its use here can be justified by the Sandwich Theorem, using g(x) h(x) 0 on the interval (0, 1). Graphical support: 19. You cannot use substitution because the expression defined at x 0. Since lim x→0 x x x x 1 and lim x→0 x x is not 1, the left- and right-hand limits are not equal and so the limit does not exist. 20. You cannot use substitution because the expression (4 (4 [ 4.7, 4.7] by [ 3.1, 3.1] 13. lim (x x→ 2 6) 2/3 (2 6) 2/3 3 ( 8) 2 3 64 4 x)2 x x)2 x 16 is not defined at x 16 x2 8x 0. Since and is equal to lim (8 x→0 8 x for all x x) x 8 0 0, the limit exists 8. Graphical support: 21. [ 4.7, 4.7] by [ 3.1, 3.1] [ 10, 10] by [ 10, 10] 14. lim x→2 x 3 2 3 5 lim 1 1 x 2 x→1 x 1 2 Algebraic confirmation: Graphical support: lim 1 1 x 2 x→1 x lim x→1 (x x1 1)(x 1) lim 1 x→1 x 1 2 1 1 1 1 22. [ 4.7, 4.7] by [ 3.1, 3.1] 15. lim (e xcos x) x→0 e 0 cos 0 Graphical support: 11 1 [ 4.7, 4.7] by [ 3.1, 3.1] lim t2 3t t2 t→2 2 4 1 4 Algebraic confirmation: lim [ 4.7, 4.7] by [ 3.1, 3.1] t→2 t2 3t t2 2 4 (t lim (t t→2 1)(t 2)(t 2) 2) lim t t→2 t 1 2 2 2 1 2 1 4 44 Section 2.1 27. 23. [ 4.7, 4.7] by [ 3.1, 3.1] 8x2 16x2 5x3 lim 4 x→0 3x [ 4.7, 4.7] by [ 3.1, 3.1] 1 2 Algebraic confirmation: 8x2 16x2 5x3 3x4 x→0 lim sin x lim 2 x→0 2x 1 x Algebraic confirmation: x2(5x 8) x2(3x2 16) x→0 5x 8 lim 2 16 x→0 3x 5(0) 8 3(0)2 16 8 1 16 2 lim sin x lim 2 x→0 2x lim x x→0 sin x lim x→0 x sin x x 1 2x 1 1 lim x→0 2x (1) 1 1 2(0) 1 28. 24. [ 4.7, 4.7] by [ 3.1, 3.1] 1 2 1 lim 2 x x→0 lim Algebraic confirmation: 1 2 1 2 lim x 2 x x→0 sin x x lim 1 x→0 sin x x lim 1 2 (2 x) x→0 x(2 x)(2) x lim x→0 x(2 x)(2) 1 lim x→0 2(2 x) 1 1 2(2 0) 4 lim x x→0 sin x x Algebraic confirmation: 1 4 x x→0 x lim [ 4.7, 4.7] by [ 3.1, 3.1] lim x→0 1 1 x→0 sin x x 2 29. 25. [ 4.7, 4.7] by [ 3.1, 3.1] sin2 x x→0 x lim [ 4.7, 4.7] by [ 5, 20] lim (2 x→0 x)3 x 8 Algebraic confirmation: sin2 x x→0 x 12 lim Algebraic confirmation: lim (2 x→0 3 x) x 8 lim lim sin x x→0 lim sin x 12x x→0 6x x lim (12 2 x 6(0) lim x→0 3 (sin 0)(1) x→0 sin x x sin x x 0 2 6x x→0 12 0 x) 2 (0) 30. 12 26. [ 2, 2] by [ 10, 10] lim 3 sin 4x x→0 sin 3x 4 [ 4.7, 4.7] by [ 3.1, 3.1] Algebraic confirmation: sin 2x lim x x→0 x→0 sin 3x 2 lim Algebraic confirmation: lim x→0 sin 2x x 2 lim x→0 sin 2x 2x 3 sin 4x 4 lim x→0 2(1) 2 4(1) 3x sin 4x sin 3x 4x sin 3x lim x→0 3x 4 lim x→0 sin 4x 4x (1) 4 1 45 Section 2.1 31. (a) True 42. Since (b) True (c) False, since lim f (x) x→0 x x 1 for x 43. (a) lim (g(x) 0. x→4 (b) lim x f (x) (d) True, since both are equal to 0. x→4 (e) True, since (d) is true. (c) lim g2(x) x→4 (f) True x x 0, lim x→0 3) 1. lim g(x) lim 3 x→4 lim x lim f (x) x→4 x→4 2 lim g(x) x→4 3 x→4 40 32 3 6 0 9 lim g(x) (g) False, since lim f (x) x→0 0. (d) (h) False, lim f (x) 1, but lim f (x) is undefined. (i) False, lim f (x) x →4 g(x) lim x→4 f (x) 1 lim f (x) 0, but lim f (x) is undefined. x→1 x→1 x→1 44. (a) lim (f (x) x→b g(x)) 0 x→4 lim f (x) 1 3 lim g(x) x→b 7 3 lim 1 x→4 x→b ( 3) 4 x→1 (j) False, since lim f (x) x→2 (b) lim ( f (x) g(x)) x→b 0. lim f (x) lim g(x) x→b x→b (7)( 3) 21 32. (a) True (b) False, since lim f (x) (c) False, since lim f (x) (c) lim 4g(x) 1. 1. x→2 x→2 (d) True x→b f (x) x→b g(x) (d) lim 4lim g(x) 4( 3) x→b lim f (x) x→b 7 7 lim g(x) x→b 12 3 3 (e) True 45. (a) (f) True, since lim f (x) x→1 lim f (x). x→1 (g) True, since both are equal to 0. (h) True (i) True, since lim f (x) x→c 33. y1 x 2 x x 2 (x 1 1 for all c in (1, 3). 1)(x 2) x1 x 2, x [ 3, 6] by [ 1, 5] (b) lim f (x) 1 x→2 2; lim f (x) x→2 1 (c) No, because the two one-sided limits are different. (c) 46. (a) 34. y1 x2 x x 2 (x 1 1)(x 2) x1 (b) 35. y1 x2 x 2x 1 1 (x x 1)2 1 x 1, x [ 3, 6] by [ 1, 5] 1 (b) lim f (x) (d) 36. y1 x→2 1; lim f (x) x→2 1 (c) Yes. The limit is 1. x2 x x 2 (x 1 1)(x 2) x1 47. (a) (a) 37. Since int x 38. Since int x 39. Since int x 40. Since int x 41. Since x x 0 for x in (0, 1), lim int x x→0 0. [ 5, 5] by [ 4, 8] 1 for x in ( 1, 0), lim int x x→0 0 for x in (0, 1), lim int x x→0.01 1 for x in (1, 2), lim int x x→2 1 for x 0, lim x0 x x 1. (b) lim f (x) x→1 0. 4; lim f (x) does not exist. x→1 (c) No, because the left-hand limit does not exist. 48. (a) 1. 1. [ 4.7, 4.7] by [ 3.1, 3.1] 46 Section 2.1 48. continued 54. f (x) (b) lim x→ 1 0; lim f (x) x→ 1 0 (c) Yes. The limit is 0. 49. (a) [ 4.7, 4.7] by [ 5, 5] lim (x 2 sin x) 0 x→0 Confirm using the Sandwich Theorem, with g(x) = h(x) x 2. x 2 sin x x 2 sin x x 2 1 x 2. 2 2 2 x x sin x x [ 2 , 2 ] by [ 2, 2] (b) ( 2 , 0) (c) c (0, 2 ) 2 (d) c Because lim ( x 2) lim x 2 x→0 2 2 x→0 gives lim (x sin x) 50. (a) x 2 and 0, the Sandwich Theorem 0 x→0 55. [ (b) , ] by [ 3, 3] , 2 2 [ 0.5, 0.5] by [ 0.25, 0.25] , (c) c lim x 2 sin (d) c x→0 51. (a) 1 x2 0 Confirm using the Sandwich Theorem, with g(x) and h(x) x 2. 1 1 x 2 sin 2 x 2 sin 2 x 2 1 x 2. x x (c) c x 1 x sin 2 x 2 x 2 Because lim ( x 2) [ 2, 4] by [ 1, 3] (b) (0, 1) 2 x→0 2 (1, 2) give lim x sin 2 (d) c x→0 0 x2 lim x 2 x→0 1 x2 0, the Sandwich Theorem 0. 56. 52. (a) [ 0.5, 0.5] by [ 0.25, 0.25] [ 4.7, 4.7] by [ 3.1, 3.1] (b) ( , 1) ( 1, 1) (1, ) (c) None lim x 2 cos x→0 (d) None 1 x2 0 Confirm using the Sandwich Theorem, with g(x) and h(x) x 2. 1 1 x2 cos 2 x2 cos 2 x2 1 x2. 53. x x [ 4.7, 4.7] by [ 3.1, 3.1] lim (x sin x) x→0 x→0 2 give lim x cos Confirm using the Sandwich Theorem, with g(x) and h(x) x . x sin x x sin x x x sin x x x→0 gives lim (x sin x) x→0 x 1 x cos 2 x 2 x lim x x→0 0. x Because lim ( x 2) 0 Because lim ( x ) 2 1 x x 0, the Sandwich Theorem x→0 x2 1 x2 2 lim x 2 x →0 0, the Sandwich Theorem 0. 57. (a) In three seconds, the ball falls 4.9(3)2 44.1 average speed is 3 14.7 m/sec. 44.1 m, so its Section 2.1 (b) The average speed over the interval from time t time 3 h is 4.9(3 h)2 (3 h) y t 29.4 4.9(3)2 3 3 to h2) 4.9(6h h 63. (a) Because the right-hand limit at zero depends only on the values of the function for positive x-values near zero. 1 1 sin (base)(height) (1)(sin ) 2 2 2 (angle)(radius)2 (1)2 Area of sector OAP 2 2 2 1 1 tan Area of OAT (base)(height) (1)(tan ) 2 2 2 (b) Area of 4.9h Since lim (29.4 4.9h) h→0 47 29.4, the instantaneous speed is 29.4 m/sec. OAP (c) This is how the areas of the three regions compare. 58. (a) y gt (d) Multiply by 2 and divide by sin . 2 g(42) (e) Take reciprocals, remembering that all of the values involved are positive. 20 16 20 (f) The limits for cos 5 or 1.25 4 20 (b) Average speed 5 m/sec 4 g sin (c) If the rock had not been stopped, its average speed over the interval from time t 4 to time t 4 h is 1.25(4 h)2 (4 h) y t 10 1.25(4)2 4 sin ( is between them, it must also have a limit of 1. ) sin sin ( ) (h) If the function is symmetric about the y-axis, and the right-hand limit at zero is 1, then the left-hand limit at zero must also be 1. h2) 1.25(8h h (g) and 1 are both equal to 1. Since 1.25h (i) The two one-sided limits both exist and are equal to 1. Since lim (10 h→0 1.25h) 10, the instantaneous speed 64. (a) The limit can be found by substitution. is 10 m/sec. lim f (x) x→2 59. (a) x 0.1 0.01 0.001 0.0001 f (x) 0.054402 0.005064 0.000827 0.000031 f (2) (b) The graphs of y1 shown. 3(2) f (x), y2 2 4 2 1.8, and y3 2.2 are (b) 0.1 x f (x) 0.01 0.054402 0.001 0.005064 0.0001 0.000827 0.000031 [1.5, 2.5] by [1.5, 2.3] The limit appears to be 0. 60. (a) x 0.1 f (x) (b) 0.8269 0.3056 0.01 0.1 f (x) 0.001 0.5064 0.5440 x 0.01 0.0001 0.001 0.0001 0.5064 0.8269 0.5440 0.3056 The intersections of y1 with y2 and y3 are at x 1.7467 and x 2.28, respectively, so we may choose any value of a in [1.7467, 2) (approximately) and any value of b in (2, 2.28]. One possible answer: a 1.75, b 2.28. (c) The graphs of y1 shown. f (x), y2 1.99, and y3 2.01 are There is no clear indication of a limit. 61. (a) x 0.1 0.01 0.001 f (x) 2.0567 2.2763 2.2999 (b) x 0.1 0.01 0.001 0.0001 2.3023 0.0001 [1.97, 2.03] by [1.98, 2.02] f (x) 2.5893 2.3293 2.3052 2.3029 The limit appears to be approximately 2.3. 62. (a) x 0.1 0.01 f (x) 0.074398 0.001 0.0001 0.009943 0.000585 0.000021 (b) x f (x) 0.1 0.074398 0.01 0.009943 The limit appears to be 0. 0.001 0.000585 0.0001 0.000021 The intersections of y1 with y2 and y3 are at x 1.9867 and x 2.0134, respectively, so we may choose any value of a in [1.9867, 2) and any value of b in (2, 2.0134] (approximately). One possible answer: a 1.99, b 2.01 48 Section 2.2 65. (a) f sin 6 s Section 2.2 Limits Involving Infinity 1 2 6 (b) The graphs of y1 shown. f (x), y2 0.3, and y3 0.7 are (pp. 65–73) Exploration 1 Exploring Theorem 5 1. Neither lim f (x) or lim g(x) exist. In this case, we can x→ x→ describe the behavior of f and g as x → lim f (x) [0, 1] by [0, 1] The intersections of y1 with y2 and y3 are at x and x 0.3047 0.7754, respectively, so we may choose any value of a in 0.3047, 6 and lim g(x) x→ One possible answer: a 0.305, b f (x), y2 sin x x sin x x lim 5 x→ 5 0 5, so the limit of the quotient exists. 2. Both f and g oscillate between 0 and 1 as x → , taking on each value infinitely often. We cannot apply the sum rule because neither limit exists. However, 0.775 0.49, and y3 5x lim approximate. (c) The graphs of y1 shown. Example 5, x→ , 0.7754 , where the interval endpoints are . We cannot apply the x→ quotient rule because both limits must exist. However, from and any value of b in 6 by writing 0.51 are lim (sin2 x cos2 x) x→ lim (1) x→ 1, so the limit of the sum exists. 3. The limt of f and g as x → apply the difference rule to f lim f (x) [0.49, 0.55] by [0.48, 0.52] The intersections of y1 with y2 and y3 are at x and x 6 6 , and any value of b in , 0.5352 , where the interval endpoints are approximate. One possible answer: a 0.513, b 0.535 x→ f (x) 0.5121 0.5352, respectively, so we may choose any value of a in 0.5121, lim g(x) x→ g(x) 66. Line segment OP has endpoints (0, 0) and (a, a ), so its midpoint is 2 a a a , 2 y y 0 a2 a0 , 2 a a2 =, and its slope is 22 2 0 a. The perpendicular bisector is the line through 0 2 a 1 with slope , so its equation is 2 a 2 1 a a x , which is equivalent to a 2 2 2 1a 1 a2 1 x . Thus the y-intercept is b . As a 2 2 P→0 a→0 a2 1 2 02 1 2 1 . 2 1) 4. The fact that the limits of f and g as x → ln 2x x 1 . do not exist does not necessarily mean that the limits of f g, f f do not exist, just that Theorem 5 cannot be applied. g 1. y 2x 3 y 3 2x y 3 2 x Interchange x and y. x 3 2 f 1 (x) y x 3 2 value of a approaches zero. Therefore, lim ln (x We can use graphs or tables to convince ourselves that this limit is equal to ln 2. the point P approaches the origin along the parabola, the lim b g. We can say that . We can write the difference as ln (2x) Quick Review 2.2 2 do not exist, so we cannot [ 12, 12] by [ 8, 8] g or Section 2.2 2. y e x ln y x Interchange x and y. ln x y f 1(x) ln x 7. (a) f ( x) 1 (b) f x cos ( x) 8. (a) f ( x) 1 (b) f x ex 1/x e 1 x ( x) e ln ( x) x 9. (a) f ( x) (b) f ln 1/x 1/x ln ( x) x x ln x [ 6, 6] by [ 4, 4] 10. (a) f ( x) 1 3. y tan x tan y x x, y 2 2 (b) f Interchange x and y. tan x 1 f y, (x) x 2 tan x, 2 x 2 1 x 1 x 1 1/x 1. [ 5, 5] by [ 1.5, 1.5] 4. y cot 1x cot y x, 0 x Interchange x and y. cot x y, 0 y f 1(x) cot x, 0 x (a) lim f (x) x→ (b) lim f (x) x→ (c) y [ 6, 6] by [ 4, 4] 1 1 1 2. 2 3 3x 3 0x2 8 x 3 10 3 5 x 3 7 3 1 [ 10, 10] by [ 1, 1] 2 3 q(x) 3x 2 r(x) 6. q(x) r(x) x 3x2 x 3x 2 2x3 3 5 2x 3 4x x 2 5 x 3 2x2 1 2x 5 2x5 2x 2 x2 2x x 7 3 2x 1 0x 4 x3 4 2x 0x3 2x4 x3 4 2x 2x3 x3 x3 1 2 (a) lim f (x) 0x 2 2x2 2x2 0x2 2x2 x2 x2 x 1 x→ (b) lim x1 2x x1 0x 1 x2 f (x) x→ (c) y 0 x ln x x 1 ( sin x) x 1 sin x x sin Section 2.2 Exercises 2 1 1 sin ( x) x x [ 6, 6] by [ 4, 4] 5. cos x 1 cos x 0 0 1 x 1 x x sin 1 x 49 50 Section 2.2 3. 6. [ 5, 5] by [ 10, 10] (a) lim f (x) x→ [ 20, 20] by [ 4, 4] (a) lim f (x) x→ 0 (b) lim f (x) 2 x→ (b) lim f (x) (c) y x→ (c) y 2 0 2, y 2 7. 4. [ 5, 5] by [ 2, 2] [ 10, 10] by [ 100, 300] (a) lim f (x) x→ 1 (b) lim f (x) (a) lim f (x) x→ 1 x→ (c) y (b) lim f (x) 1, y 1 8. x→ (c) No horizontal asymptotes. 5. [ 5, 5] by [ 2, 2] [ 20, 20] by [ 4, 4] (a) lim f (x) x→ (b) lim f (x) (a) lim f (x) x→ x→ 3 (c) y (b) lim f (x) 3 x→ (c) y 3, y 3 1 1 1 Section 2.2 9. 13. [ 2, 6] by [ 1, 5] lim x→2 [ 4, 4] by [ 3, 3] 1 x2 10. lim x→0 int x x 0 14. [ 2, 6] by [ 3, 3] lim [ 4, 4] by [ 3, 3] x lim x→2 x 2 11. x→0 int x x 15. [ 7, 1] by [ 3, 3] lim [ 3, 3] by [ 3, 3] 1 lim csc x x→ 3 x 3 12. x→0 16. [ 7, 1] by [ 3, 3] lim x x→ 3 x 3 [ lim x→( /2) , ] by [ 3, 3] sec x 51 52 Section 2.2 17. 22. [ 4, 4] by [ 3, 3] (a) x 2, x [ 2 , 2 ] by [ 3, 3] 2 (a) x (b) Left-hand limit at 2 is . Right-hand limit at 2 is Left-hand limit at 2 is . Right-hand limit at 2 is . . (b) If n is even: Left-hand limit is . Right-hand limit is 2(x 1) x x x2 x2 x1 x 1 5 x2 5 x2 2 x2 x 3 2x 2 2 3 15 x x x 2 5x 5 2 23. y x x [ 7, 5] by [ 5, 3] 2 An end behavior model for y is (b) Left-hand limit at 2 is . Right-hand limit at 2 is . lim y 19. lim 1 x→ lim y 2 x 24. y [ 6, 6] by [ 12, 6] 5x3 1 (b) Left-hand limit at 1 is . Right-hand limit at 1 is . lim 1 lim y 5x2 1 x2 2 10x x2 x3 2 1 lim 5 x→ x x 5x2 1 x2 lim y 5x3 x3 5. 5 x→ x→ 1. 1 x→ An end behavior model for y is 20. x3 x3 1 x→ x→ (a) x . If n is odd: Left-hand limit is . Right-hand limit is . 18. (a) x n , n any integer 2 lim 5 5 x→ 25. Use the method of Example 10 in the text. [ 2, 4] by [ 2, 2] 1 ,x 3 2 cos (a) x lim x→ 1 (b) Left-hand limit at is . 2 1 Right-hand limit at is 2 . Right-hand limit at 3 is cos lim x→ Left-hand limit at 3 is . . 1 1 x 1 x 1 1 x 1 x x→ lim 2x lim 2 x→ Similarly, lim y x→ [ 2 , 2 ] by [ 3, 3] (a) x k , k any integer (b) at each vertical asymptote: Left-hand limit is . Right-hand limit is . cos (0) 10 1 1 1 cos x x cos (0) 10 1 1 1 0 2. x→0 1 26. Note that y So, lim y 21. cos x x lim x→0 1 sin x x lim x→ 2. 2 sin x . x sin x x 2 Section 2.2 sin x 2x 2 x sin x lim 0 x x→ 1 lim 0 2x 1 x→ sin x x 27. Use y So, lim y lim 0. x is both a right end behavior lim ln x x 1 x→ 1 lim 0. x→ x→ x2 sin x x2 2x3 x lim x→ 1 sin x x2 2x 2. (a) 30. An end behavior model is x5 2x2 0.5x 3. (c) 31. An end behavior model is 2x4 x 2x 3. (d) x x2 32. An end behavior model is [ 4, 4] by [ 1, 3] x 2. (b) lim f (x) x→ 2 lim f (x) x→ (b) None f 1 1/x e is shown. x 0 44. 4x 3 34. (a) 1 x 1 lim f x x→0 1 lim f x x→0 The graph of y 4 (b) None 35. (a) x 2x2 (b) y 36. (a) 1 2x 3x2 x2 (b) y 1 x 1 lim f (x) lim f x x→ x→0 1 lim f (x) lim f x x→ x→0 The graph of y 3 3 3 37. (a) [ 4, 4] by [ 1, 3] 0 4x x 4x 2 x4 x2 1 1/x e is shown. x2 0 45. (b) None 38. (a) f x2 (b) None because lim ex 2x x→ e (b) The function y ex 2x 2x e x 1 0 1 1. f x ln 1 is shown. x 0 0 46. ex x2 lim 1 x→ e 2 lim (x e x→ x2 (b) The function y x→ ex 2x lim 1 x 1 lim f (x) lim f x x→ x→0 1 lim f (x) lim f x x→ x→0 The graph of y 1. x is a right end behavior model x2 x→ x→ 2x 0 2 40. (a) The function y because lim 1 2x is a left end behavior model x→ because lim 2x ex lim 1 x x→ because lim [ 3, 3] by [ 2, 2] e x is a right end behavior model 39. (a) The function y x e e 1) x 1 0 is a left end behavior model 2 x lim x x→ 0 1 x e 1. x 1 1. [ 5, 5] by [ 1.5, 1.5] 1 x 1 lim f (x) lim f x x→ x→0 1 lim f (x) lim f x x→ x→0 The graph of y f 1. x 2 is both a right end behavior 43. 29. An end behavior model is 33. (a) 3x 0 model and a left end behavior model because 0 and lim y x ln x x 42. (a, b) The function y 1 sin x xx So, lim y x x→ x→ sin x x 2 41. (a, b) The function y 1 model and a left end behavior model because 0 and lim y x→ 28. y 1 2x sin x is shown. x 1 1 1. 53 54 Section 2.2 47. (a) lim f (x) x→ (b) lim f (x) x→ (c) lim f (x) x→0 (d) lim f (x) x→0 48. (a) lim f (x) x→ (b) lim f (x) x→ (c) lim f (x) x→0 (d) lim f (x) x→0 1 x lim x→ 53. (a) Using 1980 as x y 2.2316x 3 0 lim ( 1) 0: 54.7134x 2 351.0933x 733.2224 1 x→ 1 x lim x→0 lim ( 1) x x lim x→ 2 1 1 lim 2 x→ x x x lim x→ (b) Again using 1980 as x 0: y 1.458561x 4 60.5740x 3 905.8877x 2 5706.0943x 12967.6288 1 0 x lim [0, 20] by [0, 800] 1 x→0 2 1 x→0 x 0 0 2 1 2 1 lim [0, 20] by [0, 800] 2 x→0 x (c) Cubic: approximately 2256 dollars Quartic: approximately 9979 dollars 49. One possible answer: y (d) Cubic: End behavior model is 2.2316x3. This model predicts that the grants will become negative by 1996. 5 y = f (x ) Quartic: End behavior model is 1.458561x4. This model predicts that the size of the grants will grow very rapidly after 1995. x 10 Neither of these seems reasonable. There is no reason to expect the grants to disappear (become negative) based on the data. Similarly, the data give no indication that a period of rapid growth is about to occur. 50. One possible answer: 54. (a) Note that fg f (x)g(x) 1. f→ as x → 0 , f → as x → 0 , g → 0, fg → 1 y 5 (b) Note that fg f (x)g(x) f → as x → 0 , f → y = f (x ) 8 (c) Note that fg f (x)g(x) 3(x 2)2. f→ as x → 2 , f → as x → 2 , g → 0, fg → 0 x 7 8. as x → 0 , g → 0, fg → (d) Note that fg 5 f (x)g(x) 3)2 (x . f → , g → 0, fg → (e) Nothing – you need more information to decide. 51. Note that f1(x)/f2(x) f1(x)g2(x) f1(x)/g1(x) g1(x)/g2(x) g1(x)f2(x) f1 f2 f2(x)/g2(x) As x becomes large, g1 and using the above equation, g2 f1/f2 g1/g2 . both approach 1. Therefore, 55. (a) This follow from x 1 int x x, which is true for all x. Dividing by x gives the result. (b, c) Since lim x x→ must also approach 1. 52. Yes. The limit of ( f g) will be the same as the limit of g. This is because adding numbers that are very close to a given real number L will not have a significant effect on the value of ( f g) since the values of g are becoming arbitrarily large. 1 Theorem gives lim x→ 56. For x 0, 0 e x lim 1 1, the Sandwich x→ x int x x 1, so 0 int x x lim x→ x e x 1. 1 . x 1 approach zero as x → , the Sandwich x ex Theorem states that must also approach zero. x Since both 0 and 57. This is because as x approaches infinity, sin x continues to oscillate between 1 and 1 and doesn’t approach any given real number. ln x2 x→ ln x 58. lim 2, because ln x2 ln x 2 ln x x 2. 55 Section 2.3 ln x x→ log x 59. lim ln x log x ln (10), since (b) lim f (x) ln x (ln x)/(ln 10) ln 10. (d) f (2) 1 x ln x 1 ln x ln 1 1 x ln x ln 1 1 ln x ln x (pp. 73–81) (g f )(x) x 2x (x 3)(x 3). The domain of f is ( (3, ) or all x 3. , 3) [ 3, 6] by [ 2, 8] Thus, lim (x x→3 10 3 20 6 2), x 2 1. lim 3x x3 x→–1 2. (a) lim f (x) (b) lim f (x) x→ 1 x→ 1 1)(x 2) for x x3 10 . 3 2 3( 1) 2( 1) ( 1)3 4 lim int (x) lim f (x) x x→2 lim (x 2 x→2 1 6 3 2 5 1 5) 1 9x 1)(x f (x) ( x x 5 1 1) 2 x ,x 1. 0. Squaring both sides 1 1 x2 1, x 1 x 1 0. 1 1 0 5) 0 1 ,x 2 5 0.453 The only solution to the original equation is x 22 0 4(2) 1. For x 3, f (x) 4 when x 2 6x 8 4, which gives x 2 6x 12 0. The discriminant of this equation is b 2 4ac ( 6)2 4(1)(12) 12. Since the discriminant is negative, the quadratic equation has no solution. 1 4x 1 1 5 9. For x 3, f (x) 4 when 5 x 4, which gives x (Note that this value is, in fact, 3.) right-hand limits are not equal. 3. (a) lim f (x) 1 2x x 1 . Therefore, f (x) x2 1 1 1 Solution: x x→ 1 int ( 1) 5 [ 5, 5] by [ 10, 10] 3. (c) lim f (x) does not exist, because the left- and (d) f ( 1) 1 3. 2 x→ 1 x→ 1 1 8. g(3), so g is continuous at x 2x 1 4 1 9 Quick Review 2.3 2 f (x) Solutions: x 1)(x 1)(x 2) x3 x→3 5. lim g(x) 10 . 3 (x 1 x 1 x g( f (x)) 1 for x x f (g(x)) (2x 3), so f (x) 1 5 4 ,x 1 (g f )(x) ( f g)(x) 7. 2x 2 3)(x 1 x 1 x 3)(x 3x 2x 1 1 gives f (x) 1 (x g 2x 2x 5 1 Therefore, 2. It appears that the limit of f as x → 3 exists and is a little more than 3. 6 2x x g( f (x)) 6. Note that Removing a Discontinuity 3. f (3) should be defined as 2 5. Note that sin x 2 (g f )(x) g( f (x)) g(x 2). Therefore: g(x) sin x, x 0 ( f g)(x) f (g(x)) = f (sin x) (sin x)2 or sin2 x, x s Section 2.3 Continuity Exploration 1 2 1 4. ( f g)(x) f (g(x)) f 1 x 2(1 x) x x2 ,x 0 (1 x) 5x 6x 1 1 x ln 1 4 2 1 , x But as x → , 1 1/x approaches 1, so ln(1 1/x) approaches ln (1) 0. Also, as x → , ln x approaches infinity. This means the second term above approaches 0 and the limit is 1. (x 2 right-hand limits are not equal. 1) ln (x 1) ln x 7x 2 1 Since ln (x 4. x 3 4 x→2 ln (x 1) ln x x→ 1. x 9 ( 3, 3) x) x→2 (c) lim f (x) does not exist, because the left- and 60. lim 2 lim (4 x→2 5 1 1. 56 Section 2.3 10. x is equivalent to x 7. The function y y 1, x x 1, 0 0. It has a jump discontinuity at x [ 2.7, 6.7] by [ 6, 6] A graph of f (x) is shown. The range of f (x) is ( , 1) [2, ). The values of c for which f (x) c has no solution are the values that are excluded from the range. Therefore, c can be any value in [1, 2). 1 2)2 (x is continuous because it is a cos x , a quotient sin x cot x is equivalent to y of continuous functions, so it is continuous. Its only points of discontinuity occur where it is undefined. It has infinite discontinuities at x Section 2.3 Exercises 1. The function y 8. The function y 0. k for all integers k. e 1/x is a composition ( f g)(x) of the 9. The function y 1 , so it is x quotient of polynomials, which are continuous. Its only continuous functions f (x) point of discontinuity occurs where it is undefined. There is continuous. Its only point of discontinuity occurs at x an infinite discontinuity at x where it is undefined. Since lim e 1/x 2. e x and g(x) , this may be x→0 x 2. The function y x2 1 is continuous because it is a 4x 3 quotient of polynomials, which are continuous. Its only points of discontinuity occur where it is undefined, that is, where the denominator x 2 4x 3 (x 1)(x zero. There are infinite discontinuities at x 3) is considered an infinite discontinuity. 10. The function y ln (x 1) is a composition ( f g)(x) of the continuous functions f (x) ln x and g(x) x 1, so it is continuous. Its points of discontinuity are the points not in the domain, i.e., x 1. 11. (a) Yes, f ( 1) 0. 1 and at (b) Yes, lim x 0, x→ 1 3. 0. (c) Yes 1 3. The function y x2 1 is continuous because it is a (d) Yes, since quotient of polynomials, which are continuous. lim f (x) x→ 1 Furthermore, the domain is all real numbers because the denominator, x 2 1, is never zero. Since the function is continuous and has domain ( 12. (a) Yes, f (1) 1 is a left endpoint of the domain of f and f ( 1), f is continuous at x 1. (b) Yes, lim f (x) x→1 , ), there are no points of 1. 2. (c) No discontinuity. (d) No 4. The function y x 1 is a composition ( f g)(x) of the continuous functions f (x) x and g(x) x 1, so it is continuous. Since the function is continuous and has domain ( , ), there are no points of discontinuity. 5. The function y 2x 3 is a composition ( f g)(x) of the continuous functions f (x) x and g(x) 2x 3, so it is continuous. Its points of discontinuity are the points not in the domain, i.e., all x 3 3 . 2 6. The function y 2x 1 is a composition ( f g)(x) of 3 the continuous functions f (x) x and g(x) 2x 1, so it is continuous. Since the function is continuous and has domain ( , ), there are no points of discontinuity. 13. (a) No (b) No, since x 2 is not in the domain. 14. Everywhere in [ 1, 3) except for x 0, 1, 2. 15. Since lim f (x) 0, we should assign f (2) 16. Since lim f (x) 2, we should reassign f (1) x→2 x→1 0. 2. 17. No, because the right-hand and left-hand limits are not the same at zero. 18. Yes. Assign the value 0 to f (3). Since 3 is a right endpoint of the extended function and lim f (x) x→3 function is continuous at x 3. 0, the extended Section 2.3 19. 27. Since lim x→0 sin x x 1, the extended function is x 0 1, y sin x , x x 0. [ 3, 6] by [ 1, 5] (a) x 28. Since lim 2 x→0 (b) Not removable, the one-sided limits are different. sin 4x x 4 lim x→0 29. For x f (x) sin 4x , x x 4, the extended 0 4, y (a) x 4(1) function is 20. [ 3, 6] by [ 1, 5] sin 4x 4x x 0. 4 (and x x 4 x 2 0), (x 2)( x x 2 2) x 2. 2 (b) Removable, assign the value 1 to f (2). The extended function is y 21. 30. For x f (x) [ 5, 5] by [ 4, 8] (a) x 1 (b) Not removable, it’s an infinite discontinuity. 22. 2 (and x 2. 2), x3 4x 2 11x 30 x2 4 (x 2)(x 5)(x 3) (x 2)(x 2) (x 5)(x 3) x2 x 2 2x 15 . x2 The extended function is y x2 2x x 15 2 . 31. One possible answer: Assume y x, constant functions, and the square root function are continuous. [ 4.7, 4.7] by [ 3.1, 3.1] (a) x x 1 (b) Removable, assign the value 0 to f ( 1). 23. (a) All points not in the domain along with x 0, 1 (b) x 0 is a removable discontinuity, assign f (0) 0. x 1 is not removable, the one-sided limits are different. 24. (a) All points not in the domain along with x 1, 2 (b) x 1 is not removable, the one-sided limits are different. x 2 is a removable discontinuity, assign f (2) 1. 25. For x 3, f (x) x2 x 9 3 The extended function is y x3 x2 (x (x x 3)(x 3) x3 3. 1 1 1)(x 2 x 1) (x 1)(x 1) x2 x 1 . x1 x2 x 1 The extended function is y . x1 26. For x 1, f (x) x 3. By the sum theorem, y x 2 is continuous. By the composite theorem, y By the quotient theorem, y x 2 is continuous. 1 x is continuous. 2 Domain: ( 2, ) 32. One possible answer: Assume y x, constant functions, and the cube root function are continuous. By the difference theorem, y 4 x is continuous. 3 By the composite theorem, y 4 x is continuous. By the product theorem, y x 2 x x is continuous. 3 By the sum theorem, y x 2 4 x is continuous. Domain: ( , ) 33. Possible answer: Assume y x and y x are continuous. By the product theorem, y x 2 x x is continuous. By the constant multiple theorem, y 4x is continuous. By the difference theorem, y x 2 4x is continuous. By the composite theorem, y x 2 4x is continuous. Domain: ( , ) 57 58 Section 2.3 34. One possible answer: Assume y x and y 1 are continuous. 39. Use the product, difference, and quotient theorems. One also needs to verify that the limit of this function as x approaches 1 is 2. Alternately, observe that the function is equivalent to y x 1 (for all x), which is continuous by the sum theorem. Domain: ( , ) [ 3, 3] by [ 2, 2] Solving x x 4 1, we obtain the solutions x 0.724 and x 1.221. 40. 35. One possible answer: y 5 y = f (x ) [ 6, 6] by [ 4, 4] x3 Solving x 5 2, we obtain the solution x lim (x 2 41. We require that lim 2ax x→3 32 2a(3) 36. One possible answer: 6a 1 4 3 5 42. Consider f (x) y = f (x ) x f (1) 1): x→3 8 a y 5 1.521. x 1 x 1 e e x . f is continuous, f (0) 1, and 0.5. By the Intermediate Value Theorem, for some c in (0, 1), f (c) 0 and e c c. 43. (a) Sarah’s salary is $36,500 $36,500(1.035)0 for the first year (0 t 1), $36,500(1.035) for the second year (1 t 2), $36,500(1.035)2 for the third year (2 t 3), and so on. This corresponds to y 36,500(1.035)int t. 37. One possible answer: y 5 (b) y = f (x ) 5 x [0, 4.98] by [35,000, 45,000] The function is continuous at all points in the domain [0, 5) except at t 1, 2, 3, 4. 38. One possible answer: 44. (a) We require: y 5 y = f (x ) f (x) 5 x { 0 1.10, 2.20, 3.30, 4.40, 5.50, 6.60, 7.25, x x x x x x x 0 1 2 3 4 5 6 x 0 1 2 3 4 5 6 24. This may be written more compactly as 1.10 int( x), 0 x 6 f (x) 7.25, 6 x 24 Section 2.4 59 49. For any real number a, the limit of this function as x approaches a cannot exist. This is because as x approaches a, the values of the function will continually oscillate between 0 and 1. (b) Section 2.4 Rates of Change and Tangent Lines (pp. 82–90) s [0, 24] by [0, 9] This is continuous for all values of x in the domain [0, 24] except for x 0, 1, 2, 3, 4, 5, 6. 45. (a) The function is defined when 1 1 x Quick Review 2.4 0, that is, on , 1) 1. x y 3 5 ( 5) 8 23 2. ( x y a b 1 3 (0, ). (It can be argued that the domain should also include certain values in the interval ( 1, 0), namely, those rational numbers that have odd 5 1 denominators when expressed in lowest terms.) 4. m (b) y [ 5, 5] by [ 3, 10] f ( 1) f (0) 1 1 1 1 10 0 1 0 1 y 1 (undefined) and 0 (undefined). Since f is undefined at y y points of discontinuity. (d) The discontinuity at x 0 is removable because the right-hand limit is 0. The discontinuity at x 1 is not removable because it is an infinite discontinuity. 4 (x 3 4 x 3 y y 9. Since 2x 2 . 3 2 [x 3 2 x 3 m [0, 20] by [0, 3] y y b 4 3 2 b 10. 3 The limit is about 2.718, or e. b 46. This is because lim f (a h→0 h) lim f (x). x→a 47. Suppose not. Then f would be negative somewhere in the interval and positive somewhere else in the interval. So, by the Intermediate Value Theorem, it would have to be zero somewhere in the interval, which contradicts the hypothesis. 48. Since the absolute value function is continuous, this follows from the theorem about continuity of composite functions. 2 3 3 6 1) 7 3 4 19 4 4 3 1 3/4 8. m (e) ( 2)] 3 (x 4 3 x 4 7. y these values due to division by zero, both values are 4 6 4 7 16 7 41 3 7 (x 1) 6 3 7 25 x 3 3 6. m (c) If we attempt to evaluate f (x) at these values, we obtain ( 1) ( 3) 3 [x 2 3 x 2 5. y 4 7 3 ( 2) 3 3 3. m 1) 4 8 3 3y 5 is equivalent to y ( 1)] 2 x 3 3 7 3 5 3 10 3 19 3 Section 2.4 Exercises 1. (a) f x f (3) 3 f (2) 2 (b) f x f (1) f ( 1) 1 ( 1) 2. (a) f x f (2) 2 (b) f x f (12) 12 f (0) 0 f (10) 10 28 9 19 1 2 0 1 2 3 1 1 2 7 41 2 0.298 5 , we use 3 60 Section 2.4 3. (a) f x f (0) 0 f ( 2) ( 2) 1 2 e 2 3 (b) f x f (3) 3 f (1) 1 e 4. (a) f x f (4) 4 f (1) 1 ln 4 0 3 (b) f x f (103) 103 5. (a) e f x f ( /2) ( /2) 6. (a) f x f( ) (b) f x f( ) f( ) () 1 h→0 h→0 1 103 ln 3 100 4 1.273 3 33 1 3 2 1.654 0.637 1 (b) The tangent line has slope ( 2, y( 2)) ( 2, 4). y 4[x ( 2)] 4 y 4x 4 (c) The normal line has slope 0 2 (16.5, 475), y ( 2)] 4 h 4h h2 h h) 4 4 and passes through 1 4 1 and passes 4 ( 2, 4). 4 9 2 46 50 [ 8, 7] by [ 1, 9] 50 PQ1 PQ2 10. (a) lim 43 46 PQ3 PQ4 y (1 h) h h→0 lim y (1) h)2 [(1 lim h→0 50 lim [12 4h 3 4(1)] h 1 2h h2 4 h h2 h→0 2h h lim (h 2) h→0 The appropriate units are meters per second. h)] 4(1 h→0 50 2 (b) Approximately 50 m/sec 8. We use Q1 (5, 20), Q2 (7, 38), Q3 Q4 (9.5, 72), and P (10, 80). 12 14 (8.5, 56), (b) The tangent line has slope (1, y (1)) (1, 3). y 2(x 1) 3 y 2x 1 (c) The normal line has slope 16 through (1, y (1)) 16 Secant PQ1 12 PQ2 1) 3 7 2 16 PQ4 1 (x 2 1 x 2 14 PQ3 y Slope 16 The appropriate units are meters per second. (b) Approximately 16 m/sec h2 4h ( 2)2 (d) 43 Slope Slope of PQ4: 1 [x 4 1 x 4 y (14, 375), Q3 (20, 650). 20 5 38 7 56 8.5 72 9.5 4 lim ( 4 h→0 through ( 2, y( 2)) 1 h)2 h (2 h→0 lim /3 Secant Slope of PQ3: lim 0.462 ln 100 1 0 650 225 20 10 650 375 Slope of PQ2: 20 14 650 475 slope of PQ3: 20 16.5 650 550 Slope of PQ4: 20 18 Slope of PQ2: y( 2) h→0 /2 (a) Slope of PQ1: 80 10 80 10 80 10 80 10 h) h = 7. We use Q1 (10, 225), Q2 Q4 (18, 550), and P (a) Slope of PQ1: y( 2 lim 3 f ( /6) ( /6) f (0) 0 ln 4 3 ln 103 1 ln 1.03 0.0099 3 f (3 /4) f ( /4) f (3 /4) ( /4) x (b) lim 8.684 2 f (100) 100 9. (a) 0.432 y (d) [ 6, 6] by [ 6, 2] (1, 3). 2 and passes through 1 2 1 and passes 2 61 Section 2.4 11. (a) lim y(2 h) h h→0 y(2) lim 1 h) (2 2 1 1 h (b) Near x h h→0 h) h h→0 1 1 2, f (x) f (2 lim h h→0 lim 13. (a) Near x 1 1 h→0 1 lim (h 1) h(h 1) 14. Near x h→0 h 1 1 and passes through h→0 1 1 (c) The normal line has slope through (2, y(2)) y 1(x y x 2) f (0 x (3 lim 2 f (1) lim 1 h→0 h) h 1 3 (x 2 x. (1 [2 lim 2) h)] h (2 1) h→0 1 lim 1 h→0 h) h 2 x. h→0 15. First, note that f (0) lim x f ( 3) h→0 1 (b) The tangent line has slope (2, y(2)) (2, 1). y (x 2) 1 y x3 f (1 h) h 1 1, f (x) h) h 1h lim h h→0 lim (2 lim h→0 1 h→0 1 lim f (2) h) h h→0 lim x. 3, f (x) f( 3 lim x 1 2. f (0) (2 lim h→0 h→0 lim ( 2 (2, 1). h) h→0 1 2 h h2 2h h lim 1 and passes h 2) 2h 2 1 f(0 lim h→0 (d) h) h f(0) (2h lim 2) h h→0 2 lim 2 h→0 2 No, the slope from the left is 2 and the slope from the right is 2. The two-sided limit of the difference quotient does not exist. [ 4.7, 4.7] by [ 3.1, 3.1] 12. (a) lim y(0 h) h h→0 y(0) lim ( h2 3h h→0 lim 1) ( 1) h h2 h→0 3h lim (h lim f (0 h→0 h h→0 16. First, note that f (0) 3) 3 lim f (0 h→0 lim (h h→0 (b) The tangent line has slope (0, y(0)) y 3(x y (0, 3x 3 and passes through f (0) f (0) lim 1 lim h→0 1) h) h ( h2 1 2 f (2) 1 lim h→0 lim through (0, y(0)) y y 1 (x 3 1 x 3 0) 1 1 (d) [ 6, 6] by [ 5, 3] (0, h) h 0 1. h→0 (c) The normal line has slope 1 1 1 1 3 0 h h→0 Yes. The slope is f (2 h lim h→0 17. First, note that f (2) 1). 0) h) h h) h 0. 1 and passes 3 lim h→0 lim 1). h→0 = 1 4 2 h 1 2 h 2 (2 h) 2h(2 h) h 2h(2 h) 1 2(2 h) 62 Section 2.4 (b) The slope of the tangent steadily decreases as a increases. 17. continued 4 lim f (2 h) h h→0 f (2) (2 4 lim [4 h→0 x→(3 /4) but f 19. (a) 3 4 f (x) cos lim f (a h→0 lim lim lim 3 4 2 2 f (a) sin [(a 2ah 2ah h→0 19.6 2 (a 2 2] h h2 h 24. Let f (t) 2) a2 2 2 lim f (10 h→0 lim h) 300 3t 2. h) h f (10) a lim lim h→0 lim h→0 h→0 3h 2 60h lim (60 a h 25. Let f (r) h 2a 2(a h) ah(a h) 2 a(a h) lim f (3 h→0 r 2, the area of a circle of radius r. h) h f (3) 1 h lim 1 1 a 1 1)(a h (a 43 r. 3 f (2 h) f (2) lim h h→0 1 26. Let f (r) 1) (b) The slope of the tangent is always negative. The tangents are very steep near x 1 and nearly horizontal as a moves away from the origin. [9 (a 9 a2 h→0 lim h→0 lim h→0 2ah h lim ( 2a h→0 2a h)2] (9 h 2ah h 2 h h2 h) lim (2 h→0 lim 8 h)3 h h)3 h 12h h→0 lim (12 h→0 6h 4 (2)3 3 23 6h 2 + h 3 h 8 h 2) 12 16 a 2) 9 lim 4 (2 3 h→0 4 3 4 3 4 3 4 3 1) 2 (a h→0 h) The area is changing at a rate of 6 in2/in., that is, 6 square inches of area per inch of radius. 1 lim 9 h lim (6 h h→0 f (a) h2 6 h→0 h) h (3)2 h→0 (a 1) (a h 1) 1)(a h 1) h→0 h(a f (a 9 h)2 h 6h h→0 a lim (3 lim h→0 lim lim 22. (a) lim 300 The rocket’s speed is 60 ft/sec. (b) The slope of the tangent is always negative. The tangents are very steep near x 0 and nearly horizontal as a moves away from the origin. h→0 300 3h) 2 a2 21. (a) lim h) 2 h 3(10 lim 60 h→0 f (a) 19.6 h h→0 2a h) h 100 4.9h) h→0 (b) The slope of the tangent steadily increases as a increases. 2 2 f (a 4.9h 2 h 19.6 h lim (2a f (a) 4.9(2) 2] [100 2 h2 h→0 h) h 100 lim ( 19.6 3 4 h) 2 a2 lim h→0 f (a h) 2] h 19.6h 4.9(2 The object is falling at a speed of 19.6 m/sec. lim lim f (2) h→0 . h→0 h→0 h) h [100 h→0 sin x x→(3 /4) lim 20. (a) f (2 4.9t 2. 100 h→0 3 4 h) h lim h→0 18. No. The function is discontinuous at x because lim 23. Let f (t) 2 h lim h→0 4h 1 4 1 . 4 Yes. The slope is 1 2 h (2 h)] 4h h→0 lim h) a2 The volume is changing at a rate of 16 in3/in., that is, 16 cubic inches of volume per inch of radius. 63 Section 2.4 s(1 27. lim h) h h→0 s(1) h)2 h 3.72h 1.86(1 lim h→0 1.86 lim 1.86(1)2 1.86h2 31. (a) From Exercise 21, the slope of the curve at x 1 1.86 lim (3.72 or a 3.72 The speed of the rock is 3.72 m/sec. s(2 h) h h→0 s(2) lim 11.44(2 h→0 lim 45.76 h→0 lim (45.76 h)2 11.44(2)2 h 45.76h 11.44h 2 h lim 0: y y 1(x 0) x1 4(a h) 2: y y 1(x 2) x3 1 1 1 2ah h2 2ah 1] (a 2 4a lim (2a h 4) h→0 2a 4h h 1 a2 4a 1 h) f (a) h [3 4(a h) 3 4a 4h h→0 lim ( 4 h→0 4 1(x 2) x1 1 4 0, or (a 4h a 2 12 a2 3 a 2 2a 3 (a 1)(a 3) a a. a2 2ah h At a h2 2a h)2] (3 4a a 2) h 2ah h 2 3 4a a 2 h h) 33. (a) 2a 2a(a 1) 2a 2 2a 0 0 1 or a 3 1 (or x 1), the slope is y 2(x 1) 12 y 2x 10 3 (or x y y 2.1 1995 1.5 1993 2( 1) 3), the slope is 2(3) 6(x 1) 12 6x 18 2. 6. 0.3 The rate of change was 0.3 billion dollars per year. 2a The tangent at x a is horizontal when 4 a 2. The tangent line is horizontal at ( 2, f ( 2)) ( 2, 7). a2) 12 a1 9 At a h→0 lim 2: y y (9 h→0 lim 1 There is only one such line. It is normal to the curve at two points and its equation is y x 1. f (a lim 1(x 0) x1 4 30. First, find the slope of the tangent at x h→0 0: y y 32. Consider a line that passes through (1, 12) and a point (a, 9 a 2) on the curve. Using the result of Exercise 22, this line will be tangent to the curve at a if its slope is 2a. The tangent at x a is horizontal when 2a a 2. The tangent line is horizontal at ( 2, f ( 2)) ( 2, 5). lim 1) and (2, 1), this time using slope 1. At x 1) 4h h→0 h2 h 4a h→0 lim 1, so we again need to find lines through (0, a. h a2 1) and (2, 1), 1 At x h→0 lim 1 passing through (0, At x f (a) h)2 [(a 1, so we need to find the equations of (b) The normal has slope 1 when the tangent has slope 29. First, find the slope of the tangent at x h) h 0 1 and 1 At x 45.76 The speed of the rock is 45.76 m/sec. f (a 0 1, so a respectively. 45.76 lim 1 lines of slope 11.44h) h→0 h→0 1 2 1)2 1 2. Note that y(0) y(2) 1 when 1, which gives (a 1)2 (a 1.86h) h→0 . The tangent has slope 1 h h→0 28. lim 1)2 (a a, is (b) 2a 0, or 3.1 1997 2.1 1995 0.5 The rate of change was 0.5 billion dollars per year. (c) y 0.0571x 2 0.1514x [0, 10] by [0, 4] 1.3943 64 Section 2.4 33. continued (d) y(5) 5 y(7) 7 y(3) 3 y(5) 5 0.31 0.53 According to the regression equation, the rates were 0.31 billion dollars per year and 0.53 billion dollars per year. (e) lim y(7 h→0 h) h y(7) lim h)2 [0.0571(7 0.1514(7 h→0 lim h 2) h 0.0571(14h h→0 lim [0.0571(14) 1.3943] h [0.0571(7)2 0.1514h 0.1514 h→0 h) 0.0571h ] 0.65 The funding was growing at a rate of about 0.65 billion dollars per year. 34. (a) [7, 18] by [0, 900] (b) Q from year Slope 440 225 17 8 440 289 17 9 440 270 17 10 440 493 17 11 440 684 17 12 440 763 17 13 440 651 17 14 440 600 17 15 440 296 17 16 1988 1989 1990 1991 1992 1993 1994 1995 1996 23.9 18.9 24.3 8.8 48.8 80.8 70.3 80.0 144.0 (c) As Q gets closer to 1997, the slopes do not seem to be approaching a limit value. The years 1995–97 seem to be very unusual and unpredictable. 35. (a) f (1 h) h f (1) e1 h e h (b) [ 4, 4] by [ 1, 5] Limit 2.718 (c) They’re about the same. (d) Yes, it has a tangent whose slope is about e. 0.1514(7) 13943] Section 2.4 f (1 36. (a) h) h f (1) 21 h 2 40. Let f (x) x2/3. The graph of y is shown. h (b) [ 4, 4] by [ 1, 5] f (0 h) h f (0) 65 f (h) h [ 4, 4] by [ 3, 3] The left- and right-hand limits are and , respectively. Since they are not the same, the curve does not have a vertical tangent at x 0. No. Limit 41. This function has a tangent with slope zero at the origin. It is sandwiched between two functions, y x 2 and y x 2, both of which have slope zero at the origin. 1.386 (c) They’re about the same. Looking at the difference quotient, (d) Yes, it has a tangent whose slope is about ln 4. 37. Let f (x) x 2/5. The graph of y f (0 h) h f (0) f (h) h is shown. h f (0 h) h f (0) h, so the Sandwich Theorem tells us the limit is 0. 42. This function does not have a tangent line at the origin. As the function oscillates between y x and y x infinitely often near the origin, there are an infinite number of difference quotients (secant line slopes) with a value of 1 and with a value of 1. Thus the limit of the difference quotient doesn’t exist. [ 4, 4] by [ 3, 3] The left- and right-hand limits are and , respectively. Since they are not the same, the curve does not have a vertical tangent at x 0. No. 38. Let f (x) x 3/5. The graph of y is shown. f (0 h) h f (0) f (h) h The difference quotient is oscillates between 1 and 43. Let f (x) f (1 h) h f (0 h) h f (0) sin 1 which h 1 infinitely often near zero. sin x. The difference quotient is f (1) sin (1 h) h sin (1) . A graph and table for the difference quotient are shown. [ 4, 4] by [ 3, 3] Yes, the curve has a vertical tangent at x lim f (0 h→0 h) h f (0) 0 because [ 4, 4] by [ 1.5, 1.5] . 39. Let f (x) x1/3. The graph of y is shown. f (0 h) h f (0) f (h) h Since the limit as h→0 is about 0.540, the slope of y sin x at x 1 is about 0.540. [ 4, 4] by [ 3, 3] Yes, the curve has a vertical tangent at x lim h→0 f (0 h) h f (0) . 0 because 66 Chapter 2 Review s Chapter 2 Review Exercises 13. Since lim ( e (pp. 91–93) e 1. lim (x 3 2x 2 x→ 2 x2 2. lim x→−2 3x 2( 2)2 ( 2)2 1 3( 2)2 2( 2) 1 2x 5 2 ( 2)3 1) 3. No limit, because the expression values of x near 4. 4. No limit, because the expression values of x near 5. 1 5. lim 2 1 2 x lim x→0 2 2x 5x 2 6. lim x→ 3 7 lim x→ 2x 5x 2 x→ 8. lim x→0 sin 2x 4x x 10. lim e x sin x x→0 11. Let x sin x and x cos x, lim x→ 2 9 x is undefined for 7 2 x→0 2x(2 21. Yes 23. No x3 x4 is 128 12x3 24. Yes 1 x. 12 25. (a) lim g(x) 1 1.5 (c) No, since lim g(x) x 1 (1) 2 sin x x sin x x→0 x e 0 sin 0 int 2 (d) g is discontinuous at x domain). 1 2 26. (a) lim k(x) x→1 sin x x lim 1 1 (b) lim k(x) x→1 2 (c) k(1) 10 0 1.5 0 0 (d) No, since lim k(x) because 6 int (6 2h) 1 (and at points not in the (f) No, the discontinuity at x 1 is not removable because the one-sided limits are different. 6, 27. lim 6 x→7/2 h, where h is in 7 2 k(1) (e) k is discontinuous at x domain). 1 . Then 2 2h – 1 x→7/2 int 2 3 (and at points not in the (e) Yes, the discontinuity at x 3 can be removed by assigning the value 1 to g(3). x→0 1 7 2 g(3). x→3 lim 1) 2h 6. 1 , 0 . Then 2 1 int (6 [ 4, 4] by [ 3, 3] (a) Vertical asymptote: x 2h) 5, (b) Left-hand limit 2h is in (5, 6). lim 2 x x→–2 x Right-hand limit: lim x x→–2 x Therefore, lim int (2x x→7/2 1. 20. Limit exists. 1 4 (b) g(3) Therefore, lim int (2x – 1) int (2x x x 19. Limit exists. x) 2h is in (6, 7). 7 2 lim x→ 22. No h, where h is in 0, because 6 sin x cos x 18. Limit does not exist. x lim 1 x x→ 12 1 lim x 12 x→ lim 1) x x 16. Limit exists. x→1 int (2x 12. Let x for all x, the Sandwich Theorem lim x→0 x→0 0, and 0. 2x is undefined for 9. Multiply the numerator and denominator by sin x. lim 1 x 17. Limit exists. x4 12x3 1 sin 2x lim 2 x→0 2x x csc x 1 lim x csc x x→0 e cos x x→3 x4 x3 3 128 x→ 12x x 4 x3 lim 12x3 128 x→ x lim e x→ 14. Since the expression x is an end behavior model for both 5 21 Therefore: lim ) cos x x gives lim e 15 2 5 7. An end behavior model for e x x 15. Limit exists. 4 lim 2 1 5 1 2 (2 x) x x) x→0 2x(2 1 1 2(2 x) 2(2 0) x→0 x x→ 1) lim 5 x→7/2 5 3 2 3 2 Chapter 2 Review 28. 33. (a) End behavior model: 2x 2 , or x2 x (b) Horizontal asymptote: y 34. (a) End behavior model: [ 4, 4] by [ 3, 3] (a) Vertical asymptotes: x 0, x 0 (the x-axis) 2x2 , or 2 x2 (b) Horizontal asymptote: y 2 67 2 3 (b) At x 35. (a) End behavior model: 0: (b) Since the end behavior model is quadratic, there are no horizontal asymptotes. x1 2 2) x→0 x (x x1 Right-hand limit lim 2 2) x→0 x (x Left-hand limit At x lim 36. (a) End behavior model: 2: 29. (a) At x Right-hand limit At x lim x→ x→ x x ex 1 1, a right end behavior model is e . lim f (x) x→ 1 lim f (x) x→ 1 lim f (x) x→0 Right-hand limit lim (1) 1 x→ 1 lim ( x) x→ 1 x→ 1 ex x (b) Since lim lim ex x 1 x→ x 1, a left end behavior model is x. lim f (x) x→0 lim ( x) x→0 lim ( x) x→ 1 ln x 0 x→0 lim f (x) x→1 lim f (x) x→1 lim ( x) 1 x→1 1 for all x ln x 0 and 0. lim (1) x→1 1 (c) At x 1: Continuous because f ( 1) the limit. At x 0: Discontinuous because f (0) the limit. At x 1: Discontinuous because the limit does not exist. 30. (a) Left-hand limit lim f (x) x→1 4(1) 3 Right-hand limit 2(1) lim (x 2 x→1 2x 2) 3 39. lim f (x) x→3 lim (x lim x2 2x – 15 x–3 x→3 3 5 x→0 5) lim x→3 1 31. Since f (x) is a quotient of polynomials, it is continuous and its points of discontinuity are the points where it is undefined, namely x 2 and x 2. 32. There are no points of discontinuity, since g(x) is continuous and defined for all real numbers. 0 (x – 3)(x 5) x–3 lim x→0 sin x 2x 8. 1 sin x lim 2 x→0 x 1 . 2 Assign the value k 41. One possible answer: 1 1 8. x→3 (b) No, because the two one-sided limits are different. (c) Every place except for x sin x ln x end behavior model. 40. lim f (x) lim f (x) 1 1, so ln x is both a right end behavior model and a left Assign the value k 4x 3 x→1 2 lim x 3 x→1 sin x 0. Hence ln x ln x sin x lim = lim ln x x→ x→ lim x→ (b) At x 1: Yes, the limit is 1. At x 0: Yes, the limit is 0. At x 1: No, the limit doesn’t exist because the two one-sided limits are different. (1)2 sin x ln x 1 ln x lim x→ Therefore, the Sandwich Theorem gives Right-hand limit (1)3 1 ln x 38. (a, b) Note that lim 0 1: Left-hand limit (d) At x lim ex 0: Left-hand limit At x ex x 37. (a) Since lim 1: Left-hand limit x4 , or x x3 (b) Since the end behavior model represents a nonhorizontal line, there are no horizontal asymptotes. x1 2 2) x→–2 x (x x1 Right-hand limit lim 2 2) x→–2 x (x Left-hand limit x , or x 2 x y 10 y = f (x ) 10 x 1 (1) 2 1 2 68 Chapter 2 Review 42. One possible answer: 48. At x y lim 5 a, the slope of the curve is f (a h→0 h) h f (a) lim [(a h)2 a2 2ah h2 lim (2a 3 The tangent is horizontal when 2a 3 lim 2ah h→0 2a V(a 44. lim h) h h→0 1 3 1 3 1 3 2 3 45. lim 2 2 V(a) 1 (a 3 2 lim h h2 2ah 3 3 . Since f 2 2 3 9 is , . 2 4 200 49. (a) p(0) 1 7e 0.1(0) or x 12 aH 3 h)2H h→0 a H lim 1 /2 a2 h h→0 H lim (2a h) h→0 (b) lim p(t) t→ aH h) h h→0 lim 6a 2 S(a) h)2 h 6(a lim h→0 6h 2 12ah lim (12a 6a2 6a 2 6h) h→0 y(a h) h h→0 lim h) 2 [(a (a h) 0, at a 3 2 9 , the point where this occurs 4 200 8 25 lim t→ 1 200 7e 0.1t 200 1 200 (a 2 2] a 2) h a2 2ah h2 a h a2 2 a 2 h h→0 lim 3 (b) y(a) h→0 lim 3a (Note that we cannot use the formula f (x) 3.20 1.35 int x, because it gives incorrect results when x is an integer.) 12a 46. lim a2 (c) Perhaps this is the maximum number of bears which the reserve can support due to limitations of food, space, or other resources. Or, perhaps the number is capped at 200 and excess bears are moved to other locations. 3.20 1.35 int ( x 1), 0 x 20 50. (a) f (x) 0, x0 h h→0 3a – 3h h Perhaps this is the number of bears placed in the reserve when it was established. H(2a) S(a 3a) h) x y = f (x ) f ( /2) – f (0) /2 – 0 (a 2 h2 h→0 3h h h→0 5 h)] h h→0 lim 43. 3(a h2 h h→0 lim (2a h→0 [0, 20] by [ 5, 32] h h 2ah 1) f is discontinuous at integer values of x: 0, 1, 2, ..., 19. 51. (a) Cubic: y Quartic: y 2a 1 47. (a) lim f (1 h→0 lim h) f (1) h 1 2h h 2 lim ( 1 h→0 3 3h 3(1 h h)] ( 2) 2 h h→0 h0 h)2 [(1 lim 52. Let A lim f (x) and B x→c 1 and passes through (c) The normal at P has slope 1 and passes through (1, 2). y 1(x 1) 2 yx3 B 3 2 1 (b) The tangent at P has slope (1, 2). y 1(x 1) 2 y x1 (b) Cubic: 1.644x 3, predicts spending will go to 0 Quartic: 2.009x 4, predicts spending will go to A h) 1.644x 3 42.981x 2 254.369x 300.232 2.009x 4 102.081x 3 1884.997x 2 14918.180x 43004.464 B and lim x→c lim g(x). Then A x→c B 2 and 3 1. Adding, we have 2A 3, so A , whence 2 1 3 2, which gives B . Therefore, lim f (x) 2 2 x→c 1 g(x) . 2 53. (a) [3, 12] by [ 2, 24] Section 3.1 (b) Year of Q 69 Slope of PQ 20.1 2000 20.1 2000 20.1 2000 20.1 2000 20.1 2000 1995 1996 1997 1998 1999 2.7 1995 4.8 1996 7.8 1997 11.2 1998 15.2 1999 3.48 3.825 4.1 4.45 4.9 (c) Approximately 5 billion dollars per year. (d) y 0.3214x 2 lim y(10 h→0 1.3471x h) h y(10) 1.3857 [0.3214(10 h) 2 1.3471(10 0.3214(20h h 2) h 1.3471h h) lim h→0 lim h→0 0.3214(20) 1.3857] h [0.3214(10)2 1.3471(10) 1.3857] 1.3471 5.081 The predicted rate of change in 2000 is about 5.081 billion dollars per year. Chapter 3 Derivatives s Section 3.1 Derivative of a Function (pp. 95–104) Exploration 1 Reading the Graphs 1. The graph in Figure 3.3b represents the rate of change of the depth of the water in the puddle with respect to time. Since y is measured in inches and x is measured in days, the derivative dy would be measured in inches per day. Those are the units that dx should be used along the y-axis in Figure 3.3b. 2. The water in the ditch is 1 inch deep at the start of the first day and rising rapidly. It continues to rise, at a gradually decreasing rate, until the end of the second day, when it achieves a maximum depth of 5 inches. During days 3, 4, 5, and 6, the water level goes down, until it reaches a depth of 1 inch at the end of day 6. During the seventh day it rises again, almost to a depth of 2 inches. 3. The weather appears to have been wettest at the beginning of day 1 (when the water level was rising fastest) and driest at the end of day 4 (when the water level was declining the fastest). 4. The highest point on the graph of the derivative shows where the water is rising the fastest, while the lowest point (most negative) on the graph of the derivative shows where the water is declining the fastest. 5. The y-coordinate of point C gives the maximum depth of the water level in the ditch over the 7-day period, while the x-coordinate of C gives the time during the 7-day period that the maximum depth occurred. The derivative of the function changes sign from positive to negative at C , indicating that this is when the water level stops rising and begins falling. 6. Water continues to run down sides of hills and through underground streams long after the rain has stopped falling. Depending on how much high ground is located near the ditch, water from the first day’s rain could still be flowing into the ditch several days later. Engineers responsible for flood control of major rivers must take this into consideration when they predict when floodwaters will “crest,” and at what levels. 70 Section 3.1 Quick Review 3.1 1. lim (2 h→0 h)2 4 2. f (3) 4 lim (4 h 2) 4h lim f (3 4 h 4 4 0 3 lim lim 4 h→0 f (3) 1 3 1 h h→0 h) h h→0 h h h→0 lim 3 (3 h) h) h→0 3h(3 2. lim+ x 2 3 x→2 5 2 2( 2x 8 x 2 lim 2( h→4 h→0 3(3 y 0, lim x 2)( x x→4 x 2) 3. f (3) lim x→3 2 f (x) x 2) 2( 4 2) 8 1 x 1 3 3 lim 3 lim 2 through (0, 1) and another point (h, h is 1) 1 h0 h. Since lim h x→3 (x 1) on the parabola 0, the slope of the line h→0 4. f (x) lim x→1 lim f (x) x→1 8. lim f (1 h→0 lim (x 1)2 lim (x 2) x→1 x→1 h) lim f (x) x→1 1)2 (1 1 2 0 3 f (x) [3(x h) 12] h lim 3h 5. dy dx lim h) h h) h 7(x 7h 1 because lim f (x) does not x→1 exist. 6. Let f (x) d2 (x ) dx (b) The normal line has slope (2, 3). y y 1 (x 5 1 x 5 2) 17 5 3 1 and passes through 5 7 h→0 f (x) (x h→0 1. (a) The tangent line has slope 5 and passes through (2, 3). y 5(x 2) 3 y 5x 7 7x x 2. lim Section 3.1 Exercises 12) y(x) lim 7 h→0 h (3x 3 h→0 y(x lim h→0 lim 0 lim 3 h→0 h h→0 9. No, the two one-sided limits are different (see Exercise 7). 10. No, f is discontinuous at x h) h lim h→0 1 9 f (x h→0 6. Use the graph of f in the window [ 6, 6] by [ 4, 4] to find that (0, 2) is the coordinate of the high point and (2, 2) is the coordinate of the low point. Therefore, f is increasing on ( , 0] and [2, ). x 3)(x)(3) 1 3x lim x→3 tangent to the parabola at its vertex is 0. 7. lim f (x) f (3) 3 x→3 x x 5. The vertex of the parabola is at (0, 1). The slope of the line (h 2 h) 1. y→0 y lim 1 9 1 lim 1 for y y x→4 3 2 y 3. Since 4. lim 2 lim x2 h→0 lim (2x h→0 lim f (x h) h h→0 h)2 x 2 h 2xh h 2 h h) f (x) x2 2x 7. The graph of y f1(x) is decreasing for x 0 and increasing for x 0, so its derivative is negative for x and positive for x 0. (b) 0 8. The graph of y f2(x) is always increasing, so its derivative is always 0. (a) 9. The graph of y f3(x) oscillates between increasing and decreasing, so its derivative oscillates between positive and negative. (d) 10. The graph of y f4(x) is decreasing, then increasing , then decreasing, and then increasing, so its derivative is negative, then positive, then negative, and then positive. (c) Section 3.1 11. dy dx lim h→0 lim y(x h) y(x) h [2(x h)2 13(x 15. (a) The amount of daylight is increasing at the fastest rate 5] (2x 2 h 13x 13h 5 h h) h→0 2x 2 4xh 2h 2 4xh lim (4x 2h 13) 4(3) 13 13x h→0 lim h→0 dy dx 4x slope 1 and passes through (3, y(3)) y x 12. Let f (x) 1, so the tangent line has (3, 16). x 3. lim h→0 lim h→0 lim f (1 (1 1 h→0 16 lim (3 h2 ) 3h (b) Yes, the rate of change is zero when the tangent to the graph is horizontal. This occurs near the beginning of the year and halfway through the year, around January 1 and July 1. (c) Positive: January 1 through July 1 Negative: July 1 through December 31 16. The slope of the given graph is zero at x 2 and at x 1, so the derivative graph includes ( 2, 0) and (1, 0). The slopes at x 3 and at x 2 are about 5 and the slope at x 0.5 is about 2.5, so the derivative graph includes ( 3, 5), (2, 5), and ( 0.5, 2.5). Connecting the points smoothly, we obtain the graph shown. h) f (1) h h)3 13 h 3h 3h 2 h 3 h h→0 about one-fourth of the way through the year, sometime around April 1. The rate at this time is approximately 13 f (1) 3) 5 13 3, 1(x 13x 1 4 hours or hour per day. 6 24 days At x y 2x 2 when the slope of the graph is largest. This occurs 5) 13h h→0 2h 2 h lim y 5 1 3 5 (a) The tangent line has slope 3 and passes through (1, 1). Its equation is y 3(x 1) 1, or y 3x 2. (b) The normal line has slope (1, 1). Its equation is y y 1 x 3 1 and passes through 3 1 (x 1) 1, or 3 4 . 3 13. Since the graph of y x ln x x is decreasing for 0 x 1 and increasing for x 1, its derivative is negative for 0 x 1 and positive for x 1. The only one of the given functions with this property is y ln x. Note also that y ln x is undefined for x 0, which further agrees with the given graph. (ii) 14. Each of the functions y property that y(0) sin x, y x has the 0, so none of these functions can be its own derivative. The function y derivative because y(1) 2 x 2 is not its own 1 but 2 (1 1 h→0 h) h lim (2 y (1) x, y 0 but the graph has nonzero slope (or undefined slope) at x h) h2 2. lim h→0 71 lim h→0 2h 17. (a) Using Figure 3.10a, the number of rabbits is largest after 40 days and smallest from about 130 to 200 days. Using Figure 3.10b, the derivative is 0 at these times. (b) Using Figure 3.10b, the derivative is largest after 20 days and smallest after about 63 days. Using Figure 3.10a, there were 1700 and about 1300 rabbits, respectively, at these times. 18. (a) The slope from x 4 to x The slope from x 0 to x The slope from x 1 to x The slope from x 4 to x 1 20 . 2 0 ( 4) 22 1 is 4. 10 2 ( 2) 4 is 0. 41 2 ( 2) 6 is 2. 64 0 is Note that the derivative is undefined at x 0, x 1, and x 4. (The function is differentiable at x 4 and at x 6 because these are endpoints of the domain and the one-sided derivatives exist.) The graph of the derivative is shown. h This leaves only e x, which can plausibly be its own derivative because both the function value and the slope increase from very small positive values to very large values as we move from left to right along the graph. (iv) x y 5 4 3 2 –5–4–3–2 –1 –2 –3 –4 –5 1234567 x 72 Section 3.1 18. continued (b) x 19. (b) 0, 1, 4 Midpoint of Interval (x) 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 y x 0.00 0 3.3 0 3.3 10.0 1 13.3 16.6 2 29.9 23.3 3 53.2 30.0 4 83.2 36.6 5 119.8 43.2 6 163.0 49.9 7 212.9 56.6 8 269.5 63.2 9 0.56 Slope 3.3 1 13.3 2 29.9 3 53.2 4 83.2 5 119.8 6 163.0 7 212.9 8 269.5 9 332.7 10 0.5 Midpoint of Interval (x) 0.56 2 0.92 2 0.92 1.19 2 1.19 1.30 2 1.30 1.39 2 1.39 1.57 2 1.57 1.74 2 1.74 1.98 2 1.98 2.18 2 2.18 2.41 2 A graph of the derivative data is shown. 2.41 2.64 2 2.64 3.24 2 Slope y x 0.28 1512 0.56 1577 0.00 116.07 0.74 1448 0.92 1512 0.56 177.78 1.055 1384 1.19 1448 0.92 237.04 1.245 1319 1.30 1384 1.19 590.91 1.345 1255 1.39 1319 1.30 711.11 1.48 1191 1.57 1255 1.39 355.56 1.655 1126 1.74 1191 1.57 382.35 1.86 1062 1.98 1126 1.74 266.67 2.08 998 2.18 1062 1.98 320.00 2.295 933 2.41 998 2.18 282.61 2.525 869 2.64 933 2.41 278.26 2.94 805 3.24 869 2.64 106.67 A graph of the derivative data is shown. [0, 10] by[ 10, 80] (a) The derivative represents the speed of the skier. (b) Since the distances are given in feet and the times are given in seconds, the units are feet per second. (c) The graph appears to be approximately linear and [0, 3.24] by [ 800, 100] (c) Since the elevation y is given in feet and the distance x down river is given in miles, the units of the gradiant are feet per mile. passes through (0, 0) and (9.5, 63.2), so the slope is 63.2 9.5 0 0 6.65. The equation of the derivative is approximately D (d) Since the elevation y is given in feet and the distance x 6.65t. downriver is given in miles, the units of the distance 20. (a) dy dx are feet per mile. (e) Look for the steepest part of the curve. This is where the elevation is dropping most rapidly, and therefore the most likely location for significant “rapids.” [ 0.5, 4] by [700, 1700] (f) Look for the lowest point on the graph. This is where the elevation is dropping most rapidly, and therefore the most likely location for significant “rapids.” 21. [ , ] by [ 1.5, 1.5] The cosine function could be the derivative of the sine function. The values of the cosine are positive where the sine is increasing, zero where the sine has horizontal tangents, and negative where sine is decreasing. Section 3.1 22. We show that the right-hand derivative at 1 does not exist. h) f (1) 3(1 lim h h→0 2 3h 2 lim lim 3 h h h→0 h→0 lim f (1 23. lim f (0 h→0 lim h) h f (0) lim 1 h→0 lim f (x) 2. x→1 (f) lim f (1 h→0 h h→0 (e) Yes, the one-sided limits exist and are the same, so (1)3 h) h h→0 0 h lim h→0 h h) h 2h lim lim h)2 h (1 h→0 h2 h) f (1) 2(1 lim h h→0 h→0 1 2h 1 lim lim 2 h h h→0 h→0 Thus, the right-hand derivative at 0 does not exist. lim (2 12 2 h) h (g) lim 24. Two parabolas are parallel if they have the same derivative at every value of x. This means that their tangent lines are parallel at each value of x. Two such parabolas are given by y x 2 and y x 2 4. They are graphed below. f (1) h) h→0 h h 73 h→0 h f (1 12 The right-hand derivative does not exist. (h) It does not exist because the right-hand derivative does not exist. 27. The y-intercept of the derivative is b a. 28. Since the function must be continuous at x 1, we have 2. lim (3x k) f (1) 1, so 3 k 1, or k x→1 x 3, 3x This gives f (x) slope 1, the graph of y y 2( 1) 1 1): y 2x 1, so for x 1. Then 1, the graph of f (x) must lie on a line of slope 1 that passes through ( 1, 1): y Thus f (x) 1(x 1) x 1 or y 2, 2x 1, f (1 h) h f (1) x x x 2. (1 3 h) h 3h 2 h (1) lim (3 3h h 2) lim h→0 lim 3h h→0 h) f (1) h h→0 (1 3h) 1 lim h h→0 lim f (1 1. 3 h→0 lim h→0 f (x) must lie on a line of 2 that passes through (0, y( 1) 1 1. Now we confirm that f (x) is differentiable at x [ 4, 4] by [ 5, 20] The parabolas are “everywhere equidistant,” as long as the distance between them is always measured along a vertical line. 25. For x x x 2, lim [3(1 h) lim 3 2] 3 (1)3] h h→0 h→0 h3 3 Since the right-hand derivative equals the left-hand derivative at x 1, the derivative exists (and is equal to 3) when k 2. 1 1 29. (a) 1 y 5 364 365 363 365 0.992 Alternate method: P 365 3 3 365 0.992 (b) Using the answer to part (a), the probability is about 1 0.992 0.008. x 5 (c) Let P represent the answer to part (b), P 0.008. Then the probability that three people all have different birthdays is 1 26. (a) f (x) lim h→0 lim f (x (x h→0 lim (2x h→0 (b) f (x) lim h→0 lim f (x 2(x h→0 (c) lim f (x) x→1 (d) lim f (x) x→1 h) h h)2 h x2 probability that all have different birthdays is lim h 2x h) h h) h h2 2xh h→0 h) f (x) lim 2x x→1 lim 2 x→1 f (x) 2x 2 (1 1 lim 2 h→0 2(1) P. Adding a fourth person, the 2 2 P) (1 362 , so the probability of a shared birthday is 365 362 P) 0.016. 365 (d) No. Clearly February 29 is a much less likely birth date. Furthermore, census data do not support the assumption that the other 365 birth dates are equally likely. However, this simplifying assumption may still give us some insight into this problem even if the calculated probabilities aren’t completely accurate. 74 Section 3.2 s Section 3.2 Differentiability (pp. 105–112) Quick Review 3.2 1. Yes Exploration 1 Zooming in to “See” Differentiability 2. No (The f (h) term in the numerator is incorrect.) 1. Zooming in on the graph of f at the point (0, 1) always produces a graph exactly like the one shown below, provided that a square window is used. The corner shows no sign of straightening out. 3. Yes 4. Yes 5. No (The denominator for this expression should be 2h). 6. All reals 7. [0, ) 8. [3, ) 9. The equation is equivalent to y slope is 3.2. 10. f (3 0.001) f (3 0.002 (3.2 5), so the 5(3 0.001) 5(3 0.002 5(0.002) 5 0.002 0.001) [ 0.25, 0.25] by [0.836, 1.164] 2. Zooming in on the graph of g at the point (0, 1) begins to reveal a smooth turning point. This graph shows the result of three zooms, each by a factor of 4 horizontally and vertically, starting with the window [ 4, 4] by [ 1.624, 3.624]. 3.2x 0.001) Section 3.2 Exercises 1. Left-hand derivative: lim f (0 h→0 h) h f (0) lim h2 h→0 0 lim h 0 h→0 h Right-hand derivative: lim f (0 h→0 [ 0.0625, 0.0625] by [0.959, 1.041] 3. On our grapher, the graph became horizontal after 8 zooms. Results can vary on different machines. 4. As we zoom in on the graphs of f and g together, the differentiable function gradually straightens out to resemble its tangent line, while the nondifferentiable function stubbornly retains its same shape. Since 0 point P. h) h f (0) lim h h→0 0 lim 1 1 h→0 h 1, the function is not differentiable at the 2. Left-hand derivative: lim f (1 h→0 h) h f (1) lim 2 h→0 2 lim 0 0 h→0 h Right-hand derivative: lim f (1 h→0 Since 0 point P. h) h f (1) lim 2(1 h) h h→0 2 lim 2 h→0 2 2, the function is not differentiable at the [ 0.03125, 0.03125] by [0.9795, 1.0205] 3. Left-hand derivative: Exploration 2 Looking at the Symmetric Difference Quotient Analytically 1. f (10 h) h f (10) f (10) 2 10 2 (10.01) 10 0.01 lim f (1 h→0 2 h) h f (1) 20.01 (1 2. h) f (10 h) 2h lim (10.01)2 (9.99)2 0.02 f (10 h) h f (10) 3 102 f (10) (10.01)3 103 0.01 h→0 h) f (10 lim 20 300.3001. lim f (1 h→0 2h h) h f (1) h 1 lim [2(1 1) 1) 1 lim 1 2 1 2 The symmetric difference quotient is 0.0001 away from Since f (10). point P. 1 2h h lim 2 h→0 300.0001. h) 1] h h→0 h→0 (10.01)3 (9.99)3 0.02 h h Right-hand derivative: 300 h) 1 1 h→0 The difference quotient is 0.3001 away from f (10). f (10 h( 1)( 1 h( 1 h h) 1 (1 The symmetric difference quotient exactly equals f (10). 3. h h→0 20 1 h h lim The difference quotient is 0.01 away from f (10). f (10 1 lim h→0 2 2, the function is not differentiable at the 1) 75 Section 3.2 4. Left-hand derivative: lim f (1 h) h h→0 15. Note that y f (1) (1 lim h) h h→0 Right-hand derivative: f (1 lim h) h h→0 f (1) lim lim 1 1 h→0 1 1 lim lim 1 h lim h→0 lim h→0 16. lim lim h→0 lim (c) None 7. (a) All points in [ 3, 3] except x 0 0 8. (a) All points in [ 2, 3] except x 1, 0, 2 1 2 0 0 10. (a) All points in [ 3, 3] except x 2, x 2, 2 2 (c) None 11. Since lim tan 1 x→0 x tan 1 0 0 y(0), the problem is a discontinuity. lim y(0 h) h h) h y(0) 4/5 h h h4/5 lim h h→0 1 h1/5 1 lim 1/5 h→0 h lim lim h→0 y(0) 13. Note that y 2, 2x 2, lim y(0 lim y(0 h→0 h) h h) h x x x y(0) x2 0 0. lim h→0 y(0) h→0 y(0 h→0 lim h→0 h) y(0) h 1 h2/3 lim 2 x 1 3 lim h→0 3 h h 0 h h 0 3 lim h→0 lim h→0 h h 3 lim h→0 h h 1 h2/3 h(2 h(2) lim [ 3 3(2 k) 3 3 lim 6 x lim 2 h→0 g (0 2 h (2h h→0 lim 0 2 lim 0 h→0 2) h (3 3 h) h h→0 lim 2 2 h→0 The problem is a vertical tangent. 3 5 1 2/3 k→0 k h) h lim (h 2 5] k k→0 19. Note that the sine function is odd, so sin x 1, P(x) sin ( x ) 1 sin x 1, The graph of P(x) has a corner at x differentiable for all reals except x 3 lim h→0 h h g (0 g (0) h) h lim (h h→0 2) h→0 h→0 lim lim 1 h→0 2. It is x0 x 0. 0. The function is 0. 21. The function is piecewise-defined in terms of polynomials, so it is differentiable everywhere except possibly at x 0 and at x 3. Check x 0: The problem is a corner. 14. lim 5 20. Since the cosine function is even, so Q(x) 3 cos ( x ) 3 cos x. The function is differentiable for all reals. The problem is a cusp. h→0 h) y(0) h 1 h2/3 h) y(0) h The function has a vertical tangent at x differentiable for all reals except x 2. (c) None h→0 (h lim h→0 k) k k→0 3 3k lim k→0 k lim 9. (a) All points in [ 1, 2] except x (b) x y(0) lim 5 h→0 18. The function is differentiable except possibly where 3x 6 0, that is, at x 2. We check for differentiability at x 2, using k instead of the usual h, in order to avoid confusion with the function h(x). (b) None 0, x 1) ( 1) h 1) ( 1) h h→0 0 0 17. Find the zeros of the denominator. x2 4x 5 0 (x 1)(x 5) 0 x 1 or x 5 The function is a rational function, so it is differentiable for all x in its domain: all reals except x 1, 5. (b) None (c) x (5h lim x x The problem is a cusp. 6. (a) All points in [ 2, 3] y(0 h→0 lim (c) None h→0 y(0 h→0 (b) None 12. lim y(0 h→0 1, the function is not differentiable at the (b) x y(0) 5x 1, x 1, 1 The problem is a corner. 5. (a) All points in [ 3, 2] (b) x y(0 h→0 h) h h) h 2x 1 lim (c) x y(0 h→0 h (1 h) h(1 h) h h(1 h) 1 1 1h h→0 h→0 Since 1 point P. 1 3x 1)2 h 1 1) h 1 lim h2 h→0 2h h 2 g (0) lim h→0 (2h The function is differentiable at x lim 2 h→0 0. 2 76 Section 3.2 21. continued Check x 27. 3: Since g(3) lim g(x) x→3 3)2 (4 lim (2x x→3 1 and 1) 2(3) 1 7, the function is not continuous (and hence not differentiable) at x 3. The function is differentiable for all reals except x 2 22. Note that C(x) x, x 2, xx x x lim C(0 h→0 3. y NDER (x) actually has two asymptotes for each asymptote of y C(0) lim h→0 hh 0 h tan x. The asymptotes of y occur at x 0. 0: h) h tan x Note: Due to the way NDER is defined, the graph of 0 , so it is 0 differentiable for all x except possibly at x Check x [ 2 , 2 ] by [ 4, 4] dy dx k 2 NDER (x) 0.001, where k is an integer. A good window for viewing this behavior is [1.566, 1.576] by lim h h→0 0 [ 1000, 1000]. The function is differentiable for all reals. 28. 23. (a) x 0 is not in their domains, or, they are both discontinuous at x 0. 1 1 ,0 x x 1 1 For 2 : NDER 2 , 0 x x (b) For : NDER 1,000,000 [ 2 , 2 ] by [ 20, 20] 0 (c) It returns an incorrect response because even though these functions are not defined at x defined at x 0, they are 0.001. The responses differ from each 1 other because 2 is even (which automatically makes x 1 1 0) and is odd. NDER 2 , 0 x x 24. The graph of NDER (x) does not look like the graph of any basic function. 29. (a) lim f (x) x→1 f (1) x) a(1)2 2 lim (3 x→1 a The relationship is a b(1) b b 2. (b) Since the function needs to be continuous, we may assume that a b 2 and f (1) 2. lim f (1 h→0 [ 5, 5] by [ 10, 10] dy x3 dx f (1) f (1 h) h f (1) lim h→0 lim h→0 lim h→0 dy dx sin x 26. 3 (1 h) a(1 1 h)2 b(1 h) 2 h a 2ah ah 2 b bh h 2ah ah 2 bh (a b h lim (2a h→0 2a 2 h lim ( 1) h→0 [ 2 , 2 ] by [ 1.5, 1.5] lim h→0 h→0 lim 25. h) h ah 2 2) b) b Therefore, 2a b 1. Substituting 2 a for b gives 2a (2 a) 1, so a 3. Then b 2 a 2 ( 3) 5. The values are a 3 and b 5. [ 6, 6] by [ 4, 4] dy dx abs (x) or x 30. The function f (x) does not have the intermediate value property. Choose some a in ( 1, 0) and b in (0, 1). Then f (a) 0 and f (b) 1, but f does not take on any value between 0 and 1. Therefore, by the Intermediate Value Theorem for Derivatives, f cannot be the derivative of any function on [ 1, 1]. 77 Section 3.3 31. (a) Note that x 1 x so lim x sin x→0 1 x sin 0 by the Sandwich Theorem. Therefore, f is continuous at x f (0 (b) h) h 1 h sin h f (0) x 10. (a) f (x) 0. 0 h) h 1 h h 2 sin g(0) 0 1. 1 h h sin h As noted in part (a), the limit of this as x approaches zero is 0, so g (0) lim h→0 x(x h lim 1 h→0 f (x) lim x h h lim h→0 hx(x h) 2 x x2 h) x h h→0 Section 3.3 Exercises (d) No, because the limit in part (c) does not exist. g(0 lim 1 x h h→0 f (x lim lim interval containing 0). (e) hx h f (x) h→0 h→0 0 (that is, for h in any open h) h h) h x (x h) hx(x h) (b) f (x) 1 and 1 an infinite number of times arbitrarily close to h lim x h→0 1 oscillates h (c) The limit does not exist because sin f (x h→0 lim 1 h sin h between 9. These are all constant functions, so the graph of each function is a horizontal line and the derivative of each function is 0. x, for all x, 2. 0. 3. s Section 3.3 Rules for Differentiation (pp. 112–121) dy dx d 2y dx 2 d ( x 2) dx d ( 2x) dx dy dx d 2y dx 2 d 13 x dx 3 d2 (x ) dx dy dx d 2y dx 2 d (2x) dx d (2) dx dy dx d 2y dx 2 d2 (x ) dx d (2x) dx dy dx d 13 x dx 3 d (3) dx 2x 0 2x 2 d (x) dx d (1) dx x2 2x d (1) dx 1 0 2 2x 0 2 0 Quick Review 3.3 1. (x 2 x 2. 2)(x 1 1) x 2 2x 1 1 2 x 2x 3 2x 2 3x 4 1 5. (x 6. 1 x x 2x 1 3x 4 2x 2 4 2 2)(x x 3 x x 5 x2 1) 2x 1 x2 1 x2 x 3x 2 1 3. 3x 2 4. x2 1 x x x2 x 2x 2 1 x 1 x 1 x 1 4. 21 1 x 5. 2 5x x2 x 2 2x 3 2x 2 2 2x 32 x 2 4 2x 2 x 2 x 1 x 1 2x 2x 2 dy dx 2 2 21 6. 2 dy dx 3 x (x 3 1 x 2 ) x 2 x 1 7. dy dx At x 1.173, 500x 6 1305. At x 2.394, 500x 6 94,212 After rounding, we have: At x 1, 500x 6 1305. At x 2, 500x 6 94,212. 8. (a) f (10) 8. 7 (b) f (0) 7 (c) f (x h) f (x ) (d) lim x x→a 9. 7 f (a) a 7 a lim 0 x→a 0 2x 3x 2 1 2 6x 2x d (4x 3) dx 4x 1 0 2x 1 42x 4 d3 (x ) dx 3x 2 2x 6x d (2x 2) dx 0 4x 3 d (21x 2) dx d (15) dx 21x 2 d (4x) dx 4x 12x 2 dy dx d 2y dx 2 d d (5x 3) (3x 5) 15x 2 15x 4 dx dx d d (15x 2) (15x 4) 30x 60x 3 dx dx dy dx d (4x 2) dx 2 dy dx 2 1 d (3x 2) dx d (7x 3) dx 21x 2 2x d2 (x ) dx 2 8x 7 lim x→a x d (1) dx d (x) dx d4 (x ) dx d 2y dx 2 0 d (x) dx d (2x) dx 4x 3 [0, 5] by [ 6, 6] d (x) dx d ( 1) dx 0 1 1 d (1) dx d 2y dx 2 7. d 12 x dx 2 d2 (x ) dx 0 2 d d (x) (1) 2x dx dx d (1) 2 0 2 dx 3 8 d (8x) dx 0 d ( 8x 3) dx 8x d (8) dx d (1) dx 3 8 24x 4 0 24x 4 78 10. Section 3.3 dy dx d1 4 x dx 4 d (3) dx x 5 x 5 d 2y dx 2 x 4 x 6 dy dx d1 2 x dx 2 d (x 1) dx (x 18. x 4 3 x 3 d ( x 5) dx 5x 11. (a) d1 3 x dx 3 5 4x x 2 x 0 4 d (x 3 ) dx 1)(2x) (x 2x 2 2x x2 3x 2 2x 1) (x 19. d (x dx (x dy dx d [(x dx 1)(x d3 (x dx 3x (x 2 2 x2 d 1) (x dx 1) (2x 3 3x 2 dy dx 2 1)] x 1) h lim f (x 3x 3x x( 1)2 x 2 2 2)(2x h) h h→0 lim (1) 3) 5x 6) f (x) lim h→0 1 h→0 d dx (x 2 3) d ( u) dx u(x) is a function of x. x 2 u(x lim u(x 3 x2 d x2 3 x dx 3 1 x2 d (x dx 3x 1 ) 1 2 3x f (x ) 0 dy 14. dx d dx 0 5x x2 5x x2 2 2x d (x dx d (1 dx (3x 5 2 x 1 ) 0 1) 3x d1x dx 1 x 2 x 2 2x 1 (1 x 2)2 (1 d x2 dx 1 x 3 (1 (2x 2)2 23. (a) At x 5)(3) 4 2)2 0, d f (x) dx 1 d (uv) dx u(0)v (0) ( 1)( 3) f (x) [ f (x)]2 v(0)u (0) 13 x ) 0, (c) At x 0, (5)(2) x)(2x) (d) At x 7(2) x 3)(2x) x 2( 3x 2) (1 x 3)2 d (c) dx v(0)u (0) u(0)v (0) du [v(0)]2 dx v ( 1)( 3) (5)(2) 7 ( 1)2 (b) At x 2 d x3 1 dx x3 3 x4 x 2)( 1) (1 (1 x 2)2 du dx 19 (3x d (1 5x 1 dx 5 2 x2 x3 3 1)(x 2 x x3 3 2)(2) (3x u(x) [ f (x)]2 (5)(2) d 2x dx 3x u(x) d d (c f (x)) c f (x) f (x) dx dx d d c f (x) 0 c f (x) dx dx d 1 22. dx f (x) This is equivalent to the answer in part (a). dy 13. dx h) h h→0 3) h) [ u(x)] h h) h h→0 lim (x 2 u(x lim h→0 3) (x) 21. dy dx d x2 dx x 2 2) 2) f (x) h→0 h 1 x2 17. 1 1)2 x. d (x) dx 3 dx x dx x(2x) dy dx x 12 6x 2 (x 2 3x 2)2 1 x2 16. 1) 2 d dx dy dx x( 5x 6) (2x 3 3x 2 (x 2 3x 2)2 1)(1) lim 2x x (x 2 15. x 2 2 2 2)(2x 3) (x 2 3x (x 2 3x 2)2 3x (b) Note that u dy dx x 1) 1)2 1)(x 1)(x 20. (a) Let f (x) (b) 2 1 dy (b) dx 12. (a) ( 1 1 1) d (x 2) dx 3 d [(x 1)(x 2 1)] dx d (x 1) (x 2 1) (x 2 dx (x (x x (x x 2 1 2 x( x 2x 2 x d dx 2 d (x 4) dx 3x dy dx 1 1) x 4 2x (1 x 3)2 0, u(0)v (0) v(0)u (0) [u(0)]2 dv dx u ( 1)( 3) (5)2 d (7v dx 2u) 2( 3) 20 7 25 7v (0) 2u (0) (x h) h x Section 3.3 24. (a) At x 2, d (uv) dx (3)(2) (1)( 4) u(2)v (2) v(2)u (2) 2 v(2)u (2) u(2)v (2) du [v(2)]2 dx v (1)( 4) (3)(2) 10 (1)2 (b) At x 2, 29. y (x) 2, [ 3, 3] by [ 10, 30] d (d) Use the result from part (a) for (uv). dx d At x 2, (3u 2v 2uv) dx d 3u (2) 2v (2) 2 (uv) dx 3( 4) 2(2) 30. y (x) 3x 2 y ( 2) 12 The tangent line has slope 12 and passes through ( 2, so its equation is y 2(2) x-intercept is 3 x 2 3x 3 3(2)2 3 9 y 1 and passes through (2, 3). 9 1 (x 9 1 x 9 2) 12x 3 29 9 31. y (x) (x 2 1)(4) 4x(2x) (x 2 1)2 At the origin: y (0) The tangent is y At (1, 2): y (1) 4x 2 4 (x 2 1)2 4 4x. 0 The tangent is y 2. Graphical support: Graphical support: [ 4.7, 4.7] by [ 3.1, 3.1] [ 4.7, 4.7] by [ 2.1, 4.1] 28. y (x) 3x 2 1 The slope is 4 when 3x 2 1 4, at x 1. The tangent at x 1 has slope 4 and passes through ( 1, 2), so its equation is y 4(x 1) 2, or y 4x 2. The tangent at x 1 has slope 4 and passes through (1, 2), so its equation is y 4(x 1) 2, or y 4x 2. The smallest slope occurs when 3x 2 1 is minimized, so the smallest slope is 1 and occurs at x 0. 32. y (x) y (2) x 2)(0) 8(2x) (4 x 2)2 (4 16x x 2)2 1 2 The tangent has slope equation is y 1 (x 2 Graphical support: Graphical support: [ 3, 5] by [ 2, 4] [ 4.7, 4.7] by [ 3.1, 3.1] (4 8), 16. The [ 3, 3] by [ 20, 20] 2 The tangent line has slope 9, so the perpendicular line has y 8, or y 6, so the 3 slope is . (iii) 2 slope 2) Graphical support: 25. y (x) 2x 5 y (3) 2(3) 5 11 The slope is 11. (iii) 26. The given equation is equivalent to y 12(x 4 and the y-intercept is 16. 3 12 27. y (x) y (2) 1 Graphical support: u(2)v (2) v(2)u (2) dv [u(2)]2 dx u (3)(2) (1)( 4) 10 (3)2 9 (c) At x 6x 2 6x 12 6(x 2 x 2) 6(x 1)(x 2) The tangent is parallel to the x-axis when y 0, at x and at x 2. Since y( 1) 27 and y(2) 0, the two points where this occurs are ( 1, 27) and (2, 0). 79 1 and passes through (2, 1). Its 2 1 2) 1, or y x 2. 2 80 33. Section 3.4 an 2 d nRT V2 dV V nb d (V nb) (nRT) dP dV (c) (nRT) dV (V 0 (V nR T nb)2 d (V dV nb) nb)2 2an 2V d (an 2V 2) dV 3 d (uv) dx (V 34. 35. ds dt d 2s dt 2 d (4.9t 2) dt d (9.8t) dt dR dM 2an V3 nRT nb)2 CM Exploration 1 M2 37. If the radius of a sphere is changed by a very small amount r, the change in the volume can be thought of as a very thin layer with an area given by the surface area, 4 r 2, and a thickness given by r. Therefore, the change in the volume can be thought of as (4 r 2)( r), which means that the change in the volume divided by the change in the radius is just 4 r 2. 38. Let t(x) be the number of trees and y(x) be the yield per tree x years from now. Then t(0) d (ty) dx 156, y(0) 12, t (0) 13, 1.5. The rate of increase of production is t(0)y (0) y(0)t (0) (156)(1.5) (12)(13) 39. Let m(x) be the number of members and c(x) be the pavillion cost x years from now. Then m(0) 250, m (0) 6, and c (0) of each member’s share is (65)(10) (250)(6) (65)2 dc dx m Growth Rings on a Tree 1. Figure 3.22 is a better model, as it shows rings of equal area as opposed to rings of equal width. It is not likely that a tree could sustain increased growth year after year, although climate conditions do produce some years of greater growth than others. 2. Rings of equal area suggest that the tree adds approximately the same amount of wood to its girth each year. With access to approximately the same raw materials from which to make the wood each year, this is how most trees actually grow. 3. Since change in area is constant, so also is change in area . If we denote this latter constant by k, we 2 k have r, which means that r varies change in radius inversely as the change in the radius. In other words, the change in radius must get smaller when r gets bigger, and vice-versa. Exploration 2 390 bushels of annual production per year. c(0) u du u v dv v (u du)(v) (u)(v dv) (v dv)(v) uv vdu uv udv v 2 vdv vdu udv v2 s Section 3.4 Velocity and Other Rates of Change (pp. 122–133) 36. If the radius of a circle is changed by a very small amount r, the change in the area can be thought of as a very thin strip with length given by the circumference, 2 r, and width r. Therefore, the change in the area can be thought of as (2 r)( r), which means that the change in the area divided by the change in the radius is just 2 r. and y (0) u v 9.8 M 3 13 M 3 dv dx u . This is equivalent to the product (d) Because dx is ‘infinitely small,” and this could be thought of as dividing by zero. 9.8t C d M2 2 dM dC2 M dM 2 du dx rule given in the text. (e) d 2 v Modeling Horizontal Motion 1. The particle reverses direction at about t t 2.06. 0.61 and 65, 10. The rate of change m(0)c (0) c(0)m (0) [m(0)]2 0.201 dollars per year. Each member’s share of the cost is decreasing by approximately 20 cents per year. 40. (a) It is insignificant in the limiting case and can be treated as zero (and removed from the expression). (b) It was “rejected” because it is incomparably smaller than the other terms: v du and u dv. 2. When the trace cursor is moving to the right the particle is moving to the right, and when the cursor is moving to the left the particle is moving to the left. Again we find the particle reverses direction at about t 0.61 and t 2.06. Section 3.4 3. When the trace cursor is moving upward the particle is moving to the right, and when the cursor is moving downward the particle is moving to the left. Again we find the same values of t for when the particle reverses direction. 81 Quick Review 3.4 1. The coefficient of x 2 is negative, so the parabola opens downward. Graphical support: [ 1, 9] by [ 300, 200] 4. We can represent the velocity by graphing the parametric equations x4(t) x1 (t) 12t 2 32t 15, y4(t) 2 (part 1), x5(t) x1 (t) 12t 2 32t 15, y5(t) t (part 2), x6 (t) t, y6(t) x1 (t) 12t 2 32t 15 (part 3) 2. The y-intercept is f (0) 256. See the solution to Exercise 1 for graphical support. 3. The x-intercepts occur when f (x) 0. 16x 2 160x 256 0 16(x 2 10x 16) 0 16(x 2)(x 8) 0 x 2 or x 8 The x-intercepts are 2 and 8. See the solution to Exercise 1 for graphical support. 4. Since f (x) 16(x 2 10x 16) 2 16(x 10x 25 9) 16(x 5)2 144, the range is ( , 144]. See the solution to Exercise 1 for graphical support. [ 8, 20] by [ 3, 5] (x4, y4) 5. Since f (x) 16(x 2 10x 16) 2 16(x 10x 25 9) 16(x 5)2 144, the vertex is at (5, 144). See the solution to Exercise 1 for graphical support. 6. [ 8, 20] by [ 3, 5] (x5, y5) f (x) 80 16x 2 160x 256 80 16x 2 160x 336 0 16(x 2 10x 21) 0 16(x 3)(x 7) 0 x 3 or x 7 f (x) 80 at x 3 and at x 7. See the solution to Exercise 1 for graphical support. [ 2, 5] by [ 10, 20] (x6, y6) dy 7. 100 dx For (x4, y4) and (x5, y5), the particle is moving to the right when the x-coordinate of the graph (velocity) is positive, moving to the left when the x-coordinate of the graph (velocity) is negative, and is stopped when the x-coordinate of the graph (velocity) is 0. For (x6, y6), the particle is moving to the right when the y-coordinate of the graph (velocity) is positive, moving to the left when the y-coordinate of the graph (velocity) is negative, and is stopped when the y-coordinate of the graph (velocity) is 0. 32x 160 100 60 32x x dy 100 at x dx 15 8 15 8 Graphical support: the graph of NDER f (x) is shown. Exploration 3 Seeing Motion on a Graphing Calculator 1. Let tMin 0 and tMax 10. 2. Since the rock achieves a maximum height of 400 feet, set yMax to be slightly greater than 400, for example yMax 420. 4. The grapher proceeds with constant increments of t (time), so pixels appear on the screen at regular time intervals. When the rock is moving more slowly, the pixels appear closer together. When the rock is moving faster, the pixels appear farther apart. We observe faster motion when the pixels are farther apart. [ 1, 9] by [ 200, 200] dy 8. 0 dx 32x dy 160 32x x 0 when x 0 160 5 5. dx See the solution to Exercise 7 for graphical support. 82 Section 3.4 9. Note that f (x) lim f (3 h) h h→0 32x f (3) (e) 160. f (3) 32(3) 160 64 For graphical support, use the graph shown in the solution to Exercise 7 and observe that NDER (f (x), 3) 10. f (x) f (x) At x 64. 30, so it was aloft 4. On Mars: 32x 160 32 7 (and, in fact, at any other of x), Velocity 2 dy dx 2 s(t) 0 24t 0.8t 2 0 0.8t(30 t) 0 t 0 or t 30 The rock was aloft from t 0 to t for 30 seconds. ds dt Solving 3.72t 32. d (1.86t 2) dt 3.72t 16.6, the downward velocity reaches 16.6 m/sec after about 4.462 sec. Graphical support: the graph of NDER(NDER f (x)) is shown. On Jupiter: Velocity ds dt Solving 22.88t d (11.44t 2) dt 22.88t 16.6, the downward velocity reaches 16.6 m/sec after about 0.726 sec. [ 1, 9] by [ 40, 10] Section 3.4 Exercises s 3, the instantaneous rate of change is 1. Since V 2. (a) Displacement = s(5) s(0) 10 m 5 sec (b) Average velocity 2 10 m 3 3s 2. 5 m/sec 2 m/sec (c) Velocity s (t) 2t 3 At t 4, velocity s (4) (d) Acceleration 12 dV ds 2(4) 2 m/sec 2 s (t) (e) The particle changes direction when s (t) 2t 3 0, so t 3 sec. 2 (f) Since the acceleration is always positive, the position s is at a minimum when the particle changes direction, at t 3 3 sec. Its position at this time is s 2 2 3. (a) Velocity: v(t) ds dt Acceleration: a(t) d (24t 0.8t 2) 24 dt dv d (24 1.6t) dt dt 1 m. 4 1.6t m/sec 1.6 m/sec2 (b) The rock reaches its highest point when v(t) 24 1.6t 0, at t 15. It took 15 seconds. (c) The maximum height was s(15) (d) s(t) 24t 0.8t 2 0 t 180 meters. 1 (180) 2 90 0.8t 2 24 24t 90 2 ( 24) 4(0.8)(90) 2(0.8) 4.393, 25.607 It took about 4.393 seconds to reach half its maximum height. 5. The rock reaches its maximum height when the velocity s (t) 24 9.8t 0, at t 2.449. Its maximum height is about s(2.449) 29.388 meters. 6. Moon: s(t) 0 832t 2.6t 2 0 2.6t(320 t) 0 t 0 or t 320 It takes 320 seconds to return. Earth: s(t) 0 832t 16t 2 0 16t(52 t) 0 t 0 or t 52 It takes 52 seconds to return. 7. The following is one way to simulate the problem situation. For the moon: x1(t) 3(t 160) 3.1(t 160) y1(t) 832t 2.6t 2 t-values: 0 to 320 window: [0, 6] by [ 10,000, 70,000] For the earth: x1(t) 3(t 26) 3.1(t 26) y1(t) 832t 16t 2 t-values: 0 to 52 window: [0, 6] by [ 1000, 11,000] 8. The growth rate is given by b (t) 104 2 103t 10,000 2000t. At t 0: b (0) 10,000 bacteria/hour At t 5: b (5) 0 bacteria/hour At t 10: b (10) 10,000 bacteria/hour Section 3.4 9. Q(t) 200(30 t)2 200(900 60t t 2) 180,000 12,000t 200t 2 Q (t) 12,000 400t The rate of change of the amount of water in the tank after 10 minutes is Q (10) 8000 gallons per minute. Note that Q (10) 0, so the rate at which the water is running out is positive. The water is running out at the rate of 8000 gallons per minute. 13. a(t) v (t) 6t 2 18t 12 Find when acceleration is zero. 6t 2 18t 12 0 6(t 2 3t 2) 0 6(t 1)(t 2) 0 t 1 or t 2 At t 1, the speed is v(1) 0 0 m/sec. At t 2, the speed is v(2) 1 1 m/sec. 14. (a) g (x) The average rate for the first 10 minutes is Q(10) 10 Q(0) 0 80,000 180,000 10 h (x) 10,000 gallons per t (x) minute. The water is flowing out at an average rate of 10,000 83 d3 (x ) 3x 2 dx d3 (x 2) 3x 2 dx d3 (x 3) 3x 2 dx (b) The graphs of NDER g(x), NDER h(x), and NDER t(x) are all the same, as shown. gallons per minute over the first 10 min. 10. (a) Average cost c(100) 100 (b) c (x) 100 0.2x Marginal cost c (100) 11,000 100 $110 per machine [ 4, 4] by [ 10, 20] $80 per machine (c) Actual cost of 101st machine is c(101) c(100) $79.90, which is very close to the marginal cost calculated in part (b). (c) f (x) must be of the form f (x) = x 3 constant. x3 (e) Yes. f (x) 11. (a) (d) Yes. f (x) x3 c, where c is a 3 15. (a) [0, 50] by [ 500, 2200] The values of x which make sense are the whole numbers, x 0. (b) Marginal revenue d 2000 1 dx r (x) d 2000 2000 dx x1 (x 1)(0) (2000)(1) 0 (x 1)2 (c) r (5) 2000 (5 1)2 2000 36 1 x (c) 1 2000 (x 1)2 [0, 200] by [ 0.1, 0.2] 55.56 The increase in revenue is approximately $55.56. (d) The limit is 0. This means that as x gets large, one reaches a point where very little extra revenue can be expected from selling more desks. 12. v(t) s (t) 3t 2 12t 9 a(t) v (t) 6t 12 Find when velocity is zero. 3t 2 12t 9 0 3(t 2 4t 3) 0 3(t 1)(t 3) 0 t 1 or t 3 At t 1, the acceleration is a(1) At t 3, the acceleration is a(3) [0, 200] by [ 2, 12] (b) The values of x which make sense are the whole numbers, x 0. 2 6 m/sec 6 m/sec2 P is most sensitive to changes in x when P (x) is largest. It is relatively sensitive to changes in x between approximately x 60 and x 160. (d) The marginal profit, P (x), is greatest at x 106.44. Since x must be an integer, P(106) 4.924 thousand dollars or $4924. (e) P P P P P P (50) 0.013, or $13 per package sold (100) 0.165, or $165 per package sold (125) 0.118, or $118 per package sold (150) 0.031, or $31 per package sold (175) 0.006, or $6 per package sold (300) 10 6, or $0.001 per package sold (f) The limit is 10. The maximum possible profit is $10,000 monthly. (g) Yes. In order to sell more and more packages, the company might need to lower the price to a point where they won’t make any additional profit. 84 Section 3.4 16. (a) 190 ft/sec (b) 2 seconds (c) After 8 seconds, and its velocity was 0 ft/sec then (d) After about 11 seconds, and it was falling 90 ft/sec then (e) About 3 seconds (from the rocket’s highest point) (f) The acceleration was greatest just before the engine stopped. The acceleration was constant from t 2 to t 11, while the rocket was in free fall. 17. Note that “downward velocity” is positive when McCarthy is falling downward. His downward velocity increases steadily until the parachute opens, and then decreases to a constant downward velocity. One possible sketch: (b) The particle speeds up when its speed is increasing, which occurs during the approximate intervals 1.153 t 2.167 and t 3.180. It slows down during the approximate intervals 0 t 1.153 and 2.167 t 3.180. One way to determine the endpoints of these intervals is to use a grapher to find the minimums and maximums for the speed, NDER x(t) , using function mode in the window [0, 5] by [0, 10]. (c) The particle changes direction at t t 3.180 sec. 1.153 sec and at (d) The particle is at rest “instantaneously” at t 1.153 sec and at t 3.180 sec. Downward velocity (e) The velocity starts out positive but decreasing, it becomes negative, then starts to increase, and becomes positive again and continues to increase. The speed is decreasing, reaches 0 at t 1.15 sec, then increases until t 2.17 sec, decreases until t 3.18 sec when it is 0 again, and then increases after that. Time 18. (a) We estimate the slopes at several points as follows, then connect the points to create a smooth curve. t (days) 0 10 20 30 40 (f) The particle is at (5, 2) when 2t 3 13t 2 22t 5 5 at t 0.745 sec, t 1.626 sec, and at t 4.129 sec. 21. (a) s (ft) 50 50 Slope (flies/day) 0.5 3.0 13.0 14.0 3.5 0.5 p′ (slope) 20 t (sec) 5 (b) s (1) 50 22. (a) t 18, s (2.5) 0, s (3.5) 12 s (ft) 30 Horizontal axis: Days Vertical axis: Flies per day (b) Fastest: Around the 25th day Slowest: Day 50 or day 0 19. The particle is at (5, 2) when 4t 3 occurs at t 2.83. 5 16t 2 15t t (sec) 5, which 20. The motion can be simulated in parametric mode using x1(t) 2t 3 13t 2 22t 5 and y1(t) 2 in [ 6, 8] by [ 3, 5]. (a) It begins at the point ( 5, 2) moving in the positive direction. After a little more than one second, it has moved a bit past (6, 2) and it turns back in the negative direction for approximately 2 seconds. At the end of that time, it is near ( 2, 2) and it turns back again in the positive direction. After that, it continues moving in the positive direction indefinitely, speeding up as it goes. (b) s (1) 6, s (2.5) 12, s (3.5) 24 23. (a) The body reverses direction when v changes sign, at t 2 and at t 7. (b) The body is moving at a constant speed, v between t 3 and t 6. 3 m/sec, Section 3.4 (c) The speed graph is obtained by reflecting the negative portion of the velocity graph, 2 t 7, about the x-axis. 85 Velocity graph: v (t) (cm/sec) 4 Speed(m/sec) 2 3 2 1 1 2 4 6 8 10 2 3 4 2 3 4 5 6 t (sec) –2 t (sec) –1 –4 –2 Speed graph: –3 (d) For 0 t 3 1 1: a For 1 t For 3 t 3 3 3 6 3( 8 0 10 3: a 6: a For 6 t 8: a For 8 t 10: a |v (t)| (cm/sec) 0 0 3 m/sec 2 Acceleration (m/sec2) 3 2 1 (1, 3) (6, 3) (8, 3) (3, 0) 2 (6, 0) 4 6 8 (8, –1.5) –1 –2 –3 10 t (sec) (10, –1.5) (3, –3) (1, –3) 24. (a) The particle is moving left when the graph of s has negative slope, for 2 t 3 and for 5 t 6. The particle is moving right when the graph of s has positive slope, for 0 t 1. The particle is standing still when the graph of s is horizontal, for 1 t 2 and for 3 t 5. (b) For 0 t 1: v 2 1 0 0 Speed For 1 t 2: v 2 2 Speed For 2 t 3: v For 3 t 5: v Speed For 5 t 2 cm/sec v 2 1 2 cm/sec 0 cm/sec v 22 32 Speed 6: v Speed 4 3 3 m/sec 2 1 ( 3) 0 m/sec 2 3 3) 3 m/sec 2 6 3 1.5 m/sec 2 8 v 2 ( 2) 53 v 4 ( 2) 65 v 0 cm/sec 4 cm/sec 4 cm/sec 0 cm/sec 0 cm/sec 2 cm/sec 2 cm/sec 2 1 5 6 t (sec) –2 –4 25. (a) The particle moves forward when v 0, for 0 t 1 and for 5 t 7. The particle moves backward when v 0, for 1 t 5. The particle speeds up when v is negative and decreasing, for 1 t 2, and when v is positive and increasing, for 5 t 6. The particle slows down when v is positive and decreasing, for 0 t 1 and for 6 t 7, and when v is negative and increasing, for 3 t 5. (b) Note that the acceleration a t 2, t 3, and t dv is undefined at dt 6. The acceleration is positive when v is increasing, for 3 t 6. The acceleration is negative when v is decreasing, for 0 t 2 and for 6 t 7. The acceleration is zero when v is constant, for 2 t 3 and for 7 t 9. (c) The particle moves at its greatest speed when v is maximized, at t 0 and for 2 t 3. (d) The particle stands still for more than an instant when v stays at zero, for 7 t 9. 86 Section 3.4 26. (a) To graph the velocity, we estimate the slopes at several points as follows, then connect the points to create a smooth curve. t (hours) 0 2.5 5 7.5 10 12.5 v (km/hour) 0 56 75 56 0 94 (b) v s (t) 980t a s (t) 980 15 The velocity was s 225 4 7 560 cm/sec. The acceleration was s 4 7 980 cm/sec 2. v (km/hr) (c) Since there were about 16 flashes during 100 5 15 the light was flashing at a rate of about 28 flashes per t (hours) second. –100 –200 –300 To graph the acceleration, we estimate the slope of the velocity graph at several points as follows, and then connect the points to create a smooth curve. t (hours) 0 v (km/hour ) 2.5 5 7.5 30 2 15 0 15 10 30 12.5 45 a (km/hr2) 50 15 60 28. Graph C is position, graph A is velocity, and graph B is acceleration. A is the derivative of C because it is positive, negative, and zero where C is increasing, decreasing, and has horizontal tangents, respectively. The relationship between B and A is similar. 29. Graph C is position, graph B is velocity, and graph A is acceleration. B is the derivative of C because it is negative and zero where C is decreasing and has horizontal tangents, respectively. A is the derivative of B because it is positive, negative, and zero where B is increasing, decreasing, and has horizontal tangents, respectively. 25 30. (a) 10 15 t (hours) dy dt –25 –50 (b) ds dt 4 of a second, 7 30t 3t 2 d t2 61 dt 12 d t t2 61 dt 6 144 d 12 6t t dt 24 t t 01 12 12 1 (b) The fluid level is falling fastest when negative, at t 0, when falling slowest at t dy dt 12, when dy is the most dt 1. The fluid level is dy dt 0. [0, 15] by [ 300, 100] d 2s dt 2 (c) 30 6t [0, 12] by [ 2, 6] y is decreasing and [0, 15] by [ 100, 50] interval. y decreases more rapidly early in the interval, The graphs are very similar. 27. (a) Solving 160 It took 160 cm 4 7 490t 2 gives t 4 . 7 4 of a second. The average velocity was 7 sec = 280 cm/sec. dy is negative over the entire dt and the magnitude of t dy dy is larger then. is 0 at dt dt 12, where the graph of y seems to have a horizontal tangent. Section 3.5 31. (a) dV dr d4 3 r dr 3 dV When r 2, dr 37. (a) Assume that f is even. Then, 4 r2 4 (2)2 f ( x) 16 cubic feet of 32. For t f( x h) h f (x h) f (x) , and substituting k h f (x k) f (x) lim k k→0 f (x k) f (x) lim f (x) k k→0 11.092 cubic feet. So, f is an odd function. 20 3600 sec t. Multiplying by 9 1h 1 km , 1000 m (b) Assume that f is odd. Then, we find that this is equivalent to 8t kilometers per hour. Solving 8t 200 gives t f ( x) 25 seconds. The aircraft takes lim f( x h→0 lim f (x h→0 25 seconds to become airborne, and the distance it travels during this time is D(25) h) h h) h and substituting k 694.444 meters. lim 33. Let v0 be the exit velocity of a particle of lava. Then f (x 16t 2 feet, so the velocity is v0t ds Solving dt v0 feet, is s 32 v0 0 gives t v0 32 v0 1900 gives v0 64 v0 32t. v0 2 v0 2 32 64 38. . Solving 348.712. The exit velocity was about 348.712 ft/sec. Multiplying by 3600 sec 1h 1 mi , we 5280 ft find that this is equivalent to about 237.758 mi/h. 34. By estimating the slope of the velocity graph at that point. d ( fgh) dx Rule gives revenue cost, the Sum and Difference d (profit) dx d (revenue) dx d (cost), where x is dx the number of units produced. This means that marginal profit marginal revenue marginal cost. 36. (a) It takes 135 seconds. (b) Average speed F t 5 73 0 0 5 73 k) k 2 26 f (x) Making a Conjecture with NDER 1. When the graph of sin x is increasing, the graph of NDER (sin x) is positive (above the x-axis). 2. When the graph of sin x is decreasing, the graph of NDER (sin x) is negative (below the x-axis). 3. When the graph of sin x stops increasing and starts decreasing, the graph of NDER (sin x) crosses the x-axis from above to below. 4. The slope of the graph of sin x matches the value of NDER (sin x) at these points. speed is approximately 2 33 f (x) s Section 3.5 Derivatives of Trigonometric Functions (pp. 134–141) (c) Using a symmetric difference quotient, the horse’s 4 59 k) k f (x) 5. We conjecture that NDER (sin x) cos x. The graphs coincide, supporting our conjecture. 0.068 furlongs/sec. F t , d d d [f (gh)] f (gh) gh ( f) dx dx dx dh dg df fg h gh dx dx dx dg dh df gh f h fg dx dx dx Exploration 1 35. Since profit f (x f (x ) So, f is an even function. . Then the maximum height, in 16 32 ds dt lim k→0 f ( x) h, k→0 v0 2 h, 0, the speed of the aircraft in meters per second after t seconds is s(t) f ( x) h→0 (b) The increase in the volume is 4 (2)3 3 lim h→0 lim volume per foot of radius. 4 (2.2)3 3 87 1 13 0.077 furlongs/sec. (d) The horse is running the fastest during the last furlong (between 9th and 10th furlong markers). This furlong takes only 11 seconds to run, which is the least amount of time for a furlong. (e) The horse accelerates the fastest during the first furlong (between markers 0 and 1). [ 2 , 2 ] by [ 4, 4] 88 Section 3.5 6. When the graph of cos x is increasing, the graph of NDER (cos x) is positive (above the x-axis). When the graph of cos x is decreasing, the graph of NDER (cos x) is negative (below the x-axis). When the graph of cos x stops increasing and starts decreasing, the graph of NDER (cos x) crosses the x-axis from above to below. The slope of the graph of cos x matches the value of NDER (cos x) at these points. We conjecture that NDER (cos x) sin x. The graphs coincide, supporting our conjecture. 5. d (4 dx d (4) dx x 2 sin x) x 2 cos x 6. d (3x dx d 4 dx cos x 180° 2. 1.7 3. sin 2.356 (1 97.403° 3 cot x) (cot x) (1 d dx (1 d cos x 10. dx 1 sin x sin2 a 7. If tan a 1, then a 1 sin a ( 1)2 1 3 4 0 0 8. k for some integer k, so . 1 cos 2 h h(1 cos h) cos h)(1 cos h) h(1 cos h) 9. y (x) 6x 14x y (3) 12 The tangent line has slope 12 and passes through (3, 1), so its equation is y 12(x 3) 1, or y 12x 35. v (t) 12 6t 2 14t Section 3.5 Exercises 1. d (1 dx 2. d (2 sin x dx 3. d1 dx x 4. d (x sec x) dx x cos x) tan x) 0 ( sin x) 2 cos x 1 x2 5 sin x 1 sec 2 x 5 cos x d dx x (sec x) x sec x tan x sin x) sin x) 2 1 sin x 11. y (x) y( ) d (sin x dx 3) cos 1 cos x The tangent line has slope 2 10. a(t) a(3) d dx (cos x) (1 cos x)2 sin x)( sin x) (cos x)(cos x) (1 sin x)2 1 (1 cos h h sin2 h h(1 cos h) 1 (sin x (1 sin x) (1 sin x)2 2 1 cot x) 2 sin x sin 2 x cos 2 x (1 sin x)2 Range: All reals 1 cot x) sin x) (cos x) (1 k for odd integers k 2 6. cos a d dx (cot x) (1 csc2 x sin2 x (1 cot x)2 sin2 x (1 5. Domain: x cos x) 2 cot x)( csc x) (cot x)( csc 2 x) (1 cot x)2 2 4. Domain: All reals Range: [ 1, 1] d (1 dx cos x) d dx (1 csc2 x (1 cot x)2 3 x 2 (1 306 ° tan x cos x x sin x (1 cos x)2 [ 2 , 2 ] by [ 4, 4] 3 4 d dx (tan x) (x) 4 sec x tan x cos x) (x) 1 180° d dx x (tan x) d dx (1 d cot x 9. dx 1 cot x (sin x)(2x)] x sec 2x d (4 sec x) dx x cos x d dx (sin x) (x 2) 2x sin x d (3x) dx x tan x) 3 d 8. dx 1 1. 135° [x 2 cos x 0 7. Quick Review 3.5 d dx x 2 (sin x) d dx sec x (x) sec x 1 sin x ( , sin 3) Its equation is y 1 and passes through ( , 3). 1(x ) 3, or y x 3. Section 3.5 d dx 12. y (x) 8 2 16 2 4 2 The normal line has slope 8 tan and passes through , 4 16 16 4 4 , 4 d cos x dx sin x d cot x dx 16. (a) 2 d dx 8 2 16 2 y 16 x 8 3 8 x 64 4 4 4 32 d dx (sin x) (cos x) (cos x) (sin x) (sin x)2 . (sin x)( sin x) (cos x)(cos x) sin 2 x 4 Its equation is y d (cos x) (1) (1) (cos x) d1 dx dx (cos x)2 dx cos x (cos x)(0) (1)( sin x) cos 2 x sin x sec x tan x cos2 x d sec x (b) dx 1 4 4 d tan x (x) x2 x sec2 x tan x x2 ( 2)2 y d dx x (tan x) d tan x dx x 89 , or (sin2 x cos2 x) sin2 x . 1 sin2 x csc 2 x Using decimals, this equation is approximately d csc x dx (b) y 1.081x 2.122. d1 dx sin x d dx (sin x) (1) 13. y (x) d2 (x sin x) dx x 2 cos x y (3) d dx (sin x) (x 2) (sin x)(0) (1)(cos x) sin 2 x 2x sin x 9 cos 3 6 sin 3 The tangent line has slope 9 cos 3 6 sin 3 and passes through (3, 9 sin 3). Its equation is y (9 cos 3 6 sin 3)(x 3) 9 sin 3, or y (9 cos 3 6 sin 3)x 27 cos 3 9 sin 3. Using decimals, this equation is approximately y 8.063x 25.460. 14. lim h→0 cos (x (sin x)2 d dx x 2 (sin x) h) h lim cos x sin2 x 17. sin x sin h) h cos x(cos h 1) h h→0 lim (cos x) h→0 cos h h cos h h h→0 (cos x) lim cos x sin x sin h 1 (sin x) 1 d tan x dx d cot x dx 19. y (x) sin h h sin h h→0 h y 15. (a) (sin x)(1) d d sin x tan x dx dx cos x (cos x)(cos x) (sin x)( sin x) cos 2 x 1 cos2 x sin x which is 0 at x 0, 1 , which is never 0. cos2 x 1 csc 2 x , which is never 0. sin2 x d ( dx 2 cos x) 2 sin x 2 sin 2 4 1 1 2 The tangent line has slope 1 and passes through (sin x) lim sin x d (cos x) (sin x) dx cos2 x sin2 x cos2 x 0, so the slope of sec 2 x 4 4 (cos x)(0) sec x tan x which is 0 at x so the slope of the tangent line is 0. 18. (cos x cos h d sec x dx csc x cot x d the tangent line is 0. cos x dx cos x h→0 lim d dx (1) (sin x) sec 2 x y d (sin x) (cos x) dx , 2 cos 1x 4 4 4 , 1 , so its equation is 1, or y x 4 1. The normal line has slope 1 and passes through (cos x)2 its equation is y 1x 4 1, or y x 1 4 , 1 , so 4 . 90 Section 3.5 d tan x dx 20. y (x) sec 2 x sec x y (x) 0 2 csc x cot x d (2x) dx y (x) (b) 2 csc x 0 2 cos x sin2 x sec 2 x 1 sin2 x 0 1) 0 2 2 1 ( sin2 x 2 2 cos x 1 cos x 2 On , , the solutions are x 22 curve are 4 , 1 and d (4 dx cot x 0 2 21. y (x) 2 csc 2 x 2 csc 2( 2) cot 2 2 1 and passes through 1x Speed: 2 cos 2 0 2 csc x cot x 0 Jerk: 2 cos 2 cos x sin2 x 0 1) 1 2 x 3 cot 3 1 4 2 csc 2 4 Greatest when cos t at point Q 3 3 Zero when cos t 0 3 3. 3. 2( csc x cot x) (or t Zero when sin t ( csc 2 x) 2 csc 4 4 4 2( 2)(1) 2 2 y 4 , 4 . Its equation is y 4x 4. k) Greatest (in magnitude) when cos t csc 2 4 4 and passes through 4x 0 or t 1 (or t k , k odd 2 24. (a) Velocity: s (t) cos t sin t m/sec Speed: s (t) cos t sin t m/sec Acceleration: s (t) sin t cos t m/sec2 Jerk: s (t) cos t sin t m/sec3 2)2 4 The tangent line has slope P ( 0 (or t Jerk: csc 2 x cot 1 k , k odd) 2 Zero when cos t (a) y k , k odd), at the 2 Greatest (in magnitude) when sin t 4 cot x) 2 csc x cot x 0 (or t Acceleration: ,4 The equation of the horizontal line is y 2 csc x k ), at the center of endpoints of the interval of motion. 3 3 d (1 dx 1 (or t the interval of motion. 3 4 The coordinates of Q are 22. y (x) 2 m/sec 3 Speed: 2 3 3 4 2 m/sec 2 4 (c) The body starts at 2, goes to 0 and then oscillates between 0 and 4. 0 cos x 4 2 m/sec Acceleration: 2 sin 1 (2 cos x sin2 x y 2 m/sec 4 y (x) 1 sin2 x 2. 23. (a) Velocity: s (t) 2 cos t m/sec Speed: s (t) 2 cos t m/sec Acceleration: s (t) 2 sin t m/sec 2 Jerk: s (t) 2 cos t m/sec 3 (b) Velocity: 2, or 2 2. csc 2 x 3 ,2 . 4 The coordinates of Q are 1 (b) ( 1) The equation of the horizontal line is y 2(1)(0) 2 3 4 1 2 3 at point Q 4 3 cot 4 2 , 2 . Its equation is y y 3 4 1 ,1 . The tangent line has slope 2 y 2 csc x cot x 12 P . The points on the 2 csc x cot x csc 2 2 4 4 x 2 csc x) csc x csc 2 x (a) y 1 cos x 4 4, or k) 91 Section 3.5 (b) Velocity: cos Speed: 0 sin 4 0 m/sec 4 (d) 0 m/sec Acceleration: sin Jerk: sin cos 4 cos 4 2 m/sec2 4 0 m/sec3 4 (c) The body starts at 1, goes to between 2. cos (x h) cos x d cos x lim h dx h→0 cos x cos h sin x sin h cos x lim h h→0 (cos x)(cos h 1) sin x sin h lim h h→0 lim cos x 2 and then oscillates 3 4 Greatest when t Zero when t 4 180 k (e) k 1802 3 4 4 180 180 2 d dx sin x 2 sin x 2 180 1802 180 cos x 3 1803 cos x 2 3 4 d cos x dx 2 k d dx sin x 180 180 180 cos x 2 1802 k cos x 3 d cos x dx 3 25. (a) 2 d dx 2 cos x 2 180 1802 180 sin x 3 1803 26. y [ 360, 360] by [ 0.01, 0.02] The limit is 180 y because this is the conversion factor sin x d csc x csc x cot x dx d ( csc x cot x) dx d d (csc x) (cot x) (cot x) (csc x) dx dx for changing from degrees to radians. (csc x)( csc2 x) (b) csc3 x 27. y [ 360, 360] by [ 0.02, 0.02] sec2 h) sin x h sin x cos h cos x sin h sin x lim h h→0 sin x(cos h 1) cos x sin h lim h h→0 cos h 1 lim sin x lim lim cos x h h→0 h→0 h→0 lim (sin x)(0) 180 sin (x y h→0 (cos x) 180 sin h h→0 h lim (cot x)( csc x cot x) csc x cot2 x d ( tan ) d d (tan ) d This limit is still 0. d sin x dx sin h h sin x d sin x dx 3 k k Greatest (in magnitude) when t (c) lim h→0 180 d cos x dx 180 3 Jerk: 4 lim sin x h→0 2 Greatest (in magnitude) when t Zero when t (sin x) 1 sin x d2 sin x dx 2 Acceleration: Zero when t h→0 (cos x)(0) Speed: cos h h lim h→0 (tan ) d () d tan d ( sec2 tan ) d d d d [(sec )(sec )] (sec2 ) ( ) (tan ) d d d d d (sec ) (sec ) (sec ) (sec ) sec2 d d sec2 2 sec2 (2 tan tan 2 sec2 2)(sec2 ) cos x or, writing in terms of sines and cosines, 2 2 tan cos2 2 cos 2 sin cos3 92 Section 3.6 28. Continuous: 33. Note that g(0) lim g(x) x→0 so b lim g(x) x→0 lim (x x→0 b) lim cos x cos (0) x→0 b. We require lim g(x) x→0 1. The function is continuous if b d sin 2x dx d (2 sin x cos x) dx d 2 (sin x cos x) dx d 2 (sin x) (cos x) dx 1, and g(0), 1. Differentiable: For b 1, the left-hand derivative is 1 and the right-hand derivative is sin (0) 0, so the function is not differentiable. For other values of b, the function is discontinuous at x 0 and there is no left-hand derivative. So, there is no value of b that will make the function differentiable at x 0. 2[(sin x)( sin x) 2 d5 dx 5 d6 dx 6 d7 dx 7 d8 dx 8 sin x cos x sin x cos x cos x 34. d cos 2x dx d [(cos x)(cos x) (sin x)(sin x)] dx d d (cos x) (cos x) (cos x) (cos x) dx dx d d (sin x) (sin x) (sin x) (sin x) dx dx cos x sin x when n 4k 2(2 sin x cos x) cos x 2 sin 2x 35. lim h→0 (cos h h 1) lim h→0 4(249) 3, d 999 cos x dx 999 1) cos2 h 1 1) h→0 h(cos h sin2 h lim 1) h→0 h(cos h sin x. lim h→0 sin h h 0 (1) 2 30. Observe the pattern: d sin x dx 2 d sin x dx 2 d3 sin x dx 3 4 d sin x dx 4 (cos h 1)(cos h h(cos h 1) lim 3 for any whole number k. Since 999 2(sin x)(cos x) 4 sin x cos x sin x Continuing the pattern, we see that dn cos x dx n 2(cos x)( sin x) cos x cos x sin x) 2 cos 2x sin x cos x (cos x)(cos x)] 2 2(cos x 29. Observe the pattern: d cos x dx d2 cos x dx 2 3 d cos x dx 3 4 d cos x dx 4 d dx (cos x) (sin x) sin h lim h→0 cos h 1 0 5 d dx 5 d6 dx 6 d7 dx 7 d8 dx 8 cos x sin x cos x sin x sin x cos x d (A sin x dx d (A cos x dx 36. y sin x sin x y sin x cos x B cos x) A cos x B sin x Solve: sin x B sin x) B cos x y ( A sin x sin x A sin x B cos x) 2A sin x Continuing the pattern, we see that dn sin x dx n At x cos x when n 4k 2 1 for any whole number At x , this gives 0, we have k. Since 725 4(181) 1, d 725 sin x dx 725 32. (a) Using y x, sin (0.12) 0.12. (b) sin (0.12) 0.1197122; The approximation is within 0.0003 of the actual value. 0, so B B cos x) sin x sin x 1 . 2 1, so A 2B 1 and B 2 Thus, A cos x. 31. The line is tangent to the graph of y sin x at (0, 0). Since y (0) cos (0) 1, the line has slope 1 and its equation is y x. 2A sin x 2B cos x (A sin x y 0. 0. s Section 3.6 Chain Rule (pp. 141–149) Quick Review 3.6 1. f (g(x) f (x 2 1) sin (x 2 1) 2. f (g(h(x))) f (g(7x) f ((7x )2 sin [(7x)2 1] sin (49x 2 3. (g h)(x) g(h(x)) 4. (h g)(x) 5. f g(x) h(x) h(g(x)) f x2 1 7x (7x)2 g(7x) h(x 2 sin 1) 1) 1) x2 1 7x 7(x 1 2 49x 2 1) 7x 2 1 7 Section 3.6 6. cos x 2 7. g(cos x) 3 cos2 x g(3 cos2 x) 2 9. g(h(cos x)) g(h( f (x))) dy dx d (x dx f (27x 4) cos 3x 10. cos 2 3x f (g(3x 2)) f (g(h(x))) dy dx d sin (3x dx [cos (3x 2. dy dx 1) 2 f ( 3x 2 f (h(h(x))) 10. 2) dy dx cot x) cot x) d sin (7 dx [cos (7 1 d dx 3 cos (3x 5x) [cos (7 5x)]( 5) 1) 5 cos (7 11. 5x) dy dx dy dx d cos ( dx [ sin ( 3x)]( 4. dy dx d tan (2x dx [sec2 (2x 5. 6. dy dx [ sin ( 3x)] ( 3) ( 5 sin x 3)](2 3x 2) x 3)] (2x 2 2 d 5 cot 5 csc2 x x dx 10 22 2 5 csc ( 2x ) csc2 x x2 dy dx d sin x dx 1 cos x sin x 2 1 cos x 2 2 2 1 sin x cos x 2 2 (1 cos x) sin x cos x cos x 1 sin x cos x 1 (1 2 (1 1 d dx sin x (1 dy dx d cos (sin x) dx dy dx cos x) cos x) d sec (tan x) dx 3 cos2 x sin x 5)3 (2x 5)3(2) 5)4 d dx 5)4 (x 3) (2x d dx 5)3 5)3[8x 5) 3x 2(2x 3x 2(2x 5)4(3x 2) 5)4 5)4 3(2x (2x 5)] 3 5) (14x 15) d (sin3 x tan 4x) dx d d (sin3 x) (tan 4x) (tan 4x) (sin3 x) dx dx d d (sin3 x)(sec2 4x) (4x) (tan 4x)(3 sin2 x) (sin x) dx dx (sin3 x)(sec2 4x)(4) (tan 4x)(3 sin2 x)(cos x) 2 cos x sin x cos x)2 4 sin3 x sec2 4x cos x cos x)2 14. 1 cos x dy dx d (4 dx sec x 1 4 2 sec x 2 d (sin x) dx d dx sec (tan x) tan (tan x) (tan x) sec (tan x) tan (tan x) sec2 x d dx (3 cos2 x) (cos x) 5)4] sec x 2 sec x 2 sec x dy dx d dx d3 [x (2x dx d (x 3) (2x dx x (2x 13. csc2 x) csc x csc x cot x x cos x 2 2 [ sin (sin x)] x) cot x) ( cot x csc x 6 sin (sin x) cos x 8. (csc x x) (sin x) x 2(2x 2 sin x (1 cos x)2 7. 6 8x 3(2x d (2x 1) dx 2 x cos x)(cos x) (sin x)( sin x) (1 cos x)2 1 sin x 2 1 cos x d sin x dx (1 (1 x 3) sin x d sin x cos x dx 1 cos x 1 dy dx x 3) 3x 2) sec2 (2x (2 dx (x 3)(4)(2x 12. d dx [sec2 (2x 2d (x 3)(4)(2x 3x) 3 sin ( 3x) x 3) (x cos3 x) 5x) d dx 3x) cot x)2 d (sin 5 x dx 5 sin 3. dx 1 (csc x)(cot x csc x) (csc x cot x)2 1) d 5x)] (7 dx 3d 1 1 d (csc x dx (csc x 1)] (3x x) 2x f (h(3x 2)) 2 [cos (3x 1)](3) 3 x) Section 3.6 Exercises 1. 2(x (csc x f (3(3x 2)2) 2 2 x) 2(x 3(cos x 2) 3( cos x 2)2 2) h(g(cos x)) h(g( f (x))) 8. 3 cos x 6 h( cos x 9. cos 27x 4 g( f (x)) 93 3 sin2 x cos x tan 4x tan x) d (sec x tan x dx (sec x tan x tan x sec x tan x sec x sec x tan x tan x tan x) sec2 x) 94 15. Section 3.6 dy dx 3 d dx 19. 2x 1 dy dx d (1 dx cos2 7x)3 ( 2x 2x ( d dx d dx 1)(0) 2x d (2x 1 dx 1 3 2 2x 2x 1 3 1) 20. dy dx 3 16. dy dx 1) 2x 1) 1 21. d x ) (x) dx d x( dx 1 21 21 x) x (1 x 2)( 1 (1 x2) 2 x sin x 2) 3 sin x2 (2x) 22. x2 2 (1 d (1 dx x2 1 x 1 3t x) 2 ds dt 2 3t 2 d [t cos ( dt d (t) [cos ( dt 23. d sin (3x dx 2) 2 sin (3x 2) cos (3x 2) (3x 2 sin (3x 2) cos (3x 6 sin (3x 2) cos (3x 2) 3 sin (6x 4) ds dt 2)(3) d (1 dx cos 2x)2 2) d dx 2) 4t)] d dt d dt cos ( 4t) (t) 4t)] ( 4t) 4t)]( 4) cos ( cos 2x) (1 d dx d dt 4 3 4 3 4 cos ( cos ( 4t)(1) 4t) 4t) 4 4 sin 3t cos 5t 3 5 d 4 d (cos 3t) (3t) ( sin 5t) (5t) dt 5 dt 4 (cos 3t)(3) ( sin 5t)(5) 5 4 cos 3t ds dt d sin dt 3 cos 2 3 cos 2 25. dr d d tan (2 d cos 2x) cos 2x)( sin 2x) (2x) 2(1 4t) 24. d dx 2(1 2(1 cos 2x)( sin 2x)(2) 4(1 4t)] sin 2a. 2) 2 sin (3x dy dx 3t t[ sin ( x 2) 17. The last step here uses the identity 2 sin a cos a 18. d dt 2 t[ sin ( 3/2 d sin2 (3x dx sec2 5x 3t ( 3) 2 4t sin ( dy dx 1/2 3t 2 1 x 1 x2 d cos 2 dt x) x )(1) 1 tan 5x 22 2 1 ds dt 5 2 or (tan 5x) sin x2 (1 ( 1 d tan 5x 2 tan 5x dx 1 d (sec2 5x) (5x) dx 2 tan 5x 1 2 tan 5x) 2 tan 5x 5 sec2 5x 2 (1 d ( dx 2 1 cos2 7x)2 cos 7x sin 7x (sec 5x)(5) 3/2 x d dx d dx 42(1 1 3(2x cos2 7x)2(2 cos 7x)( sin 7x)(7) d dx 1 2x (2x cos2 7x)2(2 cos 7x)( sin 7x) (7x) 1 cos2 7x) (2) 2x 2 1) cos2 7x)2(2 cos 7x) (cos 7x) 3(1 1)2 1) (3) cos2 7x)2 (1 3(1 3( 3(1 3(1 d dx ( 2x sec2 (2 sin 5t 3 7 t cos t 2 4 7 d3 d7 t t sin t t 4 dt 4 dt 2 3 7 7 t sin t 2 4 4 ) )( 1) sec2 (2 sec2 (2 ) d (2 d ) ) cos 2x)(sin 2x) 26. dr d d (sec 2 tan 2 ) d d d (sec 2 ) (tan 2 ) (tan 2 ) (sec 2 ) d d d d (sec 2 )(sec2 2 ) (2 ) (tan 2 )(sec 2 tan 2 ) (2 ) d d 2 sec3 2 2 sec 2 tan2 2 Section 3.6 27. dr d d d 1 sin sin 1 2 cos dr d 2 (1 sec ) 2 ( sec ) d (sec ) d sec 1 (2 ) ( g (x) 1 (2 ) (sec y 2 (4) sec 2 sec 2) d tan x sec2 x dx d d sec2 x (2 sec x) (sec x) dx dx 10 36. f (u) 3 csc (3x y 3[2 csc (3x 18 csc (3x 32. y y d 9 tan dx x 3 sec2 3 d 3 sec2 dx x 6 sec 3 x 2 sec2 3 (1) csc2 ud u 10 du 10 5 x 2 f (5)g (1) 5 2 2 5 2 4 2 (cos u) d 2(cos u) 3 cos u du 2 sin u cos3 u 1) 1 1) 37. f (u) d dx 1)] (3x 9 sec2 xdx 3 dx 3 3 2 sec x x tan 3 3 x tan 3 x 3 d x sec dx 3 dx dx 3 (u 2 1) 1) cot (3x 1) sec 1 g ( 1) 2 2 1 () 3 5 1) cot (3x x 3 f 2 1)][ csc (3x x 3 x) x)2 (1 f (g(1))g (1) d [u du 1)] 1) cot (3x 3[2 csc (3x (1 d ( x) dx 1 1 1 ( f g) f (g( ))g 4 4 4 1 f g 4 4 1)] [ csc (3x 2 1) (3x d 1)] csc (3x dx 3[2 csc (3x dx 1 ( 1) 1) d [ 3 csc2(3x dx 2d 2 d dx csc2 (3x 5 2 1 u2 x) 1 2 g (x) 2 csc2 x cot x 1) 1 2 u f (g( 1))g ( 1) csc2 1 ( 2 csc x)( csc x cot x) 2 (5) 1 1 d cot x csc2 x dx d d ( csc2 x) ( 2 csc x) (csc x) dx dx d cot (3x dx 1 4 10 2 sec x tan x 31. y 2 x) ( f g) (1) 2 y x) ) u d cot 10 du u csc2 10 10 d g (x) (5 x) dx (2 sec x)(sec x tan x) 30. y f (1)g (1) (1 1 35. f (u) sec sec 2 u ( f g) ( 1) d (2 ) d tan ) sec )(tan ) sec ( tan 29. y d (1 du d (1 dx 34. f (u) sec ) d ( d f (g(1))g (1) sin ) sin d (2 d (2 ) ( f g) (1) sin 2 28. g (x) sin ( cos d5 (u 1) 5u 4 du 1 d ( x) dx 2x 33. f (u) sin 2 d d (sin ) (sin ) ( ) d d 1 2 d ( sin ) d 1)](3) g (x) 2u d du u 2 1 (u 2 1) d (2u) du (u 2 (2u) 1) d2 (u du 1) 2 1)(2) (2u)(2u) 2u 2 2 (u 2 1)2 (u 2 1)2 d (10x 2 x 1) 20x 1 dx ( f g) (0) f (g(0))g (0) f (1)g (0) (0)(1) 0 95 96 Section 3.6 38. f (u) u 2 u u 2 u g (x) du du u 12 1 (u 1 1 1 (u 1 1) ( f g) ( 1) d (u du 1du 1 du u 1) (u 1) 1 1 d (u du 1) 3 2x f (g( 1))g ( 1) dy dt dx dt 2 cos t 2 cos t 2 sin t cot t and has slope cot 2, or y x dx dt d (sin 2 t) dt (cos 2 t) (2 t) dy dt 42. 2) 6 sin u d (cos 2 t) dt dy dt dx dt 2 sin 2 t 2 cos 2 t and has slope tan 1) y 2 43. 6 sin 2u dy 40. (a) dx dx dt d (sec 2 t dt 2 , cos 31 , 2 2 6 2 3. Its equation is 6 1 , or y 2 3x 2. d dt (2 sec t) (sec t) (2 sec t)(sec t tan t) 2) dy dt 1)(1) 2x 2x cos (x 2 d tan t dt dy dx d 1) (x 2) dx 2x cos (u dy dt dx dt 1) sec 2 y 1) 1 cot t. 2 1, tan 4 1 cot 2 1 (x 2 1) 44. 2 sec2 t 2 sec2 t tan t slope (cos u)(2x) 2x cos u sec 2 t The line passes through 1) dy du du dx d d (sin u) (x 2 du dx 2x cos (x 6 2 sec 2 t tan t dy du du dx d sin (u du cos (u dy (b) dx 1) 2 sin 2 t tan 2 t 2 3 3x ( sin 2u)(2) (3) 2 cos 2 t d dt 2) dy du du dx d d (cos 2u) (3x du dx 6 sin (6x 2) 2. ( sin 2 t) (2 t) d dt The line passes through sin dy dx 2 (x dy dx dy du du dx d d (cos u) (6x du dx 6 sin (6x ( 2, 4 8 ( sin u)(6) (b) , 2 sin 1. Its equation is 4 2) 4 y ( 4)(2) dy dx d (2 sin t) dt 2 sin t The line passes through 2 cos f (0)g ( 1) 39. (a) d (2 cos t) dt dy dx 1) dx dt dy dt 41. (u 1)2 (u 1) 4(u 1) 1)2 (u 1)3 1) (u d (x 2 dx u u 2 dx dt dy dt d sec t dt d tan t dt dy dx dy dt dx dt 1, or y 2x 1 . 2 sec 2 t sec t tan t The line passes through sec or y 1 x 2 sec t tan t sec 2 t sec t tan t has slope csc 1) and has 1 . Its equation is 2 4 1) (1, 4 1 sin t 6 , tan csc t 2 6 2. Its equation is y 6 1 , 3 2x 2 1 3 3. and 3 , 3 Section 3.6 dx dt d t dt dy dt d dt dy dx 45. dy dt dx dt 1 t 1/(2 t) 1 1 2 y 1 4 (b) 11 , and has slope 42 1 4 1x dx dt dy dt dy dt dx dt 1 , or 2 (2t 4t 4t 3 Then dy dx dy dt dx dt 1 sin t) 1 cos t) 1 y 1)(sin t) 2 cos t (2t 1)2 du dx du , so dx dt dx du dt dx . Therefore, dt d dy dt dx (2t dx dt 1)(sin t) 2 cos t (2t 1)2 1)(sin t) 2 cos t (2t 1)3 cos t sin t (2t 1) (d) The expression in part (c). sin t cos t 50. Since the radius passes through (0, 0) and (2 cos t, 2 sin t), it has slope given by tan t. But the slope of the tangent is sin 3 ,1 cos dy dt dx dt dy dx 3 31 , and has slope 2 3 2 cos t 2 sin t cot t, which is the negative reciprocal of tan t. This means that the radius and the 2 tangent are perpendicular. (The preceding argument breaks 3 cos du dt d dy dx dx The line passes through sin 1) (2t d (t dt d (1 dt 1) 2 dy . dx (c) Let u t2 d dt (cos t) (2t 1)( sin t) (cos t)(2) (2t 1)2 (2t 3) 4t 3 4t dx dt dy dt 3 d cos t dt 2t 1 d (2t 1) (cos t) dt The line passes through (2( 1)2 3, ( 1)4) (5, 1) and has slope ( 1)2 = 1. Its equation is y 1(x 5) 1, or y x 4. 47. d dy dt dx (2t d (2t 2 dt d4 (t ) dt 1 cos t 2t 1 1 . 4 x dy dx 46. dy dt dx dt t 1. Its equation is y 1 4 d2 (t t) 2t dt d sin t cos t dt t 2 1 The line passes through , 4 1 2 dx dt dy dt dy dx 49. (a) 1 3. Its equation is down when t 3 1 , or 2 3 3x 3 2 k , where k is an integer. At these values, 2 either the radius is horizontal and the tangent is vertical or the radius is vertical and the tangent is horizontal, so the y 3x 2 . result still holds.) 3 48. 97 dx dt dy dt d cos t sin t dt d (1 sin t) cos t dt dy dx dy dt dx dt cos t sin t cot 2 ds dt ds d d dt ( sin ) When cot t The line passes through cos has slope 51. 2 ,1 sin 0. Its equation is y (0, 2) and 2 2. 52. dy dt d d (cos ) dt d d dt 3 d and 2 dt dy dx dx dt (2x When x 7) d2 (x dx 5, ds dt 7x 5) sin 3 (5) 2 5. dx dt dx dt 1 and dy dt 1 dy , 3 dt [2(1) 7] 1 3 3. 98 53. 54. Section 3.6 dy dx 1 x cos 2 2 1 x Since the range of the function f(x) cos is 2 2 dy 1 the largest possible value of is . dx 2 d x sin dx 2 dy dx xdx 2 dx 2 cos d (sin mx) dx d dx (cos mx) (mx) The desired line has slope y (0) (e) 11 ,, 22 through (0, 0), so its equation is y At x f (g(x))g (x) 2, the derivative is f (g(2))g (2) m cos mx m cos 0 d f (g(x)) dx (f) 1 ( 3) 3 f (2)g (2) 1 d dx f (x) f (x) d f (x) f (x) dx 2 At x f (x ) 2 m and passes mx. 2, the derivative is f (2) 55. dy dx x d 2 tan 4 dx x sec 2 2 4 y (1) 2 sec 2 2 sec 2 (g) 4 2 2)2 ( The tangent line has slope 1, 2 tan y . x (x 1) (h) 1 equation is y 12 2 d1 dx g 2(x) d [g(x)] 2 dx 2(5) ( 4)3 1) 2, or y x f 2(x) g 2(x) 2. g 2(x) 2, the derivative is 2 f (2) At x (c) g(x)] g (x) g(x)] f (x)g (x) g (3) 2 g(x) f (x) 3, the derivative is f (3)g (3) g(3) f (3) 5. 5 f (1) (b) 15 ( 4)(2 ) 8. 10 3 2 At x g(x) f (x) f (x)g (x) [g(x)]2 2, the derivative is g(2) f (2) f (2)g (2) [g(2)]2 74 3 4 37 . 6 (2) 1 3 (8)( 3) (2)2 22 3 17 5 f (x) g (x) 1, the derivative is g (1) d f (x)g 3(x) dx 1 3 5 8 3 d g 3(x) f (x) dx 3 f (x)g 2(x)g (x) 1. d dx f (x) g 3(x) d dx At x (2)( 3) 5 17 g(x)] g 3(x) f (x) g 3(x) f (x) 0, the derivative is 3 f (0)g 2(0)g (0) 3(1)(1)2 d f (x) (d) dx g(x) d dx 2g(x) g(x)] 82 f (x)[3g 2(x)] g(x) (3)(5) 1 3 g 2(2) d [5 f (x) dx At x 3, the derivative is f (3) d [ f (x) dx At x f (x) (8) g(2)g (2) 68 57. (a) d f (x) dx g 2(x)] g 2(x) 10 3 2 f (x) 2 . 3 d2 [ f (x) g 2(x) dx 2, the derivative is f 2(2) 1 2 3 2g (x) [g(x)]3 g(x) g(x)g (x) f 2(x) [ 4.7, 4.7] by [ 3.1, 3.1] d (b) [ f (x) dx [2f (x) 2 f (x) f (2) f (2) At x 2 f 2(x) 1 At x d [2 f (x)] dx dx 5 . 32 1 d dx 2 Graphical support: 56. (a) 10 64 f (x) f (x) (x 1 3d 2[g(x)] 3, the derivative is and passes through (1, 2). Its 1 . 2, or 2. 1 1 6(2 2) 2g (3) [g(3)]3 and passes through The normal line has slope 1 28 At x (1, 2). Its equation is y 4 1 3 2 f (2) xd x 4 dx 4 1. 1 3 (1)3(5) 6. g 3(0) f (0) 99 Section 3.6 [g(x) d d 1] f (x) f (x) [g(x) dx d f (x) (c) dx g(x) 1 1]2 [g(x) [g(x) 58. For y 1] d dx sin 2x, y (cos 2x) (2x) dx 2 cos 2x and the slope at the origin is 2. 1] f (x) f (x)g (x) [g(x) 1]2 For y x 2 sin , y cos 1 x cos and the 2 2 xdx 2 dx 2 1 . Since the slopes of the two 2 1 tangent lines are 2 and , the lines are perpendicular and 2 slope at the origin is At x 1, the derivative is [g(1) 1] f (1) f (1)g (1) [g(1) 1]2 (4 At x and y 1 3 1 3 8 (5) 3 g (1) f (0) f (x)] 2 2[g(x) f (x)] 59. Because the symbols 40 3 3d dx 2[g (1) f (1)] [g(1) f (1)]3 8 3 (4 1 3 values. [g(x) 6 1 3)3 d dx 6. 60. Velocity: s (t) 2 bA sin (2 bt) acceleration: s (t) 4 2b 2A cos (2 bt) 33 jerk: s (t) 8 b A sin (2 bt) The velocity, amplitude, and jerk are proportional to b, b 2, and b 3, respectively. If the frequency b is doubled, then the amplitude of the velocity is doubled, the amplitude of the acceleration is quadrupled, and the amplitude of the jerk is multiplied by 8. d 2 37 sin (x dt 365 61. (a) y (t) d (25) dt 101) f (x g(x))(1 g (x)) 37 cos 2 (x 365 101) d2 (x dx 365 At x 0, the derivative is 37 cos 2 (x 365 101) 2 365 f (0 g(0))(1 g (0)) f (0 1) 1 4 f (1) 3 14 33 g(x)] f (x)] g(x))] f (x g(x)) [x dy dy du , , and are not fractions. The dx du dx individual symbols dy, dx, and du do not have numerical 1, the derivative is 2 d [ f (x dx 1 x is shown. 2 [ 4.7, 4.7] by [ 3.1, 3.1] g ( f (x)) f (x) 2[g (x) f (x)] [g(x) f (x)]3 (g) 2x 1 . 9 0, the derivative is d [g(x) dx At x the curves are orthogonal. A graph of the two curves along with the tangents y 1. f (1)g (0) g ( f (0)) f (0) (f) 1)2 0, the derivative is d g( f (x)) dx At x 8 3 f (g(x))g (x) f (g(0))g (0) (e) (3) (4 9 9 d (d) f (g(x)) dx 1 3 1) 1 3 74 2 cos (x 365 365 101) Since cos u is greatest when u 4 . 9 y (t) is greatest when x 101) 2 (x 365 0, 101) 2 , and so on, 0, or 101. The temperature is increasing the fastest on day 101 (April 11). (b) The rate of increase is y (101) 74 365 0.637 degrees per day. 0 100 Section 3.6 d dt 62. Velocity: s (t) 1 1 4t 4 2 At t d (1 4t dt 21 66. dT du dT dL dL du 4t 1 1 2 m/sec 5 4(6) 1 2 4t 2 6, the velocity is 1 4t d dt (1 4t) (2) ( 21 1 d dt 1 4t 4t)2 1 d (1 4t dt 1 2 2 4t) 4t y 4t 1 4t 4 4t)3/2 (1 L g dL (kL) dL g 1 (kL) g k d f (g(x)) dx 6, the acceleration is f (g(x)) has a horizontal tangent at x f (g(1))g (1) dv dt 63. Acceleration k 2 (k 4 4(6)]3/2 [1 dv ds ds dt dv (v) ds 4 m/sec 2 125 d (k ds 0, so either g (1) s tangent at x 1, or the graph of y tangent at u 1, then 0 or f (g(1)) 0. This g(x) has a horizontal f (u) has a horizontal g(1). s) (k s) k2 , a constant. 2 s) kT 2 f (g(x))g (x). If the graph of means that either the graph of y At t L g 67. No, this does not contradict the Chain Rule. The Chain Rule states that if two functions are differentiable at the appropriate points, then their composite must also be differentiable. It does not say: If a composite is differentiable, then the functions which make up the composite must all be differentiable. 68. Yes. Note that 4 1 2 L g 2 d dt L (kL) g d 2 dL 2 1 Acceleration: s (t) 4t) 69. For h 1: 64. Note that this Exercise concerns itself with the slowing down caused by the earth’s atmosphere, not the acceleration [ 2, 3.5] by [ 3, 3] caused by gravity. For h Given: v 0.5: k s Acceleration dv dt dv ds ds dt k dv (v) ds (v) dv ds dk s ds s d [ 2, 3.5] by [ 3, 3] s (k) k ds k d ds For h 0.2: s ( s)2 s k (2 s) k s s k2 2s 2 [ 2, 3.5] by [ 3, 3] As h → 0, the second curve (the difference quotient) ,s 0 Thus, the acceleration is inversely proportional to s 2. 65. Acceleration dv dt df (x) dt df (x) dx dx dt f (x) f (x) approaches the first (y 2 cos 2x). This is because 2 cos 2x is the derivative of sin 2x, and the second curve is the difference quotient used to define the derivative of sin 2x. As h → 0, the difference quotient expression should be approaching the derivative. 101 Section 3.7 70. For h 1: 72. dG dx d dx 1 2 2x uv 2 uv uv d dx x(x d (x 2 dx cx A G cx) d dx c) 2x 2x x2 c 2 cx x cx (x c) 2 x(x c) [ 2, 3] by [ 5, 5] For h 0.7: s Section 3.7 Implicit Differentiation (pp. 149–157) Exploration 1 1. 2x dy dx [ 2, 3] by [ 5, 5] For h 2y An Unexpected Derivative 2xy 2yy 1 (provided y 0. Solving for y , we find that x). 2. With a constant derivative of 1, the graph would seem to be a line with slope 1. 0.3: 3. Letting x 0 in the original equation, we find that y 2. This would seem to indicate that this equation defines two lines implicitly, both with slope 1. The two lines are y x 2 and y x 2. [ 2, 3] by [ 5, 5] As h → 0, the second curve (the difference quotient) approaches the first (y 2x sin (x 2)). This is because 2x sin (x 2) is the derivative of cos (x 2), and the second curve is the difference quotient used to define the derivative of cos (x 2). As h → 0, the difference quotient expression should be approaching the derivative. 71. (a) Let f (x) 4. Factoring the original equation, we have [(x y) 2][(x y) 2] 0 x y 2 0 or x y 2 0 y x 2 or y x 2. The graph is shown below. x. Then d u dx d f (u) dx f (u) du dx d u du du dx u u. u The derivative of the absolute value function is positive values, at 0. So f (u) 1 for negative values, and undefined 1, 1, u u 0 0. g (x) u evaluates. u (2x)(x 2 9) d x2 9 (x 2 9) x2 9 dx x2 9 d ( x sin x) dx d d x (sin x) (sin x) x dx dx x sin x x cos x x Note: The expression for g (x) above is undefined at x 0, but actually g (0) lim g(0 h→0 h) h g(0) lim h→0 h sin h h Therefore, we may express the derivative as x cos x g (x) 0, x sin x , x 5. At each point (x, y) on either line, x 0 x 0. 0. dy dx 1. The condition y x is true because both lines are parallel to the line y x. The derivative is surprising because it does not depend on x or y, but there are no inconsistencies. Quick Review 3.7 But this is exactly how the expression (b) f (x) 1 for [ 4.7, 4.7] by [ 3.1, 3.1] 1. x y2 x x y1 0 y2 y x, y2 [ 6, 6] by [ 4, 4] x 102 Section 3.7 2. 4x2 9y 2 36 9y 2 36 7. y sin x 4x 2 36 y2 4x 2 3 2 3 y1 4 (9 9 y xy y (sin x x)y y x cos x y x cos x x 2) x2 9 y 2 3 2 9 xy y sin x 2 9 y x cos x x , y2 9 x cos x sin x x y) 8. x(y 2 y (x 2 y) xy 2 y (x 2 y 9. [ 4.7, 4.7] by [ 3.1, 3.1] x2 4y 2 2y) 0 10. y x ,y 22 3 x2 x3 x x 2 y1 x+ x 1/2 x x 2/3 x x x 2/3 x 3/2 x 3/2 x 2 y x 1/2(x x 1/3) x 1/2x x 1/2x 1/3 x 3/2 x 5/6 x) 0 2y)(x (x x2 3 x(x x) xy2 y 3. x 2 3/2 5/6 x Section 3.7 Exercises 1. [ 6, 6] by [ 4, 4] 4. x 2 2 y y2 y y1 9 9 2. x2 9 x2 9 x2, y2 3. 9 y2 4. 5. [ 4.7, 4.7] by [ 3.1, 3.1] 5. x 2 2 y y2 y y1 2x 2x 3 3 x2 2x 3 x2 2x 3 x2, y2 6. 2x 3 dy dx dy dx dy dx dy dx dy dx dy dx x2 [ 4.7, 4.7] by [ 3.1, 3.1] 6. x 2y 2xy x 2y y 4x y 4x y 4x y x2 2xy dy dx 9 5/4 x 4 3 ( 3/5) x 5 d x 3/5 dx 3 8/5 x 5 1 d3 x dx d 1/3 x dx 1 (1/3) x 3 1 1 2/3 x 3 d4 x dx d 1/4 x dx 1 (1/4) x 4 1 1 3/4 x 4 1 d d (2x 5) 1/2 (2x 5)( 1/2) 1 (2x 2 dx dx 1 (2x 5) 3/2(2) (2x 5) 3/2 2 d (1 dx 2 (1 3 2 (1 3 4(1 7. 9 (9/4) 1 x 4 d 9/4 x dx 6x)2/3 1d 6x)(2/3) 6x) 1/3 6x) dx (1 ( 6) 1/3 d (x x 2 1) dx d x x2 1 x2 dx d x (x 2 1)1/2 (x 2 dx 12 x (x 1) 1/2(2x) 2 x 2(x 2 1) 6x) 1/2 (x 2 1 d (x) dx 1)1/2 (x 2 1)1/2 1)1/2 2xy Note: This answer is equivalent to 2x 2 1 x2 1 . 5) Section 3.7 dy x d dx dx x2 (x 2 8. 1)1/2 (x d dx 1)1/2 x x (x 2 x2 1 12 (x 2 x x2 x2 d dx (x 2 1) 1/2 1)1/2 x x2 12. d2 (x ) dx 1 y xy dx y dx x y (2x) (x 1 x 1 (x 2 (x 2x x 2y 9. d2 (x y) dx x 2 dy y(2x) dx x(2y) x xy 2 6 d (xy 2) dx dy y 2(1) dx 2 dy 2xy dx dy x d3 (x ) dx 2 d3 (y ) dx 2 dy 2 dy dx dy 3x 3y y 3 dx (3y 2 3y 18x 18x) dx dy dx dy d2 y dx dy 2y 2y x x d dx (x dx dy dx dy dx y 2) (2xy 2xy 18x 18y x(x x y)2 x y y) x y x3 x 2y x y x2 2 1) 3x 2 y2 13. dy dx 6x 14. (x 1)2 y (x x d (x) dx y)2 dy dx 3x 2 1 dx dy dy dx 1 3x d 3(2x 1/2 dx (2x x2 1/2 (2x 2 x dy dx 2xy 1 1/2 1) 1) 1) 1/3 4/3 d dx 4/3 (2x (x 3/2 (2x 1/2 1) 1/2 3/2 (x 15. 2xy 2 d (1 x 1/2)1/2 dx 1 d (1 x 1/2) 1/2 (1 x 1/2) 2 dx 1 1 1/2 (1 x 1/2) 1/2 x 2 2 1 (1 x 1/2) 1/2x 1/2 4 1)(1) 1)2 1 y(x 1)2 d (y) dx dy 1 dx 3x 2 18y y)2 y(2x) dx 2 18y dx d2 (x y) dx dy (x 3x y)2 y x x2 3y 2 18x 6y x 2 1 1 1)(1) (x y x 2(x 18y (1) dx x dx x(x dx dy y2 2xy x dx y)2 dy y dy y 2) 1 1 x x dy Alternate solution: 3x 2 dx y2 (2xy dx dy x dx 18xy dx dy 11. y y d3 (x ) dx d (18xy) dx dy 2x (x y)2 x 0 dx 10. y dx x2 x(x d (6) dx dx dy x 2) dx dy (2xy 3 dy (x 2y 3/2 1) x dy y) 1 y)2 2x 1)3/2 2 (x dx (x 1 x2 1)(x 2 1)1/2 2 dy y) 1 2x 4/3 d 3(csc x)3/2 dx 9 d (csc x)1/2 (csc x) 2 dx 9 (csc x)1/2( csc x cot x) 2 9 (csc x)3/2 cot x 2 ) 1) dy dx y y dy dx 103 104 16. Section 3.7 dy dx 17. d [sin (x dx 5 [sin (x 4 5 [sin (x 4 x (c) Differentiating both sides of the given equation 5)]5/4 5)]1/4 d sin (x dx 5)]1/4 cos (x f (x) 5) dy dx x 22. (a) If g (t) sin y d (x) dx dy dx 4 4 t 19. d (4t 1/4 dx 4) 3/4 t x 1 t 3/4 g (t) tan xy t 3/4 dy dx dy dx dy dx (c) If g (t) 0 g (t) 0 4 1 (sec2 xy)(y) 1 d (x) dx 1 sin y y sec2 xy (cos y) (cos y x) x2 x 1/3 d2 (x ) dx 2x 1 y 1 x x y2 y cos y d2 (y ) dx 2yy 0 2x x y y x y d dx 1 4/3 x 3 y x 1/3 x x y y2 . y2 y3 7, then 3 2/3 x and f (x) 2 (x)(y ) y2 x2 f (x) 1 3/4 t , which 16 d (1) dx (y)(1) which contradicts the given equation f (x) (b) If f (x) 4t 1/4 and 1 3, then and f (x) 9 5/3 x 10 1 , which matches the given 1 1/4 t , then g (t) 4 2yy y cos y t 3/4 y 1 3 2/3 x 2 21. (a) If f (x) f (x) dy dx dy dx 3/4 16 5/4 t , then g (t) 5 1 Conclusion: (a) and (c) could be true. y x d (xy) dx dy x (y)(1) dx dx 7 . t contradicts the given equation. xy d (sin y) dx dy t (d) If g (t) 23. x 4 equation. 0 x sec2 xy 1 cos2 xy x t 3 7/4 t , which is not 4 gives g (t) 0 d (tan xy) dx d 1 sec2 (xy) (xy) dx dy (sec2 xy)[x (y)(1)] dx 20. 1 , which matches the t 3/4 (b) Differentiating both sides of the given equation d (x) dx (sec2 xy)(x) , which matches 4, then consistent with g (t) 1 1/3 given equation. d (sin y) dx dy cos y dx 1 sec y cos y 1 x Conclusion: (b), (c), and (d) could be true. g (t) 18. 6, then f (x) the given equation. d (tan y) dx dy sec2 y dx 1 cos2 y sec2 y 1 1 4/3 x . 3 3 2/3 x 2 (d) If f (x) 1 4/3 x , so it must be true 3 gives f (x) that f (x) 5) tan y d (x) dx 1/3 x x 1/3 , which matches the given equation. Since our original equation was x 2 substitute 1 for x 2 y 2, giving y y2 1 . y3 1, we may Section 3.7 x2/3 24. d 2/3 (x ) dx y2/3 1 d 2/3 (y ) dx 2 1/3 x 3 0 1 /3 y 1/3 1 /3 x y y x y 1/3 d dx y x 1y 3x 2/3 d 1y 3x 2/3 xy x2 y 1/3 x Slope at (3, 3x 4/3y 1/3 substitute 1 for x y2 25. x2 d2 (y ) dy y 2/3 , giving y 3x4/3y1/3 . 2x y y (x 2)(2y) (x 1) x 1 y (y 2)(2x) 0 2 y2 d2 (y ) dx d (2x) dx y (2y y y 2y 2x 2y) 2)y 2 Slope at ( 2, 1): (a) Tangent: y (b) Normal: y (y 1) 2 (y 1) 2 1 1)3 y 1 y 1 1) 3 or y 1) 4y 1 d (1) dx dy 24 0 dx dy (2y 4) dx dy dx dy dx 1 1 1(x 6 1 x 3 8 3 0 d (0) dx 0 2 1 y 2 1 2 (x 3x 3 or y d (4y) dx 1) 1 y1 d1 dx y 1 (y 2y y x 3 2x 1 d (2x dx 2xy 2 2x 2y 1 (x 3 30. 2x, we may 2x 1) 1, 2xy 2 3(x (b) Normal: y 1)2 (x y3 dy dx 4 x 3 ( 4) or y d (9) dx (a) Tangent: y Since our original equation was y x write y 2 (x 1)2 (x 2 2x) (x 2 1 which gives y 3. d2 (y dx 25 4 9 dy dx dy dx 3 Slope at ( 1, 3): 1 2 26. y 3 x 4 3) 2x 2y y2 2 3) x 2y 2 2 y2 ( 4) or y 3 4 3 4 d 22 (x y ) dx 2x 2 x1 2y y dx 1 dx y (y)(1) (x 1)y y2 y 29 7 x y 4 (x 3 29. 4 x 7 0 (b) Normal: y d (2x) dy 1 2 3 or y 2) 3 (x 4 (a) Tangent: y 7 x 4 3 or y y x d (25) dx 4): 1, we may 1 2x d2 (x ) dy 2yy y 2/3 2x 2y 25 d2 (y ) dx dy 2x 2y dx dy dx y 2/3 Since our original equation was x 2/3 y2 d2 (x ) dx 2/3 1/3 y 1x y 3 x4/3y2/3 2/3 x2 28. y xy x 2/3 1 2) 4 (x 7 (b) Normal: y (y)(1) 4/3 2/3 3 7 (x 4 (a) Tangent: y y dx x (x) 1 y2 xy d2 d d2 d (x ) (xy) (y ) (1) dx dx dx dx dy dy 2x x (y)(1) 2y 0 dx dx dy (x 2y) 2x y dx dy 2x y dx x 2y 2(2) 3 7 Slope at (2, 3): 2(3) 2 4 d (1) dx 2 1/3 y y 3 x2 27. 2) 2) 1 or y 1 or y x x 1 3 105 106 Section 3.7 6x 2 31. d (6x 2) dx d (3xy) dx dy 12x 3x dx dy 3x dx 2y 2 3xy 17y 6 0 34. x sin 2y d d d (17y) (6) (0) dx dx dx dy dy (3y)(1) 4y 17 00 dx dx dy dy 4y 17 12x 3y dx dx dy (3x 4y 17) 12x 3y dx dy 12x 3y dx 3x 4y 17 y cos 2x d (2y 2) dx d (x sin 2y) dx (x)(cos 2y)(2) dy dx (sin 2y)(1) dy dx dy (cos 2x) dx dy (y)( sin 2x)(2) (2x cos 2y) d (y cos 2x) dx (cos 2x) dy dx 2y sin 2x sin 2y 2y sin 2x 2x cos 2y sin 2y cos 2x dx 6 (x 7 (a) Tangent: y 1) 7 (x 6 (b) Normal: y 2x 23 3 (a) Tangent: y d (sin y) dx dy cos y dx dy cos y) dx dy dx 2y(1) (2x 2 d (5) dx (b) Normal: y y 3 : 2 2 (x (x 1 2 4 0 2 cos ( x Slope at (1, 0): d (2 ) dx x 4 2 2 (x or y 2x 1 x 2 or y 2 2 sin ( x 2 cos ( x 5 8 y) d 2 sin ( x dx y) dy dx y) 2 cos ( x y) 2 cos ( x y) 1 2 cos ( x y) 2 cos 2 cos (a) Tangent: y 1 cos 2 dy dx dy dx dy y)] dx dy dx 4y [1 sin ( ) 2 cos ( ) y 2 1 1) 2 ( 1) 2( 1) 0 or y 2 2x 2 0 1 (x 2 (b) Normal: y 2y 1) x 2 0 or y 1 2 2y 2x 1) 1) sin 2x 2x x3 4 35. 2x 3 y 2 2 0 0 2( /2) 2(1) cos( /2) (a) Tangent: y (b) Normal: y 0 ( 1) (a) Tangent: y 4(2) sin y d 2 (xy) dx Slope at 1, 2 2 3 2xy dy dx 3 42 ( )(1) 7 6 : 2 (b) Normal: x 33. 23 6 7 , 5 d 2 (y 2 ) dx dy 3(y)(1) 4y dx dy ( x 3 4y) dx dy dx Slope at ( 3, 2): 2x 2y 2 3xy 2 Slope at 7 x 6 0 or y d 3 (xy) dx dy 3(x) dx 6 7 6 x 7 0 or y 1) x2 32. d2 (x ) dx 12 14 12( 1) 3(0) 3( 1) 4(0) 17 Slope at ( 1, 0): 2 2 2 2 x sin y 0 d2 (x cos2 y) dx d (sin y) dx dy (x 2)(2 cos y)( sin y) dx 2 or y or y 36. x 2 cos2 y cos y x (2x 2 cos y sin y 2 2 dy dx cos y) 2 cos y d (0) dx (cos2 y)(2x) dy dx 2x cos y 2x 2 cos y sin y Slope at (0, ): 1 2(0) cos 2(0)2 sin (a) Tangent: y (b) Normal: x 0 (cos y) 2x cos2 y 1 2x cos y 2x2 sin y 0 dy dx 0 107 Section 3.7 y4 37. (a) y2 d4 (y ) dx dy 4y3 dx dy 3 (4y 2y) dx dy dx 3 3 At , : 4 2 x2 39. (a) ( 1)3(1)2 d2 d2 (y ) x dx dx dy 2y 2x dx (x 3)(2y) 2x 2y x (2x 3y 2y3 y 2 2 2 4 3 4 3 4 At 33 4 2 3 y3 2y 1 3 4 13 2 1 2 1 4 4 4 1 5 t y (2 Tangent: y Normal: y x) dy dx dy x) dx dy dx 1 (x 2 1) 1) xy . Hence, 5 2 , and 1 xy (y)(1) 0 x)y y (3y 2 y 3y 2 x y d dx 3y 2 x (3y 2 x)(y ) (y)(6y y (3y 2 x)2 y xy 3y 2 y 2 (3y x)2 y 3x 2 2 y y2 2 3x 2y(2 1 or y 5 2 1 3 3x 2 4 2 1). The d ( 1) dx d3 (x ) dx x)] 3(1)2 (1)2 2(1)(2 1) 2(x 1 d (xy) dx 3y 2y x x)(2y) 2y(2 Slope at (1, 1): 1 y 1 = 0 are d3 (y ) dx 2 d2 [y (2 dx y y3 [ 1.8, 1.8] by [ 1.2, 1.2] 1 1)(y 2 2y + 1 as (y . (b) Parameter interval: 3 2 1 is one solution, and there are three possible y-values: 1, 2 (2 3 2 1, or (1) 4(1)( 1) 2(1) 3 1 (y2)( 1) 3x 2y 2 sin y 2y 2 1 y 3 (b) 38. (a) 2x 3y 0. Clearly, y solutions of y 2 2 3x 2y 2 1 we may factor y 3 3 4 1 2 dy dx dy dx 2, we have y 3 40. (a) When x 1 31 ,: 42 Slope sin y) dy dx 3( 1)2(1) 2( 1)3(1) sin 3 The slope of the tangent line is . 2 33 3 3 2 ( sin y)( ) Slope at ( 1, 1): 4 3 d cos ( y) dx (y 2)(3x 2) dy dx 1. cos( y) d 32 (x y ) dx 3 Slope x 3y 2 (b) 2x 4y3 cos ( ) is true since both sides equal y x) 2 2x 1 or y (b) One way is to graph the equations y Since we are working with numerical information, there is no need to write a general expression for y in terms of x and y. 1 1 x 2 3 2 x3 2 1) x . To evaluate f (2), evaluate the expression for y using x 2 and y f (2) 1: 1 3(1)2 2 1 1 1 To evaluate f (2), evaluate the expression for y using x 2, y f z (2) 1, and y (1) 2(1) [3(1)2 1: 3(1)2(1) 2]2 4 1 4 108 Section 3.7 41. Find the two points: The curve crosses the x-axis when y 0, so the equation becomes x 2 0x 0 7, or x 2 7. The solutions are x 7, so the points are ( 7, 0). Show tangents are parallel: x2 y2 xy (2x y) 0 2( 7) 2x 2 d (2x 2) dx 2 0 2 2(0) d2 (x ) dx 2x x d (xy) dx dy dx y2 xy 2y d (7) dx dy dx 0 d (3y 2) dx dy 4x 6y dx dy dx 5 d (5) dx 0 4x 6y (2x dy dx y) 3x 2 3x 2 2y the slopes are 2x y x 2y 2x 3y d3 x dx At (1, 1), the slopes are dy 2y) dx (x 3y 2 x3 d2 y dx dy 2y dx dy dx 7 d2 (y ) dx (y)(1) 7 . 3 Second curve: y2 x2 2 43. First curve: 2x y x 2y The tangents at these points are parallel because they have the same slope. The common slope is 2. 42. 7 7 and 2 , 3 3 Note that these are the same points that would be obtained by interchanging x and y in the solution to part (a). 2(0) 7 7 3 The points are 0 27 7, 0): 7 , 3 7 7 7 7 d (7) dx 7 Slope at ( ( 2y) y2 y2 y2 3y 2 y 7 d2 d d2 (x ) (xy) (y ) dx dx dx dy dy 2x x (y)(1) 2y dx dx dy (x 2y) dx dy dx Slope at ( 7, 0): x 2 xy ( 2y)(y) 4y 2 2y 2 2 2 and 3 2 3 and respectively. At (1, 3 2 1), 3 respectively. In both cases, the 2 tangents are perpendicular. To graph the curves and normal lines, we may use the following parametric equations for (a) The tangent is parallel to the x-axis when dy dx t 2x y x 2y Substituting 0, or y : 2x. First curve: x 2x for y in the original equation, we have Second curve: x 2 x2 2 x xy y (x)( 2x) ( 2x)2 x 2 2x 2 4x 2 3x 2 x The points are 7 7 7 7 7 7 ,2 and 3 3 5 cos t, y 2 32 t ,y 7 3 7 , 3 2 7 . 3 dx can be obtained by interchanging x and y dy dy dx 2y x in the expression for . That is, . The dx dy x 2y dx tangent is parallel to the y-axis when 0, or dy equation, 2y. Substituting equation, we have: t Tangents at (1, 1): x 1 3t, y x 1 2t, y 1 3t Tangents at (1, 1): x 1 3t, y x 1 2t, y 1 3t (b) Since x and y are interchangeable in the original x 5 sin t 3 2y for x in the original [ 2.4, 2.4] by [ 1.6, 1.6] 1 2t 1 2t Section 3.7 s (t) a(t) d (4 dt 9(4 44. v(t) 3 (4 2 6t)1/2 At t (6) 27(4 2, the velocity is v(2) 6t) 1/2 dv dt t) 4(s t) 4(s t) 1/2 (v 4(s t) 1/2 1] [(8(s 1/2 x 3 0 x6 27 3x 3 0 54) 0 9x x 1) t)1/2 1) 1] At x 1/2 (s 0 2 x 3 1 dt 9xy 23 x3 3 (x 27 x3 1/2 1/2 ds t) y3 x3 d [8(s dt 32(s 3y y2 x3 36 m/sec and the 27 m/sec2. 4 acceleration is a(2) 45. Acceleration x2 x2 0, or y . 3x 3 x2 Substituting for y in the original equation, we have: 3 6t)1/2] 1/2 6t) (b) The tangent is horizontal when 6t)1/2(6) dy dx d [9(4 dt v (t) 9 (4 2 6t)3/2 109 t) 3 0 or x 02 3 0, we have y 54 32 ft/sec2 y 13 (9 4) 3 3 2 0, which gives the point (0, 0), which is the origin. At x 132 (3 2) 3 3 3 3 3 3 3 2, we have 4, so the point other 3 than the origin is (3 2, 3 4) or approximately 46. y 4 4y 2 x d4 (y ) dx d (4y 2) dx dy dy 8y 4y 3 dx dx 4 9x d4 (x ) dx 4x 3 4y 3 47. (a) 2): (c) The equation x 3 9(3) 4(2) 2(3)3 2( 2)3 x3 d3 d3 (x ) (y ) dx dx dy dy 3x 2 3y 2 9x dx dx (3y 2 3(2) Slope at (4, 2): 2 (2) 3(4) Slope at (2, 4): 2 (4) 2x 3 2y 3 the line y 9x 4y 3 9(3) 4( 2) d dx x2 48. 27 8 2) or 2x d (0) dx 9(y)(1) 0 dy dx dy dx 9y 3y2 0 d d d 2 (xy) (3y2) (0) dx dx dx dy dy 2x 2(y)(1) 6y 0 dx dx dy (2x 6y) 2x 2y dx dy 2x 2y xy dx 2x 6y 3y x 0 9 (xy) 2xy d2 (x ) dx 27 8 9xy (4)2 3(4) (2)2 3(2) 3 approximately (4.762, 3.780). 27 8 9( 3) 4( 2) 9x) x and we may find the desired point by to part (b). The desired point is (3 4, 3 9( 3) 4(2) y3 9xy is not affected by interchanging the x-value and the y-value in the answer 27 8 2( 3)3 2( 2)3 y3 interchanging x and y, so its graph is symmetric about 18x 8y 2( 3)3 Slope at ( 3, 2): 2(2)3 2): d (9x 2) dx 18x 2(3)3 Slope at (3, 2): 2(2)3 Slope at (3, (3.780, 4.762). 4x 3 dy dx Slope at ( 3, 2 At (1, 1) the curve has slope normal line is y 3x 2 9y 3x 2 3y 2 9x 10 5 8 4 4 8 5 10 3y y2 x2 3x Substituting x 1(x 11 3(1) 1 1) + 1 or y 2 2 1, so the x 2. 2 for y in the original equation, we have: x 2 2xy 3y 2 0 x 2x( x 2) 3( x 2)2 0 2 x 2x 2 4x 3(x 2 4x 4) 0 4x 2 16x 12 0 4(x 1)(x 3) 0 x 1 or x 3 Since the given point (1, 1) had x 1, we choose x and so y (3) 2 1. The desired point is (3, 2 3 1). 110 Section 3.7 49. xy d (xy) dx dy x dx 2x y 0 d (2x) dx d (y) dx dy (y)(1) 2 dx dy (x 1) dx dy dx 0 b 2x 2 0 2 y 2 1 2, we wish to 2, that is, where 1 2y 1 0 2b 2x 2a 2y b 2x1 a 2y1 2y Substituting 2y d (x) dx 1 dy dx y b 2x1x But a 2y12 3 . b 2x1 y1 a 2y1 (x x1). This gives: 2 a 2y12 b 2x12 b 2x12. a 2b 2 since (x1, y1) is on the dividing by a 2b 2 gives x2 a2 (b) b 2x 2 b2 a 2b 2 d 22 (b x ) dx d 22 (a y ) dx dy 2b 2x 2a 2y dx dy dx d 22 (a b ) dx 0 2b 2x 2a 2y b 2x 1 a 2y 1 y1 a 2y1y a2y12 b2x1x b2x12 a 2y12 b 2x1x a 2y1y hyperbola. Therefore, b 2x1x 2b3 b 2b b2 1 greater than if b 2 2b(x 1 or b 2 1 12 2 2 b2 ) b. This line intersects the x-axis at 1 , which is the value of a and must be 2 b) will be perpendicular when 1, which gives 2y 1 or 1 . The corresponding value of a is 2 1 3 b2 . Thus, the two nonhorizontal 2 4 3 normals are perpendicular when a . 4 a 2y12 a 2b 2 since (x1, y1) is on the dividing by a 2b 2 gives y2 b2 y2 b2 y1y 2 b2 a y 2 y 1 x2 a2 a 2y1y x1x 52. (a) Solve for y: x2 a2 0. The two normals at (b 2, they have slopes x1). b2x12 b, or x (x This gives: line at (b 2, b) is y 2b b 2x1 a 2y1 But b 2x12 2bx b 2x a 2y . 2y. Thus, the normal y 1. 1 The normal line at (x, y) has slope 3 a 2b 2, and y1y a2 a 2y 2 The slope at (x1, y1) is b 2x1x x1x y2 b2 The tangent line is y d2 (y ) dx dy 2y dx 1 2y b 2x12 b 2x1x ellipse. Therefore, a 2y1y 3 in the original equation, we have: 3)y x a 2y12 a 2y1y xy 2x y 0 2( 2y 3) y 0 2y 2 8y 6 0 2(y 1)(y 3) 0 y 1 or y 3 At y 1, x 2y 3 2 3 1. At y 3: x 2y 3 6 3 3. The desired points are ( 1, 1) and (3, 3). Finally, we find the desired normals to the curve, which are the lines of slope 2 passing through each of these points. At ( 1, 1), the normal line is y 2(x 1) 1 or y 2x 3. At (3, 3), the normal line is y 2(x 3) 3 or y 2x 3. ( 2y y b 2x a 2y x a 2y1y x 50. d 22 (a b ) dx The tangent line is y x 4 a 2b 2 The slope at (x1, y1) is 1 2 y) a 2y 2 dy dx 1 the tangent has slope . Thus, we have 2 2(2 1 d 22 (a y ) dx dy 2b 2x 2a 2y dx y x 0 is find points where the normal has slope y x y2 b2 d 22 (b x ) dx y 2y x1 Since the slope of the line 2x 2 1 x2 a2 51. (a) 1 b2 2 (x a 2) a2 b x2 a2 a 1. a 2b 2, and Section 3.8 b f (x) x→ g(x) (b) lim x2 lim a lim x2 x→ lim x→ (c) lim x→ f (x) g(x) lim x→ x5 x dy at dx 1, which is 7. 2x 1 at the point (1, 2). The slope is x2 a2 x2 1 b a x2 7. The slope of L is the reciprocal of the slope of its reflection 1 x2 8. 1 7 a2 x 1 y gets reflected to become x since a2 b x a x→ lim y a2 x→ lim 6. The reflection of line L is the tangent line to the graph of a2 b x a x→ 111 Quick Review 3.8 2 a2 x2 1. Domain: [ 1, 1] 1 Range: s Section 3.8 Derivatives of Inverse Trigonometric Functions (pp. 157–163) Exploration 1 Finding a Derivative on an Inverse Graph Geometrically 1. The graph is shown at the right. It appears to be a one-toone function At 1: , 22 2 2. Domain: [ 1, 1] Range: [0, ] At 1: 0 3. Domain: all reals Range: At 1: , 22 4 4. Domain: ( [ 4.7, 4.7] by [ 3.1, 3.1] 4 2. f (x) 5x 2. The fact that this function is always positive enables us to conclude that f is everywhere increasing, and hence one-to-one. Range: 0, , 1] 2 2 At 1: 0 1 3. The graph of f is shown to the right, along with the graph of f. The graph of f 1 is obtained from the graph of f by reflecting it in the line y x. 5. Domain: all reals Range: all reals At 1: 1 6. f (x) y 3x y 8 3x x [ 4.7, 4.7] by [ 3.1, 3.1] 4. The line L is tangent to the graph of f 1 at the point (2, 1). y 8 8 3 Interchange x and y: y f 1 7. f (x) (x) y [ 4.7, 4.7] by [ 3.1, 3.1] y 5. The reflection of line L is tangent to the graph of f at the point (1, 2). 3 x x 8 3 x 8 3 3 x y x 5 5 3 5 Interchange x and y: y f [ 4.7, 4.7] by [ 3.1, 3.1] 1 x3 5 (x) 3 5 x [1, ) , x 1 . It is . y 7 112 Section 3.8 8. f (x) 8 x 8 y y x 5. dy ds 1 f 9. f (x) 2s 6. 2 x xy 3x 3)x 7. 2 x 3 1 4s 2s d csc 1 (x 2 dx 3 1 y (x 2 (x2 2x 1) x s 1 d (5s) 1 ds (5s)2 5s s2 1 1 dy dx x2 s 25s 2 1) d (x 2 1 dx 1)2 1) 2 4 2x 2 (x Note that the condition x x 2 x2 1) 2 0 is required in the last step. 2 3 x 8. x arctan 3 y x , 3 tan y 4s d sec 1 5s ds 2 2 (x) 10. f (x) 1 1) (2) 2 1 2 y3 y 1) dy ds Interchange x and y: 1 (2s 2 x f 1 d (2s ds 1 2 1 3x y (y 2s 8 x 8 x (x) 1) 1 Interchange x and y: y d sec 1 (2s ds 3 tan y, 2 y 2 1 d x csc 1 dx 2 dx 1 dx 2 x2 2 x 2 2 y 2 dy dx x2 x 4 2 Interchange x and y: y f 3 tan x, 1 (x) x 2 3 tan x, 2 9. 2 x dy dt 1 1 d sec 1 t dt 1 t 2 1 Section 3.8 Exercises 1. dy dx 1 d cos 1 (x 2) dx 1 x4 1 d2 (x ) dx (x ) x4 1 1 1 t2 1 dy dx 10. 1 1 d cos 1 x dx d 1 2 dx x 1 1 1 1 x 2 1 x2 0 is required in the last step. dy dt 1 d 3 sin 1 2 dt t 1 x 1 1 1 x2 x t2 1 11. 1 dy dt d cot 1 dt 3 t2 6 6 t3 9 t4 d 3 2 dt t2 1 2. 1 Note that the condition t 22 1 2x (2x) 12 t 1 t 1 t d1 dt t 2 t t4 1 t 1 ( 9 d t)2 dt t 1 3. 4. dy dt d sin 1 dt dy dt d sin 1 (1 dt 1 2t t2 1 2t ( 2 t)2 1 1 t) 1 (1 d ( dt 2 2 t) d (1 2 dt t) 1 t) 2 t (t 1) 2t 2 12. dy dt d cot 1 dt 1 1 t t 1 1 1 ( 1 12t 1 ) d dt 1) 2 (t 1 2t t 1 t 1 1 113 Section 3.8 13. dy ds d (s ds 1 (s) 2 d (cos 1s) ds s2) 1 ( 2s) 2 1 1 s (1 s) 1 14. dy ds d ds [sin 1 1 (2) 4x 2 1 (2x)]2 1 dy dx 4x 2 sec 2 x, the slope at The tangent line is given by y d sec 1 s ds 1 y 1 (2s) s2 (2x) 1 2 1 s2 1 1 sin 2 (2x)] s 19. (a) Since s2 2 s 2 1 dx 2 s2 2s 2d (2x)] 2 1 2 1 [sin [sin s 2 d [sin 1(2x)] 1 dx 1 s2 1 2 dy dx 1 s2)(1) 1 s s2 1 ( 18. 1 (b) Since s2 s 2x 1 2x 1 x2 1 , the slope at 1, s2 s dy 15. dx 1 d (tan 1 dx x 1 1 1 1 x2 2 2 1) x2 x x 2 1 1 x2 x y 1 f x Note that the condition x dy dx 1 d cot 1 x dx 5x 4 1 1) 4 1 . 2 , or 6x 2 1. Thus f (1) 3 and f (x) includes the point (1, 3) and (x) will include (3, 1) and the slope will be 1 . (We have 12 (x) is defined and differentiable at 5x 4 6x 2 1, which is never zero.) 21. (a) Note that f (x) sin x 3, which is always between 2 and 4. Thus f is differentiable at every point on the interval( , ) and f (x) is never zero on this interval, so f has a differentiable inverse by Theorem 3. d (tan 1 x) dx d1 dx x ) (3) 3. This is true by Theorem 3, because f (x) 1 is required in the last step. 1 1 and ( f 1 assumed that f 1 0 16. 1 1 + 12 . 1 . Thus, f 1(3) 12 x2 x is the slope of the graph is 12 at this point, the graph of 1 x x2 4 1 1 1 4 (b) Since the graph of y 1 1) (2x) x2 1 2 20. (a) Note that f (x) f (1) 12. d (csc 1 x) dx d ( 1)2 dx x2 ( 1 x 2 y 1 2. 4 1, or 4 1 (x 2 The tangent line is given by y ss , 1 is sec 2 1. 2 dy dx 4 1 1 x2 1 ( 1 1 x2 1 ) 1 x2 1 x2 (b) f (0) cos 0 3(0) f (0) sin 0 3 x2 1 1; 3 (c) Since the graph of y 1 f (x) includes the point (0, 1) 1 x2 1 and the slope of the graph is 3 at this point, the graph of y 1 1 0, x 1 x2 f 1 (x) will include (1, 0) and the slope will be 1 . Thus, f 1(1) 3 22. 0 The condition x 0 is required because the original function was undefined when x 0. 17. dy dx d (x sin 1 x) dx 1 (x) 1 sin 1 x x2 d ( dx (sin 1 1 x 2) [ 2 , 2 ] by [ 4, 4] 1 x)(1) 2 1 x2 ( 2x) (a) All reals (b) , 22 0 and (f 1 ) (1) 1 . 3 114 Section 3.8 22. continued (b) y (c) At the points x (d) k , where k is an odd integer. 2 y 2 (c) None, since 30. (a) y 0. x2 x 1 0 (b) y 1 d sec 1 x dx 0 2 (c) None, since 1 d csc 1 x dx 0. x2 x 1 2π 1 x is 1. (b) None, since sin –2π 31. (a) None, since sin undefined for x x 1 x is undefined for x (c) None, since –2 1 x is undefined for x (b) None, since cos 1 x is undefined for x 1 d sin x 1 sin2 x dx cos x which is (c) None, since 23. (a) v(t) (b) a(t) (c) 24. 1 which is always positive. 1 1 So cos t2 2t which is always negative. (1 t 2)2 x, x sin 1 sin (b) β 1 x x 2 . x α 2 d cos 1 (x) dx d dx sin 2 1 x 1 tan 1 So tan x, x cot cot (c) x2 1 d cot 1 x dx 1 d sin 1 (x) dx 1 0 25. 0. x cos 1 x2 1. α cos x is positive or negative. dx dt dv dt 1 1. β 1 1 depending on whether 1 d cos 1 x dx 33. (a) sin2 x 1 0. x2 1 32. (a) None, since cos d sin 1 (sin x) dx (e) f (x) 1 d sin 1 x dx 1. 1 1 x x 2 . β x α d tan 1 (x) dx 2 d 0 tan 1 (x) dx 1 1 x2 1 1 sec So sec 1 x x, csc csc 1 x 1 x 2 . 34. d 26. csc 1 (x) dx d dx 2 x (x) x 2 A B 1 C 2 (b) y 2 (c) None, since 28. (a) y 1 d sec 1 (x) dx 1 0 27. (a) y sec d tan 1 x dx 1 0. x2 1 0 (b) y (c) None, since 29. (a) y 2 d cot 1 x dx 1 1 x2 0. The “straight angle” with the arrows in it is the sum of the three angles A, B, and C. A is equal to tan 1 3 since the opposite side is 3 times as long as the adjacent side. B is equal to tan 1 2 since the side opposite it is 2 units and the adjacent side is one unit. C is equal to tan 1 1 since both the opposite and adjacent sides are one unit long. But the sum of these three angles is the “straight angle,” which has measure radians. Section 3.9 35. s 6. Fold 1 log4 x 15 15 log4 x 12 12 log4 x log4 x ln Fold 2 s 2 Fold 3 s s s tan 1, so tan 1 tan 1 ln 19 ln 3 5t ln 5 1 1 tan 2 tan 1 t ln 5 s Section 3.9 Derivatives of Exponential and Logarithmic Functions (pp. 163–171) Exploration 1 Leaving Milk on the Counter 55 and solve; 72 30(0.98)t 4. ln (x 2 ln (x dy dx d (2e x) dx 2e x dy dx d 2x (e ) dx e 2x dy dx d ex dx dy dx d e 5x dx e 5. dy dx d 2x/3 e dx e 2x/3 6. dy dx d e x/4 dx e dy dx d (xe 2) dx dx (e ) dx dy dx d 2x (x e ) dx d (xe x) dx (e x)(2x) tan x x 2e x 2 ln (x 2) ln x x 4 2 ln (x 2)(x 2) x2 2) 5. log2 (8x 5) log2 (23)x 5 log2 23x 15 2.71 Section 3.9 Exercises e x ln 7 4) ln 3 ln 2 ln 3 x d (2x) dx xd e dx 2e 2x ( x) 5x d dx x e ( 5x) d 2x dx 3 5e 5x 2 2x/3 e 3 , ln 8 ln 5 ) ln 3 (x 2)(e x) ln (0.98) Quick Review 3.9 3. ln (e ln 2) 7. 17 30 0.343 degrees/minute. tan x x ln 2 4. 30 ln (0.98) (0.98)t. At t e ln 7 1) ln 3 x(ln 3 28.114 ln (0.98) ln 2. 7x ln 2x 3. 17 30 The milk reaches a temperature of 55 F after about 28 minutes. x 1 1.50 2x ln 3x (x 1 8. ln 1. log5 8 ln (ln 5) ln 5 2. 17 ln 30 t ln (0.98) dy 5. dt dy dt ln 18 1. 55 17 30 (0.98)t t ln (ln 5) 3x 10. 2. The temperature of the room is 72 F, the limit to which y tends as t increases. 3. The milk is warming up the fastest at t 0. The second derivative y 30(ln(0.98))2(0.98)t is negative, so y (the rate at which the milk is warming) is maximized at the lowest value of t. ln 18 t 1. The temperature of the refrigerator is 42 F, the temperature of the milk at time t 0. 4. We set y 18 ln 5 18 ln ln 5 ln 5t 3. 2.68 18 5t 1 ln (12x 2) ln 19 From Exercise 34, we have tan ln 3x ln 19 x 9. ln x 3 19 x ln 3 2. 0 ln (4x 4) 1 and 2, so s 2 (x 3)(12x 2) 3x ln 3x If s is the length of a side of the square, then tan ln (12x 2) 3x 8. 5 ,x 4 ln 3x 7. 3 ln x s 15 12 3x 15 9. 10. x/4 d xe x 1 x/4 e 4 x 4 dx e2 ex [(x)(e x) (e x)(1)] ex dy dx d ex dx e dy dx d (x 2) e dx e (x ) xd dx 2 ( x) d2 (x ) dx e x 2x 2xe (x ) 2 115 116 Section 3.9 11. dy dx d (x ) dx 12. dy dx d1 (x dx 13. dy dx d x dx 14. dy dx d 1e x dx 15. dy dx dx 8 dx 16. 17. dy dx 26. 2 2x e1 (1 dy dx d ln (2 dx sin x 2 cos x dy dx d ln (ln x) dx 30. 2 d ln (2x 2) dx 1 ,x 1 x1 27. 2)x dy dx 21 e)x 1 (1 21 dy dx 29. 2)x 1 (1 dy dx 28. 2) (1 e e)x 8x ln 8 d 9x dx dy dx 25. 1 x 9 d csc x 3 dx x d dx (ln 9) ( x) 9 x ln 9 d dx 3csc x (ln 3) (csc x) 3csc x (ln 3)( csc x cot x) 18. d cot x 3 dx 3 cot x 31. d (ln 3) (cot x) dx dy dx 3cot x (ln 3)( csc 2 x) dy 32. dx 19. Use logarithmic differentiation. x ln x ln y ln y d (ln y) dx 1 dy y dx dy dx dy dx ln x ln x 33. ln x ln x d (ln x)2 dx 1 (2 ln x) x 2y ln x x 2x ln x ln x x 34. 36. dy dx ln x 1/ln x ln y 1 ln x ln x ln y 1 y dy dx dy dx x 1/ln x ln y dy dx 35. 20. Use logarithmic differentiation. y e dy dx d (e) dx 21. dy dx d ln (x 2) dx 22. dy dx d (ln x)2 dx 23. dy dx d ln (x 1) dx 24. dy dx 10 d ln x dx 37. dy dx 38. 0, x 2 ln x dy dx 39. dy dx 0 1d 2 (x ) x 2 dx 1 (2x) x2 d (ln x) dx d ( ln x) dx d (ln 10 dx ln x) 2 x 1 d2 (x 1 dx x2 1d ln x ln x dx 1 x) (x) 1 x ,x 2 2 2x 2 cos x) 2x 1) 1 x 1 ln x 2 2) 1 d (2 cos x dx 2 1) x d (2x 2 dx 2x x2 1 1 x ln x (ln x)(1) 1 ln x d d ln x 2 (log4 x 2) dx dx ln 4 1 2 2 1 ln 4 x x ln 4 x ln 2 d dx 2 (ln x) ln 4 ln x d2 dx ln 5 1 ,x 2x ln 5 d d ln x 1/2 (log5 x) dx dx ln 5 1 1 d 1 (ln x) 2 ln 5 dx 2 ln 5 x d log2 (3x dx 3 ,x (3x 1) ln 2 1) (3x 1 ,x x 40. 0 1 x 1 ,x x 0 dy dx 0 1 d (3x 1) ln 2 dx 1) 1 3 1d d log10 (x 1)1/2 log10 (x 1) 2 dx dx 1 1 d 1 (x 1) ,x 2 (x 1) ln 10 dx 2(x 1) ln 10 1 d log2 x dx d ( log2 x) dx d 1 1 d (log2 x) dx log2 x (log2 x)2 dx 1 1 1 or (log2 x)2 x ln 2 x(ln 2)(log2 x)2 d (ln 2 log2 x) dx 1 1 (ln 2) ,x x x ln 2 1 1 ,x x ln 2 0 ln 2 x(ln x)2 d dx (ln 2) (log2 x) 0 d 1 d log3 (1 x ln 3) (1 dx (1 x ln 3) ln 3 dx ln 3 1 1 ,x (1 x ln 3) ln 3 1 x ln 3 ln 3 d (log10 e x) dx d (x log10 e) dx log10 e 1 ln 10 2 ln x x 0 1 1 2) 1 3cot x (ln 3)(csc 2x) y d (x 2 dx x cos x) d (x ln x dx ln x 1 2) d ln (x 2 dx 1 3csc x (ln 3)(csc x cot x) dy dx d ln (x dx d ln 10x dx d (x ln 10) dx ln 10 ln e ln 10 x ln 3) Section 3.9 41. The line passes through (a, e a) for some value of a and has e a. Since the line also passes through the origin, slope m y ea , so a a xe x, we have y (x)(e x) (e x)(1) (x has slope m (a 1 (x 1)e a (a includes the point (0, 0), so we have: need to use a y 43. y 0) 46. 0e 0, or y ln y y ln (sin x)(1) 47. dA dt ln (sin x)] ln y (tan x)(ln x) At t d [(tan x)(ln x)] dx 1 (tan x) (ln x)(sec 2 x) x tan x y (ln x)(sec 2 x) x tan x x tan x (ln x)(sec 2 x) x dA dt 1 1 (2x) 2 x2 1 1 x 2d ln (x 3 dx 21 (1) 3x 1 x x2 1) 2 1 3(x xx 11 (x 1)2/3 x 1) x x2 2 1 3(x 1) d 1 t/140 dt 2 d 20 2 t/140 dt (2 ln (x tan x) 1d ln (x 2 2 dx 1) 20 20(2 x tan x 2 ln (x 3 1) 2 20 (2 ln y d ln y dx 1 dy y dx dy dx dy dx y dy dx ln (sin x)] (sin x)x[x cot x 1 x dy dx x(x 2 1)1/2 (x 1)2/3 1 ln (x 2 2 d ln x dx 1 dy y dx d [x ln (sin x)] dx 1 (x) (cos x) sin x y[x cot x ln x d ln y dx x ln (sin x) d ln y dx 1 dy y dx dy dx dy dx ln ln y ln (sin x)x 6 5(2x 5) x(x 2 1)1/2 (x 1)2/3 ln y x. 6 5(2x 5) 3)4(x 2 1) 1/5 (2x 5)3 x x2 1 (x 1)2/3 y t/140 )(ln 2) d dt t/140 )(ln 2) t 140 1 140 t/140 )(ln 2) 7 2 days, we have (2 1/70 )(ln 2) 7 0.098 grams/day. This means that the rate of decay is the positive rate of approximately 0.098 grams/day. 48. (a) (b) d ln (kx) dx 5)] 3 1 (2) 5 2x 5 2x 5(x 2 1) 3) 2x 5(x 2 1) 3) 3 ln (2x 5) 1 1 (2x) 5 x2 1 4 5(x (x 4 5(x (sin x)x ln y 44. dy dx 0. The equation of the normal line is 1 (x (0 1)e 0 41 5x 3 y 1) 3) 3d ln (2x 5 dx 1) dy dx ln (x 2 3) 4d ln (x 5 dx 1 dy y dx ae a. The desired normal line 1 0 (0 a) ae a (a 1)e a a 0 ae a (a 1)e a 1 0a ea (a 1)e a 1 a 0 or ea 0 (a 1)e a 1 The equation e a 0 has no solution, so we (a 1)e a 1 [4 ln (x 5 1d ln (x 2 5 dx 1 and its equation is 1)e a a) 1 (x 3)4(x 2 1) ln 5 (2x 5)3 d (ln y) dx 1)e x, so the normal line through the point (a, ae a) y 3)4(x 2 1) 1/5 (2x 5)3 (x ln ln y 1. Hence, the slope is e and the equation is 3)4(x 2 1) 1/5 (2x 5)3 (x ln y 0 and we have 0 ex. 42. For y 3)4(x 2 1) (2x 5)3 (x 5 y ln y the slope is also given by m ea ea a 45. 117 1d kx kx dx d d ln (kx) (ln k dx dx 1 d 0 ln x x dx k kx ln x) 1 x 1) 118 Chapter 3 Review 2x ln 2, f (0) 49. (a) Since f (x) (b) f (0) lim f (h) h→0 f (0) 20 ln 2 lim 2h 20 h→0 h s Chapter 3 Review Exercises ln 2. lim h→0 h (pp. 172–175) 2h 1 (c) Since quantities in parts (a) and (b) are equal, lim 2h h→0 1 dy dx d5 x dx dy dx d (3 dx 3. dy dx d (2 sin x cos x) dx d 2(sin x) (cos x) dx ln 2. h (d) By following the same procedure as above using 7x, we may see that lim g(x) 1. 2. h 7h h→0 1 ln 7. h y3 3 dy dx 4 dy dx d 2x dx 2x 1, or a dx a dx (d) y2 1. y1 dx ln a, then a will equal a x if and only if dx So if y3 ln a y2 5. ds dt 1, or a 12 x 2 d dx k x and ds dt 2 d cot t dt 2 22 csc t t2 7. dy dx a x if and d dx 8. dy dx 1 . x c) 9. dr d Therefore, at any given value of x, these two curves will 10. dr d 53. (a) Since the line passes through the origin and has slope (b) The graph of y y positive x e. Therefore, ln x x for all e 11. to see that 2d2 t dt t 1 3/2 x 2 e 1)2 x. e . Therefore, e is bigger. 1 d sec (1 d 2x 3) 1 2 t 2t) 2 t2 1/2 x ) ( 2x 1)(1) 1 1 3 ) tan (1 d tan 2 (3 d (2) 2x sec (1 3 ) tan (1 3 )(3) 3) 2 ) 2d ) 2 ) sec 2 (3 2 2 tan (3 d 2 ) )( 2 ) 2 2 ) sec (3 ) d2 (x csc 5x) dx 5x 2 csc 5x cot 5x dy dx d ln dx 13. dy dx d ln (1 dx 14. dy dx d (xe x) dx 12. 2 sin (1 csc 2 1 d (x 2x 1) (x) dx 2 x (2x 1) 3x 1 (x 2)( csc 5x cot 5x)(5) e (d) Exponentiating both sides of ln x e x, we have e e ln x e x, or x e e x for all positive x e. (e) Let x dy dx 2t)( 2) d 1/2 (x dx 1 1 2x3/2 2x 4 tan (3 e. x or ln x 4 (2x x 2 tan (3 ln x lies below the graph of the line (c) Multiplying by e, e ln x 1)(2) 1 1 2 tan (3 x . e x for all positive x e (cos 2x)(2) (2x 1)2 sin (1 csc 2 3 sec (1 have perpendicular tangent lines. 1 , its equation is y e 1)(2) (2x 2t) x 1 1/2 x 2 e. d (ln x dx d sin 2x dx (2x 1 1 2x 52. 2 cos 2 x d cos (1 dt e. a x ln a. This will equal y1 only if ln a d dx 5 We conclude that the graph of y3 is a horizontal line at y ln a. a x if and only if y3 21x 6 2(cos x) (sin x) d (2 sin x cos x) dx 4. 0.693147 1.098613 1.386295 1.609439 dx a dx 21x 2 2 cos 2x ln a 0.693147 1.098612 1.386294 1.609438 (c) 3x 7) 1 4 Alternate solution: 6. 2 1 x 4 5x 4 a. (b) The graph of y3 is always a horizontal line. a 1 x 4 7x 3 2 sin 2 x 50. Recall that a point (a, b) is on the graph of y e x if and only if the point (b, a) is on the graph of y ln x. Since there are points (x, e x) on the graph of y e x with arbitrarily large x-coordinates, there will be points (x, ln x) on the graph of y ln x with arbitrarily large y-coordinates. 51. (a) The graph y4 is a horizontal line at y 12 x 8 x 1 2x csc 5x d x dx e x) (x)(e x x 1 1 x 1 1 (csc 5x)(2x) d (1 e x dx )( 1) (e 2 e x) x 1 ,x 2x x )(1) 0 ex 1 ex xe x e x Chapter 3 Review 15. dy dx d 1 ln x) (e dx dy 16. dx d 1 ln x (e e ) dx d ln (sin x) dx d (ex) dx Alternate solution, using logarithmic differentiation: e (2x)2x y cos x sin x 1d (sin x) sin x dx x2 cot x, for 1 ln (2x) dr 17. dx 1) ), where k is even. d 1 d ln (cos 1 x) cos 1 x dx cos 1 x dx 1 1 1 cos 1 x 1 x2 cos 1x 1 x 2 18. dr d d log2 ( 2) d 19. ds dt d log5 (t dt t ln y 2 7) 1 d2 () ln 2 d (t 2 ln 2 1 d (t 7) ln 5 dt 7) (t 1 , 7) ln 5 7 23. d (8 t ) dt d 8 (ln 8) ( t) dt t 8 t dy dx 24. ds 20. dt x d tan 1x e dx e tan 1 dy du 1 u2 d sin 1 du 1 1 25. d (ln x)2 dx d 2 ln x ln x dx 2y ln x x 2x ln x ln x x dy dt x2 26. x2 dy dt 1 27. (2x)(2x) 1 2 x2 dy dz t 2) cot t 2) 2t 2 4t 2 x 2 1)(2 )(2x ln 2 2) (x 2 1)3/2 (2 2x)[(x 2 x 3 1)(x ln 2 (x 2 1)3/2 2x (2 ) 1) 2 (2 2 )(x ln 2 x x ln 2 (x 2 1)3/2 cos 28. x 2] 1 2 dy dx d (2 dx (2 t 1 1 2t)(2t) 2t z 2) 1 1 (cos 1 z)(1) 2 1 1 z z z2 1 x 1 1 csc x) 1 1) x) x 1 x) 1 2 ( x) x x)2 1 x 1 csc x x 1 1 1 csc 1 1 1 2x 1) 1 x (cot z x ( 1 2t t)(1) 2t] cos z2 (2 csc (2 2x)(x 3 ln 2 x ln 2 (x 2 1)3/2 1 z2 z 1 u2 1 1 2t 2t cot 1 1 x 1 d (z cos 1 z dz 1 1 1 1 (2) (2t)2 1 1 (x (sec (z) (2x) u u 1 d [(1 dt u2 1 ( 2u) sec t2 2 1 1 (2x)(2)] x 1 t 1[(2x)(2x)(ln 2) etan x 1 x2 1 tan dx 1 t d (2x)(2x) dx 1 xd 1 ln t) 2 t2 t x2 2 u2 d (t sec 1 t dt 1 x2 x2 1 (1 d 1 dx [(2x)2x] x2 1 (t) (2x)2x d dx x ln 2 d u 2)2 du 1 u2 2 (ln x)(ln x) d ln y dx 1 dy y dx dy dx dy dx ( 1 ln (x ln x) ln y 1 1 ln (x 2 1) 2 1 d [ln 2 ln x x ln 2 ln (x 2 1)] 2 dx 1 1 1 0 ln 2 (2x) x 2 x2 1 1 x y ln 2 x x2 1 (2x)2x 1 x ln 2 x2 1 x2 1 x ln x ln y dy 22. dx ln x x2 ln ln 8 21. Use logarithmic differentiation. y ln 2 d ln y dx 1 dy y dx dy dx dy dx 2 ln 2 2 ln (2x) ln y values of x in the intervals (k , (k 119 x x 1 2 x z2 ( 2z) 120 29. Chapter 3 Review dy dx 32. Since y d csc 1 (sec x) dx 1 d (sec x) 1 dx 2 sec x sec x 1 tan 2 x sec x dy dx sin x cos x sin x cos x sin x sin x 2 cos 2 1 x ,, 2 ,x 2 3 , we may 2 ,x (sec x) 0 x , d1 d1 x 1, sin cos sin c os (1 1 2 1 sin c os cos 1 1 sin c os cos (1 2 (2x 7) dy dx x xy 2 3 2 2,x x , x 2,x x 2 2 cos )(cos ) (1 sin )(sin ) (1 cos )2 cos2 sin (1 cos )2 all x 0. x 2x 5) 7)(1) (x 7)2 2x x2 3y 5x 4/5 d (5x 4/5) dx 4x 1/5 10y 6/5 5 is defined for all 7 5)(2) 17 , the (2x 7)2 7 . 1 d (1) dx 0 (y y x 2) 2 3 15 d (10y 6/5) dx dy 12y 1/5 dx dy dx d (15) dx 0 4x 1/5 12y 1/5 1 3(xy)1/5 37. Use implicit differentiation. xy d dx 1 [x xy dy dx 0 and 1 xy d (1) dx (y)(1)] 0 dy dx 0 y dy dx y x Alternate method: Since xy Therefore, 1, we have xy dy dx 1 . x2 1, 1. 36. Use implicit differentiation. 2 2 , the function is differentiable for x 2x d d d (xy) (2x) (3y) dx dx dx dy dy x (y)(1) 2 3 dx dx dy (x 3) dx dy dx sin2 sin 1 cos )2 ln x is defined for all x 1d 2 (x ) x 2 dx (x (2x x 2 31. Since y 1 (2x 7 dy and 2 dx 2 1 1 2 34. Since y 2 3 2 Note that the derivative exists at 0 and 2 only because these are the endpoints of the given domain; the two-sided derivative of y csc 1(sec x) does not exist at these points. dr d , which is defined only for x 35. Use implicit differentiation. x 2,x , 1, 0 dy Therefore, dx 30. x 2)3/2 x (1 2 x x, 2 1 function is differentiable for all x x x), x, 2 2x 1 and x)(2x) (cos x) 0 ( 2 x 2)( 1) (1 (1 x 2)2 (1 x x2 21 3 2 x x, 2 1 2 1 x2 (sec x) sec 2 x sin x, the x is defined for all x x2 1 dy dx rewrite the function as follows: 1 (cos x)(1) the function is differentiable for all x On the domain 0 1 1 1 33. Since y Alternate method: csc (x)( sin x) sec x tan x sign (sin x), x y cos x x cos x is defined for all real x and function is differentiable for all real x. sec x tan x sec x tan x 1 cos x 1 cos x sin x 1 and y 1 . x 121 Chapter 3 Review 38. Use implicit differentiation. y y3 41. y 2 cos x x 2 x1 d x dx x 1 (x 1)(1) (x)(1) (x 1)2 1 2y(x 1)2 d2 y dx dy 2y dx dy dx 39. x d3 (x ) dx 3x 2 3 y 3 d3 (y ) dx d (y) dx 3y 2y (3y 2 y 2 sin x 1)y 2 sin x 1 d3 (y ) dx 3y 2y 0 y x2 y2 2 sin x 3y 2 1 d dx (3y 2 1)(2 cos x) (2 sin x)(6yy ) (3y 2 1)2 (3y 2 1)(2 cos x) 2 x y2 d dx y 2 sin x 3y 2 1 y d (1) dx y d (2 cos x) dx (y 2)(2x) (x 2)(2y)(y ) y4 (y 2)(2x) (x 2)(2y) (3y 2 2xy 2x x y2 2 x 1/3 42. 4 5 y 2x(x 3 y 3) y5 2x y5 since x 3 y3 (3y 2 d 1/3 (x ) dx 1 2/3 x 3 y 1/3 d 1/3 (y ) dx 1 2/3 y y 3 y 1 y 40. y2 d2 (y ) dx 2yy y y 1 2 x d d2 (1) dx dx x 2 x2 2 1 x 2(2y) x 2y d1 dx x 2y 1 d2 (x y) (x 2y)2 dx 1 [(x 2)(y ) (x 2y)2 1 1 (x 2) 2 (x 2y)2 xy 11 2xy x 4y 2 y 1 2xy 2 x 4y 3 1)2 1)2 cos x 12y sin 2 x (3y 2 1)3 4 d (4) dx 0 x y y 2/3 x 2/3 2/3 d dx y 2/3 x 2y 3x 1/3 2y 3x 1/3 xy (x) (y)(1) x2 y 2/3 x 43. y 2x 3 3x 1, y 6x 2 3, y 12x, y (4) 12, and the rest are all zero. 2xy x4 , 24 3 x , 6 2 x , 2 44. y y y y (4) y (5) x, 1, and the rest are all zero. y x2 2 1/3 1/3 xy ( x 5/3y 2/3 3 2 4/3 1/3 2 5/3 2/3 x y x y 3 3 (y)(2x)] 2 sin x 3y2 1 2 y4 3 (12y sin x) x 2 y) 122 Chapter 3 Review 45. dy d dx dx x2 (2x x2 32 2(3) 3 1 2 32 2(3) x 2) 2x 3, we have y At x 2 dy dx and 1 2x (x 3 3 or y 46. dy dx 2 d (4 dx 3) 2 3 or y csc 2 x At x y 2 4 3 3) dy dx At t 3 y 2 csc x) 2 csc x cot x csc 2 2 csc 2 y 2 (a) Tangent: y (b) Normal: y 4 2 2 csc cot 2 1x 0 2 and 1 2(1)(0) 2 or y 2 or y 2 50. 2 2 2 1x x x 1. 2 2 dy dx At t y 2y 2 d (2y 2) dx dy 2x 4y dx dy dx Slope at (1, 2): y d (9) dx dy 51. dx 0 2x 4y x 2y (a) Tangent: y At t 1 4 1 2(2) 1 (x 4 11 5 1) dy dt dx dt 2 sin t 2 cos t tan t 3 , we have x 4 2 cos 3 4 3 4 2 sin 2, and 2, dy dx tan 3 4 1. 1 x 4 2 or y 1(x 2) 2), or y 4 cos t 3 sin t dy dt dx dt ( 4 cot t 3 3 , we have x 4 4 sin 3 4 2 3 cos 2, and x 2 3 4 2. 3 2 2 dy dx , 4 3 cot 3 4 4 . 3 The equation of the tangent line is 9 d2 (x ) dx 4 x 5 1 or y 6 2 2 47. Use implicit differentiation. x2 4) 5 x 4 1 or y The equation of the tangent line is , we have cot dy dx x 49. 3 cot x 4) . 3 (x 2 53 2 x 4 (x 5 (b) Normal: y 3 (b) Normal: y 5 (x 4 (a) Tangent: y 2x 3 2 (a) Tangent: y 1 x2 9 4 y 4 x 3 32 2 dy dt dx dt 6 2 5 sec 2 t 3 sec t tan t , we have x 5 tan 5 sec t 3 tan t 3 sec 4 2. 5 3 sin t 2 6 53 dy , and dx 3 6 4 x 3 2, or y 3, 10 . 3 5 3 sin 6 The equation of the tangent line is (b) Normal: y 4(x 1) 2 or y 4x 2 y 48. Use implicit differentiation. xy x d (x) dx 1 1 2 (x) xy d ( dx dy dx xy) (y)(1) x 2 dy xy dx dy dx 6 d (6) dx 0 y 1 2 xy x 2 1 4 1 4 2 2 y 1 2 y 2 x Slope at (4, 1): xy 2 1 4 y x 5 4 xy 10 (x 3 2 3) 53 , or y 3 10 x 3 5 3. Chapter 3 Review (c) Nowhere in its domain dy dt dx dt dy 52. dx At t 1 cos t sin t 59. y 2 4 , we have x 2 cos 4 2 y = f (x ) , –3 y dy dx 2 sin 4 1 123 4 cos 4 4 sin 2 2 2 2 4 y 1. y = f (x ) The equation of the tangent line is y (2 y (1 2 1) x 2)x 2 4 2 2 1 This is approximately y 2 4 x –2 , and 60. 2 2 1 3 , or x . 2.414x 3.200. 61. (a) iii (b) i (c) ii 53. (a) 62. The graph passes through (0, 5) and has slope x 2 and slope 0.5 for x 2. 2 for y 7 [ 1, 3] by 5 3 1, (b) Yes, because both of the one-sided limits as x → 1 are equal to f (1) 1. (c) No, because the left-hand derivative at x the right-hand derivative at x 1 is 1. 1 is 1 and 54. (a) The function is continuous for all values of m, because the right-hand limit as x → 0 is equal to f (0) 0 for any value of m. (b) The left-hand derivative at x 0 is 2 cos (2 0) 2, and the right-hand derivative at x 0 is m, so in order for the function to be differentiable at x 0, m must be 2. 55. (a) For all x 0 (b) At x y = f (x ) 7 x 63. The graph passes through ( 1, 2) and has slope 2 for x 1, slope 1 for 1 x 4, and slope 1 for 4 x 6. y 7 0 (c) Nowhere 56. (a) For all x y = f (x ) (b) Nowhere (c) Nowhere 7 57. Note that lim f (x) lim (2x x→0 lim f (x) x→0 x→0 lim (x 3) x→0 3) x 3 and 3. Since these values agree with f (0), the function is continuous at x 0. On the other 64. i. If f (x) f (x) 9 7/3 x 28 9, then f (x) 3 4/3 x and 4 x 1/3, which matches the given equation. hand, 2, 1 x 0 f (x) , so the derivative is undefined at 1, 0 x x 4 (0, 4] (b) At x 0 58. Note that the function is undefined at x 0. (c) Nowhere in its domain (a) [ 2, 0) (0, 2] 9 7/3 x 28 3 4/3 x , which 4 2, then f (x) contradicts the given equation f (x) 0. (a) [ 1, 0) ii. If f (x) (b) Nowhere iii. If f (x) 3 4/3 x 4 6, then f (x) the given equation. x 1/3. x 1/3, which matches 124 Chapter 3 Review 64. continued (d) 3 4/3 x 4 iv. If f (x) 4, then f (x) x 1/3 and f (x) 5 tan 0)( 5 sec 2 0) (e) however that i and iii could not simultaneously be true. (2 (f) [ 1, 5] by [ 10, 80] (b) t interval avg. vel. 38 0.5 58 1 70 1.5 74 2 70 2.5 58 3 38 3.5 10 4 [0, 0.5] [0.5, 1] [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] 10 0 38 0.5 58 1 70 1.5 74 2 70 2.5 58 3 38 3.5 1 2 67. (a) 8 f (1) 1 At x 2 d f( dx (b) 56 f(x) At x d dx g (x) g ( 1) 3(2) 1 5. d2 [f (x)g 3(x)] dx (c) d g(f (x)) dx At x f (x) g (f (x))f (x) 1, the derivative is x 1 ( 3) 2 13 . 10 (d) d f (g(x)) dx g (0)f ( 1) (4)(2) 8. f (g(x))g (x) 1, the derivative is f (x) f (x) f ( x) 3f (x) 2 5 ( 3)2(0) f 2(x) 3g 2(x)g (x) g 3(x) 2f (x)f (x) f (x)g 2(x)[3f (x)g (x) 2g(x)f (x)] At x 0, the derivative is f (0)g 2(0)[3f (0)g (0) 2g(0)f (0)] ( 1)( 3)2[3( 1)(4) 2( 3)( 2)] 9[ 12 12] 0. 1 5 x f (g( 1))g ( 1) 2 f (x) f (0) 0, the derivative is x) 5 f 2(1) cos 1 (1) 5 g (f ( 1))f ( 1) 2 (c) 2 1, the derivative is 3f ( 1) 40 At x 1 f(x) 2 x f (x) cos 2 2 g(x)] d [3f (x) dx At x 24 1 1 x 2 12. 1, the derivative is 1 f (1) 10f 2(x) cos 8 (d) Average velocity is a good approximation to velocity. 2 2 . 3 1, the derivative is 24 [ 1, 5] by [ 80, 80] x f (x) x2 f (x)] 2 x 10 sin (2f (x)f (x)) 2 x 20 f (x)f (x) sin 5 2 3f (0) 32 d [10 sin dx 20( 3) (c) x f (x)] 0, the derivative is 20 f (1) f (1) sin 40 cos x)( f (x)) ( f (x))( sin x) (2 cos x)2 cos 0)(f (0)) (f (0))( sin 0) (2 cos 0)2 At x 56 (2 d f (x) dx 2 cos x At x 65. (a) d dx 1 ( 5) 5 f (1)( 5) 1. Answer is D: i and iii only could be true. Note, (b) 5 tan x)( 5 sec 2 x) 0, the derivative is x 1/3. At x f (1 f (1 1 2/3 x , which contradicts the given equation 3 d [ dx 5 tan x) At x f (x) 66. (a) d f (1 dx 2 f (0) 29 f( 1 . 3 d f (x) dx g(x) 2 (g(x) 1 5 2 2. 0, the derivative is (g(0) 1 . 10 (2)(1) 2)f (x) f (x) g (x) (g(x) 2)2 At x x) 2x f ( 1) f (1) 1, the derivative is 2 21 (e) f ( 1)g ( 1) 2)f (0) f (0)g (0) (g(0) 2)2 6. (3 2)( 2) ( 1)(4) ( 3 2)2 Chapter 3 Review d g(x dx (f) f (x)) g (x f (x))(1 At x dw ds f (x)) (x 1)[1 dw dr dr ds ( 2)] d [sin ( dr 1 cos ( r f (0))[1 (1)( 1) r 2) 0, we have r dw ds cos ( 4 cos 0 4 2)] 1 4 t 0, the particle first reaches the origin at 2 . The velocity is given by v(t) 1 10 sin t 4 is 10 sin speed at t 6 t 4 4 , so the velocity at 10, and the 2 is 4 by a(t) 6 10 10. The acceleration is given 10 cos t is 10 cos , so the acceleration at 4 0. 2 4 and so 6 71. (a) 8 cos 0 6 24 183 4 2 6 f (0)] d 8 sin s ds 8 cos s 8 sin 0 2) 8 cos (d) Since cos t 2r At s f (x)) f (x)) 0, the derivative is g (0 g (0 68. d dx g (x 3 ds dt d 2s dt 2 d (64t dt d (64 dt 16t 2) 64 32t) 32t 32 (b) The maximum height is reached when 69. Solving 2 t 1 for t, we have 1 t 2 1 2 d2 ( d 12 ( 3 7) 2(2 3( 2 dr dt 7) 2 4( 2 3( ) At t and so dr dt 2 0, we may solve 2(1)4(12 3(1 2 t 7) 2) 2 (b) s(0) 10 cos t 10 cos 4 5 4 2/3 2(8) 2/3 3 1, (b) Since v(t) 1 . 6 (980) , y(t) 12.3 sec. 393.8 ft. 4 sec. 7 280 cm/sec. 980t, the velocity is 4 = 560 cm/sec. Since a(t) 7 dV dx 73. 2 d dx 10 (20x 1 dv dt 980, the 1, we have s(t) 10. 1, we have s(t) 10. 3 x2 x 3 x2 x 40 d dx 13 x 3 x 2) 9x 10x 2 32 x 20 13 x 1600 (b) The marginal revenue is r (x) 4 ds dt 5.2t, the 64 5.2 acceleration is 980 cm/sec 2. Farthest right: When cos t 64 5.2 4 7 74. (a) r(x) 4 64 The average velocity is 160 (c) Farthest left: When cos t 2.6t 2) 490t 2, it takes 72. (a) Solving 160 7) 2) 0, which 64, so the velocity is 64 ft/sec. d (64t dt ) 70. (a) One possible answer: x(t) ds dt ds dt The maximum height is s 1 to obtain 2/3 0, ) 3 ds dt 2 sec. maximum height is reached at t 1 dr (2 dt 2/3 2 (d) Since dr d ( dt d (2 ) 3 (c) When t dr dt dt d 7)1/3 2/3 occurs at t , and we may write: dr d 125 9 3 (x 2 1600 3 (x 1600 3 x 10 160x 40)(x 32 x 1600 4800) 120), which is zero when x 40 or x 120. Since the bus holds only 60 people, we require 0 x 60. The marginal revenue is 0 when there are 40 people, and the 40 2 corresponding fare is p(40) 3 $4.00. 40 126 Chapter 3 Review 74. continued 78. (a) P(0) (c) One possible answer: If the current ridership is less than 40, then the proposed plan may be good. If the current ridership is greater than or equal to 40, then the plan is not a good idea. Look at the graph of y r(x). 1 200 e5 (b) lim P(t) lim t→ t→ 1 student 1 200 e5 d 200(1 dt (c) P (t) 200(1 dx dt tan , we have (sec 2 ) d dt (b) 0.6 rad sec 2 0.6 sec 0 1 revolution 2 rad e5 t)2(200e5 t)( 1) (200e5 t)(2)(1 (1 e5 t)4 e 5 t)( 200e 5 t) 400(e 5 t)2 (1 e 5 t)3 0.6 km/sec. f (x) Since P 0 when t cos (x cos (x cos x sin x) (x sin x)(1 sin x) and P (t) cos x). This derivative is zero when t 0 (which we need not solve) or when of these values, f (x) 0) P (t) 5. To confirm that this corresponds to the maximum value of P (t), note that P (t) sin x) 1,which occurs at x sin (2k 5, the critical point of y sin x). Then d dx cos (x e5 t)(e5 t)( 1) (200e 5 t)(e 5 t 1) (1 e 5 t)3 60 sec 18 = revolutions per 1 min occurs at t sin (x e 5 t) 2(e 5 t)( 1) (1 minute or approximately 5.73 revolutions per minute. 76. Let f (x) 1 (1 P (t) 0.6 sec 2 . At point A, we have dx 0 and dt e 5 t) 200 students 200e 5 t (1 e 5 t)2 [0, 60] by [ 50, 200] 75. (a) Since x 200 1 t f (2k ) 0. Thus, f (x) sin (2k sin 2k ) f (x) 0 for x 5 5. The maximum rate occurs at 5, and this rate is 200e 0 (1 e 0)2 P (5) 2k for integers k. For each 0 for t 0 for t 200 22 50 students per day. Note: This problem can also be solved graphically. 2k , 79. which means that the graph has a horizontal tangent at each of these values of x. 77. y (r) y (l) y (d) 1 2rl y (T) 1 2rl d1 dr 2rl d1 dl 2rl d1 dd 2rl 1 T d 2 d1 dT 2rl 1 1 d2 T Since y (r) 1 T d1 1 2l d dr r 2r 2l 1 T d1 1 2r d dl l 2rl 2 1 Td (d 1/2) 2rl dd T 1 d3 4rl 1 1d ( T) 2rl d dT 1 T d T d T d 3/2 T d 4rl 0, y (l) T d T d dT 0, and y (d) 0, increasing r, l, or d would decrease the frequency. Since y (T) increasing T would increase the frequency. 0, , ] by [ 4, 4] [ (a) x (b) k , where k is an odd integer 4 , 22 (c) Where it’s not defined, at x (d) It has period 2 in this window. k , k an odd integer 4 and continues to repeat the pattern seen 127 Section 4.1 Quick Review 4.1 80. Use implicit differentiation. x2 y2 1 d2 (y ) dx 2x 2yy y y 1. f (x) d (1) dx 2. f (x) d 2(9 dx 24 0 2x x 2y y dx dx y (y)(1) (x)(y ) y2 y 12 (x 3 2 x 6. g (x) 1 y3 y2 1 ( 3) 1 3 8. h (x) x 2 → 0 . Therefore, lim f (x) . (9 12 (x 1) 4/3 3 2x 3(x 2 1)4/3 (2x) d2 (x dx d ln x e dx 2x x 2)3/2 1) d ln x dx 2x x2 1 sin (ln x) x 2e 2x 1) . d x dx 1 33 Chapter 4 Applications of Derivatives s Section 4.1 Extreme Values of Functions 9. As x → 3 , 10. As x → 9 3, x→3 x 2 → 0 . Therefore, lim 9 Finding Extreme Values 1. From the graph we can see that there are three critical points: x 1, 0, 1. Critical point values: f ( 1) 0.5, f (0) 0, f (1) 0.5 Endpoint values: f ( 2) 0.4, f (2) 0.4 Thus f has absolute maximum value of 0.5 at x 1 and x 1, absolute minimum value of 0 at x 0, and local minimum value of 0.4 at x 2 and x 2. d3 (x dx 2x) 3x 2 3(1)2 2 d (x dx 2) f (3) (b) f (x) x→ 3 f (1) 11. (a) (pp. 177– 185) Exploration 1 1) d 2x dx e 2x 3/2 d2 (x dx sin (ln x) 7. h (x) (since the given equation is x 2 dy dx 2 4/3 1 x 2) (9 1/3 1) 2 1 y3 3), x2 x x 2) 1/2 ( 2x) 1) 4 2 3/2 1 5. g (x) y3 At (2, 1 x) d (9 dx x 2) d2 (x dx 4. f (x) y2 x x 2) (9 x x y y 2 d (4 dx 3 1/4 x 4 3. f (x) d2 (x ) dx 1 . 1 2 1 1 (c) Left-hand derivative: lim f (2 h→0 lim h) h h3 f (2) lim (h 2 h)3 [(2 h) 2(2 h h)] 4 10h 6h h→0 [(2 h→0 6h 2 h h→0 lim 10) 10 [ 2, 2] by [ 1, 1] 2. The graph of f has zeros at x 1 and x 1 where the graph of f has local extreme values. The graph of f is not defined at x 0, another extreme value of the graph of f. Right-hand derivative: lim f (2 h→0 lim h→0 h) h f (2) lim h→0 2] 4 h h h lim 1 h→0 [ 2, 2] by [ 1, 1] 1 d 3. Using the chain rule and x dx x df 1 x2 . x (x 2 1)2 dx () x , we find x Since the left- and right-hand derivatives are not equal, f (2) is undefined. 128 Section 4.1 12. (a) The domain is x 2. (See the solution for 11.(c)). 3x 2 2, x 2 (b) f (x) x2 1, 14. The first derivative k (x) Section 4.1 Exercises 1. Maximum at x b, minimum at x c2; The Extreme Value Theorem applies because f is continuous on [a, b], so both the maximum and minimum exist. 0. Since k(0) 15. The first derivative f (x) x Endpoint values: 4. No maximum, no minimum; The Extreme Value Theorem does not apply, because the function is not continuous or defined on a closed interval. Maximum value is 1 at x 6. Maximum at x a, minimum at x c; The Extreme Value Theorem does not apply since the function is not continuous. 4 minimum value is 2 ln 0.5 1 4 f (4) Maximum value is 1 4 x e ( 1) e 1 at x e 13. The first derivative h (x) x ; 5 ; 4 sec x tan x has zeros at x and is undefined at x 0 and x 2 2 . Since g(x) 0 sec x is , the critical points occur only at . x x 0 g(x) g(x) 1 1 Since the range of g(x) is ( , 1] [1, ), these values must be a local minimum and local maximum, respectively. Local minimum at (0, 1); local maximum at ( , 1) 2 3/5 x is never zero but is 5 17. The first derivative f (x) undefined at x has no zeros, so we need g(1) e 1 1 e 0. 1 has no zeros, so we need only consider the endpoints. h(0) ln 1 0 h(3) ln 4 Maximum value is ln 4 at x 3; minimum value is 0 at x 0. f (x) 3 f (x) 0 ( 3)2/5 32/5 1.552 Since f (x) 0 for x 0, the critical point at x 0 is a local minimum, and since f (x) ( 3)2/5 for 3 x 1, the endpoint value at x 3 is a global maximum. Maximum value is 32/5 at x 3; minimum value is 0 at x 0. undefined at x 1. 1 0 3 2/5 x is never zero but is 5 18. The first derivative f (x) 1; x 4 4; ln 2 Maximum value is e at x minimum value is 1.307 2 0 ; Critical point value: x Endpoint value: x 12. The first derivative g (x) e only consider the endpoints. g( 1) 1 1 7 ,0 4 1. 1; 1 ,2 2 1 Critical point values: 1.636 ln 4 at x minimum value is 1 at x local maximum at ln 4 f (x) 1 2 also undefined at x Endpoint values: f (0.5) f (x) 7 4 1 f (x) 1 at x local minimum at 0, and x ln 1 0 x 9. Maximum at (0, 5) Note that there is no minimum since the endpoint (2, 0) is excluded from the graph. 1 f (x) x 8. Minima at ( 2, 0) and (2, 0), maximum at (0, 2) Critical point value: f (1) , has zeros at 4 4 5 4 16. The first derivative g (x) 10. Local maximum at ( 3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, 1) 0, 0. x local maximum at 7. Local minimum at ( 1, 0), local maximum at (1, 0) x→ 5 . 4 ,x 3. Maximum at x c, no minimum; The Extreme Value Theorem does not apply, because the function is not defined on a closed interval. 11. The first derivative f (x) 0. 1 and lim k(x) cos x Critical point values: x 5. Maximum at x c, minimum at x a; The Extreme Value Theorem does not apply, because the function is not continuous. 02 e the maximum value is 1 at x 2. Maximum at x c, minimum at x b; The Extreme Value Theorem applies because f is continuous on [a, b], so both the maximum and minimum exist. 1 has a zero at x x has a zero at x Since the domain has no endpoints, any extreme value must occur at x 1 x2 x2 2xe 0. Critical point value: x Endpoint value: x 0 3 f (x) f (x) 0 33/5 1.933 Since f (x) 0 for x 0 and f (x) 0 for x 0, the critical point is not a local minimum or maximum. The maximum value is 33/5 at x 3. Section 4.1 19. 129 24. [ 2, 6] by [ 2, 4] [ 4.7, 4.7] by [ 3.1, 3.1] Minimum value is 1 at x 2. To confirm that there are no “hidden” extrema, note that 20. y x (x 2 1) 2 (2x) 3x 2 To find the exact values, note that y 2 ,4 3 2, which is 2 . Local maximum at 3 46 9 46 9 1). 25. [ 6, 6] by [ 2, 7] 2 ,4 3 2x which is zero only at 1)2 0 and is undefined only where y is undefined. There is a local maximum at (0, zero when x (x 2 [ 1.5, 1.5] by [ 0.5, 3] ( 0.816, 5.089); local minimum at The minimum value is 1 at x 0. 26. (0.816, 2.911) 21. [ 4.7, 4.7] by [ 3.1, 3.1] The actual graph of the function has asymptotes at x 1, so there are no extrema near these values. (This is an example of grapher failure.) There is a local minimum at (0, 1). [ 6, 6] by [ 5, 20] To find the exact values, note that y 3x 2 2x 8 (3x 4)(x 2), which is zero when 27. 4 . Local maximum at ( 2, 17); 3 4 41 local minimum at , 3 27 x 2 or x 22. [ 4.7, 4.7] by [ 3.1, 3.1] Maximum value is 2 at x minimum value is 0 at x 1; 1 and at x 28. [ 6, 6] by [ 4, 4] Note that y 3x 2 6x 3 3(x 1)2, which is zero at x 1. The graph shows that the function assumes lower values to the left and higher values to the right of this point, so the function has no local or global extreme values. [ 4, 4] by [ 80, 30] 23. Minimum value is 115 at x 2 local maximum at (0, 10); local minimum at 1, 29. [ 4, 4] by [ 2, 4] Minimum value is 0 at x 13 2 1 and at x 1. [ 5, 5] by [ 0.7, 0.7] 1 at x 1; 2 1 minimum value is at x 1. 2 Maximum value is 3; 3. 130 Section 4.1 30. x3 36. Note that f (x) x 9x, 9x, 3 x 3x 2 9, 3x 2 9, Therefore, f (x) [ 5, 5] by [ 0.8, 0.6] 3 or 0 x x 0 or x 3 x 3 3. 3 or 0 x x 0 or x 3 3 3. (a) No, since the left- and right-hand derivatives at x are 9 and 9, respectively. 31. 0 (b) No, since the left- and right-hand derivatives at x are 18 and 18, respectively. 1 at x 0; 2 1 minimum value is at x 2. 2 Maximum value is 3 (c) No, since the left- and right-hand derivatives at x are 18 and 18, respectively. (d) The critical points occur when f (x) (at x [ 6, 6] by [0, 12] x Maximum value is 11 at x 5; minimum value is 5 on the interval [ 3, 2]; local maximum at ( 5, 9) 0 3) and when f (x) is undefined (at x 3). The minimum value is 0 at x and at x 3, at x ( 3, 6 3). [ 4, 4] by [ 3, 3] [ 3, 8] by [ 5, 5] Maximum value is 4 on the interval [5, 7]; minimum value is 4 on the interval [ 2, 1]. x2/3(1) y 33. crit. pt. x [ 6, 6] by [ 6, 6] Maximum value is 5 on the interval [3, ); minimum value is 5 on the interval ( , 0 5x 2) 3 4 3 x derivative extremum 4 5 x 2 1/3 x (x 3 0 local max value 12 1/3 10 25 undefined local min 2]. 34. [ 4, 4] by [ 3, 3] x2/3(2x) y [ 6, 6] by [0, 9] 2 1/3 2 x (x 3 4) 8x 2 3 8 3 x Minimum value is 4 on the interval [ 1, 3] 35. (a) No, since f (x) x 2. 2 (x 3 2) 1/3 , which is undefined at (b) The derivative is defined and nonzero for all x Also, f (2) 0 and f (x) 0 for all x 2. crit. pt. derivative extremum value x 2. (c) No, f (x) need not not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form [a, b] would have both a maximum value and a minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a. 1 x 0 x 1 0 minimum undefined local max 0 minimum 1.034 0 38. 3 0 3 0 or 0, 3, 6 3) and 3; local maxima occur at ( 37. 32. 3 131 Section 4.1 39. 42. [ 2.35, 2.35] by [ 3.5, 3.5] y 1 x x2 24 x 2 ( 2x) 2 (4 [ 4, 4] by [ 1, 6] x) 1, y 2 x x 2x, 0 0 crit. pt. derivative 2x 2 4 x2 x2 (1) 4 x2 4 extremum value x crit. pt. undefined local max undefined local min 3 1 0 local max 4 derivative extremum value x 0 x 4 2 0 minimum 2 0 maximum 2 undefined local min x 0 2 x 2 x 2 43. 0 [ 4, 6] by [ 2, 6] 40. 2x 2x y 2, 6, x x 1 1 crit. pt. derivative extremum value x 1 x2 y 23 3 crit. pt. x 3 3 x 5x 2 23 0 0 0 5 1 3 0 maximum 12x x value minimum local max [ 4, 6] by [ 5, 5] 0 144 1/2 15 125 undefined minimum 0 4.462 We begin by determining whether f (x) is defined at x where 12 x 4 f (x) x3 41. 1 x 2 6x 2 15 , 4 lim f (1 h) h h→0 f (1) lim h→0 [ 4.7, 4.7] by [0, 6.2] 2, 1, x x 1 h→0 1 1 crit. pt. derivative x lim undefined lim h→0 2 1 1 (1 4 h)2 1 (1 2 h h2 h 4h 1 (h 4 h) 4h) Right-hand derivative: lim f (1 h→0 lim h→0 lim h) f (1) h (1 h)3 h3 h→0 lim (h 2 h→0 1 Thus f (x) 3h 2 h 3h 1 x 2 3x 2 6(1 h)2 h 8(1 h) h 1) 1 , 2 12x 1, 1 1 extremum value minimum x x 8x, Left-hand derivative: y 5 44. derivative extremum 12 5 x x) x 0 x 2x x 4x(3 2 x 2 ( 1) 1 x [ 4.7, 4.7] by [ 1, 5] maximum undefined local min 1 x x 8, 1 x 1 3 15 4 3 132 Section 4.2 44. continued 1 x 2 Note that 3x 2 12x 12 8 48 2 3 crit. pt. x x 1, and 122 4(3)(8) 2(3) 12 23 . 3 0.845 3 occur at x 0 when x 0 when x 2 6 But 2 1 2 1 and x [ 3, 3] by [ 5, 5] 2 2 3 3 derivative extremum 3.155. value 0 3.155 local max 4 0 1 The function f (x) x 3 x has no critical points. (b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.) 52. (a) By the definition of local maximum value, there is an open interval containing c where f (x) f (c), so f (x) f (c) 0. 1, so the only critical points local max 3.079 (b) Because x → c ,we have (x c) 0, and the sign of the quotient must be negative (or zero). This means the limit is nonpositive. 45. Graph (c), since this is the only graph that has positive slope at c. (c) Because x → c , we have (x c) 0, and the sign of the quotient must be positive (or zero). This means the limit is nonnegative. 46. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. (d) Assuming that f (c) exists, the one-sided limits in (b) and (c) above must exist and be equal. Since one is nonpositive and one is nonnegative, the only possible common value is 0. (e) There will be an open interval containing c where f (x) f (c) 0. The difference quotient for the left-hand derivative will have to be negative (or zero), and the difference quotient for the right-hand derivative will have to be positive (or zero). Taking the limit, the left-hand derivative will be nonpositive, and the right-hand derivative will be nonnegative. Therefore, the only possible value for f (c) is 0. 47. Graph (d), since this is the only graph representing a function that is differentiable at b but not at a. 48. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b. 49. (a) V(x) 160x 52x 2 4x 3 V (x) 160 104x 12x 2 4(x 2)(3x 20) The only critical point in the interval (0, 5) is at x The maximum value of V(x) is 144 at x 2. 2. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x 2. 50. (a) P (x) 2 200x 2 The only critical point in the interval (0, ) is at x 10. The minimum value of P(x) is 40 at x 10. 53. (a) (b) The smallest possible perimeter of the rectangle is 40 units and it occurs at x 10, which makes the rectangle a 10 by 10 square. [ 0.1, 0.6] by [ 1.5, 1.5] 2 2bx c is a quadratic, so it can have 51. (a) f (x) 3ax 0, 1, or 2 zeros, which would be the critical points of f. Examples: f (0) 0 is not a local extreme value because in any open interval containing x 0, there are infinitely many points where f (x) 1 and where f (x) 1. (b) One possible answer, on the interval [0, 1]: f (x) The function f (x) x x 1 and x 1. x) cos 1 , 0 x x 1 x 1 This function has no local extreme value at x 1. Note that it is continuous on [0, 1]. [ 3, 3] by [ 5, 5] 3 1 0, (1 3x has two critical points at s Section 4.2 Mean Value Theorem (pp. 186–194) Quick Review 4.2 1. [ 3, 3] by [ 5, 5] The function f (x) x 0. x3 1 has one critical point at 2x 2 6 2x 2 x2 3 Interval: ( 0 6 3 x 3, 3 3) Section 4.2 2. 3x 2 6 3x 2 x2 x Intervals: ( 0 6 2 Since h (x) is never zero and is undefined only where h(x) is undefined, there are no critical points. Also, the domain ( , 0) (0, ) has no endpoints. Therefore, h(x) has no local extrema. 2 or x 2 2) ( 2, ) 2x 2 0 8 2x 2 4 x2 2x2 The domain is [ 2, 2]. 3. Domain: 8 (b) Since h (x) is never positive, h(x) is not increasing on any interval. (c) Since h (x) 0 on ( , 0) on ( , 0) and on (0, ). 4. f is continuous for all x in the domain, or, in the interval [ 2, 2]. 6. We require x 2 1 0, so the domain is x Since k (x) is never zero and is undefined only where k(x) is undefined, there are no critical points. Also, the domain ( , 0) (0, ) has no endpoints. Therefore, k(x) has no local extrema. 1. 7. f is continuous for all x in the domain, or, for all x 1. 10. 1 1 C 2( 2) C 1. C 5 (c) Since f (x) is never negative, f (x) is not decreasing on any interval. 2x 6. (a) f (x) 0 on f (x) 5 , 2 0 on , 5 , f (x) 2 0 at x 5 , and 2 , we know that f (x) has a local (b) Since f (x) is never positive, f (x) is not increasing on any interval. 25 maximum at x , the local 4 5 25 maximum occurs at the point , . (This is also a 24 (c) Since f (x) is always negative, f (x) is decreasing on ( , ). global maximum.) (b) Since f (x) , 0 on 2x 5 , 2 0 on g (x) 1 , 2 0 on minimum at x 1 2 1 , 2 . , 1 , g (x) 2 0 at x (c) Since y is negative on ( 2, ), y is decreasing on [ 2, ). 1 , and 2 8. (a) y , we know that g(x) has a local (c) Since g (x) 0 on 1 , 2 , g(x) is increasing on , 1 , 2 1 , g(x) is decreasing on 2 2 In the domain [ 2, ), y is never zero and is undefined only at the endpoint x 2. The function y has a local maximum at ( 2, 4). (This is also a global maximum.) (b) Since y is never positive, y is not increasing on any interval. 49 . (This is also a global minimum.) 4 0 on 1 . 2 5 , 2 1 . 2 49 , the local minimum occurs at the 4 (b) Since g (x) , , f (x) is decreasing on 1 Since g (x) point 2x 5 , f (x) is increasing on 2 5 . 2 (c) Since f (x) 1 7. (a) y 0 on 0.5x 0.5e Since f (x) is never zero or undefined, and the domain of f (x) has no endpoints, f (x) has no extrema. 5 5 . Since f 2 2 Since g 0 on (0, ), k(x) is decreasing on (0, ). 2e 2x (b) Since f (x) is always positive, f (x) is increasing on ( , ). Since f (x) 2. (a) g (x) , 0), k(x) is increasing on Since f (x) is never zero or undefined, and the domain of f (x) has no endpoints, f (x) has no extrema. C Section 4.2 Exercises , 0 on ( 5. (a) f (x) (1)2 2(1) 3C 4 1. (a) f (x) (b) Since k (x) ( , 0). (c) Since k (x) 8. f is differentiable for all x in the domain, or, for all x 4 3 (0, ), h(x) is decreasing 2 x3 4. (a) k (x) 5. f is differentiable for all x in the interior of its domain, or, in the interval ( 2, 2). 9. 7 7 C 2 x2 3. (a) h (x) , 133 . 4x 3 20x 4x(x 5)(x 5) 5, x 0, The function has critical points at x and x 5. Since y 0 on ( , 5) and (0, 5) and y 0 on ( 5, 0) and ( 5, ), the points at x 5 are local minima and the point at x 0 is a local maximum. Thus, the function has a local maximum at (0, 9) and local minima at ( 5, 16) and ( 5, 16). (These are also global minima.) (b) Since y 0 on ( 5, 0) and ( 5, ), y is increasing on [ 5, 0] and [ 5, ). (c) Since y 0 on ( , decreasing on ( , 5) and (0, 5), y is 5] and [0, 5]. 134 Section 4.2 9. (b) Since h (x) 0 on ( , 2) and (2, ), h(x) is increasing on ( , 2] and [2, ). (c) Since h (x) [ 2, 2]. 12. [ 4.7, 4.7] by [ 3.1, 3.1] (a) f (x) 1 x 2 3x 4 x 8 4 ( 1) 4 x 2 x [ 4.7, 4.7] by [ 3.1, 3.1] 8 and 3 The local extrema occur at the critical point x at the endpoint x (a) k (x) 4. There is a local (and absolute) and a local minimum at (4, 0). , 0 on , 0 on 4)(1) x(2x) (x 2 4)2 x2 (x 2 4 4)2 (b) Since k (x) is never positive, k(x) is not increasing on any interval. 8 , f (x) is increasing on 3 (c) Since k (x) is negative wherever it is defined, k(x) is decreasing on each interval of its domain: on ( , 2), ( 2, 2), and (2, ). 8 . 3 (c) Since f (x) (x 2 Since k (x) is never zero and is undefined only where k(x) is undefined, there are no critical points. Since there are no critical points and the domain includes no endpoints, k(x) has no local extrema. 8 16 maximum at , or approximately (2.67, 3.08), 33 3 (b) Since f (x) 0 on ( 2, 2), h(x) is decreasing on 8 , 4 , f (x) is decreasing on 3 13. 8 ,4 . 3 10. [ 4, 4] by [ 6, 6] (a) f (x) 3x 2 2 2 sin x Note that 3x 2 2 2 for x 1.2 and 2 sin x 2 for all x, so f (x) 0 for x 1.2. Therefore, all critical points occur in the interval ( 1.2, 1.2), as suggested by the graph. Using grapher techniques, there is a local maximum at approximately ( 1.126, 0.036), and a local minimum at approximately (0.559, 2.639). [ 5, 5] by [ 15, 15] (a) g (x) 1 2/3 x (x 3 x 1/3(1) 8) 4x 8 3x 2/3 The local extrema can occur at the critical points x 2 and x occurs at x 0, but the graph shows that no extrema minimum at ( 2, ( 2, (b) f (x) is increasing on the intervals ( , 1.126] and [0.559, ), where the interval endpoints are approximate. 0. There is a local (and absolute) 6 3 2) or approximately 7.56). (b) Since g (x) 0 on the intervals ( 2, 0) and (0, ), and g(x) is continuous at x 0, g(x) is increasing on [ 2, ). (c) Since g (x) 0 on the interval ( decreasing on ( , 2]. , (c) f (x) is decreasing on the interval [ 1.126, 0.559], where the interval endpoints are approximate. 14. 2), g(x) is 11. [ 6, 6] by [ 12, 12] (a) g (x) 2 sin x Since 1 g (x) 3 for all x, there are no critical points. Since there are no critical points and the domain has no endpoints, there are no local extrema. [ 5, 5] by [ 0.4, 0.4] (a) h (x) (x (x 2 2)(x (x 2 4)2 4)( 1) ( x)(2x) (x 2 4)2 2) x2 (x 2 (b) Since g (x) 4 4)2 and a local (and absolute) minimum at 2, , ). (c) Since g (x) is never negative, g(x) is not decreasing on any interval. The local extrema occur at the critical points, x There is a local (and absolute) maximum at 0 for all x, g(x) is increasing on ( 2. 1 2, 4 1 . 4 135 Section 4.2 15. (a) f is continuous on [0, 1] and differentiable on (0, 1). (b) f (1) f (0) 10 2 ( 1) 1 2c c The equation is y 1 2 2 (3, f (3)) 1 f (c) 2c 1 2 3 2 1/3 c 3 1/3 c c c (b) f (c) 1 c2 c2 1 1 c 1 c 4 2 f (2) 2 ln 1 2 ln 3 2 2 c f (c) 1 2 31 , . Its equation is y 2 2 1 1 x , or y f (c) 1 2/3 c 3 1 2/3 c 3 2.820 c (0.5, 2.5) 2.5. 0 1 1 f (1) 2 The tangent line has slope 0 and passes through (1, 2), so its equation is y 2. 2/3 c 1 and passes through 2 1 x 2 0.707x 3 2 1 or 2 0.354. 2 2 1 2/3 x , f is not differentiable at x 3 [ 1, 1] by [ 1, 1] (c) (b) The slope of the secant line is 0, so we need to find c such that f (c) 0. c c 1 2 0.771 19. (a) The secant line passes through (0.5, f (0.5)) and (2, f (2)) (2, 2.5), so its equation is y 1 3 2 2 2 ln 3 1 f (b) 2 c . 1 2 3 2 4 1 f (4) 4 ln 3 1 2 18. (a) f is continuous on [2, 4] and differentiable on (2, 4). f (c) 1 2 f (c) y 2 c 0.707. 2 1 21. (a) Since f (x) 1 0 0.707x The tangent line has slope 4 2 1) 1 1 c 2 1 , or y c 2 c (x 2 c 2 2 1 (b) x 1 2 1 1 2c 2 1 . 2 2 f ( 1) ( 1) 2 c 1 2 2 (b) We need to find c such that f (c) 17. (a) f is continuous on [ 1, 1] and differentiable on ( 1, 1). f (1) 1 (1, 0) and 2), so its slope is 2 f (1) f (0) 10 10 1 3 2 33 2 8 27 f (c) (3, 0 1 or y 16. (a) f is continuous on [0, 1] and differentiable on (0, 1). (b) 20. (a) The secant line passes through (1, f (1)) f (1) f ( 1) 1 ( 1) 1 ( 1) 2 1 3 3 3/2 0.192 0. 136 Section 4.2 22. (a) Since f (x) 1, 0 3 1, x x (b) 1 1 , f is not differentiable at x 1. (If f were differentiable at x 1, it would violate the Intermediate Value Theorem for Derivatives.) [ (b) , by [ 1, 2] sin x, cos x, (c) Note that f (x) f( ) We require f (c) For c [0, 3] by [ 1, 3] f (3) 3 (c) We require f (c) f (x) f (0) 0 2 1 c 1 , but 3 3 1, x 1 x 0 0 11 25. f (x) x2 2 26. f (x) 2x x2 x cos x ex ln (x f ( 1) ( 1) 0 0 2 1 2 1 for all x where f (x) is defined. Therefore, 1 C there is no such value of c. f (x) 24. (a) We test for differentiability at x 0, using the limits 32. given in Section 3.5. Left-hand derivative: f (0 h) h f (0) lim h→0 cos h h 1 0 Right-hand derivative: lim h→0 f (0 h) h f (0) lim h→0 lim h→0 (1 sin h h sin h) h 1 f (x) f (1) 11/4 C 1C C f (x) 33. ln ( 1 1 Since the left- and right-hand derivatives are not equal, f is not differentiable at x 0. 34. C 1 C 1) 1 x f (2) 0, but C C f (x) 31. [ 1, 1] by [ 1, 2] lim , so c C x 30. f (x) h→0 1 cos 1 C 3 29. f (x) f (x) sin c 2.818. For 0 1 C 28. f (x) f (1) 1 ( 1) f (x) f (0) 0C C f (x) 1 2 1 x C, x 0 1 ,x 2 0 x 1/4 2 2 2 3 1/4 x 1 2 0.324, and 1.247. 27. f (x) (b) . 1 . , so 0.324 or sin 2.818, , f is not differentiable at x 0. (If f were differentiable at x 1, it would violate the Intermediate Value Theorem for Derivatives.) (c) We require f (c) 1 possible values of c are approximately 1 1, ) 0 ) 0, this occurs when occurs when cos c there is no such value of c. f( ( 11 c 1 for all x where f (x) is defined. Therefore, 23. (a) Since f (x) sin x x 0 C 3 f (x) f ( 1) 2) C 0C C f (x) ln (x 3 3 3 3 ln (x 2) C 2) 3 x2 3 3 3 x2 x sin x C x sin x 3 c , this 1.247. The 137 Section 4.2 35. Possible answers: (a) (b) [ 2, 4] by [ 2, 4] [ 1, 4] by [0, 3.5] 42. The runner’s average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0 mph and the runner’s speed is continuous, by the Intermediate Value Theorem, the runner’s speed must have been 11 mph at least twice. 43. (a) Since v (t) 1.6, v(t) 1.6t C. But v(0) 0, so C 0 and v(t) 1.6t. Therefore, v(30) 1.6(30) 48. The rock will be going 48 m/sec. (c) (b) Let s(t) represent position. Since s (t) v(t) 1.6t, s(t) 0.8t 2 D. But s(0) 0, so D 0 and s(t) 0.8t 2. Therefore, s(30) 0.8(30)2 720. The rock travels 720 meters in the 30 seconds it takes to hit bottom, so the bottom of the crevasse is 720 meters below the point of release. [ 1, 4] by [0, 3.5] 36. Possible answers: (a) (b) [ 1, 5] by [ 2, 4] (c) [ 1, 5] by [ 1, 8] (d) [ 1, 5] by [ 1, 8] [ 1, 5] by [ 1, 8] (c) The velocity is now given by v(t) 1.6t C, where v(0) 4. (Note that the sign of the initial velocity is the same as the sign used for the acceleration, since both act in a downward direction.) Therefore, v(t) 1.6t 4, and s(t) 0.8t 2 4t D, where s(0) 0 and so D 0. Using s(t) 0.8t 2 4t and the known crevasse depth of 720 meters, we solve s(t) 720 to obtain the positive solution t 27.604, and so v(t) v(27.604) 1.6(27.604) 4 48.166. The rock will hit bottom after about 27.604 seconds, and it will be going about 48.166 m/sec. 37. One possible answer: 44. (a) We assume the diving board is located at s water at s 0 and the 10, so that downward velocities are positive. The acceleration due to gravity is 9.8 m/sec2, [ 3, 3] by [ 15, 15] 38. One possible answer: so v (t) 9.8 and v(t) have v(t) [ 3, 3] by [ 70, 70] 39. Because the trucker’s average speed was 79.5 mph, and by the Mean Value Theorem, the trucker must have been going that speed at least once during the trip. 40. Let f (t) denote the temperature indicated after t seconds. We assume that f (t) is defined and continuous for 0 t 20. The average rate of change is 10.6 F/sec. Therefore, by the Mean Value Theorem, f (c) 10.6 F/sec for some value of c in [0, 20]. Since the temperature was constant before t 0, we also know that f (0) 0 F/min. But f is continuous, so by the Intermediate Value Theorem, the rate of change f (t) must have been 10.1 F/sec at some moment during the interval. 41. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. C. Since v(0) 0, we 9.8t. Then the position is given by s(t) where s (t) s(0) 9.8t v(t) 4.9t 2 9.8t, so s(t) 0, we have s(t) 4.9t 2. Solving s(t) D. Since 10 gives 100 10 t2 , so the positive solution is t . The 49 7 10 10 velocity at this time is v 9.8 14 m/sec. 7 7 10 4.9 (b) Again v(t) 9.8t C, but this time v(0) v(t) 9.8t s(t) 4.9t 2 2t s(t) 4.9t 2 2t. Solving s(t) solution t 2. Then s (t) 2 9.8t 2, so D. Since s(0) 10 2 9.8 2 and so 0, we have 10 gives the positive 1.647 sec. The velocity at this time is v 2 10 2 9.8 9.8 about 14.142 m/sec. 2 10 2 9.8 2 10 2 m/sec or 138 Section 4.3 45. Because the function is not continuous on [0, 1]. The function does not satisfy the hypotheses of the Mean Value Theorem, and so it need not satisfy the conclusion of the Mean Value Theorem. 52. (a) Toward: 0 t 2 and 5 away: 2 t 5 (c) Regression equation: y 0.0820x 3 0.9163x 2 47. f (x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f (x) is zero twice between a and b. Then by the Mean Value Theorem, f (x) would have to be zero at least once between the two zeros of f (x), but this can’t be true since we are given that f (x) 0 on this interval. Therefore, f (x) is zero once and only once between a and b. ln (x 1 1 x 1 53. , which is never zero on [0, 3]. Now f (0) 0, so x 0 is one solution of the equation. If there were a second solution, f (x) would be zero twice in [0, 3], and by the Mean Value Theorem, f (x) would have to be zero somewhere between the two zeros of f (x). But this can’t happen, since f (x) is never zero on [0, 3]. Therefore, f (x) 0 has exactly one solution in the interval [0, 3]. 50. Consider the function k(x) f (x) g(x). k(x) is continuous and differentiable on [a, b], and since k(a) f (a) g(a) 0 and k(b) f (b) g(b) 0, by the Mean Value Theorem, there must be a point c in (a, b) where k (c) 0. But since k (c) f (c) g (c), this means that f (c) g (c), and c is a point where the graphs of f and g have parallel or identical tangent lines. 3.3779 (d) Using the unrounded values from the regression equation, we obtain f (t) 0.2459t 2 1.8324t 2.5126. According to the regression equation, Priya is moving toward the motion detector when f (t) 0 (0 t 1.81 and 5.64 t 8), and away from the detector when f (t) 0 (1.81 t 5.64). 1). Then f (x) is continuous and differentiable everywhere on [0, 3]. f (x) 2.5126x [ 0.5, 8.5] by [ 0.5, 5] 48. Let f (x) x 4 3x 1. Then f (x) is continuous and differentiable everywhere. f (x) 4x 3 3, which is never zero between x 2 and x 1. Since f ( 2) 11 and f ( 1) 1, exercise 47 applies, and f (x) has exactly one zero between x 2 and x 1. x 8; (b) A local extremum in this problem is a time/place where Priya changes the direction of her motion. 46. Because the Mean Value Theorem applies to the function y sin x on any interval, and y cos x is the derivative of sin x. So, between any two zeros of sin x, its derivative, cos x, must be zero at least once. 49. Let f (x) t f(b) b 1 b Thus, c f (b) b f (c) a 1 , so c2 f (c) 54. 1 a b f(a) a 1 ab 1 c2 1 and c 2 ab ab. ab. f (a) a b2 b 2c, so 2c a2 a b b a a and c a b 2 . 55. By the Mean Value Theorem, sin b sin a (cos c)(b a) for some c between a and b. Taking the absolute value of both sides and using cos c 1 gives the result. 56. Apply the Mean Value Theorem to f on [a, b]. Since f (b) f (a), f(b) b f(a) is negative, and hence f (x) a must be negative at some point between a and b. 57. Let f (x) be a monotonic function defined on an interval D. For any two values in D, we may let x1 be the smaller value and let x2 be the larger value, so x1 x2. Then either f (x1) f (x2) (if f is increasing), or f (x1) f (x2) (if f is decreasing), which means f (x1) f (x2). Therefore, f is one-to-one. [ 1, 1] by [ 2, 2] 51. (a) Increasing: [ 2, 1.3] and [1.3, 2]; decreasing: [ 1.3, 1.3]; local max: x 1.3 local min: x 1.3 (b) Regression equation: y 3x 2 5 s Section 4.3 Connecting f and f with the Graph of f (pp. 194–206) Exploration 1 [ 2.5, 2.5] by [ 8, 10] (c) Since f (x) 3x 2 5, we have f (x) x 3 5x But f (0) 0, so C 0. Then f (x) x 3 5x. C. Finding f from f 1. Any function f (x) x 4 4x 3 C where C is a real number. For example, let C 0, 1, 2. Their graphs are all vertical shifts of each other. 2. Their behavior is the same as the behavior of the function f of Example 8. Section 4.3 Exploration 2 Finding f from f and f Section 4.3 Exercises 1. f has an absolute maximum at x 0 and an absolute minimum of 1 at x 4. We are not given enough information to determine f (0). 2. f has a point of inflection at x 1. (a) Zero: x 1; positive: ( , 1) and (1, ); negative: ( 1, 1) 2. (b) Zero: x 0; positive: (0, ); negative: ( , 0) 3. 2. (a) Zero: x 0, 1.25; positive: ( 1.25, 0) and (1.25, ); negative: ( , 1.25) and (0, 1.25) (b) Zero: x 0.7; positive: ( , 0.7) and (0.7, ); negative: ( 0.7, 0.7) [ 3, 5] by [ 5, 20] Quick Review 4.3 1. (x x2 3)(x Intervals Sign of (x 3)(x 9 3) 0 0 3. (a) ( x 3 3 x 33 x x(x x3 2)(x 3) (b) ( 4x 2) 2 2 x 0 0 x 2 xe x f :x 2 x 5. (a) [0, 1], [3, 4], and [5.5, 6] (b) [1, 3] and [4, 5.5] (c) Local maximum: x 3; local minimum: x 0 7. y 0, since f (x) 6; 5.5 (b) Nowhere 3 2/5 x 5 (x 2)(1) (x)(1) (x 2)2 (x 2 2)2 2x 1 x Intervals 6. f: all reals f :x 2; 2 (a) [0, 3] 2 2, since f (x) 2] and [2, ) 6. If f is continuous on the interval [0, 3]: ex 4. f: all reals 5. f: x 2; (c) Local maxima: x 1, x 4 (if f is continuous at x 4), and x local minima: x 0, x 3, and x (2, ) 3. f: all reals f : all reals, since f (x) 0, since f (x) , (c) Local maximum: x local minimum: x 0 0 Sign of x(x 2)(x 2) Solution set: ( 2, 0) f :x 2 and x 0 4. (a) [ 2, 2] x Intervals 2] and [0, 2] (c) Local maxima: x local minimum: x Solution set: ( 3, 3) 2. , (b) [ 2, 0] and [2, ) 1 2 x 1 2 Sign of y 2 3/5 x 5 Behavior of y Decreasing Increasing 7. Left end behavior model: 0 Right end behavior model: x 2e x Horizontal asymptote: y 0 2 8. Left end behavior model: x e Right end behavior model: 0 Horizontal asymptote: y 0 y 2 (always positive: concave up) Graphical support: x 9. Left end behavior model: 0 Right end behavior model: 200 Horizontal asymptotes: y 0, y 10. Left end behavior model: 0 Right end behavior model: 375 Horizontal asymptotes: y 0, y [ 4, 4] by [ 3, 3] 200 (a) 375 1 , 2 1 2 (b) , (c) ( ,) (d) Nowhere (e) Local (and absolute) minimum at (f) None 1 , 2 5 4 139 140 Section 4.3 6x 2 8. y 12x 6x(x Intervals x 2) 0 0 xe 1/x( x 10. y x 2 2 2 e 1/x ) 1 x e 1/x 1 x Intervals Sign of y x 0 0 x 1 1 x Sign of y Behavior of y Decreasing Increasing Decreasing Behavior of y Increasing Decreasing Increasing y 12x 12 12(x Intervals x 1) 1 x e 1/x(x y 1 ) Behavior of y 0 x Graphical support: [ 8, 8] by [ 6, 6] (a) [0, 2] (b) ( (c) ( (a) ( , 0] and [2, ) , 1) , 0) and [1, ) (b) (0, 1] (c) (0, ) (d) ( (f) None (e) Local maximum: (2, 5); local minimum: (0, 3) 11. y 1 x 28 (f) At (1, 1) 8x 8x(x x 1)(x 1 Intervals 1) 1 x x2 00 x 1 1 x ( 2x) ( 8 8( 3x 1)( 3x 1 x Intervals Increasing x 2 2 x 1 3 x Decreasing ( x2)( 4x) 8 Increasing 3 1 3 x 3 Sign of y ( 2x 3 24x (8 x 2)3/2 Concave up Concave down Concave up Graphical support: x 8 8 x2 0 x Decreasing ( 2x) x2)2 8 2x(x 2 12) (8 x 2)3/2 Intervals Behavior of y 1 2x2) (8 2 y 1 22 Sign of y Decreasing Increasing 1) x2 8 8 Behavior of y Behavior of y Decreasing 2x2 8 x 2)(1) 8 Sign of y 24x 2 , 0) (e) Local minimum: (1, e) (d) (1, ) y 0 Behavior of y Concave down Concave up [ 4, 4] by [ 6, 6] Intervals x Sign of y Concave up Concave down Graphical support: 8x 3 e 1/x x3 1 1/x e ( x 2) x 1 Intervals Sign of y 9. y 2 8 x 0 8 Sign of y Behavior of y Concave up Concave down Graphical support: [ 4, 4] by [ 3, 3] (a) [ 1, 0] and [1, ) (b) ( , (c) , [ 3.02, 3.02] by [ 6.5, 6.5] 1] and [0, 1] (a) [ 2, 2] 1 and 3 (d) 1 , 3 1 , 1 , 3 1 9 1) and (1, 1) 8, 0) (d) (0, 3 8, (c) ( 1 (e) Local maximum: (0, 1); local (and absolute) minima: ( 1, (f) (b) [ 3 2] and [2, 8] 8) 8, 0) and (2, 4); (e) Local maxima: ( local minima: ( 2, 4) and ( 8, 0) Note that the local extrema at x 2 are also absolute extrema. (f) (0, 0) Section 4.3 2x, 2x, 12. y x x 141 0 0 Intervals x 0 x 0 Sign of y Behavior of y Increasing Increasing 2, y x x 2, 0 0 Intervals x 0 x 0 Sign of y Behavior of y Concave down Concave up Graphical support: [ 4, 4] by [ 3, 6] (a) ( , 0) and [0, ) (b) None (c) (0, ) (d) ( , 0) (e) Local minimum: (0, 1) (f) Note that (0, 1) is not an inflection point because the graph has no tangent line at this point. There are no inflection points. 12x 2 13. y 42x 36 Intervals 2)(2x 2 x 6(x 2 3) 3 2 x 3 2 x Sign of y Behavior of y y 24x 42 Increasing 6(4x Increasing 7) 7 4 x Intervals Decreasing 7 4 x Sign of y Behavior of y Concave down Concave up Graphical support: [ 4, 4] by [ 80, 20] (a) ( , (b) 2, (c) , 3 , 2 3 2 7 , 4 (d) 2] and 7 4 (e) Local maximum: ( 2, (f) 7 , 4 321 8 40); local minimum: 3 , 2 161 4 142 Section 4.3 14. y 4x 3 12x 2 4 Using grapher techniques, the zeros of y are x Intervals x 0.53 0.53 x 0.53, x 0.65 0.65 x 0.65, and x 2.88 2.88. 2.88 x Sign of y Behavior of y y 12x 2 Increasing 24x 12x(x Intervals x Decreasing Increasing Decreasing 2) 0 0 x 2 2 x Sign of y Behavior of y Concave down Concave up Concave down Graphical support: [ 2, 4] by [ 20, 20] (a) (b) (c) (d) (e) ( , 0.53] and [0.65, 2.88] [ 0.53, 0.65] and [2.88, ) (0, 2) ( , 0) and (2, ) Local maxima: ( 0.53, 2.45) and (2.88, 16.23);local minimum: (0.65, 0.68) Note that the local maximum at x 2.88 is also an absolute maximum. (f) (0,1) and (2, 9) 2 4/5 x 5 15. y Intervals x 0 0 x Sign of y Behavior of y Increasing Increasing 8 x 9/5 25 y Intervals x 0 0 x Sign of y Behavior of y Concave up Concave down Graphical support: [ 6, 6] by [ 1.5, 7.5] (a) ( ,) (b) None (c) ( , 0) (d) (0, ) (e) None (f) (0, 3) 16. y 1 2/3 x 3 Intervals x 0 0 x Sign of y Behavior of y Decreasing Decreasing Section 4.3 143 2 5/3 x 9 y Intervals x 0 0 x Sign of y Behavior of y Concave down Concave up Graphical support: [ 8, 8] by [0, 10] (a) None (b) ( ,) (c) (0, ) (d) ( , 0) (e) None (f) (0, 5) 17. This problem can be solved using either graphical or analytic methods. For a graphical solution, use NDER to obtain the graphs shown. y y y [ 10, 20] by [0, 5] [ 10, 20] by [0, 0.3] [ 10, 20] by [ 0.02, 0.02] An analytic solution follows. y (e x 3e 0.8x)(5e x) 5e x(e x (e x 3e 0.8x)2 5e 2x (e x 2.4e 0.8x) 15e 1.8x 5e 2x 12e 1.8x (e x 3e 0.8x)2 3e 1.8x 3e 0.8x)2 Since y 0 for all x, y is increasing for all x. (e x 3e 0.8x)2(5.4e 1.8x) (e x y 3e 0.8x)(5.4e 1.8x) (6e 1.8x)(e x (e x 3e 0.8x)3 ( 0.6e x (e x 0.6(3 (e x (3e 1.8x)(2)(e x (e x 3e 0.8x)4 e 0.2x)e 2.6x 3e 0.8x)3 e 0.2x 0: 3 0 ln 3 x Intervals 2.4e 0.8x) 1.8e 0.8x)e 1.8x 3e 0.8x)3 0.2x Solve y 3e 0.8x)(e x x 5 ln 3 5 ln 3 5 ln 3 x Sign of y Behavior of y Concave up Concave down 2.4e 0.8x) 144 Section 4.3 17. continued (a) ( ,) (b) None (c) ( , 5 ln 3) ( (d) (5 ln 3, ) , 5.49) (5.49, ) (e) None (f) 5 ln 3, 5 2 (5.49, 2.50) 18. This problem can be solved using either graphical or analytic methods. For a graphical solution, use NDER to obtain the graphs shown. y y y [ 6, 10] by [0, 4] [ 6, 10] by [0, 0.6] [ 6, 10] by [ 0.15, 0.15] An analytic solution follows. x 8e y 2e x (2 y 1.5x 5e 8 5e 2 0.5x 0.5x 5e 0.5x )(0) (8)( 2.5e (2 5e 0.5x)2 ) 20e 0.5x (2 5e 0.5x)2 Since y 0 for all x, y is increasing for all x. (2 y 0.5x 2 5e 0.5x ) ( 10e ) (2 0.5x (2 5e 10e 0.5x (5e 5e (2 (20e 5e )(10e 0.5x) (40e (2 5e 0.5x)3 0.5x 5e )( 2.5e 0.5x 2) ) 5e 0.5x 2 0 0.5x 2 5 0.5x 0: 0.5x )(2)(2 ) 0.5x 4 0.5x 3 e Solve y 0.5x ln 2 5 x x Intervals 2 ln 2 ln 2 5 2 ln 5 2 2 ln 5 2 x Sign of y Behavior of y (a) ( Concave up ,) (b) None (c) , 2 ln 5 2 ( , 1.83) 5 2 (1.83, ) 5 2 (1.83, 2) (d) 2 ln , (e) None (f) 2 ln , 2 Concave down 5 2 ) 0.5x )( 2.5e 0.5x ) Section 4.3 2, 19. y 2x, x x 1 1 Intervals x 1 1 x Sign of y Behavior of y 0, y 2, Increasing x x Decreasing 1 1 Intervals x Sign of y 1 1 x 0 Behavior of y Linear Concave down Graphical support: [ 2, 3] by [–5, 3] (a) ( , 1) (b) [1, ) (c) None (d) (1, ) (e) None (f) None 20. y ex y ex Since y and y are both positive on the entire domain, y is increasing and concave up on the entire domain. Graphical support: [0, 2 ] by [0, 20] (a) [0, 2 ] (b) None (c) (0, 2 ) (d) None (e) Local (and absolute) maximum: (2 , e2 ); local (and absolute) minimum: (0, 1) (f) None 21. y 2 xe 1/x ( 2x 2 e 1/x (1 Intervals 3 2x x 2 (e 1/x )(1) ) 2 ) e 1/x 2 2 x2 2 x2 2 x 00 x 2 2 x Sign of y Behavior of y Increasing Decreasing Decreasing Increasing 145 146 Section 4.3 21. continued 2 (e 1/x )(4x y (e 1/x 2 2e 3 (1 3 )(2x 1/x 2 x ) 2 2 2x 2 )(e 1/x )( 2x 3 ) 5 4x ) 2 x5 Intervals x 0 0 x Sign of y Behavior of y Concave down Concave up Graphical support: [ 12, 12] by [ 9, 9] (a) ( , 2] and [ 2, ) 2, 0) and (0, (b) [ 2] (c) (0, ) (d) ( , 0) 2, (e) Local maximum: ( 2e) ( 1.41, 2.33); local minimum: ( 2, 2e) (1.41, 2.33) (f) None 22. This problem can be solved using either graphical or analytic methods. For a graphical solution, use NDER to obtain the graphs shown. y y y [ 4.7, 4.7] by [ 3, 11] [ 4.7, 4.7] by [ 10, 10] [ 4.7, 4.7] by [ 10, 10] An analytic solution follows. 1 x2 y 2 ( 2x) 9 x2 3x 3 18x 9 x 3x(x 2 2 Intervals x 2(2x) 9 9 3 6) x 2 6 x 6 x 00 x 6 6 x 3 Sign of y Behavior of y Increasing x 2)( 9x 2 (9 Decreasing ( 3x 3 18) 1 18x) 29 y ( (9 x 2)( 9x 2 6x 4 81x 2 (9 x 2)3/2 162 x 2)2 ( 3x 3 18) (9 9 2 3/2 x) Increasing 18x)(x) x2 ( 2x) Decreasing Section 4.3 Find the zeros of y : 3(2x 4 27x 2 54) (9 x 2)3/2 2x 4 27x 2 0 54 0 x2 27 27 2 4(2)(54) 2(2) 27 3 4 27 x 27 Note that we do not use x Intervals 3 x 1.56 3 4 33 1.56 33 3 33 4 1.56 3.33, because these values are outside of the domain. x 1.56 1.56 x 3 Sign of y Behavior of y Concave down 6] and [0, (a) [ 3, and [0, 2.45] 6, 0] and [ (b) [ Concave up 6] or, 6, 3] or, [ 3, 2.45] [ 2.45, 0] and [2.45, 3] (c) Approximately ( 1.56, 1.56) (d) Approximately ( 3, 1.56) and (1.56, 3) 6, 6 3) ( 2.45, 10.39); (e) Local maxima: ( local minima: (0, 0) and ( 3, 0) (f) ( 1.56, 6.25) 1 23. y x2 1 Since y 0 for all x, y is always increasing. d (1 dx y x 2) 1 Intervals (1 x 0 x 2) 2(2x) 0 (1 x Sign of y Behavior of y Concave up Concave down Graphical support: [ 4, 4] by [ 2, 2] (a) ( ,) (b) None (c) ( , 0) (d) (0, ) (e) None (f) (0, 0) Concave down 2x x 2 )2 147 148 Section 4.3 x 3/4(5 24. y 5x 3/4 x) 15 1/4 x 4 y 7 3/4 x 4 0 Intervals x 7/4 (a) [1, ) (b) ( 15 7 x 15 7 , 2) and (0, ) (d) ( 2, 0) x (e) Local minimum: (1, 3 Sign of y (f) ( 2, 6 2) Behavior of y Increasing 15 5/4 x 16 3(7x 5) 16x 5/4 y , 1] (c) ( 15 7x 4x 1/4 Decreasing x 1/4(x 26. y 21 1/4 x 16 0, y is always increasing on its 0. 5 x 3/4 16 y Graphical support: 5x 3 4x 3/4 0 for all x domain x 0, the graph of y is concave down 3x 1/4 3 3/4 x 4 Since y Since y 0 for all x for x 0. x 5/4 3) 5 1/4 x 4 y 3) ( 2, 7.56) and (0, 0) 5x 9 16x 7/4 9 x 7/4 16 0 Intervals x 9 5 9 5 x Sign of y Behavior of y [0, 8] by [ 6, 6] (a) Concave up Graphical support: 15 0, 7 (b) Concave down 15 , 7 (c) None [0, 6] by [0, 12] (d) (0, ) (a) [0, ) (e) Local (and absolute) maximum: (b) None 15 15 3/4 , 7 7 20 7 15 , 5.06 ; 7 (c) local minimum: (0, 0) (d) 0, (f) None 25. y x 1/3(x y 4 1/3 x 3 9 , 5 9 5 (e) Local (and absolute) minimum: (0, 0) x 4/3 4) 4 2/3 x 3 Intervals x 4x 1/3 (f) 4x 4 3x 2/3 0 0 x 1 1 9 24 , 55 4 9 5 9 , 5.56 5 27. We use a combination of analytic and grapher techniques to x Sign of y Behavior of y Decreasing Decreasing Increasing y solve this problem. Depending on the viewing window chosen, graphs obtained using NDER may exhibit strange 4 2/3 x 9 8 5/3 x 9 Intervals x behavior near x 4x 8 9x 5/3 2 NDER (y, 2) 2 x 0 0 x x 2 because, for example, 1,000,000 while y is actually undefined at 2. The graph of y Sign of y Behavior of y Concave up Concave down Concave up Graphical support: [ 4,7, 4.7] by [ 5, 15] [ 4, 8] by [ 6, 8] x3 2x 2 x x2 1 is shown below. Section 4.3 2)(3x 2 (x y 2x 3 4x 8x 2 8x (x 2)2 (x 3 2)2 1) (x 2x 2 x 1)(1) 1 The graph of y is shown below. [ 4,7, 4.7] by [ 10, 10] The zeros of y are x Intervals x 0.15, x 0.15 1.40, and x 0.15 x 2.45. 1.40 1.40 x 22 x 2.45 2.45 x Sign of y Behavior of y Decreasing (x 2)2(6x 2 (x y 2)(6x 2 16x 16x 2x 3 12x 2 24x (x 2)3 2(x 1)(x 2 5x (x 2)3 8) Increasing (2x 3 8x 2 (x 2)4 8) 2(2x 3 (x 2)3 Decreasing 8x 8x 8x 2 1)(2)(x Decreasing Increasing 2) 1) 14 7) The graph of y is shown below. [ 4,7, 4.7] by [ 10, 10] Note that the discriminant of x 2 Intervals x 1 7 is ( 5)2 5x 1 x 2 4(1)(7) 2 3, so the only solution of y x Sign of y Behavior of y Concave up Concave down Concave up (a) Approximately [0.15, 1.40] and [2.45, ) (b) Approximately ( (c) ( , 0.15], [1.40, 2), and (2, 2.45] , 1) and (2, ) (d) (1, 2) (e) Local maximum: (f) (1, 1) (1.40, 1.29);local minima: (0.15, 0.48) and (2.45, 9.22) 0 is x 1. 149 150 Section 4.3 (x 2 28. y x2 1)(1) x(2x) (x 2 1)2 Intervals (x x 1 1)2 2 1 1 x 1 1 x Sign of y Behavior of y Decreasing Increasing (x 2 1)2( 2x) (x 2 1)( 2x) 4x( x 2 (x 2 1)3 2x 3 (x 2 6x 1)3 y ( x 2 1)(2)(x 2 (x 2 1)4 Decreasing 1)(2x) 1) 2x(x 2 3) (x 2 1)3 Intervals 3 x 3 x 0 0 x 3 3 x Sign of y Behavior of y Concave down Concave up Concave down Concave up Graphical support: [ 4.7, 4.7] by [ 0.7, 0.7] (a) [ 1, 1] (b) ( , 1] and [1, ) 3, 0) and ( (c) ( (d) ( , 3, ) 3) and (0, 3) 1 ; 2 1 1, 2 (e) Local maximum: 1, local minimum: (f) (0, 0), 29. y 3, 1)2(x (x Intervals 3 4 , and 3 3, 4 2) x 1 1 x 2 2 x Sign of y Behavior of y Decreasing Decreasing Increasing y (x (x (x Intervals 1)2(1) (x 2)(2)(x 1) 1)[(x 1) 2(x 2)] 1)(3x 5) x 1 1 x 5 3 5 3 x Sign of y Behavior of y Concave up Concave down Concave up (a) There are no local maxima. (b) There is a local (and absolute) minimum at x (c) There are points of inflection at x 1 and at x 2. 5 . 3 Section 4.3 30. y (x 1)2(x 2)(x Intervals x 4) 1 1 x 2 2 x 4 4 x Sign of y Behavior of y d [(x dx y Increasing 1)2(x 2 Increasing 6x 8)] (x 2 1) (2x 6) (x 1)[(x 1)(2x 6) (x 1)(4x 2 20x 22) 1)(2x 2 2(x (x 10x 2 6x 10 2 10 8)(2)(x 2(x 2 1) 6x 8)] 11) Note that the zeros of y are x x Decreasing Increasing 4(2)(11) 1 and 10 12 4 4 5 3 2 1.63 or 3.37. The zeros of y can also be found graphically, as shown. [ 3, 7] by [ 8, 4] Intervals x 1 1 x 1.63 1.63 x 3.37 3.37 x Sign of y Behavior of y Concave down (a) Local maximum at x 4 (c) Points of inflection at x 31. 1, at x y y = f ′(x ) y = f (x ) P x y = f ′′(x) 32. Concave down 2 (b) Local minimum at x Concave up y P y = f (x ) 0 y = f ′(x ) x y = f ′′(x) 1.63, and at x 3.37. Concave up 151 152 Section 4.3 33. (a) Absolute maximum at (1, 2); absolute minimum at (3, 2) 38. (a) v(t) (b) None s (t) 2 (b) a(t) v (t) 2 (c) It begins at position 6 and moves in the negative direction thereafter. (c) One possible answer: 39. (a) v(t) y = f (x ) 2 3 3t 2 v (t) 6t 3 (c) It begins at position 3 moving in a negative direction. It moves to position 1 when t 1, and then changes direction, moving in a positive direction thereafter. 1 1 s (t) (b) a(t) y 2 2t x 40. (a) v(t) –2 34. (a) Absolute maximum at (0, 2); absolute minimum at (2, 1) and ( 2, 1) (b) At (1, 0) and ( 1, 0) 6t 6t 2 (b) a(t) –1 s (t) v (t) 6 12t (c) It begins at position 0. It starts moving in the positive direction until it reaches position 1 when t 1, and then it changes direction. It moves in the negative direction thereafter. 41. (a) The velocity is zero when the tangent line is horizontal, at approximately t 2.2, t 6, and t 9.8. (c) One possible answer: (b) The acceleration is zero at the inflection points, approximately t 4, t 8, and t 11. y 2 42. (a) The velocity is zero when the tangent line is horizontal, at approximately t 0.2, t 4, and t 12. y = f (x ) –3 3 x –1 –2 (d) Since f is even, we know f (3) f ( 3). By the continuity of f, since f (x) 0 when 2 x 3, we know that f (3) 0, and since f (2) 1 and f (x) 0 when 2 x 3, we know that f (3) 1. In summary, we know that f (3) f ( 3), 1 f (3) 0, and 1 f ( 3) 0. (b) The acceleration is zero at the inflection points, approximately t 1.5, t 5.2, t 8, t 11, and t 13. 43. No. f must have a horizontal tangent at that point, but f could be increasing (or decreasing), and there would be no local extremum. For example, if f (x) x 3, f (0) 0 but there is no local extremum at x 0. 44. No. f (x) could still be positive (or negative) on both sides of x c, in which case the concavity of the function would not change at x c. For example, if f (x) x 4, then f (0) 0, but f has no inflection point at x 0. 45. One possible answer: 35. y y 4 5 3 y = f (x ) 2 1 –5 1 2 3 4 5 6 x x –1 36. 5 –5 y 46. One possible answer: 5 4 3 2 1 y 5 2 –1 –2 –3 3 x –5 37. (a) v(t) s (t) 2t (b) a(t) v (t) 2 4 (c) It begins at position 3 moving in a negative direction. It moves to position 1 when t 2, and then changes direction, moving in a positive direction thereafter. 5 –5 x Section 4.4 47. One possible answer: (1 51. (a) f (x) y (–2, 8) 10 ) bx 2 ) bx 5 (2, 0) abce , (e bx a)2 x so the sign of f (x) is the same as the sign of abc. –10 (e bx a)2(ab 2ce bx) (abce bx)2(e bx (e bx a)4 (e bx (b) f (x) a)(ab 2ce bx) (abce bx)(2be bx) (e bx a)3 48. One possible answer: y 8 (6, 7) (4, 4) 4 Since a 2 (2, 1) –2 2 4 a)(be bx) ab 2ce bx(e bx a) (e bx a)3 6 6 8 0, this changes sign when x the e bx x ln a due to b a factor in the numerator, and f (x) has a point of inflection at that location. –2 52. (a) f (x) 4ax 3 3bx 2 2cx d f (x) 12ax 2 6bx 2c Since f (x) is quadratic, it must have 0, 1, or 2 zeros. If f (x) has 0 or 1 zeros, it will not change sign and the concavity of f (x) will not change, so there is no point of inflection. If f (x) has 2 zeros, it will change sign twice, and f (x) will have 2 points of inflection. 49. (a) Regression equation: y bx bx abce (1 ae (0, 4) –5 bx )(0) (c)( abe (1 ae bx)2 ae 153 1 2161.4541 28.1336e 0.8627x [0, 8] by [ 400, 2300] (b) At approximately x 3.868 (late in 1996), when the sales are about 1081 million dollars/year (b) If f has no points of inflection, then f (x) has 0 or 1 zeros, so the discriminant of f (x) is 0. This gives (6b)2 4(12a)(2c) 0, or 3b 2 8ac. If f has 2 points of inflection, then f (x) has 2 zeros and the inequality is reversed, so 3b 2 8ac. In summary, f has 2 points of inflection if and only if 3b 2 8ac. (c) 2161.45 million dollars/year s Section 4.4 Modeling and Optimization b 7 , 3a 4 which is the x-value where the point of inflection 50. (a) In exercise 13, a 4 and b 21, so occurs. The local extrema are at x which are symmetric about x 2 and x 7 . 4 (pp. 206–220) 3 , 2 b (b) In exercise 8, a 2 and b 6, so 1, which 3a is the x-value where the point of inflection occurs. The local extrema are at x 0 and x Exploration 1 Constructing Cones 1. The circumference of the base of the cone is the circumference of the circle of radius 4 minus x, or 8 Thus, r 8 x 2 x. . Use the Pythagorean Theorem to find h, and the formula for the volume of a cone to find V. 2, which are 2. The expression under the radical must be nonnegative, that symmetric about x 1. is, 16 (c) f (x) 3ax 2 2bx c and f (x) 6ax 2b. The point of inflection will occur where b . 3a If there are local extrema, they will occur at the zeros of f (x). Since f (x) is quadratic, its graph is a parabola and any zeros will be symmetric about the vertex which will also be where f (x) 0. f (x) x2 8 2 0. Solving this inequality for x gives: 0 0, which is at x [0, 16 ] by [ 10, 40] x 16 . 154 Section 4.4 3. The circumference of the original circle of radius 4 is 8 . Thus, 0 x 8 . 9. x 2 x2 y 2 4 and y 3x ( 3x)2 4 x 2 3x 2 4 4x 2 4 x 1 Since y 3x, the solutions are: 3, or, x 1 and y 3. x 1 and y In ordered pair notation, the solutions are (1, 3) and ( 1, 3). [0, 8 ] by [ 10, 40] 4. The maximum occurs at about x volume is about V 25.80. 5. Start with dV dx 2 dr rh 3 dx 4.61. The maximum dh dh dr . Compute and , dx dx dx dV set 0, and solve for x to dx r2 3 dV substitute these values in , dx 8(3 6) obtain x 4.61. 3 128 3 Then V 25.80. 27 10. x2 4 y2 9 1 and y x x2 4 3)2 4(x 3)2 36 24x 36 36 13x 2 2. y 6x 2 6x 12 6(x 2)(x 1) y 12x 6 The critical points occur at x 2 or x 1, since y 0 at these points. Since y ( 2) 18 0, the graph has a local maximum at x 2. Since y (1) 18 0, the graph has a local minimum at x 1. In summary, there is a local maximum at ( 2, 17) and a local minimum at (1, 10). 1 (5)2(8) 3 r 2h 4. V SA 200 3 24x 0 x(13x 24) 0 2 r2 600 1000 . r2 Substituting into the surface area equation gives r 2 r2 600. Solving graphically, we have 11.14, r 4.01, or r value and using h x x 3, the solutions are: 0 and y 24 and y 13 3, or, x 15 . 13 In ordered pair notation, the solutions are (0, 3) and 24 15 , . 13 13 x, where 0 x 20. (a) The sum of the squares is given by f (x) x 2 (20 x)2 2x 2 40x 400. Then f (x) 4x 40. The critical point and endpoints occur at x 0, x 10, and x 20. Then f (0) 400, f (10) 200, and f (20) 400. The sum of the squares is as large as possible for the numbers 0 and 20, and is as small as possible for the numbers 10 and 10. Graphical support: 7.13. Discarding the negative 1000 to find the corresponding values r2 of h, the two possibilities for the dimensions of the cylinder are: [0, 20] by [0, 450] (b) The sum of one number plus the square root of the r 4.01 cm and h 19.82 cm, or, r 7.13 cm and h 6.26 cm. sin x is an odd function, sin ( 6. Since y cos x is an even function, cos ( sin ( 8. cos ( ) ) other is given by g(x) g (x) 5. Since y 7. Since y 24 13 0 or x 1. Represent the numbers by x and 20 Solving the volume equation for h gives h 2000 r x Section 4.4 Exercises cm3 1000 2 rh 1 9 4x 2 9x 2 1. y 3x 2 12x 12 3(x 2)2 Since y 0 for all x (and y 0 for x 2), y is increasing on ( , ) and there are no local extrema. 12 rh 3 3 9x 2 Quick Review 4.4 3. V (x ) sin . ) cos . 1 2 2 20 x 1 x 20 20 x. Then . The critical point occurs when x 1, so 20 x 1 and x 4 sin cos cos sin 0 cos ( 1) sin sin endpoints and critical point, we find cos cos ( 1) cos cos g(20) 79 . Testing the 4 sin sin 0 sin g(0) 20 20. 4.47, g 79 4 81 4 20.25, and Section 4.4 The sum is as large as possible when the numbers are 79 1 79 and summing + 4 4 4 155 Graphical support: 1 , and is as small as 4 possible when the numbers are 0 and 20 (summing 0 + 20). [0, 20] by [0, 40] Graphical support: 4. Let x represent the length of the rectangle in meters (0 x 4). Then the width is 4 x and the area is A(x) x(4 x) 4x x 2. Since A (x) 4 2x, the critical point occurs at x 2. Since A (x) 0 for 0 x 2 and A (x) 0 for 2 x 4, this critical point corresponds to the maximum area. The rectangle with the largest area measures 2 m by 4 2 2m, so it is a square. [0, 20] by [ 10, 25] 2. Let x and y represent the legs of the triangle, and note that 5. Then x 2 y2 0 x y 0). The area is A dA dx 1 x 22 1 2x 2 25 x2 25 2 x2 25 25, so y 1 xy 2 1 x 2 1 2 ( 2x) 25 x 2 (since x 2, so 25 x2 25 Graphical support: [0, 4] by [ 1.5, 5] 5. (a) The equation of line AB is y y-coordinate of P is x 1. . (b) A(x) The critical point occurs when 25 2x 2 0, which means 2x(1 x 1, so the x) d (2x 2x 2) 2 4x, the critical point dx 1 1 occurs at x . Since A (x) 0 for 0 x and 2 2 1 A (x) 0 for x 1, this critical point corresponds 2 (c) Since A (x) 5 x , (since x 0). This value corresponds to the 2 largest possible area, since dA dx 5 0 for 2 5 25 y 1 xy 2 A x 0 for 0 5 x and 2 5. When x 5 to the maximum area. The largest possible area is , we have 1 square unit, and the dimensions of the 2 1 rectangle are unit by 1 unit. 2 2 2 2 52 1 2 dA dx 2 5 A and 2 25 . Thus, the largest possible area is 4 1 2 Graphical support: 25 5 5 cm2, and the dimensions (legs) are cm by cm. 4 2 2 Graphical support: [0, 1] by [ 0.5, 1] [0, 5] by [ 2, 7] 3. Let x represent the length of the rectangle in inches (x 16 Then the width is and the perimeter is x 16 32 P(x) 2 x 2x . x x 2(x2 16) Since P (x) 2 32x 2 the critical point x2 occurs at x P (x) 4. Since P (x) 0 for x 0 for 0 x 0). 6. If the upper right vertex of the rectangle is located at (x, 12 x 2) for 0 x 12, then the rectangle’s dimensions are 2x by 12 x 2 and the area is A(x) 2x(12 x 2) 24x 2x 3. Then A (x) 24 6x 2 6(4 x 2), so the critical point (for 0 x 12) occurs at x 2. Since A (x) 0 for 0 x 2 and A (x) 0 for 2 x 12, this critical point corresponds to the maximum area. The largest possible area is A(2) 32, and the dimensions are 4 by 8. Graphical support: 4 and 0, this critical point corresponds to the minimum perimeter. The smallest possible perimeter is P(4) 16 in., and the rectangle’s dimensions are 4 in. by 4 in. [0, 12] by [ 10, 40] 156 Section 4.4 7. Let x be the side length of the cut-out square (0 Then the base measures 8 2x in. by 15 x 4). 2x in., and the volume is V(x) 2x)(15 x(8 2x) 4x 3 92x 120 4(3x 12x 2 V (x) 46x 2 120x. Then 5)(x 6). Then the 5 critical point (in 0 x 4) occurs at x . Since 3 5 5 V (x) 0 for 0 x and V (x) 0 for x 3 3 4, the 9. Let x be the length in meters of each side that adjoins the river. Then the side parallel to the river measures 800 2x meters and the area is A(x) x(800 2x) 800x 2x 2 for 0 x 400. Therefore, A (x) 800 4x and the critical point occurs at x 200. Since A (x) 0 for 0 x 200 and A (x) 0 for 200 x 400, the critical point corresponds to the maximum area. The largest possible area is A(200) 80,000 m2 and the dimensions are 200 m (perpendicular to the river) by 400 m (parallel to the river). Graphical support: critical point corresponds to the maximum volume. 2450 27 5 14 35 dimensions are in. by in. by in. 3 3 3 The maximum volume is V 5 3 90.74 in3, and the [0, 400] by [ 25,000, 90,000] Graphical support: 10. If the subdividing fence measures x meters, then the 216 m and the amount of fence x pea patch measures x m by needed is f (x) f (x) [0, 4] by [ 25, 100] 8. Note that the values a and b must satisfy a 2 and so b A dA da 400 a 2. Then the area is given by a 2) (400 2 400 when a 2 a 200 2 400 200. Since 0 for 200 a dA da a2 a2 . The critical point occurs 0 for 0 a 400 a2 f (x) 2 432x occurs at x 2 216 x 12. Since f (x) 0 for x 432x 1 . Then and the critical point (for x 0 for 0 x 0) 12 and 12, the critical point corresponds to the minimum total length of fence. The pea patch will measure 12 m by 18 m (with a 12-m divider), and the total amount of fence needed is f (12) 72 m. Graphical support: 200 and 200 then 200, so the maximum area occurs [0, 40] by [0, 250] 11. (a) Let x be the length in feet of each side of the square base. Then the height is when a 3x 20, this critical point corresponds to the maximum area. Furthermore, if a b 202 1 1 ab a 400 a 2 for 0 a 20, and 2 2 1 1 1 a ( 2a) 400 a 2 2 2 400 a 2 2 a2 dA da b2 3 3x b. 500 ft and the surface area x2 (not including the open top) is Graphical support: S(x) S (x) x2 2x 4x 500 x2 2000x point occurs at x [0, 20] by [ 30, 110] and S (x) x2 0 for x 2 2000x 1. Therefore, 2(x 3 1000) and the critical x2 10. Since S (x) 0 for 0 x 10 10, the critical point corresponds to the minimum amount of steel used. The dimensions should be 10 ft by 10 ft by 5 ft, where the height is 5 ft. (b) Assume that the weight is minimized when the total area of the bottom and the four sides is minimized. Section 4.4 12. (a) Note that x 2y 5(x 2 5x 2 10x 30x 33,750x 33,750x 10xy (for 0 0 x dc dx 0 for 15, the critical point 5 ft. (b) The material for the tank costs 5 dollars/sq ft and the excavation charge is 10 dollars for each square foot of the cross-sectional area of one wall of the hole. 13. Let x be the height in inches of the printed area. Then the 50 in. and the overall x 50 8 in. by 4 in. The amount of x width of the printed area is paper used is 400 2 in . Then x (x A (x) 4 (for x 0) occurs at x 10. Since A (x) 0 10 and A (x) 0 for x 400x 4 2 4x 4(x 2 82 100) x2 and the critical point 0 for 10, the critical point 8 and 50 x 4 for x 10, the overall dimensions are 18 in. high by 9 in. wide. 14. (a) s(t) v(t) At t , the critical point 2 2 (or 90 ). 16. Let the can have radius r cm and height h cm. Then 1000 . The area of material used is r2 2000 A r 2 2 rh r2 , so r dA 2 r 3 2000 2 2 r 2000r . The critical point dr r2 3 1000 dA occurs at r 10 1/3 cm. Since 0 for dr dA 0 r 10 1/3 and 0 for r 10 1/3, the critical dr r 2h 1000, so h point corresponds to the least amount of material used and hence the lightest possible can. The dimensions are r 1/3 10 6.83 cm and h 10 1/3 6.83 cm. In Example 2, because of the top of the can, the “best” design 17. Note that r 2h A dA dr 8r 2 16r dA dr 1000 . Then r2 0 for r 8r 2 5, the critical point corresponds to the least amount of aluminum used or wasted and hence the most economical can. The dimensions are r 2 16t 96t 112 s (t) 32t 96 0, the velocity is v(0) and 1000, so h 2000 , so r 3 16(r 125) 2000r 2 . The critical point r2 3 dA 125 5 cm. Since 0 for 0 r 5 dr 2 rh occurs at r corresponds to the minimum amount of paper. Using x 0 for is less big around and taller. 50 x A(x) x 8) 0 for maximizes the triangle’s area is 3375) 15. Since 0 for x 15 ft and y dimensions are x and A ( ) . Since A ( ) x2 corresponds to the minimum cost. The values of x and y are x 2 2 corresponds to the maximum area. The angle that 1 The critical point occurs at x dc 15 and dx 1 ab cos . The critical point 2 and A ( ) ) occurs at 0 1125 x2 10(x 3 2 1 ab sin 2 A( ) 30xy 5x 2 dc dx 4xy) 5x 2 c 15. We assume that a and b are held constant. Then 1125 . Then x2 1125, so y 157 h 96 ft/sec. (b) The maximum height occurs when v(t) 0, when t 3. The maximum height is s(3) 256 ft and it occurs at t 3 sec. (c) Note that s(t) 16t 2 96t 112 16(t 1)(t 7), so s 0 at t 1 or t 7. Choosing the positive value of t, the velocity when s 0 is v(7) 128 ft/sec. 40 , so the ratio of h to r is 8 to 1. 5 cm and 158 Section 4.4 18. (a) The base measures 10 2x in. by 15 2x 2 (d) V (x) 24x 2 336x The critical point is at in., so the volume formula is x(10 V(x) 2x)(15 2 2x) 2x 3 25x 2 ( 14)2 2(1) 14 x 75x. 24(x 2 864 4(1)(36) 14x 14 52 2 36) 7 13, that is, x 3.39 or x 10.61. We discard the larger value because it is not in the domain. Since V (x) 24(2x 14), which is negative when x 3.39, the critical point corresponds to the maximum volume. The maximum value occurs at x 7 13 3.39, which confirms the results in (c). (b) We require x 0, 2x 10, and 2x 15. Combining these requirements, the domain is the interval (0, 5). (e) 8x 3 168x 2 864x 1120 8(x 3 21x 2 108x 140) 0 8(x 2)(x 5)(x 14) 0 Since 14 is not in the domain, the possible values of x are x 2 in. or x 5 in. [0, 5] by [ 20, 80] (c) (f) The dimensions of the resulting box are 2x in., (24 2x) in., and (18 2x) in. Each of these measurements must be positive, so that gives the domain of (0, 9). [0, 5] by [ 20, 80] 20. 6 mi The maximum volume is approximately 66.02 in3 when x 1.96 in. x 2 mi (d) V (x) 6x 2 50x 6–x Village 4 + x2 miles 75 Jane The critical point occurs when V (x) x 50 25 5 6 ( 50)2 2(6) 7 4(6)(75) , that is, x 0, at 50 Let x be the distance from the point on the shoreline nearest 700 12 1.96 or x Jane’s boat to the point where she lands her boat. Then she 6.37. We discard needs to row x 2 mi at 2 mph and walk 6 4 x mi at the larger value because it is not in the domain. Since 5 mph. The total amount of time to reach the village is V (x) f (x) 12x 50, which is negative when x 1.96, x2 4 6 x 5 2 the critical point corresponds to the maximum volume. f (x) The maximum volume occurs when f (x) hours (0 1 1 (2x) 2 2 4 x2 x 6). Then 0, we have: x 25 57 6 x 1 5 x2 24 19. (a) The “sides” of the suitcase will measure 24 2x in. by 18 2x in. and will be 2x in. apart, so the volume formula is V(x) 2x(24 2x)(18 2x) 8x 3 168x 2 864x. 2 24 x 2 1 . Solving 5 1 5 5x 1.96, which confirms the result in (c). x 25x 2 4(4 21x 2 x2 4 16 (b) We require x 0, 2x 18, and 2x 24. Combining these requirements, the domain is the interval (0, 9). x 2) 4 x 21 We discard the negative value of x because it is not in the domain. Checking the endpoints and critical point, we have [0, 9] by [ 400, 1600] (c) f (0) 2.2, f land her boat 4 21 4 2.12, and f(6) 0.87 miles down the shoreline from 21 the point nearest her boat. [0, 9] by [ 400, 1600] The maximum volume is approximately 1309.95 in3 when x 3.39 in. 3.16. Jane should Section 4.4 21. If the upper right vertex of the rectangle is located at (x, 4 cos 0.5x) for 0 x , then the rectangle has width 2x and height 4 cos 0.5x, so the area is A(x) 8x cos 0.5x. Then A (x) 8x( 0.5 sin 0.5x) 8(cos 0.5x)(1) 4x sin 0.5x 8 cos 0.5x. Solving A (x) graphically for 0 x , we find that x 1.72. Evaluating 2x and 4 cos 0.5x for x 1.72, the dimensions of the rectangle are approximately 3.44 (width) by 2.61 (height), and the maximum area is approximately 8.98. 22. Let the radius of the cylinder be r cm, 0 2 r 2 100 1 2 r3 2 r(200 100 r 3r 2) The critical point for 0 200 3 0 10 r Cubic: A 1.74 a 3 ( 2r) (2 100 20 h 4000 3 10 2 3 1.86a 0.19 r 2)(2r) r 2) [ 0.5, 1.5] by [ 0.2, 0.6] 10 occurs at Quartic: A 1.92a 4 5.96a 3 6.87a 2 2.71a 0.12 0 for 0 for 10 2 < r < 10, the 3 critical point corresponds to the maximum volume. The dimensions are r 3.78a 2 2 2 . Since V (r) 3 2 10 and V (r) 3 r 0.34 [ 0.5, 1.5] by [ 0.2, 0.6] r2 100 r 0.54a r and the volume is 100 r 2 4 r(100 2 [0, 1.1] by [ 0.2, 0.6] (d) Quadratic: A 0.91a 2 10. Then r 2 cm3. Then 2 r2 V (r) (c) 2 the height is 2 100 V(r) r 159 8.16 cm and [ 0.5, 1.5] by [ 0.2, 0.6] (e) Quadratic: ’ 11.55 cm, and the volume is 3 2418.40 cm3. 3 23. (a) f (x) x( e x) e x(1) e x(1 x) The critical point occurs at x 1. Since f (x) 0 for 0 x 1 and f (x) 0 for x 1, the critical point corresponds to the maximum value of f. The absolute maximum of f occurs at x 1. (b) To find the values of b, use grapher techniques to solve xe x 0.1e 0.1, xe x 0.2e 0.2, and so on. To find the values of A, calculate (b a)ae a, using the unrounded values of b . (Use the list features of the grapher in order to keep track of the unrounded values for part (d).) a b A 0.1 3.71 0.33 0.2 2.86 0.44 0.3 2.36 0.46 [ 0.5, 1.5] by [ 0.2, 0.6] According to the quadratic regression equation, the maximum area occurs at a 0.30 and is approximately 0.42. Cubic: [ 0.5, 1.5] by [ 0.2, 0.6] According to the cubic regression equation, the maximum area occurs at a 0.31 and is approximately 0.45. Quartic: 0.4 2.02 0.43 0.5 1.76 0.38 0.6 1.55 0.31 0.7 1.38 0.23 0.8 1.23 0.15 0.9 1.11 0.08 1.0 1.00 0.00 [ 0.5, 1.5] by [ 0.2, 0.6] According to the quartic regression equation the maximum area occurs at a 0.30 and is approximately 0.46. 160 Section 4.4 24. (a) f (x) is a quadratic polynomial, and as such it can have 0, 1, or 2 zeros. If it has 0 or 1 zeros, then its sign never changes, so f (x) has no local extrema. If f (x) has 2 zeros, then its sign changes twice, and f (x) has 2 local extrema at those points. the radius is 36, we know that y 18 x and the height is 18 2 x. In part (a), x, and so the volume is given by x2 (18 2 r 2h radius is x and the height is 18 2 2 given by r h x (18 x). In part (b), the x, and so the volume is x). Thus, each problem requires x 2(18 us to find the value of x that maximizes f (x) in the interval 0 x x) 18, so the two problems have the same answer. To solve either problem, note that f (x) 18x 2 x 3 and so f (x) 36x 3x 2 3x(x 12). The critical point occurs at x 12. Since f (x) 0 for 0 x 12 and f (x) 0 for 12 x 18, the critical point corresponds to the maximum value of f (x). To maximize the volume in either part (a) or (b), let x 12 cm and y 6 cm. 27. Note that h 2 r2 3 and so r volume is given by V for 0 h critical point (for h 0 h dV 1 and dh h 2. Then the 3 r 2h 3 dV 3, and so dh 3 (3 h 2)h h2 h 3 h3 h 2). The (1 dV dh 0) occurs at h 1. Since 0 for 1 3, the critical point h 0 for corresponds to the maximum volume. The cone of greatest volume has radius ax , we have f (2) requirement is that 4 a 4 4 0. Therefore, a 16. To and so f (x) 2 32x 3 , so f (2) positive as expected. So, use a 2x 16x 2 6, which is 16. (b) We require f (1) 0. Since f (x) 2 2ax 3, we have f (1) 2 2a, so our requirement is that 2 2a 0. Therefore, a 1. To verify that x 1 is in fact an inflection point, note that we now have f (x) 2 2x 3, which is negative for 0 x 1 and positive for x 1. Therefore, the graph of f is concave down in the interval (0, 1) and concave up in the interval (1, ), So, use a 1. 2x 3 a , so the only sign change in x2 a 1/3 f (x) occurs at x , where the sign changes from 2 29. f (x) 2x ax 2 negative to positive. This means there is a local 12 x (18 4 x) 2x 2. Since a and so our 4 minimum, note that we now have f (x) 25. Let x be the length in inches of each edge of the square end, and let y be the length of the box. Then we require 4x y 108. Since our goal is to maximize volume, we assume 4x y 108 and so y 108 4x. The volume is V(x) x 2(108 4x) 108x 2 4x 3, where 0 x 27. Then V 216x 12x 2 12x(x 18), so the critical point occurs at x 18 in. Since V (x) 0 for 0 x 18 and V (x) 0 for 18 x 27, the critical point corresponds to the maximum volume. The dimensions of the box with the largest possible volume are 18 in. by 18 in. by 36 in. 2y f (x) 2 verify that the critical point corresponds to a local (b) Possible answers: No local extrema: y x 3; 2 local extrema: y x 3 3x 26. Since 2x 28. (a) We require f (x) to have a critical point at x 2 m, height 1 m, and volume 2 m3. 3 minimum at that point, and there are no local maxima. 30. (a) Note that f (x) 3x 2 2ax b. We require f ( 1) 0 and f (3) 0, which give 3 2a b 0 and 27 6a b 0. Subtracting the first equation from the second, we have 24 8a 0 and so a 3. Substituting into the first equation, we have 9 b 0, so b 9. Therefore, our equation for f (x) is f (x) x 3 3x 2 9x. To verify that we have a local maximum at x 1 and a local minimum at x 3, note that f (x) 3x2 6x 9 3(x 1)(x 3), which is positive for x 1, negative for 1 x 3, and positive for x 3. So, use a 3 and b 9. (b) Note that f (x) 3x 2 2ax b and f (x) 6x 2a. We require f (4) 0 and f (1) 0, which give 48 8a b 0 and 6 2a 0. By the second equation, a 3, and so the first equation becomes 48 24 b 0. Thus b 24. To verify that we have a local minimum at x 4, and an inflection point at x 1, note that we now have f (x) 6x 6. Since f changes sign at x 1 and is positive at x 4, the desired conditions are satisfied. So, use a 3 and b 24. 161 Section 4.4 31. Refer to the illustration in the problem statement. Since x2 y2 9, we have x 33. (a) Note that w 2 y 2. Then the volume of 9 dS dw 12 12 rh x (y 3) 3 3 1 2 (9 y )(y 3) 3 3 for 3 y 3y 9y 27), 3 ( 3y 2 9) 3)(y (y y 2 2y 3) 6 1), so the critical point in dV 1. Since dy the interval ( 3, 3) is y 0 for 1 6y w 2)3/2, so k(144 144 w 2)(w 2 0 for 3 y 0 for 0 w w w 2) 12) occurs at w dS 6 and dw 6. 0 for 12, the critical point corresponds to the maximum stiffness. The dimensions are 6 in. wide by 1 3, the critical point does w w2)3/2(1) 36) 144 The critical point (for 0 3. (y dV and dy 2 dS Since dw dV dy Thus (y 3 kw(144 w 2)( 3w 2 (k 144 ( 4k w 2. Then we 144 w2)1/2( 2w) 3 (144 2 kw 122, so d kwd 3 may write S the cone is given by V d2 6 3 in. deep. (b) 32 3 correspond to the maximum value, which is V(1) cubic units. 32. (a) Note that w 2 d2 kwd 2 may write S for 0 so w dS dw dS Since dw 43 w 144 w 2) kw(144 w 2. Then we 144kw The graph of S w(144 w2)3/2 is shown. The maximum stiffness shown in the graph occurs at w which agrees with the answer to part (a). kw 3 3kw 2 w 3k(w 2 48). The critical 12) occurs at w 0 for 0 w 4 dS 3 and dw 48 4 6, (c) 3. 0 for 12, the critical point corresponds to the maximum strength. The dimensions are 4 3 in. wide by 4 [0, 12] by [ 2000, 8000] 12, 144k point (for 0 122, so d 6 in. deep. (b) [0, 12] by [ 2000, 8000] The graph of S d 3 144 d 2 is shown. The maximum stiffness shown in the graph occurs at d 6 3 10.4 agrees with the answer to part (a), and its value is the same as the maximum value found in part (b), as expected. Changing the value of k changes the maximum stiffness, but not the dimensions of the stiffest beam. The graphs for different values of k look the same except that the vertical scale is different. [0, 12] by [ 100, 800] The graph of S 144w w 3 is shown. The maximum strength shown in the graph occurs at w 4 3 6.9, which agrees with the answer to part (a). (c) 34. (a) v(t) s (t) 10 sin t The speed at time t is 10 sin t . The maximum speed is 10 cm/sec and it occurs at t t 1 ,t 2 7 sec. The position at these times is s 2 3 ,t 2 5 , and 2 0 cm (rest position), and the acceleration [0, 12] by [ 100, 800] 2 a(t) v (t) 10 2 cos t is 0 cm/sec2 at these 2 The graph of S d 144 d is shown. The maximum strength shown in the graph occurs at d 4 6 9.8, which agrees with the answer to part (a), and its value is the same as the maximum value found in part (b), as expected. Changing the value of k changes the maximum strength, but not the dimensions of the strongest beam. The graphs for different values of k look the same except that the vertical scale is different. times. (b) Since a(t) 10 2 cos t, the greatest magnitude of the acceleration occurs at t 0, t 1, t 2, t 3, and t 4. At these times, the position of the cart is either s 10 cm or s 10 cm, and the speed of the cart is 0 cm/sec. 162 Section 4.4 35. Since di dt 2 sin t 2 cos t, the largest magnitude of the current occurs when 2 sin t 2 cos t x 2 is positive, it has an absolute minimum at the point 0, or cos t. Squaring both sides gives sin2 t sin t we know that sin2 t cos2 t 1, so sin2 t Thus the possible values of t are cos2 t, and 1 . 2 cos2 t 35 , , and so on. 444 , Eliminating extraneous solutions, the solutions of sin t cos t are t times i k for integers k, and at these 4 2 cos t 2 sin t 2 2. The peak current is 2 2 amps. 36. The square of the distance is D(x) 32 2 x so D (x) 2x Since D (x) ( 0)2 x x2 2 and the critical point occurs at x 0 for x 1 and D (x) where f (x) 0 for x 5 D(1) 2 1 13 ,. 24 0, and that point is (b) No. Since f (x) is continuous on [0, 2 ], its absolute minimum occurs at a critical point or endpoint. Find the critical points in [0, 2 ]: f (x) 4 sin x 2 sin 2x 0 4 sin x 4 sin x cos x 0 4(sin x)(1 cos x) 0 sin x 0 or cos x 1 x 0, , 2 The critical points (and endpoints) are (0, 8), ( , 0), and (2 , 8). Thus, f (x) has an absolute minimum at ( , 0) and it is never negative. 1. 1, the critical point corresponds to the minimum distance. The minimum distance is 2x 39. (a) Because f (x) is periodic with period 2 . 40. (a) 9 , 4 2x 38. No. Since f (x) is a quadratic function and the coefficient of 2 sin t sin 2t 2 sin t 2 sin t cos t 2(sin t)(1 cos t) 0 sin t 0 or cos t 1 t k , where k is an integer The masses pass each other whenever t is an integer multiple of seconds. (b) The vertical distance between the objects is the absolute value of f (x) sin 2t 2 sin t. Find the critical points in [0, 2 ]: . 37. Calculus method: The square of the distance from the point (1, (x, (x 1)2 ( x2 2x 1 2x D (x) 2 48 2 48 3x 2 2(2 cos2 t 3)2 x2 16 2 2 x2 16 20 2 48 3x 2 2(2 cos t 3 0 cos t 1) 0 1)(cos t 1) 0 ( 6x) The distance from the origin to (1, 6x 2 48 3x 2 . or cos t 1 t 2 , 4 3 , 0, 2 3 The critical points (and endpoints) are (0, 0), 2 , 3 33 4 33 , , , and (2 , 0) 3 2 2 2 The distance is greatest when t sec and when 3 4 33 t sec. The distance at those times is meters. 3 2 41. (a) sin t sin (t sin t sin t cos sin t 1 sin t 2 tan t 3 ) cos t sin 3 1 3 sin t cos t 2 2 3 cos t 2 3 3 3) is 2. The shortest distance from the point to the semicircle is the distance along the radius containing 3). That distance is 4 1 2 The semicircle is centered at the origin and has radius 4. the point (1, 2 cos t 1) 0 cos t Geometry method: ( 3)2 2 cos t 3x 2. Then Solving D (x) 0, we have: 6x 2 48 3x 2 36x 2 4(48 3x 2) 9x 2 48 3x 2 12x 2 48 x 2 We discard x 2 as an extraneous solution, leaving x 2. Since D (x) 0 for 4 x 2 and D (x) 0 for 2 x 4, the critical point corresponds to the minimum distance. The minimum distance is D(2) 2. 12 2 cos 2t 2(2 cos2 t x 2) is given by 16 D(x) f (x) 3) to 2 2. Solving for t, the particles meet at t t 4 sec. 3 3 sec and at 163 Section 4.4 (b) The distance between the particles is the absolute value of f (t) sin t sin t 3 3 cos t 2 1 sin t. Find 2 3 2 sin t 1 cos t 2 3 sin t 2 tan t measure 1 ft and (1 2 sin ) ft, so the volume is given by 1 (cos )(1 2 V( ) the critical points in [0, 2 ]: f (t) 42. The trapezoid has height (cos ) ft and the trapezoid bases 1 20(cos )(1 0 2 sin )(20) sin ). Find the critical points for 0 1 cos t 2 1 V( ) 20(cos )(cos ) 2 20(1 20 cos2 3 5 11 The solutions are t and t , so the critical 6 6 5 11 , 1 and , 1 , and the interval points are at 6 6 3 3 , and 2 , . The particles endpoints are at 0, 2 2 5 11 sec and at t sec, and are farthest apart at t 6 6 sin )( sin ) 20 sin2 0 20 sin 20 sin2 0 20(2 sin2 sin 0 1)(sin 20(2 sin 1) 1) 0 sin 1 or sin 2 The critical point is at 3 f (t) 2 1 sin t 2 1 sin t 2 cos t 0. 0 0 3 For the function y sec and t 6 6 and V ( ) , 15 0 for 3 . Since V ( ) 6 2 0 for , the critical point corresponds to the maximum possible trough volume. The 3 2 cos t volume is maximized when This is the same equation we solved in part (a), so the solutions are t 1 6 the maximum distance between the particles is 1 m. (c) We need to maximize f (t), so we solve f (t) 0 20 sin sin2 ) 20(1 : 4 sec. 3 43. (a) D 6 . C 8.5 R S y f (t), the critical points occur at Q 4 , 1 , and the interval endpoints are at 3 3 1 1 0, and 2 , . 2 2 4 Thus, f (t) is maximized at t and t . But 3 3 , 1 and these are the instants when the particles pass each other, so the graph of y points and f (t ) has corners at these d f (t) is undefined at these instants. We dt t or 3 4 the distance is changing faster than at any other 3 time in the interval. L x x A P B Sketch segment RS as shown, and let y be the length of segment QR. Note that PB 8.5 x, and so QB x 2 (8.5 x)2 8.5(2x 8.5). Also note that triangles QRS and PQB are similar. QR RS cannot say that the distance is changing the fastest at any particular instant, but we can say that near t y y 8.5 y2 8.52 PQ QB x 8.5(2x 8.5) x2 8.5(2x 8.5) y2 8.5x 2 2x 8.5 L2 x2 y2 L2 x2 8.5x 2 2x 8.5 L2 L2 x 2(2x 8.5) 8.5x 2 2x 8.5 2x 3 2x 8.5 164 Section 4.4 43. continued 46. (b) Note that x y 2x L2 4.25, and let f (x) 2x 8.5 11, the approximate domain of f is 5.20 x 2 3 8.5)(6x ) (2x )(2) (2x 8.5)2 x (8x 51) (2x 8.5)2 and f (x) 0 for x 0 for 5.20 x 6.375 6.375. Therefore, the value of x that minimizes L 2 is x 6.375 in. (c) The minimum value of L is 2(6.375)3 2(6.375) 8.5 M2 44. Since R dR dM f (M) CM C C 2 11.04 in. M 3 C2 M 2 2 M . Let f (M) 13 M , we have 3 CM x c–x R Let P be the foot of the perpendicular from A to the mirror, and Q be the foot of the perpendicular from B to the mirror. Suppose the light strikes the mirror at point R on the way from A to B. Let: a distance from A to P b distance from B to Q c distance from P to Q x distance from P to R To minimize the time is to minimize the total distance the light travels going from A to B. The total distance is D(x) x2 a2 (c x)2 b 2 Then 2 M . Then 1 D (x) 2 x2 x 2M, and the critical point for f occurs at C .This value corresponds to a maximum because 2 C C f (M) 0 for M and f (M) 0 for M . The value 2 2 C dR of M that maximizes is M . 2 dM M 45. The profit is given by P(x) (n)(x c) 2bx b(100 a2 x2 a b(100 2 (100 c (100 x)(x c) x c a2 50 A selling price of 50 x)2 b2 [ 2(c x)] b2 0 gives the equation x x)2 (c c)bx (a b2 x2 50 x)2 x)2 b 2 b 2] (c x)2(x 2 x 2b 2 (c x)2x 2 (c x)2a 2 x 2b 2 [c 2 2xc 0 (a 2 b 2)x 2 0 x 2(c 100 c 2 (c [(a b)x x2 x 2[(c c)b a2 x 2b 2 100bc). x)2 x)2 c , and this 2 value corresponds to the maximum profit because 0 for x (c x)2 (c x we have: 2x). The critical point occurs at x P (x) 2 c a2 Solving D (x) x2 1 (2x) which we will refer to as Equation 1. Squaring both sides, bx Then P (x) Q c 6.375 in., and this corresponds to a minimum value of f (x) because f (x) b u1 u2 P 2 5.20, the critical point occurs at 51 8 x A 8.5. a (2x For x B . Since Then f (x) Normal 3 c c and P (x) 0 for x 50 . 2 2 c will bring the maximum profit. 2 (c ac x Note that the value x because x and c a ac a b b a 2) x)2a 2 (c x 2]a 2 2a 2cx ac][(a a 2c 2 b)x ac or x a b is an extraneous solution x have opposite signs for this value. The only critical point occurs at x ac a b . ac] 165 Section 4.4 To verify that this critical point represents the minimum distance, note that x2 ( a 2)(1) D (x) x 2 x)2 ( (c (x) a ( x x2 a2 ) (c (c x) 2 (c (x 2 a 2) x2 (x 2 a 2)3/2 (x 2 a2 a 2)3/2 x) [(c b x) x)2 (c b2 2 50. (a) Since A (q) x)2 b 2] (c [(c x)2 b 2]3/2 x)2 when km q2 kmq 2 h , or q 2 h , the critical point occurs 2 2km . This corresponds to the h minimum value of A(q) because A (q) b2 , x)2 b 2]3/2 [(c [200 2(x 50)]x 2x 2 300x 6000 32x r(x) c(x) 2x 2 268x 6000, 50 x 80 r(x) c(x) p(x) Since p (x) 4x 268 4(x 67), the critical point occurs at x 67. This value represents the maximum because p (x) 4, which is negative for all x in the domain. The maximum profit occurs if 67 people go on the tour. 2 b 2)( 1) 49. Revenue: Cost: Profit: which is positive for q which is always positive. 3 , 0. (b) The new formula for average weekly cost is We now know that D(x) is minimized when Equation 1 is true, or, equivalently, 2kmq PR AR (k B(q) QR . This means that the two BR bq)m q cm bm cm km q right triangles APR and BQR are similar, which in turn A(q) hq 2 hq 2 bm implies that the two angles must be equal. 47. Since B(q) differs from A(q) by a constant, the dv dx minimum value of B(q) will occur at the same q-value ka 2kx ka 2k The critical point occurs at x as the minimum value of A(q). The most economical a , which represents a 2 quantity is again maximum value because d 2v dx 2 2k, which is negative for all x. The maximum value of v is kx 2 kax 48. (a) v dv dr ka cr0r 2 2cr0r a 2 k a2 2 ka 2 . 4 cr 3 3cr 2 cr(2r0 3r) The critical point occurs at r 2r0 3 . (Note that r 0 is not in the domain of v.) The critical point represents a maximum because d 2v dr 2 2cr0 which is negative in the domain (b) We graph v (0.5 6cr r0 2 2c(r0 r 51. The profit is given by p(x) r(x) c(x) 6x (x 3 6x 2 15x) x 3 6x 2 9x, for x 0. Then p (x) 3x 2 12x 9 3(x 1)(x 3), so the critical points occur at x 1 and x 3. Since p (x) 0 for 0 x 1, p (x) 0 for 1 x 3, and p (x) 0 for x 3, the relative maxima occur at the endpoint x 0 and at the critical point x 3. Since p(0) p(3) 0, this means that for x 0, the function p(x) has its absolute maximum value at the points (0, 0) and (3, 0). This result can also be obtained graphically, as shown. 3r), r0. [0, 5] by [ 8, 2] r)r 2, and observe that the maximum indeed occurs at v 2 0.5 3 2km . h 1 . 3 52. The average cost is given by a(x) c(x) x a (x) 2x x2 20x 20,000. Therefore, 20 and the critical value is x represents the minimum because a (x) 10, which 2, which is positive for all x. The average cost is minimized at a [0, 0.5] by [ 0.01, 0.03] production level of 10 items. 166 Section 4.4 53. (a) According to the graph, y (0) 0. (b) According to the graph, y ( L) (c) y(0) 0, so d c 2bx r h 3ax 2 bx 2 and 3aL 2 and y ( L) 2bL bL 2 H 0, so we have two linear 2. and let m y 3aL . Substituting into the first 2 3aL 3 aL 3 equation, we have aL 3 H, or H, so 2 2 H H a 2 3 . Therefore, b 3 2 and the equation for y is L L H H x3 x2 y(x) 2 3 x 3 3 2 x 2, or y(x) H 2 3 . L L L L m(0 equation gives b 2a x and so the 2 2a x2 . 2 a2 height is h r2 a2 r 2h 3 x2 2a 2 3 x2 2a 2 a2 m(x x0) x0) m(x x0) a a 3 d2 (r 3 dr f (r) = a2 a2 2 r 3 3 a2 2a 2r 3 a2 r(2a 2 2 2 a2 h r a 6 3 r 2)(2r) mx0 m x0 have A(x) y2 2 2a 2 , which gives 3 2 a6 . Then 3 a2 mx0 a m a m a2 m a2 m f(x) 2 . f (x) f (x) x To graph, let y1 3r 2) r2 mx0 mx0) (b) The domain is the open interval (0, 10). 2 The critical point occurs when r r ( a2 r2 r (a m m a. 3r 3 2 . 1 (x-intercept of line RT)(y-intercept of line RT) 2 1 mx0 a 2 2 r 2) r 2 3a 2 a 3 r a m Substituting x for x0, f (x) for m, and f (x) for a, we ( 2r) r2 2r(a a r 2, where 0 mx0 0, or x 2(Area of triangle ORT) . r 2) 1 r2 3 r2 mx0, (Area of triangle RST) m f(r) a Let O designate the origin. Then (b) To simplify the calculations, we shall consider the volume as a function of r: volume a. The y-intercept of this line is and the x-intercept is the solution of Therefore, V(x) f (x0) be the slope of line RT. Then the equation of line RT is equations in the two unknowns a and b. The second 54. (a) The base radius of the cone is r a3 , the relationship is 3 and h P has coordinates (x0, a), aL 3 2bx. Then y( L) 6 3 55. (a) Let x0 represent the fixed value of x at point P, so that c, so y (0) implies that ax 3 0. Therefore, y(x) y (x) 0. 0. 3ax 2 Now y (x) a (c) Since r f (x) y3 A(x) f (x) 5 51 x2 , 100 NDER(y1), and y2 x y1 2 y2 . The graph of the area function y3 A(x) is shown below. 2a 3 2 2 a 3 a 3 3 . Using a3 , 3 and h we may now find the values of r and h for the given values of a. [0, 10] by [ 100, 1000] When a 4: r when a 5: r when a 6: r when a 8: r 46 ,h 3 56 ,h 3 2 4 3 3 5 3 3 ; ; 6, h 6 8 3 2 3; 8 ,h 3 3 The vertical asymptotes at x 0 and x 10 correspond to horizontal or vertical tangent lines, which do not form triangles. Section 4.4 167 (c) Using our expression for the y-intercept of the tangent line, the height of the triangle is a mx f (x) f (x) x 1 2 1 2 5 5 x x2 100 2 100 x2 2 100 x2 100 x2 x x2 We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at x 8.66. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the y-coordinate of the center of the ellipse. (d) Part (a) remains unchanged. The domain is (0, C). To graph, note that f (x) f (x) x2 C2 B1 B B C2 1 C2 x2 B C B C2 x 2 and Bx ( 2x) C C2 . x2 Therefore, we have A(x) f (x) 2 f (x) f (x) x C2 C B C B Bx x x2 Bx C C2 Bx C C2 x2 1 BCx C2 BCx C2 BCx C2 x2 1 x2 1 C2 x2) Bx [Bx 2 (BC B C2 x 2)( [Bx 2 BC C 2 x2 x2 x C2 C2 [BC(C C2 x2 2 B(C 2 C2 BC(C x2 C2 x 2)]2 x 2)]2 x 2)]2 x 2)2 x2 (x C 2 A (x) B (BC x x2 2 C2 x 2)(2)(C C2 x x 2) C2 BC x 2 (C 2 C 2 x 2) BC(C 22 x (C x 2) 2x 2 C 2 x 2) BC(C 22 x (C x 2) [ BC(C C 2 x 2) 22 x (C x 2) ( C2 (C Cx2 2x 2 C2 Cx 2 C2 BC(C x 2(C 2 C 2 x 2) [Cx 2 x 2)3/2 BC 2(C x 2(C 2 C 2 x 2) (2x 2 x 2)3/2 x2 x2 C C(C 2 C2 C2 x 2) C C2 x 2) C2 (C x2 ( x2 C 2 C2 x2 x2 ) C2 C2 C2 x 2) x C2 x2 x 2) C C2 x2 x 2 )2 x x 2] x2 (C 2 ) x2 x 2) C2 x 2(1) 168 Section 4.5 55. continued To find the critical points for 0 2x 2 4x 4 C2 C 4C 2x 2 C4 C4 4x 4 3C 2x 2 3C 2) x2 C 2x 2 0 The minimum value of A(x) for 0 corresponding triangle height is a C, we solve: 0 x 2(4x 2 C2 x mx f (x) C occurs at the critical point x C3 , or x 2 2 3C 2 . The 4 f (x) x B C B x C2 x2 Bx2 C C2 x2 2 B 3C B B C B BC C2 B B 2 C2 3C 2 4 4 CC 2 3C 2 4 3BC 2 4 C2 2 3B 2 3B This shows that the triangle has minimum area when its height is 3B. s Section 4.5 Linearization and Newton’s Method (pp. 220–232) Exploration 1 1. f (x) Approximating with Tangent Lines 2x, f (1) 2, so an equation of the tangent line is y 1 2(x 1) or y 2x 1. 3. Since (y1 y2)(1) y1(1) y2(1) 1 1 0, this view shifts the action from the point (1, 1) to the point (1, 0). Also (y1 y2) (1) y1 (1) y2 (1) 2 2 0. Thus the tangent line to y1 y2 at x 1 is horizontal (the x-axis). The measure of how well y2 fits y1 at (1, 1) is the same as the measure of how well the x-axis fits y1 y2 at (1, 0). 4. These tables show that the values of y1 y2 near x x 1. Here are two tables with Table 0.0001. 1 are close to 0 so that y2 is a good approximation to y1 near Section 4.5 Exploration 2 Grapher Using Newton’s Method on the 1–3. Here are the first 11 computations. 7. (a) x x 1 169 0 1 (b) 2ex 2ex x e e 1 (e 1 0 1) 0.684 2e 8. f (x) 3x 2 4 f (1) 3(1)2 4 1 Since f (1) 2 and f (1) 1, the graph of g(x) passes through (1, 2) and has slope 1. Its equation is g(x) 1(x 1) ( 2), or g(x) x 1. x 2. dy dx (x 1)(1 1) sin x) (x (x 1)2 x x sin x 1 (x sin x 1)2 1 x 2x cos (x 2 2 2.069 2.1 2.072 2.2 2.003 2.3 9. f (x) cos x f (1.5) cos 1.5 Since f (1.5) sin 1.5 and f (1.5) cos 1.5, the tangent line passes through (1.5, sin 1.5) and has slope cos 1.5. Its equation is y (cos 1.5)(x 1.5) sin 1.5, or approximately y 0.071x 0.891 Quick Review 4.5 d2 (x dx 1.9 2 1.3 1) 1.871 1.2 cos (x 2 1.8 1.1 dy dx 1.7 1.688 1.0 1. 1.457 0.8 0.9 2. g(x) 0.7 4. Answers will vary. Here is what happens for x1 f (x) 1) cos x)(1) cos x cos x (x 1) sin x (x 1)2 [0, ] by [ 0.2, 1.3] 10. For x , and so f (4) 2 1 . Since 2 x3 1 , the tangent line passes through 2 1 1 (4, 1) and has slope . Its equation is y (x 4) 1, or 2 2 1 y x 1. 2 f (4) 3. 1 3, f (x) 1 and f (4) [ 2, 6] by [ 3, 3] x 0.567 4. [ 1, 7] by [ 2, 2] [ 4, 4] by [ 10, 10] x 0.322 5. f (x) (x)( e x) (e x)(1) e x xe x f (0) 1 The lines passes through (0, 1) and has slope 1. Its equation is y x 1. 6. f (x) (x)( e x) ( e x)(1) e x xe x f ( 1) e 1 ( e 1) 2e The lines passes through ( 1, e 1) and has slope 2e. Its equation is y 2e(x 1) ( e 1), or y 2ex e 1. Section 4.5 Exercises 1. (a) f (x) 3x 2 2 We have f (2) 7 and f (2) 10. L(x) f (2) f (2)(x 2) 7 10(x 2) 10x 13 (b) Since f (2.1) 8.061 and L(2.1) 8, the approximation differs from the true value in absolute value by less than 10 1. 170 Section 4.5 1 2. (a) f (x) 2 x 9 We have f ( 4) L(x) x 8. (a) f (x) 2 9 4 . 5 5 and f ( 4) f ( 4) 4 x 5 x)6 (1 2 (b) f (x) 1 2 9 5 (b) Since f ( 3.9) 4.9204 and L( 3.9) 4.92, the approximation differs from the true value by less than 10 3. 1 2[1 ( x)] 1 1/2 x (1 ( 4)) 4) ( x)]6 [1 1 6( x) 1 x 2 2[1 1 6x ( 1)( x)] 2x (c) f (x) f ( 4)(x 4 (x 5 5 x (2x) 2 (d) f (x) x) x2 2 21 x 2 1/2 2 x2 4 21 1 x2 22 x 2 1 21 2 3. (a) f (x) 1 x We have f (1) 2 and f (1) L(x) f (1) f (1)(x 1) 2 0(x 1) 2 0. (e) f (x) 41/3 1 (b) Since f (1.1) 2.009 and L(1.1) 2, the approximation differs from the true value by less than 10 2. 4. (a) f (x) (f) f (x) x 1 2 3 1 2x We have f (0) L(x) (b) Since f (0.1) 0.0953 and L(0.1) 0.1 the approximation differs from the true value by less than 10 2. 5. (a) f (x) sec2 x We have f ( ) 0 and f ( ) L(x) f ( ) f ( )(x ) 0 1(x ) x f (0) f (0)(x ( 1)(x 2 x x 3 3 and f (0) 2 x 2 1 x 1. 0) 0) (b) Since f (0.1) 1.47063 and L(0.1) 1.47080, the approximation differs from the true value in absolute value by less than 10 3. x 6 3x cos x 1 1 and f (0) f (0)(x 3 2 0) 3 x 2 1 x 2 1. 0.002)100 1 0.021 10 0.009)1/3 1 1.009 (1 1.009 1.003 9 10 6 (100)(0.002) 1 1 (0.009) 3 10 f (x) 5 0(x 1) 5 1 2/3 x 3 We have f (8) L(x) 1.003; 8 2 and f (8) k1 k. 2 2 11. Center 1 f (x) 4x 4 We have f ( 1) 5 and f ( 1) 0 L(x) f ( 1) f ( 1)(x ( 1)) 5 12. Center 2 7. f (x) k(1 x) We have f (0) 1 and f (0) L(x) f (0) f (0)(x 0) 1 k(x 0) 1 kx 2/3 1 1 1 10. (a) (1.002)100 (1 1.2; 1.002100 1.2 x2 2 2/3 1 The linearization is the sum of the two individual for 1 We have f (0) f (0) 1 (b) 1 41/3 1 linearizations, which are x for sin x and 1 1. (b) Since f ( 0.1) 0.10033 and L( 0.1) 0.1, the approximation differs from the true value in absolute value by less than 10 3. 6. (a) f (x) 1 3x 34 3x 1/3 4 x 4 41/3 1 1 9. f (x) 1. 3x)1/3 1 1 We have f (0) 0 and f (0) L(x) f (0) f (0)(x 0) 0 1x x L(x) (4 f (8) f (8)(x 8) 1 . 12 2 1 (x 12 8) x 12 4 3 Section 4.5 13. Center 1 (x 1)(1) (x)(1) 1 f (x) (x 1)2 (x 1)2 1 1 We have f (1) and f (1) 2 4 1 1 L(x) f (1) f (1)(x 1) (x 2 4 Then f (x) xn 1 x 4 1) 1 4 f 3 2 14. Center f 1 0 and f 2 f x3 x 4 . 25 4 x 25 1x 1. Then f (x) x 3x 2 f (xn) xn 2 xn3 xn 3xn2 1 1 2 x Solution: x 1 f (xn) x4 x 2xn 2xn 2 1 sin xn cos xn f (x) shows that f (x) xn f (xn) xn xn4 xn 4xn 3 1 and 0 has two x1 x2 x3 x4 x5 1 0 has two x4 xn 1 f (xn) xn xn4 xn f (xn) 19. (a) Since dy Solutions: x 2 4xn3 . Note that f (x) 0 clearly has two solutions, namely 4 x 2. We use Newton’s method to find the decimal equivalents. x1 1.5 x2 1.2731481 x3 1.1971498 x4 1.1892858 x5 1.1892071 x6 1.1892071 20. (a) Since [ 3, 3] by [ 4, 4] 4x 3 and 2. Then f (x) 1.189207 dy dx 3x 2 3, dy (b) At the given values, dy (3 22 3)(0.05) solutions. 1.5 1.455 1.4526332 1.4526269 1.4526269 2 1.9624598 1.9615695 1.9615690 1.9615690 0.386237, 1.961569 Solutions: x 3 f (x) shows that f (x) 2 0.3 0.3825699 0.3862295 0.3862369 0.3862369 18. Let f (x) . 4x 3 3. Then f (x) f (xn) x1 x2 x3 x4 x5 Solutions: x 0.682328 The graph of y x1 x2 x3 x4 x5 xn2 xn 1 and Note that f is cubic and f is always positive, so there is exactly one solution. We choose x1 0. x1 0 x2 1 x3 0.75 x4 0.6860465 x5 0.6823396 x6 0.6823278 x7 0.6823278 xn xn cos x and 9 25 1. 2 0 2 f (xn) xn 16. Let f (x) 2 f (xn) sin x. [ 4, 4] by [ 3, 3] 2 15. Let f (x) xn 3 2 3 2 sin x We have f L(x) 3 and f 5 4 x 25 2 f (x) 2x 1 solutions 3 3 , we have f 2 2 3 3 3 f x 2 2 5 Using center 1 2x The graph of y Alternate solution: L(x) x2 17. Let f (x) 171 (3x 2 3) dx. 9(0.05) 0.45. (1 x 2)(2) (2x)(2x) dy (1 x 2)2 dx 2 2 2x dx. (1 x2)2 2 (1 2x 2 , x2)2 (b) At the given values, x1 x2 x3 x4 x5 1.2 1.6541962 1.1640373 1.1640351 1.1640351 1.452627, 1.164035 2 [1 dy 2( 2)2 (0.1) ( 2)2]2 2 8 52 (0.1) 0.024. 21. (a) Since dy dy dx 1 x (x 2 ) (2x ln x (ln x)(2x) 2x ln x x, x) dx. (b) At the given values, dy [2(1) ln (1) 1](0.01) 1(0.01) 0.01 172 Section 4.5 dy dx x2 22. (a) Since 1 dy 2 x 1 1 (x) (1 x2 x2 1 2 ( 2x) x2 1 x 2) (1 1 2x 1 x 2x 2 1 2 29. (a) 1 x2 , f f (0.55) 20 11 f (0.5) (b) Since f (x) 2 x 2 11 2 , f (0.5) 4. 2 x 2 Therefore, df dx. (0)2 1 e sin x cos x, dy f df 30. (a) ( 0.2) 4 dx f f (1.01) 0.2. 4(0.05) 1 5 2 11 (c) 2(0)2 1 (b) At the given values, dy dy 23. (a) Since dx x 2)(1) f (1) 1 55 1.04060401 (b) Since f (x) 4x 3, f (1) 4. Therefore, df 4 dx 4(0.01) (cos x)e sin x dx. (c) f 1 5 0.2 df 0.04060401 dV dr 4 r 2, dV 0.04 1 0.04060401 0.04. 0.00060401 (b) At the given values, (cos )(e sin )( 0.1) dy ( 1)(1)( 0.1) 0.1. 31. Note that a to a dy 24. (a) Since dx x 3 csc 1 cot 1 3 x x csc 1 cot 1 , 3 3 x x dy csc 1 cot 1 dx. 3 3 1 3 x 3 25. (a) y(1 x Since dy a to a x a to a x1 1)(1) (x)(1) (x 1)2 dy (x dx dx . (x 1)2 (x dV dx dS dx dy dx 1) dV dr 36. Note that sec (x 2 1) tan (x 2 1) tan (x 2 2(1.5) sec (1.52 1) (2x), 1 ) dx. 1) tan (1.52 0.15 sec 1.25 tan 1.25 27. (a) f f (0.1) (c) f df f f (1.1) f (0) 0 0.21 0.2 f (1) 1) (0.05) 0.21 0.01 0.231 0 1, f (1) (b) Since f (x) 3x Therefore, df 2dx 2(0.1) f df dr, the change in volume is approximately 0.231 0.2 0.231 2. 0.2. 0.031 a to a dS dh 2 r, so dS 2 r dh. When h changes from dh, the change in lateral surface area is 37. (a) Note that f (0) cos 0 1. L(x) f (0) f (0)(x 0) 1 1.431663. 0.21 2 (c) 2 rh dr. When r changes approximately 2 r dh. (b) Since f (x) 2x 2, f (0) 2. Therefore, df 2 dx 2(0.1) 0.2. 28. (a) 2 rh, so dV 2 ah dr. (b) At the given values, dy 12x dx. When x changes from 0.01. 2x sec (x 2 dy 12x, so dS 2, from a to a 26. (a) Since 3x 2 dx. When x changes from dx, the change in surface area is approximately 35. Note that 0.01 (0 1)2 3x 2, so dV 12a dx. 1 (b) At the given values, dy 8 r dr. When r changes from dx, the change in volume is approximately 3a 2 dx. 34. Note that x y 8 r, so dS dr, the change in surface area is approximately 33. Note that 0 x) dS dr 8 a dr. 1 1 cot 1 (0.1) 3 3 2 2 0.1 csc cot 0.205525 3 3 xy 4 a 2 dr. a to a csc 1 y dr, the change in volume is approximately 32. Note that (b) At the given values, dy 4 r 2 dr. When r changes from (b) f (0.1) L(0.1) 1x x 1 1.1 (c) The actual value is less than 1.1. This is because the derivative is decreasing over the interval [0, 0.1], which means that the graph of f (x) is concave down and lies below its linearization in this interval. Section 4.5 dA dr r 2 and 38. (a) Note that A 2 r, so dA When r changes from a to a 2 r dr. dr, the change in area is approximately 2 a dr. Substituting 2 for a and 0.02 for 43. Let angle of elevation and h dh 30 tan , so d h want dh 2 30 sec height of building. Then 0.04h, which gives: dA (b) A 0.08 4 39. Let A D 0.08 dA D2 2 (2) C dD 1 1 and dD dC. dC C2 dA C , so and 4 dC 2 40. Let x 10 (2) 2 edge length and V 3x2 dx. With x have V 103 0.04 sin 5 radians, we have 12 5 5 0.04 sin cos 0.01 radian. The angle should 12 12 75 be measured with an error of less than 0.01 radian (or approximately 0.57 degrees), which is a percentage error of approximately 0.76%. 44. Note that dV dh 3 h 2, so dV 3 h 2 dh. We want 0.01V, which gives 3 h 2 dh dV 0.01( h 3), or 0.01h . The height should be measured with an error 3 1 of no more than %. 3 dh 2 10 in . volume. Then V x 3, and so r 2h 45. (a) Note that V 10 cm and dx 1000 cm3 and dV cos in. to 0.6366 in. and the area increases by approximately dA dV , so C2 2 2 0.04 sin cos d circumference, and 2 in. the diameter increases by 1 d For C dC. When C increases from 10 2 10 dD 2% cross section area, C Also, A 0.04 (30 tan ) 0.2513 0.02 diameter. Then D d 1 cos2 d 2 (2)(0.02) 30 sec2 d . We and dh 30 sec2 dr, the change in area is approximately 173 0.01x 10 r 2 2.5 D2, where D is the 0.1 cm, we 3(10)2(0.1) 30 cm3, interior diameter of the tank. Then dV 5 D dD. We want dV dV dD 5 D, so 0.01V, which gives so the percentage error in the volume measurement is approximately 41. Let x dA dx dV V 30 1000 side length and A 2x, so dA gives 2x dx 5 D dD 0.03 0.01(2.5 D 2), or dD 0.005D. The 3%. interior diameter should be measured with an error of area. Then A 2x dx. We want dA 0.02x 2, or dx x2 and no more than 0.5%. 0.02A, which (b) Now we let D represent the exterior diameter of the 0.01x. The side length tank, and we assume that the paint coverage rate should be measured with an error of no more than 1%. (number of square feet covered per gallon of paint) is 42. The volume of a cylinder is V fixed, we have For h dV dr 30 in., r 2 r h. When h is held 2 rh, and so dV 6 in., and dr 2 rh dr. paint within 5%, we need to calculate the lateral 0.5 in., the thickness of the shell is approximately dV 2 rh dr 2 (6)(30)(0.5) known precisely. Then, to determine the amount of surface area S with an error of no more than 5%. Note that S 180 3 565.5 in . dS 2 rh 10 D, so 10 dD. We want dS 10 dD dS dD 10 and 0.05S, which gives 0.05(10 D), or dD 0.05D. The exterior diameter should be measured with an error of no more than 5%. 174 Section 4.5 r 2h, we have 46. Since V We want dV dV dr 2 rh, and dV 52. (a) i. Q(a) f (a) implies that b0 f (a). ii. Since Q (x) b1 2b2(x implies that b1 f (a). 0.0005r. The variation of a), Q (a) iii. Since Q (x) 0.001V, which gives 0.001 r 2h, or dr 2 rh dr 2 rh dr. f (a) implies that 1 the radius should not exceed of the ideal radius, that 2000 2b2, Q (a) f (a) f (a) 2 b2 is, 0.05% of the ideal radius. In summary, b0 47. We have Then dW dg 2 bg dWmoon , so dW b(5.2) b(32) dWearth 2 2 bg 2 dg. 2 dg dg 32 5.22 37.87. The ratio is Since f (0) 2 L 1/2g L 1/2g dT 3/2 1/2 , so dT dg L 1/2g 3/2 L1/2g (100)1/2(980) 49. If f (x) h x1 f (x1) 0 f (x1) x1 f (x1) 1 and x2 2h 1/2 h, then f (x1) f (x1) x1. h h h 1/2 x x 2. As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (d) g(x) x 1 g (x) x g (x) 2x 1 2h 1/2 and x2 h h 1 2h 1/2 1, g (1) 1, and g (1) 1, b1 1, and b2 2, the 2 2 1. The quadratic approximation is Q(x) h 1/2 2 3 coefficients are bo h, then h. If x1 2 1 1. The [ 2.35, 2.35] by [ 1.25, 3.25] 979.0235. Since g(1) 1 2 2 1, and b2 3/2 x1, and all later approximations are also 2h 1, b1 2, the (c) equal to x1. 50. If x1 1, and f (0) 3/2 0, we have x2 Therefore x2 (1 x) 2 2(1 x) 3 quadratic approximation is Q(x) dg. dg dT dg 0.001 dg 0.9765 0.9765, g 980 0.9765 Since dg coefficients are b0 and (b) Note that dT and dg have opposite signs. Thus, if g increases, T decreases and the clock speeds up. (c) 1, f (0) f (a) . 2 f (a), and b2 (b) f (x) (1 x) 1 f (x) 1(1 x) 2( 1) f (x) 2(1 x) 3( 1) about 37.87 to 1. 48. (a) Note that T f (a), b1 2h 1 (x 1) (x 1)2. h [ 1.35, 3.35] by [ 1.25, 3.25] As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. [ 3, 3] by [ 0.5, 2] 51. Note that f (x) xn xn1/3 xn 2/3 1 2/3 x and so xn 1 3 xn 3xn xn 2xn. For x1 f (xn) f (xn) 1, we have 3 x2 2, x3 4, x4 [ 10, 10] by [ 3, 3] 8, and x5 16; xn 2n 1. Section 4.6 (e) h(x) h (x) h (x) Since h(0) x)1/2 (1 55. g(a) 1 (1 x) 1/2 2 1 (1 x) 3/2 4 1 1, h (1) , and h (1) 2 Then E(x) f (x) 0, then g(a) g(x) f (x) f (a) and c f (a) m(x f (a). a). f (x) f (a) E(x) m. xa xa f (x) f (a) E(x) lim f (a), so lim f (a) m. xa a x→a x→a x E(x) Therefore, if the limit of is zero, then m f (a) and xa Thus, 1 , the 4 1 4 1 , and b2 2 1, b1 coefficients are b0 c, so if E(a) 175 The quadratic approximation is Q(x) 1 . 8 2 g(x) L(x). x2 . 8 x 2 1 s Section 4.6 Related Rates (pp. 232–241) Exploration 1 Sliding Ladder 1. Here the x-axis represents the ground and the y-axis represents the wall. The curve (x1, y1) gives the position of [ 1.35, 3.35] by [ 1.25, 3.25] the bottom of the ladder (distance from the wall) at any As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (f) The linearization of any differentiable function u(x) at x a is L(x) u(a) u (a)(x a) b0 b1(x a), where b0 and b1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization for f (x) at x for g(x) at x 1 is 1 0 is 1 (x linearization for h(x) at x x; the linearization 1) or 2 13 . The curve (x2, y2) gives the position 3 13 of the top of the ladder at any time in 0 t . 3 13 2. 0 t 3 time t in 0 3. This is a snapshot at t 3.1. The top of the ladder is moving down the y-axis and the bottom of the ladder is moving to the right on the x-axis. Both axes are hidden from view. x; and the x . 2 0 is 1 53. Just multiply the corresponding derivative formulas by dx. d (a) Since (c) dx 0, d(c) du dx d (c) Since (u + v) dx d (d) Since (u dx (e) Since du dx v d (f) Since u n dx 54. tan x x→0 x lim 0. v) du v dx v2 nu n 1 du dx v) du v) , d(u ) n1 sin x/cos x x x→0 1 sin x lim x→0 cos x x sin x 1 lim lim x→0 cos x x→0 x 0.348 ft/sec , y (1) 0.712 ft/sec2, 1.107 ft/sec2, y (2) 1.561 ft/sec2. u dv v2 nu du. . v du. 1. , the speed of the top of the ladder t→(13/3) is infinite as it hits the ground. Quick Review 4.6 1. D 2. D (7 0)2 (b 2 0) (0 5)2 49 25 (0 2 2 2 a) a 3. Use implicit differentiation. lim (1)(1) 9t 2 2 dv u dv v du n 132 Since lim y (t) u v ,d 9t y (t) y (1.5) du v , d(u dx dv u dx dy dt y (0.5) c du. dv , d(u dx dv u dx [ 1, 15] by [ 1, 15] 4. du c , d(cu) dx d (b) Since (cu) dx t d (2xy dx dy 2x dx 2y(1) (2x 2y y 2) dy 2y dx dy 1) dx dy dx d (x dx (1) 1 2y 1 2x y) dy dx 2y 2y 1 b 74 176 Section 4.6 4. Use implicit differentiation. d (x sin y) dx (x)(cos y) dy dx (x (sin y)(1) x cos y) 3. (a) Since d (1 dx dy x dx dy dx dy dx dy dx y 2x dy dx dy dx dV dh dV , we have dh dt dt (b) Since dV dt dV dr dV , we have dr dt dt xy) y(1) sin y (c) y sin y x x cos y y sin y x x cos y 5. Use implicit differentiation. d2 x dx dV dt 4. (a) d tan y dx dy sec2 y dx 2x sec2 y 2x cos2 y dV dt dV dt dV dt d2 rh dt dh r2 dt dh r2 dt dP dt dP dt dP dt dP dt d ln (x y) dx dy 1 1 dx xy dy 1 dx dy dx dh . dt dr dt 2 rh . d (RI 2) dt dR d R I2 I2 dt dt dI dR R 2I I2 dt dt dI dR 2RI I2 dt dt d2 (r h) dt dr h(2r) dt dr 2 rh dt (b) If P is constant, we have 6. Use implicit differentiation. r2 2RI d (2x) dx dI dt dR dt 0, or y2 I2 dP dt 0, which means dR dt 2R dI I dt 2P dI . I 3 dt z2 2 2(x 2x 5. y) 2y 8. Using A(0, 4), we create the parametric equations x 0 at and y 4 bt, which determine a line passing through A at t 0. We now determine a and b so that the line passes through B(5, 0) at t 1. Since 5 0 a, we have a 5, and since 0 4 b, we have b 4. Thus, one parametrization for the line segment is x 5t, y 4 4t, 0 t 1. (Other answers are possible.) 10. One possible answer: 2 3 2 d dt x2 1 ds dt 7. Using A( 2, 1) we create the parametric equations x 2 at and y 1 bt, which determine a line passing through A at t 0. We determine a and b so that the line passes through B(4, 3) at t 1. Since 4 2 a, we have a 6, and since 3 1 b, we have b 4. Thus, one parametrization for the line segment is x 2 6t, y 1 4t, 0 t 1. (Other answers are possible.) 9. One possible answer: ds dt t dA dt dA dr dA , we have dr dt dt 2. Since dS dt dS dr dS , we have dr dt dt 6. 2x 2 y y 2 2 dx x dt dy y dt x2 dA dt dA dt dA dt y2 (x 2 y2 dx dt 2y z 2) z dt 1 z z 2 2x dy dt 2z dz dt dz dt z2 d1 ab sin dt 2 1 da db b sin a 2 dt dt 1 da db b sin a sin 2 dt dt dV dt 2 sin ab ab cos d sin dt d dt 1 volt/sec. (b) Since I is decreasing at the rate of dI dt Section 4.6 Exercises 1. Since ds dt 2x d 2 7. (a) Since V is increasing at the rate of 1 volt/sec, 3 2 t ds dt 1 2 dr dt 2r . dr dt 8r . 1 amp/sec. 3 (c) Differentiating both sides of V dV dt 1 amp/sec, 3 I dR dt dI dt R. IR, we have Section 4.6 (d) Note that V IR gives 12 2R, so R 6 ohms. Now substitute the known values into the equation in (c). 1 2 dR dt (b) P 1 3 6 dR 2 dt 3 ohms/sec 2 3 dR dt 3 ohms/sec. Since this 2 At the instant in question, dr dt 0.01 cm/sec, r dA . dt Step 5: 2r dr dt 52 At the instant in question, cm2/sec 2 (50)(0.01) At the instant in question, the area is increasing at the rate cm2/sec. At the instant in question, dw dt 2 cm/sec, 2 cm/sec, l 12 cm, 5 cm. dy dt y = 3 m, and z We want to find 2 m/sec, dz dt 1 m/sec, x 4 m, 2 m. dA dP dD , , and . dt dt dt dV dS ds , , and . dt dt dt Steps 4, 5, and 6: (a) V xyz xy dz dt xz (4)(3)(1) dy dt yz dx dt (4)(2)( 2) dS dt dS dt 2(xy xz dy 2x dt yz) dx y dt x dz dt z dx dt y lw 2[(4)( 2) (3)(1) (3)(1) (2)( 2)] dz dt z dy dt (4)(1) (2)(1) Steps 4, 5, and 6: 2 m3/sec (3)(2)(1) The rate of change of the volume is 2 m3/sec. (b) S Step 3: dw l dt 1 m/sec, dV dt dV dt Step 2: We want to find dx dt Step 3: 9. Step 1: l length of rectangle w width of rectangle A area of rectangle P perimeter of rectangle D length of a diagonal of the rectangle dA dt dA dt dl 2l 2 2 dt 2l w (12)( 2) (5)(2) 122 dw l w dt dt dw 2w 2 dt l w2 14 cm/sec 13 1 Step 2: Step 6: (a) A w2 10. Step 1: x, y, z edge lengths of the box V volume of the box S surface area of the box s diagonal length of the box r2 and w 0 cm/sec (d) The area is increasing, because its derivative is positive. The perimeter is not changing, because its derivative is zero. The diagonal length is decreasing, because its derivative is negative. Step 4: dl dt 2(2) 14 cm/sec. 13 50 cm. Step 3: of 2( 2) The rate of change of the length of the diameter is Step 2: dA dt dw dt dl dD dt dD dt 8. Step 1: r radius of plate A area of plate dA dt 2 l2 (c) D value is positive, R is increasing. A 2w dl 2 dt The rate of change of the perimeter is 0 cm/sec. R is changing at the rate of We want to find 2l dP dt dP dt 177 0 m2/sec The rate of change of the surface area is 0 m2/sec. dl w dt (12)(2) (5)( 2) 14 cm2/sec The rate of change of the area is 14 cm2/sec. 178 Section 4.6 13. Step 1: x distance from wall to base of ladder y height of top of ladder A area of triangle formed by the ladder, wall, and ground angle between the ladder and the ground 10. continued x2 (c) s y2 z2 1 ds dt 2 x 2 y2 2x z2 dx dt 2y dy dt 2z dz dy Step 2: dx x dt dy y dt x 2 y dz z dt 2 z (4)(1) ds dt At the instant in question, x 2 (2)(1) 32 0 22 29 Step 4, 5, and 6: 11. Step 1: s (diagonal) distance from antenna to airplane x horizontal distance from antenna to airplane ds dt Step 4: x 2 49 y 300 mph. dx . dt s2 x2 169 169 122 5. rate of 12 ft/sec. (Note that the downward rate of 49 motion is positive.) 2s s2 49 ds dt s s2 ds 49 dt 10 102 (300) 3000 49 mph 1 xy 2 1 dy x 2 dt (b) A Step 6: dx dt 0 The top of the ladder is sliding down the wall at the s 2 or x 1 2 169 dy 2y dt dy Then 2(12)(5) 2(5) 0 dt dy dy 12 ft/sec or 12 ft/sec dt dt Step 5: dx dt y2 To evaluate, note that, at the instant in question, Step 3: We want to find (a) x 2 dx 2x dt Step 2: At the instant in question, 10 mi and 5 ft/sec. dy dA d , , and . dt dt dt We want to find 0 m/sec The rate of change of the diagonal length is 0 m/sec. s dx dt Step 3: (3)( 2) 42 12 ft and dA dt 420.08 mph y dx dt Using the results from step 2 and from part (a), we 51 The speed of the airplane is about 420.08 mph. have 12. Step 1: h height (or depth) of the water in the trough V volume of water in the trough dA dt 1 [(12)( 12) 2 (5)(5)] 119 ft/sec. 2 The area of the triangle is changing at the rate of 59.5 ft2/sec. Step 2: At the instant in question, dV dt 2.5 ft3/min and h 2 ft. y x (c) tan 2 x d dt Step 3: sec dh We want to find . dt Since tan Step 4: 4 3 cos The width of the top surface of the water is h, so we have V 1 4 (h) h (15), or V 2 3 10h 2 20h y dx dt x2 5 , we have for 0 12 1 12 and so sec2 13 dh dt d dt Step 6: of dh dt 2.5 20(2) dh dt 0.0625 1 ft/min 16 The water level is increasing at the rate of 1 ft/min. 16 2 12 2 13 169 . 144 Combining this result with the results from step 2 and from part (a), we have Step 5: dV dt dy dt 169 d 144 dt (12)( 12) (5)(5) , so 122 1 radian/sec. The angle is changing at the rate 1 radian/sec. Section 4.6 14. Step 1: 179 Step 5: Kite s Inge dV dt 12 r dr dt Step 6: 300 ft dV dt x 1 3000 12 (1.900) 19 2500 0.0076 3 0.0239 in /min s x length of kite string horizontal distance from Inge to kite The volume is increasing at the rate of approximately 0.0239 in3/min. Step 2: At the instant in question, dx dt 25 ft/sec and s 500 ft 16. Step 1: Step 3: We want to find h ds . dt Step 4: x 2 r 300 2 s 2 r h V Step 5: 2x dx dt 2s ds dx or x dt dt s ds dt base radius of cone height of cone volume of cone Step 2: Step 6: At the instant in question, since x 2 x s2 3002 5002 ds dt Thus (400)(25) (500) , so 3002 3002 ds dt s 2, we have At the instant in question, h 4 m and dV dt 10 m3/min. Step 3: 400. 20 ft/sec. Inge must let the string out at the rate of 20 ft/sec. We want to find dh dr and . dt dt Step 4: 15. Step 1: Since the height is 3 of the base diameter, we have 8 4 h. 3 12 142 16 h 3 We also have V rh hh . We will 3 33 27 16 h 3 4 use the equations V and r h. 27 3 h 6 in. 3 (2r) or r 8 Steps 5 and 6: r (a) The cylinder shown represents the shape of the hole. r radius of cylinder V volume of cylinder Step 2: At the instant in question, dr dt 0.001 in. 3 min and (since the diameter is 3.800 in.), r Step 3: Step 4: r 2(6) 1.900 in. 10 dh dt 16 h 2 dh 9 dt 16 (4)2 dh 9 dt 45 m/min 128 1125 cm/min 32 The height is changing at the rate of 1125 32 11.19 cm/min. (b) Using the results from Step 4 and part (a), we have dV We want to find . dt V 1 in./min 3000 dV dt dr dt 4 dh 3 dt 4 1125 3 32 375 cm/min. 8 The radius is changing at the rate of 6 r2 375 8 14.92 cm/min. 180 Section 4.6 17. Step 1: Step 6: 45 m r 6 6 6m 144 dy dt h or r h V radius of top surface of water depth of water in reservoir volume of water in reservoir r dV dt At the instant in question, 50 m3/min and h 5 m. Step 3: We want to find 0.01326 m/min 25 6 1.326 cm/min y)2 (13 169 132, y)2 (13 26y dh dr and . dt dt Step 2: At the instant in question, Note that Then V h 6 by similar cones, so r 7.5h. r 45 12 1 rh (7.5h)2h 18.75 h3 3 3 We want to find the value of 18.75 h 3, dV dt 56.25 h 2 dh . dt From part (b), r 7.5h, dr dt dr dt dh dt 80 cm/min. The rate 3 or of change of the radius of the water’s surface is 80 3 18. (a) Step 1: y depth of water in bowl V volume of water in bowl 1 y2 26y 2 (26 2y) dy dt 13 26y y dy y 2 dt 13 8 26(8) 5 288 125 72 1 24 82 5 12 1 24 0.00553 m/min 0.553 cm/min Step 2: No numerical information is given. Step 2: dV dt y 2. 26y 19. Step 1: r radius of spherical droplet S surface area of spherical droplet V volume of spherical droplet 8.49 cm/min. At the instant in question, dr . dt Step 6: dr dt 7.5 1 m/min. 24 Step 5: falling is positive.) (b) Since r dy dt 8 m, Step 4: 56.25 (52) 50 6 m3/min, y Step 3: dh , and so dt dh 8 32 m/min cm/min. dt 225 9 32 The water level is falling by 1.13 cm/min. 9 dh (Since 0, the rate at which the water level is dt Thus dV dt and therefore (from part (a)) Step 5 and 6: (a) Since V y 2. (c) Step 1: y depth of water r radius of water surface V volume of water in bowl Step 4: y dy dt dy dt 1 24 (b) Since r 2 Step 2: (82)] [26 (8) 6 m3/min and Step 3: We want to show that 8 m. Step 3: Step 4: dy We want to find the value of . dt S Step 4: dr is constant. dt Steps 5 and 6: V 3 y 2(39 y) or V Step 5: dV dt (26 y y 2) dy dt 13 y 2 3 y3 4 r 2, V 4 3 dV r, 3 dt kS for some constant k 43 dV dr r , we have 4 r2 . 3 dt dt dV Substituting kS for and S for 4 r 2, we have kS dt dr or k. dt Differentiating V dr dt S, Section 4.6 20. Step 1: r radius of spherical balloon S surface area of spherical balloon V volume of spherical balloon (b) 16 cos l 1 d dt Step 2: At the instant in question, dV dt 100 ft3/min and r 1 d dt dr dS We want to find the values of and . dt dt 6 dl l 2 dt 62 l 5 ft. Step 3: 1 0.6 1 6 ( 2) 102 2 The rate of change of angle 3 radian/sec 20 3 radian/sec. 20 is Steps 4, 5, and 6: (a) V dV dt dr dt 4 (5)2 100 dr dt 22. Step 1: x distance from origin to bicycle y height of balloon (distance from origin to balloon) s distance from balloon to bicycle 43 r 3 4 r2 dr dt Step 2: 1 ft/min dS dt dS dt dS dt 4 r2 8r dy dx is a constant 1 ft/sec and is a constant dt dt We know that The radius is increasing at the rate of 1 ft/min. (b) S 17 ft/sec. Three seconds before the instant in question, the values of x and y are x 0 ft and y 65 ft. Therefore, at the instant in question x 51 ft and y 68 ft. Step 3: dr dt 8 (5)(1) We want to find the value of 40 ft2/min ds at the instant in question. dt Step 4: x2 The surface area is increasing at the rate of s 40 ft2/min. y2 Step 5: ds dt 21. Step 1: l length of rope x horizontal distance from boat to dock angle between the rope and a vertical line dl At the instant in question, dt 2 ft/sec and l ds dt 10 ft. 2 dx dt dx dt 2 51 dc dt l2 102 dx dt x2 y dy dt y2 11 ft/sec 682 6x 2 12x [3(2)2 15x) 15) 12(2) dx dt 15](0.1) 0.3 dr dt dp dt dl 36 dt 10 dy 2y dt (68)(1) d3 (x dt (3x 2 dx d and . dt dt 36 l x2 (51)(17) Step 4, 5, and 6: l2 dx 2x 2 dt y The distance between the balloon and the bicycle is increasing at the rate of 11 ft/sec. 23. (a) Step 3: We want to find the values of x 1 Step 6: Step 2: (a) x 181 ( 2) 2.5 ft/sec 36 The boat is approaching the dock at the rate of (b) dx d (9x) 9 9(0.1) 0.9 dt dt dr dc 0.9 0.3 0.6 dt dt dc dt d3 x dt 3x 2 2.5 ft/sec. 45 x 45 dx x 2 dt 6x 2 12x 3(1.5)2 12(1.5) 45 (0.05) 1.52 1.5625 dr dt dp dt d (70x) dt dr dc dt dt 70 dx dt 3.5 70(0.05) ( 1.5625) 3.5 5.0625 182 Section 4.6 24. (a) Note that the level of the coffee in the cone is not needed until part (b). Step 1: V1 volume of coffee in pot y depth of coffee in pot Step 2: dV1 dt Step 2: 10 in3/min At the instant in question, Q Step 3: We want to find the value of dy dt dD dt . dt y dy dt dy dt Q D Step 5: Step 6: 9 0 mL/min . Step 4: 9 10 41 mL/L, 2 dy . dt We want to find the value of 9y Step 5: dV1 233 mL/min, D dQ 2 (mL/L)/min, and dt Step 3: Step 4: V1 25. Step 1: Q rate of CO2 exhalation (mL/min) D difference between CO2 concentration in blood pumped to the lungs and CO2 concentration in blood returning from the lungs (mL/L) y cardiac output dy dt 10 9 dy dt 0.354 in./min D dQ dt Q dD dt D2 The level in the pot is increasing at the rate of Step 6: approximately 0.354 in./min. dy dt (b) Step 1: V2 volume of coffee in filter r radius of surface of coffee in filter h depth of coffee in filter (41)(0) 466 1681 0.277 L/min2 The cardiac output is increasing at the rate of approximately 0.277 L/min2. 26. Step 1: y Step 2: At the instant in question, h (233)( 2) (41)2 dV2 dt 10 in3/min and (x , y ) 5 in. Step 3: We want to find dh . dt x Step 4: Note that 3 , so r 6 r h 12 rh 3 Then V2 h . 2 h3 . 12 x y Step 5: dV2 dt h 2 dh 4 dt x-coordinate of particle’s location y-coordinate of particle’s location angle of inclination of line joining the particle to the origin. Step 2: Step 6: At the instant in question, (5)2 dh 10 4 dt dh 8 in./min dt 5 dh Note that 0, so the rate at which the level is dt dx dt Step 3: falling is positive. The level in the cone is falling at the Step 4: 8 rate of 5 Since y 0.509 in./min. 10 m/sec and x We want to find x 0, 3 m. d . dt x 2, we have tan tan 1 x. y x x2 x x and so, for Section 4.6 Step 5: d dt 1 1 28. Step 1: x x-coordinate of particle y y-coordinate of particle D distance from origin to particle dx x 2 dt Step 6: d dt 1 32 1 183 (10) Step 2: 1 radian/sec At the instant in question, x The angle of inclination is increasing at the rate of dx dt 1 radian/sec. 1 m/sec, and dy dt 5 m, y 12 m, 5 m/sec. Step 3: 27. Step 1: We want to find y dD . dt Step 4: x2 D (x , y ) y2 Step 5: θ dD dt x x 1 2 dx 2x dt y2 x2 dy 2y dt dx dt x2 y dy dt y2 Step 6: x y (5)( 1) dD dt x-coordinate of particle’s location y-coordinate of particle’s location angle of inclination of line joining the particle to the origin (12)( 5) 52 5 m/sec 122 The particle’s distance from the origin is changing at the rate of 5 m/sec. 29. Step 1: Step 2: At the instant in question, dx dt 8 m/sec and x Street light 4 m. Step 3: We want to find d . dt 16 ft Step 4: 6 ft Since y x, we have tan and so, for x y x x 0, 1 tan [ ( x) 1/2 x tan 1 ( x) ( x) 1/2 . Shadow 1/2 , x x s s distance from streetlight base to man length of shadow Step 2: Step 5: d dt 1 1 [( x) 1 1 x 1 1/2 2 x(x 1 dx ( x) 3/2( 1) 2 dt dx 1 2( x)3/2 dt 1 2 10 ft. Step 3: We want to find ds . dt By similar triangles, dx 1) dt 16s 6s 6x, or s s s 6 x 16 . This is equivalent to 3 x. 5 Step 5: 1 2 5 ft/sec and x Step 4: Step 6: d dt dx dt At the instant in question, 4( 4 ( 8) 1) 2 radian/sec 5 ds dt 3 dx 5 dt The angle of inclination is increasing at the rate of Step 6: 2 radian/sec. 5 ds dt 3 ( 5) 5 3ft/sec The shadow length is changing at the rate of 3 ft/sec. 184 Section 4.6 Step 2: 30. Step 1: s distance ball has fallen x distance from bottom of pole to shadow At the instant in question, dV dt Step 2: At the instant in question, s ds dt 1 32 2 16 12 2 4 ft and Step 4: dx . dt 43 r and S 3 We have V Step 4: x 30 By similar triangles, 50 s x 1500 50x 1 1500s x . This is equivalent to 50 sx, or sx 1500. We will use 2 ds dS dt 1500(4) 2(16) 1500 ft/sec. 31. Step 1: x position of car (x 0 when car is right in front of you) camera angle. (We assume is negative until the car passes in front of you, and then positive.) At the first instant in question, x 1 (264) 2 A half second later, x 0 ft and dx dt 264 ft/sec. 2 1/3 dV 3V 4 dt 4 4000 (10)3 . 3 3 1/3 4000 ( 8) 3 2 33. Step 1: p x-coordinate of plane’s position x x-coordinate of car’s position s distance from plane to car (line-of-sight) At the instant in question, dp dt 0, 120 mph, s d at each of the two instants. dt Step 4: (x Step 4: 1 x 132 p)2 32 s2 Step 5: 2(x Step 5: 1 1 dx 132 dt x2 132 dx dt p) dp dt 2s ds dt Step 6: Note that, at the instant in question, Step 6: 1 d 0: dt 132: 5 mi, and dx . dt We want to find We want to find When x 3 dV 4 dt Step 3: Step 3: When x 1/3 3V 4 area is decreasing at the rate of 1.6 cm2/min. p 132 ft and 264 ft/sec. 1 2 3 4 Step 2: Step 2: d dt 3V 2/3 4 16 3 1.6 cm2/min 3 4 1000 dS Since 0, the rate of decrease is positive. The surface dt 1500 ft/sec The shadow is moving at a velocity of tan 4 Step 5: Note that V dt Step 6: dx dt 3V 1/3 , so S 4 Step 6: 1500s dx dt 4 r 2. These equations can be combined by noting that r dS dt . Step 5: dx dt dS . dt We want to find Step 3: 50x 1 (20) = 10 cm. 2 Step 3: 16 ft/sec. We want to find 8 cm3/min and r 8 mL/min 02 132 1 d dt 1 (264) 132 1 1 132 132 2 32. Step 1: r radius of balls plus ice S surface area of ball plus ice V volume of ball plus ice 1 (264) 132 52 x 2 radians/sec 1 radian/sec 2(4 32 dx dt dx 8 dt dx dt 0) 4 mi. 120 2(5)( 160) 120 120 dx dt The car’s speed is 80 mph. 1600 200 80 mph ds dt 160 mph. 185 Section 4.6 Step 6: 34. Step 1: s shadow length sun’s angle of elevation d dt (20)( 2) (10)(1) 102 202 0.1 radian/sec 5.73 degrees/sec Step 2: To the nearest degree, the angle is changing at the rate At the instant in question, s 60 ft and d dt 0.27 /min of 0.0015 radian/min. 6 degrees per second. 36. Step 1: Step 3: A ds . dt We want to find Step 4: c a 80 or s s tan 80 cot 120° Step 5: ds dt O 80 csc d dt 2 a b c Step 6: 80 and 60 Note that, at the moment in question, since tan 0 2 4 and so csc 5 , we have sin distance from O to A distance from O to B distance from A to B At the instant in question, a b 5 nautical miles, da dt 14 knots, and c2 c2 3 nautical miles, a2 a2 db dt 21 knots. Step 3: 2.25 in./min We want to find 7.1 in./min ds Since dt B Step 2: 5 . 4 52 80 (0.0015 ) 4 12 in ft 0.1875 1 ft min ds dt b dc . dt Step 4: Law of Cosines: 0, the rate at which the shadow length is b2 b2 2ab cos 120 ab decreasing is positive. The shadow length is decreasing at Step 5: the rate of approximately 7.1 in./min. 2c 35. Step 1: a distance from origin to A b distance from origin to B angle shown in problem statement 10 m, and b da dt 2 m/sec, a2 db dt dc dt dc 14 dt dc dt 2(7) 1 m/sec, 20 m. Step 3: We want to find d . dt tan 37. a tan 1 b Step 5: d dt b 1 1 a2 b da dt a b2 2b db dt a db dt b da dt b2 (5)2 ab 2(5)(14) 2(3)(21) (3)2 (5)(3) (5)(21) 49 (3)(14) 413 29.5 knots The ships are moving apart at a rate of 29.5 knots. Step 4: a or b da dt Note that, at the instant in question, c At the instant in question, 2a Step 6: Step 2: a dc dt db dt b da dt a db dt a2 b2 dy dt dy dx dx 10(1 x 2) 2(2x) dx dt dt dx Since 3 cm/sec, we have dt dy 60x cm/sec. dt (1 x2)2 (a) dy dt [1 60( 2) ( 2)2]2 (b) dy dt (1 60(0) 0 2)2 (c) dy dt 60(20) (1 202)2 120 52 (1 24 cm/sec 5 0 cm/sec 0.00746 cm/sec dx 20x x 2 ) 2 dt 7 186 38. Chapter 4 Review dy dt dy dx dx dt dx Since dt (3x 2 4) 41. (a) 2 cm/sec, we have dy dt 8 (b) dy dt 8 6(1)2 (c) dy dt 8 6(4)2 (a) dx dt 6( 3)2 dy dt dx dt dy dt 2 cm/sec (b) At dx dt dy dt 2 dx dt dy dt cos 2) 32 cos d (uv) dt 1. y x2 1 x ( 1) 2 3x 4 22 32 2 2 100.531 ft/sec dx dt dy dt x)(1) x 4 3 x : 32 sin 4 3 x Endpoint values: 0 ft/sec Critical point value: 0 ft/sec 32 cos y 2 The global maximum value is 32 100.531 ft/sec minimum value is 4 at x 46 9 y 2 x 4 y 4 6 0 4 , and the global 3 at x 9 1.09 2. 2. Since y is a cubic function with a positive leading 30 sin coefficient, we have lim y and lim y x→ . There are x→ no global extrema. 3. y 2 2 (x 2)(e 1/x )( 2x 3) 2e 1/x 1 x 2 2e 1/x (x In general: At t 2 x 2 dx dt dy dt ( 22 x x 2(2 x) 71.086 ft/sec (b) Since the ferris wheel makes one revolution every 10 sec, we may let 0.2 t and we may write x 30 cos 0.2 t, y 40 30 sin 0.2 t. (This assumes that the ferris wheel revolves counterclockwise.) At t du dt v The first derivative has a zero at . 40. (a) One possible answer: x 30 cos , y 40 dx dt dy dt dv dt x 2 16 ( 2) u s Chapter 4 Review (pp. 242–245) 71.086 ft/sec : 32 sin At 16 ( dy dt u(0.03v) v( 0.02u) 0.01uv 0.01y The total production is increasing at the rate of 1% per year. y 4 0.09y, the rate of growth of total production 32 sin : 32 cos v(0.04u) is 9% per year. 88 cm/sec 4 du dt v 0.09y dy Since dt d 2 sin 2(sin )(16 ) dt d 2 cos 2(cos )(16 ) 32 dt 32 sin dv dt u(0.05v) 46 cm/sec (c) In general, assuming counterclockwise motion: 4 u 0.09uv (b) One possible answer is 16 t, where t is in seconds. (An arbitrary constant may be added to this expression, and we have assumed counterclockwise motion.) At d (uv) dt 6x 2 cm/sec. 8 39. (a) The point being plotted would correspond to a point on the edge of the wheel as the wheel turns. dx dt dy dt dy dt (e 1/x )(2x) x 1)(x 1) x 30(sin 0.2 t)(0.2 ) 6 sin 0.2 t Intervals 30(cos 0.2 t)(0.2 ) 6 cos 0.2 t 5: 6 sin 6 cos 1 0 ft/sec 6 ( 1) y 18.850 ft/sec x 00 x 1 x 1 Sign of y Increasing 2 d [2e 1/x ( x 1 dx 2 1) (x 2 2 1 2x (2e 1/x )(x 17.927 ft/sec 1/x 2 (x 4 x 2 4 2) x4 5.825 ft/sec 2 2e 1/x [(x 2 Decreasing Increasing x)] 2 (2e 1/x )(x 2e 6 cos 1.6 1 Behavior of y Decreasing 8: 6 sin 1.6 x 0.5)2 x4 1.75] 1 2 x)(2e 1/x )( 2x 3) 2x 2 ) Chapter 4 Review 5. y 1 2x 4x 3 Using grapher techniques, the zero of y is x The second derivative is always positive (where defined), so the function is concave up for all x 0. Graphical support: Intervals x 0.385 0.385 187 0.385. x Sign of y Behavior of y Increasing Decreasing (a) [ 1, 0) and [1, ) y 2 12x 2 2(1 6x 2) The second derivative is always negative so the function is concave down for all x. (b) ( , Graphical support: (c) ( , 0) and (0, ) [ 4, 4] by [ 1, 5] 1] and (0, 1] (d) None (e) Local (and absolute) minima at (1, e) and ( 1, e) (f) None [ 4, 4] by [ 4, 2] 4. Note that the domain of the function is [ 2, 2]. y x 1 ( 2x) 2 4 x2 x 2 (4 x 2 ) ( (a) Approximately ( 4 x )(1) (c) None (d) ( 4 x2 2x 2 4 4 x ,) (e) Local (and absolute) maximum at (0.385, 1.215) (f) None 2 ex 6. y Intervals , 0.385] (b) Approximately [0.385, ) 2 2 2 x 2 x 2 2 x 2 1 1 Intervals Sign of y x 1 1 x Sign of y Behavior of y ( 4 Decreasing x 2)( 4x) (4 Increasing 1 2x 2) 24 y 4 x x2 Decreasing ( 2x) 2 Behavior of y Decreasing Increasing ex y 1 The second derivative is always positive, so the function is concave up for all x. 2x(x 2 6) (4 x 2)3/2 Graphical support: 6 are not zeros of y because Note that the values x they fall outside of the domain. Intervals 2 x 0 0 x 2 Sign of y Behavior of y Concave up Concave down [ 4, 4] by [ 2, 4] (a) [1, ) Graphical support: (b) ( , 1] (c) ( ,) (d) None (e) Local (and absolute) minimum at (1, 0) (f) None [ 2.35, 2.35] by [ 3.5, 3.5] (a) [ 2, (b) [ 2, 2] 2] and [ 2, 2] (c) ( 2, 0) (d) (0, 2) (e) Local maxima: ( 2, 0), ( 2, 2) Local minima: (2, 0), ( 2, 2) Note that the extrema at x 2 are also absolute extrema. (f) (0, 0) 188 Chapter 4 Review Graphical support: 7. Note that the domain is ( 1, 1). y x 2) (1 1/4 1 (1 4 y x 2) x 5/4 ( 2x) Intervals 1 x 2)5/4 2(1 x 00 x [ 4.7, 4.7] by [ 3.1, 3.1] 1 (a) ( Sign of y , 1/3 (b) [ 2 Behavior of y Decreasing 5 (x)(2) (1 4 2 5/4 2(1 x ) (1) (1 2 1/4 y Increasing (c) ( 2 , 1) (d) ( 2 , 1) x ) ( 2x) ( , 0.794] [ 0.794, 1) and (1, ) 21/3) , ( , 1.260) and (1, ) ( 1.260, 1) (e) Local maximum at 5x 2] x ) [2 2x 4(1 x 2)5/2 1/3 2 1/4 x 2)5/2 4(1 1/3 2 2 3x 2 2 4(1 x 2)9/4 , 3 1 3 21/3, (f) The second derivative is always positive, so the function is concave up on its domain ( 1, 1). Graphical support: 1/3 2 1/3 2 21/3 ( 0.794, 0.529) ( 1.260, 0.420) 9. Note that the domain is [ 1, 1]. 1 y 1 x2 Since y is negative on ( 1, 1) and y is continuous, y is decreasing on its domain [ 1, 1]. d [ (1 x 2) 1/2] dx 1 (1 x 2) 3/2( 2x) 2 y [ 1.3, 1.3] by [ 1, 3] (a) [0, 1) (b) ( 1, 0] (c) ( 1, 1) Intervals 1 x (1 0 x x 2)3/2 0 x 1 (d) None Sign of y (e) Local minimum at (0, 1) Behavior of y (f) None (x 3 8. y 1)(1) (x)(3x 2) (x 3 1)2 Intervals x 2 2x 3 1 (x 3 1)2 1/3 2 1/3 Concave up Concave down Graphical support: x 1 1 x Sign of y Behavior of y Increasing Decreasing Decreasing [ 1.175, 1.175] by y (x 3 (x 3 1)2(6x 2) 2 1)(6x ) (x 3 (2x 3 1)(2)(x 3 (x 3 1)4 3 (2x 1)3 5 4 (b) [ 1, 1] (c) ( 1, 0) 2 1)(6x ) (d) (0, 1) (e) Local (and absolute) maximum at ( 1, ); local (and absolute) minimum at (1, 0) (f) x , (a) None 1)(3x 2) 6x 2(x 3 2) (x 3 1)3 Intervals 4 21/3 21/3 x 0 0 x 1 1 x Sign of y Behavior of y Concave up Concave down Concave down Concave up 0, 2 Chapter 4 Review 10. This problem can be solved graphically by using NDER to obtain the graphs shown below. y y y [ 4, 4] by [ 1, 0.3] [ 4, 4] by [ 0.4, 0.6] [ 4, 4] by [ 0.7, 0.8] An alternative approach using a combination of algebraic and graphical techniques follows. Note that the denominator of y is always positive because it is equivalent to (x (x 2 y 2x 3)(1) (x 2 2x 2 x 3 2x 3)2 (x 2 Intervals (x)(2x 3)2 2) 3 x 1)2 3 x 3 3 x Sign of y Behavior of y Decreasing (x 2 2x 3)2( 2x) (x 2 y 2x 3)( 2x) (x 2x 3 (x 2 2 Increasing Decreasing ( x 2 3)(2)(x 2 (x 2 2x 3)4 2)( x 2 2(2x 2x 3) 2x 3)(2x 2) 3) 3 18x 12 2x 3)3 Using graphing techniques, the zeros of 2x 3 x 2.584, x 0.706, and x 3.290. Intervals ( , 2.584) ( 2.584, 18x 12 (and hence of y ) are at 0.706) ( 0.706, 3.290) (3.290, ) Sign of y Behavior of y Concave down (a) [ (b) ( 3, , Concave up 3] 3] and [ 3, ) (c) Approximately ( 2.584, (d) Approximately ( , (e) Local maximum at 0.706) and (3.290, ) 2.584) and ( 0.706, 3.290) 3, 31 4 (1.732, 0.183); local minimum at ( 1.732, (f) Concave down 3, 3 4 1 0.683) ( 2.584, 0.573), ( 0.706, (3.290, 0.161) 0.338), and Concave up 2. 189 190 Chapter 4 Review 11. For x 0, y For x 0: y d 1 ln x dx x d ln ( x) dx 1 ( 1) x 1 x 1 for all x in the domain. x Thus y Intervals (0, 2) ( 2, 0) Sign of y Behavior of y Decreasing Increasing y x2 The second derivative is always negative, so the function is concave down on each open interval of its domain. Graphical support: [ 2.35, 2.35] by [ 3, 1.5] (a) (0, 2] (b) [ 2, 0) (c) None (d) ( 2, 0) and (0, 2) (e) Local (and absolute) maxima at ( 2, ln 2) and (2, ln 2) (f) None 12. y 3 cos 3x 4 sin 4x Using graphing techniques, the zeros of y in the domain 0 x 2 are x x 2.148, and x Intervals 0 0.176, x 2.965, x x 0.994, x 3.834, x 0.176 0.176 2 3 ,x 2 x 1.57, 5.591 0.994 0.994 x 22 x 2.148 2.148 x 2.965 Sign of y Behavior of y Intervals Increasing 2.965 x Decreasing 3.834 3.834 x Increasing 3 2 3 2 x Decreasing 5.591 5.591 Increasing x Sign of y Behavior of y Decreasing Increasing Decreasing Increasing 2 Chapter 4 Review y 9 sin 3x 16 cos 4x Using graphing techniques, the zeros of y in the domain 0 x 2 are x 0.542, x 1.266, x 1.876, x 2.600, x 3.425, x 4.281, x 5.144 and x 6.000. Intervals 0 x 0.542 0.542 x 1.266 1.266 x 1.876 1.876 x 2.600 2.600 x 3.425 Sign of y Behavior of y Concave down Intervals 3.425 x Concave up 4.281 4.281 x Concave up Concave down 5.144 5.144 x 6.000 6.000 x Concave down 2 Sign of y Behavior of y Concave up Concave down Concave up Concave down Graphical support: 4 , 9 4 by [ 2.5, 2.5] (a) Approximately [0, 0.176], 0.994, (b) Approximately [0.176, 0.994], 2 2 , [2.148, 2.965], 3.834, , 2.148 , [2.965, 3.834], and 3 , and 5.591, 2 2 3 , 5.591 2 (c) Approximately (0.542, 1.266), (1.876, 2.600), (3.425, 4.281), and (5.144, 6.000) (d) Approximately (0, 0.542), (1.266, 1.876), (2.600, 3.425), (4.281, 5.144), and (6.000, 2 ) (e) Local maxima at (0.176, 1.266), ,0 2 3 and (2.965, 1.266), , 2 , and (2 ,1); 2 local minima at (2.148, (0, 1), (0.994, 0.513), (3.834, and (5.591, 1.806), 1.806) Note that the local extrema at x and x (f) 0.513), 3.834, x 3 , 2 5.591 are also absolute extrema. (0.542, 0.437), (1.266, 0.267), (1.876, (5.144, 0.120), and (6.000, 0.329) 0.267), (2.600, 0.437), (3.425, 0.329), (4.281, 0.120), 191 192 Chapter 4 Review x e 13. y , 3x 2, 4 x x Intervals 0 0 x 0 0 2 2 x x 3 3 Sign of y Behavior of y Decreasing e y x , 6x , x x Intervals Increasing Decreasing 0 0 x 0 0 x Sign of y Behavior of y Concave up Concave down Graphical support: [ 4, 4] by [ 2, 4] (a) 0, 2 3 (b) ( 2 , 0] and , 3 (c) ( , 0) (d) (0, ) (e) Local maximum at 2 16 , 33 (1.155, 3.079) 3 (f) None. Note that there is no point of inflection at x at this point. 14. y 5x 4 7x 2 10x 4 Using graphing techniques, the zeros of y are x Intervals x 0.578 0.578 x 0 because the derivative is undefined and no tangent line exists 0.578 and x 1.692 1.692 x Sign of y Behavior of y Decreasing Increasing y 20x 3 14x 10 Using graphing techniques, the zero of y is x Intervals x 1.079 1.079 x Sign of y Behavior of y Concave up Concave down Graphical support: [ 4, 4] by [ 10, 25] Decreasing 1.079. 1.692. Chapter 4 Review (a) Approximately [ 0.578, 1.692] (b) Approximately ( , 0.578] and [1.692, ) (c) Approximately ( , 1.079) (d) Approximately (1.079, ) (e) Local maximum at (f) 2x 4/5 15. y (1.692, 20.517);local minimum at (1.079, 13.601) x 9/5 8 1/5 x 5 y 9 4/5 x 5 8 x 0 Intervals 9x 5 5 x 0 8 9 x 8 9 x Sign of y Behavior of y Decreasing 36 1/5 x 25 8 x 6/5 25 y 2 9 x Intervals Increasing Decreasing 4(2 9x) 25x 6/5 2 9 x 0 0 x Sign of y Behavior of y Concave up Concave down Concave down Graphical support: [ 4, 4] by [ 3, 3] (a) 0, 8 9 (b) ( , 0] and (c) , (d) 8 , 9 2 9 2 , 0 and (0, 9 ) (e) Local maximum at 8 10 , 99 8 4/5 9 local minimum at (0, 0) (f) 2 20 , 99 2 4/5 9 2 , 0.667 9 (0.889, 1.011); ( 0.578, 0.972) 193 194 Chapter 4 Review 16. We use a combination of analytic and grapher techniques to solve this problem. Depending on the viewing windows chosen, graphs obtained using NDER may exhibit strange behavior near x is actually undefined at x 5 2. The graph of y 4x 4x x2 2 x [ 5.875, 5.875] by [ 50, 30] (x y 2)( 4 2x 3 3x 2) (5 (x 2)2 16x 3 8x 10x2 (x 2)2 4x 2 4x x 3)(1) The graph of y is shown below. [ 5.875, 5.875] by [ 50, 30] The zero of y is x Intervals 0.215. x 0.215 0.215 x 2 2 x Sign of y Behavior of y Increasing Decreasing (x 2)2( 6x 2 20x 16) ( 2x 3 (x 2)4 (x y 2)( 6x 2 20x 16) (x Decreasing 2( 2x 3 2)3 2(x 3 6x 2 12x (x 2)3 10x 2 10x 2 16x 16x 3)(2)(x 3) 13) The graph of y is shown below. [ 5.875, 5.875] by [ 20, 20] The zero of x 3 6x 2 (and hence of y ) is x Intervals x 12x 13 3.710. 2 2 x 3.710 3.710 x Sign of y Behavior of y Concave down (a) Approximately ( Concave up , 0.215] (b) Approximately [0.215, 2) and (2, ) (c) Approximately (2, 3.710) (d) ( , 2) and approximately (3.710, ) (e) Local maximum at (f) (3.710, 3.420) (0.215, 2.417) Concave down 2 because, for example, NDER (y, 2) 3 2) is shown below. 5,000,000 while y Chapter 4 Review 17. y 6(x 2)2 1)(x Intervals f (x) 3 4/3 x 4 f (1) 0 C 0 24. x 1 1 x 2 2 x Sign of y Behavior of y Decreasing Increasing 3 4 Increasing 1 3 1 2 1 C y Intervals x f (x) 6(x 2) (1) (x 2)] 0 0 x 2 2 x Sign of y Behavior of y Concave up Concave down Concave up (a) There are no local maxima. (b) There is a local (and absolute) minimum at x 1. (c) There are points of inflection at x 2. 18. y 6(x 0 and at x 2) Intervals x 1 1 x 2 2 x Sign of y Behavior of y Increasing d 6(x 2 dx y 25. v(t) s(t) s(0) C s(t) s (t) 4.9t 2 10 10 4.9t 2 26. a(t) v(t) v(0) C1 x 2) 6(2x v(t) s(t) s(0) C2 s(t) e 20. Since d sec x dx 21. Since d 2 ln x dx L(x) f (x) 23. 1 x 5 x 5 f e , sec 4 2x 2x 28. f (x) 13 x 3 13 x 3 x 4 4 2 1 1 . 2 x f 4 4 x 4 4 1 2 sec x f (x) sec x tan x f C. x sec x 2 x x x2 C. f 4 sec 1 , 29. f (x) x cos x sin x sin x 2 (1 tan 4 4 2 2 4 tan x) 2 (sec2 x) 1 cos2 x(1 tan x)2 1 (cos x sin x)2 C cos x 4 1 tan x 1 f (x) x 2(1) x 2x x 4 sec 4 2 1, C. d 2 3/2 x 2x 1/2 dx 3 2 3/2 x 2x 1/2 C. 3 f (x) f( ) 0C C f (x) 20t tan 2. sec x tan x, f (x) 2 ln x 22. Since 20 C2 L(x) 14 x 4 f (x) 32t 20t x (c) There is a point of inflection at x f (x) 10 sec2 x 1. (b) There is a local minimum at x e 5t tan x f (x) 1 2 (a) There is a local maximum at x x 31 12 Increasing Behavior of y Concave down Concave up 14 x 4 x 9.8t 5 5t C s (t) 16t 2 5 5 16t 2 Sign of y d 19. Since dx 12 x 2 1) 1 2 x Intervals Decreasing C v (t) 32 32t C1 20 20 27. f (x) 1)(x x 13 x 3 3 4/3 x 4 12 x 2 31 12 2 6(x 1)(2)(x 2) 6(x 2)[(2x 2) 18x(x 2) 13 x 3 3 3 2 L(x) f (0) 1 f (0)(x 1(x x 1 0) 0) 4 x 4 195 196 Chapter 4 Review 30. f (x) e x sin x f (x) e x cos x L(x) f (0) f (0)(x 1 2(x 0) 2x 1 (b) f (x) Using a 0) 1 x (x) 0.5 and b 1 occurs at x 2 2. 32. (a) The values of y and y are both negative where the graph is decreasing and concave down, at T. 1 0.5 ln 0.5 2.5 ln c 0.4 ln(27 2) c 35. e c (c) The local extreme values occur only at the endpoints of the domain. A local maximum value of 1 occurs at x 3, and a local maximum value of 3 occurs at x 2. 1 2.5 (b) The function is decreasing on the interval [ 3, 0). 34. The 24th day 33 0.50.5 ln c 33. (a) The function is increasing on the interval (0, 2]. 3, we solve as follows. 3 ln 3 ln c ln x f (0.5) 0.5 ln (b) The value of y is negative and the value of y is positive where the graph is decreasing and concave up, at P. 1 f (3) 3 f (c) 31. The global minimum value of (ln x)(1) e 1 1 (27 2)0.4 15 1458 1.579 (c) The slope of the line is y m 2 f (b) b f (a) a 0.4 ln (27 2) 0.2 ln 1458, and the line passes through (3, 3 ln 3). Its equation is –3 3 x y y y = f (x ) –3 36. (a) We know that f is decreasing on [0, 1] and increasing on [1, 3], the absolute minimum value occurs at x 1 and the absolute maximum value occurs at an endpoint. Since f (0) 0, f (1) 2, and f (3) 3, the absolute minimum value is 2 at x 1 and the absolute maximum value is 3 at x 3. (b) The concavity of the graph does not change. There are no points of inflection. (c) y 3 0.2(ln 1458)(x 1.457x 1 3 4 –3 37. (a) f (x) is continuous on [0.5, 3] and differentiable on (0.5, 3). s (t) (b) a(t) –1 3 ln 3, or approximately 1.075. (d) The slope of the line is m 0.2 ln 1458, and the line passes through 5 5 (c, f (c)) (e 1 1458, e 1 1458( 1 0.2 ln 1458)) (1.579, 0.722). Its equation is y 0.2(ln 1458)(x c) f (c), 5 y 0.2 ln 1458(x e 1 1458) 5 e 1 1458( 1 0.2 ln 1458), 5 y 0.2(ln 1458)x e 1 1458, or approximately y 1.457x 1.579. 38. (a) v(t) x 3) 4 v (t) 3t 2 6t 6 6t (c) The particle starts at position 3 moving in the positive direction, but decelerating. At approximately t 0.528, it reaches position 4.128 and changes direction, beginning to move in the negative direction. After that, it continues to accelerate while moving in the negative direction. 39. (a) L(x) f (0) 1 (b) f (0.1) f (0)(x 0) 0(x 0) 1 L(0.1) 1 (c) Greater than the approximation in (b), since f (x) is actually positive over the interval (0, 0.1) and the estimate is based on the derivative being 0. 40. (a) Since dy (b) dy dy dx (2x (x 2)( e x) x 2)e x (e x)(2x) dx. [2(1) (1)2](e 1)(0.01) 0.01e 1 0.00368 (2x x 2)e x, 197 Chapter 4 Review 41. (a) Regression equation y 43. Let t represent time in seconds, where the rocket lifts off at t 0. Since a(t) v (t) 20 m/sec2 and v(0) 0 m/sec, we have v(t) 20t, and so v(60) 1200 m/sec. The speed after 1 minute (60 seconds) will be 1200 m/sec. 2701.73 17.28e 0.36x 1 44. Let t represent time in seconds, where the rock is blasted upward at t 0. Since a(t) v (t) 3.72 m/sec2 and v(0) 93 m/sec, we have v(t) 3.72t 93. Since s (t) 3.72t 93 and s(0) 0, we have s(t) 1.86t 2 93t. Solving v(t) 0, we find that the rock attains its maximum height at t 25 sec and its height at that time is s(25) 1162.5 m. [0, 20] by [ 300, 2800] (b) Note that y d 2701.73(1 dx 2701.73(1 0.36x 17.28e ) 0.36x 1 ) 2(17.28)( 0.36e 17.28e 0.36x ) 45. Note that s 0.36x 16,806.9e (1 17.28e r2 A 0.36x 2 ) 100 2r and the sector area is given by 1 rs 2 s 2r the domain of A(r) The graph of y is shown below. 0 s 1 r(100 2 50r 2r) r 2, note that r 50 2 r, which gives 12.1 A (r) 50 r 2. To find 50r 1 0 and r 50. Since 2r, the critical point occurs at r 25. This value is in the domain and corresponds to the maximum [0, 20] by [ 75, 275] area because A (r) Using graphing techniques, y has its maximum at x 7.92. This corresponds to the year 1998 and represents the inflection point of the logistic curve. The logistic regression equation predicts that the rate of increase in debit card transactions will begin to decrease in 1998, and since y(7.92) 1351, there are approximately 1351 million transactions that year. 2, which is negative for all r. The greatest area is attained when r 46. y (x, 27 – x2) –4 f (x) 2 cos x 1 For 0 21 xn 1 xn x A(x) f (xn) 2 cos xn 2 sin xn The graph of y near x 1. 1 A xn 1 2 1 0 has one solution, 1 0.8361848 0.8283814 0.8283608 0.8283608 Solution: x 0.828361 27, the triangle with vertices at (0, 0) and x 2) has an area given by 1 (2x)(27 2 27 3x 2 x 2) 3(3 27x x)(3 x 3. Since x) and A 6x, the 27) occurs at x 3 and corresponds to the maximum area because A (x) is negative in this interval. The largest possible area is A(3) x1 x2 x3 x4 x5 x critical point in the interval (0, xn f (x) shows that f (x) [ 2, 10] by [ 6, 2] x ( x, 27 f (xn) xn 4 x 1 2 sin x 50 ft. 27 (c) As x increases, the value of y will increase toward 2701.73. The logistic regression equation predicts a ceiling of approximately 2702 million transactions per year. 42. f (x) 25 ft and s 54 square units. 198 Chapter 4 Review 47. If the dimensions are x ft by x ft by h ft, then the total amount of steel used is x 2 x 2 4xh 108 and so h x 2h by V(x) V (x) 108x x 3 4 0.75x 2 27 V (x) 4xh ft2. Therefore, 27x 0.75(6 V x) and 1.5x. The critical point occurs at x 108 62 0. The corresponding height is 4(6) 12 2r. 12 rh 3 0 0.25x 3. Then x)(6 12 h , so h 12 r 6 The volume of the smaller cone is given by 108 x 2 . The volume is given 4x 12 23 r (12 2r) 4 r 2 r for 3 3 dV 6. Then 8 r 2 r 2 2 r(4 r), so the dr r critical point occurs at r 6, and it corresponds to the maximum volume because V (x) x 50. Note that, from similar cones, 4. This critical point corresponds to the maximum volume because 0 for 0 3 ft. The r 4 and dV dr 0 for 4 r dV dr 0 for 6. The smaller cone has the largest possible value when r 4 ft and h 4 ft. base measures 6 ft by 6 ft, and the height is 3 ft. 51. 48. If the dimensions are x ft by x ft by h ft, then we have x 2h Lid 32 . Neglecting the quarter-inch x2 32 and so h x thickness of the steel, the area of the steel used is A(x) x2 x2 4xh 10 in. Base 128 . We can minimize the weight x x of the vat by minimizing this quantity. Now A (x) 2x A (x) 2 128x 2 23 (x x2 x 43) and x 15 in. (a) V(x) 256x 3. The critical point occurs at x 4 and x(15 2x)(5 (b, c) Domain: 0 x x) 5 corresponds to the minimum possible area because A (x) 32 42 0 for x 0. The corresponding height is 2 ft. The base should measure 4 ft by 4 ft, and the [0, 5] by [ 20, 70] height should be 2 ft. h2 49. We have r 2 2 3, so r 2 The maximum volume is approximately 66.019 in3 and it occurs when x 1.962 in. 2 3 h . We wish to 4 minimize the cylinder’s volume V r 2h Since dV dh d 2V dh 2 3 3 h2 h 3h 4 2 3h 3 (2 4 4 h3 for 0 4 h)(2 h 2 3. h) and 3h , the critical point occurs at h 2 (d) Note that V(x) 2x 3 25x 2 so V (x) 6x 2 50x 75. Solving V (x) 0, we have 50 2 and it d 2V corresponds to the maximum value because 2 dh 22 h 0. The corresponding value of r is 3 4 0 for The largest possible cylinder has height 2 and radius ( 50)2 2(6) 50 x 2. 2. 10 7 12 4(6)(75) 75x, 50 700 12 25 5 6 7 . These solutions are approximately x 1.962 and x 6.371, so the critical point in the appropriate domain occurs at x 25 5 6 7 . 199 Chapter 4 Review 52. 54. The length of the track is given by 2x y 10 2x 2r 400 and therefore x 2 r, so we have 200 r. Then the area of the rectangle is (x, 8 cos 0.3x) A(r) 2rx 2r(200 –2π –π For 0 A(x) π 2π x 400r Therefore, A (r) 5 , the area of the rectangle is given by 3 x (2x)(8 cos 0.3x) Then A (x) 16(cos 0.3x Solving A (x) because A (r) 0.3x sin 0.3x) x 2.868 and the corresponding area is 30(20 Then C (x) y) 30 40 2 Solving C (x) x x2 600 (2x) 144 40 30 200 0 for all r. 100 100 m. 100 100 m and r x 144. 30x x2 2k dollars per hundred grade A tires. . 144 Then the profit is given by P(x) 2kx 144 9x 2 2304 4 16x 2 144 2k 2304 x2 2k 7x 2 P (x) 2k Choose the positive solution: x 2k 18.142 mi The solutions of P (x) 7 y x 2 m. 55. Assume the profit is k dollars per hundred grade B tires and 2 40 3x 48 m 0, we have: 30x x2 40x 100 4 , so the The rectangle will have the largest possible area when 53. The cost (in thousands of dollars) is given by 40x 4 r and A (r) . The corresponding value of x is approximately 29.925 square units. C(x) 400 200 r and this point corresponds to the maximum rectangle area 16(cos 0.3x)(1) 0 graphically, we find that the critical point occurs at x 2 r 2, for 0 critical point occurs at r 16x cos 0.3x. 16x( 0.3 sin 0.3x) r) 12 2 36 7 13.607 mi x 10 2 ( 10) 2(1) 40 5 k 10x x 5x) 5 x(5 x (20 20 5 x2 x (5 x)( 2x) (5 x) x2 (5 (20 x )2 x 2)( 1) 10x 20 x)2 0 are 4(1)(20) the appropriate domain is x 5 5 5, so the solution in 5 2.76. Check the profit for the critical point and endpoints: Critical point: x 2.76 P(x) 11.06k Endpoints: x0 P(x) 8k x4 P(x) 8k The highest profit is obtained when x which corresponds to 276 grade A tires and 553 grade B tires. 2.76 and y 5.53, 200 Chapter 4 Review 56. (a) The distance between the particles is f (t) where cos t f (t) cos t sin t f (t) sin t Solving f (t ) 4 . Then 4 0 graphically, we obtain t 1.178, t 4.230, and so on. [0, 2 ] by [ 2, 2] Alternatively, f (t ) f (t) t sin 8 sin t 2 sin 0 may be solved analytically as follows. cos 8 8 sin 8 cos t 8 cos t t 8 8 8 sin 8 sin t 8 8 cos 8 cos t 8 , so the critical points occur when cos t f (t) 0, or t 8 2 cos 3 8 3 8 k . At each of these values, 0.765 units, so the maximum distance between the particles is 0.765 units. (b) Solving cos t cos t 4 graphically, we obtain t 2.749, t 5.890, and so on. [0, 2 ] by [ 2, 2] Alternatively, this problem may be solved analytically as follows. cos t cos cos t 8 cos 8 t 8 sin t 8 2 sin t 8 sin t cos 8 sin sin 8 8 7 8 4 t cos t 8 8 8 cos 8 sin t 8 cos 8 0 0 8 t The particles collide when t cos t 7 8 k 2.749 (plus multiples of if they keep going.) sin 8 Chapter 4 Review 57. The dimensions will be x in. by 10 2x in. by 16 2x in., so V(x) x(10 2x)(16 2x) 4x 3 52x 2 160x for 0 x 5. Then V (x) 12x 2 104x 160 4(x 2)(3x 20), so the critical point in the correct domain is x 2. This critical point corresponds to the maximum possible volume because V (x) 0 for 0 x 2 and V (x) 0 for 2 x 5. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in3. 201 59. Step 1: x x-coordinate of particle y y-coordinate of particle D distance from origin to particle Step 2: At the instant in question, x dx dt dy 1 m/sec, and dt 5 m, y 12 m, 5 m/sec. Step 3: Graphical support: We want to find dD . dt Step 4: x2 D y2 [0, 5] by [ 40, 160] Step 5: 58. Step 1: r radius of circle A area of circle dD dt At the instant in question, dr dt 2 m/sec and r Step 3: dA . dt x2 2 dy 2y dt dx dt x2 y dy dt y2 10 m. (5)( 1) dD dt (12)( 5) 5 m/sec 52 122 dD Since is negative, the particle is approaching the origin dt at the positive rate of 5 m/sec. 60. Step 1: x edge of length of cube V volume of cube Step 4: A dx 2x 2 dt y Step 6: Step 2: We want to find x 1 r2 Step 5: Step 2: dA dt At the instant in question, dr 2r dt dV dt Step 6: dA dt 2 (10) 2 40 The area is changing at the rate of 1200 cm3/min and x 20 cm. Step 3: 2 40 m /sec. We want to find dx . dt Step 4: V x3 Step 5: dV dt 3x 2 dx dt Step 6: 1200 dx dt 3(20)2 dx dt 1 cm/min The edge length is increasing at the rate of 1 cm/min. 202 Chapter 4 Review 63. Step 1: r radius of outer layer of cable on the spool clockwise angle turned by spool s length of cable that has been unwound 61. Step 1: x x-coordinate of point y y-coordinate of point D distance from origin to point Step 2: Step 2: At the instant in question, x 3 and dD dt 11 units per sec. At the instant in question, ds dt 6 ft/sec and r 1.2 ft Step 3: Step 3: We want to find dx We want to find . dt d . dt Step 4: Step 4: s Since D 2 x2 x2 D y 2 and y x 3 for x x 3/2, we have Step 5: 0. Since r is essentially constant, 1 (2x 2 x2 x3 2x 3x 2 dx 2x 1 x dt 3x 2) 3x 2 6 dx dt d dt 2 dx 1 x dt 3(3) 2 dx 2 4 dt dx dt 4 units per sec 62. (a) Since h r 10 , we may write h 4 5r or r 2 2h . 5 (b) Step 1: h depth of water in tank r radius of surface of water V volume of water in tank 1.2 5 radians/sec 64. a(t) v (t) g 32 ft/sec2 Since v(0) 32 ft/sec, v(t) s (t) 32t 32. Since s(0) 17 ft, s(t) 16t 2 32t 17. The shovelful of dirt reaches its maximum height when v(t) 0, at t 1 sec. Since s(1) 1, the shovelful of dirt is still below ground level at this time. There was not enough speed to get the dirt out of the hole. Duck! 65. We have V 12 dV r h, so 3 dr 2 rh and dV 3 When the radius changes from a to a dV dt 5 ft3/min and h 6 ft. dS Step 3: We want to find dh . dt 6x 2, which means 12x dx. We want dS 12x dx 2 0.02(6x ) or dx surface area of dS dx 12x and 0.02S, which gives 0.01x. The edge should be measured with an error of no more than 1%. Step 4: 12 rh 3 dr, the volume edge of length of cube and S cube. Then S 2 rh dr. 3 2 ah dr. 3 change is approximately dV 66. (a) Let x At the instant in question, 4 h3 75 Step 5: dV dt d . dt d dt Step 2: V r The spool is turning at the rate of 5 radians per second. Step 6: 11 ds dt Step 6: Step 5: dD dt r dV dx dh 4 h2 dt 25 volume of cube. Then V 3x 2 and dV dh 4 (6)2 dt 25 dh 125 0.276 ft/min dt 144 dh Since is negative, the water level is dropping at the dt 5 0.276 ft/min. so dV x 3, which means 3x 2 dx. We have dx which means 3x 2 dx Step 6: positive rate of (b) Let V 3x 2(0.01x) 0.01x, 0.03V, 0.03V. The volume calculation will be accurate to within approximately 3% of the correct volume. 203 Chapter 4 Review 67. Let C circumference, r (a) Since C radius, S dC dr 2 r, we have Therefore, dC C 2 dr 2r surface area, and V 2 and so dC dr r 0.4 cm 10 cm volume. 2 dr. 0.04 The calculated radius will be within approximately 4% of the correct radius. dS dr dS dS 8 r dr. Therefore, S 2 dr 2(0.04) 0.08. The r (b) Since S 4 r 2, we have 8 r and so 8 r dr 4 r2 calculated surface area will be within approximately 8% of the correct surface area. 43 dV r , we have 3 dr dV 2 4 r dr. Therefore V 4 r 2 and so (c) Since V dV 4 r 2 dr 3 dr r 43 r 3 3(0.04) 0.12. The calculated volume will be within approximately 12% of the correct volume. 68. By similar triangles, we have approximately 6 have dh 69. dy dx 120(15) 120a 2 sin x cos x Therefore, dy dx 2 2 da 1 a 6 a 20 h , which gives ah 6a 120, or h 120a 14 ft. The estimated error in measuring a was da 120(15) 21 12 2 ft, so the estimated possible error is 45 3. Since sin x and cos x are both between 1 and 3 6 1 for all values of x. dy Since is always negative, the function decreases on every interval. dx 1 . The height of the lamp post is dh 1 ft. Since da 12 2 8 ft or in. 45 15 1 in. 120a 2 , we 1, the value of 2 sin x cos x is never greater than 2. 204 Section 5.1 Chapter 5 The Definite Integral s Section 5.1 Estimating with Finite Sums (pp. 247–257) Exploration 1 Which RAM is the Biggest? y y y 5 5 5 4 4 4 3 3 3 2 2 2 1 1 1 1 2 1. LRAM x 3 MRAM 1 2 x 3 1 2 x 3 RRAM y y y 5 5 5 4 4 4 3 3 3 2 2 2 1 2 2. MRAM x 3 RRAM 1 2 x 3 1 2 3 x LRAM 3. RRAM MRAM function. LRAM, because the heights of the rectangles increase as you move toward the right under an increasing 4. LRAM MRAM function. RRAM, because the heights of the rectangles decrease as you move toward the right under a decreasing Quick Review 5.1 1. 80 mph 5 hr 400 mi 2. 48 mph 3 hr 144 mi 3. 10 ft/sec2 10 sec 100 ft/sec 3600 sec 1 mi 68.18 mph 1h 5280 ft 3600 sec 24 hr 365 days 4. 300,000 km/sec 1 hr 1 yr 1 day 12 100 ft/sec 9.46 10 5. (6 mph)(3 h) (5 mph)(2 h) 6. 20 gal/min 1 h 60 min 1h 7. ( 1 C/h)(12 h) 18 mi (1.5 C)(6 h) 8. 300 ft3/sec 3600 sec 1h 3600 sec 1h 10 mi 28 mi 1200 gal 24 h 1 day 9. 350 people/mi2 50 mi2 10. 70 times/sec 1 yr km 3C 1 day 25,920,000 ft3 17,500 people 1 h 0.7 176,400 times Section 5.1 Section 5.1 Exercises 1. (a) y 2 R 2 (b) x y 2 2 x 1 2 LRAM: [2(0) 2. (a) x (0)2] 1 2 12 1 2 2 1 2 2 [2(1) (1)2] 1 2 2 32 1 2 2 3 2 5 4 1.25 1 2 5 4 1.25 y 2 2 RRAM: 2 (b) 1 2 12 1 2 2 x [2(1) (1)2] 1 2 2 32 1 2 2 3 2 [2(2) (2)2] y 2 2 MRAM: 2 3. 12 1 4 2 1 4 x 2 3 4 n LRAMn MRAMn RRAMn 10 1.32 1.34 1.32 50 1.3328 1.3336 1.3328 100 1.3332 1.3334 1.3332 500 1.333328 1.333336 1.333328 4. The area is 1.333 4 . 3 32 1 4 2 2 5 4 52 1 4 2 2 7 4 72 1 4 2 11 8 1.375 205 206 5. Section 5.1 n LRAMn MRAMn RRAMn 10 12.645 13.4775 14.445 RRAM: 10 (44 50 13.3218 13.4991 13.6818 Average 100 13.41045 12. LRAM: 10 (0 13.499775 13.59045 Estimate the area to be 13.5. n 1.16823 1.09714 1.03490 50 1.11206 1.09855 1.08540 100 1.10531 1.09860 1.09198 500 1.09995 1.09861 1.09728 1000 1.09928 1.09861 15 3490 ft 3840 ft 2 3490 ft 35) 3840 ft 3665 ft 1.09795 n MRAM 10 526.21677 LRAMn MRAMn RRAMn 20 524.25327 10 0.98001 0.88220 0.78367 40 523.76240 50 0.90171 0.88209 0.86244 80 523.63968 100 0.89190 0.88208 0.87226 160 523.60900 500 0.88404 0.88208 0.88012 1000 0.88306 0.88208 0.88110 n 4 (5)3 3 15. V 500 3 error n Estimate the area to be 0.8821. 523.59878 % error 10 2.61799 0.5 n LRAMn MRAMn RRAMn 20 0.65450 0.125 10 1.98352 2.00825 1.98352 40 0.16362 0.0312 50 1.99934 2.00033 1.99934 80 0.04091 0.0078 100 1.99984 2.00008 1.99984 500 1.99999 2.00000 8. … 30 ) 14. Use f (x) 25 x 2 and approximate the volume using r 2h ( 25 ni2)2 x, so for the MRAM program, use (25 x 2) on the interval [ 5, 5]. Estimate the area to be 1.0986. 7. 35 … (b) The halfway point is 0.4845 mi. The average of LRAM and RRAM is 0.4460 at 0.006 h and 0.5665 at 0.007 h. Estimate that it took 0.006 h 21.6 sec. The car was going 116 mph. LRAMn MRAMn RRAMn 10 15 13. (a) LRAM: 0.001(0 40 62 … 137) 0.898 mi RRAM: 0.001(40 62 82 … 142) 1.04 mi Average 0.969 mi 500 13.482018 13.499991 13.518018 6. 44 1.99999 160 0.01023 0.0020 16. (a) Use LRAM with (16 x 2). S8 146.08406 S8 is an overestimate because each rectangle is above the curve. Estimate the area to be 2. 9. LRAM: Area f (2) 2 f (4) 2 2 (0 0.6 1.4 44.8 (mg/L) sec RRAM: Area f (4) 2 f (6) 2 2(0.6 1.4 2.7 44.8 (mg/L) sec f (6) 2 … … 0.5) f (22) 2 (b) … f (8) 2 … 0) f (24) 2 (b) 60 sec 1 min Note that estimates for the area may vary. (b) RRAM: 1 (12 22 10 5 6 0) 87 in. 7.25 ft 11. 5 min 5 13 11 6 (b) RRAM: 300 (1.2 1.2 1.7 1.7 2.0 9% V S8 V 0.10 10% x 2) on the 6 2 (b) V S8 V 0.11 11% 11 … … 1.2) 5220 m 0) 4920 m 19. (a) (5)(6.0 8.2 9.1 … 12.7)(30) 15,465 ft3 (b) (5)(8.2 13 300 sec (a) LRAM: 300 (1 0.09 18. (a) Use LRAM with (64 interval [4, 8], n 8. S 372.27873 m3 6.7 L/min 10. (a) LRAM: 1 (0 12 22 10 2 6) 87 in. 7.25 ft S8 V 17. (a) Use RRAM with (16 x 2). S8 120.95132 S8 is an underestimate because each rectangle is below the curve. Patient’s cardiac output: 5 mg 44.8 (mg/L) sec V 9.1 9.9 … 13.0)(30) 16,515 ft3 20. Use LRAM with x on the interval [0, 5], n 5. 1(0 2 3 4 ) 10 31.41593 Section 5.1 21. Use MRAM with x on the interval [0, 5], n 1 1 2 3 2 5 2 7 2 22. (a) LRAM5: 32.00 19.41 (b) RRAM5: 19.41 11.77 9 2 25 2 5. triangles with a hypotenuse of length 1 and an acute 39.26991 7.14 7.14 4.33 4.33 2.63 74.65 ft/sec (5 sec)(32 ft/sec2) Area 16 45.28 ft/sec (c) The upper estimates for speed are 32.00 ft/sec for the first sec, 32.00 19.41 51.41 ft/sec for the second sec, and 32.00 19.41 11.77 63.18 ft/sec for the third sec. Therefore, an upper estimate for the distance fallen is 32.00 51.41 63.18 146.59 ft. 23. (a) 400 ft/sec (b) Think of the octagon as a collection of 16 right angle measuring 11.77 2 97 70 (b) Upper 70 Lower 50 97 136 190 265 369 516 720 2363 gal 70 97 136 190 265 369 516 1693 gal . 8 4 2 2.828 (c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle 2 . 32 16 1 32 sin cos 2 16 16 measuring (b) Use RRAM with 400 32x on [0, 5], n 5. 368 336 304 272 240 1520 ft 70 50 2 16 1 sin cos 2 8 8 4 sin 240 ft/sec 24. (a) Upper Lower 207 Area 136 190 265 758 gal 97 136 190 543 gal 8 sin 3.061 8 (d) Each area is less than the area of the circle, . As n increases, the area approaches . 28. The statement is false. We disprove it by presenting a (c) 25,000 2363 22,637 gal 22,637 720 31.44 h (worst case) 25,000 1693 23,307 720 23,307 gal 32.37 h (best case) 25. (a) Since the release rate of pollutants is increasing, an upper estimate is given by using the data for the end of each month (right rectangles), assuming that new scrubbers were installed before the beginning of January. Upper estimate: 30(0.20 0.25 0.27 0.34 0.45 0.52) 60.9 tons of pollutants A lower estimate is given by using the data for the end of the previous month (left rectangles). We have no data for the beginning of January, but we know that pollutants were released at the new-scrubber rate of 0.05 ton/day, so we may use this value. Lower estimate: 30(0.05 0.20 0.25 0.27 0.34 0.45) 46.8 tons of pollutants (b) Using left rectangles, the amount of pollutants released by the end of October is 30(0.05 0.20 0.25 0.27 0.34 0.45 0.52 0.63 0.70 0.81) 126.6 tons. Therefore, a total of 125 tons will have been released into the atmosphere by the end of October. 26. The area of the region is the total number of units sold, in millions, over the 10-year period. The area units are (millions of units per year)(years) (millions of units). 27. (a) The diagonal of the square has length 2, so the side length is 2. Area ( 2)2 2 counterexample, the function f (x) 0 x 1, with n LRAM1 RRAM1 1. MRAM1 1f (0) 2 x 2 over the interval 1f (0.5) 0.25 1f (1) 2 0 1 2 0.5 MRAM1 ( x)[ f (x1) f (x2) … f (xn 1) f (xn)] ( x)[ f (x0) f (x1) f (x2) … f (xn 1)] ( x)[ f (xn) f (x0)] LRAMn f ( x)[ f (xn) f (x0)] But f (a) f (b) by symmetry, so f (xn) f (x0) 0. Therefore, RRAMn f LRAMn f. 29. RRAMn f 30. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle 2 . The area of each isosceles triangle 2n n 1 1 2 2 sin cos sin . 2 n n 2 n measuring is AT (b) The area of the polygon is AP nAT lim AP n→ n 2 sin , so 2 n n 2 lim sin n n→ 2 (c) Multiply each area by r 2: AT AP 12 r sin 2 n2 r sin 2 lim AP n→ 2 n 2 n r2 208 Section 5.2 s Section 5.2 Definite Integrals (pp. 258–268) Exploration 1 1. 7. 0. (The equal areas above and below the x-axis sum to zero.) Finding Integrals by Signed Areas 2. (This is the same area as sin x dx, but below the 0 x-axis.) [ 2 , 2 ] by [ 3, 3] 8. 4. (Each rectangle in a typical Riemann sum is twice as wide as in sin x dx.) 0 [ 2 , 2 ] by [ 3, 3] 2. 0. (The equal areas above and below the x-axis sum to zero.) [ 2 , 2 ] by [ 3, 3] 9. 0. (The equal areas above and below the x-axis sum to zero.) [ 2 , 2 ] by [ 3, 3] 3. 1. (This is half the area of sin x dx.) 0 [ 2 , 2 ] by [ 3, 3] 10. 0. (The equal areas above and below the x-axis sum to zero, since sin x is an odd function.) [ 2 , 2 ] by [ 3, 3] 4. 2 2. (The same area as sin x dx sits above a rectangle 0 of area 2.) [ 2 , 2 ] by [ 3, 3] Exploration 2 More Discontinuous Integrands 1. The function has a removable discontinuity at x 2. [ 2 , 2 ] by [ 3, 3] 5. 4. (Each rectangle in a typical Riemann sum is twice as tall sin x dx.) as in [ 4.7, 4.7] by [ 1.1, 5.1] 0 2. The thin strip above x 2 has zero area, so the area under 3 the curve is the same as (x 0 [ 2 , 2 ] by [ 3, 3] 6. 2. (This is the same region as in 2 units to the right.) sin x dx, translated 0 [ 4.7, 4.7] by [ 1.1, 5.1] [ 2 , 2 ] by [ 3, 3] 2) dx, which is 10.5. Section 5.2 3. The graph has jump discontinuities at all integer values, but n the Riemann sums tend to the area of the shaded region ∑ P →0 shown. The area is the sum of the areas of 5 rectangles (one 1 tition of [0, 1]. 1 k1 4 x 2 dx where P is any par- 0 n 6. lim 5 0 ck2 xk 4 ∑ (sin3 ck) P →0 of them with height 0): int(x) dx 5. lim 2 3 4 10. 0 partition of [ sin3 x dx where P is any xk k1 , ]. 1 5 dx 7. 5[1 ( 2)] 15 ( 20)(7 3) 2 7 [ 2.7, 6.7] by [ 1.1, 5.1] 3 (1)2 ( 160) dt 9. 5 ∑ n2 (2)2 (3)2 (4)2 (5)2 ( 160)(3 0) 55 1 4 ∑ (3k 2) [3(0) 2] [3(1) 2] 480 0 n1 2. 80 3 Quick Review 5.2 1. ( 20) dx 8. [3(2) 2] 10. 2 4 d 2 [1 3 2 ( 4)] k0 [3(3) 2] [3(4) 2] 20 3.4 11. 0.5 ds 3. 0.5[3.4 ( 2.1)] ∑ 100( j 1)2 100[(1)2 (2)2 (3)2 (4)2 (5)2] j0 18 2 dr 12. 5500 2( 18 2) ∑k x 2 13. Graph the region under y k1 25 5. 4 2 99 4. 2.75 2.1 4 ∑ 2k 3 for 2 x 4. y 5 k0 500 6. ∑ 3k 2 k1 50 50 7. 2∑ x 2 3∑ x x1 x1 50 ∑ (2x 2 x x1 8 20 20 k0 8. 5 3x) k9 k0 ∑ xk ∑ xk ∑ xk n 9. ∑( 1)k 0 if n is odd. 4 k0 2 n 10. ∑( 1)k 1 (6)(2 2 5) 14. Graph the region under y Section 5.2 Exercises n ∑ ck 2 2 xk P →0 k 1 x dx where P is any partition of [0, 2]. ∑ k1 1) 2 5 (x 2 3x) dx where P is any 7 3 2 1 1 x ck k n ∑ P →0 4. lim 4 for 4 5 partition of [ 7, 5]. P →0 k 1 2x 0 3ck) xk k1 n 21 y 2 n ∑ (ck2 P →0 2. lim 3. lim 3 dx 1 if n is even. k0 1. lim x 2 1 1 dx where P is any partition of [1, 4]. x 3 1 1 4 ck xk 2 1 1 x 5 3/2 ( 2x of [2, 3]. x x=1 x=3 2 2 dx where P is any partition 1/2 4) dx 1 (1)(3 2 209 1 2 x 3 . 2 210 Section 5.2 15. Graph the region under y x 2 for 9 3 x 3. 18. Graph the region under y 1 x 1. x for 1 x 1. 2 5 5 x 2 1 This region is half of a circle of radius 3. 9 x for y y 3 1 2 x dx 3 1 (3)2 2 1 (2)(1) 2 1 19. Graph the region under y 2 (1 x ) dx 1 9 2 16. Graph the region under y x 16 2 x for 4 x 0. y 2 y 5 2 5 x 1 (2 The region is one quarter of a circle of radius 4. 0 16 x 2 dx 4 1 (4)2 4 17. Graph the region under y x x ) dx 1 1 (1)(1 2 2) 1 20. Graph the region under y 4 x for 2 x 1 (1)(1 2 1 2) x 2 for 3 1 y 1. 2 y 2 2 2 x x 1 (1 1 x 2) dx 1 (1)2 2 (2)(1) 1 1 x dx 2 1 (2)(2) 2 1 (1)(1) 2 5 2 21. Graph the region under y for 2 y 2 2 2 d 1 (2 2 )(2 ) 2 2 3 2 2 x 1. 211 Section 5.2 22. Graph the region under y r for 2 r 5 33. Observe from the graph below that the region under the graph of f (x) 1 x 3 for 0 x 1 cuts out a region R from the square identical to the region under the graph of g(x) x 3 for 0 x 1. 2. y 10 y 1 y = f (x ) √2 5 r 5√2 2 r dr 2 10 2 2)( 1 (5 2 2 5 2) 1 24 1 b 23. 1 (b)(b) 2 x dx 0 b 2s ds 1 (b 2 a)(2b 2a) b2 3t dt 1 (b 2 a)(3b 3a) 32 (b 2 b 25. a b 26. a 2a 27. 1 (2a 2 x dx a 3a 28. x dx a 1 ( 2 a)(2a a) a)( 3a x x 3 dx 1 0 1 4 3 4 34. Observe from the graph of f (x) ( x 1)3 for 1 x 2 that there are two regions below the x-axis and one region above the axis, each of whose area is equal to the area of the region under the graph of g(x) x 3 for 0 x 1. 2b 2 3a 1 0 1 (b)(4b) 2 0 1 x 3) dx (1 12 b 2 4x dx 24. R a2 y a 2) 2 3a 2 2 a) 1 (3a 2 2 a 2) a2 y = f (x ) 2 x 29. Observe that the graph of f (x) x 3 is symmetric with respect to the origin. Hence the area above and below the x-axis is equal for 1 x 1. 1 x 3 dx (area below x-axis) 1 (area above x-axis) 2 0 3 is three units higher than the 30. The graph of f (x) x graph of g(x) x 3. The extra area is (3)(1) 3. 1 (x 3 1 4 3) dx 0 13 4 3 31. Observe that the region under the graph of f (x) (x 2)3 for 2 x 3 is just the region under the graph of g(x) x 3 for 0 x 1 translated two units to the right. 3 (x 1 2)3 dx 1 4 x 3 dx 2 0 32. Observe that the graph of f (x) x 3 is symmetric with respect to the y-axis and the right half is the graph of g(x) x 3. 1 1 x 3 dx 1 2 x 3 dx 0 1 2 1)3 dx (x 3 2 1 1 4 1 4 35. Observe that the graph of f (x) 1 4 x3 for 0 2 a horizontal stretch of the graph of g(x) by a factor of 2. Thus the area under f (x) 0 x for 0 2 0 x 2 is just x 3 for 0 x 2 is twice the area under the graph of g(x) x x3 dx 2 1 x3 for 2 x3 1. 1 2 x 3 dx 0 1 2 36. Observe that the graph of f (x) x 3 is symmetric with respect to the origin. Hence the area above and below the x-axis is equal for 8 x 8. 8 8 x 3 dx (area below x-axis) (area above x-axis) 0 212 Section 5.2 37. Observe from the graph below that the region between the graph of f (x) x 3 1 and the x-axis for 0 x 1 cuts out a region R from the square identical to the region under the graph of g(x) x 3 for 0 x 1. 44. (a) The function has discontinuities at x 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5. (b) y 1 x [ 6, 5] by [ 18, 4] 5 2 int(x 3) dx ( 18) ( 12) ( 10) ( 16) ( 8) ( 14) 6 y = f (x ) R 0 –1 1 (x 3 1) dx 1 4 1 0 2 ( 6) ( 4) 88 45. (a) The function has a discontinuity at x 3 4 ( 2) 1. (b) 38. Observe from the graph below that the region between the 3 graph of f (x) x and the x-axis for 0 x 1 cuts out a [ 3, 4] by [ 4, 3] region R from the square identical to the region under the 4 x 3 for 0 graph of g(x) x 1. 3 x2 x 1 (4)(4) 2 1 dx 1 1 (3)(3) 2 46. (a) The function has a discontinuity at x y 7 2 3. (b) 1 R y = f (x ) [ 5, 6] by [ 9, 2] 6 x 1 1 x dx 1 0 x 39. NINT 40. 3 x2 4 1 4 x 2 , x, 42. NINT(x 2e x, x, 2, 2) 1, 3) 1 (9)(9) 2 3 1 n 1 dx x2 lim ∑ n→ ∑ n 10.6667 n→ 1.8719 k1 n k2 n→ (b) … 1 22 lim n 1 0. 1 n 1 k2 n k1 lim 0 4.3863 43. (a) The function has a discontinuity at x 1 . n2 1 1 … 22 n2 1 … →. n2 n and n → , so Note that n 1 1 22 n1 48. (a) [ 2, 3] by [ 2, 2] 3 2 x dx 2 3 77 2 (b) Using right endpoints we have 0.9905 2 NINT tan x, x, 0, 41. NINT(4 1 (2)(2) 2 47. (a) As x approaches 0 from the right, f (x) goes to . 3 4 , x, 0, 5 x2 dx 3 9 5x 1 ,x nk x RRAM 1 x n ∑ k1 n (b) k2 n ∑ k1 (c) n ∑ k2 1 n3 k 1 k n 12 n k2 1 n n n ∑ 1 n k1 1 n3 22 n 1 n n(n k2 n3 … 1 n n2 n 1 n n ∑ k2 1 n3 k 1)(2n 6 1 1) n(n 1)(2n 6n 3 1) Section 5.3 (d) lim n→ k2 n ∑ k1 1 n lim n lim n(n 2n 3 n→ 2 6 1)(2n 1) 6n3 3n 2 n 6n 3 1 3 Exploration 2 Integral 213 Finding the Derivative of an Pictures will vary according to the value of x chosen. (Indeed, this is the point of the exploration.) We show a typical solution here. y 1 x 2 dx equals the limit of any Riemann sum (e) Since 0 over the interval [0, 1] as n approaches , part (d) 1 1 . 3 x 2 dx proves that 0 x 0a b x s Section 5.3 Definite Integrals and Antiderivatives (pp. 268–276) Exploration 1 How Long is the Average Chord of a Circle? 1. The chord is twice as long as the leg of the right triangle in the first quadrant, which has length r2 x2 by the Pythagorean Theorem. 1. We have chosen an arbitrary x between a and b. 2. We have shaded the region using vertical line segments. x 3. The shaded region can be written as f (t) dt using the a definition of the definite integral in Section 5.2. We use t as y a dummy variable because x cannot vary between a and r2 − x2 r itself. 4. The area of the shaded region is our value of F(x). x x y 2. Average value 3. Average value 1 ( r) r 2 2r 1 r r x r r2 2 x2 dx. 0a r2 x2 dx (area of semicircle of radius r) r2 2 5. We have drawn one more vertical shading segment to represent F. 6. We have moved x a distance of x so that it rests above the new shading segment. r 2 y 4. Although we only computed the average length of chords ∆x perpendicular to a particular diameter, the same computation applies to any diameter. The average length of a chord of a circle of radius r is 5. The function y 2 x ∆F r 1 r b r 2 r x 0a r . 2 b x ∆F 2 x is continuous on [ r, r], so the Mean Value Theorem applies and there is a c in [a, b] so that y(c) is the average value r . 2 7. Now the (signed) height of the newly-added vertical segment is f (x). 8. The (signed) area of the segment is F so F (x) lim x→0 F x f (x) x f (x), 214 Section 5.3 Quick Review 5.3 dy 1. dx dy 2. dx 9 (c) 9 [2 f (x) 3h(x)] dx 7 sin x 7 cos x 1 sec x tan x sec x 4. dy dx c os x sin x 5. dy dx sec x tan x sec2 x sec x tan x (d) tan x (e) f (x) dx 7 f (x) dx 9 9 f (x) dx f (x) dx 1 7 1 ln x 5 6 7 (f) ( n 1) x n n1 f (x) dx 1 9 sec x 1 1 9 f (x) dx 1 ln x 7 [h(x) f (x)] dx 7 h(x) dx 9 xn 2 1 7 1 x x dy dx 7 3(4) 9 f (x) dx 9 cot x 9 f (x) dx 9 9 dy dx 9. dy dx xe x 10. dy dx 2 1 (2x 2x ln 2 (2x 1)2 (ln 2)2x 1)2 9 h(x) dx f (x) dx 7 8. 9 3 h(x) dx 7 2(5) dy dx 7. 3h(x) dx 7 9 2 f (x) dx 3. dy 6. dx 9 2f (x) dx 7 4 5 1 3 f (z) dz 5 2 3. (a) ex f (u) du 5 1 2 1 x 2 (b) 3f (z) dz 1 1 1 (c) Section 5.3 Exercises 3 1 2 f (t) dt f (t) dt 2 5 1 2 1. (a) g(x) dx 2 0 (d) 2 2 [ f (x)] dx f (x) dx 1 1 (b) g(x) dx g(x) dx 5 3 8 4. (a) 1 0 g(t) dt g(t) dt 0 2 (c) 5 1 5 2 3 2 3 f (x) dx 3 1 f (x) dx 3( 4) 0 12 (b) 1 g(u) du 2 3 5 (d) 1 f (x) dx 5 f (x) dx 2 2 3 5 [ f (x) g(x)] dx 1 1 8 5 [4f (x) g(x)] dx g(x) dx g(x) dx [ f (x) 7 7 5 0 3 f (t) dt 0 3 4 2 6. (a) 1 h(r) dr 1 h(x) dx 3 h(r) dr 1 h(r) dr 1 3 1 7 4 f (t) dt 0 4 f (t) dt 9 f (x) dx 4 3 4 3 h(x)] dx f (z) dz f (t) dt 16 2( 1) 9 4 0 0 f (t) dt 4 1 9 f (z) dz 0 3 7 2 f (x) dx 1 7 3 9 2f (x) dx (b) 3 (b) 1 8 4 f (z) dz 3 1 5 5 1 9 3 1 f (z) dz 4 f (x) dx 2. (a) 0 f (z) dz 5 4(6) g(r) dr 3 2 0 4f (x) dx 1 2 2 5 1 5. (a) 1 6 (f) g(x) dx 2 0 1 dr 4 5 f (x) dx g(x) dx 3 g(r) (d) 10 5 (e) 0 1 6 [ g(x)] dx 3 f (x) dx 1 0 (c) 5 f (x) dx 4 0 f (x) dx 1 2 h(r) dr 1 9 1 (b) 1 h(u) du 3 h(r) dr 1 h(u) du 3 6 1 3 h(u) du 1 h(u) du 1 1 h(u) du 1 6 Section 5.3 7. An antiderivative of 7 is F(x) 17. Divide the shaded area as follows. 7x. 1 7 dx F(1) F(3) 7 215 21 y 14 3 3 52 x 2 8. An antiderivative of 5x is F(x) 2 5x dx F(2) F(0) 10 0 R2 R1 10 y = x2 – 1 0 x2 . 16 9 1 16 x is F(x) 8 25 F(3) 16 9. An antiderivative of 5 x dx 8 3 F(5) 10. An antiderivative of 2t t2 3 is F(t) x 2 R3 Note that an antiderivative of x 2 3t. 1 is F(x) 2 (2t 3) dt F(2) F(0) 2 0 Area of R1 3(1) Area of R2 2 0 (3)(1) 11. An antiderivative of t 2 is F(t) t 2. (x 2 (t 2) dt F( 2) F(0) 3 1 z is F(z) 2 5 F(2) 4 1 z dz 2 1 2 F(1) 13. An antiderivative of 1 1 x2 11 dx 1 x2 1 F(1) 1/2 1/2 dx 1 x2 is F(x) F( 1) 1 14. An antiderivative of x2 1 1 F 2 F e x dx F(2) e2 F(0) tan 4 1 2 1 sin 6 F(0)] 2 3 2 1 6 0 3 Total shaded area x. x 3. 1) dx 5 3 0 x. 2 3 5 3 2 3 16 3 18. Divide the shaded area as follows. y 3 3 e x. 1 (x 2 Area of R3 4 is F(x) 15. An antiderivative of e x is F(x) 2 3 2 3 1 z2 . 4 7 4 z F(1)] 2 3 1 0 12. An antiderivative of 1 [F(2) [F(1) 0 3x 1) dx 1 3 2 x. 3 2 12 t 2 13 x 3 y = 3 – 3x 2 R1 x 2 6.389 0 16. An antiderivative of 3 0 3 dx x1 3 x 1 F(3) R2 1. R3 0 3 ln 4 3 ln x F(0) 3 ln 4 is F(x) 4.159 –9 3x 2 is F(x) Note that an antiderivative of 3 1 Area of R1 Area of R2 3x 2) dx (3 F(1) 0 (9)(1) 9 2 Area of R3 (9)(1) 3x 2) dx (3 1 9 [F(2) 9 (2 Total shaded area 2 F(1)] 2) 9 5 5 16 F(0) 2 216 Section 5.3 19. Divide the shaded area as follows. 2 3 (x 2 (b) Area 6x 8) dx 0 y 2.25 [F(2) y = 3x – x 2 20 3 R1 (x 2 6x 8) dx 2 F(0)] 0 [F(3) 6 F(2)] 20 3 22 3 x R2 4 22. –4 Note that an antiderivative of 3x 3 Area of R1 x 2) dx (3x 0 4 Area of R2 32 13 x x. 2 3 9 9 0 2 2 x 2 is F(x) F(3) F(0) [0, 2] by [ 5, 3] x 2) dx (3x 13 x 3 F(x) 3 [F(4) 8 3 F(3)] 9 2 2 (a) 11 6 9 2 Total shaded area x2 An antiderivative of 52 x 2 ( x2 5x 5x 4 is 4x. 4) dx F(2) 2 3 F(0) 11 6 0 1 19 3 (b) Area 2 ( x2 5x ( x2 4) dx 0 5x 4) dx 1 [F(1) 20. Divide the shaded area as follows. 2 3 0 F(0)] [F(2) F(1)] 0 2 3 11 6 3 x2 13 x. 3 11 6 y 23. 3 y= x2 – 2x R2 [0, 3] by [ 3, 2] x 3 R1 x 2 is F(x) An antiderivative of 2x Note that an antiderivative of x 2 2 Area of R1 (x 2 13 x 3 2x is F(x) 3 x 2. (a) 2x) dx [F(2) 2 (x 2 F(0) 0 3 x 2 ) dx (2x 0 0 (2x 4 3 2x) dx x 2) dx 2 [F(2) 4 3 0 F(3) 0 F(0)] 4 3 3 x 2 ) dx (b) Area 0 Area of R2 (2x 0 F(0)] 0 [F(3) 4 3 0 F(2)] 8 3 2 F(3) 0 24. F(2) 4 3 4 3 Total shaded area 4 3 4 3 8 3 [0, 5] by [ 5, 5] 21. An antiderivative of x 2 5 (a) 0 [0, 3] by [ 1, 8] An antiderivative of x 2 3 (a) 0 (x 2 6x 8) dx 6x F(3) 13 x 3 8 is F(x) F(0) 6 0 3x 2 6 8x. (x 2 4x) dx 4x is F(x) F(5) F(0) 13 x 3 25 3 2x 2. 0 25 3 Section 5.3 4 (b) Area 5 (x 2 4x) dx 0 30. The region between the graph and the x-axis is a rectangle 4x) dx 4 [F(4) F(0)] [F(5) 25 3 32 3 with a half circle of radius 1 cut out. The area of the region F(4)] 0 32 3 25. An antiderivative of x 2 x. (x 2 3 F( 3) 14 2 2 f (t) dt 1 0 2 1 2 av( f ) F(0) 0) . 2 4 4 . 31. There are equal areas above and below the x-axis. 1) dx 0 1 4 1 1 2 av( f ) 13 x 3 1 is F(x) 1 (1)2 2 is 2(1) 13 3 1 av (x 2 217 1 2 f (t) dt 0 0 0 3 1 (0 32. Since tan is an odd function, there are equal areas above 3 Find x c in [0, 3] such that c2 c2 1 c 1 Since 1 is in [0, 3], x 1. 1 and below the x-axis. 0 33. min f 2 3 x x 26. An antiderivative of is F(x) . 2 6 3 1 x2 1 1 av dx [F(3) F(0)] 30 2 3 3 c2 3 Find x c in [0, 3] such that . 2 2 c2 9 2 1 2 3 2 1 2 8 17 1 2 1 4 8 17 49 68 c 3 Since 3 is in [0, 3], x 27. An antiderivative of 1 1 av 3. 3x 2 1 is F(x) x 3 x. 1 ( 3x 2 1) dx F(1) F(0) 2 0 Find x 3c2 c in [0, 1] such that 3c2 1 2 1 c 1 0 0 1 16 17 0.5 16 17 0.5 1 x4 1 0 1 1 1 (1) 2 dx 1 2 dx x4 1 1 1 2 x4 0.5 1 1 1 dx x4 0.5 1 1 1 1 4 x4 01 1 1 dx x4 01 0 35. max sin (x 2 ) 1 3 2 2 f( ) d /4 1 and max f 2 1 1 dx 1 x4 01 34. f (0.5) 3 /4 1 /2 av( f ) 1 16 2 17 dx 8 17 dx 1 2 8 17 33 34 sin (1) on [0, 1] sin (x 2 ) dx sin (1) 1 0 1 c 36. max 3 1 Since 1 is in [0, 1]. x 3 1 3 av 3 1)2 dx (x . 22 0 3 and min x 1 (x 1)3. 3 18 1 F(0)] 33 3 37. (b 1 a) min f (x) Find x c in [0, 3] such that (c 1) 1. c1 1 c 2 or c 0. Since both are in [0, 3], x 0 or x 2. 29. The region between the graph and the x-axis is a triangle of height 3 and base 6, so the area of the region 1 2 av( f ) 1 6 9. 2 f (x) dx 4 9 6 3 2 8 8 dx 3 0 on [a, b] b 0 (b a) min f (x) f (x) dx a 2 is (3)(6) x 0 1)2 is F(x) 1 [F(3) 3 8 1 3 28. An antiderivative of (x x 38. (b a) max f (x) 0 on [a, b] b f (x) dx a (b a) max f (x) 0 2 2 on [0, 1] 218 Section 5.4 b 39. Yes, a 46. An antiderivative of F (x) is F(x) and an antiderivative of b av ( f ) dx f (x) dx. a G (x) is G(x). This is because av( f ) is a constant, so b av( f ) dx a b a b a b F (x) dx av( f ) b av( f ) a F(b) F(a) G (x) dx av ( f ) x G(b) G(a) a b (b a)av( f ) (b Since F (x) b a a b F (x) dx G (x) dx, so a b 1 a) G (x), f (x) dx F(b) F(a) G(b) a G(a). b a f (x) dx s Section 5.4 Fundamental Theorem of Calculus (pp. 277–288) 40. (a) 300 mi (b) 150 mi 30 mph 150 mi 50 mph 300 mi (c) 8h 8h Exploration 1 37.5 mph 2. The function y (d) The average speed is the total distance divided by the total time. Algebraically, 1 d1 2 t1 d2 d1 d2 t1 t2 1000 m3 100 min 10 m3/min 1000 m3 Time for second release 50 min 20 m3/min total released 2000 m3 1 Average rate 13 m3/min total time 150 min 3 41. Time for first release 1 42. 1 sin x dx x dx 0 0 1 43. 1 sec x dx 1 0 0 1 21 x 20 multiples of . There are six of these between 3. In attempting to find F( 10) 10 and 10. tan (t) dt 5, the 3 calculator must find a limit of Riemann sums for the integral, using values of tan t for t between 10 and 3. The large positive and negative values of tan t found near the asymptotes cause the sums to fluctuate erratically so that no limit is approached. (We will see in Section 8.3 that the “areas” near the asymptotes are infinite, although NINT is not designed to determine this.) 4. y tan x 1 2 x2 dx 2 x3 1 60 x 7 6 [1.6, 4.7] by [ 2, 2] 1 bh 2 44. (a) Area 2 tan x has vertical asymptotes at all odd 10 . The driver computed . The two expressions are not equal. t2 Graphing NINT f 5. The domain of this continuous function is the open interval (b) h2 x 2b C 2 b (c) y (x) dx 0 45. av(x k) Graph y1 1 k h 2b x 2b 0 k x k dx 0 hb 2 2b 1 bh 2 11 k xk 1 kk 1 0 xx 1 and y2 x(x 1) 3 . 2 6. The domain of F is the same as the domain of the kk 1 k(k 1) x on a graphing calculator and find the point of intersection for x , 1. continuous function in step 4, namely 2 , 3 . 2 7. We need to choose a closed window narrower than 2 , 3 2 to avoid the asymptotes. [1.6, 4.7] by [0, 16] [1, 3] by [0, 3] 8. The graph of F looks like the graph in step 7. It would be Thus, k 2.39838 decreasing on 2 , and increasing on vertical asymptotes at x 2 and x 3 . 2 , 3 , with 2 219 Section 5.4 Exploration 2 The Effect of Changing a in 10. x f (t) dt a dy dx 1 dx/dy 1 3x Section 5.4 Exercises 1. 3 1. 3 1 dx x 2 1/2 2x ln x 1/2 (6 [ 4.7, 4.7] by [ 3.1, 3.1] 2. ln 3) 1 ln 1 2 1 2 5 ln 3 ln 5 ln 3 ln 2 5 ln 6 3.208 [ 4.7, 4.7] by [ 3.1, 3.1] 3. Since NINT(x 2, x, 0, 0) 0, the x-intercept is 0. 2 4. Since NINT(x , x, 5, 5) 1 0, the x-intercept is 5. 5. Changing a has no effect on the graph of y It will always be the same as the graph of y d dx 1 1 3x ln 3 3x dx 2. 2 x 2 11 ln 3 3 f (t) dt. a 9 26 3 ln 3 f (x). 7.889 x 6. Changing a shifts the graph of y f (t) dt vertically in a 1 3. a dy dx dy 2. dx 3. dy dx f (t) dt. a 4. 2 5/2 x 5 x 3/2 dx 0 2 32 Quick Review 5.4 1. 5 1 13 x 3 x) dx 0 such a way that a is always the x-intercept. If we change from a1 to a2, the distance of the vertical shift is (x 2 6/5 x 5. dx 5 2 (25 5 0 1/5 5x 1 cos(x 2) 2x 1 6. 2 5 2 x 2 32 5 dx 1 dy dx 3 3x 5. dy dx 1 1/2 x 2 5 2 1 2 21 1 0 7. sin x dx cos x 1 0 ( 1) 2 0 8. (1 cos x) dx x sin x 0 2x ln 2 dy 6. dx 1 1 22.361 2 0 4. 1 5 0) 2 2(tan x)(sec2 x) 2 tan x sec2 x (0 10 1 2 x 2 2 3 0) 2 sin x cos x 2(sec x)(sec x tan x) 2 sec2 x tan x 2 dx x2 1 1 3 0 1 2x cos (x 2) 2(sin x)(cos x) 1 2 3/2 x 3 7. dy dx 8. dy dt dy dx 7 7x ( 0 (0 0) 3.142 /3 /3 dx dt dy/dt dx/dt 0 2( sin t x sin x cos x x2 2 5 /6 10. 2y) y x (y 1 2y y 2y csc2 3 3 1 x 3.464 5 /6 d cot /6 3 1) 0) /6 cot t dy 2y dx 1 2 tan 0 cos t sin t (1)y d x 2 ( sin x)(x) (cos x)(1) x2 cos t, 2 sec2 9. 1 9. Implicitly differentiate: dy x dx dy (x dx dy dx 0) 2 ( 3 3) 3.464 3 /4 3 /4 11. csc x cot x dx csc x ( /4 4 sec x tan x dx 0 ( 2) /3 /3 12. 2) /4 4 sec x 4(2 0 1) 4 0 220 Section 5.4 1 13. (r 1)2 dr 1 (r 3 u 4 1 4 14. 1 0 du 1)3 1 8 3 1 1/2 (u x3 17. Graph y 8 3 0 3x 2 2x. 1) du 0 u 4 2u1/2 u [0, 2] by [ 1, 1] 0 (4 15. Graph y 2 4) (0 0) Over [0, 1]: 0 1 x. (x 3 3x 2 14 x 4 0 x3 14 x 4 1 2 2x) dx x3 1 1 4 0 0 1 4 0 ( 4) 4 4 0 x2 0 1 4 Over [1, 2]: 2 (x 3 3x 2 2x) dx 1 2 Over [0, 2]: (2 2x 2x x) dx 12 x 2 12 x 2 0 3 Over [2, 3]: (2 x) dx 2 Total area 16. Graph y 1 2 2 3x 2 1 4 Total area [0, 3] by [ 2, 3] x3 18. Graph y 2 1 4 x2 2 1 1 4 4x. 2 0 3 3 2 2 2 1 2 5 2 [ 2, 2] by [ 4, 4] 3. Over [ 2, 0]: 0 (x 3 14 x 4 4x) dx 2 2x 2 0 2 Over [0, 2]: 2 [ 2, 2] by [ 4, 10] (x 3 14 x 4 4x) dx 0 Over [ 2, 1 (3x 2 1]: 3) dx Total area x3 1 3x 2 2 ( 2) 4 2 (b) (3x 2 3) dx x3 3x 1 1 2 2 (3x 2 Over [1, 2]: 3) dx 1 1 x x3 8 1 is discontinuous at x 1 1, x 1 2 2 ( 2) 4 9 . The area 2 between the graph of f and the x-axis over (1, 3] where 1 1 2 f is positive is (2)(2) 4 1. 1 2 3x 4 0 [ 2, 1) where f is negative is (3)(3) 1 Total area 2 The area between the graph of f and the x-axis over 4 1 2 x2 x 4 x2 x 19. (a) No, f (x) Over [ 1, 1]: 1 4 2x2 4 4 12 3 2 2 x x 20. (a) No, f (x) (b) 9 x2 3x 9 9 2 1 dx 1 2. 2 5 2 9 x2 is discontinuous at x 3x 9 3 x 3 ,x 3. 3 Note that f is negative for x in [0, 5]. f (0) 1 and 8 . The area between the graph of f and the 3 1 8 55 x-axis over [0, 5] is (5 0) 1 2 3 6 5 55 9 x2 dx 6 9 0 3x f (5) Section 5.4 21. (a) No, f (x) tan x is discontinuous at x 3 . 2 and x 2 26. First find the area under the graph of y 1 (b) The integral does not have a value. If 0 b 2 0 , then tan x dx ln cos x ln cos b since the 0 2 b ln cos b → cos b → 0 so or 2 3 0 2 2 3 x 2 dx 1 tan x dx → . 8 3 x 2. 7 3 Area of the shaded region , 13 x 3 2 1 3 0 Fundamental Theorem applies for [0, b]. As b → 1 x. Next find the area under the graph of y b b 2 3/2 x 3 x 1/2 dx 221 1 7 3 3 0 Hence the integral does not exist over a subinterval of 27. First find the area under the graph of y (1 [0, 2 ], so it doesn’t exist over [0, 2 ]. 0 x x2 b 1 dx 1 x since 2 x 0 1 1 x 1 1 x 1 dx ln x 0 x x2 1 dx → 1 b 0 x-axis for 5 /6 or The area of the rectangle is sin Area of the shaded region 31. 1, 2 2.55. The integral exists since 2x 4 x4 1 , x, 1 0.8, 0.8 cos x, x, 1 NINT( 2 1, 1) 2x 2 8 32. sin x is bounded. x the area is finite because 1 , x, 0, 10 2 sin x 3 24. (a) No, f (x) 0. 2 3 6 3 3 3.802 1.427 0.914 2x 2, x, 8.886 2, 2) t2 , t, 0, x), y2 2 0.6 in a [0, 1] by [0,1] window, then use the intersect function to find (b) NINT 1 cos x , x, x2 2, 3 2.08. The integral exists since the area is finite because 1 cos x is bounded. x2 1 x 2 dx 0 13 x 3 1 0 2 (2 1 x) dx 2x 12 x 2 Area of the shaded region 2 3 2 2 1 1 3 2 1 2 1 2 5 6 y 3 1 3 0, x x 1 3 NINT( 1 1 3 Next find the area under the graph of y 0.699. 34. When y y x 2. 25. First, find the area under the graph of y x x. 3 3 2 and x NINT(e 33. Plot y1 2 0 between x NINT( 8 cos x is discontinuous at x x2 3 2 /6 30. NINT 0. 3 cos x /6 29. NINT 1 5 . 6 sin x dx . Hence the integral does not exist sin x , x, x , 6 5 /6 1→ sin x and the 1, and the Fundamental Theorem sin x is discontinuous at x x . 28. First, find the area of the region between y ln b [0, 2]. 23. (a) No, f (x) 2 1, then over a subinterval of [0, 2], so it does not exist over (b) NINT 0 Area of the shaded region 1 applies for [0, b]. As b → 1 , ln b b sin x 1. b 1 cos x. The area of the rectangle is 2 . (b) The integral does not have a value. If 0 b x 0 1 is discontinuous at x 1 x x2 22. (a) No, f (x) cos x) dx 1 1. 3 x3 x 3, x, 0, 1) 0.883 222 Section 5.4 x 35. 43. Choose (d). x f (t) dt K f (t) dt a b x f (t) dt K dy dx x f (t) dt a b a t2 e dt b dy dx 1 (t 2 K 3t 1) dt 13 t 3 3 2 that cos 2x 0 3 sec t dt sec t dt t 8 3 ( 1) 6 dy dx 1 2 1 cos 2x. 2 sec t dt y (0) 0 1 sin (2x) 4 sec t dt 1 x d dx t2 e 1 1 t2 e x 2 48. f (x) 0 2 d dx sin 2 cos 2 2 dy dx cos (2x) x d dx x d dx 1 d ( dx 1 dt t x) d (2x) dx 2 1.189 f (0) (sin x) x cos (2t) dt 1 x 1 1/2 x 2 dt 0 0 3x cos (2x ) 2 x 0 cos x sin x cos x 10 1 t 1) 2x 2 10 dt 10 1 t dt 1 x 2 x 50. f (x) sin x 2x d f (t) dt dx 0 d (x cos x) dx x( sin x) f (4) 4 sin 4 1 cos x cos x cos 4 1 cos (2x ) 2x 2x cos (2x 2) 51. One arch of sin kx is from x /k sin kx dx 0 t 2 dt 2 sin x t 2 dt 0 t 2 dt 0 cos2 x ( sin x) sin x cos2 x 2x 10x Area d dx d dx 3 2 cos (2x ) 3x 3 3 d2 (x dx f (t) dt x sin x 0 2 x2 e 3 1 2 2 cos (2t) dt 0 2 4 10 2 cos (2x) 3 3 3 x 0 sin ( x)2 dt d 2 dx 49. f (x) f (0) dy dx 4 a 2 1 dt t 0 f (t) dt a since L (x) 1 d dx 4 a 47. x sin 2 cos 2 2 x sec x 46. Choose (a). 0 dy 37. dx 4 1 0 1 sin x cos x 2 1 4 0 1 cos (2x) dx 2 1 x 2 4 0 0 y (1) 1 x 2 0 x d dx dy dx sin t dt 2 4 45. Choose (b). 3 2 2 2 sin2 x, so sin2 x 1 1 2 sec x 2 2 0 4 1 2 dy dx 3 1 36. To find an antiderivative of sin2 x, recall from trigonometry 42. 0 x d dx y ( 1) 1 32 t 2 1 3 dy 41. dx 3 1 2 40. x2 e 44. Choose (c). f (t) dt 39. 3 b a dy 38. dx dt f (t) dt x K t2 e y( ) x f (t) dt x d dx sin2 x cos x sin2 x cos x 52. (a) (6 x x 2) dx 0 to x 1 cos kx k 6x 3 22 3 125 6 /k 0 12 x 2 27 2 k . 1 k 13 x 3 1 k 2 3 2 k Section 5.4 ( 1) 2( 1) (b) The vertex is at x ax 2 vertex of a parabola y y 1 2 1 . Recall that the 2 bx b . 2a c is at x 25 25 , so the height is . 4 4 55. (a) s (3) (b) s (3) f (3) 0 units/sec f (3) 0 so acceleration is positive. 3 (c) s (3) 1 ( 6)(3) 2 f (x) dx 0 (c) The base is 2 ( 3) 2 (base)(height) 3 5. 2 25 (5) 3 4 6 (d) s (6) 125 6 223 9 units 1 ( 6)(3) 2 f (x) dx 0 1 (6)(3) 2 particle passes through the origin at t 0, so the 6 sec. 0 53. (a) H(0) f (t) dt (e) s (t) 0 0 (b) H (x) H (x) x d dx f (t) dt f (x) 0 0 when f (x) 0. H is increasing on [0, 6]. (c) H is concave up on the open interval where H (x) f (x) 0. f (x) 0 when 9 x 12. H is concave up on (9, 12). f (t) dt 0 at t 7 sec (g) The particle is on the positive side since 9 s(9) f (x) dx 0 (the area below the x-axis is 0 smaller than the area above the x-axis). 12 (d) H(12) f (t) (f) The particle is moving away from the origin in the negative direction on (0, 3) since s (0) 0 and s (t) 0 on (0, 3). The particle is moving toward the origin on (3, 6) since s (t) 0 on (3, 6) and s (6) 0. The particle moves away from the origin in the positive direction for t 6 since s (t) 0. 0 because there is more area 0 0 x Si( x) H(12) is positive. 0 (e) H (x) f (x) 0 at x 6 and x 12. Since H (x) f (x) 0 on [0, 6), the values of H are increasing to the left of x 6, and since H (x) f (x) 0 on (6, 12], the values of H are decreasing to the right of x 6. H achieves its maximum value at x 6. (f) H(x) 0 on (0, 12]. Since H(0) minimum value at x 0. 54. (a) s (t) f (t). The velocity at t 0, H achieves its 5 is f (5) 2 units/sec. 0 (b) Si(0) 0 (c) Si (x) sin (t) dt t sin (t) dt t x x sin (t) dt t 0 x sin (t) dt t x 56. (a) f (t) is an even function so above the x-axis than below the x-axis. 0 0 sin t dt t f (t) Si(x) 0 0 when t k, k a nonzero integer. (d) (b) s (t) f (t) 0 at t 5 since the graph is decreasing, so acceleration at t 5 is negative. 3 (c) s (3) f (x) dx 0 1 (3)(3) 2 4.5 units [ 20, 20] by [ 3, 3] (d) s has its largest value at t 6 sec since s (6) f (6) 0 and s (6) f (6) 0. (e) The acceleration is zero when s (t) f (t) occurs when t 4 sec and t 7 sec. 100 57. (a) c(100) c(1) 1 100 0. This (f) Since s (0) 0 and s (t) f (t) 0 on (0, 6), the particle moves away from the origin in the positive direction on (0, 6). The particle then moves in the negative direction, towards the origin, on (6, 9) since s (t) f (t) 0 on (6, 9) and the area below the x-axis is smaller than the area above the x-axis. dc dx dx 1 1 9 s (9) f (x) dx 0 (the area below the x-axis is 0 smaller than the area above the x-axis). x x 2 10 1 400 c(100) 100 400 1 9 or $9 dc dx dx 1 400 dx 100 (g) The particle is on the positive side since 100 dx (b) c(400) sin (t) dt. t 20 2 x 10 x 100 10 or $10 224 Section 5.5 3 58. 2 2 (x 0 1) dx 2 2x 2(x 1 2 6 1 1) 9 2 2 3 64. Solving NINT 0 x 30 30 0 x2 dx 2 450 0 x3 6 1 450x 30 (b) (300 drums)($0.02 per drum) local minima at x f (x). (b) True, because h and h are both differentiable by part (a). 0. f (1) 1) , (2k 2) ], where 0, Si(x) has its 2k , where k is a positive integer. Furthermore, each arch of y f (x) is smaller in height than (2k 1) (2k 2) the previous one, so f (x) dx f (x) dx. 2k (2k 1) This means that (2k 2) Si((2k (d) True, because h (1) 1) ] and k is a nonnegative integer. Thus, for x $6 f (1) sin x , Si(x) is x f (x) decreasing on each interval [(2k 0 f (x) and therefore h (x) (c) True, because h (1) d Si(x) dx increasing on each interval [2k , (2k 30 300 drums 60. (a) True, because h (x) 1.0648397. We now argue that there are no other Exercise 56. Since The company should expect $4500. 1 1 graphically, the solution is solutions, using the functions Si(x) and f (t) as defined in 4.5 thousand 59. (a) sin t , t, 0, x t 0 and h (1) f (1) 2) ) Si(2k ) f (x) dx 0, so each 2k 0. successive minimum value is greater than the previous one. (e) False, because h (1) f (1) 0. Since f (2 ) (f) False, because h (1) f (1) 0 0 1.42 and Si(x) is 0, this means Si(x) Si(x) 2 . Now, Si(x) 1) for x 1.42 (and hence 1 has exactly one solution in the interval [0, ] because Si(x) is increasing on x f (t) dt sin x , x, 0, 2 x continuous for x (g) True, because h (x) f (x), and f is a decreasing function that includes the point (1, 0). 61. Since f (t) is odd, NINT f (t) dt because the area x 0 this interval and x 1.065 is a solution. Furthermore, between the curve and the x-axis from 0 to x is the opposite Si(x) of the area between the curve and the x-axis from 1 has no solution on the interval [ , 2 ] because x to 0, Si(x) is decreasing on this interval and Si(2 ) 1.42 1. but it is on the opposite side of the x-axis. x 0 x f (t) dt 0 f (t) dt x x Thus Thus, Si(x) x f (t) dt 1 has exactly one solution in the interval f (t) dt 0 0 [0, ). Also, there is no solution in the interval ( , 0] f (t) dt is even. 0 because Si(x) is odd by Exercise 56 (or 62), which means 0 62. Since f (t) is even, x f (t) dt f (t) dt because the area x that Si(x) 0 for x 0 (since Si(x) 0 for x 0). 0 between the curve and the x-axis from 0 to x is the same as the area between the curve and the x-axis from x 0 f (t) dt 0 Thus f (t) dt x x x to 0. f (t) dt Exploration 1 0 f (t) dt is odd. 0 x 63. If f is an even continuous function, then d dx f (t) dt is odd, but 0 x f (t) dt s Section 5.5 Trapezoidal Rule (pp. 289–297) x f (x). Therefore, f is the derivative of the odd 0 Area Under a Parabolic Arc 1. Let y f (x) Ax 2 Bx C Then y0 f ( h) Ah 2 Bh y1 f (0) A(0)2 B(0) C y2 f (h) Ah 2 Bh C. 4y1 2. y0 y2 x continuous function f (t) dt. h 0 Similarly, if f is an odd continuous function, then f is the x derivative of the even continuous function f (t) dt. 0 3. Ap 3 (Ax 2 h 2 Ah 2 Bh C 2Ah 2 6C. Bx x x B Cx 3 2 3 2 h h A B Ch 3 2 3 h 2A 2Ch 3 h (2Ah 2 6C) 3 A C, C, and 4C Ah 2 h2 2 Ch C) dx h h A h3 3 B Bh C Section 5.5 4. Substitute the expression in step 2 for the parenthetically enclosed expression in step 3: Section 5.5 Exercises 1. (a) f (x) h (2Ah 2 6C) 3 h (y 4y1 y2). 30 Ap 2 x, h x 0 Quick Review 5.5 1. y sin x y cos x y 0 on [ 1, 1], so the curve is concave down on [ 1, 1]. 2. y y y 3 4x 12 12x 2 0 on [8, 17], so the curve is concave up on [8, 17]. 3. y 12x 2 6x y 24x 6 y 0 on [ 8, 0], so the curve is concave down on [ 8, 0]. 4. y y y 2 (c) 5. y y y 2e 4e 2x 0 on [ 5, 5], so the curve is concave up on [ 5, 5]. 6. y y y y 8. y y 0 0 1 2 1 4 1 0 4 T y 2 (c) 2 2 4 2.75 1 2 3 2 9 4 1 1 1 4 2(1) 2 2 4 9 4 x ,h 0 1 2 1 8 0 1 0 4 T 0 1 2 4 0 f (x) 8 3 2 3 3. (a) f (x) 2 13 x 3 x 2 dx 0 2 3 2 27 8 1 1 1 8 2(1) 2 2 8 27 8 8 4.25 (b) f (x) 3x 2, f (x) 6x 0 on [0, 2] The approximation is an overestimate. 2 (c) 2 14 x 4 x 3 dx 0 0 on [3, 6], so the curve is concave up on [3, 6]. cos x sin x sin x cos x 0 on [1, 2], so the curve is concave down. 2 (b) f (x) 2x, f (x) 2 0 on [0, 2] The approximation is an overestimate. 4 0 1 ,h x 2 x 1 f (x) 1 5 4 4 5 4. (a) f (x) 9. y 100x 9 y 900x 8 y 0 on [10, 1010], so the curve is concave down on [10, 1010]. 10. y y y 0 4 0 on [100, 200], so the curve is concave down on csc x cot x ( csc x)( csc2 x) (csc x cot x)(cot x) csc3 x csc x cot2 x 0 on [0, ], so the curve is concave up on [0, ]. 3 2 2 f (x) 1 x2 2 x3 2 0 2 x 1 x2 2 2 2 x [100, 200]. 7. y 2(1) x 2, h 2. (a) f (x) 1 x y 1 1 2 12 x 2 x dx 0 [48 , 50 ]. 2x 2 3 2 3 2 1 (b) f (x) 1, f (x) 0 The approximation is exact. 1 x cos 2 2 1 x sin 4 2 0 on [48 , 50 ], so the curve is concave down on 1 0 4 T 1 2 1 2 1 2 0 f (x) 0 4 1 1 8 T (b) f (x) 1 1 4 4 2 4 5 3 2 2 3 2 1 , f (x) x2 2 3 2 x3 7 4 4 7 2 2 1 2 4 7 1 2 0 on [1, 2] The approximation is an overestimate. 2 (c) 1 1 dx x 2 ln x ln 2 1 0.693 0.697 225 226 Section 5.5 5. (a) f (x) 4 x, h 0 1 4 x 0 1 f (x) 0 1 1 (0 2 T 2 3 2( 4 x 3 2 f (x) 2( 2 2(1) 3) 2) 2) 1 3/2 x 4 1 4 0 6. (a) f (x) 0 f (x) 0 8 (b) f (x) 0 16 3 0 4(0) 4 (x 3 2x) dx 2 2 2 14 x 4 2 0 sin x dx cos x 7. 5 (6.0 2 1.896 1 (126 2 2 65 1 12 … 2 66 841 3.13791, S100 2(800) 1.08943, S100 15. S50 1.37066, S100 limit S50 1.37076, S100 lower limit 26,360,000 ft3 2(860) 0.0001 as lower 1.37076 using a 0.000000001 as 0.82812 17. (a) T10 1.983523538 T100 1.999835504 T1000 1.999998355 (b) ET n 10 988.5, the town can sell at most 988 licenses. 2 0.016476462 1.6476462 100 1.64496 1000 1 to change seconds 3600 Tn 1.645 (3.2 2.2)(30 (4.5 3.2)(40 50) (5.9 (7.8 5.9)(60 70) (10.2 (12.7 10.2)(80 90) 16.0)(100 20.6)(110 26.2)(120 130)) (d) b 80) 12.7)(90 a 100) ET n ET 10n 1 3600 2 10 10 ET n , h2 2 120) (37.1 7.8)(70 60) 110) (26.2 10n 4.5)(50 (16.0 (20.6 (c) ET 40) 0.9785 miles 12 n 2 3 12(10n)2 2 ,M n2 3 1 12n 2 10 2 4 10 to hours 30) 841 1.37066 using a 0.82812, S100 16. S50 19,770 fish to be caught. Since 9. Sum the trapezoids and multiply by 110) 1.08943 … 2(1000) 2 58 3.14029 14. S50 (b) You plan to start with 26,360 fish. You intend to have 1 (2.2(0 2 0. 70.08 13. S50 0)(20) 19,770 20 1 0 since f (4) 2x, Mf (4) 13.0)(30) 2(12.7) 2(520) (0.75)(26,360) 1 4 9 (b) We are approximating the area under the temperature graph. By doubling the endpoints, the error in the first and last trapezoids increases. 15,990 ft3 200 8. (a) (0 2 12. (a) 0 on [0, ] 2 … 2(9.1) x3 av( f ) 0 2 0 2(8.2) 3 11. The average of the 13 discrete temperatures gives equal weight to the low values at the end. 2 2 sin x 0 12 (d) Simpson’s Rule for cubic polynomials will always give exact values since f (4) 0 for all cubic polynomials. The approximation is an underestimate. (c) x2 21) 0 (c) For f (x) 2 cos x, f (x) 4(4) 12 4 2(1) 2 21 1 3 4 1 2 4 81 4 Es 2 3 2( 1) 5.333 4 2 1 1 0 sin x, h x T 4 2 3/2 x 3 x dx 1 0 The approximation is an underestimate. (c) ( 1) 4 1 0 1 (1 3 (b) 0 on [0, 4] 3 2x, h 1 S 5.146 3 1 1/2 x , f (x) 2 (b) f (x) x3 10. (a) f (x) ET n 6 10 2 Chapter 5 Review 22. Note that the tank cross-section is represented by the shaded area, not the entire wing cross-section. Using Simpson’s Rule, estimate the cross-section area to be 18. (a) S10 2.000109517 S100 2.000000011 S1000 2.000000000 (b) ES n 10 2 1.09517 100 10 1000 (c) ES 10 a ,h 4 ES 4 n4 ,M n ES 10n 1 h[ y0 yn 1 180(10n)4 y6] 2(2.0) 2yn 1 10.63 ft yn] h[ y1 y2 … 4y3 … 2y2n RRAMn LRAMn 4 10 ES 2 n 2x cos (x 2) 2x 2x sin (x 2) 2 cos (x 2) 4x 2 sin (x 2) 2 cos (x 2) 24. S2n h (y 30 4y1 2y2 4y2n 1 [h(y0 3 (b) y2n ) 2y2 2y1 1 … (2h)(y1 2T2n MRAMn 3 y3 2y2n … y5 b , where h 1 a 2n y2n) y2n 1)] . [ 1, 1] by [ 3, 3] (c) The graph shows that for 1 x 1. 1 (d) ET (e) For 0 1 (f) n 3 h2 2 0.1, ET ( 1) h 2 0.1 f (x) 2 so f (x) 3 s Chapter 5 Review (pp. 298–301) 1. h2 2 ( 1) 2 (h )(3) 12 h f (4)(x) y 4 0.12 2 0.005 0.01 20 2 4x2 2x cos (x 2) 8x sin (x 2) 4x sin (x 2) 8x3 cos (x 2) 12x sin (x 2) 8x3 2x sin (x 2) 24x2 cos (x 2) 12x 2x cos (x 2) 12 sin (x 2) (16x 4 12)sin (x 2) 48x 2 cos (x 2) 20. (a) f (x) 1 2. 2 1 (b) 2 x y 4 [ 1, 1] by [ 30, 10] (c) The graph shows that 30 f (4)(x) so f (4)(x) 30 for 1 x 1. 1 (d) Es (e) For 0 (f) n 21. h ( 1) 4 (h )(30) 180 h 1 24 in. 6 2 0.4 4 h 3 0.4, ES ( 1) h h4 3 2 10 4 0.4 3 0.00853 0.01 LRAM4: 5 4 in. Estimate the area to be 4 [0 3 0] 4(18.75) 466.67 in2 yn] 2 180n 4 5 19. (a) f (x) f (x) … 5 180 n 4 4(1.9) 1 1 42 lb/ft3 11.2 ft2 2y2 … 4y5 11.2 ft2 2.1] 2y1 y1 2y4 2(1.8) (5000 lb) h [y 20 23. Tn n 4 4y3 4(2.1) Length 4 ES 8 0 10n (d) b 4 10 1.1 1 [y 4y1 2y2 30 1 [1.5 4(1.6) 3 Sn 2(24) 4(26) 2(24) 4(18.75) 1 0 2 15 8 3 x 21 8 15 4 3.75 2 227 228 Chapter 5 Review 3. 2 y 5 f (x) dx 9. (a) f (x) dx 5 4 3 2 The statement is true. 5 (b) 2 [ f (x) 2 g(x)] dx 5 5 f (x) dx g(x) dx 2 2 1 MRAM4: 4. 1 63 2 64 165 64 f (x) dx x 2 2 5 195 64 4 105 64 5 f (x) dx 2 g(x) dx 2 3 2 2 9 4.125 The statement is true. y 5 (c) If f (x) 4 5 g(x) on [ 2, 5], then f (x) dx g(x) dx, 2 2 but this is not true since 5 2 5 f (x) dx 2 f (x) 2 5 2 g(x) dx f (x) 4 3 7 and 2 2. The statement is false. 2 1 1 15 RRAM4: 28 21 8 3 10. (a) Volume of one cylinder: r 2h x 2 15 4 0 Total volume: V lim n→ 3.75 y ∑ sin2(mi ) x i1 sin2 x on [0, ]. (b) Use 5. sin2 (mi ) x n NINT( sin2 x, x, 0, ) 4.9348 4 11. (a) Approximations may vary. Using Simpson’s Rule, the area under the curve is approximately 2 1 [0 3 4(0.5) 2(4.75) 1 (LRAM4 2 T4 2 6. 2 (4x x 3) dx 2x 2 7. 14 x 4 1 15 24 15 4 3.75 0] 26.5 s 4 4 0 LRAMn MRAMn RRAMn 1.78204 1.60321 1.46204 20 1.69262 1.60785 1.53262 30 1.66419 1.60873 1.55752 50 10 1.62557 1.60937 1.59357 The curve is always increasing because the velocity is always positive, and the graph is steepest when the velocity is highest, at t 6. 1000 1.61104 1.60944 1.60784 1 dx x 5 ln x ln 5 1 ln 1 ln 5 t Time (sec) 1.64195 1.60918 1.57795 100 1 4(2) 4(4.5) 30 8 10 5 (b) 2 n 8. 2(3.5) 2(3.5) The body traveled about 26.5 m. RRAM4) 0 4(4.5) 4(2) Position (m) 1 x 2(1) 1.60944 229 Chapter 5 Review 10 e x 3 dx 12. (a) 1 10 (b) 1 x sin x dx 25. 0 10 ln x 1 36(2x dx 1)3 1) 9(2x 1 2 10 (e) 1) ( 9) 0 1 13. The graph is above the x-axis for 0 x-axis for 4 x x 2 1 dx x2 x 26. x sin dx 10 2 9 8 x 2) dx (x 1 12 x 2 4 and below the 1 x 1 2 2 0 0 Total area 6 (4 x) dx (4 0 27. x) dx 2 1 3 2 6 4 sec x tan x dx sec x 1 /3 4 12 x 2 [8 0] 0 [6 8] 1 4 28. 10 x 2) cos (1 1 1 x 2 and below the 29. 1 0 2 2 y 0 1 1 1 dy 2 ln y 1 2 ln 3 0 2 ln 3 0 30. Graph y x 2 x 2) dx 2x sin (1 2 14. The graph is above the x-axis for 0 1 6 12 x 2 4x 2 /3 4 4x x-axis for dx 1 dx x2 1 2 0 1 10 3 0 2)2 dx x(3x 0 1 1 36 0 (d) 0 1 (2x 0 (c) e 1 dx x 24. 0 4 x 2 on [0, 2]. /2 Total area cos x dx cos x dx 0 /2 /2 sin x sin x 0 (1 /2 0) (0 1) 2 [ 1.35, 3.35] by [ 0.5, 2.6] The region under the curve is a quarter of a circle of radius 2. 2 2 15. 5 dx 5x 10 ( 10) 50 2 8 20 2 2 5 4x dx 16. 2x 2 2 42 0 2 31. Graph y /4 /4 cos x dx 17. 2 sin x 0 1 (3x 2 4x x3 7) dx 1 2x 2 7x 1 6 1 (8s 3 12s 2 ( 10) 2s4 5) ds 16 20. 1 5s 4 x 4 dx x2 y 4 1 4/3 8 x dx 2 ( 4) 2 32. Graph y 1 dy 3y 1/3 1 (4)(4) 2 64 1 (8)(8) 2 40 x 2 on [ 8, 8]. 27 1 ( 3) 2 1 ( 2) 1 dt tt sec2 0 3 4 4 t 3/2 dt 2t 1 1/2 4 1 d tan 3 0 1 [ 9.4, 9.4] by [ 3.2, 9.2] The region under the curve y of radius 8. /3 /3 23. 0 2 1 22. 3 0 27 21. [ 4, 8] by [0, 8] The region under the curve consists of two triangles. 1 4s3 0 2 x dx on [ 4, 8]. 2 1 19. 1 (2)2 4 2 0 2 0 18. x 2 dx 4 5 0 3 8 2 8 64 x 2 dx 8 2 64 8 64 x 2 dx x 2 is half a circle 2 1 (8)2 2 64 230 Chapter 5 Review 33. (a) Note that each interval is 1 day 24 hours Upper estimate: 24(0.020 0.021 0.023 0.025 0.028 0.031 0.035) 4.392 L Lower estimate: 24(0.019 0.020 0.021 0.023 0.025 0.028 0.031) 4.008 L 24 (b) [0.019 2 2(0.020) 0.035] 3 [5.30 2 x 2 25 2(0.031) dt 1 0 1 t2 1 dt 1 2 x2 1 1 x2 dt t 4t1/2 1 50 x 50 25 4 1.11) … 0) 2(1.11) 20 50 4 x 30 4 2500 87.15 ft … x c(2500) … 2(5.04) 2(5.25) t 0 x 1 2 1 (2x)2 1 2 4x 2 1 4.2 L 34. (a) Upper estimate: 3(5.30 5.25 5.04 Lower estimate: 3(5.25 5.04 4.71 (b) 2x d dx 43. c(x) … 2(0.021) dy dx 42. 103.05 ft 30 230 0] The total cost for printing 2500 newsletters is $230. 95.1 ft 14 44. av(I) 35. One possible answer: The dx is important because it corresponds to the actual physical quantity x in a Riemann sum. Without the x, our integral approximations would be way off. 4 36. 0 f (x) dx 4 0 0 x 2 dx 2) dx 4 13 x 3 2x 4 [0 64 3 16] 2 1 since min sin x 0) 16 3 1 1 1 x4 sin x dx sin2 x dx 1 (max f )(1 0) 2 4 1 4 0 0 2t (24 0 14 t 4 14 x 4 3) dt 4 1 16 43 0 0 3x a 1 a 0 a 1 2 3/2 ax a3 x dx 0 39. dy dx 2 4 3 x2 3x 4 dy dx 2 cos3 (7x 2) 4x 2 12x 16 dy 41. dx d dx 1 6 3 t4 x2 3x 0 0 3.09131 1.63052. f (x). f (1) (d) False, because g (1) f (1) (e) True, because g (1) f (1) f (1) 0. 0. 0 and g (1) f (1) 0. (g) True, because g (x) f (x), and f is an increasing function which includes the point (1, 0). d (7x 2) dx 14x 2 cos3 (7x 2) 1 x 4 dx 1 47. F(1) 0 x 3t (b) True, because g is differentiable. 2 3/2 a 3 0 192 0 x t2 (f) False, because g (1) a 14 0 46. (a) True, because g (x) cos3 x 40. 12t 2 4 (c) True, because g (1) (b) av( y) 1 24t 14 24t) dt x2 or x 1 2 3/2 x 43 x dx 24t Using a graphing calculator, x 0 38. (a) av( y) (t 3 14 x 4 14 x 4 2 24 14 0 0 1 0 45. 0 1 (min f )(1 x 0 0 2 since max sin2 x min f 4 sin2 x 1 1 14 Rich’s average daily holding cost is $192. We could also say (0.04)4800 192. 0 0 12 x 2 0.04 I(t) av(c) 4 (x max f c(t) f (x) dx 4 37. Let f (x) Rich’s average daily inventory is 4800 cases. 4 f (x) dx 1 (600 600t) dt 14 0 14 1 [600t 300t 2 4800 14 0 dt 6 3 x4 x 48. y (x) 5 sin t dt t 3 F(0) 0. Chapter 5 Review 49. y 1 x 1 x2 2x y 2 b (b) Let h S2n a 2n h [y 30 1 y(1) 1 1 y (1) 1 dt t 1 2 1 3 0 1 1 [h(y0 3 2 h( y2i 2t dt x t2 4 (x 2 4 1 1) x2 4 1 [2.5 2(2.4) 24 29 2.42 gal 12 4y2i 2 h(y2 4y2n 1 h. 12 54. (a) g(1) 2(2.4) 2 MRAMn 1 f (t) dt 2 1 (c) g( 1) 24.83 mi/gal (2)(1) 2704 ft. When her parachute opens, her 4296 ft. For t SA(t) 6144 6208 16 6208 388 sec. For t 45 4296 16(t so B lands at t 5224 16 53. (a) Area of the trapezoid 58) 16t, so A lands at 13 58 sec, B’s 5224 0 and the absolute minimum is 3 1 f (t) dt f (t) dt 3 1 (2)2 2 The range of g is [ 2 , 0]. 55. (a) NINT(e x 2/2 , x, 10, 10) , x, 2.506628275 20, 20) x 2/2 2.506628275 16t, (b) The area is 2. 326.5 sec. B lands first. 56. First estimate the surface area of the swamp. 1 (2h)(y1 2 (2h)y2 h(y1 h(y1 2(2hy2) (f) g (x) f (x), f (x) 0 at x 1 and f (x) is not defined at x 2. The inflection points are at x 1 and x 2. Note that g (x) f (x) is undefined at x 1 as well, but since g (x) f (x) is negative on both sides of x 1, x 1 is not an inflection point. NINT(e Area of the rectangle y3) (e) g ( 1) f ( 1) 2 The equation of the tangent line is y ( ) 2(x 1) or y 2x 2 1 position is given by SB(t) 1 (2)2 4 (d) g (x) f (x); Since f (x) 0 for 3 x 1 and f (x) 0 for 1 x 3, g(x) has a relative maximum at x 1. g( 3) 4 sec, A’s position is given by t f (t) dt 1 (g) Note that the absolute maximum is g(1) (c) Let t represent the number of seconds after A jumps. 4) 1 1 f (t) dt 1 (b) The distance B falls in 13 seconds is 16(t . 0 3 (b) g(3) 2704 Tn 3 f (t) dt 2.3] 6144 ft. altitude is 7000 … 1 … When her parachute opens, her altitude is 1 (32)(132) 2 y4) 1 12 52. (a) Using the freefall equation s gt from Section 3.4, 2 1 the distance A falls in 4 seconds is (32)(42) 256 ft. 2 256 4y3 y2n )] 1 1 6400 2 the ith of n rectangles plus the area of the ith of n trapezoids, S2n 2(2.3) 12 (b) (60 mi/h) h/gal 29 2y2n y2i ) is equal to twice the area of 1 3 1 51. (a) Each interval is 5 min … 2y4 Since each expression of the form 50. Graph (b). x y2) 2 4y3 y2n] 1 4y1 h(y2n Thus, it satisfies condition ii. y 2y2 4y2n 1 1 1 2 . 4y1 Thus, it satisfies condition i. 231 y3) 2hy2 4y2 y3) h(y1 y3) 20 [146 2 13] 2(122) 2(76) 2(54) 8030 ft2 (5 ft)(8030 ft2) 1 yd3 27 ft3 1500 yd3 2(40) 2(30) 2. 232 Section 6.1 57. (a) V 2 (V max)2 sin2 (120 t) 5. Again h 1 1 av(V ) 7 2 2 (Vmax) 2 2 sin (120 t) dt 240 21 (Vmax)2 2 2 2 1 2 Vmax ,3 1 2 ,5 1 2 , ..., 19 1 2 2 1 2 339.41 volts Chapter 6 Differential Equations and Mathematical Modeling 1 2 3 2 7 ,0 2 x 1, x 3 2 7 ,1 2 x 2, x 5 2 7 ,2 2 x 3, 1 2 y3 1 2 1 3 y2 s Section 6.1 Antiderivatives and Slope Fields (pp. 303–315) x 2 y1 Exploration 1 2 The 10 graphs are graphs of the functions. 2 2 19 . From left to right the 2 slopes of the line segments are (Vmax) 0 2 (b) Vmax 2 (Vmax) sin (120 t) dt (Vmax)2 Vrms 135 222 each line segment are , , , ..., 1 0 1 1. The y-coordinate of the midpoint of each line segment is . The x-coordinates of the midpoint of Using NINT: 2 k 5 2 1 Constructing a Slope Field 1. As i and j vary from 1 to 10, 100 ordered pairs are produced. Each ordered pair represents a distinct point in the viewing window. 2 y10 19 2 x 19 2 1 7 ,9 2 x 10. 2. The distance between the points with j fixed and i r and i r 1 is the distance between their x-coordinates. Xmin 2(r (Xmin 1) Xmin) 1 h 2 (2r Xmin 2 1 (2r 2r 1) 1) h 2 h 2 [0, 10] by [0, 10] h 3. The distance between the points with i fixed and j r and j r 1 is the distance between their y-coordinates. Ymin (2(r k 2 1) Ymin) (Ymin 1) (2r Ymin 2 1 (2r 2r 1) 1) k 2 k 2 6. For each line segment in part (5), make a column of parallel line segments as in part (4). 7. WL Quick Review 6.1 1. 100(1.06) k $106.00 2. 100 1 4. Here h k 0.06 4 4 3. 100 1 0.06 12 12 4. 100 1 0.06 365 365 1. Each line segment in the third column has 4 7 slope , because the x-coordinate of the midpoint of each line segment is 2.5. The y-coordinates are , , , …, 135 222 19 . 2 The 10 graphs are graphs of the functions y $106.14 $106.17 $106.18 n ,2 2 x 3, for n 1, 3, 5, …, 19. The length of the line segment can be increased or decreased by adjusting the restriction 2 [0, 10] by [0, 10] x dy dx d sin 3x dx (cos 3x)(3) 6. dy dx 5 d tan x 2 dx sec 2 x dy dx d Ce 2x dx (Ce 2x)(2) 8. 2.5) 5. 7. 4 (x 7 dy dx d ln (x dx 2) 5 2 3 cos 3x 5 2 5 5 sec 2 x 2 2 3. 2Ce 2x 1 x 2 Section 6.1 9. 6. 1 x dx 3 ln x 3 C Check: d [ln x dx [0.01, 5] by [ 3, 3] By setting the left endpoint at x 0.01 instead of x 7. (x 5 0, we avoid an error that occurs when our calculator attempts 1 to calculate NINT , x, 1, 0 . The graph appears to be the x same as the graph of y 3 ln x. 6x 8. ( x 3 9. x x 3 x6 6 3) dx 5 t2 e t/2 1 C] 3x 2 2 x 1) dx 2e t/2 10. [ 5, 0.01] by [ 3, 3] 11. By setting the right endpoint at x 0.01 instead of x 43 t dt 3 1 dx x3 x3 (x 3 1 , x, x to calculate NINT 1, 0 . The graph appears to be the same as the graph of y 1. (x 2x 13. x3 3 1) dx x 2 x C 3 dx dx 3 x2 x 4 2. ( 3x ) dx x2 C 3 x 2x 1 15. 17. d (x 3 dx C) 3x x) dx (x 2 4x x3 3 1/2 ) dx 8 3/2 x 3 Check: d x3 dx 3 4. (8 8 3/2 x 3 cos 1 x 1 C 1 x 1 4x e 4 2 x 2 8x 4x 1/2 x csc x 2 C dx cos 3x 3 C C C 2 ln x e cos 5x 5 tan 5r cot 7t 7 x C cos2 x dx 8 csc x cot x C 22. sin 2 x dx Check: d 1 4x e dx 4 x 2 2 sec t sin 5x 2 ln x 21. 5. e 4x dx C 3 cos x sin 20. csc 2 7t dt C C) ) dx C x dx 2 1/3 3 2/3 x 2 sin 3x) dx Check: csc x x 3 4/3 x 4 19. 5 sec 2 5r dr csc x cot x) dx d (8x dx (x 1/3 x 1/3 2 x C 1 ln x C 4 18. 4 2 C 2 1 2x 2 dx 1 2/3 x dx 3 C 2 x x 16. 2 sec t tan t dt C Check: 3. (x 3 14. (3 sin x Check: 2 1 x ln ( x). Section 6.1 Exercises 2 3 12. C x 3) dx x 4 x4 4 we avoid an error that occurs when our calculator attempts C C t 4/3 4 0, 1 5 t 4 1/3 t dt 3 x ) dt 5t t/2 C 2 5t 10. 2e x2 2 2 (e t/2 dt 3x 23. tan 2 d dx 2x e 2 C C C 1 cos 2x dx 2 1 cos 2x dx 2 2 sin 2x x C 4 2 1 1 2 e 4x 2x cos 2x dx 2 cos 2x dx 2 (sec 2 1) d x 2 tan sin 2x 4 C C 233 234 Section 6.1 24. cot 2 t dt (csc 2 t 1) dt cot t t C dy dx 29. dy dx dx 25. (a) Graph (b) (b) The slope is always positive, so (a) and (c) can be ruled out. 1 when x is positive since slope dy dx and (c) don’t show this slope pattern. y 2x 1 2 x. Graphs (a) tan 1 negative, zero slope when x is zero and negative slope dy dx dx 1 4 C C 4 C Solution: y (2x tan x 2 x2 1) dx x C , 0 by [ 8, 8] dy dx x 22 30. dy dx dx 2 y 1 x2 2/3 x 2/3 3x 1/3 dx C Initial condition: y ( 1) 5 3( 1)1/3 5 3C 2C y 3x 1/3 2 [ 4, 4] by [ 3, 3] dy dx C 1 Initial condition: y (2) 0 22 2 C 02C 2C Solution: y x 2 x 28. tan x Initial condition: y (b) The solution should have positive slope when x is dy dx sec 2 x dx y 26. (a) Graph (b) 27. sec 2 x 5 C x dy dx dx (x 2 x) dx y x 1 x2 2 y x2 2 1 x 1 1 1 2 Solution: y 2 2 3 2 [ 4, 4] by [ 8, 4] C Initial condition: y (2) 2 C 1 2 1 dy dx dy dx dx C C y C x2 2 31. 1 x [ 6, 6] by [ 4, 4] 1 2 9x 2 (9x 2 3x 3 4x 4x 2x 2 5 5) dx 5x C Initial condition: y ( 1) 0 0 3( 1)3 2( 1)2 5( 1) C 0 10 C 10 C Solution: y 3x 3 2x 2 5x 10 Section 6.1 dy dx 32. cos x dy dx dx (cos x y sin x cos x dy dt 2e dy dt dt C y t ln 2 0 2 2 0 1 y 0 C dy dx dx y t 1 1 x cos 1) d sin sin 0 C Initial condition: y(e 3) 0 0 ln (e 3) C 03C 3C Solution: y ln x 3 0 3 C2 C2 d 2y dx dx 2 2 dy dx dx 2 dy dx 1 dx x 1 C2 Solution: y 36. ln x dy d ( cos 3 dy dx C1 Initial condition: y (0) C 2e 0 C1 dy d d C 3 Solution: y C1 C1 C 1 34. cos 0 First derivative: Initial condition: y (ln 2) 2e 0 dt t d cos 1 2e 0 sin Initial condition: y (0) t 2e sin d 2y d d2 dy d sin x) dx Initial condition: y ( ) 1 1 sin cos C 11C 0C Solution: y sin x cos x 33. d 2y d2 35. sin x sin 2 3 6x (2 6x) dx 3x 2 2x C1 Initial condition: y (0) 4 4 3(0)2 2(0) 4 C1 First derivative: dy dx dx y x2 dy dx 4x 02 1 03 4) dx C2 Initial condition: y (0) 1 3x 2 2x 3x 2 (2x x3 C1 1 C2 4(0) C2 Solution: y x2 x3 4x x3 x2 4x 1 or y 1 4 235 236 37. Section 6.1 d 3y 1 dt 3 t3 d 3y dt t 3 dt dt 3 d 2y 12 t C1 dt 2 2 38. 4 1 (1) 2 2 1 C1 2 2 5 2 cos sin cos 0 dy dt dt 2 dy 11 t dt 2 3 1 2 dy dt 2 12 t 2 12 5 t 2 2 5 t C2 2 1 (1) 1 2 3 3 5 (1) 2 5 2 C1 C1 dy d d3 d 2y d2 sin 0 1 1 0 C2 2 5 t 2 C2 cos 0 2(0) d 2y d2 cos 0 1 2 cos cos 2 2 )d 2 sin 1 C3 1 2 sin 0 0 C3 dy d cos dy d d y C3 C3 First derivative: 1 4 C2 sin ( sin 1 C3 52 t 4 1 Initial condition: y (0) 1 C2 C2 dy d d2 dy cos d C3 1 ln t 2 2 2) d cos 2 dy 11 First derivative: t dt 2 dy 11 5 dt t t dt dt 2 2 1 52 y ln t t C3 2 4 sin sin sin Second derivative: 1 52 ln 1 (1) 2 4 5 C3 4 cos ( cos 1 3 C2 Solution: y d 3y d3 Initial condition: y (0) C2 Initial condition: y (1) C1 3 dt Initial condition: y (1) 3 C1 3 sin 0 Third derivative: 2 1 4 cos ) d 3 C1 Second derivative: 1 (sin Initial condition: y (3)(0) 2 1 cos 2 C1 0 sin dy d d4 d 3y d3 Initial condition: y (1) 2 d 4y d4 (cos 2 sin 2 sin 2 2) d 3 sin cos 2 3 Initial condition: y (0) 3 sin 0 3 1 4 cos 0 C4 3 03 3 C4 2(0) C4 C4 3 Solution: y 39. ds v dt ds dt dt s 4.9t sin 9.8t (9.8t 2 5t cos 5 5) dt C Initial condition: s(0) 10 10 4.9(0)2 5(0) C 10 C Solution: s 4.9t 2 5t 10 3 2 4 Section 6.1 40. ds v sin t dt ds dt sin t dt dt 1 s cos t 42. v C Initial condition: s (1) 1 0 1 0 1 cos t dt sin t C1 Initial condition: v(0) 1 C C C sin 0 1 (1 cos t s cos t) C1 ds dt v sin t (sin t ds dt dt 1 1) dt cos t t 1 32 1 cos 0 1 1 2 0 C2 C2 32 dt 32t C1 Initial condition: v(0) 20 C2 Solution: s 20 32(0) 20 C1 Velocity: C1 ds dt ds dt dt 16t 1 C2 Initial condition: s (0) dv a dt dv dt dt s 1 C1 Velocity: 1 or s v 0 cos t 1 cos Solution: s 41. dv a dt dv dt dt t 2 43. v (32t 2 cos t 32t 20 20) dt 20t [ 2, 2] by [ 3, 3] C2 Initial condition: s (0) 0 16(0)2 + 20(0) 0 44. 0 0 C2 [ 2, 3] by [ 3, 3] Solution: s 16t 2 20t 45. d (tan 1 x dx C) 46. d (sin 1 x dx C) d (sec 1 x dx C) 47. 48. dy x dx dy dx dx x2 y 2 x2 1 x2 1 1 x d ( cos 1 x dx 49. (a) 1 1 x2 1 C) 2 2 1 2 1 2 3 2 x2 1 1 x2 (x x x 1 2 ) dx x2 2 C Initial condition: y (1) 2 1 1 1 1 x C 2 C C C Solution: y x2 2 1 x 1 ,x 2 0 237 238 Section 6.1 49. continued (b) Again, y 52. (a) 1 x ( 1)2 1 2 ( 1) 1 C 2 1 3 2 (b) (c) For x d (x sin x) dx dx [ f (x)] dx f (x) dx x 2e x [ g(x)] dx g(x) dx x sin x (e) C x2 2 dy dx 0, g(x) dx (c) 1 C Solution: y d 2x (x e ) dx dx [ f (x) C. Initial condition: y ( 1) 1 f (x) dx (d) x2 2 1 x 3 ,x 2 x2 2 d1 dx x 1 x2 0 C1 g(x)] dx (f) 1 . x2 dy x2 d1 For x 0, C2 dx 2 dx x 1 x x2 1 x . x2 dy And for x 0, is undefined. dx [ f (x) g(x)] dx (g) [x (h) [g(x) 53. (a) (d) Let C1 be the value from part (b), and let C2 be the value from part (a). Thus, C1 1 . 2 f (x)] dx d 2s dt dt 2 ds kt dt 1 1 7 2 1 y ( 2) 22 2 1 2 5 2 50. r x 2 1 2 C2 (3x 3 1 and C2 2 3x 2 2 6x 03 12x 0 3(0)2 ( 2)2 1 2 ( 2) 3 C1 2 ( k)(0) 88 C1 0 0 C dc dx dx c (3x 2 x3 6x 2 12x 15x Initial condition c(0) 400 03 400 6(0)2 0 when t kt 2 2 C2 kt 88 88t 0 88 k 15) dx (c) C k 88 2 2k 400 15(0) C 88 k 88 88 k 3872 k s k x3 0 0 12x C Solution: c(x) C2 88(0) ds dt t 51. 88) dt Solution: s 3x 2 88 6x 2 15x 400 242 242 242 16 ft/sec 2 x 2e x x sin x 0 C2 (b) x3 kt 88t k2 (0) 2 C Solution: r (x) 88 when t Initial condition: s 0 4 dx C1 ( kt k2 t 2 s C 12(0) ds dt ds dt ds dt dt 7 . 2 x2 2 C1 Velocity: C1 C C1 2 12) dx Initial condition: r (0) 0 2 C2 Thus, C1 dr dx dx C2 g(x) dx k dt 88 (e) y (2) C x sin x g(x) dx Initial condition: C f (x) dx x dx 4] dx C x sin x x 2e x C g(x) dx f (x) dx x C x sin x f (x) dx x 2e x x 3 and C2 2 x 2e x C 4x C Section 6.1 54. We first solve s (0) d 2s dt 2 k with the initial conditions 2 44 and s (0) k ( k)(0) ds dt kt ( kt k2 t 2 44) dt 44t k2 (0) 2 0 kt C2 s0 44 44 , so it takes k 0 when t 44 k 44 k V 45 45 968 45 k 21.5 It requires a constant deceleration of approximately 0 ds dt 4 25 t 24 125t 48 0 5.2t 2.6t 2 2.6(0)2 s0 C2 43 h 75 10 when t h 5/2 0. 10 5/2 125t 48 125t 48 The height is given by h 105/2 105/2 2/5 125t 48 105/2 volume is given by 4 when t 0 V C2 C2 Position: s (t) v0t 2 (10)5/2 C 5 2 (10)5/2 5 2 5/2 2 h (10)5/2 5 5 h 5.2 dt Initial condition: s 4 0 when t C1 ds dt C2 122 hh 35 d4 3 h dt 75 4 2 dh h 25 dt dh h 3/2 dt 3/2 dh h dt dt 2 5/2 h C 5 h5/2 ds dt dt s C C1 C1 Velocity: 12 rh 3 dV dt 1 h 6 25 24 25 dt 24 25 t 24 25 (0) 24 5.2 dt 5.2(0) a2 t 2 Initial condition: h 21.5 ft/sec 2. Initial condition: (v0 )(0) 57. We use the method of Example 7. 45 968 k d 2s dt dt 2 ds 5.2t dt s0 44t 44 2k 0 C2 C2 s 55. v0t a2 (0) 2 44 seconds to stop, and we require: k k 44 2 v0 v0) dt Position: s ds dt at (at s0 C2 Now, ds dt 0 k2 t 2 C1 Initial condition: s (0) C2 44(0) Position: s v0 C1 ds dt dt a2 s t 2 44 Initial condition: s (0) 0 (a)(0) Velocity: ds dt dt s C1 v0: C1 v0 44 C1 Velocity: s0, and v(0) a dt v0 Initial condition: s (0) 44 a, s(0) Initial condition: s (0) C1 44 d 2s dt 2 ds dt dt 2 ds at dt 0. 2 ds dt dt 2 ds kt dt 56. Solving 2.6t 2 4 4 , so the positive 2.6 Solving s (t) 0, we have t 2 solution is t 1.240 sec. They took about 1.240 sec to fall. 43 h 75 4 75 125t 48 105/2 6/5 . 2/5 and the 239 240 Section 6.1 58. (a) y 500e 0.0475t (b) 1000 2 63. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (1, 1): y e (x 1)/2 ( 1, 2): y 2e (x 1)/2 (0, 2): y 2e x /2 ( 2, 1): y e (x 4)/2 500e 0.0475t e 2 0.0475t 2 2 ln 2 t 0.0475t 2 ln 2 0.0475 14.6 It will take approximately 14.6 years. 59. (a) y 1200e 0.0625t (b) 3600 [ 6, 6] by [ 4, 4] 1200e 0.0625t 3 ln 3 The concavity of each solution curve indicates the sign of y. e 0.0625t 0.0625t t ln 3 0.0625 64. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (0, 1): y e x (0, 2): y 2e x (0, 1): y ex 17.6 It will take approximately 17.6 years. x x 2 cos x dx 60. (a) t 2 cos t dt C 0 0 (b) We require t 2 cos t dt C 0 1, so C x The required antiderivative is 1. t 2 cos t dt 1. 0 x xe x dx 61. (a) te t dt C 0 0 (b) We require te t dt C 1, so C 0 x The required antiderivative is 65. Use differential equation graphing mode. For reference, the equations of the solution curves are as follows. (0, 1): y 3e 2x 4 (0, 4): y 4 (0, 5): y e 2x 4 1. te t dt 1. 0 62. (a) d 2y dx dx 2 dy 3x 2 dx 6x dx C1 Initial condition (horizontal tangent): y (0) 0 3(0)2 0 0 C1 C1 First derivative: dy dx dx y x3 [ 3, 3] by [ 4, 10] dy dx 3x 2 dx C2 (0)3 1 The concavity of each solution curve indicates the sign of y. 3x 2 Initial condition (contains (0, 1)): y (0) 1 1 C2 Solution: y C2 x3 1 (b) Only one function satisfies the differential equation on ( [ 4, 4] by [ 3, 3] The concavity of each solution curve indicates the sign of y. , ) and the initial conditions. 241 Section 6.2 66. Use differential equation graphing mode. For reference, the equations of the solution curves are as Section 4.2, y1 and y2 must differ by a constant. We find follows. that constant by evaluating the two functions at x 0. 1 (0, 1): y x 1 2 (0, 2): y (0, 2 . By Corollary 3 to the Mean Value Theorem of 3 y2 3. y1 2x 1 1 x1 1): y (0, 0): y 0 [ 3, 3] by [ 2, 8] 4. [ 2.35, 2.35] by [ 1.55, 1.55] [ 10, 10] by [ 30, 30] The concavity of each solution curve indicates the sign of y. 67. (a) (b) d (ln x dx d [ln ( x) dx for x 1 for x x C) 0 Exploration 2 Two Routes to the Integral 1d ( x) x dx C] 5. The derivative with respect to x of each function graphed in part (4) is equal to 1 x 2. 1 ( 1) x 1 x 0 1 3x 2 1. C ln x C, which is a solution to the differential equation, as we showed in part (a). For x 0, ln x C ln ( x) C, which is a 3x 2 2 1. 0, we have y ln ( x) 1 dx x3 0, we have y 0 5 x 1 dx (x 1 part (b). Thus, dy dx s Section 6.2 Integration by Substitution 1 1 15 (0) 5 1)1/2 dx 1 3. dy dx 2 (x 3 dy dx 3x 5. dy dx 4(x 3 2x 2 3)3(3x 2 4x) (pp. 315–323) 6. 1 x 2 2x dx u du 2 sin (4x 5) cos (4x 5) 4 5) cos (4x 5) 7. 2 3/2 u C 3 2 (1 x 2)3/2 3 2. Their derivatives are equal: dy dx 8 sin (4x Exploration 1 Supporting Indefinite Integrals Graphically 1. dy1 dy2 dx dx 1 x 2 2x. dy dx 1 cos x 8. C 0 42 3 23 (x 1)3/2 so 3 42 . 3 32 5 3x 4. 1 for all x except 0. x 1)3/2 2 3/2 2 3/2 (4) (0) 3 3 2 16 (8) 3 3 C1, which is a solution to the differential equation, as we showed 15 (2) 5 0 5 C2, which is a ln x 23 (x 3 2 solution to the differential equation, as we showed in part (a). For x 2 3/2 u 3 u du 1 dx 15 x 5 x 4 dx 2. (d) For x x 3 2 Quick Review 6.2 1 for all x except 0. x d ln x dx 2 1 solution to the differential equation, as we showed in part (b). Thus, 2 3/2 u 3 u du 0 2. 3x 0, ln x 1 dx 1 1 (c) For x 2 x3 dy dx 1 sin x sin x cos x tan x cot x 1)3/2 5 1 242 9. Section 6.2 dy dx 1 (sec x tan x sec x tan x sec x tan x sec2 x sec x tan x sec x(tan x sec x) sec x tan x 4. u sec2 x) du dy dx dx d Check: [(7x dx 1 ( csc x cot x csc x cot x csc x cot x csc2 x csc x cot x csc x(cot x csc x) csc x cot x csc2 x) 3 du dx d Check: dx 1 sin u du 3 1 cos u C 3 1 cos 3x C 3 1 cos 3x 3 C 1 ( sin 3x)(3) 3 6. u du 1 1 r 3 9 1 3 3u 1 cos u du 4 1 sin u C 4 1 sin (2x 2) C 4 1 d1 Check: sin (2x 2) C cos (2x 2)(4x) 4 dx 4 Check: x cos (2x2) du d (6 dx 1 C r3 r3 C C) 21 r3 1 du 2 dx 1 du 2 Check: 2x d1 sec 2x dx 2 1 cos t 2 1 t sin dt 2 2 t 2 du sin dt 2 t2 t 1 cos sin dt 2 2 du dx sec 2x tan 2x dx 1 6 9r 2 7. u 1 sec u tan u du 2 1 sec u C 2 1 sec 2x C 2 1 C sec 2x tan 2x 2 2 sec 2x tan x2 u 1/2 61 2x 1 9 du 3(2)u1/2 x cos (2x 2) dx 3. u 1 3 x2 3 r3 1 du r 2 dr 3 9r2 dr 1 du 4 x dx 1 1 31 3r 2 dr du 4x dx 2)3 3u 9 sin 3x 2x 2 2. u 28(7x C 9u 2 9 du 3 9 u2 1 1 du 3 u2 1 1 tan 1 u C 3 1 x tan 1 C 3 3 d1 x Check: tan 1 C dx 3 3 3 dx 2)3(7) 4(7x 2)4 (7x dx dx 3x sin 3x dx C] C 1 dx 3 du Section 6.2 Exercises 1 du 3 2)4 u4 x 3 5. u x2 du 1 28u3 du 7 2)3 dx 28(7x csc x 1. u 2 7 dx 1 du 7 sec x 10. 7x 2 u2 du 23 u 3 2 1 3 t3 d2 1 cos dx 3 2 t2 t1 2 1 cos sin 2 22 2 t t 1 cos sin 2 2 Check: C C cos t3 2 C r3 ( 3r 2) Section 6.2 y4 8. u 4y 2 14. Let u du (4y 3 8y) dy du 3 tan x du 1 sec2 x dx 2y) dy 4( y /4 1 du 4 (y3 8( y 4 2y) dy 4y 2 /4 1)2(y 3 2y) dy 8 1 4 d24 (y dx 3 4y 2 2( y 4 4y 2 1)2(4y 3 8( y 4 4y 2 1)2( y 3 1)3 13 u 3 C 4y 2 1)3 C 1 3 8y) 15. Let u du 9. Let u 1 du x dx 1 1 x du 2 sec (x 2) dx C 2) 2 3/2 u C 3 2 3/2 (cot x) 3 C 17. Let u tan x du 2 tan x sec x dx 1/2 u du 2 3/2 u C 3 2 (tan x)3/2 3 du C 2 tan 2 2 d sec u tan u du sec u ln x 1 du dx x 6 dx e x ln x ln 6 1 du u ln 6 ln u ln (ln 6) 1 2 C tan 2 u7 du 18 u 8 1 x tan8 4 2 2 C sec 13. Let u u6 du x 2 1 2x du sec dx 2 2 x 7x tan sec2 dx 2 2 d sec 1 dx x 17 u C 7 17 (ln x) 7 18. Let u 12. Let u ln x ln6 x dx x sec x dx 2 u1/2 du sec u du tan (x C C cot x 2 tan u du 1 cos u du 3 1 sin u C 3 1 sin (3z 4) 3 cot x csc2 x dx 2 2 3 csc2 x dx du dx 11. Let u 4) dz C 16. Let u x 1 3 4 C 1 1 ( 1)3 3 dz cos (3z u 10. Let u 3z du u2 x)2 (1 1 3 dz 1 du 3 dx 1 1 (1) 3 C 2y) u2 du 1 u2 du 23 u 3 24 (y 3 Check: 1 tan2 x sec2 x dx C C C 243 244 Section 6.2 s4/3 19. Let u du 3 du 4 3 /4 cot x dx /4 /4 Let u s1/3 ds dx sin2 3x Let u 23. 4 1/3 s ds 3 s1/3 cos (s4/3 20. 3 /4 8 3 cos u du 4 3 sin u C 4 3 sin(s4/3 8) 4 8) ds du cos x dx sin x sin x cos x dx 3 /4 C /4 x 3 /4 cos x dx sin x csc2 3x dx 1 /4 u x x ln u 3 /4 x 3x du /4 3 /4 du 1 du 3 ln sin x 3 dx /4 dx 24. Let u du du cos (2t sin (2t 1 du sin (2t 2 sin (2t 1) dt cos2 (2t 1) 1) 2 2 1) dt 1 u 2 du 2 x2 25. Let u ln 2 1) 1) cos t dt 1 C C du 2x dx 1 du 2 x dx x dx 2 1 1 2 10 2 1 du u 1 ln u 2 6u 2 2 1.504 C 1x 6u 0 1 du u9 ln 9 sin t 6 cos t dt (2 sin t)2 2 2 ln u 3 du 9 2 1)(2) dt 1 2 cos (2t 1 sec (2t 2 2 ln 2 dx x 0 11 u 2 22. Let u x 2 dx 7 21. Let u 2 ln 1 csc2 u du 3 1 cot u C 3 1 cot (3x) C 3 csc2 3x dx du 1 6 sin t 10 2 1 (ln 10 2 C C ln 2) 1 ln 5 2 0.805 Section 6.2 5 26. 0 40 25 5 40 dx x 2 25 30. csc x dx x2 5 0 25 2 25 5 40 25 1 dx dx x2 5 Let u 5 du csc x cot x 0 cot x csc2 x dx 1 du u csc x dx ln u dx 5 40 dx x 2 25 cot x dx cot x 1 1 dx 5 du csc x du 0 csc x csc x csc2 x csc x cot x dx csc x cot x x 5 Let u csc x 8 (5) 5 1 0 1 u2 1 C ln csc x du 1 31. Let u 8 arctan u y cot x C 1 0 du 8(arctan 1) dy 3 8 27. dx cot 3x 4 4 y 2 0 1 4 2 3/2 u 3 1 2 3/2 2 3/2 (4) (1) 3 3 2 2 14 (8) 3 3 3 sin 3x dx cos 3x Let u cos 3x du 3 sin 3x dx 1 du 3 dx cot 3x sin 3x dx 1 3 1 3 1 3 32. Let u 1 du u ln u ln cos 3x du 5x r dr 1 1 ln sec 3x 3 0 1 u1/2 du 21 0 1 2 3/2 u 23 1 1 1 (0) (1) 3 3 r 2 dr r1 C.) 0 8 5 dx 1 du 5 dx dx 5x 8 29. sec x dx Let u du 2r dr 1 du 2 C r2 1 du C (An equivalent expression is 28. Let u u1/2 du 1 dy 33. Let u sec2 x dx 0 0 tan x sec2 x dx u du /4 1 12 u 2 sec x tan x dx sec x tan x sec2 x sec x tan x dx sec x tan x sec x sec x 1 du u 0 1 1 (0) 2 1 ( 1)2 2 1 2 tan x sec x tan x sec x dx tan x du 1 u 1/2 du 5 1 2u1/2 C 5 2 5x 8 C 5 sec2 x dx ln u 34. Let u C ln sec x tan x C 1 3 du 1 du 21 1 (4 4 r2 2r dr r dr 5r dr r 2)2 5 2 5 u 5 2 du 0 245 246 Section 6.2 35. Let u 39. 3 1/2 d 2 du 2 du 3 1 1/2 dy ( y 5)(x 2) dx dy (x 2)dx y5 Integrate both sides. d 10 3/2 2 (1 0 3/2 1 ) dy 2 (10) 3 d y 2 2 u du 2 let u y 1 20 1 32 5 du 1 20 3 2) dx On the left, 1 20 1 u 3 (x 5 1 2 dy 1 du u 10 3 12 x 2 12 x 2 12 x 2 ln u 2x C 2x C 3 sin x 2x C 5 5 e (1/2)x y du 4 ln y y 36. Let u 5 e Ce(1/2)x 3 cos x dx 1 du 3 cos x dx cos x dx 1 3 1/2 u du 0 2 2x eC or C We now let C 4 2 2x C e C, depending on whether (5t 4 du 1 t5 4 2t 2) dt 2t (5t 4 u1/2 du 2) dt 2 2x C e (1/2)x 2 2x value C 3 2 3/2 u 3 2 3 27 2 3 40. dy dx dy cos 2 sin 2 d 5. y x dx y cos2 y Integrate both sides. dy x dx y cos2 sin 2 d 3 2 2x y cos2 x 2 sin 2 d /6 0 gives a solution to the original differential Ce (1/2)x y cos 2 1 du 2 5 equation), we may write the solution as 0 2 3/2 (3) 3 0 C e (1/2)x Since C represents an arbitrary constant (note that even the 0 du 5 3 0 38. Let u 5) is positive or negative. Then y t5 37. Let u (y y 3 sin x 4 y On the left, let u 1 2 1 2 1/2 u 3 du 1 11 42 1 (3) 4 1 u2 2 2 1 3 4 y 1 1/2 y dy 2 du 1/2 2 du 1/2 y dy 1 12 x C 2 12 2 sec2 u du x C 2 12 2 tan u x C 2 12 2 tan y x C 2 12 tan y x C 4 2 du cos2 u (Note: technically, C is now C y y tan tan 2 1x 4 1x C 2 2 4 C C . But C’s are generic.) 2 247 Section 6.2 41. dy dx dy dx dy ey (cos x)(e y esin x) dy dx dy cos x esin x dx Integrate both sides. (cos x)e y sin x 44. dy On the right, let u ln x dx x 4 y cos x e sin x dx du y ln x x ln x 4 dx x y Integrate both sides. dy ey 4 On the right, let u sin x du 2y1/2 4 u du 2y1/2 cos x dx ln x 1 dx x 4 u2 1 2 e y e u du e y e u e y e sin x C y (ln x) C e sin x C y [(ln x)2 C]2 y e 1/2 C y 1/2 C]2 1 (1 2 1 C 0 e sin x) x y e xe y e C [(ln e)2 y (e) ) e ln (C dy 42. dx dy dx dy ey C.) 2 sin x ln (C y 2(ln x) 2y (Note: technically C is now C C 2 C) y (ln x)4 Note: Absolute value signs are not needed because the original problem involved ln x, so we know that x 0. e x dx 45. (a) Let u x 1 Integrate both sides. dy ey y e e y e dx x ln (e C) dy 43. 2xy 2 dx dy 2x dx y2 dy 2x dx y2 y 1 y x2 x y (1) C C 3 y dy1 x x dx 1 and antiderivatives of dy2 x dx 1. x 1, so both are 1. x 0 1 2 y1 0.25 4 0 1.219 2.797 y2 4.667 y2 y1 1 2 x (c) Using NINT to find the values of y1 and y2, we have: C 1 C C C 1 1 C (b) By Part 1 of the Fundamental Theorem of Calculus, 1 2 u1/2 du 1 dx 2 3/2 u C 3 2 (x 1)3/2 3 d2 Alternatively, (x 1)3/2 dx 3 C x dx x x e dy y du e x dx 3 4.667 3.448 4.667 1.869 4.667 2 3 C 4 (d) C y1 y2 x x x 1 dx 0 x x 1 dx x 1 dx 0 3 x x 0 x 1 dx 3 3 1 dx 3 4 4.667 6.787 0 2.120 4.667 4.667 248 46. (a) Section 6.2 d [F(x) dx x4 47. Let u C] should equal f (x). 1 (b) The slope field should help you visualize the solution curve y F(x). 10 x 3 dx (a) x4 0 9 x (c) The graphs of y1 F(x) and y2 4x 3 dx. 9, du f (t) dt should 0 9 1 2 1 2 1 1/2 1 1/2 u du u 4 2 1 10 9 2 3 10 0.081 2 10 9 differ only by a vertical shift C. x3 (b) (d) A table of values for y1 y2 should show that y1 y2 C for any value of x in the appropriate domain. x 4 9 1 1/2 u 2 (e) The graph of f should be the same as the graph of NDER of F(x). x 0 x3 x 4 9 dx u1/2 x2 1 48. Let u C 9 1 2 x4 9 10 1 2 1 2 2x dx. x4 1 2 1, du 1 1/2 u du 2 dx 1 C 1 2 1 x2 (f) First, we need to find F(x). Let u x2 1 1/2 u du 4 dx 10 3 2 C 1 0 1 cos 3x, du (1 9 0.081 3 sin 3x dx. cos 3x) sin 3x dx /3 x2 Therefore, we may let F(x) (a) 1. 2 /6 a) d ( dx x2 1 1 C) 12 (2) 6 (2x) 2 x2 x x2 1 1 u du 3 1 f (x) 1 (b) (1 12 (1) 6 12 u 6 cos 3x)2 /3 (1 cos 3x) sin 3x dx /6 c) 1 (1 6 12 (2) 6 49. We show that f (x) f (x) [ 4, 4] by [ 3, 3] f (x) d) x 1 2 y1 1.000 1.414 2.236 3.162 4.123 y2 y1 0 0.000 0.414 1.236 2.162 3.123 1 1 1 y2 3 1 4 1 f (3) e) [ 4, 4] by [ 3, 3] tan x and f (3) cos 3 5. cos x cos 3 d ln 5 cos x dx d (ln cos 3 ln cos x 5) dx d ln cos x dx 1 ( sin x) tan x cos x cos 3 ln 5 (ln 1) 5 5 cos 3 ln 1 1 2 C 1 (1 6 [ 4, 4] by [ 3, 3] 2 1 u du 3 cos 3x) sin 3x dx b) 12 u 6 C /3 cos 3x)2 /6 12 (1) 6 5, where 1 2 Section 6.3 50. (a) u 2 csc2 2 d cot 2 , du 1 u du 2 1 u2 C 2 2 u2 C 4 1 cot2 2 C 4 csc2 2 cot 2 d F1( ) csc 2 , du 2 csc 2 cot 2 d 1 u du 2 1 u2 C 2 2 u2 C 4 1 csc2 2 C 4 F2 ( ) d (sin2 x C) 2 sin x cos x dx d ( cos2 x C) ( 2 cos x)( sin x) dx 1 1 d cos 2x C sin 2x (2) 2 dx 2 s Section 6.3 Integration by Parts (pp. 323–329) Exploration 1 Integrals 1. u Evaluating and Checking dx and dv x ln x ⇒ du 2 csc 2 cot 2 (d) F1( ) F2( ) b 51. (a) u sin x, du 1 (csc2 2 cot2 2 ) 4 1 1 cos2 2 1 sin2 2 sin2 2 4 4 sin2 2 2. cos x dx cos x, du 2 sin x cos x dx dx x ln x 1 4 d (x ln x dx x) x ln x 3. The slope field of 2 2u du u C 2 sin x C dy dx ( 2u) du C cos2 x C C x 1 x 1 ln x ln x shows the direction of the curve as it is graphed from left to right across the window. sin x dx u2 x. Thus, v du x ln x b dx ⇒ v u dv uv 1 csc2 2 4 2 sin x cos x dx (b) u ln x dx b 1 cot2 2 4 2 sin x cos x 2 sin x cos x 1 cot 2 ( 2 csc2 2 ) csc2 2 cot 2 2 1 csc 2 ( 2 csc 2 cot 2 ) 2 (c) F1 ( ) sin 2x dx sin 2x 1 csc2 2 4 F2( ) 2 dx 1 sin u du 2 1 cos u C 2 1 cos 2x C 2 (d) csc2 2 cot 2 du 2x, du 2 sin x cos x dx 1 cot2 2 4 (b) u (c) u 249 [0, 6] by [ 2, 5] 250 Section 6.3 4. The graph of y2 x ln x x appears to be a vertical shift of the graph of y1 1 ln t dt (down 1 unit). Thus, y2 appears to be an antiderivative of ln x which supports x ln x x dy dx x sin x dy 9. x (x sin x)dx Integrate both sides. C as the set of all antiderivatives of dy ln x. (x 12 x 2 y y (0) sin x) dx cos x 10. [0, 6] by [ 2, 5] dy dx (x 3)(cos 2x)(2) 2x 3 cos 2x 2. dy dx 1 1 (2x)2 2 1 4x 2 dy dx 5. y tan tan y 1)(2e 2x) 2 x du dv 1 3) 2. Let u du cos x dx sin x sin x C C) (cos x)( 1) cos x x2 1 2x dx 1 cos e 2x dx Integrate both sides. C sin x 2 x sin x dx 1 2[ x cos x (x 2 cos 0 (2x)( sin x) sin x] 2)sin x d Check: [2x cos x dx cos x e 2x x 2 sin x 2x cos x 1 ( 1) cos x dx v x 2 sin x (x 2 C C 2)sin x (2 cos x)(1) (sin x)(2x) 2 x 2 cos x e 2x dx dv Using the result from Exercise 1, 0 1 2x e 2 x cos x x 2 cos x dx 1 y cos x 2 3x 1 dy v dx sin x dx ( x)( sin x) (x sin x dx dy 1x e sin x 2 Section 6.3 Exercises x cos x cos 1 (x 1) x1 cos y 1 dy dx 1x e sin x 2 1 2 1x e cos x 2 cos x) e x (sin x d Check: ( x cos x dx 0 8. sin x) x sin x 1 7. cos x) x sin x dx 1 tan y 3 y cos y x 3 1) 3x x 6. cos x 1 1 3 d1x e (sin x dx 2 1x e (cos x 2 1x e cos x 2 1. Let u ln (3x 3x 1 3e 2x 2e 2x ln (3x 3x 1 dy 3. dx 4. (sin 2x)(3x 2) 3x 2 sin 2x 3 (e 2 x) 2 e x sin x Quick Review 6.3 1. 12 x 2 C C y 1 C (x 2 C] 2)(cos x) Section 6.3 3. Let u ln y dv 1 dy y du 12 y 2 6. Let u 12 y 2 v 12 y ln y 2 12 y ln y 2 12 y ln y 2 y ln y dy y dy 1 sin dv 1 du 1 dy y sin d d v 2 1 1 sin 1 1 2 1 Let w 1 y dy 2 12 y C 4 2d 1 1 1 d 1 1 sin 2 sin sin sin 12 d12 Check: y ln y y C 4 dy 2 12 1 1 y (ln y)(y) y 2 y 2 d 2 1 dw 1 u 1 dy y tan 1 dy v y y y 1 y2 du 2y dy 1 y Check: 5. Let u du dy y2 tan 1 y x dv dx v x sec2 x dx 1 y 1 y 1 y 1 2 1 2 1 2 1 ln (1 y 2) 2 1 1 (2y) 2 1 y2 d y tan 1 y dy y 1 y2 1 y tan y tan y t2 7. Let u y tan y tan [ 1, 1] by [ 0.5, 2] dy y2 1 dw dv y2 1 y dy Let w y 1 du tan 1 tan 1 dw w ln (1 y 2) Let u C tan C x tan x t cos t (2 cos t dt sin t 2 t cos t dt 2t sin t 2 sin t dt 2t sin t 2 cos t t 2) cos t 2t sin t C dv v C csc2 t dt cot t tan x dx x tan x v 2 tan x 2 (cos t)(t) dt dv t 2 cos t sec2 x dx t 2 cos t dt du y cos t t t 2 cos t 1 sin t dt v t 2 sin t dt ln w C dv 2t dt sin x dx cos x x tan x ln cos x C [ 3, 3] by [0, 8] 8. Let u du t dt t csc2 t dt [ 1.5, 1.5] by [ 1, 4] t cot t cot t dt t cot t cos t dt sin t t cot t [0, 3] by [ 4, 4] 1 1 2 y ln y 4. Let 251 ln sin t C dw w 1/2 C 1 2 w C 252 Section 6.3 9. Let u ln x 1 dx x du x 3 dx dv 14 x 4 v 14 x ln x 4 14 x ln x 4 14 x ln x 4 x 3 ln x dx f (x) and its derivatives f (x) and its derivatives 1 41 x dx 4 x 13 x dx 4 14 x C 16 3x2 (–) 6x x and g (x) (+) 6 (–) e. x4 x 3e e–x (+) 4x 3 (–) (+) e–x 24x (–) –e–x (+) e–x 24 x 4e x x xe (x 4 4x e 4x 3 x2 11. Let u du (x 2 3 x 2 12x e 12x 2 24)e dv 24e C 5x)e x x C du e y sin y dy u 2x du (x 2 (x 2 5 5x)e x (x 2 5x)e x (x 2 7x e x(2x x e 5)e x 5)e x (2x C e du Let u e y cos y 1y e (sin y 2 y e C e du 2e x dx 2e x v e y cos y sin y e y sin y sin ye y dy e y sin y cos y) C dv y e y e cos y dy y 2e y dy v sin y e y e cos y dy cos y dy cos y dy e ysin y e ycos y dy 5) dx (2x 7)ex e dx v 2 dx 5x)e x x dv cos y dy 5) dx 14. Let u Let e dy e y sin y dy e x(2x cos y e y dy dv y 2 e y sin y dy ex v (x 2 x x e x dx 5) dx 5x)e x dx 24xe 24x 5x (2x x cos y e y cos y ey C sin y dy v e y sin y dy Let u 3 2x e 8 dv e y dy du dx 4 ey 13. Let u –e–x 0 dx 1 3 2x 3 2 2x 3 xe xe xe 2x 2 4 4 3 2 x 3x 3x 3 e 2x C 2 4 4 8 –e–x 12x 2 2x . 1 –2x e 2 1 –2x e 4 1 –2x e 8 1 –2x e 16 0 g(x) and its integrals 2x e–2x (+) x e g(x) and its integrals x3 4 10. Use tabular integration with f (x) x 3 and g(x) 12. Use tabular integration with f (x) sin y y dv y y e e sin y dy v dy cos y sin y y dy sin y 1y e (sin y 2 e ycos y e y e cos y cos y) C y cos y dy Section 6.3 x 2 and g(x) 15. Use tabular integration with f (x) f (x) and its derivatives sin 2x. g(x) and its integrals x2 (+) 2x (–) 2 (+) sin 2x 1 cos 2x 2 1 sin 2x 4 1 cos 2x 8 0 12 x cos 2x 2 x 2 sin 2x dx 1 x sin 2x 2 2x 2 cos 2x 4 1 /2 1 x 2 sin 2x dx 0 1 cos 2x 4 x sin 2x 2 2x 2 cos 2x 4 C C x /2 sin 2x 2 0 2 1 2 2 ( 1) 4 2 8 1 2 0 0.734 Check: NINT x 2 sin 2x, x, 0, 0.734 2 x 3 and g(x) 16. Use tabular integration with f (x) f (x) and its derivatives 1 (1) 4 0 cos 2x. g(x) and its integrals x3 (+) 3x2 (–) 6x (+) 6 (–) cos 2x 1 sin 2x 2 1 cos 2x 4 1 sin 2x 8 1 cos 2x 16 0 x 3 cos 2x dx 13 x sin 2x 2 x3 2 /2 3x sin 2x 4 x3 2 x 3cos 2x dx 0 0 3 4 32 x cos 2x 4 3x 2 4 3 ( 1) 8 32 16 Check: NINT x 3cos 2x, x, 0, 3 cos 2x 8 3x 2 4 3x sin 2x 4 32 16 3 x sin 2x 4 0 1.101 2 1.101 3 cos 2x 8 C 3 /2 cos 2x 8 0 3 (1) 8 253 254 Section 6.3 e 2x 17. Let u dv 2e 2x dx du e 2x cos 3x dx Let 1 sin 3x 3 1 1 (e 2x) sin 3x sin 3x (2e2x dx) 3 3 1 2x 2 2x e sin 3x e sin 3x dx 3 3 v e2x u du cos 3x dx dv 2e 2x dx sin 3x dx 1 cos 3x 3 2 2x (e ) 3 v 1 2x 1 1 e sin 3x cos 3x cos 3x (2e 2x dx) 3 3 3 1 2x 4 2x e (3 sin 3x 2cos 3x) e cos 3x dx 9 9 13 2x 1 2x e cos 3x dx e (3 sin 3x 2 cos 3x) 9 9 1 2x e 2x cos 3x dx e (3 sin 3x 2 cos 3x) 13 3 1 2x 3 e 2x cos 3x dx e (3 sin 3x 2 cos 3x) 13 2 2 e 2x cos 3x dx 16 [e (3 sin 9 13 16 [e (2 cos 9 13 2 cos 9) e 4 3 sin 9) e 4 (3 sin ( 6) (2 cos 6 2 cos ( 6)] 3 sin 6)] 18.186 Check: NINT(e 2x cos 3x, x, 18. Let u e du 2x e 2x 2, 3) 18.186 dv 2x 2e sin 2x dx (e v dx 2x sin 2x dx 1 cos 2x 2 1 cos 2x 2 ) 1 cos 2x ( 2e 2x dx) 2 1 2x e cos 2x 2 Let u e du 2x dv 2x 2e 2x e cos 2x dx cos 2x dx 1 sin 2x 2 v e sin 2x dx 2x 2e 2x 2 e 1 2x e cos 2x 2 sin 2x) 1 2x e (cos 2x 2 sin 2x dx e sin 2x dx 2x (e 1 2x e (cos 2x 2 2x e 1 2 (cos 2x e sin 2x) sin 2x) 4 (cos 2x sin 2x) 4 e sin 2x dx C 4 3 sin 4) e6 [cos ( 6) 4 (cos 4 sin 4) e6 (cos 6 4 4 e 2 (cos 4 4 125.028 2x 2x C 2x e 3 Check: NINT(e 1 sin 2x ( 2e 2x dx) 2 ) sin 2x 2x 4 sin 2x dx 2x sin 2x, x, 3, 2) 125.028 sin ( 6)] sin 6) 255 Section 6.3 23. Let x 2e4x dx 19. y x2 Let u du y 1 4 1 2 4x xe 4 y y dx y v 1 1 (x) e4x 2 4 1 4x 1 4x xe e 8 32 1 4x e C 32 Let u 2 x cos x du 2 v 1 x dx 2 1 sec 2 2 Let w 2 y ) sec 2 2 y 22. y Let y y 2 sec x2 (x 2 2 ln sec x 2x v x 1)e 2 dx 1)e x e v 2 x 1)e x (2x 1)e x (x 2 1 dx dx (x 2 d dx x e x 1)e x (2x 1)e x (x 2 1 x dx x e (2x dv x x e dx. dx 1 1)e x 1)e dv x x x 1 x 1)e du 1 3x 4)e x x 2e 2e x dx C C [ 3, 3] by [ 3, 3] dv sec v sec tan d The graph shows that the two curves intersect at x where k 1.050. The area we seek is k (x 2 x 1)e k x x 2 dx 0 d tan 4 1) dx (2x Let u d sec x (x 2 12 2 x sin x dx 0 Let u C Note: In the last step, we used the result of Exercise 29 in Section 6.2. 2 x sin x dx 24. We begin by evaluating (x 2 1 w 1/2 dw 4 1 1/2 w C 2 1 2 1C 2 sec tan u du d sec 0 3 2d 1 sec ( 1) 0 d 12 2 2d 1 sec 2 0 x sin x dx (x 2 12 2 1 4 1 2 y 0 2 2 (c) 1, so no absolute value is needed 1, dw 1 0 0(1) 2 dv 1 (sec 0 x sin x dx sin x 2 (1) in the expression for du. y x sin x dx du Note that we are told y sin x 2 (b) C d 1 du x sin x dx 0 ( 1) 1 4x e 4 1 4x e dx 4 dv 1 sec C x cos x 1 13 du dx v x x 3 1 13 1 (ln x) x 3 x dx 3 3 x 13 12 x ln x x dx 3 3 13 13 x ln x x C 3 9 sec sin x 3 ln x 21. y cos x cos x dx e4x dx x ln x dx Let u y x sin x dx 0 2 20. y y (a) dv 1 2 4x xe 4 1 2 4x xe 4 x2 x 4 8 y x cos x x cos x 1 xe4xdx 2 x du v dx sin x dx 1 4x e (2x dx) 4 (x 2) e4x Let u dv x sin x dx 1 4x e 4 v x du e4xdx dv 2x dx u 0 (x 2 ( 2.888 0.726 3x 4)e 4) xk 0 (0.386 13k x 30 0) k, 256 Section 6.3 t 25. First, we evaluate e (c) Using the result from part (b): cos t dt. x3 Let u Let u e t du e t t e dt e sin t du dv t e t dt t e cos t dt or [x n e t(sin t cos t) C e tcos t dt 1t e (sin t 2 cos t) C 0 t ( 1)n n(n 1)x n 1 1)!x ( 1)n (n d nx xe dx n C 2 ( 1)n(n!)]e x C 2e tcos t for integration. Alternately, show that the derivative of the answer to 1 part (d) is x ne x: 2 1 2 Average value 2e tcos t dt d xn dx 0 2 [x n 1 e t(sin t 2 2 cos t) 0 1 [e 2 ( 1) 2 1 dx … [xn … ex 1)e x … C 2 ( 1)nn!]e x 1 1)x n n(n 2 ( 1)nn!] dv 2x dx v x2 ex 2x 1)x n n(n 2 ( 1)n 1(n!)x ( 1)nn!]e x 1)x n 1 n(n 2)x n 1)(n 2 3 ( 1)n 1n!]e x sin e x dx x dx Let u ex du w dw w sin w dw 1)e x 2)e x dx x. Then dw , so dx 2 x dw 2x 2x e x dx 2(x 1 x ne x C x2 (x 2 nx n n(n (b) Using the result from part (a): x 2e x 1)x n n(n C [nx n 27. Let w x 2 e x dx 2 ( 1)nn! e x (n!)x ( 1)n 1(n!)x nx n e x dx (x du 1 ( 1)n 1(n!)x d dx ex v xe x u 1 ex [x n e0( 1)] e x dx dv xe x nx n 1)x n n(n ( 1)n … 0.159 x xe x dx 1 2 e 2 du nx n … e tcos t dt 0 Let … C n (e) Use mathematical induction or argue based on tabular 2. 26. (a) Let u 1 C 6)e x 6x dn x dx 2 nx n 2)e x 2x 3x 2 2 dn x dx … 2 e tcos t dt Now we find the average value of y 3(x 2 (x 3 (d) x n e cos t 3x 2 e x dx x 3e x cos t ex v x3 ex dt t e sin t x 3 e x dx sin t dt v t e cos t dt t e x dx dv 3x 2 dx du sin t sin t e t e cos t dt v t cos t dt Let u dv dv v sin x dx 2 w sin w dw sin w dw cos w w cos w w cos w C C (sin w)(2w dw) cos w dw sin w C 2 w sin w dw 2w cos w 2 sin w 2 x cos x 2 sin C x C 2w dw. Section 6.3 28. Let w 2 3 dx e 3x 9. Then dw 3x 3x 9 Let u 9 dw 2 (e ) w dw 3 dx dw v w ew dw xe dx nx dx x n cos x dx cos x dx v sin x x n sin x 2 w ew dw 3 (sin x)(nx n 1dx) x n sin x 32. Let ew w ew n xn 1 dv sin x dx v cos x xn u du ew dw nx x n sin x dx ew n1 x ncos x 2 w ew dw 3 2 (w 1)ew 3 2 ( 3x 9 3 2 3 x2 (x ) e xn 33. Let u sin x dx 1)e 3x 9 C w 3 and g(w) e w. g(w) and its integrals (+) 34. Let (–) (+) ew 6 (–) 1 ax e (nx n 1 dx) a dv 0 dx n1 dx v n x n(ln x)n x x (ln x)n(x) x(ln x) ew 6w (ln x)n dx cos x dx n n 1 ax x e dx, a a n(ln x) x dx) 1 ax e a v (ln x)n u ew 3w 2 dx x ne ax a du w3 1 1 1 e ax dx (x n) e ax 2x dx. x dx n xn 1 a x ne ax dx 1 3w w e dw. 2 ( cos x)(nx n dv nx n du Use tabular integration with f (x) f (w) and its derivatives dx (x n)( cos x) 1)ew x 2. Then dw 29. Let w 7 x2 dx du 9 dv n1 e dw (w 3x 9 (3) dx, so w w ew e 2 3x w w dv du 2 w dw. 3 xn 31. Let u 1 n1 n (ln x) 1 dx dx ew 35. (a) Let y f 1 (x). Then x 1 Hence, f (b) Let u (x) dx f ( y) dy. (y)[ f ( y) dy] y dv y f ( y) dy f ( y)dy ew 0 du w 3 e w dw w3 ew (w 3 2 x 7e x dx 3w 2e w 3w 2 6w 6w e w 6)e w 6e w 1 dr, and so dr r r dy Using the result of Exercise 13, we have: sin (ln r) dr v y f ( y) f C f ( y) dy (x)(x) (x) dx 1 f e y dy. f 1 (x) xf 37. (a) Using y 1 (x) dv d f 1(x) dx dx (x) dx f ( y) dy y f ( y) dy xf 36. Let u f ( y) 1 1 Hence, f 1 3w w e dw 2 13 (w 3w 2 6w 6)e w C 2 2 (x 6 3x 4 6x 2 6)ex C 2 ln r. Then dy dy y f ( y) dy C du 30. Let y f (y), so dx 1 f 1 v dx x d x f 1(x) dx dx (x) (x) f ( y) dy. sin 1 x and f (y) sin y, y (sin y)e dy 1y e (sin y cos y) C 2 1 ln r e [sin (ln r) cos (ln r)] C 2 r [sin (ln r) cos (ln r)] C 2 y 2 sin 1 x dx 2 , we have: x sin 1 x x sin 1 x cos y x sin 1 x cos (sin sin y dy C 1 x) C 257 258 Section 6.4 37. continued 40. (a) Using y 1 sin x sin x dx 1 x 1 x d sin 1 x dx dx 1 x x sin (b) x x2 1 u x 2, du 1 1 1 f x y 2 (x) 1/2 (b) C x2 1 log2 x dx x log2 x x log2 x C x log2 x x log2 x 1 x and f (y) x tan 1 (c) 2log2 x x 1 x 2 y, we have 2 y dy 2y C ln 2 1 log2 x 2 ln 2 d log2 x dx dx 1 x dx x ln 2 dx ln 2 1 xC ln 2 x tan y, , we have: x dx log2 x and f (y) x log2 x x2 x tan tan u tan (x) x log2 x dx 1 u 1/2 du 2 x 1 1 x log2 x x x) 38. (a) Using y 1 1 x sin 2 1 x sin 1 log2 x dx 2x dx x sin (c) cos (sin f tan y dy ln sec y x s Section 6.4 Exponential Growth and Decay (pp. 330–341) C (Section 6.2, Example 7) Exploration 1 x tan u 1 x ln cos (tan x tan x dx ln cos y 1 x x tan 1 tan x x tan (b) 1 1 x x 2, du 1 1 1 x 1 x) f ( y) 1 x 1 ln (x) cos x, 0 1 1 C x cos 1 x 1 x sin y 1 x x cos x cos 1 0 2 x) v (u) du traveled by the object at time t. cos y dy C 1 x) C [0, 70] by [0, 1500] (b) cos 1 x dx x x d x cos 1 x dx dx 1 x 1 u 1 x 2, du 2x dx x cos 1 x x cos 1 x u1/2 C x cos 1 x 1 x2 (c) sin (cos 1 x) 1 x2 x2 dx Quick Review 6.4 1. a eb 2. c ln d 3. ln (x x 1 u 1/2 du 2 2x C 0, the integral gives the distance 0 3. The total distance traveled is about 200 units for m 1, about 400 units for m 2, about 800 units for m 4, and about 1200 units for m 6. sin (cos 1 distance traveled by the object over the time interval [0, t]. Since s (0) , we have: x cos x dx v (u) du gives the 0 x and cos x t 2. As we saw in Section 5.1, s (t) 1 ln (1 2 x2 x cos cos 1. As m increases the velocity of the object represented by the graph slows down more slowly. That is, the y-coordinates of the graphs decrease to 0 more slowly as m increases. [0, 20] by [0, 120] 1 u 1 du 2 1 ln u C 2 1 ln (1 x 2) 2 x 1 f C 2x dx x tan 39. (a) Using y x) dx d tan 1 x dx dx 1 x dx 1 x2 x tan (c) ln cos (tan 1 Slowing Down More Slowly x x tan 1 C 4. 100e e2x 2x x 3) 3 x 2 e2 e2 600 6 ln 6 1 ln 6 2 3 Section 6.4 0.85x 5. 5. Doubling time: 2.5 ln 0.85x x ln 0.85 A0ert A(t) ln 2.5 ln 2.5 1000e0.086t 2000 2 ln 2.5 ln 0.85 x ln 2 5.638 e0.086t 0.086t ln 2 0.086 t 2 k1 ln 2 k1 6. (k 3 k Amount in 30 years: ln 3 k A 1)ln 2 k ln 3 ln 2 k(ln 3 1.1t 7. 1.710 10 ln 1.1t ln 10 t ln 1.1 ln 10 t ln 10 ln 1.1 2t ln 24.159 1 4 2t 8. e 1 log 1.1 9. ln ( y y 10. ln y y y 1) 1 y 1 ln 4 2 2x 3 e2x 3 1 e2x 2 2 2 y 3t 1 e3t 1 e3t 1 2 e3t A(t) ln 2 2898.44 A0 1. y(t) y(t) A(t) 1200 1 2 y0e kt 200e ln 2 y0e kt 100e1.5t 2. y(t) y(t) A0e rt A0e(0.0525)(30) 2898.44 e1.575 t A0e rt 600e0.0525t e0.0525t 0.0525t ln 2 0.0525 A(t) 10,405.37 104.0537 12 104.0537 ln 12 50 20.2t A0e rt 1200e(r)(30) e30r 30r 104.0537 1 ln 12 30 r y(t) y(10) 1 2 1 ln 2 k kt e A(t) 60e 10k 2400 10k Solution: y(t) A0e rt 1200e0.072t 2 1 2 0.1 ln 2 60e (0.1 ln 2)t e0.072t ln 2 10k 0.1 ln 0.072 Doubling time: 60e kt 30 13.2 years 8. Annual rate: 05t 3. y(t) y0e kt y(t) 50e kt y(5) 100 50e 5k 2 e 5k ln 2 5k k 0.2 ln 2 Solution: y(t) 50e(0.2 ln 2)t or y(t) y0e $600.00 Doubling time: 3 Section 6.4 Exercises 4. y(t) $13,197.10 7. Initial deposit: 1 4 1 1 ln 2 4 t 1000e(0.086)(30) 6. Annual rate: A(t) A0e rt 4000 2000e(r)(15) 2 e15r ln 2 15r ln 2 r 0.0462 4.62% 15 Amount in 30 years: A(t) A0e rt A 2000e[(ln 2)/15](30) 2000e2 ln 2 2000 22 $8000 ln 2) ln 2 ln 3 ln 2 k 8.06 yr 0.072 t t or y(t) 60 2 t/10 ln 2 0.072 9.63 years 7.2% 259 260 Section 6.4 9. (a) Annually: 2 1.0475 t ln 2 t ln 1.0475 ln 2 t 14.94 years 11. (a) Since there are 48 half-hour doubling times in 24 hours, there will be 248 2.8 1014 bacteria. (b) The bacteria reproduce fast enough that even if many are destroyed there are still enough left to make the person sick. ln 1.0475 0.0475 12t 12 0.0475 12t ln 1 12 ln 2 2 t Hence e2k 14.62 years 0.0475 12 12 ln 1 y0 ln 2 ln 2 4 ln 1.011875 13. ln 2 0.0475 0.9 ln 0.9 t 14.59 years (b) e 0.05 ln 2 ln 1.0825 0.005t ln 0.05 0.005 8.74 years 15. Since y0 y (0) 5 0.0825 12 12t ln 1 k 12 ln 1 0.0825 12 8.43 years t t ln 2 0.5 ln 2.5 2 2e (0.5 ln 2.5)t y (0) 0.0825 4t 4 y 4t ln 1.020625 3 1.1e(k)( or y 2e0.4581t 1.1, we have: 1.1e kt 1 ln 2 4 ln 1.020625 8.49 years (d) Continuously: ln 2 2k ln 5 Function: y ln 3 k 2 ln 2 16. Since y0 (c) Quarterly: ln 2 2e(k)(2) ln 5 12 ln 2 2 2e 0.0825 12t 1 t 2, we have: kt y ln 2 599.15 days The sample will be useful for about 599 days. (b) Monthly: 2 0.005t e t t ln 1.0825 t 1250. 0.18t 1.0825t ln 2 23 0.18t ln 0.05 10. (a) Annually: 2 10,0000 ln 0.9 0.585 days 0.18 ln 2 ln 2 14. (a) Half-life 138.6 days k 0.005 0.0475t t y0e3ln 2, we have 1250. There were 1250 bacteria initially. population was 14.68 years e0.0475t ln 2 ln 2. Solving 10,000 population would have doubled 3 times, so the initial (d) Continuously: 2 4, or k , which gives 2 hrs, so the doubling time is 1 hr. Thus in 3 hrs the 4t ln 1.011875 t y0e3k population increased by a factor of 4, i.e. doubled twice, in 0.0475 4t 4 1 y0e5k 40,000 10,000 y0e5k. We could solve this more quickly by noticing that the (c) Quarterly: 2 y0e3k and 40,000 10,000 1 ln 2 y0e kt, we have 12. Using y (b) Monthly: e 0.0825t ln 2 0.0825 ln 1.1 1 3 8.40 years 3k (ln 1.1 Function: y 0.0825t 3) ln 3) 1.1e(ln 1.1 ln 3)t/3 or y 1.1e 0.3344t Section 6.4 3 17. At time t kt y0e k y0e 20. First, we find the value of k. , the amount remaining is k(3/k) 3 y0e 0.0499y0. This is less than 5% of the original amount, which means that over 95% has Taking “right now” as t (T0 Ts) e 35 65 (T0 65)e 65 (T0 65)e (T0 Ts)e 60e( e20k k 7 1 ln 20 6 (k)(20) Dividing the first equation by the second, we have: k (a) T e10k kt k)( 20) (k)(10) 50 Ts 7 6 kt 0, 60 above room temperature 60. Thus, we have 70 Ts 2 Ts means T0 T decayed already. 18. T 261 Ts (T0 Ts)e kt 60e( (1/20)ln (7/6))(15) 53.45 1 ln 2 10 It will be about 53.45 C above room temperature. Substituting back into the first equation, we have: 30 (b) T [(ln 2)/10](10) (T0 65)e 30 (T0 T0 5 T0 Ts)e kt 60e( (1/20)ln (7/6))(120) 23.79 1 65) 2 60 (T0 Ts It will be about 23.79 above room temperature. 65 Ts (T0 10 60e( (c) T The beam’s initial temperature is 5 F. ln 19. (a) First, we find the value of k. T Ts (T0 Ts)e 20 (90 20)e kt (1/20)ln (7/6))t 7 1 ln t 6 20 20 ln (1/6) ln (7/6) t kt 60 1 6 Ts)e 232.47 min (k)(10) 4 7 e It will take about 232.47 min or 3.9 hr. 10k 4 1 ln 7 10 k 20 15 ln 70e (4/7)]t 3 14 10 ln 4 ln 7 0.445 ln 0.445 ln 0.445 k t [(1/10) ln (4/7)]t 5700 ln 0.445 ln2 6658 years Crater Lake is about 6658 years old. 4 1 ln t 7 10 3 14 t 20)e[(1/10) ln (90 kt kt e When the soup cools to 35 , we have: 35 ln 2 (see Example 3). 5700 21. Use k ln 2 (see Example 3). 5700 22. Use k (a) e kt 0.17 27.53 min kt It takes a total of about 27.53 minutes, which is an additional 17.53 minutes after the first 10 minutes. t ln 0.17 ln 0.17 k 5700 ln 0.17 ln 2 14,571 years The animal died about 14,571 years before A.D. 2000, (b) Using the same value of k as in part (a), we have: Ts (T0 ( 15) [90 T 35 50 10 ln 21 t 105e Ts)e kt ( 15)]e[(1/10) ln (4/7)]t in 12,571 B.C. (b) e kt 0.18 [(1/10) ln (4/7)]t kt 4 1 ln t 7 10 10 10 ln 21 4 ln 7 t ln 0.18 ln 0.18 k 5700 ln 0.18 ln 2 14,101 years The animal died about 14,101 years before A.D. 2000, 13.26 It takes about 13.26 minutes in 12,101 B.C. 262 Section 6.4 22. continued 26. kt 0.16 kt ln 0.16 (c) e ln 0.16 k t 5700 ln 0.16 ln 2 15,070 years 0.2 e 0.1t ln 0.2 0.1t t 10 ln 0.2 16.09 yr It will take about 16.09 years. 27. (a) The animal died about 15,070 years before A.D. 2000, in 13,070 B.C. 23. Note that the total mass is 66 v v0e (k/m)t v 9e 3.9t/73 7 73 kg. 2190 3.9t/73 e C 13 2190 Since s (0) 0 we have C and 13 2190 2190 lim s(t) lim (1 e 3.9t/73) 13 13 t→ t→ (a) s(t) 3.9t/73 9e ln p 1 p 168.5 Ae kh p0 p 1013 millibars, so: 1013e kh v 9e (59,000/51,000,000)t 90 v 9e 59t/51,000 90 1013 459,000 59t/51,000 e 59 459,000 Since s (0) 0, we have C and 59 459,000 lim s (t) lim (1 e 59t/51,000) 59 t→ t→ 459,000 7780 m 59 59t/51,000 9e dt The ship will coast about 7780 m, or 7.78 km. (b) 1 9e 59t 51,000 C 1013e(k)(20) e20k 1 90 ln 20 1013 k Thus, we have p 1013e 1013e((1/20) ln (90/1013))(50) (c) 1 0.121h 2.383 millibars. 1013e kh 900 e kh 1 900 ln k 1013 h 51,000 ln 9 59 0.121 km (b) At 50 km, the pressure is 900 1013 59t/51,000 ln 9 t p0e kh Using the giving altitude-pressure data, we have (k/m)t (a) s (t) 0 p0 Solution: p 41.13 sec p0 when h Ae0 A 73 ln 9 3.9 v0e C eCe kh p0 It will take about 41.13 seconds. 24. v C Initial condition: p ln 9 t e kh p p 3.9t/73 9e 3.9t 73 kh eln dt The cyclist will coast about 168.5 meters. (b) dp kp dn dp k dh p dp k dh p 20 ln (900/1013) ln (90/1013) 0.977 km The pressure is 900 millibars at an altitude of about 1899.3 sec 0.977 km. It will take about 31.65 minutes. 25. y 800 0.8 k At t y y0e 28. By the Law of Exponential Change, y 100e t 1 hour, the amount remaining will be 100e 0.6(1) 54.88 grams. kt (k)(10) 1000e ln 0.8 10 10 (b) 0.1 14 1000e 1000e . At 29. (a) By the Law of Exponential Change, the solution is V V0e (1/40)t. 10k e 0.6t 24 h: ln 0.1 ( ln 0.8/10)24 2.4 ln 0.8 585.4 kg About 585.4 kg will remain. t e (1/40)t t 40 40 ln 0.1 92.1 sec It will take about 92.1 seconds. Section 6.4 30. (a) A(t) A0e t It grows by a factor of e each year. 34. (a) T Ts 79.47(0.932)t (b) T 10 263 79.47(0.932)t t (b) 3 e ln 3 t It will take ln 3 1.1 yr. (c) In one year your account grows from A0 to A0e, so you can earn A0e A0, or (e 1) times your initial amount. This represents an increase of about 172%. (k/m)t v0e 31. (a) s (t) v0m dt k Initial condition: s (0) v0m 0 (k/m)t (c) Solving T 12 and using the exact values from the regression equation, we obtain t 52.5 sec. C 0 (d) Substituting t 0 into the equation we found in part (b), the temperature was approximately 89.47 C. C k v0m e [0, 35] by [0, 90] k(T dT Ts k dt T dT Ts k dt 35. (a) v0m s (t) dT dt T C k k v0m k (b) lim s (t) (k/m)t e v0m k 1 e (k/m)t v0 m lim 1 e v0m (k/m)t ln T e(r)(100) T Ts ln 90 100r T Ts r 32. (a) ln 90 100 T Ts 90 (b) t→ k k 0.045 or 4.5% e(r)(100) 131 ln 131 k (0.80)(49.90) k k We know that v0m Ae Ts Ae T0 coasting distance Ts kt kt T0 when t 0 (k)(0) A Solution: T (T0 Ts Ts)e kt 1.32 (b) lim T 998 33 v0m lim [Ts t→ 1.32 and k (k/m)t k m 998 33(49.9) e 1.32(1 e 36. (a) 2y0 20t/33 1.32(1 e 0.606 t ) (T0 kt Ts Ts y0e rt rt ln 2 r t ) Ts)e e rt ln 2 ) A graph of the model is shown superimposed on a graph of the data. t→ Horizontal asymptote: T 20 . 33 2 (1 k eCe 0.049 or 4.9% Using Equation 3, we have: s (t) kt C e T0 ln 131 100 v0m C Initial condition: T 100r r 33. kt Ts t→ Ts) (b) [0, 0.1] by [0, 100] [0, 4.7] by [0, 1.4] (c) ln 2 0.69, so the doubling time is almost the same as the rules. (d) 70 5 14 years or 72 = 14.4 years 5 0.69 which is r 264 Section 6.4 36. continued r 0.5 x 0.5 x x 1 (e) 3y0 y0e rt 3 ln 3 rt t 10 100 1000 10,000 100,000 ert ln 3 r Since ln 3 1.099, a suitable rule is 108 108 or . i 100r (We choose 108 instead of 110 because 108 has more factors.) 37. (a) 1 10 100 1000 10,000 100,000 Graphical support: 2.5937 2.7048 2.7169 2.7181 2.7183 Graphical support: 1x , y2 x 1 [0, 10] by [0, 3] e 38. (a) To simplify calculations somewhat, we may write: [0, 50] by [0, 4] 2 x 1 10 100 1000 10,000 100,000 e2 2x x 6.1917 7.2446 7.3743 7.3876 7.3889 mg k mg k mg k y1 1 [0, 500] by [0, 10] e 2 2 e2at 1 The left side of the differential equation is: m dv dt m mg (2)(e 2at k 4ma mg 2at (e k 4m Graphical support: e at e at eat eat e at eat e2at 1 e2at 1 (e2at 1) 2 e2at 1 mg 1 k 7.389 2x , y2 x e0.5 (c) As we compound more times, the increment of time between compounding approaches 0. Continuous compounding is based on an instantaneous rate of change which is a limit of average rates as the increment in time approaches 0. v(t) (b) r 0.5 x , y2 x 1 1x x 2.7183 y1 1.6487 y1 x e e 0.5 1.6289 1.6467 1.6485 1.6487 1.6487 gk m 4mge 2at (e 2at 1)2 2 1) 1) mg 2at (e k 2 (2ae 2at) (e 2at) 1) 2 (e 2at) Section 6.5 The right side of the differential equation is: mg kv 2 mg k mg 1 1 mg 1 2 2 e 2at 1 mg 1 k 2 2 e 2at 1 4 e2at 1 1 4 (e 2at 1)2 2at mg 4(e 1) 4 (e 2at 1)2 4 mg e2at (e2at 1)2 Since the left and right sides are equal, the differential equation is satisfied. mg e 0 k e0 And v (0) e0 e0 0, so the initial condition is also satisfied. mg eat e at e k ea t e a t e t→ t→ mg 1 e 2 a t lim k 1 e 2at t→ mg 1 0 mg k1 0 k mg The limiting velocity is . k (b) lim v (t) (c) lim mg k 160 0.005 at at 179 ft/sec The limiting velocity is about 179 ft/sec, or about 122 mi/hr. s Section 6.5 Population Growth (pp. 342–349) Quick Review 6.5 1. All real numbers 2. lim f (x) x→ lim f (x) x→ 3. y 0, y 50 1 0 50 0 50 4. In both f and f , the denominator will be a power of 1 5e 0.1x, which is never 0. Thus, the domains of both are all real numbers. 5. [ 30, 70] by [ 10, 60] f (x) has no zeros. 265 266 Section 6.5 6. Use NDER f (x), or calculate the derivative as follows. d 50 dx 1 5e 0.1x (1 5e 0.1x)(0) (50)(5e (1 5e 0.1x)2 0.1x 25e (1 5e 0.1x)2 f (x) 0.1x )( 0.1) [ 30, 70] by [ 0.5, 2] (a) ( ,) (b) None 7. Use NDER(NDER f (x)), or calculate the second derivatives as follows. d 25e 0.1x dx (1 5e 0.1x)2 (1 5e 0.1x)2(25e f (x) 0.1x )( 0.1) (25e (1 5e 2.5e 0.1x[(1 5e 0.1x) 2(5e 0.1x] (1 5e 0.1x)3 12.5e 0.2x 2.5e 0.1x (1 5e 0.1x)3 0.1x )(2)(1 ) 5e 0.1x )(5e 0.1x )( 0.1) 0.1x 4 [ 30, 70] by [ 0.08, 0.08] Locate the inflection point using graphical methods, or analytically as follows. f (x) 12.5e (1 2.5e 0.2x 0.1x 2.5e ) 0 0.1x 3 5e (5e 0 0.1x 0.1x 1) e 0 0.1x 1 5 0.1x ln 5 x 10 ln 5 16.094 (a) Since f (x) 0 for x 10 ln 5, the graph of f is concave up on the interval ( (b) Since f (x) 0 for x 10 ln 5, the graph of f is concave down on the interval (10 ln 5, ), or approximately (16.094, ). 8. Using the result of the previous exercise, the inflection point occurs at x Since f (10 ln 5) 1 50 5e ln 5 25, the point of inflection is (10 ln 5, 25), or approximately (16.094, 25). 9. x x2 12 4x A x x 12 A( x 4) Bx x 12 (A B)x 4A Since A B 2. B B x 4 1 and 4A 12, we have A 3 and 10 ln 5. , 10 ln 5), or approximately ( , 16.094). Section 6.5 10. 2x 16 x2 x 6 2x A x 16 A( x When x B 3 x 2) 4. (a) 2 B(x 3) 3, the equation becomes 10 x 2, the equation becomes 20 B 5A, and when 5B. Thus, A dP dt dP dt dP dt 4. Section 6.5 Exercises M Ae kt 150 Ae 0.02t 1 P dP 1. (a) dt k P(M P) m 0.02 P(150 P) 150 1 P(150 P) 7500 2 and (b) P 1 Initial condition: P(0) 0.025P 1 A A (c) 15 150 1 Ae0 150 10 15 15 (b) Using the Law of Exponential Change from Section 6.4, the formula is P 75,000e 0.025t 267 9 Formula: P 150 9e 0.02t 1 (c) [0, 100] by [0, 1,000,000] 2. (a) dP dt 0.019P [0, 200] by [0, 200] (b) Using the Law of Exponential Change from Section 6.4, the formula is P 110,000e0.019t. 5. The growth rate is 0.3 or (c) 7. dP dt 0.0004P2 0.04P 0.0004P(100 dP dt dP dt dP dt (b) P P k P(M P) M 0.05 P(200 P) 200 0.00025P(200 1 1 10 1 A A Thus, k P) M Ae kt 200 Ae 0.05t Initial condition: P(0) 200 19e 50 P 1 P dP dt dP dt Thus, k 0.04 and the carrying capacity is M 2 100. P 250 1 2 50 1 1 25 P 250 P 500 P 0.04 1 500 P k1 M 0.04 and the carrying capacity is M 500. 9. Choose the slope field that shows slopes that increase as y increases. (d) 19 1 8. 10 200 1 Ae0 200 20 10 Formula: P P) 0.04 P(100 P) 100 k P(M P) M [0, 100] by [0, 1,000,000] 3. (a) 30%. 6. The growth rate is 0.075 or 7.5%. 0.05t 10. Choose the slope field that matches a logistic differential equation with M 100. (b) 11. Choose the only slope field whose slopes vary with x as well as with y. (c) (c) 12. Choose the slope field that matches a logistic differential equation with M 150. (a) [0, 100] by [0, 250] 268 Section 6.5 1000 e4.8 0.7t 1000 1 e4.8e 0.7t M 1 Ae kt 13. (a) P (t) 17. (a) 1 (b) P (0) 0.7 and Thus, k 1 6 1 1 25 A 24 Formula: P 1 and 1 dP dT 1 person 14 sec 0.225t 0.225t e 365 24 3600 sec 1 yr 24e 24e Initially 1 student has the measles. 15. (a) Note that 0.225t 0.225t 2,252,571 people per year. t 125 The relative growth rate is dP dT 2,252,571 257,313,431 P 1 100 1 0.00875 or 0.875% (b) The population after 8 years will be approximately P0e rt 257,313,431e 8r 275,980,017, where r is the unrounded rate from part (a). 24e 0.225t 24e 0.225t e 1 0.225t 0.225t t 16. (a) Let t be the number of years. 1000 0.1 ln 0.1 t 0.8 t t ln 0.8 10.32 It will take about 10.32 years. 10,000(0.8)t. So that f (t) will round to less (b) Let f (t) than 1, we actually require f (t) 0.5 0.00005 ln 0.00005 t 10,000(0.8)t 0.8t t ln 0.8 ln 0.00005 ln 0.8 150 24e 0.225t 150 1 24e 0.225t 3 2 1 2 1 48 ln 48 ln 48 17.21 weeks 0.225 150 1 24e 0.225t 6 5 1 5 1 120 ln 120 ln 120 0.225 21.28 It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies. 10,000(0.8)t ln 0.1 ln 0.8 6 150 Ae0 A (b) 200 e 5.3 150. Initial condition: P(0) This is a logistic growth model with k M 200. 1 P) M Ae kt 150 Ae 0.225t 1 8 200 1 e 5.3 t 200 1 e 5.3e t M 1 Ae kt (b) P (0) P) 0.225 and M P Initially there are 8 rabbits. 14. (a) P (t) 0.0015P(150 0.225 P(150 150 k P(M P) M This is a logistic growth model with k M 1000. 1000 1 e4.8 dP dt 44.38 It will take about 44.4 years. 0.5. Section 6.5 18. (a) dP dt 0.0004P(250 Now find the time to grow from 10,000 bees to 25,000 bees. P) 0.1 P(250 P) 250 k P(M P) M Thus, k P 1 1 dP dt M Ae 250 Ae 10,000e t/12 2.5 ln 2.5 0.1t Initial condition: P(0) year 1970. 28(1 25,000 250. 10,000e t/12 kt 28 1 28, where t 0 represents the 250 28 t 12 t dx dt 21. (a) 250 A 12 ln 2.5 e t/12 The total time required is 4 ln 2 250 Ae0 A) 1 P 12 P(t) 0.1 and M Formula: P(t) 111 14 1 1000 7.9286 1 249.5 10 ln 1000 0.1x t ln 1000 0.1x 0.1(t 1000 0.1x e0.1(t 0.1x 111e 0.1t 14 (249.5)(111e 0.1t) 14 249.5 1 0.1t e 0.1t t (b) y0e 0.01(t/1000) Solution: x ln 14) 82.8 0.00001t (c) y y0e (0.00001)(20,000) 0.819y0 The tooth size will be about 81.9% of our present tooth size. 20. First find the time to grow from 5000 bees to 10,000 bees. P(t) 5000e0.25t 10,000 5000e0.25t 2 e0.25t ln 2 0.25t 4 ln 2 t 10,000 Ae0.1t 11,000 14 55,389 14 ln 55,389 0.9 e ln 0.9 0.00001t t 100,000 ln 0.9 10,536 It will take about 10,536 years. 1 P 4 e0.1C e0.1t 10e0.1C e0.1t 1000 1000 0.5 0.00001t dP dt C) 10,000 Initial condition: x(0) 250 y0e C) x It will take about 83 years. 19. (a) y C x 250 111e 0.1t/14 10(ln 55,389 0.10x 1000 249.5. 1 13.8 years. dt 0.1x (b) The population P(t) will round to 250 when P(t) 12 ln 2.5 1000 dx 250 , or approximately 111e 0.1t/14 250 7.9286e 0.1t 1 P(t) 269 10,000 Ae0 A 11,000e0.1t 10,000 (b) 100,000 11,000e0.1t 10,000 10 e0.1t ln 10 0.1t t 10 ln 10 23 yr It will take about 23 years. 22. (a) Using the Law of Exponential Change in Section 6.4, the solution is p(x) p0e x/100, where p0 p(0). Initial condition: p(100) 20.09 20.09 p0e 1 20.09e p0 Solution: p(x) (20.09e)e x/100 or p(x) 20.09e1 0.01x (b) p(10) p(90) 20.09e0.9 20.09e0.1 $49.41 $22.20 (c) r(x) x p(x) 20.09x e1 0.01x r (x) 20.09[(x)(e1 0.01x)( 0.01) (e1 0.01x)(1)] 20.09e1 0.01x(1 0.01x) The derivative is zero at x 100, positive for x 100, and negative for x 100, so r (x) has its maximum value at x 100. 270 Section 6.5 23. (a) Note that the given years correspond to x x y 50, x 1 70, and x 0, x 20, 26. (a) y 16.90 5.132e 0.0666x 1 80. (b) Carrying capacity 18.70 1.075e 0.0422x lim y x→ 16.90 representing 16.9 million people. 27. dy dx (cos x)e sin x dy dy [0, 100] by [0, 20] y 18.70, representing (c) Using NDER twice and solving graphically, we find that y 0 when x 1.7, corresponding to the year 1912. The population at this time was about y (1.7) 9.35 million. 1 eu du eu y (b) Carrying capacity lim y 18.7 million people. x→ 24. (a) y (cos x)e sin x dx e sin x 0 esin 0 1 C dy dx dy y3 ln y [0, 50] by [0, 15] y lim y 24.76, representing x→ ln P ln M ln ln P M M P P k P(M M M P P M P M P 5 Ae0 3 A Solution: y dy dx 2 y 2 3 2e 2x x y x dx 2 x 2 C Initial condition: kt C kt y (0) C Ae APe C This gives kt e 02 2 2 C e 2 2 2 C kt kt C y2 2 x2 2 2, or y 2 x2 4. 4. The solution of the initial value problem is the function that satisfies the initial condition, namely y Ae M Ae x2 But this equation represents two functions, y kt P(1 1 2x Ae y dy k dt P M 3 C 2x C e 2 29. k dt P 2x 5 P) k dt P 3) 2 dx 3 2 P 2( y 1 Initial condition: y (0) (c) Using NDER twice and solving graphically, we find that y 0 when x 38.44, corresponding to the year 1988. The population at this time was about y(38.44) 12.38 million. dP dt M dP P(M P) (M P) P dP P(M P) 1 1 dP P MP e sin x 3 y 24.76 million people. 0 C Solution: y 28. 25. C Initial value: y(0) 24.76 7.195e 0.0513x (b) Carrying capacity C kt kt ) x2 4. Section 6.5 30. dy dx dy y y x (c) x1/2 dx ln y 2 3/2 x 3 y A) 1200A 100 300 100 1200A 300A 200 900A e(2/3)x 3/2 C 3/2 eCe(2/3)x y Ae(2/3)x y 3/2 P(t) 1 P(t) Ae0 1 A P(t) e(2/3)x Solution: y 31. (a) Note that k (d) 3/2 0 and M 0, so the sign of same as the sign of (M both M P and P positive. For P and P P)(P dP is the dt m). For m P M, [0, 75] by [0, 1500] m are positive, so the product is m or P M, the expressions M Note that the slope field is given by P dP dt m have opposite signs, so the product is negative. (b) 2 9 1200(2/9)e11kt/12 100 1 (2/9)e11kt/12 1200(2)e11kt/12 100(9) 9 2e11k/12 300(8e11kt/12 3) 9 2e11kt/12 A Initial condition: y(0) 1 1200Ae0 100 1 Ae0 300 300(1 C 0.1 (1200 1200 P)(P k (M M dP dt k (1200 1200 P)(P M 1200 P)(P (1200 1100 P)(P M 100) dP 100) dt 1 11 k 12 P dP 1 P 100 dt P P P ln P 11 k 12 1 dP P 100 P ln M (P 100) (1200 P) dP (1200 P)(P 100) dt P M 11 k 12 P m 11 k dt 12 (1200 1 1200 1 1200 ln 1200 ln P P 100 1200 P P 100 1200 P ln P 100 11 kt 12 1 1 m ln 11 kt 12 C P(1 Ae (M dP dt M dP dt M m ) P C 11kt/12 ee P(0) P 100 1200 P P P(1 Ae11kt/12 1200Ae11kt/12 100 Ae11kt/12) P 1200Ae11kt/12 m M M 1200Ae11kt/12 100 1 Ae11kt/12 C C m)kt/M Ae (M m)kt/M (M P)Ae (M AMe (M m)kt/M m)kt/M m AMe (M m)kt/M m 1 Ae (M m)kt/M AM e 0 m 1 Ae 0 AM M) m A k dt e C e (M A(P(0) 100 m kt A) m P(0) P(0) M k M Mm kt M P(0)(1 APe11kt/12 k M m M m m)kt/M m M dP Pm MP Pm MP Pm MP P C m) k 1 M dP 100) dt dP dt P)(P k M m m (M P)(P m) (P m) (M P) (M P)(P m) k k (M M dP m) dt M P)(P (M m) P)(P 100). dP dt (e) dP dt 271 AM m 1A m P(0) P(0) m M P(0) Therefore, the solution to the differential equation is P AMe (M m)kt/M m where A 1 Ae (M m)kt/M P(0) m . M P(0) 272 Section 6.6 dp dt dp p 32. (a) s Section 6.6 Numerical Methods k(t)p (pp. 350–356) k(t) dt Quick Review 6.6 t ln p k(u) du 1. f (x) f (2) t k(u) du eCe 0 t p Ae 0 p0 . 0 Ae 0 p0 Ae A f Solution: p(t) 9 (b) p0e 9 k(u) du 0 0 sec2 4 4. L(x) t f k(u) du 0 f 4 1 x 2 4 4 2 5. f (x) f (4) 0 ln 1) 0.2x 5x 2 0.2(4) 5(4) 6. L(x) 9 u) 0.04(ln 10 4 2x 2x 0.04 ln (1 2) ( 2)2 4 1 0.04 du 1u 9 sec2 x 3. f (x) k(u) du 0 p0 f (2) f (2)(x 2 9(x 2) 9x 16 k(u) du Initial condition: p(0) p0 3x 2 3 3(2)2 3 2. L(x) 0 p C f (4) f (4)(x 4) 2.85 0.4875(x 4) 0.4875x 0.9 0.04 ln 10 2 0.4875 9 p(9) 7. L(4.1) k(u) du p0 e 0 100e0.04 ln 10 f (4.1) 109.65 0.4875(4.1) 0.1(4.1)2 After 9 years during which the inflation rate is (a) L(4.1) 0.04 per year, the price of an item which originally 1t (b) f (4.1) L(4.1) f (4.1) f (4.1) 0.9 5 4.1 2.89875 2.900512 0.001762 0.00061 0.061% cost $100 will be increased to $109.65. 8. L(4.2) 9 (c) p(9) p0e 0.04 du 0 0.04(9) 100e 143.33 The price will be $143.33. 9 0 (0.04 0.004u) du 0.002u2 k(u) du 0 p(9) p0 e dP dt P 2 dP P 1 P 100e 0.522 168.54 9. L(4.5) f (4.5) (b) k dt kt C 0.4875(4.5) 2 0.1(4.5) f (4.5) L(4.5) f (4.5) f (4.5) 10. L(3.5) 0.4875(3.5) 1 kt 1 C 1 P0 Solution: P L(4.2) f (4.2) f (4.2) (a) L(4.5) f (3.5) C Initial condition: P(0) C 0.522 kP2 33. (a) f (4.2) P0 1 (1/P0) 0.1(3.5)2 (a) L(3.5) (b) kt 0.9 5 4.2 2.9475 2.954476 0.006976 0.00236 0.236% 9 0 9 P0 (b) 0 0.04u 0.1(4.2)2 (a) L(4.2) 9 k(u) du (d) f (4.2) 0.4875(4.2) P0 1 (b) There is a vertical asymptote at t kP0t 1 kP0 f (3.5) L(3.5) f (3.5) f (3.5) 0.9 5 4.5 3.09375 3.136111 0.042361 0.01351 0.9 5 3.5 1.351% 2.60625 2.653571 0.047321 0.01783 1.783% Section 6.6 Section 6.6 Exercises y x y 2e x) 1 x (x Therefore, y 2e x( 1) 1 x 1 2e ) x 1 1 x 2e 2e 0 1 y. dy dx dy 1y 6. (0) 2e 1 2 1 1 Initial condition: y (0) 1 1 Ae0 1 2A Solution: y 2 e x 1 x Check the initial condition: y (0) Ae x y 1. Check the differential equation: d (x dx eC ex y 1 ln 1 x(1 y) x dx 12 x 2 y 2. Check the differential equation: C (x2/2) y x 1 x (x Therefore, y e x( 1) e) x 1 e 1 e y e (x2/2) C y x 1 y 1 e x) 1 e C 1 d (x dx y e C x x e x2/2 y. y Ae x2/2 1 1 Check the initial condition: y (0) 0 1 e (0) 1 1 Initial condition: y ( 2) 0 2 0 Ae ( 2) /2 1 2 0 Ae 1 e2 A 2 Solution: y e2 e x /2 1 or y 2 3. Check the differential equation: 2x y de dx 2y 2 sin x 5 sin x e2x 2 2e2x 2 e2x cos x 2e 2x 2 sin x cos x 5 4 sin x 2 cos x 5 2 cos x sin x 5 2 cos x 5 sin x dy dx dy y dy y 7. sin x 5 sin x 2y(x 1) 2(x 1) dx (2x x2 ln y 2) dx 2x C y 2 e x 2x C sin x y eC e x Check the initial condition: y Therefore, y 2y e2(0) y (0) 2 sin 0 5 cos 0 1 1 4. Check the differential equation: y dx (e dx y 2x e e2x 1 (e Therefore, y ex 1) x y 2x e e 2x 1) 2e2x 2x e e x y (0) e e 2(0) 1 2e y 1. 1 2x) 2 dy (1 2x) dx 1 x2 x 1 1 x x2 ( 1)2 1 C y C x 1 y ex 1 y C C ex C 1 C 1 Solution: y y C 1 ( 1) dx ln 1 C Initial condition: y ( 1) 1 1 2 y 2(1 y 5. Note that we are finding an exact solution to the initial value problem discussed in Examples 1–4. dy dx dy 1y 2x y 2x Check the initial condition: 0 2 dy dx 8. 1 A ex 2x Initial condition: y ( 2) 2 2 Ae( 2) 2( 2) 2A 2 Solution: y 2e x 2x 0 5 2 x2 1 x 1 e (x2/2) 2 1 273 274 Section 6.6 9. To find the approximate values, set y1 2y sin x and use EULERT with initial values x 0 and y 0 and step size 0.1 for 10 points. The exact values are given by 1 2x (e 5 y x 2 sin x cos x). y (Euler) y (exact) Error 11. To find the approximate values, set y1 2y(x 1) and use IMPEULT with initial values x 2 and y 2 and step size 0.1 for 20 points. The exact values are given by 2 y 2e x 2x. x y improved Euler y (exact) Error 0 0 0 0 0.1 0 0.0053 0.0053 2 2 2 0 0.2 0.0100 0.0229 0.0129 1.9 1.6560 1.6539 0.0021 1.3983 1.3954 0.0030 0.3 0.0318 0.0551 0.0233 1.8 0.4 0.0678 0.1051 0.0374 1.7 1.2042 1.2010 0.0032 1.0578 1.0546 0.0032 0.5 0.1203 0.1764 0.0561 1.6 0.6 0.1923 0.2731 0.0808 1.5 0.9478 0.9447 0.0031 0.8663 0.8634 0.0029 0.7 0.2872 0.4004 0.1132 1.4 0.8 0.4090 0.5643 0.1553 1.3 0.8077 0.8050 0.0027 0.7683 0.7658 0.0025 0.9 0.5626 0.7723 0.2097 1.2 1.0 0.7534 1.0332 0.2797 1.1 0.7456 0.7432 0.0024 1.0 0.7381 0.7358 0.0023 0.9 0.7455 0.7432 0.0023 0.8 0.7682 0.7658 0.0024 0.7 0.8075 0.8050 0.0024 Error 0.6 0.8659 0.8634 0.0025 0 0.5 0.9473 0.9447 0.0026 1.8048 0.0048 0.4 1.0572 1.0546 0.0026 1.2036 1.2010 0.0026 10. To find the approximate values, set y1 x y and use EULERT with initial values x 0 and y 2 and step size 0.1 for 10 points. The exact values are given by y x 1 e x. x y (Euler) y (exact) 0 2 0.1 1.8000 2 0.2 1.6100 1.6187 0.0087 0.3 0.3 1.4290 1.4408 0.0118 0.2 1.3976 1.3954 0.0022 1.6553 1.6539 0.0014 1.9996 2 0.0004 0.4 1.2561 1.2703 0.0142 0.1 0.5 1.0905 1.1065 0.0160 0 0.6 0.9314 0.9488 0.0174 0.7 0.7783 0.7966 0.0183 0.8 0.6305 0.6493 0.0189 0.9 0.4874 0.5066 0.0191 1.0 0.3487 0.3679 0.0192 Section 6.6 12. To find the approximate values, set y1 x(1 y) and use IMPEULT with initial values x 2 step size 0.1 for 20 points. The exact values are given by y e (x /2) 2 1. x 2 y improved Euler y (exact) Error 0 0 0 1.9 0.2140 0.2153 0.0013 1.8 0.4593 0.4623 0.0029 1.7 0.7371 0.7419 0.0049 1.6 1.0473 1.0544 0.0071 1.5 1.3892 1.3989 0.0097 1.4 1.7607 1.7732 0.0125 1.3 2.1585 2.1740 0.0155 1.2 2.5780 2.5966 0.0186 1.1 3.0131 3.0350 0.0219 1.0 3.4565 3.4817 0.0252 0.9 3.9000 3.9283 0.0284 0.8 4.3341 4.3656 0.0315 0.7 4.7491 4.7834 0.0344 0.6 5.1348 5.1719 0.0370 0.5 5.4815 5.5208 0.0394 0.4 5.7796 5.8210 0.0413 0.3 6.0210 6.0639 0.0430 0.2 6.1986 6.2427 0.0441 0.1 6.3073 6.3522 0.0449 0 6.3438 6.3891 0.0452 2 and y 0 and 275 276 Section 6.6 13. To find the approximate values, set y1 x y and use EULERT and IMPEULT with initial values x y 1 and step size 0.1 for 20 points. The exact values are given by y x 1 2e x. x y (Euler) y improved Euler y (exact) Error (Euler) Error improved Euler 0 1 1 1 0 0 0.1 0.9000 0.9100 0.9097 0.0097 0.0003 0.2 0.8200 0.8381 0.8375 0.0175 0.0006 0.3 0.7580 0.7824 0.7816 0.0236 0.0008 0.4 0.7122 0.7416 0.7406 0.0284 0.0010 0.5 0.6810 0.7142 0.7131 0.0321 0.0011 0.6 0.6629 0.6988 0.6976 0.0347 0.0012 0.7 0.6566 0.6944 0.6932 0.0366 0.0012 0.8 0.6609 0.7000 0.6987 0.0377 0.0013 0.9 0.6748 0.7145 0.7131 0.0383 0.0013 1.0 0.6974 0.7371 0.7358 0.0384 0.0013 1.1 0.7276 0.7671 0.7657 0.0381 0.0013 1.2 0.7649 0.8037 0.8024 0.0375 0.0013 1.3 0.8084 0.8463 0.8451 0.0367 0.0013 1.4 0.8575 0.8944 0.8932 0.0357 0.0012 1.5 0.9118 0.9475 0.9463 0.0345 0.0012 1.6 0.9706 1.0050 1.0038 0.0332 0.0012 1.7 1.0335 1.0665 1.0654 0.0318 0.0011 1.8 1.1002 1.1317 1.1306 0.0304 0.0011 1.9 1.1702 1.2002 1.1991 0.0290 0.0010 2.0 1.2432 1.2716 1.2707 0.0275 0.0010 0 and Section 6.6 14. To find the approximate values, set y1 y e2x 1 and use EULERT and IMPEULT with initial values x y 1 and step size 0.1 for 20 points. The exact values are given by y e x e2x 1. y (Euler) x improved Euler y y (exact) Error (Euler) Error 277 0 and improved Euler 0 1 1 1 0 0 0.1 1.1000 1.1161 1.1162 0.0162 0.0002 0.2 1.2321 1.2700 1.2704 0.0383 0.0004 0.3 1.4045 1.4715 1.4723 0.0677 0.0007 0.4 1.6272 1.7325 1.7337 0.1065 0.0012 0.5 1.9125 2.0678 2.0696 0.1571 0.0018 0.6 2.2756 2.4954 2.4980 0.2224 0.0026 0.7 2.7351 3.0378 3.0414 0.3063 0.0037 0.8 3.3142 3.7224 3.7275 0.4133 0.0050 0.9 4.0409 4.5832 4.5900 0.5492 0.0068 1.0 4.9499 5.6616 5.6708 0.7209 0.0092 1.1 6.0838 7.0087 7.0208 0.9370 0.0121 1.2 7.4947 8.6872 8.7031 1.2084 0.0159 1.3 9.2465 10.7738 10.7944 1.5480 0.0206 1.4 11.4175 13.3628 13.3894 1.9719 0.0267 1.5 14.1037 16.5696 16.6038 2.5001 0.0342 1.6 17.4227 20.5358 20.5795 3.1568 0.0437 1.7 21.5182 25.4345 25.4902 3.9720 0.0556 1.8 26.5664 31.4781 31.5486 4.9822 0.0705 1.9 32.7829 38.9262 39.0153 6.2324 0.0891 2.0 40.4313 48.0970 48.2091 7.7778 0.1121 dy dx dy y2 2y 2(x 2 dy (2x y 1 15. (a) y 1) 2(x 1)dx x2 2) dx 2x C 1 2 Initial value: y (2) 2 22 2 C 2(2) Solution: y (3) y 32 C 1 1 2(3) x2 2x 2 2 or y 1 5 1 2x 2 0.2 (b) To find the approximation, set y1 5 points. This gives y (3) x2 2y 2(x 0.1851; error (c) Use step size 0.1 for 10 points. This gives y (3) 0.1929; error 0.0071. (d) Use step size 0.05 for 20 points. This gives y (3) 0.1965; error 0.0035. 1) and use EULERT with initial values x 0.0149. 2 and y 1 and step size 0.2 for 2 278 Section 6.6 dy dx dy y1 16. (a) y dx 5 ln y 1 x y 1 ex y 1 y 19. Set y1 2y sin x and use EULERG with initial values x 0 and y 0 and step size 0.1. The exact solution is 1 2x y (e 2 sin x cos x). 1 C C eCe x Ae x 1 [ 0.1, 1.1] by [ 0.13, 0.88] Initial condition: y (0) 3 Ae0 2 20. Set y1 x y and use EULERG with initial values x 0 and y 2 and step size 0.1. The exact solution is y x 1 e x. 3 A 1 Solution: y y (1) 2e 2e x 1 1 6.4366 (b) To find the approximation, set y1 y 1 and use EULERT with initial values x 0 and y 3 and step size 0.2 for 5 points. This gives y (1) 5.9766; error 0.4599. (c) Use step size 0.1 for 10 points. This gives y (1) 6.1875; error [ 0.1, 1.1] by [ 2.3, 0.3] 21. Set y1 2y(x x 2 and y 2 y 2e x 2x. 1) and use IMPEULG with initial values 2 and step size 0.1. The exact solution is 0.2491. (d) Use step size 0.05 for 20 points. This gives y (1) 6.3066; error 0.1300. 17. The exact solution is y x2 1 2x (a) To find the approximation, set y1 IMPEULT with initial values x 2 , so y (3) 2y 2(x 2 and y step size 0.2 for 5 points. This gives y (3) error 0.2. 1) and use 1 and 2 [ 2.2, 0.2] by [ 0.2, 2.2] 22. Set y1 x(1 y) and use IMPEULG with initial values x 2 and y 0 and step size 0.1. The exact solution is 2 y e (x /2) 2 1. 0.2024; 0.0024. (b) Use step size 0.1 for 10 points. This gives y (3) 0.2005; error 0.0005. (c) Use step size 0.05 for 20 points. This gives y (3) 0.2001; error 0.0001. (d) As the step size decreases, the accuracy of the method increases and so the error decreases. 18. The exact solution is y 2e x y (1) 2e 1 6.4366. 1, so [ 2.2, 0.2] by [ 7.3, 1.1] 23. To find the approximate values, set y1 x y and use EULERT with initial values x 0 and y 1 and step size 0.1 for 10 points. The exact values are given by y 2e x x 1. x Error 1 1.0 0 0.1 0.9000 0.9097 0.0097 0.2 0.8200 0.8375 0.0175 0.3 0.7580 0.7816 0.0236 (b) Use step size 0.1 for 10 points. This gives y (1) 6.4282; error 0.0084 0.4 0.7122 0.7406 0.0284 (c) Use step size 0.05 for 20 points. This gives y1 6.4344; error 0.0022. 0.5 0.6810 0.7131 0.0321 0.6 0.6629 0.6976 0.0347 0.7 0.6566 0.6932 0.0366 0.8 0.6609 0.6987 0.0377 0.9 0.6748 0.7131 0.0383 1.0 0.6974 0.7358 0.0384 (a) To find the approximation, set y1 y 1 and use IMPEULT with initial values x 0 and y 3 and step size 0.2 for 5 points. This gives y (1) 6.4054; error 0.0311. (d) As the step size decreases, the accuracy of the method increases and so the error decreases. 0 y (Euler) y (exact) Section 6.6 24. To find the approximate values, set y1 x y and use IMPEULT with initial values x 0 and y 1 and step size 0.1 for 10 points. The exact values are given by y 2e x x 1. x y improved Euler 0 y (exact) (b) [0, 10] by [0, 3] Error 1 1.0 0 0.1 0.9100 0.9097 0.0003 0.2 0.8381 0.8375 0.7824 0.7816 0.0008 0.4 0.7416 0.7406 0.0010 0.5 0.7142 0.7131 0.0011 0.6 0.6988 0.6976 0.0012 0.7 0.6944 0.6932 0.0012 0.8 0.7000 0.6987 0.0013 0.9 0.7145 0.7131 0.0013 1.0 0.7371 0.7358 28. Set y sin (2x y) and use IMPEULG with initial values x 0 and y 1 and step sizes 0.1 and 0.05. 0.0006 0.3 279 0.0013 x 25. Set y1 y e 2 and EULERG, with initial values x 0 and y 2 and step sizes 0.1 and 0.05. (a) (a) [0, 10] by [0, 5] (b) [0, 10] by [0, 5] 29. To find the approximate values, let y1 y and use EULERT with initial values x 0 and y 1 and step size 0.05 for 20 points. This gives y (1) 2.6533. Since the exact solution to the initial value problem is y e x, the exact value of y (1) is e. [0, 4.7] by [0, 100] 30. To find the approximate values, let y1 3y and use IMPEULT with initial values x 0 and y 1 and step size 0.05 for 20 points. This gives y (1) 19.8845. Since the exact solution to the initial value problem is y e3x, the exact value of y (1) is e3. (b) [0, 4.7] by [0, 100] 26. Set y1 cos (2x y) and use EULERG with initial values x 0 and y 2 and step sizes 0.1 and 0.05. 31. To find the approximate values, let y1 1 y and use RUNKUTT with initial values x 0 and y 1 and step size 0.1 for 10 points. The exact values are given by y 2e x 1. x y (Runge-Kutta) y (exact) Error 0 and y 1 and step size 0.1 and 0.05. 3 (a) [0, 10] by [0, 3] 0.0000006 1.9836 1.9836 0.0000009 0.5 2.2974 2.2974 0.0000013 2.6442 2.6442 0.0000017 3.0275 3.0275 0.0000022 3.4511 3.4511 0.0000027 0.9 x 0.0000004 1.6997 0.8 ln y and use IMPEULG with initial values 1.4428 1.6997 0.7 27. Set y1 1.4428 0.6 1 y 2 0 0.0000002 0.4 [0, 10] by [0, 6] 1 1.2103 0.3 (b) 1 1.2103 0.2 [0, 10] by [0, 6] 0 0.1 (a) 3.9192 3.9192 0.0000034 1.0 4.4366 4.4366 0.0000042 280 Chapter 6 Review 32. (a) Set y1 x y and use RUNKUTT with initial values x 0 and y 1 and step size 0.1. 5. Let u du sin x cos x dx /2 1 5 sin 3/2 x cos x dx 5u 3/2 du 0 0 2 5/2 u 5 5 2(1 [0, 10] by [ 3, 10] (b) Use RUNKUTT with initial values x and step size 0.1. 0 and y 2 2 4 x2 6. 3x x 1/2 4 dx (x 3) dx (x 4 12 x 2 11 24 3(4) s Chapter 6 Review Exercises 20 d tan tan 0 0 2 2. 12 x 2 1 dx x2 x 1 x 1 3 tan 0 13 8 7. Let u du 1 2 1 2 1 tan x sec 2 x dx /4 1 e tan x sec 2 x dx e u du 0 3 2 3. Let u du 1 du 2 1 eu 2 e 1 8. Let u dx du 3 3 1) 1 du u3 1 u2 18 2 1 9 1 9 8 9 9 dx 18 e 1 3 1 du du 1 ln r dr r u 1/2 du 0 2 3/2 u 3 2 (1 3 9. Let u du du 2x dx 2x sin (1 1 x 2) dx 0 sin u du 0 1 0 0) 2 3 2x dx 1 1 1 dr r 1 x2 1 e0 ln r 8 4. Let u 1 0 e1 2 dx 36 (2x 0 2x 0 1 2 4 2 12 8 147 8 2 1 1 (4) 2 3 1 8 20 /3 sec 2 1/2 3 2 (pp. 358 – 361) /3 3x 1 (16) 2 [0, 10] by [ 3, 10] 1. 0) 1/2 0 2 2 sin x cos x dx cos x dx cos x dx sin x 1 du u ln u C ln 2 sin x C 0) 1 0 Chapter 6 Review 10. Let u 3x du 1 dx x 1 du u dx x ln x dx ln u dx 3x ln x du 3 dx 1 du 3 3 15. Let u 4 1 u 1/3 du 3 1 3 2/3 u C 32 1 (3x 4)2/3 C 2 4 dt du 1 du 2 t2 3/2 12. Let u 1 2 sec 2 1 du u 1 ln u 2 ln t 2 5 ln (t 2 5) f (x) and its derivatives tan (+) cos x 3x2 (–) sin x 6x (+) –cos x 6 1 2 (–) –sin x d 1 d sec u tan u du 0 C 1 C x 3 sin x 1 dy y 18. Let u du sin u du 1 dw w ln w C ln cos u C ln cos (ln y) 14. Let u du C x e e x dx e x sec (e x) dx 6x sin x ln x dv 1 dx x v x 4 ln x dx cos u dw 3x 2 cos x 6 cos x x 4 dx 15 x 5 tan u du sin u du cos u Let w cos x x 3 cos x dx ln y tan (ln y) dy y cos x. g(x) and its integrals x3 1 x 3 and g (x) C sec du C 17. Use tabular integration with f (x) C C sec u 13. Let u C t 1 2 1 2 1 2 1 1/2 2t t dt du dt 5 2t dt t dt t2 5 C dt t 3/2 tt t 11. Let u C ln ln x 16. 281 sec u du ln sec u tan u C ln sec (e x) tan (e x) C 15 x ln x 5 15 x ln x 5 15 x ln x 5 151 x dx 5 x 14 x dx 5 15 x C 25 C 282 Chapter 6 Review e 3x 19. Let u e sin x dx e 3x sin x dx v 3e 3x dx du 3x dv cos x 12 x 12 x dx x 2 1 x dx x 1 x2 2 dx x2 13 x 2x x 1 C 3 1 21C1 3 dy dx 22. x dy 3 cos x e 3x dx cos x dy Integrate by parts again. 3e 3x Let u dv 9e 3x dx du v y cos x dx y sin x y (1) e 3x 10 e 3x e e sin x dx sin x dx 3x x2 20. Let u du cos x cos x dv 2x dx dx 1 2 3x xe 3 dv Let u du e sin x 3e 3x sin x 9e 3x 3x sin x dx 4 3 1 1 3 C C x3 3 y 2x 1 x Graphical support: dx 2 e 3xx dx 3 e 3x dx [ 2, 2] by [ 10, 10] 23. 1 3x e 3 1 2 3x 2 1 1 xe xe 3x e 3x dx 3 3 3 3 1 2 3x 2 2 xe xe 3x e 3x dx 3 9 9 1 2 3x 2 2 xe xe 3x e 3x C 3 9 27 x2 2x 2 e 3x C 3 9 27 dx C C 1 3x e 3 v x 3x 3e 3x 1 [ e 3x cos x 3e 3x sin x] 10 3 sin x cos x 3x e C 10 10 e 3x sin x dx x 2e 3x dy dt 1 t 4 1 dy v t dy y dt 4 1 t ln t 4 dt 4 y ( 3) C ln (1) C ln (t 2 2 y C 4) 2 Graphical Support: x2 1x 2 x2 dy 1x dx 2 2 x dy 1x dx 2 12 13 yx x x C 2 6 dy 21. dx y (0) y C x3 6 x 24. dy d csc 2 cot 2 dy 1 x2 2 [ 4.5, 5] by [ 2, 5] csc 2 cot 2 d 1 dy csc 2 cot 2 d Graphical support: 1 csc 2 C 2 1 C1 2 y y C [ 4, 4] by [ 3, 3] y 4 3 2 1 csc 2 2 3 2 [0, 1.57] by [ 5, 3] 1 3 283 Chapter 6 Review d(y ) dx 25. d (y ) d (y ) x2 y y (1) x 2 1 C 26. d(r ) dt cos t d (r ) 1 x2 1 2x dx x2 1 2x dx x2 2x cos t dt d (r ) cos t dt r C r (0) 1 sin t C r C sin t d(r ) y (1) 2 3 C C y x (x 2 dy y 1 1 1 x r 1) dx 13 x ln x 3 1 01 3 r (0) x3 3 t C C C 0 r cos t r ln x 1) dt 1 2 dr (cos t sin t 2 t t2 2 2) dt C 2t 1 We first show the graph of y r r C sin t 1 t2 2 Graphical support: 2 . 3 x t 2t r (0) 2 3 x ln x 1 C 0 2 3 x3 3 cos t C We first show the graph of y x 1 ( sin t x Graphical support: Let f (x) 1 1 x2 y C f (x) 0, along with the slope field for y x2 x f (x) 1 1, 2x with the slope field for y sin t r 1 along cos t. 1 . x2 [ 6, 4] by [ 3, 3] [ 0.5, 4.21] by [ 9, 21] Next, we show the graph of y We now show the graph of y for y f (x) x 2 x f (x) along with the slope field 1 with the slope field for y r cos t r sin t t 2 along 1. 1. [ 6, 4] by [ 3, 3] [ 0.5, 4.21] by [ 9, 21] Finally we show the graph of y r along with the slope field for y r [ 6, 4] by [ 8, 2] sin t cos t t2 2 2t t 2. 1 284 27. Chapter 6 Review dy y2 dx dy dx y2 dy dx y2 ln y 2 2 y Since 1 C C x x2 2 C g (x)] dx x C. 2 f (x) dx 2(1 2 g (x) dx x) (x x C 2 4e x C C is an arbitrary constant, we may write the 31. [2f (x) 2 x) 1 4 y f (x) dx (1 indefinite integral as 2 C x dx x 2 x2 2 x Ce x f (x)] dx 2 Ce x y y (0) 30. [x 2 32. [g (x) 4] dx x g (x) dx Graphical support: C 4 dx (x 2) 4x 2 Since 2 2) 3x C C C is an arbitrary constant, we may write the indefinite integral as 3x C. [ 5, 5] by [ 5, 20] 28. dy dx dy y1 dy y1 33. We seek the graph of a function whose derivative is (2x 1)(y 1) Graph (b) is increasing on [ (2x 1) dx (2x ln y 1 x2 y 1 Ce x 2 sin x is positive, x and oscillates slightly outside of this interval. This is the 1) dx x , ], where sin x . x correct choice, and this can be verified by graphing C NINT x sin x , x, 0, x . x 34. We seek the graph of a function whose derivative is e x . Since e x 0 for all x, the desired graph is increasing for all x. Thus, the only possibility is graph (d), and we may verify that this is correct by graphing NINT(e x , x, 0, x). 2 2 y Ce x x 1 1 2 1 2 y ( 1) C C 2 y 2e x 35. (iv) The given graph looks like the graph of y 2 x satisfies 1 36. Yes, y Graphical support: 37. (a) dy dx 2 dv [ 3, 3] by [ 10, 40] 29. 2t 1 Since 1 4 x) x C x C. 0 4 C C is an arbitrary constant, we may write the indefinite integral as 6t) dt 3t 2 C Initial condition: v f (x) dx (1 6t (2 2t 4 when t C v f (x) dx 1. x is a solution. dv dt v 2x and y (1) C 3t 2 4 1 1 v (t) dt (b) 0 3t 2 (2t 0 t2 t3 4) dt 1 4t 0 6 0 6 The particle moves 6 m. 0 x 2, which Chapter 6 Review 38. 285 40. Set y1 (2 y)(2x 3) and use IMPEULT with intial values x 3 and y 1 and step size 0.1 for 20 points. x y 3 2.9 x 0.6680 2.8 [ 10, 10] by [ 10, 10] 39. Set y1 y cos x and use EULERT with initial values x 0 and y 0 and step size 0.1 for 20 points. 1 0.2599 2.7 0.2294 y 2.6 0.8011 0 0 2.5 1.4509 0.1 0.1000 2.4 2.1687 0.2 0.2095 2.3 2.9374 0.3 0.3285 2.2 3.7333 0.4 0.4568 2.1 4.5268 0.5 0.5946 2.0 5.2840 0.6 0.7418 1.9 5.9686 0.7 0.8986 1.8 6.5456 0.8 1.0649 1.7 6.9831 0.9 1.2411 1.6 7.2562 1.0 1.4273 1.5 7.3488 1.1 1.6241 1.4 7.2553 1.2 1.8319 1.3 6.9813 1.3 2.0513 1.2 6.5430 1.4 2.2832 1.1 5.9655 1.5 2.5285 1.0 5.2805 1.6 2.7884 1.7 3.0643 1.8 3.3579 1.9 3.6709 2.0 4.0057 x x 41. To estimate y (3), set y1 initial values x 0 and y points. This gives y (3) 1 and step size 0.05 for 60 0.9063. 42. To estimate y (4), set y1 with initial values x x2 2y x 1 and y 60 points. This gives y (4) 43. Set y1 e (x x 0 and y 2y and use IMPEULT with 1 y 2) 1 and use EULERT 1 and step size 0.05 for 4.4974. and use EULERG with initial values 2 and step sizes 0.1 and 0.1. (a) [ 0.2, 4.5] by [ 2.5, 0.5] 286 Chapter 6 Review 43. continued (b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x 1. (This occurs because the analytic solution is y 2 ln (2 e x ), which has an asymptote at x ln 2 0.69. Obviously, the Euler approximations are misleading for x 0.7.) kt 47. T Ts (T0 Ts)e We have the system: 39 Ts (46 Ts )e 10k 33 Ts (46 Ts )e 20k Thus, 39 Ts 46 Ts Since (e e 10k 2 ) 10k 20k e and 33 Ts 46 Ts (39 Ts)2 (33 78Ts Ts2 1518 Ts 3 39 x2 ey 44. Set y1 values x Ts 2 33 Ts 46 1521 Ts 46 Ts y and use IMPEULG with initial x 0 and y 0 and step sizes 0.1 and 0.1. (a) . , this means: () [ 1, 0.2] by [ 10, 2] 20k e Ts)(46 Ts) 79Ts Ts2 The refrigerator temperature was 3 C. 48. Use the method of Example 3 in Section 6.4. kt 0.995 kt ln 0.995 e [ 0.2, 4.5] by [ 5, 1] 1 ln 0.995 k t (b) 5700 ln 0.995 ln 2 41.2 The painting is about 41.2 years old. 49. Use the method of Example 3 in Section 6.4. Since 90% of the carbon-14 has decayed, 10% remains. [ 4.5, 0.2] by [ 1, 5] 2.645 k (b) Mean life 46. T T Ts 40 (T0 1 k 180 50. Use t kt 250 e rt kt e rt 180 and t (220 15k 180 140 1 ln 15 k (k)(15) 9 7 (220 40)e ((1/15) ln (9/7))t 70 40 (220 40)e ((1/15) ln (9/7))t ln 6 t 15 ln 6 ln (9/7) 6 107 min It took a total of about 107 minutes to cool from 220 F to 70 F. Therefore, the time to cool from 180 F to 70 F was about 92 minutes. 18,935 1924 64 years. 7500 30 rt ln 30 ln 30 t ln 30 64 0.053 The rate of appreciation is about 0.053, or 5.3%. 9 7 180 30 1988 r 40)e 40 9 1 ln t 7 15 5700 ln 0.1 ln 2 15 to find k. T e((1/15) ln (9/7))t 1 ln 0.1 k 3.81593 years 40 e ln 0.1 The charcoal sample is about 18,935 years old. 40)e Use the fact that T 0.1 t 0.262059 Ts)e (220 kt kt e ln 2 k ln 2 k ln 2 2.645 45. (a) Half-life Chapter 6 Review 51. Using the Law of Exponential Change in Section 6.4 with (c) 1 appropriate changes of variables, the solution to the differential equation is L(x) kx L0e the surface intensity. We know 0.5 , where L0 e 150 e 4.3 125 t 6 5 1 5 L(0) is 18k , so 1 e 4.3 e 4.3 t t ln 5 4.3 t t ln 0.5 k and our equation becomes 18 1 x/18 L(x) L0e (ln 0.5)(x/18) L0 . We now find the depth 2 4.3 ln 5 5.9 days It took about 6 days. where the intensity is one-tenth of the surface value. 54. Use the Fundamental Theorem of Calculus. 1 x/18 2 1 x ln 2 18 18 ln 0.1 ln 0.5 0.1 ln 0.1 x ln c kA (c y) V kA dt V kA tC V kA tC V y ln c y c y c y c e y c De Initial condition y c Solution: y k y(0) c 150 1 e 4.3 1 3(0)2 sin(t 2) dt kt P), or 1 150 e 4.3 0.002P P 800 800 P 800 0.002 dt (800 P) P dP P(800 P) 0 0.002 dt 1 dP 800 P 0.002 dt ln 800 0.002t C c)e 0.002t C P (kA/V)t P 800 P 800 P ln P ln (kA/V)t c)e 150 e 4.3e c 800 P P t where M 150, A 4.3 e , and dP dt 1 P(150 150 P). The 1 P Initial condition: P(0) 50 2 Initially there were 2 infected students. 0.002t 1 50 800 1 Ae 0 A 16 A 15 Solution: P 1 800 15e 0.002t e Ae 1 C 0.002t C e 800 P P 800 P carrying capacity is 150. (b) P(0) 0.002P 1 1 P (y0 1 2 800 dP P(800 P) differential equation. k P(M M 2 dP dt 1. Therefore, it is a solution of the logistic dP dt 0 03 dP dt (kA/V)t (y0 M Ae 1 55. ln P t 1 0 D t→ This is P 0 y0 when t lim [c t→ (sin 02) y (0) D y0 (b) lim y(t) Verify the initial conditions: (kA/V)t C y c 6x (kA/V)t C e y0 1) Thus, the differential equation is satisfied. (kA/V)t C e 2) 6x 2x cos (x 2) dy dt dy cy x 1) 3x 2 (cos x 2)(2x) 59.8 feet. 53. (a) P(t) (3x 2 d (sin x 2 dx y d3 (x dx sin t 2 dt 0 (sin x 2) 59.8 ft You can work without artificial light to a depth of about 52. (a) x d dx y C e 0.002t 0.002t 800 Ae 0.002t 287 288 Section 7.1 56. Method 1–Compare graph of y1 y2 x 3 ln x 3 NDER x 3 ln x 3 79.961(0.9273)t 60. (a) T x3 . The graphs should be the same. 9 Method 2–Compare graph of y1 y2 x 2 ln x with NINT(x 2 ln x) with x3 . The graphs should be the same or differ 9 [ 1, 33] by [ 5, 90] only by a vertical translation. (b) Solving T(t) 40 graphically, we obtain t 9.2 sec. The temperature will reach 40 after about 9.2 seconds. 10,000(1.063)t 57. (a) 20,000 (c) When the probe was removed, the temperature was about T(0) 79.96 C. 1.063t 2 ln 2 t ln 1.063 ln 2 ln 1.063 t 61. 11.345 v0m It will take about 11.3 years. (b) 20,000 k ln 2 0.063 11.002 It will take about 11.0 years. d dx d dx 58. (a) f (x) g (x) v0m (1 e 0.97(1 e (27.343/30.84)t s(t) 0.063t t 27.343 s(t) 0.063t e 0.063t ln 2 0.97 s(t) 10,000e 2 coasting distance k (0.86)(30.84) k 0.97(1 e 0.8866t k (k/m)t ) ) ) A graph of the model is shown superimposed on a graph of the data. x u(t) dt u(x) u(t) dt u(x) 0 x 3 [0, 3] by [0,1] (b) C f (x) g(x) x x u(t) dt 0 x u(t) dt 3 3 u(t) dt 0 3 u(t) dt x u(t) dt 0 59. (a) y 1 Chapter 7 Applications of Definite Integrals 56.0716 5.894e 0.0205x s Section 7.1 Integral as Net Change (pp. 363–374) Exploration 1 1. s(t) s(0) [ 20, 200] by [ 10, 60] (b) The carrying capacity is about 56.0716 million people. (c) Use NDER twice to solve y 0. The solution is x 86.52, representing (approximately) the year 1887. The population at this time was approximately P(86.52) 28.0 million people. 8 t2 03 3 Thus, s(t) 2. s(1) Revisiting Example 2 13 3 (t 1) 8 0 C 1 t3 3 8 t 8 1 1 t3 3 dt 2 1 1 9⇒C 8 t 1 C 1 1. 16 . This is the same as the 3 answer we found in Example 2a. 3. s(5) 53 3 8 5 1 found in Example 2b. 1 44. This is the same answer we Section 7.1 Quick Review 7.1 7. 1. On the interval, sin 2x 0 when x 2 4 , sin 2x 3 , sin 2x 4 x and 2 1; for x 4 – – 3x + – 0 (x 1)(x 2 f (x ) 32 ,x 2 for x 3x 3x 2 2 x2 x2 0 when x x2 x2 x2 x2 x2 x2 1 or 2. Test 0, x 2 3x 2 2; 3, –2 1 17 x2 2 ; for x 1.9, 2 4.13; for x 0, 9 x 4 2 1 x 2 5 ; for x 1.9, 2 4.13; and for x , 2 x 4 2 17 . The function changes sign at 2, 2, 2 9 2 4 2 4 2 4 – + – f (x ) 1)2(2x 1) 3x 2 1 0, 2x 3 4; for x 1 or 1. 2 0 when x 3x 2 1 0 when x function changes sign at – 0 4 x 2 ; and for x + 6. xe 0 when x always positive. – + + 3 4 , x cos 2x 4, x cos 2x 4 , 3 5 0, , , or . 44 4 k or 2.1783 + – 32.7984. The 8 2 k , where k is an f (x ) x –2.1783 –0.9633 0.9633 2.1783 1 x sin 1 x 1 x 0.15, sin 1 1 or . Test one 3 2 1 0.1, sin 0.54; for x 0 when x 0.37; and for x ; for x , The graph is + 3 f (x ) x 54 4 0. On the rest of the interval, xe – 0.1 1 0.58. The x is 0.2, 0.96. The graph changes sign at , 3 5 , and . The graph is 4 4 – sin2 1) 1 – x Test one point on each subinterval: for x ; for x 2.4030. point on each subinterval: for x x 2 5. On the interval, x cos 2x x cos 2x 1: sec (1 is undefined when k for any integer k. Test for sin2 0) 1 changes sign at 0.9633 1; and f (x ) + 1 16 0: sec (1 10. On the interval, sin –1 2 2 x 1. The function changes sign at + x cos 2x k or 2.1783 cos x ) integer. The graph is 1, 1 . The graph is 2 – 0.9633 cos (1 sign alternates over successive subintervals. The function Test one point on each subinterval: for x 1 x Test for x x (x x 3 1 sin2 x) 1 9. sec (1 4 2 1 f (x ) + √2 2 x 2 –4 –2 5 , 2 f (x ) + + 3 , 2x 3 2 2 and is undefined when and 2. The graph is 3. x 2 2x 3 0 has no real solutions, since b 2 4ac ( 2)2 4(1)(3) 8 0. The function is always positive. The graph is x x 0 when x + – 3x 2 1 . The 2 2. The function changes sign at 1 and 2. + 2x 3 1 30 –3 –2 –√2 3x 2 x x2 2. Test one point on each subinterval: for x The graph is 4. 2x 3 1, f (x ) + 2 4 x 1 ; and for x 4 1 ; for x 2 1 –5 0 x 2) x2 – , 0, 2 2 1, 1; and for 2 2 0. Test one point on each subinterval: x function changes sign at 0. The graph is 1. The function changes sign at one point on each subinterval: for x x 1; , sin 2x 0 when x 1 for x 8. + 2 . Test one . The graph is –3 2. x 2 2 3 , sin 2x 4 point on each subinterval: for x for x , 0, or x x2 289 – 1 2 f (x ) x 0.2 1 1 , and . 3 2 290 Section 7.1 Section 7.1 Exercises 1. (a) Right when v(t) when 0 0, which is when cos t or 2 2 i.e., when t 6t 2 v(t) 0, i.e., 3 t 2 . Left when cos t 2 3 t . Stopped when cos t 0, 2 3 or . 2 2 t i.e., when 4. (a) Right when 18t 12 6(t 1)(t 2) i.e., when 0 t 1. Left when 6(t i.e., when 1 0, t 0, 2. Stopped when 6(t 1)(t 2) 1)(t 0, i.e., when x 2) 0, 1, or 2. 2 0 2 2 5 cos t dt 5 sin t 5[sin 2 0 sin 0] 2t 3 0 9t 2 12t 2 3 /2 5 cos t dt 0 10 5 t 3 2 t 3 36 . Left when sin 3t . Stopped when sin 3t 18t 0, i.e., when 0, i.e., when t /2 6 sin 3t dt 1 cos 3t 3 6 0 3 2 cos 0 0 ( 6t2 12) dt 1 /2 cos t 0, which is when sin t 0, i.e., when 0 when sin t 0 and cos t 2 0 (b) Displacement 2 3 or 2 2 t 0] 0 and 0, i.e., when 2 . Left t 2 or 0 or cos t 5 sin2 t cos t dt 5 0, 1 sin3 t 3 2 6 sin 3t dt 4 2 3. (a) Right when v(t) Left when 49 /2 49 9.8t 9.8t Stopped when 49 0, i.e., when 0 0, i.e., when 5 9.8t t (49 0, i.e., when t 5. 5 sin2 t cos t dt 0 2 49[(10 10) 0] 0 0 5 sin2 t cos t dt 5 sin2 t cos t dt 3 /2 5 3 10 3 when 0 10 3 /2 /2 5 3 20 3 6. (a) Right when v(t) 9.8t) dt 0 4.9t 2 5. 0 10. 10 (b) Displacement t 5 sin2 t cos t dt (c) Distance 6 /3 t 0, which is when 4 4. Left: never, since negative. Stopped when 4 t 4 t 0, i.e., t cannot be 0, i.e., when t 4. 10 (c) Distance (49 49 0 9.8t dt ( 49 5 122.5 4 (b) Displacement 10 9.8t) dt 0 122.5 0 0 /2 0 5 2 6 sin 3t dt 6 sin 3t dt 49t 12) dt t 0 5[0 0 /3 18t 6 3 . Stopped when sin t 2 3 i.e., when t 0, , , , or 2 . 2 2 /2 (c) Distance 4 12 dt t 2 cos 0] 2 18t 5. (a) Right when v(t) 0, . (b) Displacement 24) 1 5 0, which is when sin 3t 6t 2 0 3 /2 20 2. (a) Right when v(t) i.e., when 0 (6t 2 5 cos t dt /2 5 [(16 0 1 2 5 cos t dt 3 (c) Distance 5 cos t dt /2 12) dt 0 2 0 18t 2 0 (c) Distance or (6t 2 (b) Displacement (b) Displacement 245 9.8t) dt 4 t dt 0 2 [0 3 8] 16 3 4 (c) Distance 4 0 t dt 16 3 2 (4 3 t)3/2 4 0 291 Section 7.1 7. (a) Right when v(t) i.e., when 0 0, which is when cos t 3 or 2 2 t cos t 0, i.e., when cos t e sin t cos t dt e 0 [e 0 0 e] 16t /2 0 e sin t cos t dt e sin t cos t dt C1, where 6 ft/sec. 90t 0. Solve s(t) 90t when t 90 16t2 2t( 8t 45 8 0 or t 0 2 /2 esin t cos t dt 32(3) s(0) 2 32t 90. v(t) dt C2 0 esin t cos t dt (c) Distance 3 /2 (b) s(t) 0 2 v(0) 32 dt Then v(3) 2 sin t a(t) dt C1 3 . Stopped when 2 3 or . 2 2 0, i.e., when t (b) Displacement 11. (a) v(t) 2 . Left when t 2 2 t 0, C2, where 0: 45) 0 5.625 sec. The projectile hits the ground at 5.625 sec. 3 /2 (e 1) 1 e e 1 e 1 2 e 2e 4.7 (c) Since starting height 8. (a) Right when v(t ) 0, which is when 0 t 3. Left: never, since v(t ) is never negative. Stopped when t 0. 3 (b) Displacement 1 0 1 [ln (10) 2 t t 3 t 0 t 2) 0. 3 (d) Max. Height 0 1.15 ln 10 2 dt t2 1 1 ln (1 2 dt ln 10 2 ln (1)] (c) Distance 2 Displacement 16 s 5.625 2 2 Distance 1.15 ending height, 5.625 2 5.625 90 2 126.5625, and 2(Max. Height) 253.125 ft. c 9. (a) v(t) a(t) dt v(t) t 2t 3/2 C, and since v(0) 2t 3/2. Then v(9) t 9 2(27) 0, 63 mph. 12. Displacement v(t) dt 4 5 24 23 cm 0 c 13. Total distance v(t) dt 4 5 24 33 cm v(t) dt 15 4 11. v(t) dt 15 4 5 16. v(t) dt 15 4 5 24 0 a (b) First convert units: 14. At t t t 3/2 t 2t 3/2 mph mi/sec. Then 3600 1800 9 3/2 t t Distance dt 1800 0 3600 2 5/2 t t 9 27 9 0 0.06525 mi 7200 4500 0 800 500 344.52 ft. a, s s(0) 0 b At t b, s s(0) 0 c At t c, s s(0) 8. 0 15. At t a, where dv is at a maximum (the graph is steepest dt upward). 4 10. (a) Displacement (t 2) sin t dt 0 sin t 16. At t 4 t cos t 2 cos t 4 cos 4 2 cos 4) 2] 1.44952 m 17. Distance (b) Because the velocity is negative for 0 for 2 t , and negative for t t 2, positive 4, 2 Distance (t 0 2) sin t dt [(2 sin 2) ( ( 2 2 cos 4 (t 2) sin t dt 2 4 (t 2) sin t dt sin 2 2 cos 4 2 sin 2 dv is at a maximum (the graph is steepest dt upward). 0 [(sin 4 c, where sin 4 2)] 2 1.91411 m. 4 1 2 (a) Final position Initial position 2 4 6; ends at x 6. 12 4 Distance (b) 4 meters 18. (a) Positive and negative velocities cancel: the sum of signed areas is zero. Starts and ends at x 2. (b) Distance meters 2) sin 4 Area under curve Sum of positive areas 4(1 1) 4 292 Section 7.1 25. (Answers may vary.) Plot the speeds vs. time. Connect the points and find the area under the line graph. The definite integral also gives the area under the curve. 7 19. (a) Final position 2 v(t) dt 0 1 (1)(2) 2 1 (2)(2) 2 2 1 (1)(2) 2 1 (2)(1) 2 1(2) 26. (a) Sum of numbers in Sales column (b) Enter the table in a graphing calculator and use QuadReg: B(x) 1.6x 2 2.3x 5.0. 5; ends at x 5. 11 (1.6x 2 (b) 1 1 (1)(2) (1)(2) 2 2 1 (2)(1) 2 v(t) dt 0 v(t) dt 0 1 1 (2)(3) (1)(3) 2 2 1 (3)(3) 2 2 1 (1)(3) 2 (3)(3) (1.6x 2 1.6 3 x 3 10 v(t) dt 1 (1)(3) 2 3) 1 (1)(3) 2 3(3) 1 (3)(3) 2 21. 27.08 e t/25 b dt 27.08 25e 10 t/25 27.08[25e 0 0.4 25] 332.965 billion barrels 3.9 22. 0 93.6 t 2.4 sin dt 12 28.8 28.8 23. (a) Solve 10,000(2 (b) Width 3.9t r) 0: r 20,000 29. F(x) r2 20,000 80,000 3 13 r 3 20,000 4 r ) dr 8 3 30. F(x) 2 r: Area 3 (c) 8(10 (2 r) r in2 flow in 396 in3 sec x2 2x dx 0 kx; 10,000 2x. 12 kx 2 kx dx 9 81 N cm 0 k(1), so k (b) For total distance: W 16 (10r 3 16 0 1244.07 in3 sec 45 81 4 W in3 sec r 3) dr 0 14 r 4 2 and F(x) 9 0 2rr 3 r2)(2 r) dr 5r 2 2(110) 18N F(x) dx (a) W Cross section area 0 16 2(112) d 10,000. 12 kd 2 0 1 (10,000)(0.5)2 2 1 (10,000)(1)2 2 5000 For second half of distance: Inches per second in. sec 2(119) 1250 inch-pounds r, Length r 2) 2(115) 2(115) k(3), so k d 0 (b) Volume per second 8(10 2(120) 0 2 83,776 24. (a) Width 2(115) 2(9) (b) W (2r 0 kx; 6 (a) F(9) 0 2 2(110) 9 2 r)(2 r) dr f (xn) 121] Area 0 ∑ 2 f (xi) 1156.5 2rr Population density 10,000(2 798.97 thousand 0.5 i1 2(120) 0 2 miles. 2 r: Area 2 (d) 5.0x n1 f (x0) 24 t cos 12 93.6 kilowatt-hours r, Length (c) Population 28.8 a 2n 18 8 [120 2(10) 0 24 10.5 2.3 2 x 2 28. Treat 6 P.M. as 18 o’clock: 19.5 meters 10 5.0) dx (b) The answer in (a) corresponds to the area of midpoint rectangles. The curve now gives a better approximation since part of each rectangle is above the curve and part is below. 0 1 (2 2 2.3x 0.5 2.5. (b) Distance 0 10.5 27. (a) 2.5; ends at x 11 (d) The answer in (a) corresponds to the area of left hand rectangles. These rectangles lie under the curve B(x). The answer in (c) corresponds to the area under the curve. This area is greater than the area of rectangles. 10 2 5.0x 904.02 thousand 7 meters 20. (a) Final position 5.0) dx 2.3 2 x 2 0 1 (2)(2) 2 1(2) 2.3x 1.6 3 x 3 (c) 7 797.5 thousand 31. 5000 (12 0) [0.04 2(12) 1250 3750 inch-pounds 2(0.04) 2(0.05) 2(0.06) 2(0.05) 2(0.04) 2(0.04) 2(0.05) 2(0.04) 2(0.06) 2(0.06) 2(0.05) 0.05] 0 0.585 The overall rate, then, is 12 0.04875. 0.585 Section 7.2 32. (12 0) [3.6 2(12) 2(4.0) 2(3.1) 2(2.8) 2(2.8) 2(3.2) s Section 7.2 Areas in the Plane (pp. 374–382) 2(3.3) 2(3.1) 2(3.9) 4.0] 2(3.2) 2(3.4) 2(3.4) 40 thousandths or 0.040 Exploration 1 1. For k ∑mkxk . Taking dm ∑mk My 33. (a) x 293 M A Family of Butterflies 1: [(2 sin x) sin x] dx (2 0 dA as mk and letting 2 sin x) dx 0 2x 2 cos x 2 4 0 dA → 0, k → x dm yields . For k dm 2: /2 ∑mky . Taking dm ∑mk My (b) y M [(4 0 (4 dA as mk and letting (2 sin 2x)] dx 4 sin 2x) dx 0 /2 4x dA → 0, k → yields 2 sin 2x) /2 2 cos 2x y dm 2 4 0 . dm 2. It appears that the areas for k 2 4. 3 will continue to be /k 34. By symmetry, x 0. For y, use horizontal strips: 3. Ak [(2k k sin kx) k sin kx] dx 0 /k y dm y dA (2k y dA 2k sin kx) dx 0 y dm dA dA If we make the substitution u kx, then du k dx and the 4 y(2 y) dy u-limits become 0 to . Thus, 0 4 2 /k y dy 2 5 2k sin kx) dx 0 /k 4 0 0 2 3/2 4 y 3 0 0 2 y 5/2 2 (2k Ak 0 (2 (2 4. 2 12 5 2 sin kx)k dx 2 sin u) du. 4 5. Because the amplitudes of the sine curves are k, the kth butterfly stands 2k units tall. The vertical edges alone have lengths (2k) that increase without bound, so the perimeters are tending to infinity. 35. By symmetry, y 0. For x, use vertical strips: Quick Review 7.2 x dm x dA x dA 1. x dm dA sin x dx 0 x(2x) dx [1 1] 2 0 1 2 cos x 0 dA 2. 1 2x e 2 e2x dx 0 1 0 12 (e 2 1) 3.195 2 0 2x dx /4 3. sec2 x dx tan x /4 1 ( 1) 2 /4 2 32 x 30 x2 2 0 /4 2 4. x 3) dx (4x 0 3 4 3 5. 9 3 radius 3.) x 2 dx 2x 2 14 x 4 2 (8 4) 0 4 0 9 (This is half the area of a circle of 2 294 Section 7.2 6. Solve x 2 4x x 6. x 2 5x 6 0 (x 6)(x 1) 0 x 6 or x 1 y 6 6 12 or y 1 (6, 12) and ( 1, 5) 7. Solve e x 1 3. 1 6 [(12y 2 (x 1)] 0. Test: e0 e 0 x ( 12y 3 (x 10 3 1) is at a minimum. x2 2y)] dy 2y) dy 1 y2 0 4 3 1 1, i.e., when 5. Use the region’s symmetry: 1. So the solution is (0, 1). 2 2x 10y 2 10 3 y 3 3y 4 3 8. Inspection of the graphs shows two intersection points: (0, 0), and ( , 0). Check: 02 0 sin 0 0 and 2 2 sin 0. 9. Solve (2y 2 2 [2x 2 (x 4 2 2x 2)] dx 2 ( x4 0 2 15 x 5 43 x 3 2 32 5 2 2 x4 x4 x 1 2 (x 2 x2 x2 2 0 0 25 x 5 2 x3 1 2 0 1 3 2 5 2 or 1 1 Throw out the negative solution. 0 (2 y 4 3/2 y 3 y) dy 12 y 2 1 0 1 x3 y 4 3 1 (0, 0), ( 1, 1 2 5 6 0 8. Integrate with respect to y: 1) and (1, 1) 1 10. Use the intersect function on a graphing calculator: 0 [(2 y) y] dy 12 y 2 2y 2 3/2 y 3 1 1 2 2 0 2 3 0 9. Integrate in two parts: 0 [(2x 3 x2 [( x 2 3x) [ 2, 2] by [ 2, 2] ( 0.9286, 0.8008), (0, 0), and (0.9286, 0.8008) 0 (1 2 cos x) dx 0 1 x 2 1 sin 2x 4 /3 0 1 sec2 t 2 /3 (sec2 t 8 sin2 t) dt 0 /3 tan t 4t 2 sin 2t 0 3 4 3 4 3 3 0 x2 2 8x) dx 3x)] dx 5x)] dx ( 2x 3 8x) dx 0 14 x 2 2 0 [0 4 sin2 t dt (2x 3 (2x 3 2 2. Use symmetry: 2 0 ( x2 5x) 2 2 Section 7.2 Exercises 1. 128 15 22 15 7. Integrate with respect to y: 8 1 2 x 1 3 2x 4) dx 0 1 2 0 6. Use the region’s symmetry: x2 1 2 32 3 (0, 0) is a solution. Now divide by x. x2 4x 2) dx 0 x 3. 1 1 12 0 0 1, so that if they are 1 is zero when e x 1 12y 3) 1 14 y 4 0 1 always greater than or equal to x x 4. 5 1. From the graphs, it appears that e is ever equal, this is when e x 13 y 3 y 3) dy 0 x x dx [e dx ( y2 4x 2 (8 0 2 16)] 14 x 2 [( 8 4x 2 16) 2 0 0] 16 5 6 Section 7.2 10. Integrate in three parts: x 1 [( x 2) 2x 2 x2 4: x 2 2x 2) (x 2 4)] dx 13. Solve 7 1, so the curves intersect at 1. 2 (4 x )] dx 1 2 [(7 1 1 2 x 2) [(4 295 (x ( 3x 2 3) dx 1 2)] dx 1 1 3 x 2) dx (1 1 3 [( x 2) x 2)] dx (4 1 (x 2 2 x 2) dx (x 2 2 x 2) dx (x 2 x 2 3 3 1 3 13 x 3 3x 2 1 1 2 3 4 14. 2) dx 2 13 x 3 1 12 x 2 1 x3 3 2 1 2 2 1 x2 2 2x 1 [ 3, 3] by [ 1, 5] 3 1 x2 2 1 3 13 x 3 2x 2x The curves intersect at x 2 8 3 2 2 1 2 1 3 4 4x 2 9 2 49 6 8 2 1 2 (x 4 2 (x 4 4x 2 4)] dx 5x 2 2 2 ( x4 4) dx 1 5 1 2 5 5x 2 4) dx 1 53 x 3 2 x5 4 1 6 5 3 4 1 4x 15 x 5 2 0 32 5 2 40 3 2 53 x 3 4x 1 1 5 8 5 3 4 8 15. 11. Solve x 2 x 8 3 6 2 [x 2 1 0 9 2 x 2] dx 4) 0 1 2 2. Use the region’s symmetry: 4 2 [(x 4 8 3 1 and x 2 2: x 2 4, so the curves intersect at 2. 2 [2 2 (x 2 2)] dx x 2) dx (4 2 33 a, a by [ a 2, a 2] 22 2 4x 1 32 x 3 2 8 x2 3: x 2 12. Solve 2x 8 3 8 3 8 2x 3 (x 32 3 3)(x 10 1) 2 3 The curves intersect at x 0, 3 (2x x 2 3) dx x 1 and x 13 x 3 2 1 (9 9 32 3 10 9) 2 3 2 x a2 3. 3x 1 x 2 dx 2 0 3 1 20 1 3 3 a. Use the region’s symmetry: a so the curves intersect at x 0 and x 23 a 3 12 (a 3 x 2)3/2 a 0 13 a 3 296 Section 7.2 18. Solve y 2 16. 2: y2 y y curves intersect at y 2 (y [ 2, 12] by [0, 3.5] 12 y 2 1 The curves intersect at three points: x 1, x 4 and x 9. Because of the absolute value sign, break the integral up at x 0 also: 0 x 4 6 5 1 x x dx 9 x x 6 12 x 2 2 ( x)3/2 3 5 6x 5 1 y 2 3/2 x 3 20 5 6x y 4 4 5 18 189 10 16 15 13 30 32 5 16 3 1 6 5 3 1 16 3 (y 2 1 3 4. y 4 4) 5)( y 0 y2 4 y3 12 32 5 y2 4 4 4 2 3 y 4 y2 4 0. 4: The curves intersect at y 5 11 10 y 4 1 1 2 1 2 1 and x 1 2 13 y 3 5 0, 4 and y 5. 0 4 0 y2 4 0, so the 4 9 12 x 2 2 3/2 x 3 y2 12 x 2 y2 4 4 1) 2. 8 3 4 9 2 19. Solve for x: x 2)(y 2y 2 Now solve 0 6x x dx dx 5 4 6 5 0 (y 1 and y y 2) dy 2 2 y 4 2 3 5 5y 4 25 8 20. Solve for x: x dy 5 dy y2 8 125 12 1 16 3 25 y 2 and x 2 243 8 20 30 3 8 2y 2. Now solve 3 17. y2 2y 2: y 2 3 1, so the curves intersect at y 1. Use the region’s symmetry: 1 [ 5, 5] by [ 1, 14] 2 0 2y 2 (3 1 y 2) dy 2 The curves intersect at x 0 and x 4. Because of the absolute value sign, break the integral up at x 2 also (where x 2 4 turns the corner). Use the graph’s symmetry: 2 2 0 x2 2 2 2 0 2 4 (4 4 2 2 4 2 3x dx 2 x3 2 20 2[4] x 2) dx 2 2 2 2 x3 6 32 3 x2 2 4 (x 2 3y 2) dy 0 (1 y 2) dy 13 y 3 6y 4) dx 1 0 1 3 61 0 4 8 dx 4 y 2 and x 21. Solve for x: x 8x 2 32 (3 1 6 2 x 2 0 4 3 16 64 3 21 1 3 y2 Now solve at y 0 3y 2 (2 1 4 3y 2: y 2 2 1, so the curves intersect 1. Use the region’s symmetry: 1 2 3y 2. 2 0 (1 y 2) dy y 2) dy 4y 1 2 0 2y 2) dy (2 13 y 3 1 41 0 1 3 0 8 3 Section 7.2 22. Solve for y: y 4x 2 Now solve 4 x4 4x 2 4x 2 and y 4 x4 (x 2 5 x4 297 27. 1. 1: 1)(x 2 5) 0. [ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x = 1. The curves intersect at x 0 and x 1. Use the area’s Use the region’s symmetry: symmetry: 1 2 0 4x 2) [(4 1)] dx 1 0 ( x4 2 4 3 6 4 2 y and x y2 2 :y 4 2 0 2 2. 0 2 3 0 y 2 /4 2 y2 dy 4 2 y3 4 2(6 (2 sin x 0.273 [sec2 x (sec2 x 1)] dx /4 dx 2x 2 0 0 (tan2 y tan2 y) dy /4 4 /4 0 0 8 41 1 cos 2x 2 2 cos x 1 2 2 4 /2 3 sin y 30. cos y dy 2 (8 cos x 0 sec x) dx 2 8 sin x tan x 0.858 2 (cos y)3/2 3 3 2 3 30 /3 y 3 and x 31. Solve for x: x 0 4 4 0 1 2 0 2 y 0 25. Use the region’s symmetry: /3 tan2 y dy 0 4 tan y 0 2 tan2 x) dx 2 2) sin 2x) dx 2 3y 2 dy 4 3 0 2 3y 24. 0 29. Use the region’s symmetry: Use the region’s symmetry: 2 (sec2 x /4 4, so the curves intersect at y 0 /4 y2 . 4 2 y2 Now solve 3 1 28. Use the region’s symmetry, and simplify before integrating: 14 15 3 12 x 2 2 0 0 x 2 cos 1 2 2 5x 5 23. Solve for x: x 2 2 1 /4 104 15 x dx 5) dx 43 x 3 1 5 2 0 4x 2 15 x 5 x 2 sin 2 1 2 (x 4 /2 0 2 y. 0 2[(4 3 3) 0] 6 3 26. [ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x [ 1.1, 1.1] by [ 0.1, 1.1] 1 The curves intersect at x cross at x 1 2 cos 2x 13 x 3 2 21 1 3 1 1, but they do not 0. x2 0 0 and x 2 x 2 sin 0 dx x 2 1 0 4 3 4 0.0601 symmetry: 2 ( y 0 y 3) dy 0 and x 1 2 2 y2 1. Use the area’s 14 y 4 1 0 1 2 298 Section 7.2 32. (b) The two areas in Quadrant I, where x c y, are equal: 4 y dy y dy 0 c 4 2 3/2 c 2 3/2 y y 3 3 0 c 2 3/2 2 3/2 2 3/2 c 4 c 3 3 3 [ 0.5, 2.5] by [ 0.5, 1.5] y 1 intersect at x x2 x and y 1 2 x dx 0 1 12 x 2 1 dx x2 1. Integrate in two parts: 1 1 x 0 1 2 1 2 8 c 3/2 1 ( 1) 33. The curves intersect when sin x 2c 3/2 2 1 4 42/3 c cos x, i.e., at x 4 . 24/3 (c) Divide the upper right section into a (4 c)-by- c /4 /4 (cos x sin x) dx sin x rectangle and a leftover portion: cos x 0 0 2 1 0.414 c 2 x 2) dx (c 34. (4 c) c c 13 x 3 cx c c [ 3, 3] by [ 2, 4] c c3/2 4c 2 2 0 1) dx 2 2 3/2 c 3 0 13 x 3 2 4x 8 3 (b) Solve y x 2 for x: x 3 y-intercepts are 3 y dy 16 3 4 c3/2 c 4 2 1 20 35. (a) 0 3 32 3 y. The c 42/3 5 y=4 2 (3 3 36. 3 y)3/2 1 16 3 32 3 y = x2 (2, 4) (–2, 4) [ 1, 5] by [ 1, 3] The key intersection points are at x 0, x 1 and x Integrate in two parts: 1 1 (–√c, c) 0 (√ c , c ) 3 ( x2 c, then x c, c) and ( c, c). x dx 4 x x If y 1 3/2 c 3 24/3 y y=c c 1 and 3. 3 2 1 3/2 c 3 4c 0 8 3 28 c 16 3 c3/2 2 4 4 3/2 c 3 x 2) dx (4 2 c 2. 2 x2 (3 4x 0 8 Use the region’s symmetry: 13 x 3 3/2 4 1 3/2 c 3 3/2 (a) The curves intersect at x x 2) dx (4 0 1 2 3 x c. So the points are x2 8 2 3/2 x 3 1 8 4 1 2 x dx 4 x 1 4 x2 8 x 0 (8 2) 4 4 1 1 8 11 3 4. 299 Section 7.2 37. First find the two areas. 43. First graph y 1 2 For the triangle, (2a)(a 2) a For the parabola, 2 (a 2 The ratio, then, is 2 a 2x 13 x 3 a 0 43 a 3 3 , which remains 4 a 43 a 3 a3 x 2) dx 0 3 x 2. cos x and y [ 1.5, 1.5] by [ 0.5, 1.5] constant as a The curves intersect at x 0.8241 approaches zero. 2 0.8241. Use NINT to find x 2) dx (cos x 1.0948. Multiplying both 0 b 38. functions by k will not change the x-value of any b [2f (x) f (x)] dx a f (x) dx, which we already know a intersection point, so the area condition to be met is equals 4. 0.8241 2 2 kx 2) dx (k cos x 0 39. Neither; both integrals come out as zero because the 1-to-0 and 0-to-1 portions of the integrals cancel each other. ⇒2 ⇒2 k(1.0948) 40. Sometimes true, namely when dA [ f (x) g(x)] dx is always nonnegative. This happens when f (x) g(x) over the entire interval. ⇒k 1.8269. 0.8241 k2 x 2) dx (cos x 0 44. (a) Solve for y: 41. x2 a2 y2 The curves intersect at x symmetry: 2 0 2x x2 1 x 3 b2 1 y [ 1.5, 1.5] by [ 1.5, 1.5] 1 y2 b2 dx 0 and x ln 4 14 x 4 1 1 2 2 ln 2 1 2 x2 a2 1. Use the area’s a 2 ln x x2 a2 b1 b1 1 x2 dx or 4 a2 a a 2 0 0.886 x2 a2 b1 (b) 2 1 a dx or a 0 x2 dx a2 b1 (c) Answers may vary. a (d, e) 2 42. x2 a2 b1 b1 a x2 dx a2 2b x 2 1 x2 a2 a 2b sin 1 (1) 2 a a x sin 1 2 a a a sin 1 ( 1) 2 ab [ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x 0.9286 to find 2 (sin x 0 and x x 3) dx 0.9286. Use NINT 0.4303. 45. By hypothesis, f (x) g(x) is the same for each region, where f (x) and g(x) represent the upper and lower edges. 0 b But then Area [f (x) a each. g(x)] dx will be the same for 300 Section 7.3 Quick Review 7.3 1 : 2 46. The curves are shown for m 1. x 2 x 2. s x2 . 2 s2 , so Area 2 3. 12 x2 r or 2 2 4. 1 2 [ 1.5, 1.5] by [ 1, 1] In general, the intersection points are where which is where x x x2 mx, 1 (1/m) 1 x x2 0 1) 1 m 1 m m 1 1 0 2x m ln (m) 2x 1. s Section 7.3 Volumes x (pp. 383–394) b Exploration 1 x and h 1 bh 2 12 x 2 (2x)2 xk2. 3xk Area 2. Unrolling the cylinder, the circumference becomes one dimension of a rectangle, and the height becomes the other. The thickness x is the third dimension of a slab with dimensions 2 (xk 1) by 3xk xk2 by x. The volume is obtained by multiplying the dimensions together. 3 3. The limit is the definite integral 2 (x 1)(3x x 2) dx. 15 x, so 2 15 2 x. 4 Volume by Cylindrical Shells 1. Its height is f (xk) x2 . 4 8. 1 ln 1 bh 2 , so Area 32 x. 4 x2 . 2 1 bh 2 x, so Area x 1 bh 2 x, so Area 2 (1/m) 1 12 mx 2 h 2 1. Then, mx dx 1 1 2 ln (x 2 2 h 3 x and h 7. b 0 or else x 5. b 6. b 1 m because of symmetry, the area is 2 d2 x2 or 2 8 9. This is a 3-4-5 right triangle. b Area 1 bh 2 4x, h 3x, and 2 6x . 10. The hexagon contains six equilateral triangles with sides of 0 4. length x, so from Exercise 5, Area 45 2 Exploration 2 Surface Area b 1. 2y 1 (a) A r 2, where r [a, b]. (b) A sin x, so dy dx b a 2 sin x x, so (d) A cos2 x dx 1 dy dx 1 2x 12 4 2 0 s 2, where s w2 2 w , so A(x) 2 w2 w, so A(x) w , so A(x) 2 14.424. 0 3. y s 2, where s (c) A cos x and dy 2 dx dx 2y 1 x1 2 x 3 32 x. 2 1. In each case, the width of the cross section is w 2 1 x 2. The limit will exist if f and f are continuous on the interval 2. y 32 x 4 Section 7.3 Exercises dy 2 dx dx a 6 and dx 36.177. A(x) 4(1 w2 (1 x 2). 2(1 2 32 w (see Quick Review Exercise 5), so 4 3 4 (2 1 x 2)2 3(1 x 2). x 2). x 2). Section 7.3 2. In each case, the width of the cross section is w w , so A(x) 2 r 2, where r (a) A (b) A s 2, where s (c) A s 2, where s w, so A(x) w 2 x. w2 x. 2 w 2 8. A cross section has width w 4x. /3 w2 , so A(x) 2 V 2x. /3 4 2 /3 32 w (see Quick Review Exercise 5), so (d) A 3 4 x)2 (2 4 4 x2 2x dx 0 4 x and area 2 2 2 w s2 (sec2 x tan x 2 sec x tan x r x 2) (2 1 x ) dx (x 2x 2 2 2 2x x ) . The volume is 2 23 x 3 w2 21 1 4 x 2) dx (1 1 1 2 x 2 and area 4x 2 x ) dx 1 2 2 (1 1 1 1 16 . 3 x 2 and area x ) dx 2x 1 7. A cross section has width w 32 w (a) A(x) 13 x 3 13 x 3 1 1 8 . 3 2 sin x. 54 y dy 4 y5 4 8. 0 10. A cross section has width w 12 s 21 12 w 2 1 y 2 and area y 2). The volume is 2(1 y 2) dy 2 2y 13 y 3 1 1 8 . 3 11. (a) The volume is the same as if the square had moved without twisting: V Ah s 2h. (b) Still s 2h: the lateral distribution of the square cross sections doesn’t affect the volume. That’s Cavalieri’s Volume Theorem. from y 3 sin x dx 12 2 6 to y 12, for a diameter of 6 and a radius of 3, the solid has the same cross sections as the right 0 3 54 y . The volume is 42 2 12. Since the diameter of the circular base of the solid extends 3 sin x, and 4 V w2 2(1 x 2). The volume is 2(1 2 5y 2 and area 9. A cross section has width w 1 2 w 21 2 . 3 x 1 6. A cross section has width w tan x)2 dx, which by same method as (sec x x 2). The volume is 4(1 x 2) dx 2(1 6 tan x)2, and (sec x in part (a) equals 4 3 1 0 1 2 /3 r2 5. A cross section has width w w2 /3 V 1) dx 16 . 15 s2 3 6 . 6 s2 22 (1 4 15 x 5 x2 (b) A(x) w2 2 2 1 A(x) /3 2 1 4(1 /3 1 x 2 sec x 3 2 16. 22 s2 x /3 2 and area A(x) 1 tan x 2x. The volume is 4. A cross section has width w A(x) tan2 x) dx 2 sec x tan x /3 3 (1 tan x)2, and (sec x tan x)2 dx (sec x 0 1 4 tan x. /3 3x. 3. A cross section has width w A(x) 4 4 A(x) w2 2 r2 (a) A(x) sec x 301 sin x dx circular cone. The volumes are equal by Cavalieri’s 0 3 Theorem. cos x 0 2 s2 (b) A(x) V 13. The solid is a right circular cone of radius 1 and height 2. 3. w2 4 sin x dx 0 V 4 sin x, and 4 sin x dx 0 4 cos x 8. 1 Bh 3 1 ( r 2)h 3 1 ( 12)2 3 2 3 14. The solid is a right circular cone of radius 3 and height 2. 0 V 1 Bh 3 1 ( r 2)h 3 1 ( 32)2 3 6 302 Section 7.3 15. A cross section has radius r r2 A( y ) 1 tan2 4 0 tan2 y dy 4 4 tan 4 y and area 19. y . The volume is 1 tan 4 4 y y [ 6, 6] by [ 4, 4] 0 1 The solid is a sphere of radius r 4 . 43 r 3 16. A cross section has radius r sin x cos x and area A(x) r2 from x 3. The volume is 36 . 20. sin2 x cos2 x. The shaded region extends 0 to where sin x cos x drops back to 0, i.e., where [ 0.5, 1.5] by [ 0.5, 0.5] x 2 2 cos2 x . Now, since cos 2x 1, we know The parabola crosses the line y 1 cos 2x cos x and since cos 2x 1 2 sin2 x, we 2 /2 1 cos 2x know sin2 x . sin2 x cos2 x dx 0 2 /2 1 cos 2x 1 cos 2x dx 0 2 2 2 /2 4 (1 0 /2 4 8 1 0 x /2 cos2 2x) dx cos 4x dx 2 1 sin 4x 4 4 0 0 x(1 x) r2 (x 2 2x 3 x 4). 14 x 2 15 x 5 The volume is 1 cos 4x) dx 1. A cross x and area (x 2 2x 3 13 x 3 x 4) dx 0 (1 0 or x 2 x 2)2 (x 0 when 0, i.e., when x x A(x) /2 8 x section has radius r sin2 2x dx /2 0 x 2 1 0 30 . 21. 2 8 2 0 0 16 . 17. [ 1, 2] by [ 1, 2] Use cylindrical shells: A shell has radius y and height y. The volume is 1 2 ( y)( y) dy [ 2, 4] by [ 1, 5] 2 0 A cross section has radius r A(x) 2 r2 0 1 2 . 3 0 22. x4. The volume is 15 x 5 x 4 dx x 2 and area 13 y 3 2 0 32 . 5 [ 1, 3] by [ 1, 3] 18. Use washer cross sections: A washer has inner radius r outer radius R 1 3 x 2 dx The volume is [ 4, 6] by [ 1, 9] 0 A cross section has radius r A(x) 2 0 x 6 dx r2 x 3 and area x 6. The volume is 17 x 7 2 0 128 . 7 (R 2 2x, and area A(x) 3 13 x 3 r 2) 1 . 0 3 x 2. x, Section 7.3 23. 303 26. [ 2, 3] by [ 1, 6] [ 1, 5] by [ 3, 1] The curves intersect when x 2 1 x 3, which is when x 2 x 2 (x 2)(x 1) 0, i.e., when x 1 or x 2. Use washer cross sections: a washer has inner radius r x 2 1, outer radius R x 3, and area A(x) (R 2 r 2) [(x 3)2 (x2 1)2] ( x 4 x 2 6x 8). The volume is 2 ( x4 x2 6x The curves intersect where x 2, which is where x 4. Use washer cross sections: a washer has inner radius r x, outer radius R 2, and area A(x) (R 2 r 2) (4 x). 4 The volume is (4 x) dx 4x 0 12 x 2 4 8 0 27. 8) dx 1 15 x 5 2 13 x 3 32 5 8 3 3x 2 8x 1 12 1 5 16 1 3 3 8 [ 0.5, 1.5] by [ 0.5, 2] 117 . 5 The curves intersect at x 24. radius r 2 r2 A(x) The curves intersect when 4 x 2 x, which is when x2 x 2 (x 2)(x 1) = 0, i.e., when x 1 or x 2. Use washer cross sections: a washer has inner radius r 2 x, outer radius R 4 x 2, and area A(x) (R 2 r 2) [(4 x 2)2 (2 x)2] (12 4x 9x 2 x 4). 4x [ 1, 3] by [ 1, 3] The curve and horizontal line intersect at x 9x 2 section has radius 2 x 4) dx 1 r2 A(x) 12x 24 2x 2 8 2.301. 28. The volume is (12 sec x tan x)2. Use NINT to find 0 2 2 2 sec x tan x)2 dx (2 [ 2, 3] by [ 1, 5] sec x tan x and area ( 0.7854 0.7854. A cross section has 15 x 5 3x 3 24 sin x)2 4 (1 4 (1 2 sin x The volume is 32 5 /2 12 2 3 1 5 2 sin x sin2 x) dx 2 cos x 4 (1 108 . 5 1 sin 2x 4 0 4 25. 4 3 x 2 3 4 2 (3 /2 0 8) 29. 3 , 3 by [ 0.5, 2] Use washer cross sections: a washer has inner radius r sec x, outer radius R 2, and area A(x) (R 2 r 2) (2 sec2 x). The volume is /4 (2 sec2 x) dx [ 1, 3] by [ 1.5, 1.5] A cross section has radius r /4 2x A(y) tan x /4 2 2 2. 2 The volume is 1 5y 2 and area 5 y 4. 1 /4 1 r2 5 y 4 dy 1 . A cross 2 sin x and area 2 1 2 y5 1 2. 1 sin2 x). 304 Section 7.3 30. 35. [ 1, 4] by [ 1, 3] [ 1, 5] by [ 1, 3] A cross section has radius r r2 A(y) 2 The curved and horizontal line intersect at (4, 2). y 3/2 and area (a) Use washer cross sections: a washer has inner radius y 3. The volume is 14 y 4 y 3 dy 0 2 r 4. 0 x, outer radius R A(x) 31. (R 2 2, and area 2 r) (4 4 (4 0 x) dx x). The volume is 12 x 2 4x 4 8 0 y 2 and area (b) A cross section has radius r [ 1.2, 3.5] by [ 1, 2.1] r2 A(y) y 4. 2 Use washer cross sections. A washer has inner radius r outer radius R y (R 2 A(y) 1 ( y2 0 2 32 . 5 0 1, and area r 2) volume is 15 y 5 y 4 dy The volume is 1, 1)2 [(y 13 y 3 2y) dy 0 (c) A cross section has radius r (y2 1] y2 1 0 2 x and area 2y). The r2 A(x) 4 . 3 x)2 (2 (4 4 x x). The volume is 32. 4 (4 4 x x) dx 8 3/2 x 3 4x 0 12 x 2 4 0 8 . 3 (d) Use washer cross sections: a washer has inner radius [ 1.7, 3] by [ 1, 2.1] r Use cylindrical shells: a shell has radius x and height x. The 1 volume is 2 (x)(x) dx 2 0 13 x 3 1 0 y 2, outer radius R 4 (R 2 A( y ) 2 . 3 (8y 2 r 2) 4, and area [16 83 y 3 15 y 5 y 2)2] (4 y4). 33. The volume is 2 0 (8y 2 y 4) dy 2 224 15 0 36. [ 2, 4] by [ 1, 5] Use cylindrical shells: A shell has radius x and height x 2. 2 2 (x)(x 2) dx The volume is 2 0 14 x 4 2 8. 0 [ 1, 3] by [ 1, 3] 34. The slanted and vertical lines intersect at (1, 2) (a) The solid is a right circular cone of radius 1 and height 2. The volume is [ 0.5, 1.5] by [ 0.5, 1.5] 1 Bh 3 The curves intersect at x 0 and x shells: a shell has radius x and height 1 is 2 (x)( 0 x x) dx 2 2 5/2 x 5 1. Use cylindrical x 13 x 3 x. The volume 1 0 2 . 15 1 ( r 2)h 3 1 ( 12)2 3 2 . 3 (b) Use cylindrical shells: a shell has radius 2 height 2x. The volume is 1 1 2 (2 x)(2x) dx 4 0 (2x x 2) dx 0 4 x2 13 x 3 1 0 8 . 3 x and 305 Section 7.3 40. 37. [ 2, 2] by [ 1, 2] [ 2, 2] by [ 1, 3] The curves intersect at ( 1, 1). x2 (a) A cross section has radius r r2 A(x) 1 x and area (1 x 2)2 (1 2x 2 2 x 4). 2x 2 (1 1. A shell has radius x and height x 2. The volume is x 2 (x)(2 x 2) dx x 13 x 3 x2 2 0 x 4) dx 23 x 3 x 1 1 5 . 6 0 16 . 15 1 y and [ 1, 5] by [ 1, 3] height 2 y. The volume is 1 1 y)(2 y) dy 14 x 4 41. 1 15 x 5 (b) Use cylindrical shells: a shell has radius 2 2 (2 x at x 1 The volume is 1 2 2 4 (2 0 y y 3/2) dy 0 4 3/2 y 3 4 2 5/2 y 5 1 A shell has radius x and height 4 56 . 15 0 2 (x)( x) dx 2 5/2 x 5 2 0 x. The volume is 4 128 . 5 0 42. (c) Use cylindrical shells: a shell has radius y 1 and height 2 y. The volume is 1 1 2 (y 1)(2 y) dy ( y3/2 4 0 2 5/2 y 5 4 38. (a) A cross section has radius r r2 A(x) b y) dy 0 2 h1 0 h2 1 x2 dx b 2 3/2 y 3 1 0 [ 2, 2] by [ 2, 2] 64 . 15 x and area b h1 x2 . The volume is b b b x3 h2 1 bh 2. 3 3 b0 The functions intersect where 2x x . The volume is b b x 2 (x)h 1 dx 2 h b 0 x (2x 1) x 2x x2 dx b b x3 b 2h. 3 3b 0 x 0 1 2 2 h x2 1. The volume is 1 1 2 (x)( x 2x 1) dx (x 3/2 2 0 2x 2 x) dx 23 x 3 12 x 2 0 2 5/2 x 5 2 7 . 15 43. A shell has height 12( y 2 b x, i.e., at x A shell has radius x and height (b) Use cylindrical shells: a shell has radius x and height h1 1 (a) A shell has radius y. The volume is 1 2 ( y)12( y 2 y 3) dy 1 ( y3 24 0 y 4) dy 0 24 (b) A shell has radius 1 1 y)12( y 2 2 (1 0 1 ( y4 24 14 y 4 15 y 5 24 2 3 2 2 (x) x dx The volume is 0 2y 3 y. The volume is y 3) dy y 2) dy 0 [ 2, 3] by [ 2, 3] x 2 x3 3 x. 2 2 8. 0 0 y 3). 39. A shell has radius x and height x 1 15 y 5 14 y 2 13 y 3 1 0 4 . 5 1 0 6 . 5 1. 306 Section 7.3 43. continued 45. (c) A shell has radius 1 2 0 8 5 8 5 y. The volume is y 12( y 2 1 24 y4 0 15 y 5 24 y 3) dy 2 y 0 1 24 0 24 The functions intersect at (2, 8). (a) Use washer cross sections: a washer has inner radius A(x) 2 12( y 2 y 3) dy 5 22 33 y4 y y dx 5 5 1 15 34 23 y y y 2. 5 20 15 0 y2 44. A shell has height 2 y2 y4 y2 . 2 4 2 2 ( y) y 2 0 (b) A shell has radius 2 2 2 (2 2 0 2 16 y 24 16 y 24 y4 y 3 2y 2 dy 2 14 23 15 y y y 4 3 10 2 0 2 0 2 y5 4 16 y 24 2 y 2 2 0 2 5 8 y5 4 16 y 24 512 . 21 0 y and 8 . 3 46. The functions intersect at (0, 0) and (1, 1). 2 8 . 5 0 y. The volume is 5y 4 4 15 y 4 y3 (a) Use cylindrical shells: a shell has radius x and height x2 1 2 (x)(x x x 2. The volume is x x 2) dx 2 0 53 y 3 13 x 3 y2 height 2x 2 8. x 2 1 2 (1 0 5 . The volume is 8 5y 4 32 x)(x x x 15 y 32 14 y 4 5y 2 dy 8 53 y 24 0 6 . x 2) dx 2 4. 0 x and x . The volume is 1 (x 3 2 2x 2 x) dx 23 x 3 12 x 2 0 6 dy y3 1 2 14 x 4 2 4 y 4 14 x 4 (b) Use cylindrical shells: a shell has radius 1 5y 2 dy 14 y 4 (d) A shell has radius y 0 2 y . The volume is 4 8 y 2 (8 y) y 1/3 dy 4 0 8 y2 2 8y 1/3 2y y 4/3 dy 4 0 8 3 7/3 832 13 2 6y 4/3 y 2 y y . 7 21 12 0 0 2 17 x 7 [ 0.5, 1.5] by [ 0.5, 1.5] y4 dy 4 y) y 2 0 2 16 3 x 3 height y 1/3 2x 2 (5 x 6). The volume is dy (c) A shell has radius 5 2 (16x 2 y. The volume is y 4 y) y 2 y5 4 r) 4 0 2 14 y 4 2 (16x 2 2 (b) Use cylindrical shells: a shell has a radius 8 y4 4 y4 dy 4 4x, and area x 6) dx (R 2 0 (a) A shell has radius y. The volume is 2 x 3, outer radius R r 2 . The volume is 5 (d) A shell has radius y 1 [ 1, 3] by [ 1.4, 9.1] 13 3 82 y y dy 5 5 1 13 4 83 y y 2. 20 15 0 . 1 0 307 Section 7.3 47. (a) Solve d dc 2 22 c 4c 2 2 0 2 c 4 0 [ 0.5, 2.5] by [ 0.5, 2.5] 1 1 , 1 , , 2 , and (1, 1). 4 4 The intersection points are and area (R 2 1 y4 1 1 y4 1 3y 3 2 r) 1 dy 16 1 (b) A shell has radius x and height 1. The volume is 2 0 2 c 1 1 , outer radius R , 4 y2 1 . The volume is 16 2 11 1 y . 48 16 1 (a) A washer has inner radius r 2 c This value of c gives a minimum for V because a 2V 2 dc 2 2 0. 2 22 Then the volume is 1 2 (x) 1/4 1 dx 2 3/2 x 3 2 x 48. (a) For 0 x x(sin x) x , x f (x) For x 0, x f (x) 01 x f (x) sin x for 0 1 1/4 2 anywhere else besides c 11 . 48 2 4 , the maximum must occur 2 at c sin x. sin 0 x 12 x 2 2 2 (b) Since the derivative with respect to c is not zero x 1 2 2 4 0 or c and for c sin x. So 1. The volume for c (3 1 it is 8) 0 is 2.238. c 2 2 4.935, 0 maximizes the volume. . (c) (b) Use cylindrical shells: a shell has radius x and height y. The volume is 2 xy dx, which from part (a) is 0 2 sin x dx 2 cos x 0 49. (a) A cross section has radius r r2 A(x) 6 144 0 (b) [0, 2] by [0, 6] 4. The volume gets large without limit. This makes sense, since the curve is sweeping out space in an ever-increasing radius. 0 144 (36x 2 (36x 2 x 4) dx x 12 36 x 2 and area x 4). The volume is 144 36 cm3 (8.5 g/cm3) 5 15 x 5 12x 3 6 0 51. (a) Using d 36 cm3. 5 C d2 2 , and A C2 yields the 4 following areas (in square inches, rounded to the nearest tenth): 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 192.3 g 10.7, 9.3, 6.4, 3.2. 50. A cross section has radius r c sin x and area (b) If C( y) is the circumference as a function of y, then the r2 A(x) sin x)2 (c (c 2 2c sin x sin2 x). area of a cross section is The volume is C(y)/ 2 A(y) (c2 sin2 x) dx 2c sin x and the volume is 0 1 x 2 c2x 2c cos x 2 1 2 c 22 c 2c 1 sin 2x 4 2c 0 (c) 1 4 6 2 . 0 5.12 11.62 C 2( y) , 4 6 1 4 C 2( y) dy. 0 6 A( y) dy 2 4c 2 1 C 2( y) dy 40 16 0 [5.42 24 4 6.32 10.82 7.82 9.42 9.02) 2(4.52 10.82 6.32] 4.42 11.62 34.7 in.3 308 Section 7.3 52. (a) A cross section has radius r 5 r2 2 y. The volume is 2 y dy 0 dV dt dh so dt For h dh so dt 1 8 0 and x a. For revolution about the x-axis, a cross section has radius x 2 and area ax A(h). r2 A(x) dh , dt x 2)2 (ax (a 2x 2 2ax 3 x 4). The volume is a 8, units3 3 3 = sec 8 53. (a) 25 . 0 4, the area is 2 (4) 0: This is true at x 5 r dV A(h) dh, so dh dV dh A(h) dh dt dV 1 A(h) dt (b) V(h) y2 x2 55. Solve ax 2y and area (a 2x 2 2ax 3 123 ax 3 x 4) dx 0 units3 . sec 14 ax 2 15 x 5 a 0 1 a 5. 30 y For revolution about the y-axis, a cylindrical shell has 2 r x 2. The volume is radius x and height ax x a x 2) dx 2 (x)(ax 13 ax 3 2 0 The remaining solid is that swept out by the shaded region in revolution. Use cylindrical shells: a shell has r2 radius x and height 2 . x 2. The volume is 1 a5 30 14 1 a yields a 6 30 2 (x)(2 r 2 2 x 2) dx 22 1 , so a 6 5. 56. The slant height s of a tiny horizontal slice can be x2 written as s r2 0 14 a. 6 Setting the two volumes equal, –2 .r a 14 x 4 y2 ( g ( y))2 y. So the 1 surface area is approximated by the Riemann sum n 22 (r 3 4 ( 8) 3 Σ2 k1 r 2 3/2 x) r2 4 32 . 3 integral. dx dy 57. g ( y) (b) The answer is independent of r. (g ( y))2 y. The limit of that is the g( yk) 1 1 2 2 y1 0 , and 2y 1 2 2 dy 4y and measure the radius r (x) of the shadow region at these 6 points. Then use an approximation such as the trapezoidal (4y 13 3 b rule to estimate the integral a r 2(x) dx. dx dy 58. g ( y) 1 2 0 y3 3 3 2 1 dx dy y1/2 1 3 2 1 1)3/2 2 0 13.614 y 2, and 2 3 ( y 2)2 dy 9 59. g ( y) 1 dy 0 2y 54. Partition the appropriate interval in the axis of revolution y1/2 1 (1 6 (2 2 y 4)3/2 0 1) 1 1/2 y , and 2 1 3/2 3 1 3/2 3 1 1/2 2 y dy 2 1 1 1 dy. 4y Using NINT, this evaluates to 1 16.110 0.638. Section 7.3 309 66. Use washer cross sections: a washer has inner radius dx dy 2y 2y 11 60. g ( y) 1 , and 1 2 r 1 5/8 area (R2 2 1 dy 2y 1 2y dy 2 4 2 3/2 y 3 2 2 dy dx 61. f (x) 2 2 x2 a 1 a2 y 2)2 5/8 y 2)2] 4 b a2 a2 a y 2 dy 4b y 2 dy a a2 2 2.997. 22 2 ab 2x, and 67. (a) Put the bottom of the bowl at (0, 2 (2x)2 dx 1 2 x2 0 4x2 dx evaluates, 1 using NINT, to (a2 y 2) dy 13 y 3 a2y a 2 2 (3x x) 1 2 (3 2x) dx evaluates, using NINT, (52 0 to dh dt 44.877. dy dx 63. f (x) 1 2 x x2 2x 1.5 1 0.5 dx x2 2 1 dx 0.5 0.5 1 2 2 5 3/2 4 2 x 3 2 4 3 a 6.283 25 3/2 x Since A1 (a2 x 2) dx 49 3 x 2 and area ( a2 13 x 3 a2x x2 r2 1 1 x 2). The a a 13 a 3 a3 13 a 3 h 2 volume of cone y and area h r1 y2 h r2 1 2y h y2 . The h2 volume is 51.313 22 h) A1. A2 23 R. 3 a2 x 2)2 a A( y ) theorem. Thus, volume of hemisphere R3 a2 5 volume of cylinder m/sec. 43 a. 3 A2, the two volumes are equal by Cavalieri’s 13 R 3 ( dx Right circular cylinder with cone removed cross sectional h2 r2 1 65. Hemisphere cross sectional area: ( R area: R2 1 120 (b) A cross section has radius x 9 3/2 4 4 1 (0.2) 24 2 5 dx 4 x 1 1 dV A dt a3 1 11 5 . 24 . The rate of rise is 1 1 2 a h) 1 and the area of a cross section is , and x 5 x h2(3a 3 x 2 2 ha volume is 1.5 2 dy dx 12) A(x) 1.5 2 x 2x 64. f (x) 4, y 68. (a) A cross section has radius r , and x2 1 2x y 2). 2x, and (b) For h 3 ( a2 The volume for height h is 53.226. 3 y 2)2 horizontal cross section is ( a ha dy dx a). The area of a 2 0 62. f (x) y 2, and y 2. The volume is 4b 5 2 a2 b a2 (b a 5 16 1 3 a2 4b 5/8 y 2, outer radius R r 2) [(b 1 2 a2 b 0 r2 1 2y h y2 dy h2 r2 y 12 r h. 3 y2 h y3 3h2 h 0 310 Section 7.4 s Section 7.4 Lengths of Curves 3. (a) x (pp. 395–401) cos2 y dy. 1 cos y, so Length 0 (b) Quick Review 7.4 1. 1 1 x 1 2. x)2, which, since x (1 x2 4 1, equals x2 , which, since x 2 1 1 (tan x)2 equals sec x. [ 1, 2] by [ 1, 4] 2, equals (c) Length 2x x or . 2 2 1 3. x2 1. 2x x or x 4. (a) x (sec x)2, which, since 0 x 2 3.820 y 2) y(1 , 1/2 , so 1/2 Length y2 1 1/2 4. 12 x x 4 1 since x 5. 1 1 2 12 x 16 x2 0, equals 4 4x (x 2 1 4 1 x2 which, x 2 2 10 3 3 [ 1, 2] by [ 1, 1] (c) Length 6. f (x) has a corner at x d 7. (5x 2/3) dx (b) , 2 cos x. equals 8. d ( dx is undefined at x 2y 2x 2 2y 1 y 0. f (x) has a cusp x2 1 is undefined for x 5(x 3)4/5 10. d 1 dx 4x 3 1)2 (y 4 x 2 Length 3. 2x 2, and 1 1. Then y 2x 1 1 2x 1 2 , and 2 dx. (b) 2 has a corner at x 2. cos x is undefined for x 3(sin x)2/3 sin x 2x 7 f (x) has a vertical tangent there. 9. 1, so x y x 3) 1.047 5. (a) y 2 4. there. 5 dy. . 2 cos2 x, which, since cos 2x 4)2 x2 y2 1 k, [ 1, 7] by [ 2, 4] where k is any integer. f (x) has vertical tangents at these (c) Length 9.294 values of x. 6. (a) y Section 7.4 Exercises cos x x sin x Length cos x x sin x, so x 2 sin2 x dx. 1 0 1. (a) y 2x, so 2 Length 1 (b) 2 (2x)2 dx 4x 2 dx. 1 1 1 (b) [0, by [ 1, 4] (c) Length 7. (a) y [ 1, 2] by [ 1, 5] (c) Length 2. (a) y 6.126 2 4.698 /6 0 (b) y 0 sec x, so Length 1 tan x dx ln ( sec x ) 4 sec x dx. /3 (b) 0, 3 , 0 by [ 3, 1] (c) Length 2.057 1 tan x, so Length 6 (c) Length by [ 0.1, 0.2] 0.549 tan2 x dx. Section 7.4 /4 sec2 y 8. (a) x 1, so Length sec y dy. /3 (e x 10. y e x) 2 3 , so Length ex 1 x2 e dx. 2 3 (b) sec2 y (b) x 1 tan y , ln (cos y) ln (cos y) y 3 so x 0 y 0 4 [ 3, 3] by [ 2, 12] (c) Length 12 (x 2 11. y 3 1 0 [ 2.4, 2.4] by (c) Length 2 , (x x 2 sec x tan x, so /3 1 2 sec x tan x dx. 3 2 3 2 1 0 , (c) Length 3 x 12. 4 dx 9x dx 4 9x 3/2 4 8 1 0 8 1 27 80 10 27 y2 by [ 1, 3] 3 1 3.139 2 1 4 0 . 3 1 , so the length is 4y 2 12 13 y2 dy y 3 4 y2 12 dy 4 y2 y2 1 1 1 4y 3 53 . 6 1 1 , so the length is 4y 3 y3 14. x 1 dx 0 2 x (b) 13. x 2x 2 x, so the length is /3 3 x4 3 13 x 3 1) dx 4 2 2, so the length is 3 0 (x 2 2 x x2 2)2 dx 0 2.198 Length 2)1/2(2x) 3 12. y 9. (a) y 20.036 2 12 dy 4 y3 y3 1 12 dy 4 y3 y3 1 14 y 4 1 8y 2 2 1 123 . 32 y2 2 15. x 1 , so the length is 2y 2 2 1 2 12 dy 2 y2 y2 2 1 1 13 y 6 x2 16. y 2x 1 1 2y 4 (x 4)2 (4x so the length is 2 1 (x 2 4(x 0 1 (x 3 1)3 2 4(x 1)2 2 1 1) dx 1)2 4(x 1 1)2 (x 2 1 1)2 0 0 12 dy 2y 2 y2 2 dx 53 . 6 2 1 1)2 17 . 12 1 4(x 1)2 311 312 Section 7.4 sec4 y 17. x /4 /4 (sec4 y 1 24. The area is 300 times the length of the arch. 1, so the length is 1) dy /4 y sec2 y dy /4 x , so the length is 50 2 x sin2 dx, which evaluates, using NINT, 2 50 sin 2 25 /4 tan y 1 2. 25 /4 to 3x 4 18. y 1 the cost (rounded to the nearest dollar): $38,422. 1 (3x 4 1 73.185. Multiply that by 300, then by $1.75 to obtain 1, so the length is 3x 2 dx 1) dx 2 2 1 1 3 3 x3 73 . 3 2 dy 2 dy 1 1 corresponds to here, so take as . Then dx dx 4x 2x 19. (a) y x C, and, since (1, 1) lies on the curve, C 0.2x y1 10 1 0. x. dx 2 dx 1 1 corresponds to 4 here, so take as 2 . dy dy y y 1 Then x C and, since (0, 1) lies on the curve, y 1 C 1. So y . 1x (b) Only one. We know the derivative of the function and the value of the function at one value of x. 21. y cos 2x, so the length is /4 5 22.3607, and . NINT fails to evaluate (y1 )2 dx because of the undefined slope at 22.36 (y1 )2 dx 1 2 0 52.548. Then, pretending the last little stretch at each end is a straight line, add 0.2(22.36)2 2 100 track 1 as 0.156 to get the total length of 52.704. Using a similar strategy, find the length of the right half of track 2 to be 2 cos x dx 0 2 sin x 32.274 + NINT Y2 /4 1. 32.274. Now 52.704 and 2 cos 2x dx 0 from the upper limit a little, and find enter Y1 /4 1 5 10 the limits, so use the track’s symmetry, and “back away” (b) Only one. We know the derivative of the function and the value of the function at one value of x. 20. (a) 0.2x 2 100 5 10 So y 0 at x 25. For track 1: y1 2 0.2t 1 150 0.2t 2 , t, x, 0 and graph in a [ 30, 0] by [0, 60] window to see the effect of 0 the x-coordinate of the lane-2 starting position on the length 22. y x 2/3)1/2x (1 1/3 , so the length is of lane 2. (Be patient!) Solve graphically to find the 1 8 1 x 2/3)x (1 2/3 intersection at x dx 2/4 coordinates ( 19.909, 8.410). 1 8 2/3 x dx 24 1 2/3 x 3 26. f (x) 1 8 x 1/3 dx 2 1 2/4 1 3 2 31 22 x 0. So, instead solve for x in terms of y using the quadratic formula. (x1/3)2 6. 23. Find the length of the curve y 20 0 to (f (x))2 dx because of the undefined slope at 2/4 1 x1/3 y 2 1/3 x , but NINT fails on 3 0 3 2 8 x 2/3 8 19.909, which leads to starting point 3 sin x for 0 20 x 20. 3 3 cos x, so the length is 20 20 2 3 3 1 cos x dx, which evaluates, using NINT, 20 20 21.07 inches. 1 (1 8 3 (1 8 x x 2 1 2 1/3 4y 4y 2/3 2 y 0, and . Using the positive values, 1)3. Then, 2 1)2 1 1 0 4y x1/3 (x )2 dy , and 4y 3.6142. 313 Section 7.5 (4x 2 27. f (x) 1) (8x 2 (4x 2 1)2 1 1/2 using NINT, to partition goes to zero will always be the length (b 4x 2 8x 1 2 dx which evaluates, (4x 2 1)2 1 length is 31. Because the limit of the sum Σ xk as the norm of the 4x 2 8x 1 , so the (4x 1)2 8x) a) of the interval (a, b). 2.1089. 32. No; the curve can be indefinitely long. Consider, for 28. There is a corner at x 0: 1 3 example, the curve sin 1 + 0.5 for 0 x x 1. 33. (a) The fin is the hypotenuse of a right triangle with leg lengths xk and [ 2, 2] by [ 1, 5] Σ k1 (b) lim n→ x3 x3 ( xk)2 2 0 x x n 0 and 1 Σ k1 lim n→ xk 1 ( f (x))2 dx b 2 3x 3x 2 y 5 5 x x 1 The length is 0 . 1 (y )2 dy 2 0 (3x 2 1 34. Yes. Any curve of the form y 5)2 dx a 1 (3x 2 5)2 dx, x)2, 0 (1 x (y )2 dx 1 c, where c is a 1, so that a 2 dx 0 0 which evaluates, using NINT for each part, to 29. y x constant, has constant slope 1 2 ( f (xk 1))2 a 2 0 1 f (xk 1) xk. ( f (xk 1) xk)2 1 5x 5x xk 1 n Break the function into two smooth segments: y df dx x xk a 2. 0 13.132. s Section 7.5 Applications from Science and Statistics 1 (pp. 401–411) Quick Review 7.5 1 1. (a) [ 0.5, 2.5] by [ 0.5, 1.5] x y 1 e into two equal segments by solving 1 . The total length is 2 4 which evaluates, using NINT, to x y 1 1 2. (a) 12 x 1/4 ex 1 e 1 1.718 /2 /2 dx, 1 e 1 0 (b) 3. (a) sin x dx x (b) 2 cos x /4 1.623. 2 /4 0.707 3 30. e x dx 0 1 with 1 0.632 1 0. So, split the curve x 0 (b) , but NINT may fail using y over the entire interval because y is undefined at x x: x dx 0 x y x e 4. (a) (x 2 13 x 3 2) dx 0 3 2x 15 0 (b) 15 2 5. (a) 1 x2 x 3 1 [0, 16] by [0, 2] 1 3/4 x , but NINT may fail using y over the entire 4 y interval, because y is undefined at x 4 x y ,0 2 1 y 2: x 0. So, use (b) 3 4y and the length is 6. 0.501 2 (x 2) sin x dx 0 0 7 16.647. 1 ln (x 3 1) 3 1 [ln 9 ln 2] 3 1 9 ln 3 2 7 (4y 3)2 dy, which evaluates, using NINT, to dx 7. (1 0 x 2)(2 x) dx 2 1 314 Section 7.5 7 cos2 x dx 8. 7 y2 (10 2 9. 0 7 3 10. 4 0 ks, so 150 k 1 and k 16 1 ,F 8 1 8 300 lb. 10. (a) F 0 s y) dy 2400 1/8 sin2 x dx 1/8 1200x 2 2400x dx (b) 2400 lb/in. Then for 18.75 in.-lb 0 0 Section 7.5 Exercises 5 1. xe x/3 dx 5 x/3 3e (3 0 0 3 2. x dx 4 x sin 0 44 x 4 sin 4 22 11. When the end of the rope is x m from its starting point, the 24 e 5/3 x) 9 4.4670 J (50 x 4 x cos F(x) 0 3. 3 2 (0.624)(50 x 2 dx x9 (esin x cos x 4. 9J (b) p1V11.4 10 esin x 19 19.5804 J 109,350 F(x) 490 1 490 20 W (490 24.5x) dx 12.25x 2 490x 0 0.4 V 4900 J. 20 4x/5 20 x 25 490 1 490 W (490 19.6x) dx 490x 9.8x 0 k1 F(x) ∑ 124.8yk 144 x/2 x 36 144 1 18 W (144 4x) dx 144x 2x (b) ks, so 800 k(14 10) and k 2 (b) F(x) 200x, and 100x 2 200x dx (c) F 9. (a) F ks, so 21,714 400 in.-lb. n Σ (62.4yk) 6 k1 7238 lb/in. (b) 374.4y 1 0 1/2 (b) F(x) 7238x. W 3619x 2 7238x dx 0 1/2 0 1 904.75 905 in.-lb, and W 7238x dx 1/2 3619x 2 1 2714.25 1/2 1, it follows that a thin y2 y, where y is 64 force is approximately 8 5) and k y2 82 distance from the top, and pressure 62.4y. The total 8 in. k(8 3 2 0 1600 200 200s, so s x2 32 horizontal rectangle has area 6 1 200 lb/in. 0 y 2)3/2 0 14. (a) From the equation 0 8. (a) F 41.6(9 1123.2 lb 1944 ft-lb 0 9 yk2) y yk2 y. 0 4x lb. 18 2 y 2 dy 3 Then 18 124.8y 5880 J. 144 9 k1 0 18 y 2 y, where y is n 20 7. When the bag is x ft off the ground, the sand weighs 32, it follows that a thin force is approximately ∑ (62.4yk)(2 9 n 2 y2 distance from the top, and pressure 62.4y. The total 19.6x N. Then 20 243 horizontal rectangle has area 2 9 0 490 109,350 and V1.4 32 V 6. When the bucket is x m off the ground, the water weighs F(x) p dV (p , V ) 11 dV 2.5V 13. (a) From the equation x 2 20 (p , V ) 22 pA dx 37,968.75 in.-lb 24.5x N. Then 780 J. 0 109,350, so p 5. When the bucket is x m off the ground, the water weighs x 20 50 (p , V ) 11 (p , V ) 2 2 109,350 V1.4 (p , V ) 11 Work 0 esin 10 (p , V ) 22 F(x) dx (50)(243)1.4 2x 0 20 x 490 20 12 x 2 0.624 50x (p , V ) 11 0 2) dx x) dx (p , V ) 22 3 x 2)3/2 10 x) N. The total work is 0 12. (a) Work 1 (9 3 0 (0.624)(50 50 3.8473 J 3 x) m of rope still to go weigh 3 2714 in.-lb. 1 yk2 64 y2 dy 64 n y Σ 374.4yk k1 7987.2 1 7987.2 lb yk2 1 y 2 3/2 64 64 8 0 y. 315 Section 7.5 3 y, it follows that a thin 8 3 horizontal rectangle has area y y, where y is the 4 18. The work needed to raise a thin disk is (10)2(51.2)y y, 15. (a) From the equation x distance from the top of the triangle, the pressure is 62.4(y ∑ 62.4(yk ∑ 46.8(yk y 2 3yk) y. k1 46.8(y 22 3y) dy 3494.4 3 ( 210.6) 15.6y 3 70.2y 2 8 5120 0 12 y 2 30 7,238,229 ft-lb 0 4 (62.4)( y 12 y 2 6 15y 0 84,687.3 ft-lb x2 , it follows that a thin 2 for the whole tank. Work to pump over the rim is y). The total force is approximately (2)2(62.4)(6 15) y for a thin disk and 6 0 n 4 (62.4)(21) dy 4 (62.4)(21)(6) 98,801.8 ft-lb for the whole tank. Through a hose attached to a valve in the yk)(2 2yk) y n 2(4 249.6 0 distance from the bottom, and pressure 62.4(4 Σ 124.8 k1 15) dy 3705 lb horizontal rectangle has area 2 2y y, where y is Σ 62.4(4 k1 15) y for a thin disk and 6 3 16. (a) From the equation y bottom is faster, because it takes more time for a pump with yk3/2) y. yk a given power output to do more work. 4 (b) 100 (51.2)y dy 19. Work to pump through the valve is (2)2(62.4)( y n 3 3) yk 4 k1 (b) 30 3). The total force is approximately n 8 where y is height up from the bottom. The total work is 3/2 124.8 2(4 y y ) dy 124.8 2 y3/2 0 8 3 1064.96 2 2 5/2 y 5 4 20. The work is the same as if the straw were initially an inch 0 long and just touched the surface, and lengthened as the 1506.1 lb 17. (a) Work to raise a thin slice 62.4(10 20 Total work 12)( y)y. 62.4 60y 2 62.4(120)y dy 0 liquid level dropped. For a thin disk, the volume is 20 0 1,497,600 ft-lb (b) (1,497,600 ft-lb) 100 min (250 ft-lb/sec) 5990.4 sec 10 62.4 60y 2 7 0 (c) Work to empty half the tank 17.5 2 14 y 17.5 2 14 y 17.5 14 y to y and the work to raise it is 4 (8 y) y. The total work is 9 24 (8 y) dy, which using NINT evaluates 9 91.3244 in.-oz. 62.4(120)y dy 0 10 374,400 ft-lb, and 21. The work is that already calculated (to pump the oil to the 0 374,400 250 1497.6 sec 25 min rim) plus the work needed to raise the entire amount 3 ft higher. The latter comes to (d) The weight per ft3 of water is a simple multiplicative factor in the answers. So divide by 62.4 and multiply 12 r h (57)(3) 3 total is 22,921.06 57 (4)2(8) 30,561.41 22,921.06 ft-lb, and the 53,482.5 ft-lb. by the appropriate weight-density For 62.26: 62.26 1,494,240 ft-lb and 62.4 62.26 5990.4 5976.96 sec 100 min. 62.4 1,497,600 For 62.5: 62.5 1,500,000 ft-lb and 62.4 62.5 5990.4 6000 sec 100 min. 62.4 1,497,600 22. The weight density is a simple multiplicative factor: Divide by 57 and multiply by 64.5. 30,561.41 64.5 57 34,582.65 ft-lb. 316 Section 7.5 (d) 0 if we assume a continuous distribution. Between 23. The work to raise a thin disk is r 2(56)h ( 102 56 (12 y 2)2(56)(10 2 59.5 in. and 60.5 in., the proportion is y) y 60.5 y 2) y. The total work is y)(100 f (x) dx 56 (12 y 2) dy, which evaluates using NINT y)(100 0 to 0.071 (7.1%) 59.5 10 1 28. Use f (x) (x 498)2/(2 1002) e 100 2 967,611 ft-lb. This will come to 500 (967,611)($0.005) (a) $4838, so yes, there’s enough money f (x) dx 0.34 (34%) 400 to hire the firm. (b) Take 1000 as a conveniently high upper limit: 1000 24. Pipe radius 1 ft; 6 700 360 12 (62.4)y dy 6 Work to fill pipe 0 385 0.217(300) 58,110,000 ft-lb. 30. The proportion of lightbulbs that last between 100 and 800 hours. 58,222,320 ft-lb, which will take 35,780,000 31. 58,222,320 1650 110,855 sec 6.5 people 29. Integration is a good approximation to the area (which represents the probability), since the area is a kind of Riemann sum. 360 Total work 0.217, which means about 112,320 ft-lb. (10)2(62.4)y dy Work to fill tank f (x) dx 6,370,000 31 hr. M 25. (a) The pressure at depth y is 62.4y, and the area of a thin 1000MG dr r2 1024, G 5.975 1 r 1000 MG 6.6726 10 35,780,000 , which for 6,370,000 11 evaluates to 1010 J. 5.1446 11 ft, so horizontal strip is 2 y. The depth of water is 6 32. (a) The distance goes from 2 m to 1 m. The work by an the total force on an end is 11/6 (62.4y)(2 dy) external force equals the work done by repulsion in 209.73 lb. 0 moving the electrons from a 1-m distance to a 2-m (b) On the sides, which are twice as long as the ends, the initial total force is doubled to 419.47 lb. When the tank is upended, the depth is doubled to 11/3 force on a side becomes 11 ft, and the 3 (62.4y)(2 dy) distance: 2 Work 23 dr 2 1 r 23 838.93 lb, 29 10 10 0 1.15 which means that the fluid force doubles. 26. 3.75 in. 5 ft, and 7.75 in. 16 31 ft. 48 31/48 Force on a side p dA (64.5y) 0 5 dy 16 4.2 lb. 27. (a) 0.5 (50%), since half of a normal distribution lies below the mean. 65 (b) Use NINT to find f (x) dx, where 63 f (x) 1 e 2 2 (x 63.4) /(2 3.2 ) . The result is 3.2 2 0.24 (24%). (c) 6 ft 72 in. Pick 82 in. as a conveniently high upper 82 limit and with NINT, find f (x) dx. The result is 72 0.0036 (0.36%). 28 10 2 1 r 29 1 J (b) Again, find the work done by the fixed electrons in pushing the third one away. The total work is the sum of the work by each fixed electron. The changes in distance are 4 m to 6 m and 2 m to 4 m, respectively. 6 Work 23 10 r2 4 23 10 7.6667 4 29 dr 29 10 1 r 29 6 4 J. 2 23 10 r2 4 1 r2 29 dr 317 Chapter 7 Review 33. F dv dt m x 2 W x 1 x 2 x 1 v 2 mv s Chapter 7 Review Exercises dv , so dx (pp. 413–415) F(x) dx 5 1. 0.2t 3) dt 0 13 t 3 mv dv 7 v 1 2. 1 mv 2 22 (t 2 0 dv dx dx mv 5 v (t) dt 0.001t 4) dt (4 0 1 mv 2 21 10.417 ft 0 7 c(t) dt 5 0.05t 4 0 0.0002t 5 4t 7 31.361 gal 0 34. Work Change in kinetic energy 12 mv . 2 100 3. 0.3125 lb 0.009765625 slug, and 32 ft/sec2 5280 ft 1 hr 90 mph 90 132 ft/sec, so 1 mi 3600 sec 1 Work change in kinetic energy (0.009765625)(132)2 2 0 1.6 oz 1 (0.003125)(280)2 2 Work 37. 2 oz 2 2 4. (x) dx (11 0 so Work 24 Work 39. 6.5 oz Work cos t 12 sin t 12 0 12 dt 24 14,400 0 6. 122.5 ft-lb. [ 1, 3] by [ 1, 2] The curves intersect at x 2 x dx 0.02832 slug, so 0 1 12 x 2 1 dx x2 1 2 109.7 ft-lb. 1 lb /(32 ft/sec2) 16 oz 1 (0.01270)(132)2 2 14 g 0.003125 slug, so 1 lb 14.5 oz /(32 ft/sec2) 16 oz 6.5 oz 2 0 300 2 0 1 lb 1 /(32 ft/sec2) slug, and 16 oz 256 5280 ft 1 hr 124 mph 181.867 ft/sec, 1 mi 3600 sec 11 (181.867)2 64.6 ft-lb. 2 256 1 (0.02832)(88)2 2 2x 2 11x 24 E(t) dt 5. 1 38. 14.5 oz 4x) dx 0 2 oz 124 mph 1464 0 300 2t 1 lb /(32 ft/sec2) 16 oz 100 33.333e0.03x 21x 85.1 ft-lb. 36. 1.6 oz e0.03x) dx (21 0 2 oz 1/8 lb 1 m slug, so 32 ft/sec2 32 ft/sec2 256 11 Work (160)2 50 ft-lb. 2 256 35. 0.3125 lb 100 B(x) dx 0.01270 slug, so 1. The area is 1 1 x 0 1 2 2 1 1 1. 7. 110.6 ft-lb. 1 1 lb slug. Compression energy of spring 8 256 12 1 12 ks (18) 0.5625 ft-lb, and final height is 2 2 4 0.5625 given by mgh 0.5625 ft-lb, so h 4.5 ft. (1/256)(32) 40. 2 oz [ 4, 4] by [ 4, 4] The curves intersect at x 1 [3 x2 2 and x 1 (x 1)] dx 2 ( x2 1. The area is x 2) dx 12 x 2 2x 2 13 x 3 1 3 9 . 2 1 2 2 1 2 8 3 2 4 318 Chapter 7 Review 8. x y 1 implies y x)2 (1 1 2 x x. 12. 2 [ 0.5, 2] by [ 0.5, 1] 1 The area is (1 2 x x) dx 4 3/2 x 3 x 0 12 x 2 1 . 6 by [ 3, 3] The area is 1 (2 sin x 0 sin 2x) dx 2 cos x 0 1 cos 2x 2 4. 0 13. x . 2 2y 2 implies y 9. x 3 2 , [ 5, 5] by [ 5, 5] The curves intersect at x [ 1, 19] by [ 1, 4] 2.1281 (4 x 2.1281 The curves intersect at x 18 x dx 2 3 0 3 or 0 y2 10. 4x 1 y 4 x 18 4 x 3/2 3x 23 y 3 2y 2 dy 18. The area is 32 2 2.1281. The area is cos x) dx, which using NINT evaluates to 18, 8.9023. 14. 0 3 18. 0 12 y 4 4 implies x 1, and 4x y 16 implies [ 4, 4] by [ 4, 4] 4. The curves intersect at x 0.8256 (3 0.8256. The area is sec2 x) dx, x 0.8256 which using NINT evaluates to 2.1043. [ 6, 6] by [ 6, 6] 15. Solve 1 The curves intersect at (3, cos x 2 cos x for the x-values at the two 4) and (5.25, 5). The area is ends of the region: x 5 1 y 4 4 5 12 y 4 4 12 y 4 4 12 y 8 425 24 38 3 3 , i.e., symmetry of the area: dy 7 /3 2 1 y 4 13 y 12 1 2 [(1 2 5 dy cos x) 2 cos x)] dx (2 cos x 1) dx 2 5 5y 2 2 sin x x 4 243 8 (2 7 /3 2 30.375. 2 3 3 7 /3 2 1.370. 5 /3 11. 16. [(2 /3 cos x) (1 5 /3 (1 2 cos x) dx /3 x [ 0.1, 1] by [ 0.1, 1] /4 The area is (x 0 sin x) dx 2 /4 12 x 2 cos x 2 2 32 0 2 0.0155. 1 2 sin x 3 4 3 5 /3 /3 7.653 cos x)] dx 5 7 or . Use the 3 3 Chapter 7 Review 17. Solve x 3 x x x x2 to find the intersection points at 1 (c) Use cylindrical shells. A shell has radius 4 height 2 x 21/4. Then use the area’s symmetry: 0 and x 2 (4 0 2 0 (x3 1 1 ln (x 2 2 2 x)(2 x x) dx (8 4x 2x 3/2 x 2) dx 2x 2 x 4 5/2 x 5 13 x 3 0 x x2 x. The total volume is 4 1/4 2 ln ( 2 x) dx 14 x 4 1) 1) 2 1 16 3/2 x 3 2 12 x 2 x and 4 the area is 2 21/4 4 64 . 5 0 (d) Use cylindrical shells. A shell has radius 4 0 319 1.2956. y and y2 . The total volume is 4 4 y2 2 (4 y) y dy 4 0 4 y3 2 4y 2y 2 dy 4 0 4 23 32 14 2 2y 2 y y . 3 3 16 0 height y 18. Use the intersect function on a graphing calculator to determine that the curves intersect at x 1.8933. The area is 1.8933 31 x2 3 dx, 10 x2 1.8933 which using NINT evaluates to 5.7312. 22. (a) Use disks. The volume is 2 19. Use the x- and y-axis symmetries of the area: 4 x sin x dx 4 sin x x cos x 0 4. 0 k (b) 0 0 A(x) r 1 V 1 k, 9 x. 9 x dx x 9 When k 1 y2 2y dy 2 4. 0 k2 0 8 8 k 2 dV (c) Since V 2 0 y2 2y dy 3x 4 and area 20. A cross section has radius r 2 2y)2 dy ( 1, 2. dt dk dt dk dt 1 dV 2 k dt 2k . 1 (2) 2 1 1 so the depth is increasing at the rate of 1 21. , unit per second. 23. The football is a solid of revolution about the x-axis. A [ 5, 5] by [ 5, 5] cross section has radius The graphs intersect at (0, 0) and (4, 4). r2 (a) Use cylindrical shells. A shell has radius y and height 2 y . The total volume is 4 4 4 y2 y3 2 ( y) y dy 2 y2 dy 4 4 0 0 4 13 14 2 y y 3 16 0 32 . 3 y 12 1 11/2 2 12 0 1 4x 2 and area 121 12 1 4x 2 . The volume is, given the symmetry, 121 11/2 4x 2 4x 2 dx 24 1 dx 121 121 0 11/2 21 2 24 x x3 3 11 0 11 11 24 2 6 88 276 in3. 24. The width of a cross section is 2 sin x, and the area is (b) Use cylindrical shells. A shell has radius x and height 2x x. The total volume is 4 4 2 (x)(2 x x) dx 2 0 (2x 3/2 2 x ) dx 0 2 4 5/2 x 5 128 . 15 13 x 3 4 0 12 1 r sin2 x. The volume is 2 2 1 1 1 sin2 x dx x sin 2x 22 4 02 2 0 4 . 320 Chapter 7 Review 29. 25. [ 1, 2] by [ 1, 2] [ 1, 3] by [ 3, 3] Use washer cross sections. A washer has inner radius r e x/2, and area (R2 outer radius R r 2) (e x 1, 1). 3x 2 y is at x 6x equals zero when x 0, the minimum at x = 2. The distance between 2 them along the curve is The volume is ln 3 (e x ex 1) dx using NINT evaluates to x 0 (3 ln 3 (2 4.5920 2 ln 3). 1) 26. Use cylindrical shells. Taking the hole to be vertical, a shell has radius x and height 2 22 3 2x 0 4 2 (4 3 2 x 2 dx x4 27. The curve crosses the x-axis at x 3 ( 2x)2 dx 1 3 1 2 x 4 dx 2 5 x 2 dx 2 2x, so the 13 x 3 5 39. 2 32. (a) (100 N)(40 m) = 4000 J (b) When the end has traveled a distance y, the weight of the remaining portion is (40 4x 2 dx, which 3 using NINT evaluates to 5 (F (x))2 dx 1 0 3. y 3 1, so 3 x2)3/2 4 (1 8) 3 28 29.3215 ft3. 3 length is 30. If (b) were true, then the curve y k sin x would have to get vanishingly short as k approached zero. Since in fact the curve’s length approaches 2 instead, (b) is false and (a) is true. 5 3 2 0 4.5920. The time taken is about 2.296 sec. 31. F (x) x 2) dx y)(0.8) 40 28. 0.8y) dy 0.4y 2 32y 0 The curves intersect at x 0 and x 1. Use the graphs’ 1 4 1 0 (3x 2 640 is (800)(8) 4750 x- and y-axis symmetry: 3x 40 640 J. 4640 J 33. The weight of the water at elevation x (starting from x [ 2, 2] by [ 2, 2] x) 0.8y. 0 (c) 4000 2 32 The total work to lift the rope is 19.4942. (32 d3 (x dx 6x)2 dx, which x 2. The volume of the piece cut out is 2 (x)(2 22 (3x 2 1 0 ln 3 0 0 or 2. The maximum is 0 4750 x/2 4750 128 4750 95 128 4750 95 1 x dx 2 1, and the total perimeter is 0) 1 x . The total work 2 128 4750x 95 12 x 4 4750 0 22,800,000 ft-lb. 1)2 dx, which using NINT evaluates to 34. F ks, so k 5.2454. 0.3 Work 0 F 800 80 N/m. Then s 3 0.3 0.3 800 800 2 x dx x 12 J. 3 6 0 To stretch the additional meter, 1.3 Work 0.3 800 x dx 3 800 2 x 6 1.3 213.3 J. 0.3 35. The work is positive going uphill, since the force pushes in the direction of travel. The work is negative going downhill. Chapter 7 Review 82 36. The radius of a horizontal cross section is y 2, where 1 42. Use f (x) 321 x 2/2 e . 2 y 2), the y is distance below the rim. The area is (64 1 (a) y )( y) y. The total work is y ) dy 0.04 3 (64y 2 y ) dy 2 36 320y(2 y) dy 320 ( y 0 2 13 y 3 1280 3 23 ft, 3.5 in. 48 38. 5.75 in. f (x) dx 1 A shell has radius x and height 2x x 2 4y, and the [ 1, 3] by [ 1, 3] 2y) dy 2 y2 0 and 44. 0 320 0.9973 (99.73%) 43. Because f (x) 113.097 in.-lb. 2 f (x) dx 3 y)4y y. The total force is 2 0.9545 (95.45%) 3 2 37. The width of a thin horizontal strip is 2(2y) force against it is 80(2 f (x) dx (c) 8 14 y 4 32y 2 0.04 (b) 2 8 2 0.04 y(64 0.6827 (68.27%). 2 2 the rim is 0.04 (64 8 f (x) dx evaluates, using NINT, to 1 y 2) y, and the work to lift it over weight is 0.04 (64 1 volume is 2 (x) 0 0 3 x dx 2 1 x3 3 x. The total 2 . 0 45. 426.67 lb. 5 ft. 6 7 ft, and 10 in. 24 For the base, Force 57 23 48 [ 3, 3] by [ 3, 3] 5 6 7 24 6.6385 lb. 5/6 Force 0 2 5/6 399 1 2 y 24 2 7 57 y dy 24 5/6 57 0 23 y dy 48 39. A square’s height is y y2 ( 0 1/2 2 1 dx x 2 dx 2 2x 1/2 3. 1/2 46. 5/6 1311 1 2 y 48 2 9.4835 lb. 0 x)2, and its area is (6 x)4. The volume is 6 2 (x) 5.7726 lb. For the sides, Force 1 x A shell has radius x and height . The total volume is For the front and back, 6 ( 6 x)4 dx, 2 , 3 2 by [ 2, 2] 0 A shell has radius x and height sin x. The total volume is which using NINT evaluates to exactly 14.4. 2 (x)(sin x) dx 40. Choose 50 cm as a conveniently large upper limit. 50 20 1 e (x 17.2)2/(2 3.42) x cos x 2 2 . 0 47. dx, evaluates, 0.2051 (20.5%). [ 1, 4] by [ 4, 1] 41. Answers will vary. Find , then use the fact that 68% of the class is within of to find , and then choose a conveniently large number b and calculate 10 sin x 3.4 2 using NINT to b 2 0 1 2 e (x )2/(2 2 ) dx. The curves intersect at x x and height x 3 (x 2 1 and x 3. A shell has radius 3x) 2 x 4x 3. The total volume is 3 2 (x)( x 2 3 4x 3) dx ( x3 2 1 4x 2 3x) dx 14 x 4 43 x 3 32 x 2 1 2 16 . 3 3 1 322 Section 8.1 48. Use the intersect function on a graphing calculator to determine that the curves intersect at x has radius x and height 3 1x x 2 2 3 10 1.8933. A shell that l’Hôpital’s Rule does not say that lim . The volume, which lim 0 evaluates to 9.7717. 5 (x 4 49. (a) y . x2 3 dx, which using NINT 10 x2 2 (x) 31 x→0 y2 y1 x→0 y2 is calculated using the right half of the area, is 1.8933 x cos x sin x . The graphs of y3 and y5 clearly show x2 y1 3. y5 2)(x 2) 52 x 4 5 [ 3, 3] by [ 2, 2] Quick Review 8.1 (b) Revolve about the line x 4, using cylindrical shells. 1. x A shell has radius 4 52 x . The total 4 x and height 5 1 2 2 (4 2 10 2 320 3 1 100 x 12 x 2 4x 2 dL dx 1 x f (x) must equal 1 x 2 2 f (x) x 1 x2 2. 12 x 4 51. y 12 x 4 1 ln x 2 3 4 sec x, so the area is 2 (tan x) 2 2.7183 1 2.7183 2.7183 2.7183 0.00001 2.7183 As x→0 , x1/(ln x) approaches 2.7183. 2 (sec x) dx, 0 which using NINT evaluates to x1/(ln x) 0.0001 3 . The 4 2 ln x 3 . 4 x2 x 0.001 C, and the /4 2 0.1 x approaches 1.1052. x 0.01 1 ln x 2 requirement to pass through (1, 1) means that C function is f (x) 1.1052 0.1 ( f (x))2, and 1 . Then f (x) 2x 1.1052 As x→ , 1 ( f (x))2, 1 1.1052 10,000 2 1.1051 1,000,000 4 dx 335.1032 in3. ( f (x))2 f (x) x 2 13 x 3 14 x 16 10 50. Since 13 x 4 1.1046 1000 52 x dx 4 x) 5 1.1000 10 volume is 2 0.1 x x 1 3. x 3.84. 1 1x x 1 1 52. x and x y 2 1 2 1 y 1 0.5 0.1 1 , so the area is y2 12 dy, y2 0.78679 Chapter 8 L’Hôpital’s Rule, Improper Integrals, and Partial Fractions 0.99312 0.0001 5.02. 0.95490 0.99908 0.00001 which using NINT evaluates to 0.01 0.001 0.99988 0.000001 0.99999 1x approaches 1. x As x→0 , 1 4. 1x x x s Section 8.1 L’Hôpital’s Rule (pp. 417–425) Exploration 1 Exploring L’Hôpital’s Rule Graphically 1. lim x→0 sin x x lim x→0 cos x 1 1 1.1 13.981 2. The two graphs suggest that lim y1 x→0 y2 lim y1 x→0 y2 . 105.77 1007.9 1.0001 1 1.01 1.001 10010 As x→ 1 , 1 1x goes to x . is equal to Section 8.1 5. 323 3. [0, 2] by [0, 3] t As t→1, 1 t [0, 3] by [0, 3] From the graph, the limit appears to be 1. The limit leads to the indeterminate form 0. approaches 2. 1 6. ln 1 ln 1 [0, 500] by [0, 3] lim As x→ , 1 x x ln 1 1 lim 1 x2 1 lim x→0 lim 1 x2 1/x x→0 7. 1 x 1 x x→0 x [ 5, 5] by [ 1, 4] 1 x 1 1 x 1 x x→0 4x 2 1 approaches 2. x1 1 x ln 1 1x x 0 1 Therefore, sin 3x As x→0, approaches 3. x 1x x lim 1 x→0 8. lim f (x) x→0 lim e ln f (x) e0 x→0 1. 4. [0, ] by [ 1, 2] tan →, 2 2 tan As [0, 1000] by [0, 1] approaches 1. From the graph, the limit appears to be about 0.714. 5x 2 2 x→ 7x lim 1 sin h h 9. y 10. y lim x→ x3 1 4x 3 x 3 x→1 h)1/h (1 3x 1 10x 3 14x lim x→ 3x 2 12x 2 1 x→1 5. lim lim 10 14 5 7 3 11 Section 8.1 Exercises 1. [0, 2] by [0, 1] [0, 2] by [0, 1] 1 4 The graph supports the answer. From the graph, the limit appears to be . x 2 x→2 x lim 2 4 1 x→2 2x lim 1 4 6. lim x→0 1 cos x x2 lim x→0 sin x 2x lim x→0 2. [ 2, 2] by [ 2, 6] [ 5, 5] by [ 1, 1] From the graph, the limit appears to be 5. lim x→0 sin 5x x lim x→0 5 cos 5x 1 5 The graph supports the answer. cos x 2 1 2 0.71429 324 Section 8.1 7. The limit leads to the indeterminate form 1 . ln(e x Let ln f (x) ln(e x x x→0 lim lim (e x x) ex ex lim 1 x x) . t ex x x→0 e 1 x lim 1 x→0 x)1/x x→0 ln(e x x x)1/x 2 1 lim e ln f (x) lim f (x) x→0 1 t 1 sin t 12. Let f (t) 2 e2 x→0 100 f (t) 0.1884 t sin t . t sin t 1 10 10 0.0167 2 10 0.0017 3 0.00017 Estimate the limit to be 0. 1 t 1 lim t→0 sin t lim t t→0 lim t→0 lim t→0 [0, 5] by [0, 10] 13. Let f (x) sin t t sin t 1 cos t t cos t sin t sin t t sin t cos t 0 cos t x)1/x. (1 The graph supports the answer. 8. lim x→ 3x 2x x3 x 1 3 1 4x 3x 2 lim x→ lim x→ 4 6x 0 102 10 x 2 103 104 105 f (x) 1.2710 1.0472 1.0069 1.0009 1.0001 Estimate the limit to be 1. ln (1 x) 1 x 1x ln (1 x) lim lim 1 x x→ x→ ln f (x) [ 5, 25] by [ 1, 2] The graph supports the answer. 9. (a) 102 10 x x)1/x lim (1 x→ 103 104 105 0 1 0 lim e ln f (x) e0 102 103 104 105 0.6525 0.6652 0.6665 0.6667 lim f (x) x→ x→ 1 x 2x2 . 3x2 5x 14. Let f (x) f (x) 1.1513 0.2303 0.0345 0.00461 0.00058 10 x Estimate the limit to be 0. ln x 5 x x→ 5 ln x x x→ 10. (a) 100 x 5/x 1 x→ lim (b) lim 1 10 10 2 f (x) 0 1 lim 10 3 10 4 f (x) 0.1585 0.1666 0.1667 0.1667 0.1667 1 6 Estimate the limit to be . (b) lim x→0 x sin x x3 0.5429 0 x 2x 2 2 5x x→ 3x lim 15. lim sin 2 2 cos 1 →0 x→0 lim x→0 1 16. lim → /2 1 sin cos 2 1 6 sin 3 . sin 4 17. lim 10 f( ) 0 10 1 10 3 4 sin 3 →0 sin 4 2 10 3 10 4 0.1865 0.7589 0.7501 0.7500 0.7500 Estimate the limit to be . lim lim 3 cos 3 →0 4 cos 4 4 6 (2)(0) cos (0)2 2 . 3 0 1 lim x→0 lim 11. Let f ( ) 2 lim →0 cos x 3x 2 sin x 6x cos x 6 2 . 3 1 4x lim lim 5 x→ x→ 6x Estimate the limit to be 3 4 cos t t t t→0 e 1 1 cos 2 sin 2 sin lim → /2 4 cos 2 sin /2 4 cos 1 4 lim → /2 lim t t→0 e sin t 1 lim t→0 c os t et 1 Section 8.1 18. lim t t→1 ln t 1 sin t 1 lim t→1 1 cos t t y→ /2 1 1 22. lim y tan y 2 ( 1) lim x→ 2 x ( 1)(1) (1) lim ln 2 x→0 x→0 ln 2 log3 (x 3) lim x→ lim (x lim lim x→0 tan 1 24. lim x tan x x→ lim x→ ln 3 ln 2 ln (y 2 2y) ln y y→0 21. lim x x→0 x ln 3 3 ln 3 x ln 2 lim x→ y→0 lim y→0 2 2y 1 1 sec2 x2 x 1 x2 lim sec2 1 x sec2 0 1 y(2y 2) y 2 2y 4y y→0 2y 4(0) 2(0) 1 x x→ 1 y 2y 2 lim 2 y→0 y lim 0 1 x lim 2y y2 1 x2 x lim 3) ln 3 x ln 2 ln 3 x→ ln 2 2 1 x lim lim ( 1) sin 2 1 x 1 x2 x→0 1 x ln 2 1 (x 3) ln 3 x→ x→ cos lim ln x 23. lim x ln x x→ x→ 2 sin 1 x ln 2 ( 1) sin y sin y 1 x ln 2 1 x→ x 20. lim y cos y y→ /2 lim log2 x 2 lim 1 y sin y cos y y→ /2 1 +1 ln (x 1) 19. lim log2 x x→ 2 lim 2 2 2y 2y 2 2 25. lim (csc x cot x x→0 2 2 1 lim x→0 lim x→0 lim x→0 1 sin x 1 cos x sin x cos x) cos x cos x cos x sin x sin x sin x cos x cos x cos x sin x sin x 1 2 325 326 Section 8.1 26. lim (ln 2x ln (x x→ 2x Let f (x) x 1 x→ x lim 1 x→ x 1 ln (x x→ 27. lim (ln x 1)) ln sin x) x→0 x lim x→0 sin x lim ln f (x) ln (ln x) lim x x→ ln 2 x→ x→0 Let f (x) 1 ln sin x) 1 x x→0 1 lim ln f (x) x→0 1 lim ln 1 ln (e lim x x x)1/x. x) ln (e x→0 x x) lim x)1/x x→0 ln lim x→0 x→0 1x x2 x→0 31. lim x→ 32. lim 2x 2 sin 7x x→0 tan 11x 5 x 2 lim (x 2 ln 2/x 3 1/x 2 2x x lim x→ 2 x 1 lim eln f (x) lim x→ 2x e1/2 x→ lim 1)x e 1 1 1) ln (x 2 (x ln (x 2 2x x e2 0 . 2x 1) 1) 2x 1) lim x2 x→1 1 4x 0 7 cos 7x 2 x→0 11 sec 11x 1)2 2(x x→1 lim (x 2 x→1 2x 1) x1 lim e 1) ln f (x) x→1 0 e 0 1 36. The limit leads to the indeterminate form 00. 1 Let f (x) 1 1 2(x 1) (x 1)2 lim 1 x→1 (x 1)2 x→0 3 2 2x 1 (x 1 lim e0 1 2x 2 lim ln (x 1 x2 1 x lim 2x 1/x 2 x→ 2x 1)x 2x x x→0 3x lim 1 x→ 2x)1/(2 ln x) x→1 lim e ln f (x) lim ln (1 2x) 2 ln x 1 x→0 lim . 2 1 lim e ln f (x) 1 x ln 2 x 1 x2 1 x lim (1 x→ ln (x 2 1 x 1x . x2 1x ln 2 x 2x) 2 Let f (x) 30. The limit leads to the indeterminate form Let f (x) . x) ex ex x→0 x lim (e x 0 1 35. The limit leads to the indeterminate form 00. ln (e x x 1/x (1 ln (1 2x) lim 2 ln x x→ 29. The limit leads to the indeterminate form 1 . (e x e0 0 1/(2 ln x) 0 x x→0 x Let f (x) x→ x→ 2x)1/(2 ln x) ln (1 28. lim 1 x ln x lim 1 lim eln f (x) x→ Therefore, lim (ln x lim 34. The limit leads to the indeterminate form 1 lim x→0 cos x x→0 1/x ln x x→ lim (ln x)1/x x sin x lim ln x . sin x Let f (x) ln (ln x) x ln (ln x)1/x 2 . (ln x)1/x. Let f (x) Therefore, lim (ln 2x 0 33. The limit leads to the indeterminate form 2x lim ln . 2 x→ 1 2x lim 1)) 0 7 11 (cos x)cos x. ln (cos x)cos x ln (cos x) sec x (cos x) ln (cos x) sin x ln (cos x) cos x lim sec x x→ /2 x→ /2 sec x tan x lim lim tan x x→ /2 sec x tan x lim x→ /2 lim (cos x)cos x x→ /2 cos x lim e ln f (x) x→ /2 0 e0 1 1 2 1 2 Section 8.1 37. The limit leads to the indeterminate form 1 . Let f (x) 1/x (1 x x) . x) lim x 1 1 lim eln f (x) e1 x→0 1/(x 1) ln x1/(x lim x e ln x 1 x→1 x lim x1/(x lim . x→ 1) e1 x→1 lim x→0 (c) lim ln (sin x) x→ 1 x x 2 cos x sin x x→0 x 2 sin x 2x cos x lim cos x x→0 lim e ln f (x) x→0 x→ 1 1/2 1) (1/2)(x 1) lim 1/2 x→ e0 x→0 9x 1 x 1 1 lim x→ 1 x 1 x 9 lim sec x x→ /2 tan x 0 ln (sin x) lim x→ /2 sec x tan x sec2 x tan x lim x→ /2 sec x (b) lim (sin x) [0, ] by [ 1, 5] ln (sin x) cot x tan x ln (sin x) cos x sin x ln (sin x) lim cot x x→0 lim csc x 1 x→0 ln f (x) 0 lim e x→0 The limit appears to be 1. lim ( sin x cos x) 2 x→0 tan x e x→0 0 sec x x→ /2 tan x (c) lim ln x1/(1 ln x lim x x→1 1 x→1 lim x→1 x) lim (b) f (x) 1 x 1 lim e ln f (x) x→1 . 7(x f (x) x→3 g(x) . ln x 1x x) lim x1/(1 x) (x f (x) lim x→3 g(x) 1 e 1 1 lim x→ /2 sin x 1 x→ /2 sin x 45. Possible answers: 1 (a) f (x) x1/(1 lim cos x cos x 41. The limit leads to the indeterminate form 1 Let f (x) 1 1 3 1 (sin x)tan x tan x x 9x 1 40. The limit leads to the indeterminate form 00. Let f (x) 9 44. (a) L’Hôpital’s Rule does not help because applying L’Hôpital’s Rule to this quotient essentially “inverts” the problem by interchanging the numerator and denominator (see below). It is still essentially the same problem and one is no closer to a solution. Applying L’Hôpital’s Rule a second time returns to the original problem. 1 x2 lim lim (sin x)x (9/2)(9x lim 9 cos x sin x 1 x 1 [0, 100] by [0, 4] x ln (sin x) lim ln (sin x) 9x x (sin x)x. x→0 ln 2 The limit appears to be 3. 39. The limit leads to the indeterminate form 0 . ln (sin x)x lim ln 2 x→ e 0 Let f (x) 2x x lim ln x→ 1 lim e ln f (x) x→1 dt t x 2x x ln (b) ln x x1 1/x lim x→1 1 1) 2x ln x 43. (a) L’Hôpital’s Rule does not help because applying L’Hôpital’s Rule to this quotient essentially “inverts” the problem by interchanging the numerator and denominator (see below). It is still essentially the same problem and one is no closer to a solution. Applying L’Hôpital’s Rule a second time returns to the original problem. 38. The limit leads to the indeterminate form 1 . Let f (x) ln 2x x x→ 1 x→0 x)1/x lim (1 ln t lim 1 ln (1 x) lim x x→0 x→0 2x dt t ln (1 x) x 1/x ln (1 2x 42. 327 1 e (c) f (x) lim x f (x) x→3 g(x) 3); g(x) lim x→3 7(x x x 3) 3 3)2; g(x) 3 lim 7 7 x→3 1 x 3 (x lim x→3 x 3)2 3 x→3 3; g(x) (x 3)3 x 3 3)3 lim x→3 (x lim lim 2(x x→3 3(x 3) 1 1 3)2 0 328 Section 8.1 46. Answers may vary. (a) f (x) 3x 1; g(x) f (x) lim x→ g(x) (b) f (x) lim x lim x→ 1 lim x→ x 1; g(x) f (x) lim x→ g(x) x x lim x x 0 lim x→ 1 1 1 f (x) g (x) 0, f (x ) 1. 1 x→0 g (x) 2 1 f (x) lim x→0 g (x) 2 (b) This does not contradict L’Hôpital’s Rule since lim f (x) 2 and lim g (x) 1. 1 x lim x→ 50. (a) For x lim 2 f (x) g(x) 3 1 lim x→ 2x x2 x→ 3 1 2 1 x 2; g(x) (c) f (x) x 3x x→ (b) Part (a) shows that as the number of compoundings per year increases toward infinity, the limit of interest compounded k times per year is interest compounded continuously. x→0 2x 1 x→0 t 51. (a) A (t) e x dx 0 47. Find c such that lim f (x) lim f (x) 3 sin 3x 5x3 9 9 cos 3x 15x2 27 sin 3x 30x 81 cos 3x 81 30 30 x→0 lim x→0 lim x→0 lim x→0 1) t 27 10 e c f (0), so f is lim x→0 x→0 ln x lim x x x→0 lim e x→0 lim x→0 x ln x e x→0 0 x 0 V(t) t→0 A (t) (c) lim lim (e lim 2 r k lim 1 lim A0 1 k→ lim k→ rt 1 rt lim k→ t 1 k r k r kt k 0.1 1 0.00495 0.00050 0.0001 0.00005 The limit appears to be 0. (b) lim sin x 2x x→0 1 rt ) t 0.04542 0.001 r r 1 k2 k 1 k2 1) 1 f (x) 0.01 1 k 2 (2) x r k t ln 1 1 2t t (1) 2t (2e e 1 2 1 52. (a) r kt . k r kt ln 1 k t ln 1 k→ e t→0 1) t e t→0 49. (a) The limit leads to the indeterminate form 1 . 1 2t (e t→ 2 ln f (k) 2 lim 2 0. Extend the 1 2 1) 1 Thus, f has a removable discontinuity at x definition of f by letting f (0) 1. Let f (k) dx 2t (e V(t) lim t→ A (t) 1 x lim 1 x 2x 1 2t e 2 0. lim f (x) leads to the indeterminate 1 x 1 x2 1 1 2x t e 2 0 2 ln x t 0 form 0 . x ln x 1 lim ( e t→ (e x)2 dx t 0 ln x x 1 0 continuous. 48. f (x) is defined at x t→ 1 et lim t→ (b) V (t) 27 . This works since lim f (x) 10 x→0 Thus, c t lim A (t) 9x lim x→0 e 0 c. x→0 t x e 0 1 0 L’Hôpital’s Rule is not applied here because the limit is A0 lim 1 k→ r kt k A0 lim e ln f (k) k→ A0 e rt not of the form limit 1. 0 or 0 , since the denominator has Section 8.2 e x ln (1 53. (a) f (x) 1 x 1 1/x) (b) f (x) 0 when x Domain: ( , 1 or x 1) f 0 sin x, g (x) f (0) 2 tan c 1 x (c) lim x ln 1 x→− lim 56. (a) ln f (x)g(x) 1 x lim e ln f (x) x→c 1 lim f (x) x→− lim e x ln (1 ( 1/x) x→− e x→0 cos x 6 x12 Exploration 1 x→ ax 2 x→ x 1. lim x→0 lim x→0 lim x→0 lim x→0 2. lim ax x x→ b 3. lim ln x 1. lim x e x→ ex 3 x→ x 2. lim 3. lim x→ 55. (a) f (x) 3x , g (x) f (1) f ( 1) 2x 2, g(1) 3c 2 2c 1 (3c c 2c 1)(c 1 or c 3 2c 1 x2 2x x→ e 2 5. 1 (ln a)2 a x 2 x→ , so a x grows lim lim 1.5x x→ lim x→ ax b a b because 0 f (x) x→ g(x) 8. lim 1 The value of c that satisfies the property is c 1 . 3 lim x→ 1 x 1. 0 ex ex 2 x→ 3x ex x→ 6x lim ex x→ 6 lim lim lim x→ 2x 2e2x lim x→ 2 4e2x 3x 4 f (x) 7. lim x→ g(x) 0 1) (ln a)(a x) 2x x→ lim x2 e2x 4. lim 2 2 3c 2 3c 2 1 g( 1) Comparing Rates of Growth as Quick Review 8.2 (d) The graph and/or table on a grapher show the value of the function to be 0 for x-values moderately close to 0, but the limit is 1/2. The calculator is giving unreliable information because there is significant round-off error in computing values of this function on a limited precision device. 2 e faster than x 2 as x→ . 3x 2x x→ 6x 5 sin x 6 12x 11 sin x 6 2x 6 6x 5 cos x 6 12x 5 cos x 6 1 2 2 lim lim eln f (x) x→c (pp. 425–433) (b) Same reason as in part (a) applies. 1 ) s Section 8.2 Relative Rates of Growth 54. (a) Because the difference in the numerator is so small compared to the values being subtracted, any calculator or computer with limited precision will give the incorrect result that 1 cos x 6 is 0 for even moderately small values of x. For example, at x 0.1, cos x 6 0.9999999999995 (13 places), so on a 10-place calculator, cos x 6 1 and 1 cos x 6 0. (c) lim x→c )( g(x) lim f (x) g(x) x→c 0 lim g(x) lim ln f (x) x→c 1 1 x 1 e x→c x→c x→− x→c ) g(x) lim f (x) g(x) (b) lim (g(x) ln f (x)) lim 2 lim g(x) lim ln f (x) x→c 1 1 x on 0, 4 ( 1 x2 x→− 1 g(x) ln f (x) x→c 1 1 x2 lim 1 tan lim (g(x) ln f (x)) 1 x x→− 1 1 c ln 1 g(0) 2 1 1 (b) The form is 0 1, so lim f (x) x→ 1 cos x 1, g sin c cos c (0, ) 329 0 6. lim x→ lim x→ x ln x x 4x 2 5x 2x 1 x 1 lim x→ lim x→ 2x 3 x 1 1 2x 2 1 5 4x 1 330 Section 8.2 ex 9. (a) f (x) f (x) 2x x x x2 ex 2x 1 2x xe e2 x 2 x x ln x 6. lim ex x→ 2 ex x) f (x) 0 or x 2 1; f (2) 4 e2 1 ecos x ex x→ ecos x grows slower than e x as x→ . (2, 1.541) and has a local xe x x x→ e 8. lim x1000 ex x→ 9. lim , 0] and [2, ). sin x sin x 1 ,x 0 x x sin x 1 since sin x x for x Observe that x sin x 112 lim f (x) 1 lim x→0 x→0 x gets lim x x→ 0. 10. lim ex doesn’t have an absolute maximum value. f is not defined at 0. 11. lim x 3x 3x 2 3 ex x→ lim x→ x 6x ex lim x→ 6 ex 0 1x x2 lim x→ x 4 grows at the same rate as x 2. 1 1 lim x 4 lim x4 x→ 2 x→ 1 x4 1 x2 x x→ e lim 2x x x→ e lim 2 x x→ e lim than e as x→ . lim x→ 4x e since 4 e (5/2)x 5x 5 lim 0 since x e 2e 2e x→ x→ 5x grows slower than e x as x→ . 2 4. lim ex 1 ex x→ 1 lim e x→ 3 x2 lim x x→ 15x 3 x2 15. lim x→ 15 x lim x→ 3 x2 0 3 grows slower than x 2 as x→ . 15x x 4 5x x2 lim x4 5x lim x4 x→ x→ 5 x3 1 1 5x grows at the same rate as x 2 as x→ . ln x x2 2x 2 x→ x 1. 16. lim 4x grows faster than e x as x→ . 5. lim 1 lim x→ 1/x 2x lim x→ 1 2x 2 0 ln x grows slower than x 2 as x→ . x 4x x x→ e x→ x4 x 4 grows slower 4 x lim 1 x→ 3 0 1 1 2 1 2e 2x 3 grows faster than x 2 as x→ . x x→ x 2 grows slower than e x as x→ , so 3. lim 4x x2 14. lim Next compare x with e . 1 2 lim x→ grows at the same rate as e x as x→ . x3 1 x 0. x1000 grows slower than e x as x→ . e x)/2 ex x2 x→ 3 13. lim 4 1000! ex 4x grows at the same rate as x 2 as x→ . 1 grows slower than e as x→ . 2. First observe that 0 Repeated application of L’Hôpital’s Rule x2 x→ 2 12. lim lim x e 2 x Section 8.2 Exercises ex (e x x→ Thus the values of f get close to 2 as x gets close to 0, so f x→ 3 lim x x→ xe xgrows faster than e x as x→ . (c) f is decreasing on ( 1 e for all x. 1.541 (b) f is increasing on [0, 2] 3x ex 0 0 since ecos x 7. lim minimum at (0, 1). x3 1 x x grows slower than e x as x→ . x ln x f has a local maximum at 1. lim e x→ 0 for x ln x ln x ex 1/x ex lim 0 2 1 x x→ lim The graph decreases, increases, and then decreases. 10. f (x) lim x→ 0 or x f (0) x x 0 ex x(2 x2 ex 2xe x (ln 2)2 x 2x x→ lim (ln 2)2 2 x 2 x→ lim . 2 x grows faster than x 2 as x→ . 1. e grows at the same rate as e x as x→ . 17. lim x→ log2 x 2 ln x lim x→ 2 log2 x ln x lim x→ 2(ln x)/(ln 2) ln x 2 ln 2 log2 x 2 grows at the same rate as ln x as x→ . 18. lim x→ log log x ln x lim x→ log x 2 ln x lim x→ (ln x)/(ln 10) 2 ln x x grows at the same rate as ln x as x→ . 1 2 ln 10 Section 8.2 1/ x ln x x→ 1 19. lim 1 lim x→ 25. Compare f1 to f2. 0 x ln x lim grows slower than ln x as x→ . x→ x ex x→ ln x x e lim 0 x 2 ln x ln x x→ lim (x lim x ln x lim x→ 2 lim x→ 1 1/x 22. lim x→ 5 ln x ln x lim x→ e xx ex lim x→ lim x→ Compare e x to (ln x)x. x lim e (ln x)x 10 f3(x) f1(x) lim x x→ 1 lim 1 x 1 x→ x 1 lim x→ f1(x) lim x→ x4 x x2 lim x→ 1 1 x3 1 e ln x f3(x) f1(x) lim x→ x4 x3 x2 lim x→ 1 1 x 1 Thus f1 and f3 grow at the same rate. x 0 By transitivity, f2 and f3 grow at the same rate, so all three e grows slower than (ln x)x. x f2(x) Compare f1 to f3. 0 x e grows slower than x x. x→ x 1 x 10 Thus f1 and f2 grow at the same rate. 23. Compare e x to x x. lim lim x→ 26. Compare f1 to f2. 5 ln x grows at the same rate as ln x as x→ . x→ x 1 x functions grow at the same rate as x→ . 5 x 10x By transitivity, f2 and f3 grow at the same rate, so all three 2 ln x grows faster than ln x as x→ . x lim x→ Thus f1 and f3 grow at the same rate. 2 2) x→ f1(x) Compare f1 to f3. grows slower than ln x as x→ . x→ 21. lim f2(x) Thus f1 and f2 grow at the same rate. 1 x x→ e ln x 20. lim 331 functions grow at the same rate as x→ . x/2 Compare e to e . ex x/2 x→ e lim e x/2 lim 27. Compare f1 to f2. x→ x x/2 e grows faster than e . lim x→ Compare x x to (ln x)x. x x x x→ (ln x) lim lim x→ x ln x x x x→ ln x since lim 1 x→ 1/x lim f2(x) f1(x) lim 24. Compare 2x to x 2. (ln 2)2 x 2x x→ x→ lim x→ Compare 2 x to (ln 2)x. 2x (ln 2)x x→ lim x→ 2x ln 2 since 2 ln 2 lim x→ x 2 ex lim x→ e 0 since since lim x 2 x→ lim x→ lim x→ 1 2 e 9x 4x 3x 9x 4x 9 1 x 4x 9 1 Thus f1 and f3 grow at the same rate. By transitivity, f2 and f3 grow at the same rate, so all three functions grow at the same rate as x→ . 1. Compare x 2 to (ln 2)x. x2 (ln 2)x x→ 2x 9 1 x→ 2 grows slower than e x. lim 9 1. Compare 2 x to e x. 2x f3(x) f1(x) lim 2 x grows faster than (ln 2)x. x 2x x Compare f1 to f3. (ln 2)2 2 x 2 x→ lim 2 x grows faster than x 2. lim 9x Thus f1 and f2 grow at the same rate. get e x/2, e x, (ln x)x, x x. lim 9x 2x 3x x→ Thus, in order from slowest-growing to fastest-growing, we 2x 2 x→ x x→ lim . x x grows faster than (ln x)x. lim lim and lim (ln 2)x x→ 0. x 2 grows faster than (ln 2)x. Thus, in order from slowest-growing to fastest-growing, we get (ln 2)x, x 2, 2 x, e x. 332 Section 8.2 28. Compare f1 to f2. x4 lim x→ f2(x) f1(x) 2x 2 x lim x 2x x→ x→ 1 1 x lim f1(x) lim x→ x→ lim x→ 2x 5 1 5 lim 1 x→ 1 x5 1 (c) True, since lim x→ x2 x 1 x 5 1. ex e2x x→ lim x 1 x 1 f (x) g(x) , so lim x→ g(x) f (x) 1 2 0. 0. lim 5 1 1 x→ lim x2 x x f (x) g(x) 1 and not equal to zero. Thus, f 1 lim lim x→ 1 x 1. 1 x2 5 x2 1 3 1 1 x2 1 x 1 x lim 1 1 x 1 lim x→ ex . xn x→ … ex x→ n! lim 1. (b) The nth derivative of a x, a 1 1, is (ln a)n a x. We can ax n. x→ x apply L’Hôpital’s Rule n times to find lim 1. ax n x→ x 1 x 1 x→ 1 x 3 and not equal to zero. Thus, f Thus e x grows faster than x n as x→ for any positive 0. 1 x x ex n x→ x lim lim 1 x f (x) g(x) integer n. x→ 3 o(g), so i is true. L’Hôpital’s Rule n times to find lim 1 x→ 2x x2 35. (a) The nth derivative of x n is n!, a constant. We can apply 0. 1/x x→ 1/x ln x x→ ln 2x (g) False, since lim (c) False, since lim 1/x lim x→ 1. and g grow at the same rate, so iii is true. 1 ln x x x→ x→ 0 0. Thus f 34. From the graph, lim 1 ex x→ 1 ln x f (x) g(x) x→ (e) True, since lim (b) True, since lim 1 x and g grow at the same rate, so iii is true. 1. 30. (a) True, since lim 1 x ln x o( f ), so ii is true. x→ 0. 1 5 x x→ Thus g 0. x (h) True, since lim x→ 0. 33. From the graph, lim 1 1 2 (f) True, since lim 31. From the graph, lim x→ x 2x x→ 1/x 1 lim x→ 32. From the graph, lim (d) True, since lim x→ 1. x2 1 lim 2x 2 x→ 2 x 29. (a) False, since lim x→ x x→ ln x ln (x 2 1) (h) False, since lim x3 Thus f1 and f3 grow at the same rate. By transitivity, f2 and f3 grow at the same rate, so all three functions grow at the same rate. (b) False, since lim ln x x lim 1 1 1 x→ x3 x 1 ex 1 x2 2 x→ lim x→ ln (ln x) ln x x→ x→ lim x ln x x2 (f) True, since lim (g) True, since lim 5 1 ex x→ x→ 1 Compare f1 and f3. x→ e lim 1 x4 1 2x 3 . 2 ex lim 1 Thus f1 and f2 grow at the same rate. f3(x) x x x3 2 x2 1 ex (e) True, since lim 2 x4 lim 3 2 lim x→ 1 4 x→ cos x 2 x→ x3 x→ lim 2 (d) True, since lim 1 … (ln a)n ax n! x→ lim Thus a x grows faster than x n as x→ for any positive 1 1. 1 0. integer n. Section 8.2 39. Compare n log2 n to n3/2 as n→ . 36. (a) Apply L’Hôpital’s Rule n times to find ex lim anx x→ n an 1x … n1 ex lim anx n x→ an 1x n a1 x … 1 a0 a1 x lim . n log2 n n3/2 n→ ex x→ ann! lim a0 log2 n lim n1/2 n→ (ln n) (ln 2) lim n→ an 1x … 2 n1/2(ln 2) n→ a0 as x→ . a1x 1 n ln 2 1 2n1/2 lim anx n1 n1/2 lim Thus e x grows faster than n n→ x→ lim x→ ax anx n an 1x … n1 ax anx n an 1x a1 x … n1 a0 a1 x Compare n log2 n to n(log2 n)2 . lim … a0 n→ Thus ax grows faster than an 1x n ln x 37. (a) lim 1/n x→ x … 1 1 x lim x→ lim x→ 1 (1/n) 1 x n n 1 x lim ax a x→ 38. lim x→ 1 a x→ ax lim 1 0 ln x an 1x n 1 x→ nanx n 1 x→ f (x) g(x) 1 M for some integer M. lim … a1 x a0 1)an 1x n (n (n 1 L 1 N for x5 2 x→ x 43. (a) lim 1)an 1 … 2 a1 xn 1 … a1 x 0 Thus ln x grows slower than any nonconstant polynomial as x→ . lim x 3 x→ x 5 grows faster than x 2. 1 nanx n g(x) f (x) 42. (a) The limit will be the ratio of the leading coefficients of the polynomials since the polynomials must have the same degree. 5x 3 3 x→ 2x (b) lim x→ L L. Then for sufficiently (b) By the same reason as in (a), the limit will be the ratio of the leading coefficients of the polynomial. 1 x lim number L such that lim some integer N. 0. anx n 19.9; it might take 20 binary Similarly, for sufficiently large x, Thus ln x grows slower than x a as x→ for any number a 0 f (x) large x, g(x) positive integer n. ln x xa x→ 1 log2 n 41. Since f and g grow at the same rate, there exists a nonzero Thus ln x grows slower than x1/n as x→ for any (b) lim n→ (b) log2 1,000,000 searches. 0 x1/n lim n(log2 n)2 40. (a) It might take 1,000,000 searches if it is the last item in the search. a0 as x→ . a1x n log2 n Thus n log2 n grows slower than n(log2 n)2 as n→ . The algorithm of order of n log2 n is likely the most efficient because of the three functions, it grows the most slowly as n→ . (ln a)na x lim ann! x→ anx n 0 Thus n log2 n grows slower than n3/2 as n→ . (b) Apply L’Hôpital’s Rule n times to find lim 333 5 2 lim x→ 3 5 2 5x 3 and 2x have the same rate of growth. (c) m n since lim xm xn x→ (d) m n since lim xm xn x→ x→ x→ lim x m n lim x m n . is nonzero and finite. (e) Degree of g degree of f (m n) since lim g(x) f (x) (f) Degree of g degree of f (m n) since lim g(x) is f (x) nonzero and finite. x→ x→ . 334 Section 8.3 44. (a) f f (x) x→a g(x) o(g) as x→a if lim 3. If p 0 1 Suppose f and g are both positive in some open interval containing a. Then f 0 1, then 1 dx xp lim c→0 O(g) as x→a if there is a positive integer M for which f (x) g(x) lim c→0 where M is a bound for the absolute value of f h (b 4 as h→0, so ES M a) 180 4. If 0 b a4 hM 180 (b) From Section 5.5, we know that ES [a, b]. Thus, int (b (4) 1 on M a) 180 0 b (c) From Section 5.6, we know that ET 12 f(x) 45. (a) lim x→ (b h2 h→0, so ET c→0 a) M 12 int (b a) x→ (b) lim x→ lim x→ g(x) f (x) g(x) 1. 1 as f (x) x→ g(x) ln x 3 1 x 1 2 1 x 1 4 4 L 4. f ( x) g( x) x→ lim x→ Thus f ( x) grows faster than g( x) by definition. f ( x) (b) lim x→ g( x) f (x) lim g(x) x→ L 1 cos x 1 Investigating 0 1 0 dx x f (x) x→ g(x) 1 lim c 0. 10. lim x→ dx x lim ln x c→0 C 1 c x2 1. 1 x2 x2 1 x for x 1 4 3 4 3 1 x 4e x x x→ 3e 4e x x x→ 3e 5 7 lim f (x) g(x) lim x→ lim ( ln c) c→0 C 3 lim lim x→ Thus f and g grow at the same rate as x→ . dx xp 1 has an infinite discontinuity at x xp c→0 1 9. lim (pp. 433–444) 2. 1 x 2 so 1 1. Because dx x2 2 x2 1 x2 s Section 8.3 Improper Integrals 1 ln 2 2 1 1, so cos x cos x cos x x2 8. x 2 definition. ln 2 6. x 1 0 for x 1 The domain is (1, ). 7. Thus f ( x) grows at the same rate as g(x) by Exploration 1 1 1 x tan 1 C 2 2 13 x 4 dx x 3 dx x4 ln 3 1 5. 9 x 2 0 for 3 x The domain is ( 3, 3). f (x) g(x) p ln 6 1 ln x 2 2 dx 2 3 1 x 2 tan 1 4 2 Thus f grows at the same rate as g as x→ by 46. (a) lim 0 1 1 0 x dx 2. 3. definition. 1 ln 2 2 p1 3 dx x 0 lim f (x) lim x→ g(x) 0. dx xp 1 p 1c 1 cp1 p1 c→0 Thus f grows faster than g as x→ by definition. f(x) 1) Quick Review 8.3 O(h ). Thus as h→0, ET→0. f (x) g(x) x lim 3 M 12 c lim 2 lim g(x) lim c→0 a2 hM because ( p 1, then 1 dx xp 1 where M is a bound for the absolute value of f on [a, b]. Thus p O(h 4). Thus as h→0, ES→0. ET p1 1 p 1c 1 cp1 p1 c→0 M for x sufficiently dx xp x lim close to a. ES c lim x→ 2x 1 x 3 2x x 1 3 2 lim x→ 1 1 x 3 x 2 Section 8.3 Section 8.3 Exercises 1. (a) The integral is improper because of an infinite limit of integration. x2 0 lim b→ 1 0 4. (a) The integral is improper because of two infinite limits of integration. 0 2x dx 2x dx 2x dx 2 1)2 (x2 1)2 (x 2 1)2 0 (x 0 2x dx 2x dx lim 2 1)2 (x 2 1)2 b→ b (x (b) b dx (b) dx 0 x2 1 1 lim tan b→ b x 1 lim (tan b (x 2 lim 0 b→ b→ 0) b b→ 2x dx (x 2 1)2 0 The integral converges. b 1) 2x dx (x 2 1)2 lim b→ 0 lim ( x 2 1) 1 lim [ (b2 1) 2x dx (x 2 1)2 1 0 b→ 2. (a) The integral is improper because the integrand has an infinite discontinuity at x 0. 1 b 1 b→ 2 0 1 1) (b 2 lim [ 1 2 (c) 335 1 1 1] 1 0 The integral converges. 1 dx (b) 0 x 1 dx lim b→0 b (c) 0 x 5. (a) The integral is improper because the integrand has an infinite discontinuity at 0. 1 lim b→0 2 x b lim (2 2 b→0 b) ln 2 2 (b) x 2 1/x e ln 2 dx 0 x b→0 b e1/x lim The integral converges. b→0 2 1/x e b 1 (b) 8 0 8 0 dx x1/3 dx x1/3 8 b b→0 b→0 lim b→0 0 dx x1/3 0 lim lim 1 1 dx x1/3 8 lim b→0 lim b→0 lim b→0 b (c) No value dx x1/3 (b) b 8 dx x1/3 d b→0 6 6 3 2 /2 b lim b→0 dx x1/3 3 2/3 x 2 cot b lim b→0 b→0 1 cos d sin /2 b ln sin b ) The integral diverges. b (c) No value 3 2/3 b 2 dx 7. 1.001 x b lim b→ lim 3 2 d ln sin lim (0 1 6 lim 8 The integral converges. 9 2 /2 cot 0 3 2 1 6. (a) The integral is improper because the integrand has an infinite discontinuity at x 0. /2 3 2/3 b 2 1 The integral diverges. dx x1/3 3 2/3 x 2 e1/b] b→0 3. (a) The integral involves improper integrals because the integrand has an infinite discontinuity at x 0. dx ln 2 lim [ e1/ln 2 (c) 2 (c) lim 9 2 b→ 1 dx x1.001 1000 x lim ( 1000b b→ 0.001 b 1 0.001 1000) 1000 336 Section 8.3 1 0 dx x 2/3 8. 1 0 1 0 b dx x 2/3 1 1 dx x 2/3 1 2 b→ 1 b→0 b lim 1 4 0 4 b 2 b→4 0 dr lim b→0 r b lim b→0 1 11. 0 1 x 2 b 4) r 0.001 1000b0.001) x 1 sin 1 b→1 lim b→ lim b→ lim b→ 4 b tan 4 2 2 tan 3 4 2 b s 4 s 2 ds b b→2 1b b b 1 2b) 3 0 b→2 0 lim 4 b→2 4 2 s2 1 sin 2 lim 4 b→2 sin 1 1 2 1 s 1 s2 4 s lim s lim b 3 2 2 0 2 2 1 2 b→0 16. (1/2) dx (x/2)2 1 1x 2) d 2 2 lim ( 0) 2 2 1 (2 b 1 2 3 ln 2 b→0 x b 3 ln b lim b 0 1 1 3 ln 2 b→0 2 2 b lim 2 3 ln t b t 1 1 2 0 dx 0 1 1 2 3 ln 1 1000 15. lim (sin 12. 1 b→ 1 lim 2 dx x2 4 t lim 3 ln 1 b b→1 2 1 b→ 4 dr lim b b b 1000r b→1 b 3 dt t 1 lim 3 ln t 0.999 b dx b t 2 b→ b→0 2 ln 3 3 b→ r 4 lim (1000 1 (t 1)] dt t(t 1) 2 lim 3 ln 1 0.999 3[t lim 0 b→4 ln 1 lim r 4 lim ( 2 10. ln b→ dr 4 lim 1 3 dt t2 t 14. 0 ln x 3 6 b→4 ln 3 ln 3 3b1/3) lim r dx 1 b b→ 2 dr 9. 3 1 1 lim 1 b 3 x ln x b→ 3x b→0 dx x 2/3 x 1 3 dx x 2/3 lim (3 1)] dx 1) 1 1 ln x b→ 3) 1/3 b→0 1 1 x b lim 1 lim (x 1)(x 2 b→0 b 1) (x lim lim (3b1/3 dx x 2/3 [(x b lim b→0 1 2 2 dx lim x 2 1 b→ b 3x1/3 lim 13. dx x 2/3 lim b→0 2 dx x 2/3 b2 2 sin 2 (s/2)2 1s 2 1b 2 2 4 ds b 0 2 s2 ds Section 8.3 dx 17. First integrate (1 x) 20. 1 x b d 5 2 b→ 6 dx. b→ (1 x) 1 2 tan 1 2 2 ln lim ln 2 3 2 3 ln C x C dx 0 x) x 1 b b b (1 lim 2 tan lim (2 tan 1 b→0 b→0 1 lim (2 tan c→ 2 1 x x2 1 x 1) b 16 tan 1 x dx 1 x2 0 1 x2 1 b→ lim 1 b→1 b x x2 1 8 4 x 1 22. 1 2 1 sec b) 0 2 sec b 2 1 lim (sec dx x 3 b→0 3 1 2 x 1 lim ( 2 x 4 b sec x x b b→0 b 1 2) 0 4 dx lim b→0 x 2) b 4 dx b x 2 b→0 6 6 4 b ds s s2 sec 1 lim (sec 1 lim b→1 1 1 2 s b b→1 1 2 2 sec sec 1 1 1 b) 3 dx x 2 x b 4 b→0 2 2 lim sec 1 lim lim 3 b→1 dx lim x dx 1 2 4 2 0 dx 0 b lim ds b)2 b 1 2 1 4 b→0 x x2 b→ 1 1 2 2 dx 1 x dx 1 0 dx 1 1 b 1 b b→ x)2 0 2 1 2 1 sec lim 1 2 x2 x lim sec s s2 0 b→ dx x b→ 1 16 tan 1 x dx 1 x2 lim dx 2 1 dx 2 C lim 8(tan dx sec 19. x)2 b→ lim (sec x x2 1 8(tan C b) b→1 1 8u2 16u du 2 2 tan . 1 1 b→1 x x2 c 1 x2 1 16 tan 1 x dx 1 x2 x c→ b lim 2 x) lim 2 tan 2 tan c (1 dx du x) x lim 8 (tan dx x x2 1 2 tan 1 dx 1 1 x 0 2 18. c→ x 1 lim (1 dx c x) 16 tan 1 x dx by letting u 1 x2 1 2 dx 1 x) x dx lim b→0 dx (1 21. Integrate 3 1 lim ln Now evaluate the improper integral. Note that the integrand is infinite at x 0. (1 d b 1 b→ u b→ 0 3 b 1 1 1 1 b→ 2 tan 2) d 2) lim ln x 2 du 1 u2 x 3) ( 3)( ( 1 lim 2 dx ( lim b 1 x, so du u by letting 4 2 6 2 ln 2 1 x, so 337 338 Section 8.3 23. Integrate e d by parts. x e 25. 0 e x dx dx x e dx 0 u du d 0 ed dv v x e ed 0 e e b→− lim b→ 0 ed lim b→− dx b b lim (1 e b→ 1) b→ 0 b lim 0 lim ( e ed b→ x e ex lim b b dx 0 C e x dx lim e e ed 0 e x dx b→− x eb) b 0 1 0 e lim b→ e be b lim ( 1 b→ Note that lim be b lim b→ lim c→ 1 ec u eb) ce c→ 0 and lim e b dx 1 1 sin 2 sin c 26. Integrate x ln x dx by parts. c ec lim c→ c u 0. c→ e dv 1 dx x v x ln x dx d 2 cos d v e sin d 2e Integrate 2e cos sin 2e cos d lim 2 cos du 2 sin d e v 12 x ln x 2 b→0 1 4 lim d cos d cos 2e sin b→0 d lim b→0 Thus, 2e sin b2 2 27. 2e cos 2e sin b tan 0 sin d lim sin d 2e 2e sin d e sin sin 2e sin d lim b→ 2e e lim b→ 2e cos lim cos C1 The integral diverges. d b lim ( e b→ sin sin e sin b e cos 0 b b cos b sin 1) lim b→0 d ln cos lim [ ln cos b b→ /2 C 0 e 12 b 4 b d b 0 b ln b 1/b 2 lim b→0 b→ /2 0 cos b→ /2 2 2e 1 0. /2 d 2e 12 x 4 1 4 e 2e 12 x 4 12 b ln b 2 Note that lim b 2 ln b 2e 12 x ln x 2 b lim d by parts. dv 1 x dx 2 x ln x dx b→0 b→0 u 12 x 2 1 x ln x 0 2e x dx 12 x ln x 2 du d by parts. dv ln x 1 du 2 1 lim e b→ 24. Integrate 2e x e b 1 0 0] 1/b 2/b 3 C 1 Section 8.3 sin 28. On [0, ], 0 1 , so 1 32. 0 1 x sin 1 d 0 4 0 b 1 d lim b→ 0 1 b lim b→ x 4 b lim 2 x b→ x lim [2 b b→ 4 4] Since this integral diverges, the given integral diverges. d 0 on [4, ) 1 dx x dx d 339 b lim 2 b→ 1 33. 0 x3 0 lim ( 2 b b→ 2 ) 1 1 on [1, x3 1 b dx x3 lim b→ 1 lim 2 12 x 2 2 b→ Since this integral converges, the given integral converges. 2xe x2 dx x2 2xe 0 dx 2xe x2 dx 0 x2 2xe b dx lim b→ 0 x2 2xe e lim [ e dx 2xe x2 lim e x 0 x lim b→1 b→1 b→0 b2 1 dx 2 lim [ 2e 2 35. 1 ln 1 2 x 1 ln 1 2 1 1 ln 2 1 x x 1 1 ln 2 1 b b 0 1 0 b] x dt 0 x t lim b→0 b 2 lim [2 b→0 x lim b→1 b→1 2 0 1 dx b x 0 b 0 0 dx 1 x dx 1 x ln 1 lim ( ln 1 x 0 b 0) Since this integral diverges, the given integral diverges. on (0, ] since sin t 0 on [0, ]. dt lim b→0 x) b t sin t 1 lim The integral converges. t 2 dx b dx 1 0 b→1 1 1 2(1 1 dx 1 2e b→0 1 1 x) b→1 b 2 x) (1 x)] dx x)(1 x) 2(1 b→1 x 2e 2e x2 lim dx x b→0 1/2[(1 (1 0 0 e x b lim dx 1 1 Since this integral diverges, the given integral diverges. e lim 2 lim 1 4 31. 0 0 b→1 4 dx x lim The integral converges. e 1 1 0 2 dx b b b→ 1 2 lim x2 x2 lim [ 1 4 x 2 dx b lim 1 2 b 1] 2xe b→ b→ 30. x 1 2 0 0 dx 0 1 dx 1 b b→ 0 2 34. 0 b2 lim b→ 1 Since this integral converges, the given integral converges. 0 x2 b 12 b 2 b→ 29. dx x3 lim 20 ) t t b 2 b] 2 Since this integral converges, the given integral converges. 340 Section 8.3 1 1 36. ln x dx 2 ln x dx by symmetry of ln x about the 1 0 40. 1 dx 0 y-axis. Integrate ln x dx by parts. 1 dx 0 x dx dx 1 x b lim b→ x x dx 1 x b u ln x dv 1 dx x du ln x dx v lim 2 lim (2 x ln x dx x ln x x 2 lim b→0 0 b→0 1 x ln x x b 2 lim [ 1 b ln b b→0 Note that lim b ln b b] b→0 1/b 1/b 2 b→0 1 42. 0 0. 2 1 on [2, ) x2 b lim b→ 2 2x 1 2b 2 1 dx x lim (ln b b→ dx b 2 2 b lim ln x b→ lim (ln b b→ ln 2) 1 ex e 1 x. ex e x(e2x 1) 1 (e x)2 e x dx Integrate by letting u e x so du 1 (e x)2 dx e x dx x x e e 1 (e x)2 du 1 u2 e x tan 1 u tan 2 1 ex C Since this integral diverges, the given integral diverges. 39. Let f (x) x1 and g(x) x2 1 . Both are continuous on x3/2 [1, ). x1 f (x) lim lim x→ g(x) x→ x→ x b 1 dx lim x 3/2 dx x3/2 1 b→ 1 b→ lim ( 2b b→ 1/2 x dx e ex x 0 x 1 lim b→ tan 1 1 b→ 4 2) 2 Since this integral converges, the given integral converges. 0 dx ex e 0 b x lim b→ x 0 ex 0 4 dx ex e 1 ex lim [tan 1 b b→ tan 1 lim tan 1 e b] tan 1 1] x b 0 b→ 2 dx e ex b b→ b 0 dx e ex b lim 1 1/2 dx e dx ex e C lim [tan 1 lim 2x 1 x 0 ex 0 lim ln ) ) 1 1 ex Since this integral converges, the given integral converges. dx x 2x b→ 43. First rewrite 1 e 1 x b Since this integral converges, the given integral converges. e 1] b→ 38. 0 b lim lim 1 b 2 on [ , x2 b→ b lim [ e sin x x2 b→ e b→ cos x on [ , ) x b dx lim lim ln x x b→ b→ lim 1 1 on [1, ) 1e e b 1 d lim ed e 1 b→ lim 2 2 dx x2 The integral converges. 1 2) Since this integral diverges, the given integral diverges. 2 lim ln b 1/b lim b→0 b 1 x dx x 41. 0 ln x dx b 2 lim 37. 0 1 Since this integral diverges, the given integral diverges. C 1 lim b b→ ln x dx b→0 x b→ x 1 2 dx 4 e 4 Thus, the given integral converges. x e x dx. 341 Section 8.3 dx 44. 1 x4 1 dx 1 0 x4 1 1 on [1, x2 1 u lim 2 x b→ dx b 1 x 1 b lim b→ dv ln x x2 Area 1 1 ln x dx x2 dx x2 v 1 dx x 1 lim 1 1 x ). 1 b→ b→ ln x du b 1 dx x2 1 b lim ln x dx by parts. x2 exists on [0, 1]. x4 1 1 Integrate 1 0 on [1, ). ln x dx x2 1 x4 1 1 exists because x4 0 0, y Area dx x4 47. For x 1 dx 0 1 by symmetry about the y-axis x4 0 1 dx 0 dx 2 x4 ln x x dx x2 ln x x b→ Note that lim b→ C b ln x x lim 1 x 1 x ln b b b→ ln b b lim b→ 1 lim 1/b 1 1 b 1 1 0. Since this integral converges, the given integral converges. 48. For x 45. Integrate (1 dy y )(1 tan 2 1 y) by letting u 1 y2 dy y 2)(1 tan 1 (1 du 1 y) 1 u ln 1 (1 C ln 1 dy y 2)(1 tan u tan 1 b 1 lim y) b→ 0 (1 y C ln x ln x dx lim dx x 1 b→ 1 x ln x dx Integrate dx by letting u ln x so du . x x ln x 12 1 dx u du u C (ln x)2 C x 2 2 b 1 1 Area lim (ln x)2 lim (ln b)2 2 b→ b→ 2 1 Area 49. (a) The integral in Example 1 gives the area of region R. dy y 2)(1 tan 1 y) Area 1 lim ln 1 b→ tan 1 tan 1 lim (ln 1 b→ b 1 x2 y 0) The surface area of the solid is given by the following integral. 2 The integral converges. 2 1 e y dy y2 1 46. 0 lim y→ 0 e y dy y2 1 0 y 1 lim 2 y→ y ey 1 1 x 1 12 dx x2 1 ey 2y y→ lim ey y→ 2 lim 1 x 1 x4 1 on [1, x3 1 x4 dx x4 1 dx x3 2 Since 0 x4 1 x 2 e y dy y2 1 e y dy diverges since y2 1 e y2 dx x (b) Refer to Exploration 2 of Section 7.3. b y 0 ln 1 0 on [1, ). b y so dy du 0 tan 0, y ), the direct comparison test shows that the integral for the surface area diverges. The surface area is . Thus the given integral diverges. (c) Volume 1 12 dx x 1 1 dx x2 b lim b→ lim b→ lim b→ 1 1 dx x2 1 x 1 b b 1 1 (d) Gabriel’s horn has finite volume so it could only hold a finite amount of paint, but it has infinite surface area so it would require an infinite amount of paint to cover itself. 342 Section 8.3 1 50. (a) f (x) x 2/2 e 6, x 2 51. (a) For x x2 6x, so e 6x e 2 e x2 dx 6x e 6 dx 6 b lim , 0]. f is decreasing on [0, ). 1 (b) NINT e 2 1 NINT x 2/2 1, 1 x 2/2 2, 2 1 36 e 6 0.954 3, 3 b→ 0.683 e 2 1 NINT 0, 0.997 , x, , x, x 2/2 e , x, (b) x2 6 dx 4 x2 e 1 1 6 1 e 36 6 17 10 x2 dx 10 dx 17 e 6 x2 e dx 4 1 2 Thus, from part (a) we have shown that the error is (c) Part (b) suggests that as b increases, the integral approaches 1. We can make 1 large enough. Also, we can x2 NINT(e (c) . e dx x2 , x, 1, 6) 0.1394027926 1 (This agrees with Figure 8.16.) b f (x) dx and 10 f (x) dx as close to 1 as b we want by choosing b 17 bounded by 4 b make e 6 1 e 6b 6 lim 1 2 f has a local maximum at (0, f (0)) b 1 6x e 6 b→ f is increasing on ( dx 6 lim [ 3, 3] by [0, 0.5] 6x e b→ f (x) dx as small as we want b by choosing b large enough. This is because 0 f (x) 0 e f (x) e x/2 x/2 x/2 dx b for x 3 dx x2 e e x2 c lim c→ e x/2 dx b lim 2e x/2 lim [ 2e c/2 e e 3x 3 c dx 3. b dx 3x 3 3x for x lim e b→ 3x dx 3 b lim 2e 1 e 3x 3 lim 1 3b e 3 19 e 3 b c→ 2e since x 2 dx dx 0 dx x2 3 3 1) Thus, x/2 e dx 0 1. (Likewise, for x c→ As b→ , 2e x2 0 b e e e f (x) dx b (d) 0.000041 b→ b/2 b/2 b→ →0, so for large enough b, b/2 f (x) dx b 3 19 e 3 0.000042 is as small as we want. Likewise, for large enough b, 52. (a) Since f is even, f ( x) b f (x) dx is as small as we want. f (x). Let u x, du dx. 0 f (x) dx f (x) dx f (x) dx 0 0 f ( u)( 1) du f (x) dx 0 f (u) du f (x) dx 0 0 2 f (x) dx 0 (b) Since f is odd, f ( x) f (x). Let u 0 x, du 0 f (x) dx f (x) dx f (x) dx 0 f ( u)( 1) du f (x) dx 0 f (u) du 0 f (x) dx 0 0 dx Section 8.3 b 2x dx x2 1 53. (a) 0 lim b→ 0 55. Suppose 0 2x dx x2 1 1) b→ b Thus the integral diverges. 0 2x dx 2x dx and must converge in order x2 1 x2 1 2x dx to converge. x2 1 (b) Both 0 for b (c) lim b→ b b 2x dx x2 1 lim ln (x 2 1) b→ lim [ln (b 1) b→ lim 0 2x Note that x2 ln (b If the infinite integral of g converges, then taking the limit in the above inequality as b→ shows that the infinite integral of f is bounded above by the infinite integral of g. Therefore, the infinite integral of f must be finite and it converges. If the infinite integral of f diverges, it must grow to infinity. So taking the limit in the above inequality as b→ shows that the infinite integral of g must also diverge to infinity. 1)] 0: b x dx e x e x lim [ e e b lim b→ 0 b b b→ 2x dx = 0. x2 1 (d) Because the determination of convergence is not made using the method in part (c). In order for the integral to converge, there must be finite areas in both directions (toward and toward ). In this case, there are infinite areas in both directions, but when one computes the integral over an interval [ b, b], there is cancellation which gives 0 as the result. For n 1: ux du dx xe x b dx 0 lim b→ xe y 2/3 xe y 1 x dy dx 3 (1 2 x dx x 1/3 (1 x 2/3)1/2 x2 e 2/3 x 2/3) (1 2/3 (x x b dx lim b→ 1) b→ dy 2 1/3 2/3 x dx 1/3 lim b→0 lim b→0 lim b→0 3 2/3 x 2 Thus, the perimeter is 4 1 x2 e x dx 0 x2 e x b b b b 2 eb 2 3 2/3 b 2 3 2 3 2 6. b→ x 2x e dx 0 0 lim 1 dx e x dx ex 2(1) b→ x 1 2b eb dx dx 0 lim 1/3 b 3 2 e b→ 2 lim b→ x b lim b2 eb lim 1 dx x x 0 dv v lim 1 0 x 0 dy 2 1 For n 2: u x2 du 2x dx 2 1/3 x 3 b e 1 eb lim x 2/3)1/2 b x b eb b→ x 2/3)3/2 (1 dx 0 b→ 2/3 x 0 b→ 0. 1 e x dx ex dv v lim 1, y 1] b→ 54. By symmetry, find the perimeter of one side, say for x b 0 = lim 0 dx 0 lim is an odd function so 1 2 0. b→ a 56. (a) For n b 2 a, g(x) dx. a 1) b→ a. b f (x) 0 lim ln (b 2 g(x) for all x From the properties of integrals, for any b b lim ln (x 2 f (x) 343 b→ xe 0 2 x dx 344 Section 8.4 56. continued c n x (b) Evaluate x e u xn du nx n 1 (b) dx using integration by parts dv e x dx v ex 0 f (x) dx c f (x) dx f (x) dx 0 0 f (x) dx f (x) dx c c f (x) dx 0 Thus, n x xe n dx x xe nx n1 x e dx c f (x) dx f (x) dx c f (n x ne 1) x 0 dx b→ x ne b x nx n 1e 0 lim b→ x n 1e (c) Since f (n f (n x n xe 0 0 nf (n), 1) … f (1) n!; thus dx converges for all integers n 0. 0 sin x , x, 0, x or create a table x 57. (a) On a grapher, plot NINT of values. For large values of x, f (x) appears to approach approximately 1.57. (b) Yes, the integral appears to converge. 1 58. (a) 1 dx 1 x 2 dx lim b→ x2 1 b tan lim (tan 1 b→ 4 1 1 1 lim b→ 1 lim [tan 1 b x b→ 1 b→ 2 1 x2 b) x2 1 3 4 1 dx 1 lim tan dx tan 3 4 2 b x x b b→ 2 1 1 lim dx 0 c f (x) dx c c f (x) dx 0 f (x) dx 0. 0 s Section 8.4 Partial Fractions and Integral Tables (pp. 444–453) Quick Review 8.4 n(n x f (x) dx, because f (x) dx dx 0. 1) 1) f (x) dx c bn eb f (x) dx 0 0 Note: apply L’Hôpital’s Rule n times to show that b→ f (x) dx c 0 dx 0 n x nf (n) bn eb 0 f (x) dx 0 lim lim c f (x) dx 0 4 b 4 4 tan 1 1] 1. Solving the first equation for B yields B 3A Substitute into the second equation. 2A 3( 3A 5) 7 2A 9A 15 7 11A 22 A 2 Substituting A 2 into B 3A 5 gives B solution is A 2, B 1. 5. 1. The 2. Solve by Gaussian elimination. Multiply first equation by 3 and add to second equation. Multiply first equation by 1 and add to third equation. A 2B C 0 7B 5C 1 B 2C 4 Multiply third equation by 7 and add to second equation. A 2B C 0 9C 27 B 2C 4 Solve the second equation for C to get C 3. Solve for B by substituting C 3 into the third equation. B 2(3) 4 B 2 B2 Solve for A by substituting B 2 and C 3 into the first equation. A 2(2) 3 0 A10 A 1 The solution is A 1, B 2, C 3. 2x 1 3. x 2 3x 4 2x 3 5x 2 10x 7 2x 3 6x 2 8x x2 2x 7 x2 3x 4 x3 2x 1 x x2 3 3x 4 Section 8.4 4. x 2 3x 2 5. x x2 3 11x 8x 3x x 6. y 7. 8. 5y 2 3 4 2 x x (x (y (y x 2 3t t2 2 3) 2 3)(x 2 (x 3) 1) 4)(y 1) 2)(y 2)(y A 1)(y x2 3. 2 (x 2)(x A A x 1) B(x 4. s 3 B)x (A x 7. 2 x B 0, B 2, B s2 C(t (B C 1) C)t 1, and 2, C C C 1. 1. 2 t 2 t 1 s(s 2 6s A s 4 s2 A( s 3)(s s 6s (A 1 Solving the system simultaneously yields A 2 1) B)t 2 A( s 2 Equating coefficients of like terms gives 3 C t2 Bt(t At 2 4 2B) x 1)2 (x B t 1 s3 2) 2 1 A t1 t 2(t 1) 1) A( x 7 4 x t 2, B Solving the system simultaneously yields s3 7 3x B) 2. 2 1 1 1 B B C 2 ,B 3 A 2 5x B 1 t2 s 6) s(s B s 2) C)s 2 2) C 3 s 2 B(s)(s B(s 2 6) 3)(s (A 2) 2s) 2B C(s)(s C(s 2 3) 3s) 3C)s 0, A 2B 3C 0, 6A Solving the system simultaneously yields (A x2 B 2 2x t1 t 2(t 1) A 3x 2B 1)2 (x Equating coefficients of like terms gives B 5 and A B 1 6A Equating coefficients of like terms gives A 2 A (A 3 7 B (A t 1 5x A 1) 2x Section 8.4 Exercises 3x A( x 2 and x2 1 (x 1) (x 1) 2(x 1)2 3(x 1) 1 (x 1)3 (x 1)3 (x 1)3 2x 2 4x 2 3x 3 1 (x 1)3 2 2x 7x 6 (x 1)3 5x 2 A Solving the system simultaneously yields A 1) (3t 4)(t 2 2) (t 1)(t 2 1) (t 2 2)(t 2 1) (t 2 2)(t 2 1) 3 2 t t t 1 3t 3 4t 2 6t 8 (t 2 2)(t 2 1) 2t 3 5t 2 5t 9 (t 2 2)(t 2 1) 2t 3 5t 2 5t 9 (t 2 2)(t 2 1) 4 1 3 1. x 2 x 2 Equating coefficients of like terms gives 2 2 x 1 2x 1 2 4x 5 x5 (x 1)(x 5) 2(x 2 4x 5) (x 5)(x 2 4x 5) (x 5)(x 2 4x 5) x 2 4x 5 2x 2 8x 10 (x 5)(x 2 4x 5) x2 12x 15 (x 5)(x 2 4x 5) 1 2 2x 2x x x2 (x x2 3(x 3) 2(x 2) (x 2)(x 3) (x 2)(x 3) 2x 4 3x 9 (x 2)(x 3) x 13 (x 2)(x 3) x 13 x2 x 6 3 3 t 9. 2 t 10. 2 1)2 1 5 2 (x 4 2x 4 4x 3x 2. x 2 2 06 10 04 Ax 5 2x 2 2x 2 4x 345 3, B 2. 4 s2 6s 4 ,C 15 2 3s 2 . 5 4 15(s 3) 2 5(s 2) 4 4. 346 Section 8.4 5. x 2 6 x2 x2 5x x 2 8 x2 x2 5x 6 (x 5x 6 x 5x 2 A( x (A 3)(x x A 2 B(x B 5 and 2A 3B 17, B x2 x 2 x2 0 and 2A 5x 6. y 2 y3 y2 6 4 y3 y3 x 0y 2 0y 2 9. y 2 (1 x2 1 1 1 x)(1 1 B y B( y 1 3) B)y (A 3B) 3B 0. 3 ,B 4 Solving the system simultaneously yields A y dy 2y 3 y 3/4 dy 3 3 ln y 4 1 1/4 dy y1 1 ln y 1 4 3 x) B(1 B)x y2 (A A y 4 A( y 1) 1 . 4 C By (A B) y(y 4 y y x) y y y2 10. x 0 and A B 1. B)y A 1 . 2 1/2 1/2 dx dx 1x 1x 1 1 ln 1 x ln 1 2 2 1 1x ln C 2 1x 1) B y 1 Equating coefficients of like terms gives Solving the system simultaneously yields 1 A ,B 2 dx 1 x2 1) 1 and A Equating coefficients of like terms gives A 3 B x A(1 (A y A( y x) A 1) B 3 B y2 x2 3)(y Equating coefficients of like terms gives The rational function cannot be decomposed any further. 7. 1 1 . 2 C y 4y 1 y2 4 y (y A 2 y 1 0 1 0y 4y 4y 3 y 2y y2 A 1 4 2y 2 (A x 1 ,B 2 1/2 dx 2 x 1 1 ln x ln x 2 2 1 x ln C 2 x2 12 3 1 1/2 dx x 2 17 1 2A 2x 12. 8 B)x dx 3B) Solving the system simultaneously yields A Bx Solving the system simultaneously yields A 3) ( 2A B Equating coefficients of like terms gives A 2 2) Equating coefficients of like terms gives B 3 B x (A 6 2) A 2) A( x 2 B)x 2 A x 1 2) 5x x2 x(x 2x x2 5x 6 2x 1 5x 1 5x x2 8. x 2 1 8 6 2 0x 5x 5x A B 1 and A 4. Solving the system simultaneously yields A x C y y2 4 dy y 4 dy y 4 ln y y 3 dy 1 3 ln y 1 C 4, B 3. Section 8.4 11. t 3 t2 1 t2 t3 1 t(t 2 2t A t 2t A( t 2)(t A( t 2 t (A t 2) t(t B 1) 2 t 1 5x 2 5x 2 x 1 B(t)(t 1) C(t)(t B( t 2 t) C(t 2 2t) C)t 2 (A B 2C)t 2A x 5x 2 x 2 5x 5 x2 x 1 5 1 5 5 5 5 5x 5x 5x 2) 2) B 15. x 2 C t 1) 2)(t 5x 2 dx x x1 5 dx 2 x 5 5x B C 0, A B 2C 0, and 2A 1 1 1 ,B ,C . 2 6 3 1/2 dt dt t t 2 2t 1 ln t 2 t 3 12. 2t 3 2t(t 2 8t t3 2t 3 8t t A t 3 4) 1/6 dt t2 1 ln t 6 t t 2A(t 2)(t 2A(t 2 4) (2A x2 1/3 dt t1 1 ln t 1 3 2 x C 2)(t 2) 2B 2C 2B(t 2 2) 0, 4B 3 ,B 8 3 dt 8t t 2t 3 13. s 2 4 s 3s 3 s3 2C(t 2 2t) (4B 4C)t 2t) 8A 4C 1, 8A s2 s2 5 1 ,C . 16 16 3/8 5/16 dt dt t t2 3 5 ln t ln t 2 8 16 s 4 s3 (x 12 2 1 2 (x x x x2 4 ds 1/16 dt t2 1 ln t 16 2 1)(x (s 2 13 s 3 2 x 2 ln (s ds (A 4) 2s s2 1 1 s 2 1) ds s ln (s 2 1/2 3/2 2x 1 tan 1 3 5 1) tan 2x 1 1 3 2x 1) (x 3 C 1) 1)2 2 x 2x x C (x 1 1)2 1)(x 1) B)x 2 B 1 B(x 1)2 1)(x A 2x A( x A( x 4 ( 3/2)2 x 1 tan 3 1)(x 2 (x 2 1/2)2 3 1 1) x2 s2 2 dx 3/4 dx C 16. (x s ds 1 1 2 1) 5 ln (x 2 2 5x 1) C(x 1) 2 1) C(x 1) A B B(x (2A C)x C Equating coefficients of like terms gives s2 1 4 3 2 1 s + 0s + 0s 2s + 0 s 4 + s3 s 2+ s + 0 s 2 2s + 0 s2 + s 1 2s 1 s 4 2s s2 s2 1 s 4 2s ds s2 1 3 4 1/2 1/2)2 5x 2 dx x1 4s s2 4 12 s 2 14. s 2 x 1/2 dx 1/2)2 3/4 The second integral was evaluated by using Formula 16 from the Brief Table of Integrals. 3 s 4s 4s 4s 4s 3 4 dx 1 ln x 2 2) 12 2 x x1 dx 1/2)2 3/4 x2 s3 1 3 4 1 ln (x 2 2 2C(t)(t Solving the system simultaneously yields A x 1 4 x 1 ln (x 2 2 2B(t)(t dx 1 (x 2) 2 2C)t 2 2B x2 1 1 x2 Equating coefficients of like terms gives 2A x C 2 x (x 2t(t B 1 x2 dx 1 To evaluate the second integral, complete the square in the denominator. Solving the system simultaneously yields A x x 5 1. 1 x2 Equating coefficients of like terms gives A A B 1, 2A C 0, and A B C 0. Solving the system simultaneously yields 1 ,B 4 A 1 2s s2 2s ds s2 1 1) 347 s2 1 ds s tan 1 2 1 C 1 . 2 x 2 dx 1)(x 2 2x (x 1 x 1 s 3 ,C 4 1/4 dx 1 1 ln x 4 1 1) x 3/4 dx 1 3 ln x 4 (x 1 1/2 dx 1)2 1 2(x 1) C C 348 Section 8.4 17. (x 2 1)2 1)2(x (x 1 1)2 A (x 2 1)2 21. x 3 B C 1)2 1)(x 1)2 B(x 1)2 1)(x 2 x2 2x 2 A 1)2 x A(x A(x 3 1 x2 C)x x2 x 2B B C 2x 1)2 x3 1)2(x x2 1) C 2D)x 2x C A 0, 2B A B C C 2D D) x (A B D 18. x 2 5x x x2 x 6 4 5x 6 4 1 ,D 4 C 1, B 2 6)(x A x B C D 1 A( x 1) 2 1 dx B)x 1 (x A 4x 4 x3 x2 1 4x r 2 2 dr 2r A 1, r2 4r 5 3 dr 4r 5 1) C x C x 1 1) B)x 2 A 3, B x2 B (Bx (A C 2, C 4x x3 C)(x B 1) C)x 4, and A 4 (A C) C 4. 1 1. 3 dx x 1 dx 3 ln x 3x 2 2x 12 (x 2 4)2 3x 2 4. 2x Ax x2 x2 1 2x 1 dx x1 ln (x 2 x 1) B)(x 2 4) (Cx D) Bx 2 (4A C)x 4B C D 4)2 (Ax 12 Cx (x 2 Ax 3 23. B 4 D Equating coefficients of like terms gives 5/7 dx x1 5 ln x 1 7 6 A C 0, B 2r 1 2 dr (r 1)2 1 2 tan 1 1)2 (r 1 (r A 0, B r2 4r (r 3 dr 2)2 4 1 1 2)2 (r 3 tan 1 (r 2 (x 2 1 1) 3, 4A C 2, and 4B D 12 Solving the system simultaneously yields C 3, C 2x 12 4)2 dx 2, D 0. 3 x2 4 dx 3 x tan 1 2 2 2x dx (x 2 4)2 1 C x2 4 The first integral was evaluated by using Formula 16 from the Brief Table of Integrals. 20. Complete the square in the denominator. r2 x 1) x2 A(x 2 4 B A 2/7 dx x6 r2 2 2. Equating coefficients of like terms gives 3x 2 C) Solving the system simultaneously yields 19. Complete the square in the denominator. 2r x 1 6B) 2 ln x 7 r2 (A C ln (x 2 Bx x Solving the system simultaneously yields 2 5 A ,B . 7 7 x4 dx x 2 5x 6 C)x 1) 2x 1 dx x2 x 1 1 A x 6) 6B B 1 dx 1 1)(x 2 1 . 4 (A (A C)(x 2, and A x 2 1 B(x (Bx 1. 22. x 3 1) x 1 1) C ln x B 6 x 2, C 2x x3 1/4 1/4 1/4 dx dx dx (x 1)2 x1 (x 1)2 1 1 1 ln x 1 C 4 4(x 1) 4(x 1) (x 1 and 1, A A D) Equating coefficients of like terms gives B B x B C x B)x 2 (A (A A A( x 2 2 0, 0, and A x2 1 Solving the system simultaneously yields Solving the system simultaneously yields 1 1 A ,B ,C 4 4 dx (x 2 1)2 1/4 dx x1 1 ln x 1 4 x 2x A 1) 2 Equating coefficients of like terms gives A 1 1) Equating coefficients of like terms gives 1) D(x 2 1) (A (A (x C(x B(x 2 1) 3 1 x Bx (A x C(x 3 (A (x D x D(x 1 (x 1 1 2) C Section 8.4 24. x 3 2x 2 2 (x 2 1)2 x3 2x 2 Ax (x 2 Cx (x 2 B 1) (Ax B)(x 2 Ax 3 2 Bx 2 D 1)2 1) (Cx (A 1 dy y2 y 1 dy y(y 1) 1 y(y 1) 27. C)x D) (B D) Equating coefficients of like terms gives A 1, B 2, A C 0, B D e x dx e x dx A y 1 ex B y 1 A( y 1) (A 2 C B(y) B)y A Solving the system simultaneously yields Equating coefficients of like terms gives A A 1, B 2, C 1, D 0. x 3 2x 2 2 dx (x 2 1)2 x x2 1 ln (x 2 2 25. x (x 2 1)2 2 x2 1) 1 y(y dx dx 1 2 tan (x 2 1 1 1 2(x 2 1) x 1 1 d 0 0 1 1 2 0 1 1 1 y ex 1 1, B 1. dy 1 C2 C 2. C or C ln y 1 1 ln 2 dy ex 1 sin 1 ln 2. d 1)2 1 dy sin d ( y 1)2 1 cos C y1 Substitute x 0 1 0 2 ,y 0. C or C 1 The solution to the initial value problem is 1 y 2 1 d 2 d 0 2 0 2 tan 0 2 1 1 1 1 tan 2 1 1 cos 1 1 y d cos 1 1 2 x2 dx 3x 2 3x 2 (x 29. dy 0 1 cos 1 y 1 2 1 1 ln y 1 0, y 0 (y ln 2 2 2 ln 2 1 1 +1 1 1 2 2 ln y Substitute x C d 1 0 1 2 ln y 1 ln 1 dx 28. d 2 1 dy y dy 1) ln y 1 0 1 The solution to the initial value problem is 1 2 1)2 1 1 A ln y x 1 +1 1 1 1 0 and 1 dx 1 B Solving the system simultaneously yields A 2 dx 1 x x2 26. 349 x 2 2)(x x2 1 3x 1 A( x 1) B(x 1 (A B)x A 1) A 2 B x 2 x 1 2) 2B Equating coefficients of like terms gives A B 0, A 2B 1 Solving the system simultaneously yields A dy y dx 3x x2 ln x 2 Substitute x 0 0 dx 2 x 2 ln x 1 C 3, y ln 2 dx x 1 0. C or C ln 2 The solution to the initial value problem is y ln x 2 ln x 1 ln 2 1, B 1. 350 Section 8.4 ds 2 dt t 2 2t 1 ds 2s 1 2t t(t 2) 2t A t t 1 A( t 2) Bt Let u 1 30. (A B)t 2A with x 2s 2 ds 2s t2 1 t2 32. (a) Complete the square in the denominator. x2 C1 B 0 and 2A 2t 1 ,B 2 1/2 dt 2 t ln s 1 ln t ln t Substitute t 1, s 0 ln 3 1 ln s 2 C or C s ln t 1 ln 2 1 t ln t t 6t 2 ln 3 2 ln 6 . x with x (x 2 1 (x 2 4x (x 2)2 2 so du B 0.5 5 (b) x2 x d1 ln x dx 2a a a 1 1 2a x a 1 2a (x a 1 x2 C C a a (x 1 a2 3 3 1 5 ln x xa a)(x a) a2 x2 dx x2 3x 0.5 x x) Bx B)x 0 and 3A 3A 1 2.5 dx 3x x 2 3 0.5 3 1 x x 9 2.5 dx x 0.5 dx 3 x 2.5 C 1d ln x 2a dx 2.5 dx 2 Solving the system simultaneously yields A 9 du 9 u2 1 u ln 6 u 1 x ln 6 x xa a)(x a) 1 . 3 x 2)2 B 3 A(3 2.5 3. 1 ,B 3 1 Equating coefficients of like terms gives A dx, and then use Formula 18 x2 x) (A 4) C 1 1/a 2a 3 1 (x/a)2 (a 2 3x A x x2 C 1 a2 9 x(3 4x) u and a dx 4x 2 1 3x 9 x x2 x 2)2 33. Volume 1) a2 1 2a 4 a 2 x 2 2a 2 1 2a 2 (a 2 x 2)2 3x 5 1 x tan 1 2a 3 a x 2) 0.5 9 Let u (1 2.5 ln 6 2 x2 4x dx and then use Formula 17 du u 2)2 1 u tan 1 u C 2 2(1 u 2) x1 1 tan 1 (x 2 2(x 2 2x 2) 2)2 1 a2 2a 2 (a 2 31. (a) Complete the square in the denominator. 5 1)2 1. a2 x2 2a 2(a 2 x 2)2 C 6t ln 1 1 1 (a 2 x 2) x(2x)] (a 2 x 2)2 2a 2 The solution to the initial value problem is ln s u and a x d dx 2a 2(a 2 1. ln 2 (x 1 so du 1 . 2 (b) 1 1 ln t ln t 2 C2 2 2 1 1 1 ln t ln t 2 C3 2 2 1 ln s 2 dx 2x (x 2 x 1 1/2 dt t dt 1)2 1) 2 Solving the system simultaneously yields A t2 2 B Equating coefficients of like terms gives A 2x 2x 1 1 (x 2 (x 1 ln s 2 ln x 2.5 ln 3 x 0.5 3 (ln 2.5 ln 0.5 3 ln 25 6 ln 5 0.5 ln 0.5 ln 2.5) 351 Section 8.4 1 34. Volume 2x 2 1)(2 (x 0 1 4 0 x 1)(2 (x x) B 2x A x) x x 1 A(2 x) (A 5 sin , dt 25 x dx 1)(2 (x 36. t dx x) 25 25 sin t 2 dt 25 B(x B)x t 2 1) (2A B 1 and 2A B B) 1 x dx 1)(2 (x 0 4 3 1 0 1 0 ln x 35. y y2 9 25 sin 2 C cos C 25 t 25 t 25 2 25 5 t2 Use Figure 8.18(b) from the text with a 0 get 3 sec2 d, 9 tan2 t2 C C 5 and x t to x 4 ln 2 3 2 ln 2) 2 t2 25 7 sec , dx 2 37. x 2 9 sec2 3 sec y 2 4x 2 5 cos . 7 sec 2 49 sec 2 49 d dx 3 sec sec ln 4x 2 d ln sec tan 49 d ,0 2 49 tan 2 tan 1 2x ln 2 7 4x 2 7 1 ln 2x 2 4x 2 49 1 ln 7 2 1 ln 2x 2 4x 2 49 C C1 9 y2 y ln 3 ln 9 y2 y C Use Formula 88 for sec d with x Figure 8.18(a) from the text with a y2 3 sec . d 1 ln sec 2 y 3 3 7/2 sec tan 7 tan 1 sec 2 y2 9 49 C1 tan ln 9 25 sin 2 4 2 dy 9 cos 2 d 2 1 2 ln 2 0 3 tan , dy 9 2 dx 2x 1 4 ( ln 2 3 1 d 25 2 2 . 3 2 (5 cos ) d 25 t sin 1 2 5 25 t sin 1 2 5 1 4 3 1 ,B 3 2 25 cos 25 2 0 x) dx x1 2 2 5 cos 25 Solving the system simultaneously yields A 4 d, 25 cos 2 Equating coefficients of like terms gives A 5 cos C1 and a 1. Use 3 to get d C1 49 Use Figure 8.18(c) from the text with a x2 49 4 7 tan . 2 C1 7 to get 2 C1 352 Section 8.4 38. x sec 2 tan , dx x2 tan 2 1 x2 2 40. x 2 x sec 2 1 sec , dx 2 1 1 sec sec 2 x3 x2 tan 2 cos 2 1 2 1 csc cos 2 d 1 sin 2 2 C x2 1 sec Use Figure 8.18(a) from the text with a 39. x 1 sin , dx x2 1 1 . 41. z cos d, sin 2 2 2 3 cos sin 3 (1 d d cos 2 ) sin d 1 x x2 1 3 C 1 x2 1 (1 3 2 C x) 1 x2 3 Use Figure 18.8(b) from the text with a 1 x2 cos . C C 1 4 ln 4 sin ) d cot 4 cos 4 z 16 z z2 4 16 z z2 from the text with a 4 2 sin , dw w2 4 4 z2 16 4 4 z2 16 4 sin 2 z2 C . Use Figure 8.18(b) 16 4 and cos d, 2 z2 . 2 4 cos 2 8(2 cos ) d 4 sin 2 2 cos w2 2 csc 2 d 2 cot 2 4 w C w 2 C Use Figure 8.18(b) from the text with a get cot C 4 to get 2 cos 8 dw w2 C 1 and x 16 z 4 , cot z 42. w 16 cos 2 d Use Formula 89 with a csc 2 d 4 sin2 sin 4 ln d ,0 C (4 cos ) d 4 sin 4 ln csc 2 x2 1 1 x x x2 1 C x2 16 sin 2 (4 csc 1 cos 2 3 1 2 4 4 cos C dz 4 cos 2 sin 1 cos 3 3 cos z2 16 z cos x 16 4 cos cos to get z2 16 cos x2 1 C x 4 sin , dz cos 2 sin x 3 dx 1 to get 1 sec C x x2 d C cos sin x d d (sin ) 1 2 2 2d sec 2 (sin ) csc d ,0 2 sec tan d sec 3 tan 1 sec d tan 2 cos sin 2 tan 1 2 dx sec 2 d tan 2 sec dx x2 d, 4 w w 2 . 2 and x w to Section 8.4 x x x 45. For x dx 43. dy 2 9 9 9 sec 3 sec 2 tan 9 d ,0 9 tan 3 sin , dx 3 cos d ,0 9 x 2 9 3 d 9 0, x 2 9 cos /2 dx 0 /2 0 3 cos2 x2 3 x 3 ln Substitute x 5, y 5 ln 3 ln 3 9 C 1 cos 2 d 2 /2 sin 2 4 0 2 C or C 0 1 0 x x2 1 2 x sec 2 tan , dx 1 2.356 2 46. Volume 4 3 d 1 3 4 ln 3. dx 2 4 0 d ,0 4 The solution to the initial value problem is y x2 3 x 3 ln 9 1 44. (x dy x x2 1) 2 dy x dx dx (x 2 1)3/2 tan , dx 1 1 0 0, cos d, 2 2 x C 0, y 1 0 47. (a) 4 1. dx x(1000 1 x(1000 1 The solution to the initial value problem is x 1. 1) 1, 4 . d 8 x) x) B)x 1 2 1 dt 250 A B x 1000 x) A(1000 (A cos 2 d 2 /4 sin 2 4 0 1 4 1 4 8 C x2 cos 2 1 2 1 Substitute x x d /4 sec 2 1 sec 2 sec 4 0 C x2 (since 0 0 Volume y /4 dx (1 x 2)2 d sin 1 0 and when x dx (x 2 1)3/2 2 sec d sec 3 y dx x 2)2 sec 2 /4 sec 2 tan 2 1 2 tan 2 (1 . When x 2 . d /2 3 2 3 cos 0 C 3, 3 cos 3 3 tan 2 0 and when x 2 3 0 9 sin d ln sec 2 2 When x x2 9 3 sec tan 3 tan sec x dx 3 0 2 x2 9 Area dx y 0 on [0, 3] 3 3 sec , dx 2 0, y 353 8.076 x Bx 1000A Equating the coefficients and solving for A and B gives 1 A 1 ,B 1000 dx x(1000 x) 1 1000 (1/1000) dx (1/1000) dx x 1000 x 1 1 ln x ln (1000 x) 1000 1000 1 x ln C1 1000 1000 x t C2 x 250 1 x ln 1000 1000 x ln 4t C 1000 x x e 4t C Ae 4t 1000 x C1 354 Section 8.4 47. continued When t 49. (a) From the figure, tan 0, x 2 998 (b) 500 4t 499e cos x) sin x cos x)2 sin 2 x z 2(1 cos x)2 1 z 2(1 cos x)2 1 cos x 1 ds 1 sin 2 x 1 cos 2 x 1 x(1000 (1 (1 500 which occurs when z 2)2 z 2)2 z 2)2 (1 z 2)2 (1 z 2)2 (1 x) is greatest. 1 2z 2 (1 Only sin x (d) z 1 dx for 0 1 4z 2 z 2 )2 A 1 x where A x x2 1 dx 1 dz 1 1 2 tan 2 1 (1 2 z 2) dx 1 dx 2 x2 1 B x1 1 and B 2 dz 1 1 1 2 1 dx 1 1 1/2 1 1 x 0 x 1 1 ln x 1 1 2 ln 1 2 ln ln 3 1 2 x ln x 3 2 1 dx 1/2 1 0 1 1 x sec 2 dx 2 2 1 2 dz z2 z2 makes sense in this case. x 2 1 2 x x2 x2 0 1 z2 dz 1/2 Arc length 1 1 tan 2z 1 z 4 1 2z 2 (1 z 2)2 2z sin x 2x 2 dx 1 x2 x2 1 dx 1 x2 x2 x2 2 z2 z2 z2 1 1 x2 1 x2 x2 1 1 does not make sense in this case. 2x x2 x2 0 z2 z2 1 1 cos x 0 1 1.553 x4 2x2 1 (1 x 2)2 0 1) cos x 1.553 as shown in part (b). dy 48. dx 1/2 (z 2 or cos x) (c) From part (b), cos x This occurs when x x2 1 cos x) cos x 4t dx 1 (c) x(1000 x) dt 250 dx will be greatest when y dt x2 cos x)(1 z 2cos x 0 Half the population will have heard the rumor in about 1.553 days. t (1 cos x)(z 2 (1 cos 2 x 2 1 1 ln 4 499 t sin x cos x z 2(1 1 499 4t e z(1 1 cos x 1000 1 499e 1 (b) From part (a), z 1 499 1 4t e x 499 1000 4t x 4t x e e 499 499 4t 4t 1000e e 499 499 1000e 4t x 499 e 4t 1000 or x 1 499e 4t x1 sin x . cos x 1 2. A or A x 1000 x 2 x dx 2 z4 355 Chapter 8 Review 50. s Chapter 8 Review Exercises 2 dz 1 z2 dx 1 sin x (pp. 454–455) 2z 1 z2 1 z2 (z 1. lim 2 dz 2z 2 dz 1)2 1 2 z 2 3. lim x→0 1 1 C x cos x sin x sin x x sin x cos x lim cos x x→0 x sin x cos x lim x→0 x1/(1 cos x 2 2 dz 2z z lim x 1 e 1 e 5. The limit leads to the indeterminate form f (x) . ln x x ln x x lim x→ 1/x 1/x 1 0 ln f (x) lim x C 0 x1/x lim e0 lim e x→ 2 tan lim e x→1 ln f (x) 1 1 ln f (x) x→1 1 2 x) ln x 1x 1/x lim x→1 1 ln x lim 2 dz 1 z2 2z 1 1 z2 x→ 1 C 1 2 2 6. The limit leads to the indeterminate form 1 . C tan 2 1 cos t for t→0 and 3 5 2 t→0 5 sec 5t x x→1 1 1/(1 x) x→ 1 3 sec2 3t lim ln f (x) C x tan 2 2 dz (z 1)2 53. 2t 2t t→0 f (x) z2 dt sin t lim 4. The limit leads to the indeterminate form 1 . 1 z 1 tan 3t t→0 tan 5t dz z2 52. 1 The limit does not exist. C 1 C x 1 2 2 dz 1 z2 1 z2 1 1 z2 d 1 sin 2 1 2t) for t→0 tan dx 1 cos x ln (1 t2 t→0 2. lim 51. t 2 dz z2 1 z z2 z2 1 1 2z 1 f (x) 2 3x x 1 ln f (x) x ln 1 1 x dz z 1 lim ln z 3 x ln 1 1 t ln tan 2 C 1 C lim 1 x→ 7. lim r→ 8. lim → /2 lim 1 x x→ x→ 3x x cos r ln r 1 3/x 2 3/x 1 x2 lim eln f (x) x→ 0 since cos r 2 sec 3 x ln 1 3 x lim lim x→ 3x x 3 3 e3 1 and ln r→ as r→ . → /2 cos 2 lim → /2 1 sin 1 356 Chapter 8 Review 1 9. lim x→1 x 1 ln x 1 ln x x 1 (x 1) ln x lim x→1 1 x lim x 1 x→1 x x→ ln x 1 2 ln x 19. lim x→ 0 10. The limit leads to the indeterminate form ln f (x) lim x→0 . f (x) x→ g(x) 20. lim ln (1 1/x) 1/x 1x x lim x→ 1 x x x→ csc 1 x 1/x x→ lim 1 /x 2 1 1/x lim x→ lim 1/x 2 x→0 x→0 lim eln f (x) e0 x→0 lim x2 1 lim x x 0 1 x lim x2 1 x2 x2 1 lim 1 1 x2 x→ x→ x→ 11. The limit leads to the indeterminate form 0 . x→0 since x and lim x 2 lim 1 1/x 2 1 1 f grows at the same rate as g. (tan ) ln f ( ) 1 tan x ln (1 1/x) 1/x 0 f( ) lim 1 1 x x ln 1 lim 1 1 f grows faster than g. 1x x x→0 1 1/x 2 1 ex x→ 100 lim f (x) g(x) lim tan 1 lim x→ 1/x f grows faster than g. x→ f (x) x x x/100 x x→ xe f (x) x→ g(x) x x ln x x/x lim x→ 18. lim 1 lim x→1 1 f (x) g(x) f grows at the same rate as g. 1 1 1 lim x→1 x 17. lim ln (tan ) 1/ f (x) x→ g(x) ln (tan ) 1/ ln (tan ) 21. lim lim x ln x lim x→ lim x (ln x)(1 2 x→0 (ln x)/ln 2 x→ 2 lim log2x lim xln x sec2 tan 1 x→0 xln x log x x→ x 2 lim sin x→ cos lim x→ 2 lim sin2 x→0 cos2 Note that 1 0 1/ln 2) 1 (ln x)(1/ln 2 x 1 ln 2 1) 0 0 since ln 2 1. f grows slower than g. ln f ( ) lim (tan ) lim e x→0 12. lim → 2 x 3 3x 2 2 x x→ 2x 1 3 3x 2 x 4 x3 x→ x 1 2 14. lim x→ f (x) g(x) lim x→ 1 f (x) x→ g(x) 22. lim sin t lim 2 t→0 t 13. lim 15. lim e x→0 1 sin 0 cos t lim 2t t→0 3x 2 4x x→ 6x 1 6x 3 x→ 4x 1 3x 2 lim lim x 5x lim x→ lim x→ 3 x→ 2 lim 6x 6 23. lim 4 x→ 6 12x 2 6x 0 lim x→ lim x→ log2 x log3 x f (x) x→ g(x) lim x→ lim lim x→ x→ lim f grows at the same rate as g. (ln x)/(ln 2) (ln x)/(ln 3) ln 3 ln 2 ln 2x ln x 2 lim x→ lim x→ 2x 3 ln x ln 2 2 ln x f grows at the same rate as g. 24. lim 1 5 f (x) g(x) x→ f (x) x→ g(x) 2x x x→ 3 lim x 0 since 2 3 1/x 2/x 1 2 f grows slower than g. f grows at the same rate as g. 16. lim x lim x→ 10x 3 2x 2 ex 30x2 4x ex 60x 4 ex 60 0 ex f grows slower than g. lim x→ 1. Chapter 8 Review f (x) x→ g(x) 25. lim 1 (1/x) 1/x tan lim x→ 1 1 lim 34. lim 2( 1 (1/x) ) 1 2 x→ 2 (x 2 x lim 2 sec 1 x 1 x→ 28. (a) lim f (x) sin x 1 ex x→0 (b) Define f (0) lim x→0 1 (ln 2)(cos x)2 ex ln 2. ln 2 39. x 9 ln x x→0 1/x 1/x lim 1/x 2 x→0 lim ( x) 2 2 3 cos x 2 0 sin 1 x2 lim 1 d 1 3 1 0 x 3 b dx 9 1 lim x 2 b→3 0 C dx x2 9 True 1 x2 30. lim x→ 1 x4 1 x4 lim lim (x 2 b→3 1) x→ lim False 31. lim x→ x b→3 x ln x 1 lim x→ 1 x 1 1 0 lim 1 ln x False 32. lim x→ ln (ln x) ln x lim 1 x ln x x→ 1 x True tan 1 x 1 x→ 33. lim True 2 2 2 C 0. x→ d, 3 cos 1 x2 1 d x→0 2 4 29. lim x 1 x x→ 1 3 cos dx lim 1 1 x 1 x 3 sin , dx x→0 (b) Define f (0) 0 1 True lim x ln x x→0 lim x→ 38. lim 2 lim lim True sin x x→0 0 1 (1/x)2 1 f grows faster than g. 27. (a) lim f (x) x2 1 x x→ ln 2x 37. lim x→ ln x 3 2x x→ 1 lim ) (1/x ) x→ 1 True 1 1 1 x2 ln x 36. lim 1 x→ x lim lim x→ 0 1 True sin 1 (1/x) (1/x 2) x→ f (x) x→ g(x) 1 x4 x2 x→ 1 x4 35. lim f grows at the same rate as g. 26. lim 1 lim 1 x4 1 x2 True 1 x→ x→ 2 x lim 2 x (1/x) x→ 1 x4 x→ 0 sin sin 1x 3 b 0 1b 3 2 2 357 358 Chapter 8 Review 40. u ln x dv ln x dx 43. 3 1 dx x du dx v x ln x x dx 1 x lim ln x dx b→0 1 x ln x b→0 b ln b 1 1 dy y 2/3 0 b→0 1 lim 3y 1/3 3 3x dy 3 b→ 0 where A 3 1 6 1 d 2 1 2 1)3/5 ( 2 1)3/5 1 B x 2 5 ( 2 lim 5 (b 2 b→ 1 0 1 0 d 1)3/5 ( lim b→ 1 lim b→ 1 lim b→ 1 0 2 d ( 1)3/5 5 2 1) 4B)x 2 1 dx x2 3x 4x 3 ln 3 C(4x (B 1, C 1) 4C)x C 1)3/5 ( b b→ b 1)2/5 ln 2 1) 5 2 2/5 5 2 du 5 ( 2 1)2/5 0 u x 1 dv dx 2x x e v x 2e b 5 2 5 (b 2 5 2 5 2 du 2 dx 0 1)2/5 x 2e x x x e v e x 2e 0 x dx 1 dx x x 2e x 2x e x x 2x e x b x 2e x lim x 2e x lim b2 eb lim b→ b→ b→ 1 dx x 2e dx 1 dx 2x e dv 3 4 b x e x ln 3 ln 1 dx x2 1 x ln x 1 b 1 ln 3 2x dx x 2e 1)3/5 4b x2 45. u d ( 1 4 1 1 b lim ln 1 x 4 4x 1 ln 4x b→ b 1 lim b→ d 1 1)3/5 ( lim b→ 1 1 3 C x2 Bx(4x 4, B d lim 1)3/5 b→ 1 ( 0 d ( b d ln 3x 1 dx x 2(4x 1) 1 lim 42. 2 b A (A 3 b 1 b b→0 2 lim 4x Ax 2 1 2 b x b 1 dx x2 3x 4x 3 3x 1 x 2(4x 1) 3b1/3] lim [3 3 1 3] 2/3 y lim 1 44. dy ln x x b→ b b→0 dy y 2/3 1 1 b→0 1 lim ln b lim [3b1/3 dx 2 3 b→ 1 3y1/3 1 x ln x b→ dy y 2/3 2/3 y 1 x 3 lim ln 1 lim b→0 b→ b lim ( b) b dy y 2/3 2A 1. lim b→0 b→0 0 B)x lim ln b 1/b 1/b lim 1/b 2 b→0 lim 1 (A b 2 dx x(x 2) lim 1 dy y 2/3 2 1, B b→0 1 1 0 b) 3 1 dy y 2/3 Bx B b b→0 41. 2) where A x lim ( 1 0 x b lim 1 3 A( x 2 1 ln x dx 0 b→ A x x(x C 2 dx x(x 2) lim 2) 2 x ln x b 2 dx 2x x 2 x 2e 2e dx x C dx 0 2x e 2b eb 2 eb x 2e x b 0 2 2 359 Chapter 8 Review 46. u x du e 3x dx dv dx 0 xe 1 3x e 3 v 1 x 3 1 x 3 x e3x dx dx has infinite discontinuities at t 1 3x e dx 3 e3x 1 3x e 9 3x e 0 3x Note that the integral must be broken up since the integrand lim xe b→ 0 b 3x lim b→0 dx lim b→ B t2 b→ C ln (t 1)2 t2 1 tan lim 1 b (b 1)2 ln 2 b 1 1 3b e 9 1 9 tan b→0 b 4t 3 t 1 47. 2 t (t 1)(t 2 1) t 1 A t t 1)(t 2 At(t 1) 22 Ct (t B E)t (t B E)t 3 D (A B)t 0 4 dx x 2 16 A D B 0, C A E B 0, D A E B 1x tan 2 0 lim 1. 0 b 4 dx x 2 16 Solving the system simultaneously yields 0, B 1, C 2, D 2, E 0 4 dt t2 2 dt t1 2t t dt t2 2 dt t1 4 dx x 2 16 2t dt t2 1 2 1 dt 1 1 ln t (t 1) t2 1 ln 1 tan 1 t C lim 2 1 t tan 1 t → C t1 4t dt t 2(t 1)(t 2 1) 1 t1 4t dt t 2(t 1)(t 2 1) 2 1 4t 3 t 1 dt t (t 1)(t 2 1) 2 1 1/2 3 t1 4t dt t 2(t 1)(t 2 1) 1 2 1 lim lim → g( ) d → 1 1 b→ 3 1 t1 4t dt t 2(t 1)(t 2 1) and 1 1 1 1 2 d 1 1 d b lim ln b→ 4t 3 t 1 dt t (t 1)(t 2 1) 1 lim ln b 2 b→ 3 2 1 2 . Both are positive continuous functions on [1, ). b 0 2 2 lim 3 1/2 f( ) g( ) Since 3 0 1 g( ) 2 ln t 4 49. Use the limit comparison test with f( ) 1 dt t2 1 2 2 0 1b lim tan t1 4t dt t 2(t 1)(t 2 1) 1 t 4 b→ 3 2 b 1x lim tan 1. b 1b b→ A 4 4 dx x 2 16 lim b→ 4 tan b→ B 4 dx x 2 16 0 1x tan b→ 4, 0 4 dx x 2 16 b lim 1 and 1 C using Formula 16 with a 4 4 dx lim 16 b→ x B Equating coefficients of like terms gives C 4 dx x 2 16 2 0 A 1 4 4 dx 16 x 48. 1) E) t 2 C ln 2 t 1) 2 (A 1 1 Since this limit diverges, the given integral diverges. 1)(t 2 (Dt D)t 4 (A E 1 B( t 1) C (A Dt t2 1 b 1 t b→0 b 1 3b be 3 4t 3 t 1 dt t (t 1)(t 2 1) 2 lim 0 1 3x e 9 1 9 lim 4t 3 1 b 1 3x xe 3 1. 4t 3 t 1 dt t 2(t 1)(t 2 1) 1 C 0 and t t1 4t dt t 2(t 1)(t 2 1) , we know that 1 g( ) d diverges and so 1 f( ) d diverges. This means that the given integral diverges. 360 Chapter 8 Review x 50. Evaluate e u cos x dx using integration by parts. cos x du e dv sin x dx x v cos x dx x Evaluate sin x e u cos x x sin x e ex dx dv cos x dx v x x x e x dx e 0 ex e dx x x lim b→ sin x e dx e sin x e cos x dx lim e x cos x dx x e cos x x e sin x e x b→ cos x dx 0 x 2e cos x dx x e sin x x e cos x dx ex e b→ C1 0 x u e x e cos x dx sin x e x cos x b cos u du lim e b→ 0 x 1 eb cos x dx lim x 2 e x cos x b 0 e sin x b cos b 2 b→ e lim b sin b 2 b→ 1 2 1 2 4 x eb 4 4 4 Since these two integrals converge, the given integral converges. 54. The integral has an infinite discontinuity at x Note that we cannot use a comparison test since e x cos x 0 for some values on [0, ). 1 dx x (1 e x ) ln z on [e, ) z b dz b lim lim ln z lim (ln b 1) e b→ b→ e z b→ 1 dz diverges, so the given Since this integral diverges, ez 2 1 4x 2 1 0 dx 4x 2 1 x 2(1 ex ) 1 dx 4x 2 lim b→0 b 1 11 4x b lim b→0 1 0 0. dx x 2(1 e x ) dx x 2(1 e x ) ex on (0, 1] since 1 Since this integral diverges, integral diverges. 1 dx x 2(1 e x ) 0 0 0 dx x (1 e x ) 2 1 1 51. 0 z dz z e C b 0 ex b→ ex x 1 0 e 1 lim tan C 2 dx e ex 1 0 b lim tan b→ e ex dx ex e lim x tan x tan 4 b C 0 dx e 1 tan b→ u x ex b lim x dx e ex 0 dx e 1 tan 0 dx e e dx 1) du u2 1 x e x dx (e x)2 1 x e , du x x e dx e (e 2 x x x Let u dx using integration by parts. sin x du x dx x e x e x e dx e e 53. 4 on (0, 1]. 1 4 lim b→0 1 4b dx diverges, so the x 2(1 e x ) given integral diverges. t e 52. 0 e t on [1, ) t e 1 t 55. x 2 b dt lim b→ e e lim e b→ dt 1 lim b→ t x2 tb 2x 1 b 1 e 7x 2x 1 7x 12 1 (x 4)(x A x 3) B 4 x 3 A( x 3) B(x 4) (A 1 e Since this integral converges, the given integral converges. 12 B)x 3A 4B Equating coefficients of like terms gives A B 2 and 3A 4B 1. Solving the system simultaneously yields A 2x 1 dx x 2 7x 12 9 ln x 9, B 7 dx x3 9 dx x4 4 7 ln x 3 C 7. 361 Chapter 8 Review 56. 2 B x C x2 Bx 2(x 2) Cx(x 8 A x 3(x 2) x Ax 3 8 B)x 3 (A D x3 2) D(x D)x B 0, 2B C 0, 2C x3 x3 1 x1 1 x x3 x A x1 x x(x 1)(x 1) D 8 1, B 1, C 8 dx x3(x 2) 2, D dx x dx x2 ln x 57. t 3 3t 2 4t t 3t 3 A t 4 4t 4 Bt t2 A( t 2 (A 4 2 dx x2 2 ln x x 4 dx x3 2 x2 A C B A Bx(x C)x 2 B C 0, B (Bt 3, C (B C C)t 1, C 1) C)x Cx(x 1 1)(x 1) 1) A 1, and A 1. 0 dx dx x x B)t 2 4 and A 1, B 1 dx x x3 x3 Ct ln x 3 x3 x3 4x 2 4x 2 dx x 1 ln x 1 C A Equating coefficients of like terms gives B 1)(x x(x Equating coefficients of like terms gives C 1 1) (A A A( x x 1 1) t 2 1) 1 1 Solving the system simultaneously yields t(t 2 t 2 x1 x(x 2 1) B C x1 x1 x 0, and 2D Solving the system simultaneously yields A 1 x1 x+0 x1 0x2 0x2 2D Equating coefficients of like terms gives A x x3 x3 2) (2C C)x 2 (2B 59. x 3 60. x 2 4x 4. x 0x 3x 3x Solving the system simultaneously yields x3 A 4, B 3t 2 1, C 4t t3 4 t 4 x 4 dt t 4 dt t dt t4 dt t2 1 t dt 4 dt t2 1 t2 1 1 ln (t 2 1) 4 tan 1 t 2 4 ln t 58. t 4 (t 2 4t 2 (t 2 3 1 1)(t 2 3) (At C)t 3 At t2 B)(t 2 (A 1 3)(t 2 B 1 (B C 0, B D D 3 D)t 2 (3A C 0, B (t 2 dt 3)(t 2 1 ,C 2 1) 0, D 1 tan 1 t 2 C)t 3B 0, and 3B D D 1/2 dt t2 3 1 1t tan 2 3 1 C 3 Evaluate the integrals using Formula 16, with x in the first integral and a (A B)x 3A (x 3x 1)(x 3 in the second. t, a 1 3) 1) B Equating coefficients of like terms gives B x 1 2 1/2 dt t2 1 B(x x3 1) Solving the system simultaneously yields A 3) x 3, 3A B 0 3 ,B 2 Solving the system simultaneously yields A D)(t 2 0, 3A A( x 3x C Equating coefficients of like terms gives A 4x 2 3x x 4x 3 x 2 4x 3 3x A B 1)(x 3) x1 x3 3x A (Ct 3) (x 1) Ct t2 2 2 4x 2 dx 4x 3 x dx x2 2 x 3/2 dx 1 3 ln x 2 1 x 9 . 2 9/2 dx 3 9 ln x 2 3 C 362 61. Chapter 8 Review dy y(500 y) 1 y(500 y) 1 0.002 dx A y A(500 63. y B 500 y y) 1 1 tan , dy 3 9y 2 A)y By 9y 2 1/500 dy 500 y 1 ln 500 500 1 y ln 500 500 y C1 y 500 y C1 y 500 y 20 480 0.002x 0, y y 500 y 20. 1x e 24 24y 500e x (e x 24)y C2 64. t yex 500e x 500 e x e x 24 y y 1 dy y2 65. x 500 24 e x y2 x 1 dx 1 x tan 1 y C1 tan 1 y ln x Substitute x tan 1 tan 1 y tan ln x 4 y 9t 2 9t 2 dt 1 C 0, y C2 1 1 4 1 and x . 1 . 3 2 2 1 cos 3 9 sec2 9 2 3 sec 5 d tan , 0 1 4 2 9 tan2 9 3 sec tan d 3 tan 9 sec ln tan tan C cos2 d ln sec C ln x 3y d, cos 3 sec , dx 5 25x 1 1 1 cos 3 sin2 1 5 dx ln x 9y 2 1 1 sin , dt 3 25x 2 dx 1 dy C 1 cos2 d 3 1 cos 2 d 6 sin 2 C 6 12 sin cos C 6 6 sin 1 3t 3t 1 9t 2 C 6 6 1 1 sin 1 3t t 1 9t 2 C 6 2 1 Use Figure 8.18(b) with a and x t. 3 1 24 ke0 or k tan C1 1 ke x Substitute x d Use Figure 8.18(a) from the text with a 1 x sec Integrate by using Formula 88 with a C ln 62. ln y 2 d ln sec 1 ln y 500 2 sec 1 . 500 1 y ln 500 500 y d, 2 sec 1/500 dy y dy y(500 y) 2 sec2 500A 1 ,B 500 where A tan 3 dy 1 (B 1 1 sec2 3 5x 3 ln (5x tan 25x 2 3 25x 2 C1 9 C1 9) C 1 4 Integrate by using Formula 88 with a Use Figure 8.18(c) with a 3 . 5 1 and x . Chapter 8 Review 68. For x 66. x sin , dx cos d, 2 0, y Area 2 0 on [0, ). b x xe dx lim x2 1 (1 cos2 4x 2 dx x 2)3/2 4 sin2 cos u d 3 cos 4(1 cos2 ) d cos2 2 (4 sec du dv dx x xe 4 tan 4 x C 0 1 1. 69. (a) 1 b→0 b 2 u (ln x) dv 2 ln x dx x du (ln x)2 dx 2 v dx x (ln x) 1 a 2 ln x dx 2 ln x du dv 2 dx x v dx x 2x ln x 2 dx (ln x)2 dx x (ln x)2 2x ln x lim b→0 x(ln x)2 b→0 2 2 2 2 2x ln x 2x 2x C C 1 2x ln x 2x b lim [2 lim b→0 b(ln b)2 (ln b)2 1/b 2b ln b b→0 2 (ln b)(1/b) 1/b2 lim 2 (ln b) 1/b lim 2 /b 1/b2 b→0 b→0 /b 1/b2 2 lim b→0 2 lim ( b→0 2 2b] ln b 1/b 2 lim lim b→0 a x 1 kt 1/a x 2 ln x dx Area kt C1 1 b) lim 2 b b→0 2 1 a 1 C 0, t C 1 Evaluate 2 ln x dx by using integration by parts. u C1 b k dt x)2 dx (a x)2 1 C2 ax 1 kt ax Substitute x x 2 (ln x) dx kt xe x)2 k(a dx (ln x)2 dx by using integration by parts. Evaluate dx dt (a (ln x) dx 0 k dt b eb 1 lim b b→ e lim 2 lim 1] dx 0 b→ ( ln x)2 dx Volume e b b b→ 0 on (0, 1]. 1 x lim [ be 1 4 sin x2 Use Figure 8.18(b) with a 0, y e x b→ C x e x xe lim 1 a kt a x 1 kt dx dx e xe 4) d 4x 1 x x e v dx x 0 dx by using integration by parts. x Area 67. For x x Evaluate xe xe b→ 0 1/a 0 x e x C 363 364 Section 9.1 Chapter 9 Infinite Series 69. continued dx (b) k dt (a x)(b x) dx k dt (a x)(b x) 1 A (a x)(b x) ax 1 A( b x) (A B(a B)x kt C1 s Section 9.1 Power Series (pp. 457–468) B b x Exploration 1 x) bA 1. 1 aB 2. x 3. 1 Equating coefficients of like terms gives A B 0 and bA aB 4. 1 5. A a b dx x)(b (a ,B a 1/(a b) dx ax x) a a a b x x b C2 a b a b ab (x 3 x 4x 8x 1/(a b) dx bx ln b x ab a b x x C2 1n (x 3 C2 abe (a b)kt ab(e (a ae (a b)kt 1) 1 (x 3 1)2 … 3 … 1)3 …. 1)n x 1 x 1 1, 2. The interval of conver- Exploration 2 1. 1 2. tan x 2 1 x A Power Series for tan 4 x x x 1 01 x 0 … 6 t2 t3 3 x3 3 t 0. x …. n 2n ( 1) x dt t2 (1 1 t4 t5 5 x5 5 … t6 t7 7 x7 7 … … ( 1)nt 2n …) dt t 2n 1 … 2n 1 2n 1 x …. ( 1)n 2n 1 ( 1)n x 0 3. The graphs of the first four partial sums appear to be converging on the interval ( 1, 1). b)kt axe (a ab(e (a b) b)kt (2x) gence is (0, 2). C a (a b)kt e b bx 1 (x 3 1) …. n 2 which is equivalent to 0 C1 D x x ( 1) x … 3 …. nn1 (x 1) (x 1) ( 1)n(x 1)n …. 1 (x 3 …. ( x)n … 4 2 x x(ae (a x 0, x ln kt b ax ln (a b)kt bx ax De(a b)kt bx Substitute t 2x … x3 This geometric series converges for b 1 ln 1 3 x 1 ln a x ab 1 2 1 Solving the system simultaneously yields 1 x Power Series for Other Functions x2 x b)kt b)kt 1) 1) b [ 5, 5] by [ 3, 3] 4. When x e bkt Multiply the rational expression by bkt . e ab(e akt e bkt) x akt bkt ae be 1 1 3 1, the series becomes 1 5 1 7 … ( 1)n 2n 1 …. This series does appear to converge. The terms are getting smaller, and because they alternate in sign they cause the partial sums to oscillate above and below a limit. The two calculator statements shown below will cause the successive partial sums to appear on the calculator each time the ENTER button is pushed. The partial sums will appear to be approaching a limit of /4 (which is tan 1(1)), although very slowly. Section 9.1 Exploration 3 A Series with a Curious Property 2 1. f (x) 1 2. f (0) x 2! x 1 3 0 x 3! … 0 … n x n! …. u2 1. u3 3. Since this function is its own derivative and takes on the value 1 at x 0, we suspect that it must be e x. 4. If y dy f (x), then dx y and y ( 1)1 1 ( 1)2 2 ( 1)3 3 ( 1)4 4 ( 1)30 30 2. u1 1 when x 0. u4 u30 3. (a) Since 5. The differential equation is separable. dy dx y dy dx y ln y x (c) an 1 Ke0 ⇒ K y e x. 1 30 18 6 1 54 18 3, the common ratio is 3. ) 4 8 (c) an 1 1 4 8 1 2 2 4 19 2 k Ke 1 3 2(3n (b) 8 y 1 2 39,366 4. (a) Since C 1 6 2 (b) 2(39) 1 , the common ratio is 2 1 . 2 1 64 1n1 2 8( 0.5)n 5. (a) We graph the points n, 6. The first three partial sums are shown in the graph below. It is risky to draw any conclusions about the interval of convergence from just three partial sums, but so far the convergence to the graph of y e x only looks good on ( 1, 1). Your answer might differ. 365 1 1 n n2 for n 1, 2, 3, … . (Note that there is a point at (1, 0) that does not show in the graph.) [0, 25] by [ 0.5, 0.5] [ 5, 5] by [ 3, 3] 7. The next three partial sums show that the convergence extends outside the interval ( 1, 1) in both directions, so ( 1, 1) was apparently an underestimate. Your answer in #6 might have been better, but unless you guessed “all real numbers,” you still underestimated! (See Example 3 in Section 9.3.) (b) lim an n→ lim n→ 1 n n2 0 6. (a) We graph the points n, 1 1n for n n 1, 2, 3, … . [0, 23.5] by [ 1, 4] (b) lim an [ 5, 5] by [ 3, 3] n→ Quick Review 9.1 1. u1 1 2 4 u2 2 u3 3 u4 u30 4 3 4 4 4 5 4 6 4 2 4 2 4 4 30 2 4 2 lim 1 n→ 1n n e 7. (a) We graph the points (n, ( 1)n) for n 1, 2, 3, … . 1 2 3 4 32 [0, 23.5] by [ 2, 2] 1 8 (b) lim an does not exist because the values of an n→ oscillate between 1 and 1. 366 Section 9.1 8. (a) We graph the points n, 1 1 2n for n 2n 1, 2, 3, … . 2. (a) Note that a0 1 ,a 32 1, a1 1n . 3 an 1 , and so on. Thus 9 1, a2 1 ,a 23 (c) Note that a0 5, a1 0.5, a2 5(0.1)n 5 . 10n (b) Note that a1 ( 1)n n an 1 1 , and so on. Thus 3 . [0, 23.5] by [ 2, 2] 1 n→ 1 (b) lim an 2n 2n lim n→ 2 2 1 an 1 for n n 9. (a) We graph the points n, 2 1, 2, 3, … . 0.05, and so on. Thus 1n1 alternate 2 3. Different, since the terms of ∑ n1 between positive and negative, while the terms of 1n1 are all negative. 2 ∑ n1 4. The same, since both series can be represented as [0, 23.5] by [ 1, 3] (b) lim an 1 n lim 2 n→ n→ 10. (a) We graph the points n, 1 2 1 2 ln (n 1) for n n 1, 2, 3, … . 1 4 …. 1 8 5. The same, since both series can be represented as 1 2 1 1 4 …. 1 8 6. Different, since ∑ ∑ n [0, 23.5] by [ 1, 1] 1 ln (n 1) lim n n→ (b) lim an n→ lim n→ (n 1) 1 u4 and 1 u1 1, 1 , so 16 u1 1, u2 un n2, or * ( 1) 2n 1 1 2n 3 1 1 8 4, u3 11 , 4 u3 1 u2 9, and u4 1 , 9 9. Converges; ∑ n0 5 4 2n 3 5 4 1 5 4 3 2 3 15 4 10. Diverges, because the common ratio is 1 and the terms do not approach zero. 11. Diverges, because the terms alternate between 1 and and do not approach zero. 16. We may write n2. 12. Converges; ∑ 3( 0.1)n 1 n0 3 ( 0.1) 30 11 3 … (b) Let un represent the value of * in the nth term, starting with n 0. Then 1 u3 u3 1 u0 1, 1 , so u0 16 1, u1 16. We may write un (c) If * ( 1)3 ( 1)6 11 , 4 u2 1 u1 4, u2 (n 1 1 1 4 2 ( 1)4 n0 1)2, or * (n 1)2. 1 2 ( 1)5 1 3 3. 1 1 1 2 2( (2 2 1 2 1 1 2 n0 1 2 n 1 2 1 2 ∑ 2 n 4 9, and … , which is the same as the desired series. Thus let * 13. Converges; ∑ sin n 1 , and 9 3, the series is 2 … but 3 2 3 1 1 4 8. Diverges, because the terms do not approach zero. 1. (a) Let un represent the value of * in the n th term, starting 1. Then 1 n0 Section 9.1 Exercises with n n 7. Converges; ∑ 0 1 n1 1n1 1 1 2 2 1 1 1 …. 2 4 8 2 2 2 1)( 2 1) 2 1) 2 2 1 2 2 1 14. Diverges, because the terms do not approach zero. 1 Section 9.1 e 15. Converges; since 1, ∑ 0.865 16. Converges; ∑ n0 6n 17. Since ∑ 2 x n0 n0 1 1 function f (x) 1 2x 18. Since ∑ ( 1)n(x 1 2x ∑[ 1)n n0 (c) The partial sums alternate between positive and negative while their magnitude increases toward infinity. 1 . 2 x 1)]n, the series (x 24. Since ∑ n0 converges when (x 1) 1 1 , the series represents the 1 tan x 1 , k x k. 4 4 tan x (b) The partial sums are alternately 1 and 0. , the series represents the 1 2 , k , where k is any integer. Since the 4 23. (a) Since the terms are all positive and do not approach zero, the partial sums tend toward infinity. 11 , . Since 22 1 and the interval of convergence is the sum of the series is x function f (x) n n0 k sum of the series is 1 5 6 1 n0 1. Thus, the series converges for 4 ∑ (2x) , the series converges when nn 2x 1 tan x 1 6 15n 66 ∑ 5n ∑ (tan x)n, the series converges when n0 e 1 n1 e 22. Since ∑ tan n x 1 en ∑ en ne n0 1 and the interval of n e e , this is a geometric series with n0 e common ratio r 1.03, which is greater than one. e convergence is ( 2, 0). Since the sum of the series is 1 [ (x 1 1 1)] 1 f (x) x 2 x , 2 2 25. , the series represents the function x ∑ xn 20 n0 1 0. 1 20, x x 1 1n (x 2 19. Since ∑ n0 converges when ∑ 3)n 2 n0 3 x 2 xn 3 20 20x 19 20 1 and the interval of convergence 1 x 1 . 2 2 the series represents the function f (x) x) x 1 ,1 x 5. For any real number a use a 2 a 4 1 2 use 1 20. For ∑ 3 n0 x 1n x 2 a 16 1 8 ∑ 2r n 1… . 32 1 1 3 x 5r , the series 3 x , 1 x Series: ∑ 2 3. n1 21. Since ∑ sin x n n0 sin x ∑ (sin x) , the series converges when n n0 1. Thus, the series converges for all values of x except odd integer multiples of 2 , that is, x for integers k. Since the sum of the series is series represents the function f (x) (2k 1) . 2 3 5 1 1 1 , sin x (2k 1) (b) ∑ n1 3n1 5 13 2 13 n 1 r 2 1 r 13 2 1 5 5r 5r r Series: n1 13 2 5r 5, r 2 1 , the sin x 1 3 2 6 represents the function f (x) 1) 5 r 6 3 (x 1: 5, r r 2 Since the sum of the series is x … . To get 0, a 32 1 16 2 1 n1 1 and the interval of convergence is ( 1, 3). 2 a 8 1 4 0, 27. Assuming the series begins at n , the series converges when (a) 1 20x 26. One possible answer: 2 1 (3 1 19 x , the series is (1, 5). Since the sum of the series is 367 3n1 10 3 2 3 10 368 Section 9.1 28. Let a 0.21 21 and r 100 0.21 1 , giving 100 33. 1.414 0.21(0.01)2 0.21(0.01) 1 0.414 0.414(0.001)2 0.414(0.001) ∑ 0.414(0.001)n 1 n0 … 0.21(0.01)3 ∑ 0.21(0.01)n 1 n0 157 111 0.21 1 0.01 0.21 0.99 7 33 29. Let a 0.234 34. 1.24123 1 , giving 1000 0.234(0.001) 0.234(0.001)3 0.234(0.001)2 … 124 100 41,333 33,300 0.234 1 0.001 0.234 0.999 26 111 0.7(0.1) 35. 3.142857 0.7(0.1)2 0.7(0.1)3 3 3 3 36. Total distance …) 0.142857 ∑ 0.000001n (0.142857) 1 1 0.000001 0.142857 0.999999 1 7 0.06(0.1) ∑ 0.06(.1)n 4(0.6)2 2[4(0.6) 4 2 4(0.6)3 …] 2 ∑ 2.4(0.6)n 4 …] 4 4 0.13 26 n0 ∑ 0.06 1 0.1 0.06 0.9 1 15 0.000001 22 7 d [1 0.1 0.12 10 d (0.1)n 10 n 0 d 1 10 1 0.1 d1 10 0.9 d 9 n0 0.142857(1 n0 … 0.7 1 0.1 0.7 0.9 7 9 0.06 3 3 ∑ 0.7(0.1)n 32. 0.06 … 0.00123 1 0.001 n1 0.00123 0.999 41 33,300 0.0000012 n0 31. 0.d 0.00123(0.001) ∑ 1.24 ∑ 0.234(0.001)n 0.7 0.00123 1.24 n0 30. 0.7 1.24 0.00123(0.001)2 234 and r 1000 0.234 0.414 1 0.001 46 111 1 1 2.4 0.6 16m 37. Total time 0.06(0.1)2 0.06(0.1)3 4 4.9 … 4(0.6)2 4.9 4(0.6) 4. 9 2 4(0.6)3 4.9 … 4 4.9 2 4(0.6) [1 4. 9 4 4.9 2 4(0.6) 4.9 7.113 sec 0.6 1 1 0.6 ( 0.6)2 …] 369 Section 9.1 38. The area of each square is half of the area of the preceding square, so the total of all the areas is ∑ 4 1 8 m 2. 1 2 1n 2 n1 n1 ∑ 2 ∑ 45. Comparing a a S a ar n 1r 41. Using the notation Sn ar a 1 4 1, then lim r n n→ ar 2 ar ar 3 1 r 4) , the first term is a (x 4). … 4)2 (x a ar n 1r n→ a 1 r ∑ ar n 4 1 3x a 1 r the common ratio is r 3x 9x 1 … 4)n 1, so the 1) with a 1 r , the leading term is a ( 3x) . 1 and (x 1 (x 4 1) … 1)2 x 2x with the common ratio is r 2x 2 4x 3 3x 1 2 x as (x 1 (x 1 1) a r , the first term is a 11 ,. 22 1, so the 2n 1 x n a 1 1 and the common ratio is r … r x , … 1) (x 1) Rewriting 1 2 2 (x 1) 1 n 1. 1, so the 1 1 x 21 and comparing with x 2 1 and the common ratio is r 2 the first is a 1 2 as 1 x 4 12 x 8 … a 1 r , x . 2 … 1n x 2n 1 x and 48. 1 eb e 2b e 3b … x 2 1, so the interval ∑ (e b)n n0 1 Interval: The series converges when 2x of convergence is 1 … of convergence is ( 2, 2). 2x. … x. 1)n and comparing with 1, so the 11 ,. 33 1 1 1 ( 1)n(x 4 Interval: The series converges when 1 1) interval of convergence is (0, 2). Series: … n Interval: The series converges when interval of convergence is , the first term is Alternate solution: 3x. … 2 1 (x Interval: The series converges when x 1 1, then the nth partial sum is na, which goes to with 1 (x 4 Series: 1 . a ar n has no finite limit and 1r n 1 1 41 The first term is a 1 n1 diverges. Series: x ( 1)n(x 1 and the common ratio is r 4 47. Rewriting 1, then r n has no finite limit as n → , so the expression 43. Comparing 1 ar n 1, ... a ar . 1r 0 and so ∑ ar n lim 1 or r Series: 1 a with interval of convergence is (0, 2). the formula from Exercise 40 is Sn 42. Comparing 4) Interval: The series converges when x n If r 1, so the interval Series: n rS S (1 n→ (x 46. Comparing r: ar n r) S If r … 3x 3n interval of convergence is (3, 5). (a ar ar 2 ar 3 ... ar n 2 ar n 1) (ar ar 2 ar 3 ar 4 ... ar n 1 ar n) a ar n a lim Sn … Interval: The series converges when x (b) Just factor and divide by 1 If r 3 and x 3. 3x 6 1 (x 1 Series: 1 1 2 2 rS , the first term is a and the common ratio is r 4 40. (a) S r of convergence is ( 1, 1). 1n 42 1 a 1 Interval: The series converges when x 3 12 2n 1 2 n0 3x 3 Series: 3 ∑ 2n 39. Total area with x3 1 the common ratio is r 2 n0 4 3 44. Comparing 1n … 1, so the interval 1 eb 9 1 9 9e b 8 eb 8 9 b ln 9e b 8 9 ln 8 ln 9 370 Section 9.1 49. (a) When t n0 1 1 t. x (1 1 t t 1 t, which is true when t For t 1 . 2 (1 ∑ n0 2(1 n t 1 10 when t 1 1 t t 1 1 (1 t t) 1 t t, so 4 1 t2 with a 1 r , the first term First four terms: 4 4t n 4t (1 4t 2n 0, so the constant term of the power series for G (x) will be 0. Integrate the terms for f (x) to obtain the terms for G (x). 43 45 47 x x x 3 5 7 4 General term: ( 1)n x 2n 1 2n 1 First four terms: 4x (c) The series in part (a) converges when 2 x) C. 4x 3 … nx n x) d (1 dx 2 2x 3x 2 4x 3 … 3 x) 0 2 2 2 …) 12x 2 6x n (n 1)x n ... … 12x 2 6x 1)x n n (n (n …. 1)x n 2)(n 1, so the interval (b) No, because if you integrate it again, you would have the original series for f, but by Theorem 2, that would have to converge for 2 x 2, which contradicts the assumption that the original series converges only for 1 x 1. t2 1, so the lim an. Then by definition of convergence, for n→ there corresponds an N such that for all m and n, n, m N ⇒ am L 2 and an L 2 Now, 1, which result in the am an 4 4 3 4 5 4 7 … ( 1)n 4 2n 1 … and G ( 1) 4 4 3 respectively. 4 5 4 7 … ( 1)n 4 1 2n 1 …, whenever m am L am convergent series G (1) …. 53. (a) No, because if you differentiate it again, you would have the original series for f, but by Theorem 1, that would have to converge for 2 x 2, which contradicts the assumption that the original series converges only for 1 x 1. 54. Let L interval of convergence is ( 1, 1). (d) The two numbers are x 2 … 2 2, this may be written as 12x 2 6x 1 … 1 of convergence is ( 1, 1). 6 General term: ( 1) (4t ) (b) Note that G (0) dx Interval: The series converges when x t 2. 4 and the common ratio is r 4 3 3x 2 Replacing n by n t 2 x) 2x 1 Thus, f (x) f (x) is a 2 x) 9. 50. (a) Comparing f (t) … 4 nx n 1 , we have 2 (c) For t 2(1 d (1 dx 1 . 2 Thus, S converges for all t 1)4 (x 1 dt t 1 Using the result from Example 4, we have: t, which is always true. S 1)2 (x 1)3 2 3 ( 1)n(x 1)n n (x 1 f (x) dx 1 (x 52. To determine our starting point, we note that 0, the inequality is equivalent to 0, the inequality is equivalent to t x 1) 1)2 … 1)3 (x 1)n t), which is always false. For (x … , we may write ln x 1 1, this inequality is equivalent to t S 1 x ( 1)n(x 1, or t t 51. Since 2. 1 2 1 t (b) S converges when For t 1 1n 2 ∑ 1, S L L N and n an an L N. 2 2 = . 2 Section 9.2 55. Given an 0, by definition of convergence there and corresponds an N such that for all n N, L1 an L 2 an . (There is one such number for each series, and we may let N be the larger of the two numbers.) Now L2 L1 L2 an an L1 an L1 L 2 an 1 L1, lim ai(n) L2, and L1 n→ ak(n) L1 , and an N2 such that for i(n) ai(n) L2 0 N1, max {N1, N2}. Then for n an L1 and an implies that lim an L2 N2, N, we have that for infinitely many n. This L1 and lim an n→ n→ L2 where L1 (b) The line y L2. 1. f (x) f (x) f (x) f (x) f (n)(x) 1 (x f (x) 3x x 1 , which means lim f (x) 1 x→ 3. an for all positive integers n, n→ s Section 9.2 Taylor Series (pp. 469–479) Exploration 1 Designing a Polynomial to Specifications 1, we know that the constant coefficient is 1. Since P (0) 1) 4 ( 1)nn!(x 1) 1)! x f (x) nx n f (x) n(n 1)x n (The 2 in the denominator is needed to cancel the factor of f (x) n(n 1)(n 2 that results from differentiating x 2.) Similarly, we find the f (k)(x) 4 5 coefficients of x 3 and x 4 to be and . 6 24 32 43 54 Thus, P(x) 1 2x x x x. 2 6 24 n for n 1 xn 5. f (x) 3 3, we know that the coefficient of x is . 2 (n 1) 3x 3x ln 3 3x(ln 3)2 3x(ln 3)3 3x(ln 3)n 4. f (x) ln x f (x) x 1 f (x) x2 f (x) 2x 3 f (4)(x) 6x 4 f (n)(x) ( 1)n 1(n 2, we know that the coefficient of x is 2. Since P (0) 3. f (x) f (x) f (x) f (x) f (n)(x) 3 1) 6(x f (n)(x) 2 1) 2(x f (x) it follows that lim an must also be 3. 1. Since P(0) x f (x) 3 is a horizontal asymptote of the graph of Because f (n) e2x 2e2x 4e2x 8e2x 2ne2x 1 2. f (x) 3 the function f (x) Approximating sin 13 4. 20 terms. an does not converge and hence diverges. 1 1 x6 6! x6 , and the Taylor series is 6! 2n … ( 1)n x …. (2n)! 1. 0.4201670368… Since the limit of a sequence is unique (by Exercise 55), 3n 57. (a) lim n→ n x4 4! x4 4! Quick Review 9.2 . Assume an converges. Let N x2 2 x2 2 Exploration 3 L2. Given an there corresponds an N1 such that for k(n) 1 2. A clever shortcut is simply to differentiate the previouslydiscovered series for sin x term-by-term! 56. Consider the two subsequences ak(n) and ai(n), where lim ak(n) A Power Series for the Cosine 1. cos (0) 1 cos (0) sin (0) 0 cos (0) cos (0) 1 cos(3) (0) sin (0) 0 etc. The pattern 1, 0, 1, 0 will repeat forever. Therefore, P6(x) 2. L2 L1 2 says that the difference between two fixed values is smaller than any positive number 2 . The only nonnegative number smaller than every positive number is 0, so L2 L1 0 or L1 L2. n→ Exploration 2 371 2 f (n)(x) 1 2 2)x n 3 n! xn k (n k)! n! 0 x n! 0! 6. dy dx d xn dx n! nx n n! 7. dy dx d 2n(x a)n n! dx 1 xn 1 (n 1)! 2nn(x a)n n! 1 2n(x (n a)n 1) ! 1 372 8. 9. 10. Section 9.2 d x 2n 1 ( 1)n dx (2n 1)! dy dx 2n dy dx d (x a) dx (2n)! dy dx 2n(x d (1 x)n n! dx n(1 ( 1)n (2n 1)x 2n (2n 1)! 2n 1 ( 1)nx 2n (2n)! 5. cos (x (x a) (2n 1)! x)n 1( 1) n! (1 x)n 1 ( n 1) ! (cos 2) at the end of Section 9.2. (2x) 3! 4x 3 3 2x 2x 5 2n 1 (2x) … ( 1)n (2x) 5! (2n 1)! 4x 5 ( 1)n(2x)2n 1 … … 15 (2n 1)! … x for x in the Maclaurin series for ln (1 x) x) x5 … 5! (cos 2)x 2 2! (sin 2)x ( x)2 ( x)3 … 2 3 n ( x) … ( 1)n 1 n 2 3 n x x …x … x 2 3 n 1 x 1, so the interval … ( 1)n(cos 2) where 2n (2n)! 2n 1 k. Thus the coefficient is B sin 2 if n is even and cos 2 if n is odd. 1 cos 2 sin 2 x , 2 cos 2 cos (2 and so on, so the general term is 2 2n 1 (x 2)3 (x 2)5 … ( 1)n (x ) 3 5 2n 1 n 4n 2 x6 x 10 … ( 1) x … 3 5 2n 1 x2 x2 This series converges when x 2 7x(1 7x x 7x 2 … n x2 …x …) 2! n! 3 n1 7x … 7x 2! n! This series converges for all real x. ), n 2 1 cos 2 n! x n. The series converges for all real x. 6. x 2 cos x x2 2! x2 1 x4 2 x2 1, so the interval of convergence is [ 1, 1]. 4. 7xe x k. Thus the coefficient is shown at the end of Section 9.2. x2 x 2n 1 … (2n 1)! (sin 2)x 3 (cos 2)x 4 3! 4! ( 1)n Another way to handle the general term is to observe that 3. Substitute x 2 for x in the Maclaurin series for tan 1 … We need to write an expression for the coefficient of x k. and B of convergence is [ 1, 1). tan x 2n (2n)! ( 1)int[(k 1)/2](cos 2) ( 1)(k 1)/2(sin 2) , which is the same as . k! (2n 1)! An ( 1) Bx n1 , where A int , Hence the general term is n! 2 ( x) This series converges when ( 1)n ( 1)int[(k 1)/2](cos 2) ( 1)k/2(cos 2) , which is the same as . k! k! n1 ( 1) (sin 2) where If k is odd, the coefficient is (2n 1)! shown at the end of Section 9.2. ln (1 (sin 2)(sin x) … If k is even, the coefficient is This series converges for all real x. 2. Substitute x4 4! x3 3! (sin 2)x 5 5! 1. Substitute 2x for x in the Maclaurin series for sin x shown sin 2x x2 2! (sin 2) x Section 9.2 Exercises 3 (cos 2)(cos x) (cos 2) 1 2n a) (2n)! 2) x6 24 … x4 … 4! ( 1)nx 2n (2n)! ( 1)n 1 x 2n (2n)! … … The series converges for all real x. 7. Factor out x and substitute x 3 for x in the Maclaurin series for 1 1 x shown at the end of Section 9.2. x 1 1 x3 x[1 1 x3 x x3 x x4 (x 3)2 x7 … The series converges for x 3 convergence is ( 1, 1). … x 3n (x 3)n 1 …] … 1, so the interval of Section 9.2 2x for x in the Maclaurin series for e x shown at 8. Substitute x1/2 12. f(4) the end of Section 9.2. 2x 1 2x ( 2x)2 … 2! nnn … ( 1) 2 x n! ( 2x) 1 e 2x 2 ( 2x)n n! … f (4) 9. f (2) 1 xx 2 f (2) f (2) 2x f (2) P0(x) P1(x) P2(x) P3(x) 10. f f f 4 1 3 x2 4 4 6x x2 1 2 1 2 1 2 1 2 x f (2) 1 2! 8 3 f (2) , so 8 3! 4 2 8 x 2 (x 4 2)3 16 (x (b) f (1) 2 /4 2 x cos x 2 /4 f 2 x 2 /4 4 , so 2 2! f 4 , so 2 12 3! 2 P1(x) 2 P2(x) P3(x) 2 2 x 2 2 2 2 2 2 4 f f f cos x 4 4 4 P0(x) 4 x3 2x 6x f (1) 6 P3(x) 3 4 4 2 1 x1 f (1) 2! 3 6, so f (1) 3! 1 1) 3(x 1)2 6, so x1 x1 4 (x 4 2x 3 x2 3x 6x 2 2x 12x 2 f (1) 12 P3(x) 3 2 8 14, so f (1) 2! f (1) 12, so 3! 2 x1 x1 11(x 1) 2 /4 2 x 2 /4 2 /4 2 P2(x) 2 P3(x) 2 7 1)2 7(x 2(x f , so 4 2 2! 4 (b) f (1) x4 1 x1 f , so f (1) 3! 12x 2 24x 1 4(x 1) 16. (a) P3(x) 2 12 4x 3 P3(x) 4 f (1) 4 5x 82 x 2! 4 5x x1 x1 4 12, so f (1) 2! 2 2 2 2 2 2 2 2 2 x 12 x 4 2 x 4 4 4 4 2 x 3 4 x1 6 24, so f (1) 3! 4 2 x 4 2 x 6(x 4x 2 4 f (0.2) P3(0.2) 4.848 1)2 63 x 3! x3 1)3 0, the Taylor 2 x 2 2 11 x1 2 P1(x) 2 x1 15. (a) Since f (0) f (0) f (0) f (0) polynomial of order 3 is P3(0) 0. 2 1)3 (x 14. (a) Since f is a cubic polynomial, it is its own Taylor polynomial of order 3. P3(x) 2x 3 x 2 3x 8 or 8 3x x 2 2x 3 f (1) 2 x 4 2 x 4)3 512 3 f (1) sin x (x x1 4 3 /4 cos x 4 4 f (1) 2 x sin x 4 x 2 x 2 2 2 2 x 12 11. f 4 x 3x 2 (b) f (1) 4 x 2 2 (x 4 2 P0(x) 4)2 64 (x 4)2 64 4 f (1) 2 sin x x 2 /4 x 4 4 f (1) 2 cos x 4 2)2 8 x 2 x 1 512 13. (a) Since f is a cubic polynomial, it is its own Taylor polynomial of order 3. P3(x) x 3 2x 4 or 4 2x x 3 2)2 (x 4 4 1 16 2 P3(x) , so 2 x sin x 4 f x2 2 P2(x) 1 4 2 x f (4) 3 , so 3! 256 1 64 2 P1(x) 1 2 f (4) 1 , so 2! 32 3 5/2 x 8 x4 P0(x) The series converges for all real x. 1 4 1 3/2 x 4 x4 f (4) … 2 x4 1 1/2 x 2 x4 f (4) 373 4(x 1)3 374 Section 9.2 16. continued (x)(e x) d ex 1 x dx xe x e x 1 x2 ee1 1 1 (c) g (x) (b) Since the Taylor series of f (x) can be obtained by differentiating the terms of the Taylor series of f (x), the g (1) second order Taylor polynomial of f (x) is given by 5 3x 2. Evaluating at x 8x f (0.2) g (x) 3 (x 2! 4 ( 1)(x 1) 4 (x 3 (x 2 f (1.2) 1) P3(1.2) 2 (x 3! 1)2 1 (x 3 1)2 1)3 3.863 ∑ second order Taylor polynomial of f (x) is given by 1) f (1.2) 1)2. Evaluating at x (x 1.2, n1 n1 n (n x ex g(x) (c) g(x) x2 2! x3 3! x4 4! f (t) t2 … … x for x in the Maclaurin series for e x shown 2 x 2 x 2 1 ex (b) g(x) x2 8 3 n! … 2 1 xn 2 … 2 G(x) … x 2! … (t 2)3 2t 4 2x 3 3 x)1/2 (1 f (0) f (0) xn 2n n! 1 (1 2 … 2t 6 P4(x) 2x 5 5 (t 2)n 2t 2n …] … x) 5/2 x0 x2 8 x 2 2x 2n 1 2n 1 1 f (0) , so 4 2! 3/2 x0 x) … … 1 2 1/2 x0 3 (1 8 1 2x 7 7 1 x0 x) 1 (1 4 3 f (0) , so 8 3! 1 8 1 16 x3 16 x x2 2! x 2! x3 3! x2 3! x2 3! f (x 2), the first four terms are (b) Since g(x) n x 3! ... x n! … xn n! … xn 1 n! … … … xn (n 1)! … 1 1 x2 2 x4 8 (c) Since h (0) x6 . 16 5, the constant term is 5. The next three terms are obtained by integrating the first three terms of This can also be written as 1 , which means 1 x 2! 1 1)! 0, the constant term is zero and we may 2x f (0) x 1 1 x 1 x x … find G(x) by integrating the terms of the series for f (x). xn 1 (n 1)! at the end of Section 9.2 e x/2 n (n (t 2)2 2t 2 (b) Since G(0) 21. (a) f (0) x2 2 nx n 1 (n 1)! … … 1)! 1 at the end of Section 9.2. 1x 2 1 t2 1 2 1 t2 1 19. (a) Substitute (n series for 2 (b) Multiply each term of f (x) by x. xn … 1. 1)! 2[1 1 . 11 x3 4! 20. (a) Factor out 2 and substitute t 2 for x in the Maclaurin 0.36 1 x 1 18. (a) Since f (0)x , f (0) . 2 2! 2! f (10)(0) 10 10! x 10 (10) Since x , f (0) 11! 10 ! 11! ∑ Therefore, g (1) differentiating the terms of the Taylor series of f (x), the 3(x x2 d x 1 3! dx 2! 2x 3x 2 1 3! 4! 2! nxn 1 1)! n 1 (n ∑ 1)3 (b) Since the Taylor series of f (x) can be obtained by 1 1)(1) From the series, 0.2, 3.52 17. (a) P3(x) (e x x2 the answer to part (b). The first four terms of the series …. for h(x) are 5 x x3 6 x5 . 40 375 Section 9.2 22. (a) a0 1 3 a 10 3 a 21 3 a 32 a1 a2 a3 31 3 3 2 9 2 3 (x 2 a2 we have f (0) 3 n 3 . n! 1 92 x 2 3x … 93 x 2 (b) Since the series can be written as ∑ n0 3 and 3e 3x 3n x n! … L2(x) 1 (x 2 3 n (3x) , it represents n! 1 x 2 3) 3 2 4x and 3, respectively. e. 3e 3 x1 23. First, note that cos 18 ∑( n 3x the function f (x) Using cos x 3) 4, f ( 1 , so the linearizations are L1(x) 2 f ( 3) n0 (c) f (1) 0, f (0) n term by , an n 1)(4) (4x)(2x) (x 2 1)2 4 4x 2 , (x 2 1)2 9 2 Since each term is obtained by multiplying the previous ∑ anx 4x d dx x 2 1 29. (a) Since f (x) 0.6603. 1)n n0 x 2n , enter the following two-step (2n)! commands on your home screen and continue to hit [ 2, 4] by [ 3, 3] (b) f (a) must be 0 because of the inflection point, so the second degree term in the Taylor series of f at x a is zero. ENTER. 30. The series represents tan tan 1 1 4 . When x tan 1( 1) The sum corresponding to N 25 is about 0.6582 (not withing 0.001 of exact value), and the sum corresponding to N 26 is about 0.6606, which is within 0.001 of the exact value. Since we began with N 0, it takes a total of 27 terms (or, up to and including the 52nd degree term). 24. One possible answer: Because the end behavior of a polynomial must be unbounded and sin x is not unbounded. Another: Because sin x has an infinite number of local extrema, but a polynomial can only have a finite number. 25. (1) sin x is odd and cos x is even (2) sin 0 0 and cos 0 1 (3x)5 35 81 so = . 5! 5! 40 x. When x 1, it converges to 1, it converges to . 1 (sin x) x 2n 1 1 x3 x5 … ( 1)n x …) (x x 3! 5! (2n 1)! 2 4 2n x x … ( 1)n x … 1 3! 5! (2n 1)! 31. (a) f (x) (b) Because f is undefined at x (c) k 0. 1 32. Note that the Maclaurin series for 1 26. Replace x by 3x in series for sin x. Therefore, we have 4 1 … x2 x 1 1 x is …. If we differentiate this series xn and multiply by x, we obtain the desired Maclaurin series x 2x 2 … 3x 3 …. Therefore, the desired nx n function is 3 27. Since is 1 4 3! d ln x dx 3 2x 3 , which is 1 at x 4 2, the coefficient 1 . 24 28. The linearization of f at a is the first order Taylor polynomial generated by f at x a. f (x) x d 1 dx 1 x 33. (a) f (x) (1 x 1 (x 1)2 . x)m f (x) m(1 x)m f (x) m( m 1)(1 f (x) x x)2 (1 m(m 1 1)(m x)m 2 2)(1 x)m (b) Differentiating f (x) k times gives f (k)(x) m(m 1)(m 2) … (m Substituting 0 for x, we have f (k)(0) m(m 1)(m 2) … (m 3 k 1)(1 k 1). x)m k. 376 Section 9.3 33. continued 3. Since f (x) is increasing and positive on [ 3, 0], M f (0) 1. (c) The coefficient is f (k)(0) k! m( m (d) f (0) 2) … (m k! 1)(m 1, f (0) k 1) m, and we’re done by part (c). 1 ,M 2 maximum value of f (x) is f (1) m 34. Because f (x) (1 x) is a polynomial of degree m. Alternately, observe that f (k)(0) 0 for k m 1. s Section 9.3 Taylor’s Theorem (pp. 480–487) Exploration 1 1. We need to consider what happens to Rn(x) as n → . By Taylor’s Theorem, Rn(x) f (n 1) (c) is the (n 1) (c) (x 1)! 1) 0) , where 1 and 1 inclusive. Therefore, no Rn(x) 1) (c) (x 1)! 0) 1 n (n 1)! n x f (x) xn → 0 for all x. This means (n 1)! that Rn(x) → 0 for all x, which completes the proof. 1. e ix 1 1 ix ix … f (0) (i) n n! … 2. If we isolate the terms in the series that have i as a factor, we get: e ix 1. f (0) f (0) nx f (4)(0) P4(x) f (0.2) 1 ix 1 x2 2! ix cos x x2 2! x4 4! x3 3! x3 i 3! x6 6! x5 5! x4 4! x5 i 5! … … ( 1)n x 2n (2n)! … x7 7! … ( 1)n x 2n 1 (2n 1)! (i) nx n n! 2x e 1 x0 2e 4e 2x 2. f (0) … f (0) f (0) i sin x. 2x 1 0 2 cos (3x) 2 on [ 2 , 2 ] and f (0) 2. Since f (x) is increasing and positive on [1, 2], M f (2) 7. f (0) 2! 2 f (0) 3! f (4)(0) 16e 2x 16, so x0 4! 43 24 2 1 2x 2x x x 3 3 8e 2x P4(0.2) 4 3 8, so x0 2 3 0.6704 f (0) f (4)(0) x 1 2x0 x sin 0 2 2x0 2 2 2 x f (0) cos , so x0 4 2 4 2! 8 3 x f (0) sin 0, so 0 8 2x0 3! 4 4 4 x f (4)(0) cos , so 16 2x0 16 4! 384 cos 2 x2 P4(x) 1 Quick Review 9.3 1. Since f (x) M 2. 4, so x0 2, 1 f (0.2) i sin 10 2 x0 … (We are assuming here that we can rearrange the terms of a convergent series without affecting the sum. It happens to be true in this case, but we will see in Section 9.5 that it is not always true.) 3. ei cos Thus, e i 3 1/2 x , so f (0) is undefined. 4 f (0) Euler’s Formula n (ix)2 (ix)3 … (ix) 2! 3! n! x2 x3 x4 x5 … i i 2! 3! 4! 5! 3 1/2 x and 2 Section 9.3 Exercises Example 3, eventually outstrips the power growth in the Exploration 2 x 3/2, we have f (x) 10. No, since f (x) xn . (n 1)! The factorial growth in the denominator, as noted in numerator, and we have 4 has a corner at x 9. Yes, since the function f (x) e x has derivatives of the form f (n)(x) e x for odd values of n and f (n)(x) e for even values of n, and both of these expressions are defined for all values of x. matter what x is, we have f (n (n x2 2. 8. Yes, since the derivatives of all orders for sin x and cos x are defined for all values of x. 1)st derivative of cos x evaluated at (c) lies between 5. On [ 3, 1], the minimum value of f (x) is f ( 3) 7 and the maximum value of f (x) is f (0) 2. On (1, 3], f is increasing and positive, so the maximum value of f is f (3) 5. Thus f (x) 7 on [ 3, 3] and M 7. 7. No, since the function f (x) n some c between x and 0. As with sin x, we can say that f (n 1 . 2 6. Yes, since each expression for an nth derivative given by the Quotient Rule will be a rational function whose denominator is a power of x 1. Your Turn f (n (n 1 and the 2 4. Since the minimum value of f (x) is f ( 1) P4(0.2) 8 4 384 x4 0.9511 x 377 Section 9.3 5 sin ( x) x 3. f(0) 0 5 sin x x 5 cos x x 0 0 5 sin x x f (0) f (4)(0) P4(x) f (0.2) 4x 2 2 2! 1 x2 (2x)2 2! 1 1 2 1 0, so 0 1 cos (2x) 2 1 2 f (0) 0 2! f (0) 5 5 cos x x 0 5, so 3! 6 (4) f (0) 5 sin x x 0 0, so 0 4! 53 5x x 6 149 P4(0.2) 0.9933 150 f (0) 1 2 8. cos2 x 5 f (0) 0 16x 4 2 4! x4 3 … 4x 2 2 2! x 2) ( 1)n 2n (x 2)2 (x 2)3 … ( 1)n 1 (x ) … 2 3 n 2n x4 x6 … ( 1)n 1 x … 2 3 n x4 x2 and f (0.2) P(0.2) 0.0392. 2 Therefore, P4(x) 5. f (0) (1 2 x) f (0) 2(1 f (0) x) 6(1 3 f (0) f (4)(0) 4 3x 2 P4(x) 1 f (0.2) P4(0.2) 6. xe x 2x x1 x 7. sin x x x5 5! x x2 x3 x 3! x3 x5 3! 5! x7 x9 7! 9! x3 2! x2 1 2x 4x 3 … (2x)6 6! (2x)2n (2n)! … … … … 2 (2n)! … ( 1)n 2n 1 2n x 12 1, the general term can be 1 2x 2x 2x 3 … (2x)2 … 4x 4 sin x. Then P4(x) …] (2x)n 2n x n P3(x) 2 … x x3 , so we use 6 the Remainder Estimation Theorem with n f (5)(x) 5x 4 … (2n)! 22n 1 x 2n 2 . (2n 2)! 1 11. Let f (x) 5 … cos x 4. Since 1 for all x, we may use M r 1, x5 giving R4(x) , so we may assure that 5! x5 4 R4(x) 5 10 by requiring 5 10 4, or 5! xn n! … … x2 x2 1.56 x2 2! written as ( 1)n 6, so x0 (2x)4 4! 2n 2n 12 x 2x 6 45 x 2[1 f (0) 3 2! f (0) 4 24(1 x) 5 x 0 24, so 3! (4) f (0) 120(1 x) 6 x 0 120, so 4! x) 22n 1x 2n (2n)! Note: By replacing n with n 2 x0 x4 3 x2 10. 1 x0 … 22nx 2n (2n)! 64x 6 2 6! 16x 4 2 4! x) at the end of Section 9.2, we have x2 (2x)2 2! ( 1)n ln (1 2 (2x)2n (2n)! 1 cos (2x) 2 1 1 2 ( 1)n ( 1)n ( 1)n 1 2 ln (1 … … 1 2 9. sin2 x 4. Substituting x 2 for x in the Maclaurin series given for x2 (2x)4 4! … xn 1 n! 5 x 0.06 0.5697. Thus, the absolute error is no greater than 5 … 0.56 x 10 4 when 0.56 (approximately). Alternate method: Using graphing techniques, … … x 2n 1 (2n 1)! 2n 1 nx ( 1) (2n 1)! ( 1)n Note: By replacing n with n … x x3 3! … 2, the general term can be sin x x3 6 x 12. Let f (x) 5 10 cos x. Then P3(x) 4 when 0.57 P2(x) 1 x 0.57. x2 , so we may 2 use the Remainder Estimation Theorem with n 3. Since 2n 5 x written as ( 1)n (2n 5)! f (4)(x) cos x 1 for all x, we may use M r 1, 4 giving R3(x) than (0.5)4 4! x . For x 4! 0.5, the absolute error is less 0.0026 (approximately). 378 Section 9.3 Alternate method: 12. continued Using graphing techniques, we find that when x Alternate method: Using graphing techniques, we find that when x error 1 cos x x2 2 1 cos 0.5 0.52 2 1 x 1 error 0.5, 0.01 1.26 15. Note that 1 0.002583. x2 tends to be too small, as shown by the 2 x2 cos x and y 1 . 2 x 2 1 0.01 2 1 5 10 . x2 is the second order Taylor polynomial 2 x e x at x for f (x) 0, so we may use the Remainder The quantity 1 graphs of y Estimation Theorem with n is less than e0.1 when x 0.1 e 0.01, 2. Since f (x) 0.1 and r e x, which 1, giving 3 x . Thus, for x 0.1, the maximum possible 3! e0.1(0.1)3 error is about 1.842 10 4. 3! R2(x) [ , 16. Note that e x by [ 1.5, 1.5] 13. Let f (x) sin x. Then P2(x) P1(x) the Remainder Estimation Theorem with n f (x) cos x e x, so we may use r Alternate method: error sin x x The inequality x by graphing y sin 10 10 sin x is true for x sin x 3 1.67 … … x 2! 1x (e e 2 1x for cosh x (e 2 x3 sinh x x 3! x2 cosh x 1 2! x3 . Thus, for x 10 3, the maximum 3! (10 3)3 possible error is about 1.67 10 10. 3! 3 x sinh x 1, giving R2(x) Using graphing techniques, we find that when x 2 1 x2 2! x ( 1)n n xn n! … and … x n! Thus the terms with n even will cancel for 2. Since 1 for all x, we may use M x 1 10 3, 10 10 0, as we may see . x ), and the terms with n odd will cancel e x). 5 x 5! x4 4! … … x 2n 1 … (2n 1)! 2n x … (2n)! 17. All of the derivatives of cosh x are either cosh x or sinh x. For any real x, cosh x and sinh x are both bounded by e x . e x and r So for any real x, let M 1 in the Remainder Estimation Theorem. This gives Rn(x) x. ex x n 1 1)! n → (n But for any fixed value of x, lim ex x n 1 (n 1)! 0. It follows that the series converges to cosh x for all real values of x. 18. For n 3 [ 10 , 10 3 by [ 2 14. Let f (x) 1 10 10 ,2 10 x. Then P1(x) 10 has derivatives of all orders in an open interval I containing 1 x , so we may use 2 the Remainder Estimation Theorem with n f (x) x 1 (1 4 x) 3/2 a, then for each x in I, f (x) f (a) R(x), where R(x) f (x) f (a) f (c)(x Letting b possible absolute error is about 1, giving 0.01 the maximum x this equation is f (b) 10 5 . a), so f (b) b f (a) f (c)(b a), f (a) for some c a between a and b. Thus, for the class of functions that have derivatives of all orders in an open interval containing a and b, the Mean Value Theorem can be considered a special case of Taylor’s Theorem. 1.27 f (c)(x a) for some c between a and x. which is equivalent to f (c) 0.2538 and r 0.2538 x 2 . Thus, for x 2! 0.2538(0.01)2 2! 1. Since , which is less than 0.2538 for 0.01, we may use M R1(x) 0, Taylor’s Theorem with Remainder says that if f Section 9.3 19. f (0) ln (cos x) ln 1 x0 1 ( sin x) x 0 cos x f (0) sec2 x x f (0) (a) L(x) tan x x 1 so 0 22. f (0) 0 f (0) 2! sec x tan x x f (0) 1 2 2 so 0 (c) The graphs of the linear and quadratic approximations fit the graph of the function near x 0. 0 0 (sec x)(sec x) f (0) 2! (tan x)(sec x tan x) x 1, 0 1 2 (a) L(x) 12 x 2 (b) P2(x) 1 0 f (0) 0 0 sec x x 379 1 (b) P2(x) x2 2 1 (c) The graphs of the linear and quadratic approximations fit the graph of the function near x 0. [ 3, 3] by [ 3, 1] [ 3, 3] by [ 1, 3] 20. f (0) sin x e 0 e x0 1 23. f (0) sin x f (0) e cos x x 1 0 (esin x)( sin x) f (0) 1 (b) P2(x) sec2 x x f (0) (2 sec x)(sec x tan x) x 1, 1 2 (a) L(x) (a) L(x) x 1 x 0 0 f (0) (cos x)(esin x cos x) x0 f (0) so 2! tan x x (c) The graphs of the linear and quadratic approximations fit the graph of the function near x 0. 0 0, so f (0) 2! 0 x (b) P2(x) x2 2 1 0 x (c) The graphs of the linear and quadratic approximations fit the graph of the function near x 0. [ 3, 3] by [ 2, 2] [ 3, 3] by [ 1, 3] 24. f (0) 21. f (0) (1 x) 1 (1 2 f (0) f (0) so 2 (a) L(x) (b) P2(x) 1 x0 2 x) 3 (1 2 (x) f (0) 2! 1/2 x2) 5/2 0 x(1 ( 2x) (1 2 x) x 2) 3/2 x0 3/2 0 1, k(1 x) f (0) k(k 1)(1 so x0 1 2 1 0 k1 f (0) 3/2 ( 2x) x x)k x (1 f (0) 2! k(k x)k 1 (c) The graphs of the linear and quadratic approximations fit the graph of the function near x 0. k(k x0 1), 1) k(k 1 For k x2 2 2 2 P2(x) 1 k x0 kx 1) 2 x 3, we have f (x) 2 x)3 and f (x) (1 6. We may use the Remainder Estimation Theorem with n 1, giving R2(x) and r 6x3 3! 2, M x . (In this particular case it is actually true that R2(x) x 3, since f (x) is a cubic polynomial.) Thus the absolute error is less than whenever x 3 when 0 x 1 100 0.01. In the interval [0, 1], this occurs 3 0.01 0.215. Alternate method: [ 3, 3] by [ 1, 3] Note that P2(x) 1 3x 3x 2. Using graphing techniques, (1 x)3 (1 3x 3x 2) 6, 3 1 when x 100 0.215. 380 Section 9.3 e x. Then P3(x) 25. Let f (x) 1 x2 2 x x3 , so we may 6 use the Remainder Estimation Theorem with n f (4)(x) 3. Since e x, which is no more than e0.1 when x 0.1 may use M for x e and r 1, giving R3(x) 27. (a) No (b) Yes, since dy dx 0.1, we 0.1 10 0.1, y x 6 0.01 2 x2 … P3(x) 1 x x2 x 3. Since f (4) 24(0.9) 5 24(1 when x …, xn x x3 3 … 0.1, we may use M 24(0.9) 4! 5 x4 0.14 0.95 1.694 (2n x 2n 1 1)(n Using graphing techniques, when x x 1 1 0.1 (1 x 1.111 1)! . x for x in the Maclaurin series for ln (1 24(0.9) 5 and x) (b) ln 1 1 x x x2 2 x ln (1 x) 2 x x 2 x 2x 10 4. Rounding up x3 3 2x 3 3 ln (1 3 x 2 29. (a) 0.1, 2 3 x) 1.11 , 2 by [ 2, 2] The series approximates tan x. 10 4. (b) 2 , 2 … xn n … x) n x … ( 1)n 1 x … 3 n n x2 x3 …x 2 3 n 2n 1 2x 5 … 2x … 5 2n 1 Alternate method: 1 … x) given at the end of Section 9.2. x4 . Thus, for 0.95 10 4. to be safe, an upper bound is 1.70 error x 2n 1 (2n 1)n! 1 0.1, an upper bound for the magnitude of the 1 ( 1)n ln (1 x) , which is no more than approximation error is x5 10 1 for n, the general term may also 28. (a) Substitute 5 1, giving R3(x) x 2, and we may obtain (c) The power series equals the function y for all real 2 values of x. This is because the series for e x converges for all real values of x, so Theorem 2 of Section 9.1 implies that the new series also converges for all x. . x r …. x n! be written as ( 1)n 0.001 6 26. Since the Maclaurin series is (x) … By substituting n 6 1 2 3 x 2 x 1 4.251 2n series. 1 e0.1 ( x 2 )n n! the remaining terms of y by integrating the above 2 ex x ( 1)n x 2! The constant term of y is y (0) 10 6. 4.605 Using graphing techniques, when x 1 … … 4 x2 1 Alternate method: error ( x 2)2 2! ( x 2) 1 e0.1 x 4 . Thus, 4! 0.1, the maximum possible absolute error is about e0.1(0.1)4 24 1 x2 e by [ 1, 4] The series approximates sec x. 381 Section 9.4 1 (1 cos 2x) 2 1 1 (2x)2 1 2 2 2! 30. (a) sin2 x (b) e ax cos bx dx (2x)4 (2x)6 … 4! 6! (2x)2n … ( 1)n (2n)! 256x 8 4x 2 16x 4 64x 6 2 8! 2 2! 2 4! 2 6! 1024x10 … 2 10! x4 2x 6 x8 2x10 … x2 3 45 315 14,175 (b) derivative (c) part (b) 4x 3 (2x)3 3! 2x 2x 3 5 4x 15 (2x)5 5! e(a a2 sin P e b2 [(a cos bx b sin bx) i(a sin bx b cos bx)] Separating the real and imaginary parts gives sin 2x e ax (a cos bx a2 b2 ax e (a sin bx a2 b2 e ax cos bx dx e ax sin bx dx b sin bx) and b cos bx) (pp. 487–496) x) sin ( x) x sin x sin x Exploration 1 Test x3 . 6 ei e x3 is less than . 6 sin P and i (cos i sin ) cos i sin )) 2 2 cos 2 (cos i sin ) (b) (cos ( 2i (cos ) i sin ( )) d 33. [eax(cos bx dx (eax)[(bi2 sin bx bi cos bx) a(cos bx i sin bx) bi)(e )(cos bx (a n n 1. 1 1 dx x 1 n→ 1 dx x2 1)2 (n lim ln x k→ lim n→ k 1 x k→ n2 lim 1 n2 (n lim ln k . k→ 1k 1 k lim k→ 1 1. 1)2 1 1 1. 1 dx. Since x 1 is less than 1 n2 1 1 dx. x2 Since the integral converges, so must the series. (aeax)(cos bx (a 1 lim Figure 9.14b shows that ∑ bi cos bx) ax lim n→ the integral diverges, so must the series. sin (e )( b sin bx (eax)[bi(cos bx 1 1 n 1 n i sin bx)] ax lim 3. Figure 9.14a shows that ∑ is greater than i sin ) 2i 2i sin 2i n n→ 1 n 2. (a) i sin ) 1 For ∑ 2 : L cos (cos i e 2i 1. For ∑ 1 :L n Finishing the Proof of the Ratio 1 (cos ( ) i sin ( 2 cos i sin 2 a(cos bx i sin bx) i sin bx)] i sin bx)] (a bi)x i sin bx) bi)e 34. (a) The derivative of the right-hand side is a a2 bi cos bx s Section 9.4 Radius of Convergence ( P (b) b sin bx ax a2 Therefore, the difference between the new estimate ei (a cos bx x where x is the error in the original But by the Remainder Theorem, x 32. (a) b2 i sin bx) ai sin bx) estimate. Then P ea x 7 8x … 315 (2x)7 … 7! dx bi (a bi)x e b2 bi ax e (cos bx b2 a a2 a a2 31. (a) It works. For example, let n 2. Then P 3.14 and P sin P 3.141592653, which is accurate to more than 6 decimal places. (b) Let P bi)x i e ax sin bx dx bi (a bi)e(a bi)x b2 a 2 (bi)2 (a bi)x e a2 b2 a 2 b 2 (a bi)x e e(a bi)x, a2 b2 which confirms the antiderivative formula. 4. These two examples prove that L 1 can be true for either a divergent series or a convergent series. The Ratio Test itself is therefore inconclusive when L 1. Exploration 2 1. L lim n→ x n Revisiting a Maclaurin Series n1 n xn 1 lim n→ n n converges absolutely when x 1 x x . The series 1, so the radius of convergence is 1. 2. When x 1 1 2 1 3 1, the series becomes … 1 …. n Each term in this series is the negative of the corresponding term in the divergent series of Figure 9.14a. Just as ∑ diverges to , this series diverges to . 1 n 382 Section 9.4 3. Geometrically, we chart the progress of the partial sums as in the figure below: Section 9.4 Exercises 1. Diverges by the nth-Term Test, since lim 1 –2 n→ n n 1 +3 n→ n 0. . (The 1 Ratio Test can also be used.) 1 L 1 1–2 2n 2. Diverges by the nth-Term Test, since lim 1 –4 1 +5 1 1 3. Converges by the Ratio Test, since etc. lim 1 1 1– 2 + 3 1 1 1 1– 2 + 3 – 4 n→ 4. The series converges at the right-hand endpoint. As shown in the picture above, the partial sums are closing in on some limit L as they oscillate left and right by constantly decreasing amounts. an 1 lim n→ an 1) 2 (n 2 2n 1 n1 n 2 1 2 1 1. 4. Converges, because it is a geometric series with r so r 1 , 8 1. 5. Converges by the Ratio Test, since 5. We know that the series does not converge absolutely at the right-hand endpoint, because ∑ 1 diverges (Exploration 1 n of this section). n→ nx n n2 x n→ n(n 3. lim n→ n→ 3 1) 2. lim xn n! 1 2n lim n n→ (3 an 1 1 1) 2n n x ∑ Comparison Test. n x lim 1 an ∑ Quick Review 9.4 1. lim 3n 1 2 1. 3 2n n→ n 2 Alternately, note that n for all n. 3 3 1 n n 2 2 Since converges, converges by the Direct n 1 n 13 n 13 lim x 1 6. Diverges by the nth-Term Test, since n2 3 lim n→ n 2 x n 3 lim n sin n→ 0 1 n 1 0 7. Converges by the Ratio Test, since (Note: This limit is similar to the limit which is discussed lim n→ at the end of Example 3 in Section 9.3.) an 1 lim (n n→ an 1)2e n 2e n n1 1 1. 1 10 1. e 8. Converges by the Ratio Test, since 4. lim n→ (n 1) 4 x 2 (2n)4 2 x lim 21 6. Since n 2 5n for n 2 1n lim 2x 2 6, an ln n for n n 2, bn n 2, an n→ 1 lim n→ 5n, and N 5 5 , bn 1, an n and N n, bn an 1 an (n 1)10 10 n 1 n→ 10 n n10 lim an 1 (n 4)! 3!(n 1)!3n n4 lim 1) n→ 3(n 1 1. 3 lim n→ an 6. 6. ln n, and 1 3!n! 3n (n 3)! 10. Diverges by the nth-Term Test, since 1 n! and hence n 9. Since 10 10 1 1 an ,b , and N 25. 10 n n n! n lim 2 1 for n n! n 10. Since n 2 1 9. Converges by the Ratio Test, since 1 6, an n for n 8. Since n N 1. 2x n→ 5 4n x 16 16 2x 1 5. lim n 1 2x n→ 2 6n 2 16n 4 2 n1n 7. Since 5 4n 3 n→ x n n4 n 3 and hence 1 ,b n2 n n 3 1 n2 25, 1n n lim 1 n→ e 0. 2 , 3 11. Converges, because it is a geometric series with r n , and N 3 for so r 1. 2. 12. Diverges by the Ratio Test, since lim n→ an an 1 lim n→ (n 1)!e n!e n n1 lim (n n→ (The nth-Term Test can also be used.) 1)e 1 . 383 Section 9.4 13. Diverges by the Ratio Test, since lim an n→ 3n 1 lim 1)3 2n n→ (n 3n 3 lim (n 1)3(2) n→ 3 1. 2 1 an (The nth 22. lim n32 n 3n 1 an n→ 1 diverges for x lim 1 (n lim n→ an 1 lim n→ 2 1 1. 2 x an n→ lim lim n→ lim n→ 2 1)ln (n 1) 2 2n 1 n ln n n 1 ln (n 1) n ln n 24. lim n→ lim n→ 1, or x n→ lim n→ lim n→ lim n→ 2 10, so the radius of convergence is an 1 lim (n n n→ an 1) x n 3 1 n2 n xn lim x x n→ 1 and diverges for x 1, so 25. lim n→ (2n 1)! 1)! n! 3)! n1 (2n 3)(2n 2) 1 0 1. 2(2n 3) an 1 xn lim n→ an (n x lim n→ 3 1 n n1 1) n x 3 13 The series converges for x n 3n xn 3 and diverges for x 3, so the radius of convergence is 3. lim an 2 3 1, or x the radius of convergence is 1. 26. lim n 1 1 , and 3 1 1 , so the radius of convergence is . 3 3 2 The series converges for x 16. Converges by the Ratio Test, since an 2 2 3 10. (n n→ (2n 1 an 3x 2n n 15. Converges by the Ratio Test, since lim n 3x 23. This is a geometric series which converges only for Term Test can also be used.) 14. Converges by the Ratio Test, since n→ 1 The series converges for 3x 10 an 3x 2 n n1 lim n→ an (n 1)! n (n 1)n 1 n! (n 1)n n (n 1)(n 1)n nn n1 1 (1 1/n)n 1 e n→ an 1 an x 2n 3 1)! n→ (n n! lim x lim 2n 1 n→ x2 n 0 1 The series converges for all values of x, so the radius of convergence is . 27. lim n→ an 1 lim (n 1) x 5n n→ an lim 1 n→ x 3 3n 1 5n 1 nx x 5 3n 3 5 The series converges for x 3 5 and diverges for 17. One possible answer: x ∑ 1 diverges (see Exploration 1 in this section) even n 1n 1 though lim 0. n→ n 28. lim n→ 3 an 5, so the radius of convergence is 5. 1 an 18. One possible answer: Let an 2 n and bn n 3 a n 20. This is a geometric series which converges only for x5 1, so the radius of convergence is 1. 1] 4 and diverges for x 1, 29. lim n→ an an 1 lim n→ 1xn n 1 3n n1 3 The series converges for x nx (4x 1) 1, or x 1 4 convergence is . 1 4 1 , so the radius of 4 n lim n→ x 3 3 and diverges for x the radius of convergence is 3. 21. This is a geometric series which converges only for 4n(n2 1) nxn 4, so the radius of convergence is 4. 3n is a divergent geometric series. 2 19. This is a geometric series which converges only for x so the radius of convergence is 1. 1 n1 The series converges for x Then ∑an and ∑bn are convergent geometric series, but ∑ bn ∑ (n 1) x n [(n 1)2 n→ 4 x x lim 4 n→ 4 lim x 3 3, so 384 Section 9.4 30. lim an n→ 1 lim 1)! x 4 n n! x 4 n (n n→ an lim (n 1) x n→ 35. This is a geometric series with first term a 1 common ratio r 4 (x 4) 1) 1 The series converges only for x 4 2 x 1 a 1 r 31. lim n→ 4 n1 lim ( 2) (n 2) x 1 2n (n 1) x 1 n lim 2 x 1 1 and diverges for 2 1 1 , so the radius of convergence is . 2 2 The series converges for x x 1 an n→ 1 4x (n 5 1)3/2 n→ n 3 4 2x 3 1)2 (x 4 1 1)2 (x common ratio r 9 1 and . It converges only when 1, so the interval of convergence is 9 5 2n 4x 4 2x 36. This is a geometric series with first term a 3/2 5)2 lim n→ an 1 2n 3 lim (4x 32. lim 1)2 x2 1 n→ 2x 4 (x 4 n1 x2 n→ an 1)2 (x 1 convergence is 0. 1 . It converges only when 3. 4, so the radius of Sum an 1 and 1, so the interval of convergence is 4 (x 1)2 (x x 2. a Sum 1 r 5)2 (4x 1 The series converges for (4x to 4x x 33. lim n→ 5 5 4 an 1 1, which is equivalent 1 and diverges for 4 1 1 . The radius of convergence is . 4 4 lim n1 x n x 0 n→ lim n→ 2 2n x 2n 3 1 Sum 2n 1 (x 2)2 n→ 2 1 (x 2)2 2 1 The series converges for (x 2 x x equivalent to x x 2 2 1 and 1. It converges only when 2 1, so the interval of convergence is 16. 1 a 1 r 2 x 1 1 2 4 x 1 and 1 common ratio r ln x. It converges only when ln x lim so the interval of convergence is Sum 2)2 8 38. This is a geometric series with first term a 2 2n x 9 2x x2 8 1 and diverges for 1, so the radius of convergence is 1. 1 1 2 The series converges for x an 9 2x common ratio r x an 1)2 37. This is a geometric series with first term a n n→ 34. lim 9 (x x2 lim x x 9 9 n x 1 1)2 (x 1 5 4 1, or x n→ an 5) 2 1, which is 2, and diverges for 2. The radius of convergence is a 1 r x e. 1 ln x 39. This is a geometric series with first term a common ratio 2. 1 1 e x2 1 3 a 1 1 r 1 1 and x2 . It converges only when so the interval of convergence is Sum 1, x2 1 3 3 2 x 3 (x 2 1) 1 3 2. 3 4 x2 1, Section 9.4 40. This is a geometric series with first term a sin x sin x common ratio . Since 2 2 1 and 46. 1 for all x, the interval ∑ n 1 (2n 6 1)(2n 3 3 1) s1 3 s2 (3 1) 1 s3 (3 1) 1 41. Almost, but the Ratio Test won’t determine whether there is convergence or divergence at the endpoints of the interval. sn 3 42. (a) For k S of convergence is x a Sum 1 1 r 2 sin x 2 1 2 sin x N, it’s obvious that … a1 . ak … a1 For all k ak ∑ nN1 cn. 47. N, … aN a1 … a1 a1 … aN a1 … aN aN … cN 1 ∑ aN … ak cn (b) Since all of the an are nonnegative, the partial sums of the series form a nondecreasing sequence of real numbers. Part (a) shows that the sequence is bounded above, so it must converge to a limit. 43. (a) For k N, it’s obvious that … dk For all k … dk d1 d1 d1 an. dN d1 … dN 1 … dN d1 … dN … ∑ ak dN sn S 48. N, d1 … nN1 ∑ s1 … 1 1 an sn (b) If ∑an converged, that would imply that ∑dn was also S nN1 convergent. 49. s1 44. Answers will vary. S lim sn n→ 1 1) ∑ n1 4n 1 3 4n s2 1 s3 1 5 1 5 1 9 1 9 1 1 9 1 9 sn 1 13 1 s2 ∑ ∑ 2n (n 2 s3 dk 1 1 lim sn 1 1 13 S 50. s1 s2 s3 sn S 1 ∑ 1 1)2 1 4 1 4 1 4 n1 ∑ n1 3 7 3 7 3 5 5 1)2 (2n 5 25 5 25 1 n2 1 4 1 4 1)2 (2n 5 25 5 5 25 5 49 5 49 5 1 9 1 9 1 1)2 (n 1 9 1 1 9 1 16 1 16 1 1 1)2 (n lim sn 1 n→ 1 1 1 2 1 1 1 2 1 1 1 2 1 lim sn 3 1 2 1 1 2 1 n n→ 3 5 3 5 1 5 n→ n 1n aN 4 45. 3)(4n n 1 (4n 1 s1 1 5 1 s2 1 5 1 s3 1 5 1 sn 1 4n 1 40n 1)2(2n 1)2 5 5 9 5 5 5 9 9 5 5 5 9 9 5 5 (2n 1)2 s3 nN1 3 3 5 3 5 2n 3 n 1 (2n s2 3 1 1 lim sn s1 ck 1 3 2n 3 2n n→ ∑ ∑ n1 385 3 1 1 3 3 4 1 ln 3 1 ln 3 1 ln 4 1 ln 5 1 ln 2 1 ln 4 1 1 1 1 ln 3 ln 2 1 1 1 ln 3 ln 2 ln 4 1 1 1 ln 3 ln 2 ln 4 1 1 ln 5 ln 2 1 1 ln(n 2) ln 2 1 lim sn ln 2 n→ 1 1 4 386 Section 9.5 1 51. s1 tan s2 (tan s3 tan 1 tan 1 (tan 1 sn 4 S 52. x 1 (tan 2) 1 tan 1 1 2 2 tan 1 3) tan 1 (tan 1 2 tan 1 3) 1. We first note that the Integral Test applies to any series of lim tan n→ 1 n 4 2 4 If p 1 1 dx xp n0 k→ k→ n1 ∑ nx x)2 1 is negative for all x 1 dx xp 1 k→ kp 1p 1 p1 1 p1 n lim 1 1 0 Multiply by x: (1 k lim n0 x is continuous and positive for all x xp p = lim ∑ nx x)2 p x 0, 0. 1: Differentiate: (1 1 where p is positive. This is because the np and f (x) 1) ∑ xn 1 The p-Series Test function f (x) (n 4 s Section 9.5 Testing Convergence at Endpoints (pp. 496–508) the form ∑ 4) 4 lim sn 1 2) 1 tan 4 Exploration 1 3 tan n→ 1 1 tan 1 2 3 1 (tan 4 1 tan 4 1 1 1 x p1k p1 1 1 (since p 1 0) . The series converges by the Intergral Test. n0 Differentiate: (1 d x dx (1 x)2 (1 x)2(1) x) (1 x ) x1 (1 x)3 x (1 ∑ n2x n 1 x)3 (1 2x (x)(2)(1 x)4 x)( 1) 2. If 0 p 1 dx xp 3 1 13 2 lim ∑ n 2x n 1 : 2 ∑ n2 n0 ∑ nn n 02 The sum is 6. 1n 2 1 dx xp 1 p1k x k→ 1 k→ n0 2 6 k→ p1 n0 x(x 1) (1 x)3 13 22 k lim lim Multiply by x: Let x 1: 1 1 (k 1 1p p 1) (since 1 p 0). The series diverges by the Integral Test. If p 0, the series diverges by the nth Term Test. This completes the proof for p 1. 3. If p 1 1: 1 dx xp k lim k→ 1 1 dx x k lim ln x k→ 1 lim ln k k→ . The series diverges by the Integral Test. Exploration 2 The Maclaurin Series of a Strange Function 1. Since f (n)(0) 0 for all n, the Maclaurin Series for f has all zero coefficients! The series is simply ∑ 0 x n 0. n0 2. The series converges (to 0) for all values of x. 3. Since f (x) 0 only at x 0, the only place that this series actually converges to its f-value is at x 0. 387 Section 9.5 Quick Review 9.5 10. Converges by the Limit Comparison Test, since 1 p dx with p 1x 1 2. Diverges, limit comparison test with integral of x 1. Converges, since it is of the form 1 lim n→ 1, and ∑ n1 1 n2 2. 11. Diverges by the nth-Term Test, since 2 4. Converges, comparison test with integral of 2 x 3n lim 1 1 1 3 3n n→ 0. 1 5. Diverges, limit comparison test with integral of 0 22 {un } is a decreasing sequence of positive terms with 8. No, neither positive nor decreasing for x lim un 3 0. n→ 9. No, oscillates 10. No, not positive for x 13. Diverges by the nth-Term Test, since lim 1 n→ Section 9.5 Exercises 5 x 1 3∑ 3 n n1 n1 1 dx diverges. 1 , which diverges by n1/2 3. Diverges by the Direct Comparison Test, since 2 and ∑ n2 ln n n 1 for n 1 diverges. n lim un n→ 1 2x 1 x1 and observe that x1 1 (x 1) ( x 1)(1) f (x) 1 dx diverges. (x 1)2 6. Converges, since it is a geometric series with 1 n n→ Then an 1 n 0 and bn e 0, and ∑ e for n n converges as a n0 geometric series with r e 1 an lim 1. 8. Converges by the Direct Comparison Test, since n ln n 2 ln n 1 , 2 1 . 2 0.37. n→ lim n→ bn 1 . n 1 and bn n2 0.91. 7. Diverges by the nth-Term Test, since lim n sin en 1 e 2n ln n ln n2 16. Diverges by the Limit Comparison Test. 1.44. Let an 1 ln 3 , 1.) 15. Diverges by the nth-Term Test since 5. Diverges, since it is a geometric series with r 1 1 1 x 2x , 2(x 1)2 x 2x f (x) which means each term is 1 ln 2 n n 0. (To show that un is decreasing, let which is negative, at least for x 4. Diverges by the Integral Test, since r . then {un} is a decreasing sequence of positive terms with the p-series Test. n 10 n n10 14. Converges by the Alternating Series Test. If un 1. Diverges by the Integral Test, since 2. Diverges because ∑ 1 , then ln n 12. Converges by the Alternating Series Test. If un x 7. Yes, for N 5n 3 3n 2)(n 2 5) 1 n2 converges as a p-series with p 1 3. Diverges, comparison test with integral of x 6. Yes, for N n 2(n 1 n2 1 n 1 n 0 for n lim n→ 2 and n 1 n lim 1 n→ 1 n 1. Since ∑ bn diverges, ∑ an also diverges. n1 n1 17. Converges absolutely, because, absolutely, it is a geometric series with r 0.1. 18. Converges conditionally: 9. Converges by the Direct Comparison Test, since n n2 1 with p 1 for n n3/2 3 . 2 1, and ∑ n0 1 converges as a p-series n3/2 1 n 1 n 1 , then {un} is a decreasing sequence n2 1n of positive terms with lim un 0, so ( 1)n 1 2 n n→ n1 If un n2 ∑ converges by the Alternating Series Test. But ∑ n1 1 n n2 1 n diverges by the Direct Comparison Test, since n2 1 1 for n 1 and diverges. n n 1n ∑ 388 Section 9.5 2n converges by the 3 19. Converges absolutely, since ∑ n2 n1 Ratio Test: an lim n→ (n 1) n1 22 1 3n n2 2 3 2 3 (c) None 1. 28. This is a geometric series which converges only for x5 1, or 6 x 4. 20. Converges conditionally. (a) ( 6, ∑ 29. This is a geometric series which converges only for 4x 1 diverges by the integral test, since n ln n lim ln ln x b→ n→ n! 2n and so the terms do not approach 0. n1 n1 sin n converges by direct n2 1 , which converges as a p-series with n2 23. Converges conditionally: n 1 n1 n ( 1)n 1 converges 24. Converges absolutely, since ∑ n1 n1 1 . 2 1 , n1/2 ∑ nn n1 n1 ∑( cos n n n1 1)n+1 . n x ∑( 1 : 3n 1. Check x 1 n 1 3x 2n 1 1: ∑ 2 2 1, or n 1) converges n 1n 1 diverges. n n 1n ∑ 1 ,1 3 1 ,1 3 1 3 (c) At x x 2 1, or 10 8 x 12. (b) ( 8, 12) (c) None n→ an 1 lim (n n→ an n 1) x n 3 1 n2 nxn The series converges absolutely when x , then {un } is a decreasing sequence of positive terms with lim un n→ 0, so ∑ ( 1) n1 n n 1 n lim n x 1. For x 1/2 n Since ∑ vn diverges as a p-series with p nth-Term Test. (a) ( 1, 1) But ∑ un diverges by the Limit Comparison Test: n→ 1 n converges by the Alternating Series Test. u 1 , then lim n n1/2 n → vn (a) 32. lim 26. Converges conditionally: diverges. n 3x The series converges absolutely when 3x 1 , which n3/2 (See Examples 2 and 4.) n1 1 (a) ( 8, 12) 25. Converges conditionally, since ∑ If vn n→ an 3x 2 n n1 31. This is a geometric series which converges only for cos n converges as a p-series. n lim n diverges by direct comparison to ∑ which diverges as a p-series with p If un 1 (b) 0, so ∑ n→ n1 n→ , then {un} is a decreasing sequence of by the Alternating Series Test. 1 an conditionally. Check x positive terms with lim un n1 30. lim n1 1 But ∑ 1 ,0 2 1 3 2. 1 0. 1 ,0 2 (b) x (c) None 22. Converges absolutely, since ∑ comparison to ∑ 1 2 1, or (a) . 2 21. Diverges by the nth-Term Test, since lim If un 1 b 1 dx x ln x p 4) (c) None converges by the Alternating Series Test. 2 4) (b) ( 6, 1 , then {un} is a decreasing sequence of n ln n 1 positive terms with lim un 0, so ( 1)n 1 n ln n n→ n2 If un n2 (a) ( 1, 1) (b) ( 1, 1) 1 an But ∑ 27. This is a geometric series which converges only for x 1 . 2 1 1 , u also 2n 1 n ∑ (b) ( 1, 1) (c) None x 1, or 1, the series diverges by the 1. 389 Section 9.5 33. lim an n→ 1 xn lim n→ an (n 1 1) n 3n 1 3, Σ when x n1 x n3 3 . 2 n x n 38. lim n→ 3 (a) Only at x (b) At x 1) x 1 4 4 4 4 4 (c) None an n→ 1 n→ an 0, x ,x 39. lim lim n→ an 1)! x 4 n n! x 4 n (n lim (n (c) None an lim n→ (b) [ 3, 3] n→ 1 3. Furthermore, (a) [ 3, 3] 34. lim an Σ1 n1 , which also converges 3/2 n n as a p-series with p x 1 The series converges absolutely for x n n 3n n x (n 2n 3 n! x 2n 1)! 1 lim n→ x 2 n 2 n 1(n 2 n (n lim n→ 1n 1n 2) x 1) x 1 2(x The series converges absolutely for 2(x 0 1 1 an The series converges absolutely for all real numbers. 1 2 (a) All real numbers 1) 1, or nth-Term Test. x (b) All real numbers (a) x3 5n 5 nx 3n n→ n→ x3 The series converges absolutely for 1, 5 x3 or 8 x 2. For 1, the series diverges by 5 an 1 lim an (n 1) x 5n 3n (c) None 40. lim n→ (n 1) x n 4n 1[(n 1)2 n→ lim 1 Check x 4n(n2 1) nxn 1] x 4 1, or 4 x 4. 4: n ∑ (n2 1) 1 converges by the Alternating Series Test. (a) 1, ∑ n2 4: an n→ diverges by the Limit Comparison Test 1 lim n→ an 1 (a) [ 4, 4) (b) ( 4, 4) ∑( 37. lim n→ 4 1 lim n→ an 1 xn 3n 1 1 n xn x 3 ∑ 3 3, the series diverges by the nth-Term Test. (a) ( 3, 3) (b) ( 3, 3) (c) None 1 n x n x x 3. 1, or 1. 1: converges by Alternating Series Test. n 1 n1 3, or 1 n Check x 3n The series converges absolutely for x For x 1) n1 n n x Check x an ∑ The series converges absolutely for x with ∑ (c) At x n1 x 1: diverges as a p-series with p n (a) [ 1, 1) (b) ( 1, 1) (c) At x 1 1 . 2 5)2 1, or 1 n3/2 3 : 2 3 2 41. lim 1 1 . 1n n 5)2 2n 1 (c) None n0 n0 ∑ (4x 1 3 2 n n n 3/2 5 2n 4x ∑ x 4 (b) 1, The series converges absolutely for Check x n→ 5 2n 3 1)3/2 4x (n ( 1) 1x 1: n3/2 n1 n 3 converges as a p-series with p . Check x 2 12n 1 3 . 3/2 converges as a p-series with p 2 n1n (c) None 1 lim 3 . Check x 2 (b) ( 8, 2) an 1 an The series converges absolutely for (4x (a) ( 8, 2) an an n→ the nth-Term Test. 36. lim 1, the series diverges by the 13 , 22 1 1 1) 13 , 22 (b) (c) None 35. lim 3 . For 2(x 2 1) 390 Section 9.5 46. (a) Diverges by the Limit Comparison Test. 42. This is a geometric series which converges only for ln x 1, or (a) x 1 . Then ak 0 and k1/2 ak k1/2 1 bk 0 for k 1 and lim lim . bk k→ k→ 2 2k 7 1 Since bk diverges as a p-series with p , 2 k 1 1 Let ak e. 1 ,e e (b) 1 e 1 ,e e 2k and bk 7 ∑ ∑ ak also diverges. k1 (c) None 43. n 13 109 365 24 3600 ln (n 1) sum 1 ln n ln (4.09968 1017 1) sum 1017) 40.5548... sum 41.5548... 40.554 sum 41.555 1 f (x) dx 1 a1 k→ e 0. (c) Converges absolutely by the Comparison Test, since cos k k2 1, 1 and ∑ 1 for k k2 k p-series with p k1 1 converges as a k2 2. n … an a1 f (x) dx. 1 If the integral diverges, it must go to infinity, and the first inequality forces the partial sums of the series to go to infinity as well, so the series is divergent. If the integral converges, then the second inequality puts an upper bound on the partial sums of the series, and since they are a nondecreasing sequence, they must converge to a finite sum for the series. (See the explanation preceding Exercise 42 in Section 9.4.) (d) Diverges by the integral test, since 3 b 18 dx x ln x lim 18 ln ln x b→ 3 47. One possible answer: ∑ n3 1 n ln n This series diverges by the integral test, since 3 45. 1k k lim 1 ln (4.09968 44. Comparing areas in the figures, we have for all n n1 (b) Diverges by the nth-Term Test, since 1017 4.09968 b 1 dx x ln x lim ln ln x . Its partial sums are b→ 3 roughly ln (ln n), so they are much smaller than the partial y sums for the harmonic series, which are about ln n. y = f (x ) an – 1 0 ( 1)k 48. (a) ak a N aN + 1 N N+1 N+2 n an n+1 x 1/k 1 6(kx)2 dx 0 ( 1)k 1 2k 2x 3 1/k 0 y ( 1) y = f (x ) (c) The first few partial sums are: an N–1 N N+1 N+2 n–1 x n Comparing areas in the figures, we have for all n n1 f (x) dx N aN … S1 N, n an k (b) The series converges by the Alternating Series Test. a N a N + 1 aN + 2 0 k12 aN f (x) dx. N If the integral diverges, it must go to infinity, and the first inequality forces the partial sums of the series to go to infinity as well, so the series is divergent. If the integral converges, then the second inequality puts an upper bound on the partial sums of the series, and since they are a nondecreasing sequence, they must converge to a finite sum for the series. (See the explanation preceding Exercise 42 in Section 9.4.) 2, S2 S7 319 ,S 210 8 5 ,S 34 1, S3 533 ,S 420 9 7 47 37 ,S ,S , 65 30 6 30 1879 . For an alternating 1260 series, the sum is between any two adjacent partial sums, so 1 S8 sum S9 3 . 2 49. (a) Diverges by the Limit Comparison Test. Let an n n and bn 3n2 1 an 1, and lim n→ bn 1 . Then an n n2 3n2 1 n→ lim ∑ bn diverges, ∑ an diverges. n1 n1 0 and bn 1 . Since 3 0 for 391 Section 9.5 ∑ 3n2n (b) S 1 n1 ∑ 3n23 3 n n1 53. ln (1 . 1 Σ This series converges by the Direct Comparison Test, since n1 ∑ n12 is convergent as a 3 3n2 1 1 and n2 n p-series with p Σ( n1 x) ( 1) n n1 (b) 1 x 1, the series is Test. 2. ∑( 54. arctan x x2 x3 2 x , so at x . This series converges by the Alternating Series 1)n n0 x) n 1 At x 50. (a) From the list of Maclaurin series in Section 9.2, ln (1 n 1x 1)n 3 … ( 1)n 1x n ∑( …. n n0 x 2n 1 2n 1 1, the sequence is 1)n( 1)2n 2n 1 ∑ 2(n 1) 1 , which converges by the 1 n n0 Alternating Series Test. At x 1 1, the sequence is ∑ 2(n 1) 1 , which converges by the Alternating Series Test. n 3 2 1 2 (c) To estimate ln , we would let x n0 55. (a) It fails to satisfy un The truncation error is less than the magnitude of the x6 6 1 26 6 ∑ (b) The sum is sixth nonzero term, or 1 384 un 1 for all n ∑ 21n 1 n n 13 N. 1 1 2 n1 0.002605 n1 51. lim k→ ak ∑ ( 1)n 1x 2n 2n 1 2n k1 1 ( 1)n 1(x 2)n n k1 2x lim 3) k→ ln (k 1 ak ln (k 2) 2k x k 1 ln (1 2 1 2 Check x k (b) lim 1 , 2 (c) p-series when p (b) For p 1, the series is ∑ Integral Test, since 2 ∑ n2 p 1 converges as a np 1 , which diverges by the n ln n 1 dx x ln x 1 1, we have p n ln n b lim ln (ln x) b→ . 2 1 , so n ln n 1 diverges by the Direct Comparison Test with n p ln n 1 from part (b). n ln n n ( lim n→ nn 2n 1 n lim n→ n) 2 2 1 2 n lim n→ 2n n an n lim n→ , n odd 1 1 2 n 2n lim n→ , n odd n lim n→ , n even Thus, lim n→ n an n an n 1 2 n 2 lim n→ , n even n x Check x 1 2 1 2n 1 , so the series converges. 2 an lim 1n nx n→ n 4 x 1 4 The series converges absolutely if 3 1. n2 (c) For 0 n3 n→ n→ , n odd n→ 52. (a) The series converges by the Direct Comparison Test, 3, and ∑ n2 2n n lim an lim 58. (a) lim 1 for n np an The series converges. ∑ 1 n ln n n n→ ∑ p 1 The series converges. ( 1) converges by the Alternating Series Test. 2) k 0 ln (k 1 Check x : 2 1 diverges by the Direct Comparison Test, since 2) k 0 ln (k 1 1 1 for k 2 and diverges. The original k ln (k 2) k 2k 1 1 series converges for x . 2 2 since n n→ x) 1 . 2 1 : 2 x ∑ 57. (a) lim 2 2x The series converges absolutely for x or 1/2 1/2 56. Answers will vary. 0.002605. ∑ 1 1 . 2 Thus, a bound for the (absolute) truncation error is (d) 1/3 1/3 x 1 4 5. 3: ∑ ( 1)n diverges. n0 Check x 5: ∑ 1n diverges. n0 The interval of convergence is ( 3, 5). 1, or 392 Chapter 9 Review 58. continued n (b) lim n→ an 2n nx lim x lim n 3n n→ 2 x x n n→ n3 x2 The series converges absolutely if 3 1 x4 n3 n an 3 x 4n n→ x4 The series converges absolutely for 1, 3 an 2. lim 2 3 1, or or 7 lim x 1. Check x 7: ∑ Check x 5. 1: ∑ n1 1: ∑ Check x n1 5: ∑ Check x ( 1)n converges. n n1 1 diverges. n n1 n n→ an n lim (b) [ 7, n→ 1) (c) ( 7, 1) (d) At x 2n x n ( 1)n converges n 1 diverges. n (a) 3 The interval of convergence is [ 1, 5). (c) lim 4n 1 1)3 n 1 x n → (n 1 7 2x 3. This is a geometric series, so it converges absolutely when The series converges absolutely if 2x r ∑ 1 diverges. (a) ∑ Check x 1 : 2n Check x 1 n n→ an x n→ ln x n 1) 1, or 1, or ∑ ln 0 1n e e: ∑ (ln e) n n0 ∑( 1)n diverges. (d) None an n→ 1 an x 1 2n (2n 1)! 1)! x 1 2n 2 n→ (2n x 12 lim 0 1)(2n) n→ (2n lim The series converges absolutely for all x. n0 The interval of convergence is 5 . 2 15 , 22 n0 ∑ 1n diverges. x 15 , 22 4. lim 1 : en 1 2 3 2 ln x e. Check x 1), the series converges absolutely when (c) n lim Check: x 2 (x 3 (b) The series converges absolutely if ln x 1 e 1 and diverges for all other values of x. Since 2 (x 3 11 ,. 22 The interval of convergence is (d) lim r 1 1 x . 2 2 1 : ( 1)n diverges. 2n 1 1, or (a) 1 ,e . e (b) All real numbers (c) All real numbers s Chapter 9 Review Exercises (d) None (pp. 509–511) 1. lim n→ an 1 an lim n→ x (n n1 1)! 5. lim n! xn lim n→ x n 1 0 an n→ 1 an 3x (d) None n2 3x 1n 3x 1 3x 2 . Furthermore, when 3 1 1, we have an and n2 1 1, or 0 1 (a) (c) All real numbers n→ 1n 1 1)2 3x (n The series converges absolutely for The series converges absolutely for all x. (b) All real numbers lim ∑ n1 x 1 converges as a p-series with p n2 2, so ∑ an also n1 converges absolutely at the interval endpoints. (a) 1 3 (b) 0, 2 3 (c) 0, 2 3 (d) None 393 Chapter 9 Review an 6. lim n→ (n 2) x 3n 3 (n 1) x 3n n→ 1 x3 lim an n→ The series converges absolutely for x 3 1 x 1. When x an 9. lim 1 xn lim n→ an n 1 n xn 1 x The series converges absolutely for x 1, or 1, the series diverges by the nth Check x ∑ Term Test. n1 ∑ (b) ( 1, 1) (c) ( 1, 1) 1 x 1. 1: ( 1)n converges by the Alternating Series Test. n Check x (a) 1 1, or 1 1: 1 . 2 diverges as a p-series with p n n1 (a) 1 (d) None (b) [ 1, 1) an 7. lim n→ 1 lim (n n→ an 2x 2) 2x 1 n (2n 3)2n 1 1 (2n 1)2n (n 1) 2x 1 n (c) ( 1, 1) (d) At x 1 2 The series converges absolutely for 3 2 1 2x 1 . When 2 2 x 2x 1 10. lim n→ 1, or 2 1, the series diverges by the lim 1 e 1 an x lim n→ (n 1, we have an ∑ 1 and ne n 1 1 ne e, so ∑ an also converges n1 (a) (d) None an 1, absolutely at the interval endpoints. 31 , 22 (c) ex 1 . e x converges as a p-series with p 31 , 22 (b) ne e xn n The series converges absolutely for e x Furthermore, when e x (a) 1 n→ en 1 x n 1 (n 1)e n→ 1 an or nth-Term Test. 8. lim an 1 xn 1 1)n n→ (n 1 lim 1) 1 1n n 1 nn xn x lim n→ (n x 1 lim e n→ n 1 nn 1)(n 1 e (b) 0 1) 11 , ee (c) 11 , ee n (d) None The series converges absolutely for all x. Another way to see that the series must converge is to observe that for n 2x, we have xn nn 1n , so the terms 2 3n an (n 1) x 2n 1 n→ n→ x2 The series converges absolutely when 1, 3 11. lim an 1 lim are (eventually) bounded by the terms of a convergent or geometric series. When x A third way to solve this exercise is to use the nth Root (a) Test (see Exercises 57–58 in Section 9.5). (b) ( 3, 3) (c) ( 3, 3) (a) (b) All real numbers (c) All real numbers (d) None 3 x (n 2) x 2 n 3n 1 1 x2 3 3. 3, the series diverges by the nth Term Test. 3 (d) None 394 Chapter 9 Review 12. lim an n→ 1 lim x n→ an 1 2n 2n 3 3 1 2n x 1 12 The series converges absolutely when x or 0 x 12 x 2n 1 1 1, or 2 n0 n ∑ 2n 1 2: ∑ n0 ( 1) ( 1) 2n 1 n0 ( 1) converges 2n 1 (a) ( 1)n converges conditionally by the 2n 1 (b) ( 3, 3, 3) (d) None 1 1 (a) 1 1 x n→ 1 lim 1)! x 2n 2n 1 (n n→ (n lim n→ 18. f (x) 2 1)x 2 2 0, x ,x 2 2n n! x 2n ln (1 x) 19. f (x) sin x 20. f (x) 0 only (c) x 10x n 1 ln (n 1) n→ 1 lim an 1 10 ln n 10x n x ∑ 1 : 10 n 2 ∑ 1 : 10 n 1 Test, since ln n 2 1x n n , 5 . 3 x 2n 1 (2n 1)! …, 0. tan 1 x cos x2 2! ( 1)n n ( 1) converges by the Alternating ln n xn n! x5 5 … . Sum …, evaluated at tan 1 ( 1)n ∑ x 2n 1 2n 1 1 …, . (Note that 6 3 when n is replaced by n 1 diverges by the Direct Comparison ln n 1 1 for n 2 and diverges. n n 2n …, 1 . 2 3 … x3 3 n 1x x 2n (2n)! 2. x 1 … 3 1, the general term of tan 1 x 2n 1 2n 1 , which matches the general term given in the exercise.) 23. Replace x by 6x in the Maclaurin series for 1 1 x given at the end of Section 9.2. 1 10 1 (b) (c) 11 , 10 10 1 n→ The series converges only at x 0. 1 an lim (a) 0 (b) x 0 only (c) x 0 (d) None 6x (6x)2 … … 36x 2 … (6x)n (6x)n … 1 1 x given at the end of Section 9.2. (n 2)! x n 1 (n 1)! x n n→ an (6x) 24. Replace x by x 3 in the Maclaurin series for 1 10 (d) At x 6x 1 1 1 11 , 10 10 n ln ( 1)n sin x4 4! e ln 2 ln 2. Sum becomes ( 1) Check n 2 3 … . Sum x evaluated at x Series Test. 15. lim 3 1 22. f (x) 1, x2 2! 1 x 10x 1 . 10 Check n (a) ex 21. f (x) The series converges absolutely for 10x or cos x 0 an n→ . Sum evaluated at x (d) None 14. lim x5 5! ( 1)n 0. (a) 0 (b) x … ln 1 x3 3! x evaluated at x The series converges only at x x3 3 2 . Sum 3 evaluated at x 0 0 x2 2 x 4 . 5 1 4 1 …, ( 1)nx n 1 (c) (0, 2) 0 and x … x2 x 1 . Sum 4 evaluated at x (b) [0, 2] an 3. It 3) (c) ( 17. f (x) an x 3 Alternating Series Test. 13. lim 3 , so it converges n conditionally by the Alternating Series Test. (d) At x 1 2 diverges for all other values of x. 0: ∑ Check x x2 absolutely when 1, 2. Check x x2 16. This is a geometric series with r lim (n 2) x (x 0) x3 1 (x 3) 1 1 x3 (x 3 ) 2 x6 … … ( x 3)n ( 1)nx 3n … … 25. The Maclaurin series for a polynomial is the polynomial itself: 1 2x 2 x 9. 395 Chapter 9 Review 26. 4x 1 4x x 33. Use the Maclaurin series for e x given at the end of 1 1 x 4x(1 x 4x 2 4x … x2 4x 3 …) xn … 4x n Section 9.2. … 1 x2 xe 5 27. Replace x by x in the Maclaurin series for sin x given at ( x)3 3! ( x)5 5! … ( 1)n ( x)2n 1 (2n 1)! … 2x in the Maclaurin series for sin x given at 3 the end of Section 9.2 ( 2x sin 3 30. x e x 2 5! 4x 5 3645 … ( ( 1)n … 2x 2n+1 1)n 3 (2n 1 (2n x2 2! x 1 1 2 x2 2! 1 3x ) ln (1 x x4 4! ( 5x)2 2! … xn n! 1 32. Replace x by end of Section 9.2. e x/2 1 1 (3x)5 5 … ( 1)n (3x)2n 1 2n 1 ( 2x)2 ( 2x)3 … 2 3 n ( 2x) … ( 1)n 1 n n 8x 3 … (2x) 2x 2x 2 3 n 36. Use the Maclaurin series for ln (1 … x) …. x) given at the end of Section 9.2. x ln (1 … x x2 2! … … x 2n (2n)! x) ( 1)n xn n! … 37. f (2) x2 2 x3 2 (3 2! 22 x 8 … n! xn 1 n! 2 ( x)] ( x) 3 x4 3 1 x) (3 2(3 x) 6(3 x) 4 n!(3 1 x 38. f ( 1) x2 2)n f ( 1) (6x f ( 1) 6x (x 2x 2 (3x 2 1 5) x 1 1 … 2) 2 1 1 1 (x 2) 3 1 … 0 for n 10, so 5 7(x f ( 1) 2! 5 1 4. x3 2 7 f ( 1) 6, so 3! f (n)( 1) … … … 4x) x 4) x n f (n)(2) n!, so n! n1 2) f ( 1) n 1 ( x) … f (2) 2! f (2) 6, so 3! x2 (x (x 3 1 2, so x2 x) ( 1)n 1 x2 (x 3 … xn n 3 f (2) … 1 x2 2 x) 3 … f (2) ( … 2 f (2) … 1 xn 2 x ln [1 ( x) 2 x x2 x in the Maclaurin series for e x given at the 2 x 2 x 2 x given at 2x 1)! … 5x)4 … 4! ( 5x)2n … ( 1)n (2n)! n 5x (5x)2 … ( 1)n (5x) 2! 4! (2n)! 1 1 2x in the Maclaurin series for ln (1 2x) f (n)(2) 5x … n! given at the end of Section 9.2. the end of Section 9.2. cos (3x)3 3 3x 35. Replace x by 5x in the Maclaurin series for cos x given at 31. Replace x by x … x x 1 1 2 1)! 1 2x 2n 1 3 x3 x5 x7 … 3! 5! 7! x 2n 1 … ( 1)n (2n 1)! 2n 1 x3 x5 x7 … ( 1)n x 3! 5! 7! (2n 1)! sin x ex 2x 5 3 4x 3 81 2x 3 29. 2x 3 3 3! 2x 3 ( 1)n 2n 1 the end of Section 9.2. tan 28. Replace x by ( x 2 )n n! … 34. Replace x by 3x in the Maclaurin series for tan the end of Section 9.2. x … x 2! x3 x sin x ( x 2)2 2! ( x 2) x1 2x 2 2 1) 5(x 1)2 (x This is a finite series and the general term for n 1)3 4 is 0. 396 Chapter 9 Review 1 xx 3 39. f (3) f (3) 2x f (n)(3) n! 1 1 x 3 40. f ( ) x3 f (3) 2 1 , so 2! 27 27 f (3) 2 1 6x 4 x 3 , so 3! 27 81 ( 1)n 3n+1 1 1 1 (x 3) (x 3)2 (x 3)3 9 27 81 n (x 3) ( 1)n n+1 3 f (3) ∑ 3 x3 sin x x f( ) 1 , then {un} is a decreasing sequence of ln (n 1) ( 1)n positive terms with lim un 0, so converges ln(n 1) n→ n1 If un 1 9 2 x f (3) 45. Converges conditionally: 1 3 by the Alternating Series Test. The convergence is conditional because … 1 cos x x 0, so (x 1 (x 3! ) lim 1 (x 7! ( 1)n (2n 3n 1 1)! n → (n 1 n! 3n lim an lim n→ 3 n 1 0. 1n for n 2 2n3n nn )5 12 and ∑ n1 1n is a convergent 2 geometric series. Alternately, we may use the Ratio Test or the nth-Root Test (see Exercise 57 and 58 in Section 9.5). 1 1 1 . ln 2 48. Converges absolutely by the Direct Comparison Test, since … )7 an n→ 1 (x 5! )3 1b ln x 2 lim b→ 47. Converges absolutely the the Ratio Test, because 0, if k is even 1, if k 2n 1, n even 1, if k 2n 1, n odd sin x 1 dx x(ln x)2 2 f( ) 0 2! f() 1 1, so 3! 6 sin x x f (k)( ) n 46. Converges absolutely by the Integral Test, because cos x x f() ∑ 1 diverges, so ∑ n 1n Comparison Test. 0 f( ) 1 1 for n 1 and n ln (n 1) 1 diverges by the Direct 1) 1 ln (n 1)! )2n (x 1 … 49. Diverges by the nth-Term Test, since 41. Diverges, because it is ∑ n1 5∑ 5 n 5 times the harmonic series: 1 ( 1)n(n2 2n2 n n→ 1) does not exist. 1 lim n 1n 50. Converges absolutely by the Direct Comparison Test, since 42. Converges conditionally. ∑ n1 converges as a p-series 3/2 1 1 If un n , then {un} is a decreasing sequence of positive terms with lim un n→ 0, so ∑ ( 1)n converges by the n n1 1)(n 3 . with p 2 n(n 1 and n3/2 n 2) 1 Alternating Series Test. The convergence is conditional because ∑ n1 1 is a divergent p-series p n 1 . 2 51. Converges absolutely by the Limit Comparison Test. 43. Converges absolutely by the Direct Comparison Test, since 0 ln n n3 with p 1 for n n2 1 and ∑ n 1 converges as a p-series n2 1 2. 1 Let an Then lim n n2 an n→ bn 1 lim n→ 1 . n2 and bn 1 and ∑ n2 n n2 n2 lim n→ an an 1 n lim n→ (n 2 1)! n! n 1 n lim n→ (n 52. Diverges by the nth-Term Test, since 2 1)2 0. lim n→ n n n 1 2 2). Therefore ∑ an converges. p-series ( p 44. Converges absolutely by the Ratio Test, since 1 n 2n 1 lim 1 n→ 1 n n 1 e 0. converges as a 397 Chapter 9 Review 53. This is a telescoping series. ∑ (2n n3 s1 s2 s3 sn S (c) The fourth order Taylor polynomial for g(x) at x x ∑ 1 3)(2n 1) n 1 2(2 3 3) 2(2 1 1 1 6 10 10 1 1 1 6 10 10 1 1 6 2(2n 1) 1 lim sn 6 n→ 1 1 3) 2(2n 1) 3 2(2n 1 1 1 3 1) 6 10 1 1 1 6 14 14 1 1 1 1 6 14 14 18 [7 3(t 4) 4)2 5(t 4 is 4)3] dx 2(t 4 7t 4)2 5 (t 3 4)3 1 (t 2 4)4 7(x 1 18 3 (t 2 4) 3 (x 2 4)2 5 (x 3 4)3 1 (x 2 x 4 4)4 (d) No. One would need the entire Taylor series for f (x), and it would have to converge to f (x) at x 3. 57. (a) Use the Maclaurin series for sin x given at the end of Section 9.2. 54. This is a telescoping series. ∑ n(n 1) n2 2 2 s1 s2 s3 1 sn 2 n n2 2 n 2 2 1 3 3 2 2 3 3 2 2 3 3 2 n2 1 S ∑ 2 1 lim sn (x/2)3 (x/2)5 3! 5! 5x 3 x5 … 48 768 x 2 5 1 5x 2 2 4 2 4 1 2 4 2 4 2 5 1 2 5 f (3) an lim n→ 1 5 lim f (3) (x 3! 1 f (3.2) 4(x 3) P3(3.2) f (3) (x 2! 3) 3) 3(x 3)2 (2n 5 1)! 2 2n+1 x 0 3)3 5 2n 1, 2, 3, …. for all x and all n 2(x (2n (c) Note that the absolute value of f (n)(x) is bounded by 3 3)2 x 2n 3 3)! 2 x/2 2 3)(2n 2) n→ (2n an lim f (3)(x ( 1)n the Ratio Test: n→ 55. (a) P3(x) (x/2)2n 1 … (2n 1)! 5 x 2n+1 … ( 1)n (2n 1)! 2 … (b) The series converges for all real numbers, according to 1 n→ x 2 5 sin We may use the Remainder Estimation Theorem with 1.936 M So if (b) Since the Taylor series for f can be obtained by term- 1 . 2 5 and r 2 x 2, the truncation error using by-term differentiation of the Taylor Series for f, the Pn is bounded by second order Taylor polynomial for f at x 5 2n 4 6(x f (2.7) 3) 6(x 3)2. Evaluated at x 3 is 1 2n 1 (n 1)! . 4. So, two terms (up through degree 4) are needed. means the graph of f is concave up near x 6, which 58. (a) Substitute 2x for x in the Maclaurin series for 1 1 x given at the end of Section 9.2. 3. 1 1 7. Since 1 2x 2x (2x)2 1 56. (a) Since the constant term is f (4), f (4) f (4) , f (4) 3! 1)! To make this less than 0.1 requires n 2.7, 2.74. (c) It underestimates the values, since f (3) 2 5 (n 2x 4x 2 … (2x)3 … 8x 3 … (2x)n (2x)n … 12. (b) Note that P4 (x) 3 10(x 4) 6(x 4)2 24(x 4)3. The second degree polynomial for f at x 4 is given by the first three terms of this expression, namely 3 10(x 4) 6(x 4)2. Evaluating at x 4.3, f (4.3) 0.54. (b) 11 1 , . The series for is known to converge for 22 1t 1 t 1, so by substituting t resulting series converges for 1 2x, we find the 2x 1. 398 Chapter 9 Review 58. continued (c) f (b) Integrate term by term. 1 4 2 , so one percent is approximately 0.0067. It 3 x ln x 1 (t 2 t 1 , the series is the 4 n0 18 2 through degree 7) are sufficient since dt 1 (t 3 1 (t 4 3)3 0.0067. … 3)4 x 3)n+1 … n1 3 (x 3)2 (x 3)3 3) 2 3 3)n 1 n (x … ( 1) n1 (x Series Estimation Theorem, it shows that 8 terms (up 2 3)2 ( 1)n 1n . If you use the Alternating 2 alternating series ∑ 1 t 3 takes 7 terms (up through degree 6). This can be found by trial and error. Also, for x 2 (t 3)4 (x … 4 It is also a geometric series, and you could use the (c) Evaluate at x remainder formula for a geometric series to determine the number of terms needed. (See Example 2 in 1 2 3.5. This is the alternating series 1 22 2 1 23 3 … ( 1)n 2n 1 1 (n 1) … By the Alternating Series Estimation Theorem, since Section 9.3.) the size of the third term is an 59. (a) lim n→ 1 (n 1)n 1 n! (n 1)! x nnn x (n 1)n 1 lim (n 1)nn n→ n 1n x lim xe n n→ 1 xn lim terms will suffice. The estimate for ln given at the end of Section 9.2. 1 , so the e 1, or x e 2x 2 ( 2x 2)2 ( 2x 2)3 2! 3! 2n … ( 2x ) … n! 4x 6 … 2x 2 2x 4 3 n 2n 2x … ( 1)n n! ( 2x 2) 1 1 e 1 11 12 31 3 1 2 1 3 9 6 5 0.278 18 1 3 13 3 22 2! 33 3! (b) Use the Ratio Test: lim n→ (c) By the Alternating Series Estimation Theorem the error is no more than the magnitude of the next term, which 14 3 44 4! (x 2) 60. (a) f (3) f (3) f (3) (x 2(x f (3) f (x) 1 2 2) 1 (x 3) ( 1)n(x f (3) 2! x3 6, so 4 ( 1)nn!, so The series converges for all real numbers, so the f (n) (3) n! (x 3)n , ). bounded by the magnitude of the fifth term, which is = 2, so 3 x lim n→ (c) This is an alternating series. The difference will be 1 x3 3 2n 1x 2n 2 n! (n 1)! 2nx 2n 2x 2 lim 0 n1 n→ 1 an interval of convergence is ( x3 2) an 0.132. 1 2) 6(x f (n)(3) 32 243 3 is 0.375. 2 2x 2 for x in the Maclaurin series for e x 61. (a) Substitute radius of convergence is . is 0.05, the first two n→ an The series converges for x e (b) f 1 24 1 f (3) 3! 1 ( 1)n 3)2 … (x 3)3 … (2x 2)4 2x 8 . Since 0.6 x 0.6, this term is less 4! 3 2(0.6)8 than which is less than 0.02. 3 62. (a) f (x) x2 1 1 x x 2(1 x x2 (b) No. At x x3 x2 x4 … … ( x)n ( 1)nx n+2 …) … 1, the series is ∑ ( 1)n and the partial n0 sums form the sequence 1, 0, 1, 0, 1, 0, …, which has no limit. Chapter 9 Review 63. (a) Substituting x 2 for x in the Maclaurin series for sin x given at the end of Section 9.2, sin x 2 6 10 x 3! x2 … x 5! du ( 1)n 4n 2 x (2n 1)! . Integrating term-by-term and observing that the constant term is 0, x x3 3 sin t 2 dt 0 x7 7(3!) x1 1 11(5!) ( 1)n 1 (4n … x 4n 3 3)(2n 1)! x2 Let u du e x dx dv 2x dx x 2e x dx x 2e x ex v x 2e x 2x 2x e x dx e x dx dv 2 dx ex v 2xe x dx x 2e x 2xe x x 2e x … 1 7(3!) suffices to use the first two nonzero terms (through degree 7). 1 x 2e x dx (x 2 2)e x 0 2e x C 2)e x 2x 2x 2e x dx 2xe x (x 2 1 … (b) sin x dx 11(5!) 0 1 …. ( 1)n (4n 3)(2n 1)! 1 1 Since the third term is 0.001, it 11(5!) 1320 1 3 (e) Let u 399 C e 2 1 0 65. (a) Because [$1000(1.08) n](1.08)n available after n years. 0.71828 $1000 will be (b) Assume that the first payment goes to the charity at the end of the first year. 1000(1.08) 1 1000(1.08) 2 1000(1.08) 3 … (c) This is a geometric series with sum equal to (c) NINT(sin x 2, x, 0, 1) (d) 1 3 1 7(3!) 0.31026830 1 11(5!) 258,019 831,600 1 15(7!) This is within 1.5 10 7 1000 0.08 1000/1.08 1 (1/1.08) 0.31026816 of the answer in (c). 12,500. This means that $12,500 should be invested today in order to completely fund the perpetuity forever. 66. We again assume that the first payment occurs at the end of 64. (a) Let f (x) 1 x 2e x dx. x 2e x dx 0 the year. 1 f (x) dx Present value 0 h f (0) 2 2f (0.5) 1 0 4 2 e0.5 8 e0.5 4 f (1) e 4 x2 1 x x2 2! 1 x3 1000(1.06) … xn n! … … 0.68333 (c) Since f is concave up, the trapezoids used to estimate the area lie above the curve, and the estimate is too large. (d) Since all the derivatives are positive (and x 0), the remainder, Rn(x), must be positive. This means that Pn(x) is smaller than f (x). 3 1000 1.06 1 The present value is $16,666.67. n2 x4 …x 2! n! x4 2 3 x P4(x) x 2 1 3 x x4 41 x5 1 P4(x) 3 4 60 10 0 0 x2 1 1000(1.06) 1000/1.06 (1/1.06) e 0.88566 (b) x 2e x 1000(1.06) 2 … 16,666.67 400 Section 10.1 67. (a) Sequence of Tosses Payoff ($) Probability Term of Series T 0 1 2 0 70. (a) Note that ∑ x n 1 12 2 2 13 2 2 13 2 14 2 3 identity ∑ x n 12 1 2 HHT 14 2 HHHT 3 is a geometric series with first term x 2 and common ratio r a 1 2 HT 1 n1 x 1 1 n1 x, which explains the 2 x (for x 1). Differentiate. ∑ (n x)(2x) (x 2)( 1) (1 x)2 (1 1)x n n1 2x (1 x2 x)2 Differentiate again. ∑ n(n 1)x n 1 n1 Expected payoff 1 2 0 (b) (c) 1 1 13 3 2 … 2 … 3x 2 2x 14 (2x x 2)(2)(1 (1 x)4 x)(2 2x) 2(2x x 2) (1 x)3 nx n (1 … 1 x 2(1 x)2 x2 2x 3 … 3x 2 2x … 3x4 nx n nx n …) 1 x)3 ∑ n(n 2x 1)x n (1 n1 1 , the formula in part (c) matches the 2 1 x (d) If x nonzero terms of the series in part (a). Since 2 (1/2) 1, the expected payoff is $1. 2 [1 ∑ n(nxn 1) 68. (a) The area of an equilateral triangle whose sides have 1 2 s2 3s 2 3 4 x)3 Replace x by . (1/2)] length s is (s) x)( 1) Multiply by x. … 1 2x) 2 (1 x2 (1 2 2 1 x)2 (1 x)2(2 (1 12 1 n1 . The sequence of areas (b) Solve x 2 x (x 13 x (x 2x 2 to get x 1)3 2x 2 ,x 1)3 1 2.769 for x 1. removed from the original triangle is b2 3 4 3 3n b2 3 4 b2 3 2 4 9 b2 3 2n 4 3b2 3 42 b2 3 4 4 Chapter 10 Vectors … … or 32b2 3 43 … 3nb2 4n …. 3 1 (b) This is a geometric series with initial term a and common ratio r b2 3 b , so the sum is 4 1 2 b2 3 4 3/4 (3/4) s Section 10.1 Parametric Functions (pp. 513–520) Exploration 1 Investigating Cycloids 1. 3, which is the same as the area of the original triangle. [0, 20] by [ 1, 8] (c) No, not every point is removed. For example, the vertices of the original triangle are not removed. But the remaining points are “isolated” enough that there are no regions and hence no area remaining. 69. 1 1 x 1 x x2 x3 … 1 x)2 1 Substitute x 2x 3x 2 4x 3 2na for any integer n. 3. a 0 and 1 cos t 0 so y 0. 4. An arch is produced by one complete turn of the wheel. Thus, they are congruent. 5. The maximum value of y is 2a and occurs when x (2n 1)a for any integer n. Differentiate both sides. (1 2. x 5x 4 1 to get the desired result. 2 … 6. The function represented by the cycloid is periodic with period 2a , and each arch represents one period of the graph. In each arch, the graph is concave down, has an absolute maximum of 2a at the midpoint, and an absolute minimum of 0 at the two endpoints. Section 10.1 Quick Review 10.1 1. (cos (0), sin (0)) 2. (a) (1, 0) 3 3 , sin 2 2 2. cos 3. x 2 y2 (0, 1 (since cos2 t (b) 1) sin2 t 1) 4. The portion in the first three quadrants, moving counterclockwise as t increases. 5. x t2 t, y 1, 1 t 3. (a) dy dx d 2y dx 2 dy dx dy/dt dx/dt dy dx which at t 3t 3 t 3 (b) 3 . 2 2/2) 0 3/(2 3t) 1/(2 t 1) 3t 1 3 cos t , 2 sin t 2/2) 0 sin t 3 dy/dt dx/dt y 3 3 3( equals 4 2( 3 sin t sin t dy /dt dx/dt 6. The graph is a circle with radius 2 centered at (2, 3). Modify the x cos t, y sin t parameterization correspondingly: x 2 cos t 2, y 2 sin t 3, 0 t 2 . 7. dy/dt dx/dt y 401 d 2y dx 2 3/(2t 2 3 1/(2 t dy /dt dx/dt 3 1/t 1/t 2 3t 3/t) 1) 1 t t2 3 3/t t 3/2 3 8. y x C. For t 2 32 3 ( 2) 2 2 3 Thus, y x3 2 2 x 3 9. y 32 2 10. y 3 3 2 2 , so 4. (a) 2. (b) 2. 3 ,x 4 2) 2 x 3 Thus, y 2 and y C and C C. For t 2 ( 3 3 2 3 ,x 4 C and C 2 and y 5 2 6 32 , so 2 5. (a) . (b) 52 . 6 dy dx d 2y dx 2 dy dx 3 d 2y dx 2 3 2 1 9 x dx 4 0 8 1 27 x 9 3/2 x 4 dx 6. (a) 3 0 1 1/t 2 dy/dt dx/dt dy /dt dx/dt [(2t 3)(6t) 12t 2 18t (2t 3)3 6t 2 18t (2t 3)3 2 1 0 3 dy /dt dx/dt y x, so Length dy/dt dx/dt y 313/2 8 . 27 (b) dy dx dy/dt dx/dt y d 2y dx 2 3t 2 2t 3 (3t 2)(2)]/(2t 2t 3 6t 2 2t 2t dy /dt dx/dt [(2t 1)(2) (b) dy dx d 2y dx 2 y dy/dt dx/dt dy /dt dx/dt 2 sin t 4 cos t 1 sec2 t 2 4 cos t 1 1 1)2 4 1 tan t 2 1 sec3 t 8 3)2 (2t 1)(2)]/(2t 2t 1 Section 10.1 Exercises 1. (a) t2 1)3 (2t 7. dy dx dy/dt dx/dt cos t sin t (a) cot 0 when t (x, y) (2, (b) 2 1 cos cot t k (k any integer). Then 2 k 2 , (2 sin k 2 1). The points are (2, 0) and (2, cot is undefined when t (x, y) 1 cos (k ), The points are (1, 2). k (k any integer). Then 1 sin (k )) 1) and (3, 1). (2 1, 1). 402 8. Section 10.1 dy dx sec2 t sec t tan t dy/dt dx/dt 13. x csc t t 2, y t, so 1 (a) Nowhere, since csc t never equals zero. 9. dy dx dy/dt dx/dt 3t 3t 2 (a) 4 4 2 2 2 2 , 12 (t 3 3 , 8 33 (0.845, 1 1) 0 1 0.609 2 4 8 1)3/2 22 3 . 3 3 3 which evaluates to (3.155, 3.079). 3 2 3 2 1 dt 1 3/2 (2 3 4 3 0 when t Then (x, y) t t2 0 3t 2 4 1 t 2 dt 0 1 (b) csc t is undefined when t k (k any integer). Then (x, y) (sec (k ), tan (k )) ( 1, 0). The points are (1, 0) and ( 1, 0). 2 (t 2)2 Length 14. x 8t cos t, y , 8t sin t, so /2 3 (8t cos t)2 Length 0 3.079) and (8t sin t)2 dt /2 8t dt 0 2 (b) Nowhere, since 4 /2 4t 2 3t is never undefined. 2 0 dy dx dy/dt dx/dt 3 cos t 3 sin t (a) 10. cot t 0 when t cot t 15. x y k (k any integer). 2 /3 ( 2, 1 2 k ,1 3 sin 2 k 3). The points are ( 2, 4) and ( 2, sec2 t /3 0 cos t)2 cot t is undefined when t (x, y) 1 dt 0 /3 ln 2 0 k (k any integer). Then 3 cos (k ), 1 3 sin (k )) (2 3, 1). The points are (1, 1) and ( 5, 1). 16. x et 2t, y e t, so 1 2 (e t Length 2t )2 (1 e sin t, y 1 cos t)2 dt (1 17. x sin t, y 2(1 4 cos2 Area cos t) dt t dt 2 0 3 Length ( 2t 2 0 3 (t 0 2) dt 12 t 2 sin t) dt 2 2t cos t 8 2 0 18. x 1 t, y , so t 2 2 3/2 t 3 2 0 t, so 3)2 (2 0 Area 1 cos2 t dt 2 2 t 2 cos dt 2 0 t 2 2 sin 4 20 3, y sin t) ( sin t)2 2 (2 0 0 2t 4.497. cos t, so 2 0 12. x ) dt , cos t, so ( sin t)2 length t2 1 which using NINT evaluates to 11. x ( sin t)2 dt tan t dt 2). (2 cos t, /3 ln sec t (b) sec t (sec t 0 3 cos cos t sin t, so Length Then (x, y) 2 sec t tan t sec2 t sec t tan t 4 3 t)2 dt (1 3 2t 0 2 t t2 1 dt 1)3/2 1) 1, y dt 0 4 (5 19. x 12 t 4 12 (t 33 21 2 t)2 ( 2 0 5 9 14.214 2t, so 3 Area 2 (t 1) 1 (2t)2 dt, 0 which using NINT evaluates to 178.561. Section 10.1 20. x sec t cos t (see Ex. 15), y sin t, so /3 Length cos t)2 2 (cos t) (sec t 0 /3 26. Parameterize the curve as x sin t dt 0 /3 2 cos t (b) x 2t, y (t) 2, y t 1, 0 Area 2 (t formula. Then 1 27. x 22 1) 0 dy dy becomes dt dy 2t t, y 5 (t 12 dt 2 3 (1 2) ( (t 1)2 dt 2t 4 12 t 2 0 12 5 1. 1) dt t 22 3 12 m 2 5 r x, h 4 12. 0 t 0 6, or m2 m 2 m t m 12 t 2 Now solve 5, so 22. (a) Because these values for x(t) and y (t) satisfy y 48 2 12 for m: 2 2m 12 solution. The midpoint is at t (h, r), and this range of t-values gives the correct initial (x, y) and terminal points. h, y 13 1)2 1 , (2 3 2 ( 13 13 1, which gives 1)3/2 (3.394, 5.160). 28. r, so 1 2 (rt) h2 Area r 2 dt 0 r h2 [ 4, 4] by [0, 10] 1 r2 t 2 Use the right half of the curve, 0 0 r2 r h2. r2 (c) Slant height x h2, so Area r r2 h2 3 cos t, y Area 2 sin 2t, y 2 sin 2t)2 ( 0 . 2 (3 sin t) (3 cos t)2 (6 cos 2t)2 dt, 0 159.485. 2 cos 2t, so /2 Length t 6 cos 2t, so which using NINT evaluates to 23. (a) x 0, and 13. Take the positive 1 which is the equation of the line through the origin and (b) x d. The 0 5 (c) Slant height y 0 4 1 1 5 t2 2 t2 Total length 1) dt 0 y, c 1, so 4 1 2 Area t 1, so 1 g( y), y parameter is y itself, so replace t with y in the general . 0 21. (a) x(t) 1. tan t dt /3 2 dx dx becomes 2 2 cos t 0 dx dt 25. In the first integral, replace t with x. Then ( sin t)2 dx 403 (2 cos 2t)2 dt 29. /2 2 dt . 0 (b) x cos t, y 1/2 Length [0, 9] by [ 3, 3] sin t, so ( cos t)2 ( sin t)2 dt 1/2 1/2 dt Use the top half of the curve, and make use of the shape’s symmetry. . 1/2 x 3 cos t, y /2 24. x 3 sin t, y 4 cos t, so Area 2 6 cos 2t, so 2 (3 sin 2t) (3 cos t)2 (6 cos 2t)2 dt 0 2 Length ( 3 sin t)2 (4 cos t)2 dt 0 which using NINT evaluates to 22.103. which using NINT, evaluates to 144.513. 404 30. y Section 10.1 0 for t 0 and t 2 .x a a cos t, y (b) x a sin t, t cos t, y t sin t, so 2 2 2 [a(1 t dt 0 2 2 a2 0 2 2 8 a2 8a cos t) 2 t 2 sin2 2 t sin3 dt 2 cos2 1 0 t 2 cos 2 2 64 a 3 dx dt 0 2 2 2 (a sin t)2 dt 0 34. All distances are a times as big as before. (1 2a 8a a cos t)2 cos t)] (a 0 2 2 2 2 cos t dt (a) x a(cos t (b) Length t 4 sin2 dt 2 t sin t), y t 2 y t dt 2 2 2 t cos3 3 20 sin 0 for t v sin /16 0 0 v0 sin . To find the path length, evaluate 32 (v0 cos )2 32t, and . The maximum height is attained 16 v0 sin in mid-flight at t t cos t) and y v0 cos v0 sin 0 or t a(sin t 2 2a For exercises 35–38, x 32t)2 dt using NINT. To (v0 sin find the maximum height, calculate a(1 cos t) (Note: integrate with respect to x from 0 to 2a ; integrate ymax (v0 sin ) 2a Area y y dx 0 0 a 2 cos t)a(1 0 or t x (1 cos t) dt a2 t 1 sin 2t 4 t 2 2 sin t 2 . 0. 0 3.206 150 sin 20 (150 cos 20 )2 Length 0 16t) 150 cos 20 , y (75 sin 20 )/8 2 2 cos t 32 75 sin 20 8 cos t) dt 2 v0 sin 16 32 t(150 sin 20 t 2a a(1 v0 sin 35. (a) The projectile hits the ground when y with respect to t from 0 to 2 .) dx 32. dt 2 (t sin t)2 dt 0 Area 31. (t cos t)2 Length so 32t 32t)2 (150 sin 20 0 2 3 a2 dt which, using NINT, evaluates to 461.749 ft 0 (b) The maximum height of the projectile occurs when a(1 cos t), so y 2 Volume [a(1 0 2 cos t)] a(1 cos t) dt 2 3 a (1 3 cos t 2 3 cos t so t 3 cos t) dt 0 3 at 3 t 2 1 sin3 3 3 sin t sin t 5 3 sin 2t 4 36. (a) (b) 2 t 0 a (b) 33. (a) QP has length t, so P can be obtained by starting at Q and moving t sin t units right and t cos t units downward. (If either quantity is negative, the corresponding direction is reversed.) Since Q (cos t, sin t), the coordinates of P are x cos t t sin t and y sin t t cos t. 75 75 sin 20 , y sin 20 16 16 5625 64 87.891 ft 840.421 ft 16,875 64 263.672 ft 38. (a) It is not necessary to use NINT. 75/8 Length 5625 16 (150 0 5625 8 (b) t t t t cos t P (x , y ) (1, 0) x 16t 2 75/8 0 2y a dx 2 dt and y with f (x). Then 150t 351.5625 39. In the integral t sin t 32t) dt 703.125 ft b Q 41.125 ft 641.236 ft 37. (a) 23 y 0, dx dx becomes dt dx 1. dy 2 dt, replace t with x dt 405 Section 10.2 40. dy dx 3 e x, so Area 2 ex 1 (e x)2 dx which using 0 NINT evaluates to 1273.371. 5. (a) 2u 3v 2u (b) dy 41. dx 4 1 , so Area x2 1 x 2 1 using NINT, evaluates to 12 dx, which, x2 1 6. (a) 9.417. (b) dy 42. dx (ln 2)2 2 x (ln 2)2 2 (2 x Area x 2 x) (ln 2)(2 x 2 ), so 7. (a) 116.687. (pp. 520–529) 2. 1)2 3 5 2 1 3. Solve 2)2 (3 8. (a) 17 1 4 3 5 b 3 4: b 0 0 1 1 4 4 ( 2), (5) 5 5 4 v 5 9. 2 2 4 32 14. cos 52 2(3)(5) cos (30 ) 34 10. 242 15 272 c 3 192 34 15 1, 2(27)(19) cos , so 272 192 257 and 2(27)(19) 513 257 cos 1 1.046 radians or 59.935 . 513 242 Section 10.2 Exercises (b) 2. (a) (b) 92 2( 2), 42 3. (a) 3 (b) ( 6)2 ( 2), 12 4. (a) 3 ( 2), 52 2 3 4, 2(5) 32 6 117 ( 10) 2 10 116 5 13 2 1, 3 10 2 ( 7)2 5 5, 74 12 v 13 15 10 , 13 13 5 ( 2) 13 ( 4) 2 15 13 3 0, 1 3 0 2 3 1, 1 3 1, ( 1), 1 2 2 , sin 3 3 3 , sin 4 1, 1 ( 1) ( 1), 2 1 60 13 4 2, 1, 0 CD E 24 10 , 13 13 6421 13 1, 3 2 24 60 , 13 13 70 2 13 1 1 14 , 55 4 ( 1) 1 0, 0 1 3 , 22 3 4 1 , 2 1 2 15. This is the unit vector which makes an angle of 2.832 1. (a) 3(3), 3( 2) = 9, 3, so 6 5 197 5 5 (3), 13 ( 3)2 2 13. cos 8 , 5 12 12 ( 2), (5) 13 13 12. E AB 45 6 5 8 ,4 5 14 2 5 5 u 13 2, 0 AB E 9 , 5 9 5 12 5 11. 0 6 (b) 16, 29 1097 3 3 (3), ( 2) 5 5 CD E 8. (a) (b) (b) 10. 30 (b) cos 0 and a 4. a b and b 6. 8 2 3 7. (a) 9. c2 3 5 2 6 120 (b) 29 3 u 5 5 u 13 11. 3 1 5 5. Slope of AB = Slope of CD, so 3 19 505 2 ( 16) 12 v 13 4. Slope of AB = Slope of CD, so 6. (a) ( 19) 2 (b) Quick Review 10.2 (5 12 12, 2u 2(3), 2( 2) 6, 4 5v 5( 2), 5(5) 10, 25 2u 5v 6 ( 10), 4 25 3 u 5 s Section 10.2 Vectors in the Plane 1. 2 4 v 5 2 x)2 dx which 2 using NINT evaluates to 2 x (ln 2)2(2 x 1 2(3), 2( 2) 6, 4 3( 2), 3(5) 6, 15 3v 6 ( 6), 4 15 7 29 120 90 210 with the positive x-axis; cos 210 , sin 210 16. cos 135 , sin 135 3 2 1 , 2 1 2 , 1 2 3, 70 13 406 Section 10.2 17. The vector v is horizontal and 1 in. long. The vectors u and 11 w are in. long. w is vertical and u makes a 45 angle 16 with the horizontal. All vectors must be drawn to scale. (a) (d) w v v u u+v u u+v+w=0 (b) 1 3, 4 5 19. 42 42 ( 3)2 21. ( 15)2 82 22. ( 5)2 u –v 34 , 55 ( 2)2 5; w v (c) 32 20. u+v+w u 1 u–v 5, 1 4, 5 5; 1 17 17; 1 5 2 –w ,y (x )2 (a) 1 ,2 17 2 2 1 2, 2 17 u 1 t 1 ,y (b) 5 (y )2 Tangent: u Normal 2 –v 25. x y 1, 4 sin t, y 5 , and 2 2, and , 17 1 , 17 . 17 1 ,y 2 3, x 1, and . 1 ,1 52 2 4 , 17 4 2 5 u–v+w 1 ,y 2 1, x 1 1; for t u–v (x )2 ; for t 2 Normal: 24. x 29 17 . 2 (y )2 Tangent: –v 2 , t u–w 18. The angles between the vectors is 120 and vector u is horizontal. They are all 1 in. long. Draw to scale. 1 1 t 2 15 8 , 17 17 15, 8 29 23. x u 3 5 29; 29 (d) 4 , 5 3 1 2 , 5 1 2 , 5 2 1 , 5 5 cos t; for t (x )2 . 5 3 ,x 2 3, 73 . 2 (y )2 w Tangent 2 2 3, 73 (c) 12 , 21 9 0.811, 0.585 , 2u Normal –v 2u – v 5 2 2 5 ,2 73 2 3 0.585, 0.811 . 5 73 , 12 219 5 73 Section 10.2 26. x 3 sin t, y 3 y 3 cos t; for t (x )2 , and (y )2 4 3 ,x , 2 3. 2 32. Since u and v are nonzero, we know that u ≠ 0 and v ≠ 0. Therefore, the dot product u v u v cos is 0 if and only if cos 0, which occurs if and only if u and v are . orthogonal 2 1 3 3 1 3 3 Tangent: Normal: 3 , 2 1 , 2 , 2 3 2 1 1 2 33. , 2 (r1, r2) 1 , 2 v . 2 u 27. AB E 3, 1 , BC E BA E AB E BA E AC E CA E 3 , and AC E 1, 1 , CB E 3, 2, CB E 2. 1 cos 1 cos 1 v 10, and r1 r2 AB AC EE AB AC EE 3(2) 1( 2) ( u 2, 2 . 2 2. cos (u1, u2) u+v (v1, v2) 1, 3 , and CA E 10, BC E Angle at A v1 v2 u1 so r1 u2 so r2 u1 u2 v1 v2 34. (a) To find u v, place both vectors with their initial points at the origin. The vector drawn from the terminal point of v to the terminal point of u is u v. Or, add u and v according to the parallelogram law. 10)(2 2) (b) 1 63.435 , 5 BC E E BA cos 1 BC BA EE Angle at B 1 cos (u1, u2) ( 1)( 3) 1 3 5 10) 1 cos 53.130 , and v 1 cos u –v (–v1, –v2) 1( 2) ( u–v u CB CA EE cos 1 CB CA EE Angle at C –v u–v ( 3)( 1) ( 10)( cos 3(2) 10)(2 2) 1 r1 r2 63.435 . ( v1) ( v2) u1 → r1 u2 → r2 u1 u2 v1 v2 5 28. AC E 2, 4 and BD E AC BD EE 2(4) 4, 35. (a) Let P 2. 4( 2) 0, so the angle measures 90 . 29. (a) u (v w) u1(v1 w1) u2(v2 (u1v1 u1w1) (u2v2 u2w2) (u1v1 u2v2) (u1w1 u2w2) uv uw (b) (u u2 2 v) (u 31. (u v) ( u1 u1 2 (u12 2 u w2) v) w (u1 v1)w1 (u2 v2)w2 (u1w1 v1w1) (u2w2 v2w2) (u1w1 u2w2) (v1w1 v2w2) uw vw u12 30. u u 407 v1)(u1 v1 2 u22) v2 ( u12 v1) u22 (v12 u22)2 u2 (u2 v2)(u2 v2) v22 v22) 1 OP E 2 (a (a, b) and Q 1 OQ E 2 c) (b d) , 2 2 (c, d ). Then 1 a, b 2 OM E (b) OM E 2 OP E 3 1 OQ E 3 (c) OM E 1 OP E 3 2 OQ E 3 1 c, d 2 ( r 1, r2) 408 Section 10.2 35. continued 40. The indicated diagonal is (u (d) M is a fraction of the way from P to Q. Let d be this d OQ E Proof: PM E so PQ E d)OP. E (1 dPQ and MQ E E 1 PM and PQ E E d 1 d 1 Therefore, PM E But PM E 1 OM E d OM E 1 d 1 1 d d)PQ, E (1 1 1 d MQ. E MQ. E OQ E OQ E 1 1 d OM, so E OM. E Therefore, 1 OM E d ⇒ OM E 1 1 d 1 d(1 ⇒ OM E 1 OP E d OM E d) 1 OP E d d) OP E (1 (u v) v uv v2 . u vv u vv If u and v are the same length then these two quantities are equal, and the diagonal makes the same angle with both sides. 41. The slopes are the same. OP and MQ E E 1 OP E d between the diagonal and u is (u v) u u2 v u . u vu u vu But the cosine of the angle between the diagonal and v is fraction. Then OM E 1 OQ. E 1d 1 OQ E 1d 42. v 0 since any other vector has positive magnitude. 43. 25 west of north is 90 800 cos 115 , sin 115 25 44. 10 east of south is 270 600 cos 280 , sin 280 10 280 “north” of east. 104.189, 590.885 v (u u v and CB E 325 cos 70 , sin 70 111.157, 305.400 . Wind velocity is 130 north of east: dOQ. E u v. Since u , these vectors are orthogonal, as v) (u v) v2 u2 115 north of east. 338.095, 725.046 45. Initial velocity is 70 north of east: 40 cos 130 , sin 130 36. CA E v). The cosine of the angle 0. 37. Two adjacent sides of the rhombus can be given by two vectors of the same length, u and v. Then the diagonals of the rhombus are (u v) and (u v). These two vectors are orthogonal since u v so (u v) (u v) u2 v 2 0. 38. Two adjacent sides of a rectangle can be given by two vectors u and v. The diagonals are then (u v) and (u v). These two vectors will be orthogonal if and only if u and v are the same length, since (u v) (u v) u2 v 2. 39. Let two adjacent sides of the parallelogram be given by two vectors u and v. The diagonals are then (u v) and (u v). So the lengths of the diagonals satisfy u v 2 (u v) (u v) u 2 2u v v2 2 and u v (u v) (u v) u 2 2u v v 2. The two lengths will be the same if and only if u v 0, which means that u and v are perpendicular and the parallelogram is a rectangle. 25.712, 30.642 . Add the two vectors to get 85.445, 336.042 . The speed is the magnitude, 346.735 mph. 336.042 The direction is tan 1 85.445 75.734 north of east, or 14.266 east of north. 46. w cos (33 15 ) 2.5 lb, so w 2.5 lb . cos 18 2.5 lb cos 33 , sin 33 cos 18 Then w 2.205, 1.432 . 47. Juana’s pull 23 cos 18 , sin 18 21.874, 7.107 ; Diego’s pull 18 cos ( 15 ), sin ( 15 ) 17.387, 4.659 . Add to get the combined pull of the children: 39.261, 2.449 . The puppy pulls with an opposite force of the same magnitude: 39.2612 2.4492 39.337 lb. 48. (a) 7 cos 45 , sin 45 has its terminal point at (4.950, 4.950). (b) 7 cos 45 , sin 45 terminal point at 49. AB E 3 0, 4 50. AB E 2 ( 4), 2, 3 5 1, 4 8 cos 210 , sin 210 has its ( 1.978, 0.950). 0 3, 4 2 1 3 CD E 1 4, 5 1 CD E Section 10.3 51. u u1, u2 , v (iii) (iv) (v) (vi) (vii) (viii) (ix) 3. dy/dt dx/dt 5 cos t , which for t 4 sin t dy dx dy/dt dx/dt 5 cos t , which for t 4 sin t v 52. Write the two vectors as a 1, 1 and b 1, sum is a b, a 7 ,b 2 a dy dx 4. w1, w2 u1 v1, u2 v2 v1 u1, v2 u2 vu (u v) w u1 v1, u2 v2 w1, w2 (u1 v1) w1, (u2 v2) w2 u1 (v1 w1), u2 (v2 w2) u (v w) u0 u1, u2 0, 0 u1 0, u2 0 u1, u2 u u ( u) u1 , u 2 u1, u2 u1 u1, u2 u2 0, 0 0 0u 0 u1, u2 0u1, 0u2 0, 0 0 1u 1 u1, u2 1u1, 1u2 u1, u2 u a(bu) a(b u1, u2 ) a bu1, bu2 abu1, abu2 ab u1, u2 (ab)u a(u v) a u1 v1, u2 v2 au1 av1, au2 av2 au1, au2 av1, av2 a u1, u2 a v1, v2 au av (a b)u (a b) u1, u2 (a b)u1, (a b)u2 au1 bu1, au2 bu2 au1, au2 bu1, bu2 au bu (i) u (ii) v1, v2 , w b , so solve a 1 . So 3, 4 2 b 77 , 22 b 5. dy dx dy/dt dx/dt 1 1 y 1 1, so y Also, at t 6 ,x 2 53 x 4 y 6. dy dx dy/dt dx/dt 6 7. lim 3 ,x 2 2 4 x 2 x→2 x (x (b) Slope y1 m(x x 1, so y y1 m(x x 2 or y x 3. 54. The slopes of the lines are 4 3 x1) becomes cos cos 7 2 10 52 5 . The equation for the 2 9 . 10 (x 2 3), or x2 2)(x 2) lim 2x, and Length 1 using NINT evaluates to t cos t sin t, and y Length t2 ( t sin t 10. y 2 y xe x e x C (use integration by parts), so 0 e0 C and C 3: xe x e x 3 Section 10.3 Exercises ( 1)]i 3)i 3. E AB CD E [0 (0 (1 [0 4)j 6i ( 4)] j 3j 3i 4j ( 3)]i (2 0)j 4)i ( 3 0)j 3i 4i (a) [3 ( 4)]i [2 ( 3)] j ( 4)]i [2 ( 3)] j 2j and 3j. i 7i 1 f (x) . Then y 1, f (x) 3 3 y 1 (x 4 x 4 1 x2 1) or . , so for x . Then y 3 1, f (x) 3 3(x 3 and 1) or y 4] j 8i 4]j 2i 3x. j 5j 2j [( 2) 6j (c) 3(5)i 3 2. f (x) [( 2) 3)i (d) [2(5)i 3 x 3 f (x) 1 3 and 3)i (b) (5 , so for x cos t)2 dt 2.958. 4. (a) (5 Quick Review 10.3 x2 cos t, and 1 dt, (b) [3 x 2x)2 dx, which (3 t sin t sin t)2 (t cos t 0 2 2. (0 8.130 . (pp. 529–539) 4 2 3.400. which using NINT evaluates to s Section 10.3 Vector-valued Functions 1. f (x) 1 4 1 lim x→2 x 0 x1) becomes 1 53 . 4 equals 0 9. x 3 and 1, which means that 4 31 6 53 x→2 (x 1. [5 41 3), or 4 to get vectors 4, 3 and 1, 1 are parallel to the respective lines. 1 2 2 8. y 1. 2) or y (x 3 and y 5 2 x 15 53 . 4 equals 10. normal line is y y 3 4 5 cos t , which for t 4 sin t Also, at t 4 5 6 5 . The equation for the 2 3 and y 5 2 tangent line is y 11 ,. 22 y1 is undefined: 5 cos t , which for t 4 sin t 2 53. (a) Slope equals zero. the tangent line is vertical. 1 . Then their 3, a 2 409 3( 2)j 2( 2)j] 15i 6j [3(3)i 3(4)j] i 16j 410 Section 10.3 5. (a) 8. (a) [ 6, 6] by [ 4, 4] [ 6, 6] by [ 3, 5] a(t) d (2 cos t)i dt d (3 sin t)j dt ( 2 sin t)i (b) v(t) (3 cos t)j d ( 2 sin t)i dt d (3 cos t)j dt ( 2 cos t)i (c) v 2 direction (d) Velocity 1 2 2 a(t) (3 sin t)j ( 2)2 2, 0 ; speed 2, 0 d d2 (2 ln (t 1))i (t )j dt dt 2 i (2t)j t1 d 2 d 2 i (2t)j i dt t 1 dt (t 1)2 (b) v(t) 02 1, 2 ; speed (c) v(1) 2, 12 22 1 1 2 direction 1, 0 1, 2 5 1, 0 (d) Velocity 1 5 , (2t 5, 5 2 5 (cos t)i , 5 6. (a) 9. v(t) 2j 5 sin t)j, r(0) j and v(0) i. So the slope is zero (the velocity vector is horizontal). (a) The horizontal line through (0, [ 4.5, 4.5] by [ 3, 3] a(t) d (cos 2t)i dt (2 cos t)j d ( 2 sin 2t)i dt ( 4 cos t)i (c) v(0) (b) The vertical line through (0, d (2 sin t)j dt ( 2 sin 2t)i (b) v(t) d (2 cos t)j dt (d) Velocity r (2 sin t)j v 02 0, 2 ; speed direction 10. v(t) 1 0, 2 2 22 2, ( 2 sin t)i (2 4 ( 3 3)i 3 3 3 x 2 y 2 [x 3 1 2 x 3 y (sec t tan2 t sec3 t)i 22 42 3 3 1 2 24 , ; speed 33 3 24 , direction 2 533 (c) v 6 (d) Velocity 25 3 1 , 5 , 5 2 5 5 25 , 3 6t) dt i 2 3)] or 3 t dt j 1 2 2 3t 2 i 6t 2t 3/2 j 1 3i 1 2 (4 2)j /4 12. ( 2 (6 1 (2 sec2 t tan t)j 3)] or 5 2 18 6 2 (sec2 t)j (2 7 2 d d (sec t)i (tan t)i (sec t tan t)i dt dt d d (sec t tan t)i (sec2 t)j dt dt 3/ 2 2 3 [x 2 62 2 3 (b) y a(t) j. So the slope is 1 0, 1 7. (a) (b) v(t) 1 j and 2 2 11. 0. (3 cos t)j, 2)i (a) y [ 6, 6] by [ 4, 4] 1): x 1. 2 4 2 0, 1 1): y /4 sin t dt i (1 /4 cos t) dt j /4 /4 /4 cos t i /4 2 2 j t sin t j /4 3 . 2 411 Section 10.3 sec t tan t dt i (sec t)i 1 5 (ln t )i t 4 t4 4 r(t) ( 32t)j C2 r(t) 18. C t j 2 100)i ( 2 cos t)i ( sin t)j. So v C, and r(0) t2 2 ti tj i a j r(0) C2 dr dt t 0 C i j, so ( 16t 2)j C1 8j)t C1t C2 . 100i 2 8i 8j. So 2)i j. 2 5 , 2 a 3 , and 2 va cos 1 va 1 cos 3 5 53.130 . 24. v(t) 3i (2t)j, and a(t) 2j. So v(0) 3i, and a(0) 2j. These are perpendicular, i.e., the angle between them measures 90 . 25. (a) Both components are continuous at t 8t)j. is 3i t2 j 2 3 32 2 9 j 3(3) 10i C1 C1t C2 10j, and 2i sin t cos t ( sin t)i. ( 3 sin t)i 0 or 0, i.e., t 3 3 ( 3 cos t)i 2i. (c) Discontinuous when t ( 1] {0}. dx dt (sin t)i (1 sin t, and Distance 1 2 0 or cos t 0, i.e., for k , k any nonnegative integer. 2 1 0, i.e., 0, i.e., cos t)2 dt (1 2 cos t dt t dt 2 2 sin 0, is 1 cos t (sin t)2 0 2 /3 (16 sin t cos t) 0 or t 0 2 /3 ( 4 sin t)j. 0: (9 sin t cos t) 0 and t cos t)j, i.e., dy dt 2 /3 (4 cos t)j, and a(t) 0j (b) Continuous so long as t ( 1, 0) (0, ). 27. v(t) 0, which is true for 21. v(t) true when sin t 0, sin 2t i lim(ln (t 1))j t t→0 2 cos 2t i lim(ln (t 1))j lim 1 t→0 t→0 k , k any nonnegative integer. 2 Solve v a 3t 0, i.e., t t→0 j, and a(t) 0: 3t lim t t2 i j (10i 10j) 2 2 t2 t2 10 i 10 j 2 2 Solve v a 2 26. (a) Use L’Hôpital’s Rule for the i-component: 0, so (cos t)i 3, so the limit 3i. (c) Discontinuous when t 19. v(t) (1 cos t)i (sin t)j and a(t) (sin t)i (cos t)j. Solve v a 0: (sin t sin t cos t) (sin t cos t) 0 implies sin t 0, which is true for t 0, , or 2 . t j, and 2 1 va 2 t 1 2)i ( 4 Then v C1, and t2 i 2 r(t) 20. v(t) ( 4 1 j. dr dt t 0 ( 16t 0, (cos t)j, and 2 (8i ( 25sin t cos t) ( 2 sin t)i (b) Continuous so long as t 2 dr dt r(t) 0: (25 sin t cos t) a(t) C2)j e j C, and 0, so C (i j) 1)i (e t 1)j 100i, and (8t t 1i ( 16t2)j ( 5 sin t)j. which is true for all values of t. C1 and r(t) r(0) ( 5 cos t)i t 2t 2 i 2t 2 (5 cos t)j, and 23. v(t) t )j (t 1) i ijC ((t 1)3/2 16. r(t) ( 5 sin t)i Solve v a C ( ln 5 (ln 5 4 C2)j dt j 3/2 15. r(t) r(0) r(t) dr dt t C1)i (ln t 17. (ln sec t (ln sec t )j 1 dt i t 14. tan t dt j C1)i (sec t 22. v(t) a(t) 13. 0 4 cos t 2 2 /3 2 0 412 Section 10.3 10 e 4 28. (a) r(0) 1 i 4 4t (b) v(t) dy dt (e 18 e 4 (4 v(0) e 2 2t 2 4t (e Length 0 2 1) 4t 4t 6 0 i C1 and C2. r(0) C2 i 2j, 1) (1 2 2)j 2)2 (1 j 10 3 10 i 5 32 t So r(t) 2 (2e ) dt 1) dt (e 4 t 1 4t e 4 e8 7 4 32. (a) 1) dt (cos t)i (b) v(t) at t So v(t) 0 (b) 746.989 d 2y dx 2 3 ; sin 2t 0 at t 2 3 0 at t ,. 22 (2 0, 2 ,, 0. cos t 12 6t 2 t2 2t 2 1 2 j. 2, 6). 12, t 2. dy dx 6t , which for t 2/t 2 2 equals 3 (40)(40 64 (b) 0 3 , and 2 . 2 4 2t (b) Horizontal tangents: t 4 0 for t 2. Vertical tangents: 2t 2 2t 0 for t 0, 1. Plugging the t-values into x 2t 3 3t 2 and y t 3 12t produces the x- and y-coordinates of the critical points. t 2: horizontal tangent at ( 28, 16) t 0: vertical tangent at (0, 0) t 1: vertical tangent at ( 1, 11) t 2: horizontal tangent at (4, 16) 160) 0 and t 160: 225 m 15 , which for t 2 3 3 t(t 160) t 64 32 15 equals m per second. 4 3 t 32 6. 24. d dt (d) v(t) 1 equals 2/t 2)6 (4/t 3)6t , (1 2/t 2)3 (1 dy /dt dx/dt 33. (a) The j-component is zero at t 160 seconds. (c) x sin t, y cos 2t. Relate the two using the identity cos 2u 1 2 sin2 u: y 1 2x 2, where as t ranges over all possible values, 1 x 1. When t increases from 0 to 2 , the particle starts at (0, 1), goes to (1, 1), then goes to ( 1, 1), and then goes to (0, 1), tracing the curve twice. 3t 2 6t 2 10 t 5 2. That corresponds to , 3( 2)2 dy/dt dx/dt y which for t (c) 0 and sin 2t dy dx (c) When y (2 sin 2t)j and dy/dt dx/dt 0 when t 2 2 12 t 2 1i 2 0 when both cos t 2 2 t2 1 point 2 1 4t e t (e4t 1)2 (2e2t)2 dt 4 2 1 4t e t (e4t 1) dt 2 04 2 1 8t 1 4t e e te4t t dt 2 4 04 122 1 8t 1 4t 1 e e (4t 1)e4t t 2 20 32 16 16 e16 12e8 69 1,737,746.456 16 29. (a) v(t) dx dt 10 j 5 3 10 t 5 t 2 dy 30. (a) dx C1 t 1)i (4 2 10 0 2 2 2 C1 1, and 2 (e 0 12 tj 2 tj 2e2t. 2 (c) 32 ti 2 (3t)i with magnitude 2, 2, e4 dx (2e )j; dt 2t 1)i j, so v(t) and since v(0) must point directly from (1, 2) toward (4, 1) (e4)j 2i 1 , 1 , terminal 4 Initial 3i r(t) j, 18 e 4 r(2) 31. a(t) (e0)j 0i 15 equals 0 at t 2 40 80 seconds (and is negative after that time). 34. (a) Solve t (t 3)2 3 3t 2 3t 2 4: t 2 for t (b) First particle: v1(t) i 2. Then check that 2: it does. 2(t 3)j, so v1(2) and the direction unit vector v1 is Second particle: v2(t) 3 i 2 the direction unit vector is 1 , 2 i 2j . 5 5 3 j, which is constant, and 2 1 1 , 2 . 2 Section 10.3 35. (a) Referring to the figure, look at the circular arc from the point where t 0 to the point “m”. On one hand, this arc has length given by r0 , but it also has length given by vt. Setting those two quantities equal gives the result. (b) v(t) v sin vt i r0 v cos v2 vt cos i r0 r0 a(t) v2 r0 vt j, and r0 c (b) vt sin j r0 r0 So, by Newton’s second law, F fu 41. u (a) 2 r0 2 T r02 2 d (u dt T (b) d (u dt , c dt du dt fu v1, v2 v1, u2 423 r GM 0 u1 v2 ) d (u dt 2 v1), v 1 , u2 v1, u2 v 1 , u2 u1 , u 2 du dt dv dt v2 ) d (u dt 2 v1), v2) v2 v1 , v2 d (u dt 1 d (u dt 1 v) GM 4 2r03 T dt f u1, f u2 u1 , u 2 GM r0 2 f u2 u1 GM r0 2 du1 du2 dt d (u dt 1 d (u dt 1 v) GM and solve for T 2: r0 for v in v 2 1 T2 T u 1 , u2 , v 2 and solve for vT. 2 r0 du2 fu1 , fu2 v2 m r. r0 result. (e) Substitute ,c c f u1, fu2 dt d fu , fu dt 1 2 r(t). Substituting for F in the law of gravitation gives the 4 d ( fu) dt du1 fu1 v2 (c) From part (b) above, a(t) vT (d) Set r0 d cu (t), cu2(t) dt 1 d d (cu (t)), (cu2(t)) dt 1 dt v2 vt sin j r0 r0 vt cos i r0 u1(t), u2(t) . 40. (a) Suppose u d (cu) dt 413 v2) v2 v1 , v2 x 36. Solve both equations for t: t e 1 and t y 1. Now eliminate the t and solve for y: ex 1 y 1, y (e x 1)2 1, or y e 2x 2e x, where t 0 so x 0. 37. (a) Apply Corollary 3 to each component separately. If the components all differ by scalar constants, the difference vector is a constant vector. (b) Follows immediately from (a) since any two anti-derivatives of r(t) must have identical derivatives, namely r(t). d 38. v 2 dt d (v v) dt v v+v v 2v v Therefore, v is constant. 39. Let C dC C1, C2 . dt dC1 dC2 dt , dt 0, 0 . 0. du dt 42. dr dt dr dt df i dt dt ds dv dt dg j dt df i dt dg j dt df dt dt i ds df i ds dt ds dg dt dg j ds dt j ds dr ds 43. f (t) and g(t) differentiable at c ⇒ f (t) and g(t) continuous at c ⇒ r (t) f (t)i g(t)j is continuous at c. 414 Section 10.4 44. (a) Let r(t) s Section 10.4 Modeling Projectile Motion x(t), y(t) . b b b kr(t) dt kx(t), ky(t) dt kx(t) dt, a a b a b a b k x(t) dt, k y(t) dt a b k a b x(t) dt, b y(t) dt a b k x(t), y(t) dt a a k r(t) dt a (b) Let r1(t) x1(t), y1(t) and r2(t) b a (pp. 539–552) ky(t) dt Exploration 1 Hitting a Home Run 1. The graphs of the parametric equations x (152 cos 20 8.8)t, y 3 (152 sin 20 )t 16t 2 and the fence are shown in the window [0, 450] by [ 20, 60]. The fence was obtained using the line command “Line(“. You can zoom in as shown in the second figure to see that the ball does just clear the fence. x2(t), y2(t) . b (r1(t) r2(t)) dt a ( x1(t), y1(t) x2(t), y2(t) ) dt b x1(t) a x2(t), y1(t) (x1(t) x2(t)) dt, y2(t) dt b a b a b a b b a (y1(t) b x1(t) dt a b x1(t) dt, a y2(t)) dt b x2(t) dt, a b y1(t) dt y2(t) dt a b y1(t) dt b x2(t) dt, a a y2(t) dt 2. angle (degrees) b r1(t) dt You can also use algebraic methods to show that t 2.984 when x 400, and that y 15.647 for this value of t. r2(t) dt (c) Let C a 25 30 45 C1, C2 , r(t) b range (ft) x(t), y(t) . C r(t) dt a (C1x(t) C2y(t)) dt b C1 588.279 665.629 4.061 4.789 6.745 3. Using the same window of part (1) we can see that the ball clears the fence. b a 523.707 flight time (sec) a b x(t) dt C2 a y(t) dt a b C1, C2 b x(t) dt, y(t) dt a a 4. angle (degrees) b C r(t) dt a 25 30 45 45. (a) Let r(t) d dt f (t)i t 559.444 630.424 724.988 4.061 4.789 6.745 g(t)j. Then d dt r(q) dq a range (ft) flight time (sec) Exploration 2 Hitting a Baseball t [ f (q)i g(q)j] dq a t d dt f (q) dq i g(q) dq j a a t d dt d dt f (q) dq i a f (t)i 1. x t g(t)j t g(q) dq j a y 152 (1 e 0.05t ) cos 20 0.05 152 3 (1 e 0.05t ) sin 20 0.05 32 (1 0.05t e 0.05t ) 0.052 r(t). t r(q) dq. Then part (a) shows that S(t) is an (b) Let S(t) a antiderivative of r(t). Let R(t) be any antiderivative of r(t). Then according to 37(b), S(t) Letting t a, we have 0 Therefore, C S(a) C. R(a) b. C. R(t) R(a) and S(t) The result follows by letting t R(t) R(a). [0, 450] by [–20, 60] 2. The ball reaches a maximum height of about 43.07 ft when t is about 1.56 sec. 3. The range is about 425.47 ft and the flight time is about 3.23 sec. Section 10.4 Quick Review 10.4 Section 10.4 Exercises 1. 50 cos 25 , 50 sin 25 45.315, 21.131 1. Solve vxt (840 cos 60°) t 40, 40 3 2. 80 cos 120 , 80 sin 120 3. To find the x-intercepts, solve 2x 2 11x 40 0 using 112 4(2)( 40) 2(2) 11 the quadratic formula: x for v0: v0 5 8. The x-intercepts are , 0 and ( 8, 0). For the 2 2(0)2 y-intercept is (0, 2v0 sin 4x 11 2 11 11 4 11 . Then the 4 6. At the vertex, g (x) 20 2x 0 and x vertex is (10, 20(10) 102) (10, 100). 7. y cos x C. y y cos x cos 10. Then the C C 2 8. y t C1 and y 2 y ( 1) ( 1) y ( 1) 1 ( 1)3 3 t3 3 16 y y y(0) 16 16 4 (so the ground is t 32 y 2 y(0) 2 k 1 (32)t 2 2 2 seconds (v0 cos ) t 5, so 5. Use y 2 (v0 sin ) t 22 2t 11 2 6.5 346 16 (32 cos 30°)2 12 gt 2 32 3 55.426 6.5. 0 2.135 seconds (by the quadratic (44 sin 45°) t to obtain x (v0 cos ) t 66.4206. 66.421 feet from the stopboard. 4 6. With the origin at the launch point, use y t 12 x 2 k x2 e 2 k 2 1 so k 2 (v0 sin )t (44 sin 40 ) At t (44 cos 40 )t 66.5193 ft 0.0987 feet farther. 2 1.974, x 12 gt . 6.5 (44 sin 40 )t 16t 2 2 (44 sin 40 )2 416 1.974 sec 32 Thus the shot would have gone C x2 2 32), use t2 t ke 6377.551 m feet away (horizontally). C 20 so k 2y x (32 sin 30°) t t t x dx e C2 3. 2 when y 12 gt . 2 (v0 sin ) t y 16t t 4e 2y y C2 2g Then x 4, so C1 (500 sin 45°)2 2(9.8) 4020 m 4. With the origin at the launch point 2 10 3 (v0 sin )2 (c) ymax 2, so C2 (tan ) x formula). Substitute that into x t ke dy 4 2y 1 ln 4 2 10. 3( 1) C1 dt ln 16 y 1 25 3 3t dy 16 y 9. C1 C1t 25 3 C2 y 13 t 3 x2 9.8 50002 + (tan 45°)5000 2(500)2 cos2 45° t 2 g 2v02 cos2 (b) y 2. 2 25,510 m 25.510 km downrange 40 5. To find the x-intercepts, solve 20x x 2 0: x 0 or 20. The x-intercepts are (0, 0) and (20, 0). For the y-intercept, find y (0): it is already known to be 0. So the y-intercept is (0, 0). sin 90° 72.154 seconds; 5002 sin 90° 9.8 v0 g sin 2 40. The 0 and x 11 2 4 11 ,2 4 441 . 8 11 , 4 40 2(500)sin 45° 9.8 g R 9.8 50 seconds. 490 m/sec. 40). 4. At the vertex, f (x) vertex is 11(0) v02 v0 sin 2 ; solve 24,500 g 3. (a) t y-intercept, find f (0) 21,000 for t: t 2 2. Use R 5 or 2 415 1.18 inches 416 Section 10.4 v02 sin 2 ; solve 10 g 7. (a) Use R v0 7 2 1 2 sin 2 9.8 v02 sin 2 : sin 2 g 1 sin 0.98 39.261 or 2 15. Use R sin 90° for v0: 9.899 m/sec. (7 (b) Solve 6 v0 2 2)2 sin 2 for : sin 2 9.8 180° 0.6 36.870 and 1 sin 40 10 2 m 5 106 m/sec 1 positive) is gt 2 2 8. t 0.6 8 10 8 1 (9.8)(8 2 meters or 3.136 12 10 71.565 . sec. Then y (taking down as 10 ) 3.136 14 10 cm. sin 18° 32 ft/sec 2 is 189.556 mph. 80 10 2 3 v02 sin 2 ⇒ 200 g 32 1 sin 1 0.9 2 10 2 3 ymax and c2 31.339, which is well below y x 90 ft/sec, and x 135 ft, (tan ) x evaluates to d r(t) dt c1i ( gt c2)j. The initial v0 cos v0 sin . Integrating again, c3)i 12 gt 2 (v0 sin )t c4 j. The initial condition on the position gives x0 and c4 y0. 20. With the origin at the launch point, ymax 29.942 . Substituting g , which is three-fourths of the maximum height. ((v0 cos )t r(t) c3 2 and it occurs when t condition on the velocity gives c1 sin 2 32.079 30°, v0 32 2v02 cos 2 8g 46.597 ft/sec. into the equation for y(t ) gives a height of 19. Integrating, the ceiling height. 11. No. For 2g v0 sin 0.9. 32.079 . Then 2(32) (v0 sin )2 5 6 73 and 12 gt , and we know the maximum height 2 v0 sin v0(sin )t 2g 3(v0 sin )2 sin 2 ⇒ sin 2 (tan 40°) x. Setting x 6.5 and solving for v0 yields v0 t Taking the smaller of the two possible angles, 80 32 x2 2v02 cos 2 40° 18. y(t) 278.016 ft/sec or 10. R 17. With the origin at the launch point, y v02 (248.8 yd)(3 ft/yd) v0 (b) To increase the range (and height) by a factor of 2, increase v0 by a factor of 2 1.41. That is an increase of 41%. y v02 sin 2 g 9. R 0.98 2 16. (a) Substitute 2v0 for v0 in the formula for range. 18.435 or 82 90 0.98; 1 50.739 . 0.6, so 143.130 and (9.8)(16,000) 4002 sin 2ymaxg v0 2(68)(32) sin 56.505° v0 sin 68 ft. Then 79.107 ft/sec. feet above the ground, which is not quite high enough. 21. The horizontal distance is 30 yd 32 x2 2(116)2 cos 2 45° 12. Use y 22 x 841 x. Set y (tan 45°) x 84 45, then solve for x using the quadratic formula and taking the larger of the two values: 481 841 1 x 4 841 369.255 ft, which is 0.255 ft y 32 x2 2v02 cos 2 20° y 37 3 (tan 20°) x. Set x 315 and 34, then solve to find cos 20° 315 tan 20° (b) Solve v0(cos 20°) t t 34 149.307 ft/sec. 149.307(cos 20°) t 315 to find 2.245 seconds. 14. In the formula for range, sin 2 sin 2(90 ). 68 45 56.5° and 1 17 ) = 84 v0 cos 1.924 seconds. Then 15 17 12 gt 2 119 15 1260 v0 4 17 (v0 sin ) t (16 13. (a) With the origin at the launch point, use tan 84 ft. Then 16 17 (from Exercise 20). So t sin 68 21 45 84 tan 119 16 17 3.059 inches beyond the pin. y v0 (v0 cos ) t, where 6 ft 17 1 119 2 (32) 2 15 17 67.698 ft. The height above the ground is 6 ft more than that, 73.698, and the height above the rim is about 73.698 70 3.698 feet. 22. The projectile rises straight up and then falls straight down, returning to the firing point. Section 10.4 23. Angle is 62° (measurements may vary slightly). For 2v0 sin g flight time t (v0 sin ) 1 sec, ymax 2 (b) v0 12 gt 8 2g 1 (32)(1)2 4 ft (independent of the measured angle). 8 gt gt v0 , so speed of engine v0 cos 2 sin 2 tan 32(1) 8.507 ft/sec (changes with the angle). 2 tan (62°) 24. The height of A is given by yA height of B is given by yB (Rcos , Rsin ) 12 gt and the 2 (v sin )t 12 gt . The second 2 R tan 417 12 gt are equal for any value of t. But 2 R A moves R units horizontally to B’s line of fall in v cos R terms in yA and yB time units, and the first terms in yA and yB are also equal at R that time: (v sin ) v cos R tan . Therefore, A and B will always be at the same height when A reaches B’s line 2v02 R cos g cos2 tan sin ( cot ( so ) is maximum when ), 90 . The initial velocity vector bisects the angle between the hill and the vertical for max range. of fall. 26. (a) r(t) x(t) y(t) 25. (a) (x(t))i (y(t))j, where (145 cos 23 14)t and 2.5 (145 sin 23 )t 16t 2. (v0 sin )2 (b) ymax (145 sin 23°)2 64 2.5 2g 52.655 feet , which is reached at t v0 145 sin 23° 32 2.5 v0 sin g 1.771 seconds. (c) For the time, solve R (Rcos , –Rsin ) y 2.5 (145 sin 23°) t 16t 2 0 for t, using the quadratic formula: x (v0 cos )t y (v0 sin )t x R cos ⇒t ⇒ ⇒R t 1 2 x (v0 cos )t R cos . Then y R sin v0 cos (v0 sin ) R cos g R 2 cos2 R sin v0 cos 2 v02 cos2 2v02 Let f ( ) cos cos sin ( sin ( cos f() 0 ⇒ tan tan ( cot ( ⇒ 90 ) ). velocity bisects angle AOR. 14)(3.585) 2.5 428.262 feet. 16t 2 (145 sin 23°) t t 145 sin 23° (145 sin 23°)2 32 1120 0.342 and 3.199 seconds. At those times the ball is about sin ) sin ( ) 1 x (0.342) (145 cos 23° and x (3.199) 14)(0.342) (145 cos 23° 40.847 feet 14)(3.199) 382.208 feet from home plate. . Note that f ( ) maximum when (145 cos 23° 20 for t, using the quadratic formula: ) ⇒ tan cos ( 3.585 sec. 3.585 is about (d) For the time, solve y ). f() 160 32 Then the range at t gt 2 ⇒ R cos g cos2 (145 sin 23°)2 145 sin 23° 90 0, so R is . Thus the initial (e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate. 418 Section 10.4 27. (a) (Assuming that “x” is zero at the point of impact.) r(t) (x(t))i (y(t))j, where x(t) (35 cos 27 )t and y(t) 4 (35 sin 27 )t 16t 2. (b) ymax (v0 sin )2 4 2g which is reached at t (d) Graph Y2 30 and use the intersect function to find that y 30 for t 0.689 and 2.305 seconds, at which times the ball is about x (0.689) 94.513 feet and x (2.305) 287.628 feet from home plate. (35 sin 27°)2 4 7.945 feet, 64 v0 sin 35 sin 27° = g 32 0.497 seconds. 29. (a) r(t) (c) For the time, solve y 4 16t 2 (35 sin 27°) t 0 ( 35 sin 27°)2 32 35 sin 27° (x(t))i 256 1.201 seconds. Then the range is about x (1.201) (35 cos 27°)(1.201) 37.406 feet. (y(t))j, where 1 (1 e 0.08t)(152 cos 20 17.6) and 0.08 152 3 (1 e 0.08t)(sin 20 ) 0.08 32 (1 0.08t e 0.08t) 0.082 x(t) y(t) for t, using the quadratic formula: t (e) Yes, the batter has hit a home run since a graph in parametric mode shows that the ball is more than 14 feet above the ground when it passes over the fence. (b) Solve graphically: enter y (t) for Y1 (where X stands in for t), then use the maximum function to find that at t 1.527 seconds the ball reaches a maximum height of about 41.893 feet. (c) Use the zero function to find that y (d) For the time, solve y 4 16t 2 (35 sin 27°) t 7 for t, using the quadratic formula: ( 35 sin 27°)2 32 35 sin 27° t 192 has traveled for 0 when the ball 3.181 seconds. The range is about x (3.181) 0.254 and 0.740 seconds. At those times the ball is about x (0.254) (35 cos 27°)(0.254) (35 cos 27°)(0.740) 23.064 feet from the 0.08(3.181) e )(152 cos 20° 17.6) 351.734 feet 7.906 feet and x (0.740) 1 (1 0.08 impact point, or about 37.460 and 37.460 23.064 7.906 29.554 feet 14.396 feet from the landing (d) Graph Y2 that y 35 and use the interesect function to find 35 for t times the ball is about x (2.190) spot. (e) Yes. It changes things because the ball won’t clear the net (ymax 7.945 ft). 0.877 and 2.190 seconds, at which x(0.877) 106.028 feet and 251.530 feet from home plate. (e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) 28. (a) r(t) x(t) y(t) (x(t))i (y(t))j, where to find that y 152 (1 e 0.12t)(cos 20 ) and 0.12 152 (1 e 0.12t)(sin 20 ) 3 0.12 32 (1 0.12t e 0.12t) 0.122 (b) Solve graphically: enter y (t) for Y1 (where X stands in for t), then use the maximum function to find that at t 1.484 seconds the ball reaches a maximum height of about 40.435 feet. (c) Use the zero function to find that y has traveled for x (3.126) 3.126 seconds. The range is about 152 (1 0.12 372.323 feet. 0 when the ball e 0.12(3.126) )(cos 20°) 20 at t 0.376 and 2.716 seconds. Then define x (w) 1 (1 0.08 and solve x (w) e 0.08(2.716) )(152 cos 20° 380 to find w w), 12.846 ft/sec. This is the speed of a wind gust needed in the direction of the hit for the ball to clear the fence for a home run. Section 10.5 419 30. (a) To save time, enter one expression for x (t) in the main window, then use the ENTRY function to repeat it six times in the parametric Y menu. Do the same for y (t), then make appropriate modifications. 32. dr dt d 2r dt (v0e kt cos )i kt ( kv0e gj k kt (v0e cos )i g kt e k sin ( kv0e kt sin g )j k kt ge )j dr dt The initial conditions are also satisfied, since r(0) v0 k e0)(cos )i (1 v0 [0, 500] by [0, 50] k Now replace T with X in all the y (t) expressions in the parametric Y menu, then change to function mode and enter Y1T for Y1, Y2T for Y2, and so on. Use the functions in the CALC menu to fill in the tables for parts (b) and (c). (b) drag coeff time at max ht 0.01 0.02 0.10 0.15 0.20 0.25 t t t t t t 3.289 3.273 3.153 3.088 3.028 2.974 462.152 452.478 386.274 352.983 324.410 299.661 lim 1 − e t, and kt e k2 kt R ,y . So, 2 max 31. The points in question are (x, y) x v02 sin cos , and y g v0 v02 sin cos g2 v0 4 g2 v0 4 g2 2g 2 4 (v0 sin )2 sin4 4 sin2 4 cos sin2 cos2 4 sin2 cos2 (sin2 )(1 v04 4g2 4g 12 4 sin 2 v02 2 2g sin2 4 2 2 v0 4 1 g2 4 . Then 4g g v0 4 (v0 sin )2 22 4y x2 (v0e0 cos )i (v0 cos )i v0e0 sin g0 e k g j k (v0 sin )j cos2 ) , so the point (x, y) lies on the ellipse. 1 16 sin2 s Section 10.5 Polar Coordinates and Polar Graphs (pp. 552–559) Exploration 1 Investigating Polar Graphs 1. The graph is drawn in the decimal window [ 4.7, 4.7] by [ 3.1, 3.1] with 2 ≤ ≤ 2 2 . . ) 4. If (r, ) is a solution of r 2 4 cos , then so is (r, because cos ( ) cos . Thus, the graph is symmetric about the x-axis. If (r, ) is a solution of r 2 4 cos , then so is ( r, ) because ( r)2 r 2 and cos ( ) cos . Thus, the graph is symmetric about the y-axis. The graph is symmetric about the origin because it is symmetric about both the x- and y-axes. You can also give a direct proof by showing that ( r, ) lies on the graph if (r, ) does. t2 . 2 k→0 As k → 0, the air resistance approaches 0. lim 1 e0) j 3. kt k dr dt t 0 0 0, 2. r1 and r2 are 0 for (d) This follows from the following two limits (as k → 0): k→0 and 0j g (1 k2 max ht k 0.01 t 1.612 44.777 k 0.02 t 1.599 44.336 k 0.10 t 1.505 41.149 k 0.15 t 1.454 39.419 k 0.20 t 1.407 37.854 k 0.25 t 1.363 36.431 (c) After flight times using the zero function, plug the x-intercepts into T and read the ranges out as X1T, X2T, etc. drag coeff flight time range k k k k k k 0i e0)(sin ) (1 1 4 420 Section 10.5 Exploration 2 Graphing Rose Curves 4. (a) No; y is a function of x and is not the zero function. All graphs are drawn in the window [ 4.7, 4.7] by [ 3.1, 3.1]. (b) No; y( x) 1. The graphs are rose curves with 4 petals when n 2, 8 petals when n 4, and 12 petals when n 6. ( x)3 x3 ( x) (c) Yes; y( x) (x 3 x x) y(x) y(x) (see part (b)) 5. (a) No; y is a function of x and is not the zero function. ( x)2 (b) No; y( x) (c) No; y( x) ( x) x2 x y(x) y(x) (see part (b)) 6. (a) No; y is a function of x and is not the zero function. n 2 (b) Yes; y( x) cos ( x) (c) No; y( x) cos x y(x) y(x) (see part (b)) 7. (a) Yes; Substitute y for y in the equation to get the original equation. (b) Yes; Substitute x for x in the equation to get the original equation. n 4 n (c) Yes; since the curve is symmetric with respect to both the x-axis and y-axis, it is symmetric with respect to the origin. (Also, substitute x for x and y for y in the equation to get the original equation.) 8. Solve for y : y (x 2)1/2 or (x 2)1/2. Enter the first expression for Y1, the second for Y2. 6 3. The graph is a rose curve with 2 n petals. 4. The graphs are rose curves with 3 petals when n 5 petals when n 5, and 7 petals when n 7. x 2 1/2 4 9. Solve for y : y 2. 2 3 or x 2 1/2 4 3 . Enter the first expression for Y1, the second for Y2. 3, 10. (x 2 4x) ( y 2 6y 9) 0 (x 2 4x 4) ( y 2 6y 9) (x 2)2 (x 3)2 22. Center (2, 3), Radius 2. 4 Section 10.5 Exercises n For Exercises 1 and 2, two pairs of polar coordinates label the same point if the r-coordinates are the same and the -coordinates differ by an even multiple of , or if the r-coordinates are opposites and the -coordinates differ by an odd multiple of . 3 n 1. (a) and (e) are the same. (b) and (g) are the same. (c) and (h) are the same. (d) and (f) are the same. 5 2. (a) and (f) are the same. (b) and (h) are the same. (c) and (g) are the same. (d) and (e) are the same. n 7 3. 5. 6. The graph is a rose cuve with n petals. Quick Review 10.5 1 1. Slope so y 3 4 ( 2) 4 1[x 2. (x 0)2 3. [x ( 2)]2 (y 0)2 (y [ 3, 3] by [ 2, 2] 1, ( 2)] or y 32, or x 2 4)2 x y2 (a) 2. 9. 22, or (x 2)2 2 cos 4 , 2 sin (b) (1 cos 0, 1 sin 0) (y 4)2 4. (c) (d) 0 cos 2 , 0 sin 2 cos 4 , 2 4 = (1, 1) (1, 0) = (0, 0) 2 sin 4 ( 1, 1) 421 Section 10.5 6. 4. [ 9, 9] by [ 6, 6] (a) 3 cos 5 , 6 3 sin 1 (b) 5 cos tan (c) ( 1 cos 7 , (d) 2 3 cos [ 9, 9] by [ 6, 6] 4 3 5 6 33 , 2 , 5 sin tan 1 sin 7 ) 2 ,2 3 3 sin 1 3 2 (a) r with (3, 4) 2, ( ( 1)2 2, tan 1 1 3 4 3 (1, 0) 2 3 3)2 ( 3, 3) in quadrant III. The coordinates are 2, 6 3 7 . 6 also works, since r has the opposite sign and differs by . 5. 4 with in quadrant I. 3 4 4 The coordinates are 5, tan 1 . 5, tan 1 3 3 32 (b) r 42 5, tan also works, since r has the opposite sign and differs [ 6, 6] by [ 4, 4] ( 1)2 (a) r 12 quadrant II. The coordinates are 5 4 2, by . 1 1 2, tan 2, 1 with in 2 is undefined with 0 3 on the negative y-axis. The coordinates are 2, . 2 3 . 4 (c) r also works, since r is the same and 2, differs by 2 . 12 (b) r ( 3)2 2 3 2 ( 2)2 also works, since r is the same and 3 with 1 (d) r 3 22 02 . 3, tan 3 is undefined with 0 the positive y-axis. The coordinates are 3, also works, since r is the same and (d) r 0 2 works, since r is the same and 32 ( 1)2 02 1, tan 0 1 2 on . 3, 5 2 differs by 2 . [ 6, 6] by [ 4, 4] 8. 0 with on the negative x-axis. The coordinates are (1, ). ( 1, 0) also works, since r has the opposite sign and differs 0 with on the positive x-axis. The coordinates are (2, 0). (2, 2 ) also 7. 02 2, tan also works, since r has the opposite sign and differs by . (c) r 2, tan by 2 . 3 2, tan in quadrant IV. The coordinates are 2, 2, 0 differs by . [ 6, 6] by [ 4, 4] 9. [ 3, 3] by [ 2, 2] 10. [ 3, 3] by [ 2, 2] differs by 2 . 422 Section 10.5 25. x 2 y 2 x2 y2 (center [ 9, 9] by [ 6, 6] r 2 and y r sin , so the equation is 4y ⇒ x 2 ( y 2)2 4, a circle (0, 2), radius 2). 26. r 11. 5 2 cos sin r sin 12. x 2r cos r cos and y a line (slope 27. r 2 sin 2 [ 6, 6] by [ 4, 4] r sin , so the equation is y 2, y-intercept 2r sin cos 2 1 x r sin , so the equation is xy r cos or, y [ 3, 3] by [ 2, 2] 28. r y and y 1 1 , a hyperbola. x cot r sin [ 3, 3] by [ 2, 2] 5, 5). (r sin )(r cos ) 14. 2x 2 2 13. 5 csc cot r sin and x y cot , so the equation is y 2 x, a parabola. 15. 29. r csc ercos r sin ercos x r cos and y r sin , so the equation is y exponential curve. e x, the sin2 30. cos2 2 (r cos ) (r sin )2 x r cos and y r sin , so the equation is x 2 y x, the union of two lines. [ 3, 3] by [ 2, 2] 16. y 2 or 31. r sin ln r ln cos r sin ln (r cos ) y ln x, the logarithmic curve. [ 3, 3] by [ 2, 2] 17. 1 32. r 2 2r 2 cos sin r 2 2(r cos )(r sin ) 1 x 2 y 2 2xy 1 (x y)2 1 xy 1, the union of two lines. [ 1.8, 1.8] by [ 1.2, 1.2] 33. r 2 x2 (x 18. [ 3, 3] by [ 2, 2] 19. y r sin , so the equation is y 0, which is the x-axis. 20. x r cos , so the equation is x 0, which is the y-axis. 21. r 4 csc r sin 4 y r sin , so the equation is y 22. r 3 sec r cos 3 x r cos , so the equation is x 4, a horizontal line. 3, a vertical line. 23. x r cos and y r sin , so the equation is x line (slope 1, y-intercept 1). 24. x 2 y 2 (center r 2, so the equation is x 2 (0, 0), radius 1). y2 y 1, a circle 1, a 4r cos y2 4x 2)2 y 2 4, a circle (center 34. r 8 sin r 2 8r sin x 2 y 2 8y x2 ( y 4)2 16, a circle (center 35. r 2 cos 2 sin 2r sin r 2 2r cos x 2 y 2 2x 2y (x 1)2 ( y 1)2 2, a circle (center (1, 1), radius ( 2, 0), radius 2). (0, 4), radius 4). 2). Section 10.5 36. r sin 6 r sin 3 2 3 2 cos 1 x 2 y 3y x cos sin 1 r cos 2 r sin 42. x 2 y 2 r 2 cos2 r 2(cos2 2 6 1 1 2 4, a line slope 1 , y-intercept 3 4 [ 4.7, 4.7] by [ 3.1, 3.1] . 3 43. 37. x 7 r cos r 2 sin2 sin2 ) 2 6 2 1 7. The graph is a vertical line. x2 y2 9 4 r 2 cos2 9 2 1 r 2 sin2 4 2 r (4 cos 9 sin 1 2 ) 36 [ 9.4, 9.4] by [ 6.2, 6.2] 38. y 1 r sin 1 The graph is a horiztonal line. [ 4.7, 4.7] by [ 3.1, 3.1] 44. xy 2 (r cos )(r sin ) 2 r 2 cos sin 2 r 2 2 cos sin 4 r 2 sin 2 4 [ 4.7, 4.7] by [ 3.1, 3.1] 39. x y ⇒ r cos r sin More generally, 4 ⇒ tan 1⇒ 4 2k for any integer k. The graph is a slanted line. [ 4.7, 4.7] by [ 3.1, 3.1] 45. y 2 4x r 2 sin2 r sin2 4r cos 4 cos [ 4.7, 4.7] by [ 3.1, 3.1] 40. x y r cos 3 r sin [ 4.7, 4.7] by [ 3.1, 3.1] 3 46. x 2 xy y 2 1 (r cos )2 (r cos )(r sin ) r 2(1 cos sin ) 1 [ 9.4, 9.4] by [ 6.2, 6.2] 41. x 2 r2 y2 4 4 or r 2 (or r 2) [ 3, 3] by [ 2, 2] [ 4.7, 4.7] by [ 3.1, 3.1] (r sin )2 1 423 424 Section 10.5 47. x 2 ( y 2)2 4 r 2 cos2 (r sin 2)2 4 r 2 cos2 r 2 sin2 4r sin 44 r 2 4r sin 0 r 4 sin . The graph is a circle centered at (0, 2) with radius 2. 52. (a) [ 3, 3] by [ 2, 2] (b) Length of interval 2 53. (a) [ 4.7, 4.7] by [ 1.1, 5.1] [ 3.75, 3.75] by [ 2, 3] 48. (x 3)2 1)2 3)2 (r cos (y (r sin r 2 cos2 r 2 r r 6r cos 6r cos 6 cos 3 cos 4 (b) Length of interval 1)2 2r sin 2 sin sin 6 (6 cos 2 (3 cos 54. (a) 4 r 2 sin2 9 2 2r sin 1 4 0 2 sin )2 24 [ 1.5, 1.5] by [ 1, 1] (b) Length of interval sin )2 6 4 55. (a) [ 15, 15] by [ 10, 10] (b) Required interval [ 6, 6] by [ 4, 4] In Exercises 49–58, find the minimum -interval by trying different intervals on a graphing calculator. ( ,) 56. (a) 49. (a) [ 3, 3] by [ 2, 2] (b) Length of interval [ 3, 3] by [ 2, 2] (b) Length of interval 2 57. (a) 2 50. (a) [ 3, 3] by [ 2, 2] (b) Length of interval 58. (a) [ 6, 6] by [ 4, 4] (b) Length of interval 2 51. (a) [ 3, 3] by [ 1, 3] (b) Length of interval [ 1.5, 1.5] by [ 1, 1] (b) Length of interval 2 2 59. If (r, ) is a solution, so is ( r, ). Therefore, the curve is symmetric about the origin. And if (r, ) is a solution, so is (r, ). Therefore, the curve is symmetric about the x-axis. And since any curve with x-axis and origin symmetry also has y-axis symmetry, the curve is symmetric about the y-axis. 60. If (r, ) is a solution, so is ( r, ). Therefore, the curve is symmetric about the origin. The curve does not have x-axis or y-axis symmetry. 425 Section 10.5 61. If (r, ) is a solution, so is (r, ). Therefore, the curve is symmetric about the y-axis. The curve does not have x-axis or origin symmetry. 62. If (r, ) is a solution, so is ( r, ). Therefore, the curve is symmetric about the origin. And if (r, ) is a solution, so is (r, ). Therefore, the curve is symmetric about the x-axis. And since any curve with x-axis and origin symmetry also has y-axis symmetry, the curve is symmetric about the y-axis. 66. (a) [ 9, 9] by [ 6, 6] The graph of r2 is the graph of r1 rotated by angle counterclockwise about the origin. (b) 63. (a) Because r a sec is equivalent to r cos a, which is equivalent to the Cartesian equation x a. (b) r a csc is equivalent to y a. 64. (a) The graph is the same for n 2 and n 2, and in general, it’s the same for n 2k and n 2k. The graphs for n 2, 4, and 6 are roses with 4, 8, and 12 “petals” respectively. The graphs for n 2 and n 6 are shown below. n 2 [ 9, 9] by [ 6, 6] The graph of r2 is the graph of r1 rotated by angle clockwise about the origin. (c) The graph of r2 is the graph of r1 rotated counterclockwise about the origin by the angle . 67. d [(r2 cos [ 3, 3] by [ 2, 2] n 6 [r22 cos2 r22 r12 [ 3, 3] by [ 2, 2] x1)2 (x2 r1 cos 2 y2)2 ( y1 1 )2 2r2 r1 cos 2 2 sin 2 r22 2r1r2 cos ( (r2 sin 2 cos 2r2r1 sin 1 2 2 2 1 r12 cos2 1 sin r1 sin 1 r12 )2]1/2 1 2 sin 1/2 1 ) 68. (a) (b) 2 (c) The graph is a rose with 2 n “petals”. (d) The graphs are roses with 3, 5, and 7 “petals” respectively. The “center petal” points upward if n 3, 5, or 7. The graphs for n 3 and n 3 are shown below. [ 9, 9] by [ 6, 6] The graphs are ellipses. (b) Graphs for 0 k 1 are ellipses. As k → 0 , the graph approaches the circle of radius 2 centered at the origin. 69. (a) [ 3, 3] by [ 2, 2] [ 5, 25] by [ 10, 10] The graphs are hyperbolas. [ 3, 3] by [ 2, 2] (e) (f) The graph is a rose with n “petals”. 65. (a) We have x we have r r cos and y r sin . By taking t , f (t), so x f (t) cos t and y f (t) sin t. (b) x 3 cos t, y (c) x (1 3 sin t (d) x (3 sin 2t) cos t, y cos t) cos t, y (1 cos t) sin t (3 sin 2t) sin t (b) Graphs for k 1 are hyperbolas. As k → 1 , the right branch of the hyperbola goes to infinity and “disappears”. The left branch approaches the parabola y 2 4 4x. 426 Section 10.6 Section 10.6 Exercises 70. (a) dy dx [ 9, 9] by [ 6, 6] The graphs are parabolas. f ( ) sin f ( ) cos f ( ) cos f ( ) sin cos cos 1. (1 (1 (b) As k → 0 , the limit of the graph is the negative x-axis. sin cos 2 sin cos cos2 s Section 10.6 Calculus of Polar Curves dy dx (pp. 559–568) Quick Review 10.6 2. dy dy/dt 5 cos t dx dx/dt 3 sin t 5 2. cot 2 0.763 3 5 cot t 3 1. 3. Solve cot t 0: t 2 or dy dx 3 ; 2 and the corresponding points are 3 cos 3 3 and 3 cos , 5 sin 2 2 (0, 5 cot t is undefined when t 3 4. dy dx 2 , 5 sin 2 3. 5. Length 0 9 sin2 t 1, 1 1 1 2 sin 2 sin 2 sin 2 cos f ( ) cos f ( ) sin dy dx /2 1 , which is undefined. 0 dy dx f ( ) sin f ( ) cos f ( ) cos f ( ) sin 3 cos 3 cos sin cos 2 cos 2 sin (2 (2 6 sin 3(cos2 3 sin ) cos 3 sin ) sin cos sin2 ) (3, 0) and dy dx (2, 0) dy dx dy dx (2, ) 25 cos2 t dt, 0 dy dx 2 , 3 0 1 /2 dy dx dy dx (5, 3 /2) 12.763. 2 3 0 dy dx ( 1, /2) dy 2 dt dt which using NINT evaluates to dy dx 1 dy , which is undefined; 0 dx 0 0, , or 2 ; ( 3, 0). dx 2 dt sin (0, 5) 5) (3 cos 2 , 5 sin 2 ) (3 cos , 5 sin ) 1 1 f ( ) sin f ( ) cos the corrresponding points are (3 cos 0, 5 sin 0) cos sin2 0 sin )cos sin )sin 0, 2 , and 3 dy dx 0 5 3 /2 0. For questions 6–8, the graph is: dy dx f ( ) sin f ( ) cos f ( ) cos f ( ) sin 3 sin2 3 sin cos 4. 3 cos (1 cos ) 3 sin (1 cos ) 3(cos2 sin2 ) cos 3 sin 3 cos 6 sin [ 2, 4] by [ 2, 2] 6. The upper half of the outer loop 7. The inner loop dy dx (1.5, /3) 8. The lower half of the outer loop 1 2 1 2 3 3 2 2 9. y 0 for x 6 (6x Area 0 or 6. x 2) dx 3x 2 0 136 x 3 0 1 2 dy dx (4.5, 2 /3) 36 3 2 , which is undefined; 1 2 3 2 0; 10. Use a graphing calculator’s intersect function to find that the curves cross at x 0.270 and x 2.248, then use NINT to find 2.248 Area [2 sin x 0.270 (x 2 2x 1)] dx 2.403. dy dx (6, ) dy dx (3, 3 /2) 11 , which is undefined; and 00 0 0 ( 1) ( 1) 1. cos 2 cos . cos 2 sin 0 1 0; Section 10.6 427 7. 5. [ 3.8, 3.8] by [ 2.5, 2.5] [ 1.5, 1.5] by [ 1, 1] The graph passes through the pole when r which occurs when 2 3 cos 0, The polar solutions are 0, 3 . Since the 2 and when only consider x k . This can be confirmed analytically by 5 dy noting that the slope of the curve, , equals the slope of dx k the line, tan . So the tangent lines are 0 [ y 0], 5 2 2 y tan x , y tan x, 5 5 5 5 3 3 4 4 y tan x , and y tan x. 5 5 5 5 (or x 2 curve at 0, 0). y 3 cos2 (3 cos ) cos (3 cos ) sin ( 3 sin )sin 3(cos2 (3 cos )cos sin2 ) 6 cos ( sin ). 8. dx , 0, and At 0, /2 2 d dy dx 3(02 12) 3. So at 0, , 0 /2 2d d dy dy 0, so is undefined and the tangent line is and dx d [ 3, 3] by [ 2, 2] The polar solutions are 0, vertical. a given k, the line 6. [ 3, 3] by [ 2, 2] A trace of the graph suggests three tangent lines, one with positive slope for 6 , a vertical one for 2 5 . 6 with negative slope for , and one Confirm analytically: dy d 6 sin 3 sin 2 cos 3 cos dx d 6 sin 3 cos 2 cos 3 sin . dy dx dy dx 6 , 0, 2 , and 0, are all solutions. dy/d , and so dx/d 6(1)(1/2) 2(0)( 3/2) 6(1)( 3/2) 2(0)(1/2) /6 dy dx dy dx 6( 1)(1) 6( 1)(0) /2 5 /6 1 6(1)(1/2) 2(0)( 3/2) 6(1)( 3/2) 2(0)(1/2) 5 6 y ; 3 2(0)(0) , which is undefined; and 2(0)(1) lines have equations and 5 6 and 6 1 x. 3 y 1 1 x, 3 . The tangent 3 2 [x 0], k 2 for k 0, 1, 2, 3, 4, and for k appears to be tangent to the 2 k . This can be confirmed analytically by 2 dy noting that the slope of the curve, , equals the slope of dx k the line, tan . So the tangent lines are 0 [ y 0] and 2 3 [x 0]. , and 2 are duplicate 2 2 curve at 0, 0, k appears to be tangent to the 5 . At this point, there appears to be a 2 Confirm analytically: dx d 0, 1, 2, 3, 4, and for a given k, the line vertical tangent line with equation and for k 1 produces the entire graph, we need -interval 0 dy d k 5 solutions. 428 9. Section 10.6 dy d cos sin (1 cos (2 sin (1 dy d 3 (cos 22 , 5 (2 sin 66 1 1 9 0). the line y the line y 1 sin 2 6 2 , for 7 11 , , or 6 6 3 ,r 2 6 1 2 dy , 2d dx d d dy dd d dx dd 4 sin 7 ,r 6 sin 1 for cos 2 ,r or cos 3 3 4 2 0 for 0 [the line x 1 , 26 3 y 3 2 y 2, Vertical at: 0, 2 [x . and 2 0]. 1 , 4 2 0], 37 , 26 x 3 11 , 26 x 5 dx . 3d 5 ,r 3 33 , 4 33 4 1 2 1 or , i.e., when 0 when 0, , 2 24 , 33 0). So there is a horizontal tangent line 3 the line y 2 ,r 3 the line y 2 2 cos 0 2 cos 2 3 sin 2 3 33 and for 4 3 5 sin 2 3 0, r 3 , so by . There is a 2 2], for 2 ,r 2 ,r 3 2], for 3 4 2 [again, the 1 2 1 2 1 4 1 cos and for ,r 2 3 4 3 2 1 2 1 again, the line x cos . 2 3 4 dy dx For , 0, but d d d dy 4 cos sin sin 0 for , and dd d dx 2 cos 2 cos 1 for , dd dy so by L’Hôpital’s rule 0 and the tangent line is horidx zontal at ,r 0 [the line y 0]. This information can be summarized as follows. Horizontal at: 3 , 23 3 y 35 , 23 Vertical at: 2 3 4 3 , 0], 33 4 y (2, 0) [x 1 , 2 1 , 2 3 4 (0, ) [y 1 , 4 y 15 , 26 9 4 0) or when This information can be summarized as follows. Horizontal at: 1 [the line x 3 2 dy is undefined and the tangent line is L’Hôpital’s rule dx vertical at 2 cos ) 2 cos line x 0, but 2 cos 2 3 [the line x 1 . 4 33 and for 4 3 3 11 the line x cos 2 2 6 11 ,r 6 cos )sin vertical tangent line for 5 ,r 6 3 7 cos 2 6 the line x for 2 1 2 ,r 1 and for 4 There is a vertical tangent line for For , (then 1 1 5 sin 2 6 again, the line y (1 0 when . So there is a horizontal tangent line for 3 2 sin 2 1 sin (then sin dx d sin2 0 when cos 0) or when 1 or 1, i.e., when 2 4 cos sin 2 1 cos )cos cos sin (1 sin sin 0 when (1 cos sin 2 , 2 dx d sin )sin sin 2 sin2 sin sin2 2 cos2 cos2 2 dy d cos2 cos cos dy d 10. 1) sin 2 dx d sin )cos 2], 1 , 4 1 , 4 x x (2, 2 ) [x 2] 429 Section 10.6 11. y dy d 2 sin2 12. 4 sin dy d cos 4 sin2 4(sin 2 sin 2 x 2 sin dx d cos 0, dx , and d , 2 , at (0, 0) [ y (0, ) [ y 0 sin 0 0 sin 0], 2, 2, [x 4 2 sin 2 , and 2 3 , the curve has 44 0]. For 3 2 cos 4 x y 2 2 cos 4 1. This information can be summarized as follows. Horizontal at: (0, 0) [y 2, 2 (0, ) [y Vertical at: 3) 2, 4 3 2, 4 1] and (then sin 2 [x 3 sin 3 137 16 dx d 0 when 0) or when 3 8 1 cos [the line y cos 5.097 (then 8 cos r 5.176 [the line y 5.176 sin 2.146 0, r 2 2.146, 4.343], for 0.267]. There is a vertical 1 [the line x 7 [the line x 1 cos 0 7 cos 1 [again, the line x cos 0.676 0.676 tangent for 1 0). So there 4.343], and for 0.676 sin 5.878 r 3 0.267], for 5.176 sin 4.137 ,r 1.186 or 0.405, r [the line y for or 2 5.176 5.878, r 1] 0, 3 8 1 0.676 sin 0.405 [the line y 1], , i.e., when is a horizontal tangent line for 0] [x 4 cos )sin 0 when cos 4.137, r 2], 4 (3 graphing calculator). 0], [y 3 cos 0.405, 2.146, 4.137, or 5.878 (values solved for with a the curve has horizontal asymptotes vertical asymptotes at 3 2, 4 cos dy d 0 when 3 , . They are never both zero. 44 2 4 sin ) 3 cos 4 sin 2 0 when 0, cos sin (8 cos 2 cos 2 For 4 cos )cos 2 8 cos2 sin 2 dy d dy d (3 2 7], for 1 cos 2 1], for 3 ,r 8 3 9 the line x , and for 2 16 3 3 9 cos 1 ,r again, the line x . 8 2 16 This information can be summarized as follows. Horizontal at: ( 0.676, 0.405) [y (5.176, 2.146) [y 0.267], 4.343], (5.176, 4.137) [y ( 0.676, 5.878) [y Vertical at: 4.343], 0.267] ( 1, 0) [x 1], (1.5, 1.186) x (7, ) [x 7], (1.5, 5.097) x 9 , 16 ( 1, 2 ) [x 9 , 16 1] 1], 2, 430 Section 10.6 13. The curve is complete for 0 2 (as can be verified 19. by graphing). The area is 2 1 (4 2 0 2 cos )2 d 2 2 (4 [ 2.5, 5.2] by [ 2, 3.1] cos2 ) d 4 cos 0 24 1 2 4 sin 2 1 sin 2 4 The circles intersect at (x, y) coordinates (0, 0) and 18 0 2 (as can be verified 14. The curve is complete for 0 (1, 1). The area shared is twice the area inside the circle r 2 sin between by graphing). The area is 2 0 12 a (1 cos )2 d 2 2 12 a (1 2 cos 20 12 1 a 2 sin 2 2 15. Use r 4 2 2 0 /4 cos2 ) d 4 2 1 sin 2 4 32 a 2 0 4 4 . The total area is 2 2 2a cos 2 ) d 2a 1 (2 sin )2 d 2 4 sin2 d 1 2 1 4 8 /4 [ 3, 3] by [ 2, 2] The circles intersect at 1, 1 cos2 2 d /4 2 17. Use r . Its area is 4 4 1 sin 4 16 1 4 /6 is 2 /4 8 /4 4 sin 2 . One loop is complete for 0 /2 Its area is 0 1 ( 2 /2 4 sin 2 )2 d . 0 0 2 . 4 0 2 /2 2. 0 d 1 sin 2 4 3 2 2 3 and 1, /6 5 . The shared area 6 12 (1) d 2 /2 sin2 1 2 6 /2 1 (2 sin )2 d 2 /6 4 2 sin 2 d cos 2 18. Use r 1. 2 /4 1 sin 2 2 2a2 (Integrating from 0 to 2 will not work, because r is not defined over the entire interval.) /4 0 cos 2 d /4 16. One leaf covers /4 1 sin 2 4 20. /4 2 2a2 . 4 0 2a2 cos 2 . One lobe is complete for /4 1 ( /4 2 /4 Shared area 0 and /6 d /6 /2 0 3 3 /6 2 3 21. 2 sin 3 . One leaf is complete for 0 3 . The total area is /3 6 0 1 ( 2 2 sin 3 )2 d /3 6 sin 3 d 0 2 [ 4.7, 4.7] by [ 3.1, 3.1] /3 cos 3 4. 0 The shared area is half the circle plus two lobelike regions: /2 1 (2)2 2 2 0 /2 2 (4 1 [2(1 2 cos )]2 d 4 cos2 ) d 8 cos 0 2 4 5 8 2 sin 1 2 1 sin 2 4 /2 0 Section 10.6 22. 431 25. [ 4.7, 4.7] by [ 3.1, 3.1] [ 6, 6] by [ 4, 4] The area in question is half the circle minus two lobelike regions: Use the symmetries of the graphs: the shared area is /2 4 0 1 [2(1 2/2 8 cos )]2 d (1 2 2 cos cos )d 0 8 1 2 2 sin 23. For a 1 sin 2 4 /2 1 (2)2 2 2 0 /2 2 /2 6 1 [2(1 2 (4 sin )]2 d 4 sin2 ) d 8 sin 0 16 0 2 1: 4 2 cos 1 2 1 sin 2 4 /2 8 0 26. [ 3, 3] by [ 2, 2] [ 3, 3] by [ 2, 2] (a) To find the integration limits, solve The curves intersect at the origin and when 3a cos a(1 2 cos 2 cos cos ) 0 Because of the curve’s symmetry, the area inside the outer loop is 1 3 1 2 . 3 . 2 /3 2 Use the symmetries of the curves: the area in question is 0 1 (2 cos 2 1)2 d 2 /3 /3 2 0 1 [(3a cos )2 2 a (1 cos ) ] d 0 /3 (8 cos 1 2 2 cos 1) d 1) d 2 sin sin 2 4 sin 0 3 3 2 /3 2 sin 2 2 cos2 ) d 2 cos 0 a2 4 4 cos 0 2 /3 (9 cos2 2 (4 cos2 2 /3 a2 a 2 2 (b) Again, use the curve’s symmetry. The inner loop’s area is a2 0 24. 2 1 2 /3 2 2 (2 cos sin 2 1)2 d 4 sin 2 /3 Subtract this from the answer in (a) to get 3 3 [ 3, 3] by [ 2, 2] The curves intersect when 6 cos 2 6 3 or 5 . 6 Use the symmetries of the curves. The area in question is /6 4 0 1 (6 cos 2 2 /6 3) d 33 . 2 6 sin 2 3 0 3 . . 432 Section 10.6 27. 30. The integral given is incorrect because r out the circle twice as cos sweeps goes from 0 to 2 . Or, you can’t use equation (2) from the text on the interval [0, 2 ] because r [ 9, 9] by [ 6, 6] To find the integration limits, solve 3 csc cos is negative for 3 5 . The correct area is , which can be found 2 2 4 3 by computing the areas of the cardioid and the circle 2 4 6 separately and subtracting. 5 . The area in question is 66 5 /6 12 (6 32 csc2 ) d /6 2 5 /6 1 36 9 cot 2 /6 , 12 9 31. dr d 2 , so 5 ( 2)2 Length (2 )2 d 0 3 5 2 28. 4d 0 12 ( 3 5 4)3/2 0 1 (27 3 19 3 8) [ 3, 3] by [-2, 2] 32. To find the intersection points, solve 9 sec2 4 6 cos 2 48 cos4 , so 2 9 6 2 e 2 0 d 2 ed 0 e e . 1 0 By the symmetry of the curves, the area in question is /6 2 e 0 3 4 cos 0 e Length 24 cos2 2 2 dr d 1 6 cos 2 2 9 sec2 4 /6 9 tan 4 3 sin 2 33. dr d sin , so d 2 3 3 4 0 0 1 tan2 2 /2 d /4 /4 1 tan 2 1 1 2 1 (0 4 4 11 csc2 22 2 2 cos y sin lim → lim x → /2 x 1, /2 1, 3 2 lim lim y /2 8 4 sin sin2 cos → → /2 2 cos d, 2d 2 d 2 0 cos 4 3 4 1) ⇒y tan d . 8 sin . 34. dr d a sin 2 2 cos 8 0 2 a2 sin4 Length sin2 0 2a cos , so a2 sin2 2 0 a y d 2 0 /4 ⇒x tan 4 cos2 0 2 cot Total area is twice that, or (b) Yes. x 2 cos 0 2 /2 1 4 0 ( sin )2 d 0 2 . 29. (a) Find the area of the right half in two parts, then double the result: Right half area /4 cos )2 (1 Length 2 2 d 2a 0 2 cos2 2 d 433 Section 10.6 35. dr d 6 sin , so cos )2 (1 41. /2 (1 62 cos )2 NINT evaluates to 6.887. Length 0 62 sin2 d , which using (1 cos )4 [ 1.5, 1.5] by [ 1, 1] 3 (Note: the integrand can simplify to 3 sec dr 36. d 2 .) sin 2 22 sin2 d , which using (1 cos )4 0 2 2.296. (Note: the integrand can simplify to csc 37. 2 dr d cos2 sin 3 /4 3 cos6 2 sin d 0 /4 4 /4 cos4 cos (4 2 2) 3.681 3 sin2 d 3 3 /4 8 0 [ 3, 3] by [ 2, 2] . dr d , so 2 cos2 2 1 sin 2 0 2 2 16a2 43. 2 2d 2. 2 sin 2 ( sin 2 )(2) 2 0 cos 0 ( f ( ))2 2 2 e 44. (a) , so surface area 2 2e /2 sin ( 2e (b) 0 /2 2 e sin 5d 0 25 5 (e 1 e (sin 2 /2 1) /2 cos ) 0 40.818 22 /2 2 ) 2 (e ( f ( ) sin )2 sin2 ) 2 0 0 /2 2 )d (c) /2 dr 2 d r2 a(1 0 1 ( a 2 ad 0 /2) sin a 0 2 a 0 /2 a cos 2 a 2 cos ) d 2 1 2 sin2 ) 2 1 /2 /2 ( f ( ) sin )2 ( f ( ))2 4.443. 0 2 f ( ) cos )2 ( f ( ))2(cos2 /4 2 f ( ) sin )2 ( f ( ))2(cos2 sin2 2 d d sin dr d d 4a2 0 ( f ( ) cos )2 2 /4 2 2 cos2 2 cos 0 sin 2 c os 2 /4 2 /2 1 sin 2 4 ( f ( ) cos )2 ( cos 2 )2 cos 2 cos d d ( f ( ) sin c os 2 cos 2 4a2 sin2 dy 2 d , so Surface area /4 dx 2 d 1 2 ( f ( ) cos 0 1 cos2 0 (1 sin 2 ) d sin 2 1 /2 16a2 sin 2 d 2 cos 2 0 1 4a2 cos2 2 (2a cos ) cos 0 Length dr d 2a sin , so surface area /2 2 sin 2 1 1: d 3 2 sin 4 3 cos 2 dr 38. d 42. For a cos4 3 0 1 2 , so 3 0 40. sin2 2 d cos 2 cos 2 /4 2 ). cos 2 sin 0 Length 39. 2 0 3 : Surface area 4 /4 NINT evalutes to . cos 2 Use the curve’s symmetry and note that r is defined for 22 cos )2 (1 /2 2 sin 2 dr d 2 sin , so (1 cos )2 Length dr d 2r d /2 45. If g( ) 2f ( ), then (g( )2 g ( )2) 2 ( f ( )2 g is 2 times the length of f. a /2 2a sin /2 f ( )2), so the length of 434 Chapter 10 Review 46. If g( ) 2f ( ), then (g( )2 g ( )2) 2 g( ) sin 4[2 f ( ) sin ( f ( )2 f ( )2)], so the area generated by g is 4 times that of f. 47. (a) Let r 0.06 2 1.75 0 evaluates to 741.420 cm 2 b 2 1/2 2 0.06 2 d 2 7.414 m. (pp. 569–572) b since 2. (a) (b) 2 741.420 cm (from part (c)), 80 La 1.75 0 0.03 2 1.75 0.06 2 d 236 741.416 cm 0 bn 2 the integral, La bL r02 b 1 2 r0n. Solving for n gives (c) Since L is proportional to time, the formula in part (a) shows that n will grow roughly as the square root of time. 3 3 a (1 5. 6 2 8 2 23 a 3 cos ) cos 25 a2 15 sin 8 2 cos 1 7. 2 42 8. 9. (a) 4, 12 (1 cos4 ) d 15 8 cos2 sin (b) 2 So x 3 2 53 a 2 3 a2 By symmetry, y 1 2 , dy dx dy/dt dx/dt 3 d 2y dx 2 3 2 2 , dy dx 3, (1/2)sec t tan t (1/2)sec2 t 3 ;x x 17 34 , 55 (4/5)2 which for t 10. (a) cos )2 d 1 sin 2 4 2 sin 3 2 2 ,y 1 sin t dy dx 1, and 3 2 4 3 2 3 x 2 . So the or 1 . 4 dy /dt dx/dt cos t (1/2)sec2 t 1 equals . 3 4 2 cos3 t, 0 0 a2 29 17 (3/5)2 y cos2 ) d 2 cos 8 1 1 5 0 2 5 2 sin a2(1 25 725 tangent line is y 3 cos3 1 cos3 4 53 a , and 2 10, 31 , 2 6. 0 23 a 3 sin 3 10 2 10 8 [assuming counterclockwise]. d 3 cos2 (cos 6, radians below the negative x-axis: 0 2 1 2 2(4) For t 2 49. 3 1, 1 6 1. (b) The take up reel slows down as time progresses. 2 20 2 80 made n complete turns, then the angle is 2 n. So from n 2 2 (b) 8, 12 1313 2( 3), (b) 9 5 4. (a) 5(2), 5( 5) 48. (a) Use the approximation, La, from 47(e). If the reel has r0 2, 4 1 3. (a) (e) L 5 322 3 is a very small quantity squared. 2r 3, 4 4 2, 17, 32 172 (b) 2 0, so the centroid is s Chapter 10 Review Exercises r 2r 0 4a . By symmetry, x 3 a2 4a 0, . 3 2 1/2 b r1 43 a , and 3 cos a2. So 1. (a) 3 (d) r 2 23 a 3 d 43 a 3 y length of the curve. 80 a3 sin 0 a2 d b , this is just Equation 4 for the 2 (c) Using NINT, 2 3 0 0.06 . 2 1.75 dr (b) Since d 50. 2 3 a2. 0 (b) 5a . 6 0, so the centroid is 5a ,0 . 6 dy dx 3/t 2 2/t 3 5 ,y 4 3 t 2 1 dy For t 2: x , and 2 dx 1 So the tangent line is y 3x 2 13 y 3x . 4 d 2y dx 2 dy/dt dx/dt dy dx equals 6. dy /dt dx/dt 3/2 2/t 3 3. 5 or 4 3t 3 , which for t 4 2 Chapter 10 Review 11. dy dt 1 tan t sec t equals zero for t k , where k is any 2 dx 1 integer. sec2 t never equals zero. dt 2 1 1 tan 0, sec 0 2 2 (a) Horizontal tangents at 1 tan 2 , 1 sec 2 0, 17. (a) 1 and 2 [ 1.5, 1.5] by [ 1, 1] (b) 2 1 . 2 0, 435 18. (a) (b) There are no vertical tangents, since dx never equals dt zero. 12. dy dt integer. dx dt [ 3, 3] by [ 2, 2] k , where k is any odd 2 2 cos t equals zero for 2 sin t equals zero for t k , where k is any (b) 19. (a) integer. 2 cos (a) Horizontal tangents at 2 cos and 3 3 , 2 sin 2 2 2 , 2 sin (0, [ 1.5, 1.5] by [ 1, 1] 2). (b) (b) Vertical tangents at ( 2 cos 0, 2 sin 0) ( 2 cos , 2 sin ) (2, 0). 13. dy dt 2 sin t cos t where k is any integer. (0, 2) 2 ( 2, 0) and k , 2 sin 2t equals zero for t dx dt 2 20. (a) sin t equals zero for t k, [ 1.5, 1.5] by [ 1, 1] where k is any integer. Where they are both zero, use (b) L’Hôpital’s rule: dy/dt t→k dx/dt lim lim t→k sin 2t sin t (a) Horizontal tangent at 2 cos 2t cos t lim t→k cos 2 , cos 21. 2. 2 (0, 0). 2 dy dt 9 cos t equals zero for t dx integer. dt 0, 3 3 , 9 sin 2 2 2 , 9 sin 2 , 0, cos 2 cos cos 2 sin 3 5 , 0, 4 4 /4 dy dx k , where k is any integer. (a) Horizontal tangents at 4 cos 4 dy dx k , where k is any odd 2 4 sin t equals zero for t f ( ) cos f ( ) sin 2 sin 2 sin 2 sin 2 cos (b) There are no vertical tangents. 14. f ( ) sin f ( ) cos dy dx 2/ 2 2/ 2 2/ 2 5 /4 2/ 2 and 0, 1, 1, dy dx dy dx 7 4 2/ 2 3 /4 7 /4 The Cartesian equations are y (0, 9) and are polar solutions. 2/ 2 22 2 (0, 9). (b) Vertical tangents at (4 cos 0, 9 sin 0) (4 cos , 9 sin ) ( 4, 0). 22. x. f ( ) sin f ( ) cos f ( ) cos f ( ) sin 2 sin 2 sin 2 sin 2 cos (1 (1 cos 2 cos3 cos sin 2 cos2 sin sin dy dx (4, 0) and 15. 4 sin2 6 cos cos 2 )cos cos 2 ) sin 2 cos2 sin 4 sin2 2 cos2 . 3 sin 2 [ 7.5, 7.5] by [ 5, 5] 16. 0, dy dx [ 7.5, 7.5] by [ 5, 5] 1. 2 4 sin2 cos 4 cos2 sin 4 cos 1, 2 and 0, /2 dy dx 3 2 are polar solutions. 3 /2 4 is undefined, so the tangent lines 0 are vertical with equation x 0. 436 23. Chapter 10 Review dy d d 1 cos sin 2 d 1 sin sin cos cos cos 2 2 2 dx d 1 cos cos 2 d d 1 sin cos sin cos sin 2 2 2 dy Solve 0 for with a graphing calculator: the solutions d are 0, 2.243, 4.892, 7.675, 25. [ 1.5, 1.5] by [ 1, 1] 1 , 1 2 r sin horizontal tangent lines at y 0.443 and y Solve dx d are 0, 0 for 1.070, reveals 9.035, 1.739. 1 Where r cos reveals 0.067, and x dy dx and both equal zero ( dt dt 1 y 0, 4 ), close y y 0. (This can be confirmed 2 cos y 4 sin2 2 sin y , 62 , 5 3 , and . 6 2 dx d reveals horizontal tangent lines at y 0 for 1 and 2 (by first using the quadratic formula to find sin ): the solutions are two solutions to find x 3 lines at x zero x 1 x or 2 2 [ 3, 3] by [ 2, 2] 4. Solve 2, and 2 Using the first, third, and fourth solutions to find r sin x 1 or 2 0 for : the solutions are y 1 26. 2 sin2 2 sin or or x 2 sin )cos ] 2 cos2 dy Solve d cos 1, and 1, respectively. 2, 1 y 4 sin 1, 1, 2 2 dy d dx d sin )sin ] 1 x x y d [2(1 d d [2(1 d . From the 2 2, 2 using L’Hôpital’s rule.) 24. , 2 2 x 1.104. inspection of the plot shows that the tangent lines are horizontal, with equation y 1 , and 2 1 x 2 y 2, x 1 , 2 1 So the equations of the tangent lines are y vertical tangent lines at x , 2 those points have slopes of 11.497, and 4 . Using the middle five solutions to find x 1 , 2 1 curve’s symmetries, it is evident that the tangent lines at with a graphing calculator: the solutions 3.531, 2 , , 2 10.323, and 4 . Using the middle four solutions to y 1 The tips have Cartesian coordinates 3 2 2 , r cos 7 11 , and . Using the last 6 6 reveals vertical tangent 2.598. Where dy dx and both equal dt dt , inspection of the plot shows that the tangent As the plot shows, the curve crosses the x-axis at (x, y)-coordinates ( 1, 0) and (1, 0), with slope 1 and 1, respectively. (This can be confirmed analytically.) So the equations of the tangent lines are y0 (x 1) y x 1 and y0x1 y x 1. 27. r cos r sin x y, a line 28. r 3 cos line is vertical, with equation x r2 3r cos x2 2 y2 3x x2 3x 9 4 0. (This can be confirmed using L’Hôpital’s rule.) x 32 2 y2 a circle center y2 9 4 32 2 3 , 0 , radius 2 3 2 437 Chapter 10 Review 29. r 4 tan sec 37. r sin 4 r cos r cos y 4 or x 2 x x 30. r cos dr d sin , so 2 2 2 0 2 3 2 cos cos 1 r cos 2 1 3 x y 2 2 x r sin 3 3 r sin 2 2 2 3 2 3 38. 3 dr d 31. x r2 r 3 4 0 x 3 or y 2 y 5y 5r sin 5 sin 4, a line 32. x 2 y 2 2y r 2 2r sin r 2 sin 4y 39. 8 sin2 cos 3 /4 /4 2 4r 2 sin2 dx = 2e2t dt 16 16, or r 2 ln 2 8 sin2 1 2t e 2 4t 4e 2e2t 8 1 2 cos 2 1 cos2 2 cos 2 sin2 2 2d 41. 2 dx dt 2 sin t, /2 dy dt 0 24 3.087. 2t, so ( 2 sin t)2 Length 8 t /2 2 t2 (2t)2 dt sin2 t dt, 0 2 1, so 42. 3 Length d /2 ln 2 dy 2t, dt d cos 2 sin2 2 d 1 cos 2 which using NINT evaluates to dx 36. dt 2 sin 2 /2 1 ln 2 8 3 0 cos 2 )2 cos 2 )2 cos 2 /2 dt d 3 , so Length (1 1 /2 e dt (2t) 2 (t 2 2 1) dt 3 dx dt 3 cos t, 3 dy dt Length 3 3.183. t, so (3 cos t)2 (3 4t 2 t4 2t 2 1 dt 3 t)2 dt 0 3 3 t 0 which using NINT evaluates to 12.363. 3 2 3 (t 2 43. Area 2 1) dt 0 3 3 3 t 3 d /4 1 /2 2t cos2 3 3 c os 2 1 /2 (et )2 dt 0 ln 2 8 sin 2 dr d ln 2 t 8 e 40. cos 3 d 3 3 2 sin 4 3 (1 1 64 12 0 2t 16 sin2 /2 12 8 0 0 ln 2 cos2 2 8 sin2 3 3 8 sin2 1 2 8 16 4 sin2 e t, so 2e2t Length d 2 8 sin3 /2 1 dy , 8 dt ln 2 0 5)2 16 (r sin 5)2 (y 2)2 2 sin )2 d (2 cos , so 3 0 r 2 cos2 3 dr d /4 4(r sin ) 4 2. Length (r cos ) 35. 2d 16 2 34. (x 2)2 (r cos 8 cos2 0 2 0 /2 2 0 0 33. x 2 cos )2 8 sin2 /2 8. 2 0 0 0 2 4 cos 2 sin , so Length /2 0 3 2 2 cos d (2 sin 2 3y sin d 2 4 sin2 0 r cos ( sin )2 d 0 2 4y, a parabola 3 cos )2 (1 Length t 4 3 3. 1 (2 2 2 cos )2 d 1 (4 4 cos 20 1 4 4 sin 2 cos2 ) d 1 2 1 sin 2 4 2 0 9 2 cos2 t dt, 438 Chapter 10 Review /3 1 sin2 3 d 2 44. Area 0 11 22 1 sin (6 ) 12 50. /3 12 0 45. [ 1.5, 1.5] by [ 1, 1] r [ 3, 3] by [ 2, 2] cos 2 dr d sin 2 and /2 The curves cross where cos 2 0, such as 4 . Using Area 1 [(1 2 /4 4 0 cos 2 )2 2 2 (cos 2 4 0 /4 1 2 cos2 2 d sin 2 sin 2 d sin 4 d [(4 cos t)i dt 51. (a) v(t) sin 2 0 ( 2 sin t)j] ( 4 sin t)i 2 4 cos 0 2 cos 2 ) d 1 sin 4 8 , so /2 1] d 0 2 2 /2 4 /4 sin 2 cos 0 the curves’ symmetry, Length 2 2 , where 0 sin 2 d [( 4 sin t)i dt a(t) 46. ( 2 cos t)j ( 2 cos t)j] ( 4 cos t)i ( 2 sin t)j 2 (b) v 4 sin 4 [ 4.5, 4.5] by [ 2, 4] 8 2 2 cos 4 1 3 2 2i 4 Since the two curves are covered over different -intervals, (c) At t find the two areas separately. Then 2 Area 0 1 [2(1 2 2 2 sin )]2 d (1 r2 2 2 sin sin 47. dx dt t, 1 2 2 cos dy dt cos 1 sin 2 4 cos 2 5 0 5 2 (2t) t 2 Area ( 22 dt 0 4 (t 2 3 48. dx dt 2t 1 4) 2 t 1 2t 2 2 1/ 2t sin 2 cos 2 sin cos 2 0 (c) At t sin2 2 d cos 2 /4 0 2 sin d /4 2 cos (2 0 2) ( 3 sec2 t)j] 3 sec t tan t)i 0, v sec3 t)i 3j, a ( , so 2 ( 3 sec2 t)j 3 sec t tan t)i 3 sec2 t tan t)j (2 3 sec4 0 0 3 3 4 dt, 10.110. /4 3 tan t)j] 2 cos 2 Area ( va cos 1 va which using NINT evaluates to dr 49. d 3 sec t)i 3 sec2 0 tan2 0 (b) v(0) 12 2t 2 j, and va 3(sec t tan2 t 4, so 2 2i 18 d [( dt a(t) 76 3 0 1 dy , 2t 2 dt Area 5 3/2 j, a 1 va d [( dt 52. (a) v(t) 2, so ,v 1 (3)(3) 7 cos 1 38.94 . 9 )d 0 2 4 1.840 53. v(t) dr dt (1 v(t) t i t 2)3/2 3i 0 3)( 0 1 cos 0 90 3) 1 j (1 t 2)3/2 2 t (1 t 2)3/2 which is at a maximum of 1 when t (1 2 1 t 2)3/2 0. 1 1 t2 , 439 Chapter 10 Review dr = (e t cos t dt dx (e t cos t dt 54. v(t) a (t) t e t sin t)i e t sin t t (e sin t (e t sin t ( 2e t sin t)i e t sin t t e cos t dx dy , dt dt 61. (a) v(t) e t cos t)i v(3) (e t sin t)(2e t cos t) 0 for all t. The angle between r and a is always 90 . 1 6t) dt i 6 cos t dt j 0 0 3t 3t 1 2 6 sin t j 0 2 e 56. e 2 e e 1 dt j t ln t 2 e e i r(0) j 3i x 3 x2 9 cos dx dt 4 2 5 j, so C (sin t)j Length i (cos t 1)i r(0) j r(t) (tan C 1 1 (tan j, and 10 i t j, so C 100 3 t 3 10 i dr dt d 2r dt dt 2 dr dt dt r(t) dr dt t 1 t 2j C1 t C2 C2. And r(0) C2 2i r(t) ( t2 r(1) 5i r(t) ( t2 t 2i 2j 6t )i j C2 6t ( 2t)j t 2j C1 3i 2)i C1t 6i 2j and C2 3j, so C2 ( t2 (b) 2t e t sin t e t cos t dy/dt dx/dt dy dx t C2 2t)j dy dx 1 1 12 2 1250 3 t)2 dt, which (5 2)j 2j, and 1040.728. e t cos t e t sin t cos t cos t sin t sin t 1 dy dx e t(sin t cos t), e t(cos t sin t) dt dt dy 2 e 2t(sin2 t 2 sin t cos t cos2 t) dt e 2t(1 2i dx 2 dt 2 1 or t 4) dt 1 5 10 t 50 5t 4 using NINT evaluates to C1 , 4i, so C1 ( t2 20t 3 t(10 t) 2 2 0 63. (a) ( 2t)i 5 i, so t 2j r(t) 60. (c) Area dr dt dt 3 2 t)2 dt, which using NINT (5 (100t 2 8 1j dr d 2r 59. dt 2tj C1, r(t) dt dt 2 dr C1 0, so r(t) t 2j dt t 0 y2 0 C i and t2 1)i 0 10 8 1j 16 x2 dx dt dt t(10 t) 2 dt 2 10 t2 3 4 2 5 y2 0 1)j t)i 2 25.874. 2 dr dt dt 58. r(t) (sin t sin 3 16 2 t, so 4 3 3.238 t so 2 evaluates to C 5 16 12 (b) Volume r(t) 32 cos 16 4 sin 1 dy and 2 dt 10 (ln 2)j 17 4 2 1. 0 (cos t)i C y 5 t and y 25 e dr dt dt i 62. (a) 2 ln (ln t) e 57. r(t) 6i 0 2 ln t dt i t ln2 t (c) 1 i 34 4 d 2y y-component: 2 dt t 3 1 (3 , and 42 25 2 32 d 2x dt 2 t 3 (b) x-component: 55. 5 , 42 92 32 e sin t)j (2e t cos t)j 3 5 sin t, cos t , 4 44 4 3 v(3) t e cos t (e t cos t)( 2e t sin t) r(t) a(t) e t cos t )j 2 sin t cos t) e 2t(cos2 t e 2(1 2 cos t sin t 2 cos t sin t) v(t ) et 2 v(3) e3 2 3 (c) Distance v(t) dt 0 3 et 2 0 2 et (e3 3 0 1) 2 sin2 t) 440 Chapter 10 Review dx dy , dt dt 64. (a) v(t) 6 5 2t, t 2 , v(4) 96 2 5 (2t)2 4 (b) Distance 69. (a) 96 , and 5 104 5 622 t dt 5 82 v(4) 8, 0 4 [ 2, 10] by [ 2, 6] dx dy , dt dt d 2x d 2y , dt 2 dt 2 (b) v(t) 2 t 25 9t 2 dt 5 2 4 (25 9t 2)3/2 135 0 0 (c) t x 2, so dy dx 6t 2/5 2t dy/dt dx/dt 65. x degrees east of north is (90 a(t) 4144 135 3 t 5 3 5 x 540 cos 10 , 540 sin 10 595 cos 10 , 485 sin 10 585.961, 84.219 . 585.9612 Speed Direction 84.2192 585.961 tan 1 84.219 v(1) a(1) 2 ,0 0, 2 0, 0 0, 2 v(3) a(3) 2 ,0 0, 2 (c) Topmost point: 2 ft/sec center of wheel: ft/sec Reasons: Since the wheel rolls half a circumference, or feet every second, the center of the wheel will move feet every second. Since the rim of the wheel is turning at a rate of ft/sec about the center, the velocity of the topmost point relative to the center is ft/sec, giving it a total velocity of 2 ft/sec. Rg , where sin 2 70. v0 R 66. Add the vectors: 45 , g 32, and range for 4325 yds 120 cos 20 , 120 sin 20 300 cos ( 5 ), 300 sin ( 5 ) 411.622, 14.896 . 14.896 411.622 1 Direction tan Length 411.6222 12,975 ft: v0 644.360 ft/sec for 4752 yds 14,256 ft: v0 675.420 ft/sec Rg sin 2 71. (a) v0 14.8962 411.891 lbs (b) The cork lands at y Solve y 45 : v0 7 4, x 59.195 ft/sec g 2v02 cos2 Solve y 177.75. g x 2 (tan )x for v0, with 2v02 cos2 gx 2 74.584 ft/sec yx 72. (a) The javelin lands at y 2 (80 sin 45 ) 2(32) (109.5)(32) 2.073 67. Taking the launch point as the origin, y (44 sin 45 )t 16t 2 equals 6.5 when t 2.135 sec (as can be determined graphically or using the quadratic formula). Then x (44 cos 45 )(2.135) 66.421 horizontal feet from where it left the thrower’s hand. Assuming it doesn’t bounce or roll, it will still be there 3 seconds after it was thrown. 68. ymax cos t 0, 0 0, 2 591.982 mph. 81.821 east of north 2 sin t, v(2) a(2) x) degrees north of east. 55 cos ( 10 ), 55 sin ( 10 ) 2 sin t v(0) a(0) 2 Add the vectors: cos t, 6.5, x x2 262 5 . 12 (tan )x for v0, with 40 : 57 feet gx 2 (2 cos 40 )( y x tan 40 ) v0 2 (b) ymax 73. We have x gt 2 2 y x 2 y (v0 sin )2 6.5 2g (91.008 sin 40 )2 64 (v0t) cos 91.008 ft/sec 6.5 59.970 ft and (v0t) sin . Squaring and adding gives gt 2 2 2 (v0t)2(cos2 sin2 ) v02t 2. Cumulative Review 74. (a) r(t) x(t) y(t) (b) ymax tmax (155 cos 18 11.7)ti (4 155 sin 18 t (155 cos 18 11.7)t 4 155 sin 18 t 16t 2 Cumulative Review Exercises (pp. 573–576) 16t 2)j 1. Since the function has no discontinuity at x (155 sin 18 )2 4 39.847 feet, reached at 2(32) 155 sin 18 1.497 sec 32 (c) y (t) 0 when t 3.075 sec (found using the quadratic formula), and then x (155 cos 18 11.7)(3.075) 417.307 ft. (d) Solve y (t) 25 using the quadratic formula: 1552 sin2 18 32 155 sin 18 t 72.406 and 11.7)t equals 333.867 feet from home plate. (155 cos 18 11.7) 1 (1 0.09 155 sin 18 (1 0.09 4 32 (1 0.092 x(t) (155 cos 18 y(t) 4 0.09t e e )i e 0.09t ) (c) AC E AB E 1 ex ex 1 x→ 1 lim 1. t (1 t 5. By l’Hôpital’s Rule, lim t→0 t sin t (1 cos t) 1 cos t t sin t 3 cos t lim 3 cos t t→0 lim t→0 cos t) sin t t cos t 2 sin t lim sin t t→0 ex ln (e x 1) 6. By l’Hôpital’s Rule, lim ln x x→0 xe x lim x lim xe x x→0 1 (e x 0.09t ln (e x x x→0 ) 8. lim 3x x→0 1 AD E 2 AD, so by part (b), AP E E x) lim ex e lim (e x 1) x→0 1 x ln (e x x x) 1 x x)1/x. Then ln f (x) 1 AC. E 2 77. The widths between the successive turns are constant and are given by 2 a. lim x)1/x 1 x (3x x→0 lim x→0 lim (e x 1)/(e x 1 x→0 x→0 ) AB E 1 1 BD = AB E E 2 2 ex ex So lim (e x 0.09t (e) No, the batter has not hit a home run. If the drag coefficient k is less than 0.011, the hit will be a home run. (This result can be found by trying different k-values until the parametrically plotted curve has y 10 for x 380.) AB E ex ex 1)2 (x x→0 )j (d) Plot y (t) and find that y (t) 30 at t 0.753 and 2.068 seconds. At those times, x 98.799 and 256.138 feet (from home plate). (b) AP E x x→ x lim x 4. By l’Hôpital’s Rule, lim 7. Use f (x) (c) Plot y (t) and find that y (t) 0 at t 2.959, then plug this into the expression for x(t) to find x(2.959) 352.520 ft. AD E x→0 3 . 4 1 1 1 ) (b) Plot y (t) and use the maximum function to find y 36.921 feet at t 1.404 seconds. 76. (a) BD E 1 x 3. By l’Hôpital’s Rule, lim 3 cos 3x 4 lim x→0 0.09t 1 (1 11.7) 0.09 155 sin 18 (1 0.09 32 (1 0.09t 0.092 0.09t sin 3x 4x x→0 x→0 e e 0.09t e e 0. 2. By l’Hôpital’s Rule, lim x→ (e) Yes, the batter has hit a home run. When the ball is 380 feet from home plate (at t 2.800 seconds), it is approximately 12.673 feet off the ground and therefore clears the fence by at least two feet. 75. (a) r(t) 11 2(1) 12 1 12 lim (155 cos 18 1, the limit is 2 4(16)(21) 0.534 and 2.460 seconds. At those times, x 441 (3x 1 sin x 1) cos x x cos x lim e ln f(x) (3x x→0 3 sin x sin x 1) sin x 6 cos x x sin x 2 cos x 2. e 2. x→0 lim x) 1) sin x x sin x 1 3 x , and 1. 442 Cumulative Review 12 9. (a) 2(1) (b) 2 1 1 19. y 20. y (d) Yes, since lim f (x) f (1) x→1 f (1 h) h h→0 21. y f (1) lim h) h→0 2 h)2 (1 h 2h 1 2h 3 lim h2 lim h h→0 2 lim (1 h→0 h) h h 1 1 cos x 2 and x , 2x x 0 to find x x) 8 , x 3 1 . 2 ,x x [ 10, 10] by [ 4, 4] 0. 23. 14. y (x 15. y (x 2)2 sin ( 1 3 sin 2 16. y 4 1 5 5 2)(1) (x 1 1 3x) y tan x cos x 18. y 1 x 2 (e x 2 1 x )(2x x 1) 2 (2x y2 2 x 1 x y) x sin (xy) 1 x 2yy y sin (xy) 0 1 xy sin (xy) 2xy x 2 sin (xy) 2y 1 d x 1/2 x 2 dx dy/dt dx/dt d (0) dx ln x] 1 x 2 cos t sin t x x x cot t 2x x 1 x 1 y ln[(cos x)x] x ln (cos x) 1 dy y dx dy dx x 1 ( sin x) cos x ln cos x x sin x cos x x sin x x (cos x) ln (cos x) cos x y ln (cos x) (cos x)x 1 [cos x ln (cos x) x sin x] sin x 27. By the Fundamental theorem of Calculus, y 1 x 3. cos t cos (x 2) 2x 1)e x tan x2 ln y 28. y 1 cos x)(1 2 cos x) sin 2 x 1 x2 2x (2x) 2 cot 2 x) cos x 2 cos2 x sin2 x d dx 2 x ln y 3x (sin x)(1 cos 2 x) cos 2 x 17. y dy dx ( 3) sin x cos 2 x 1 1 2 x 2 2)2 1/2 sin 1 x 3 (x cot x csc x cot x csc x (1 sin (xy)(xy 3x sin x sec x csc 2 x 1 d [cos (xy) dx 26. 1)(1) 1 3x) (1 2 2 csc 2 x 2 25. 9 3 csc2 x (1 cos x)4 d dx 2 22. y 24. y f (0) 0 cot 2 x) while 0, x (2 12. One possible function is y f (5) 5 cot x csc x 1 1 3 13. 3 csc2 x (1 cos x)4 1 x→ 2 csc x sin x cos x csc x cot x) 2. 1, the end behavior at both ends is y Vertical: solve 2x 2 )e x 3(1 2 cos x) (sin 4 x)(1 cos x)3 Since the left- and right-hand derivatives are not equal, f is not differentiable at x 1. 11. Horizontal: since as x → 4 csc x cot x 3 (sin 2 x)(1 cos x)4 1 h lim 0: all x 3x cos x)( csc x cot x) (1 cos x)2 3 (sin x)(1 cos x)4 h→0 x2 2 (1 csc x 1 cos x 2 f (1) 3 (x 3 csc2 x (csc 2 x (1 cos x)4 0 Right-hand derivative: 10. Solve 4 3 (1 1 h2 h h→0 h) h e xx e 3 csc 2 x (1 (1 cos x)4 1 h h→0 f (1 4x (1 2(1 lim h→0 x2 1 1 (e) No. Left-hand derivative: lim x2 x 1 (c) 1 [from (a) and (b)] lim 3x 1 2x tan y 2x sin (x 2) 2 sin (2x) cos (2x); cot 2 x) Cumulative Review 29. d2 (y dx 2yy 2y y (1 3)( 1) (2)(3) (1 3)2 a 3t 2 12t 6t (b) Solve v t 3. 3 12 36. At t 1)(t 3) 0; t t 1 or 4 r :r dy dx 5 32. For x dy dx 1, y 6(1) (a) y 1 (b) y 1 12(1) 33. For x 3 dy dx ,y 3 sin 3 3 x 1 34. (2x) 4 1, y (a) y y (b) y or y 3 2 33 2 3 x 2 33 2 2 3 2 2 3 3 6 (x (x x 2 j and j 2i 2j, so that 1 (x 0.707x 1 1.414x 1 2) or 2 2x 1 2(3) C2 C1 x 2x 4, x 5, x 3 , choose C1, C2 so that 3 1. 3 . 3 . 0, 2 6 x 2 2y 9 (b) x 2.458x 9x . 4y 2 (d) Absolute maximum of 2 at x absolute minimum of 0 at x 0: 2, 2, 3 39. According to the Mean Value Theorem the driver’s speed at 3 2 some time was . 1) or 0.866x 2.050 0 (c) x 1.155x 3 2 [ 3, 6] by [ 1, 5] 3.464 1) 5 x 0.950 or 3 3 3 3 1 2 0.407x , the slope is y 2 2i or 6 0; y 3 4 j 2) or y C1, x C2, x f (x) 3 3 3 1 (2yy ) 9 x 3 x 3 4 sec 2 i 4 3 and 6 x 6 tan x 2x 38. (a) x 3 y 6 3 6 6 i 4 37. With f (x) 1 3 2 3 3 6 (b) y 2x 3 6 3 1.443 2 1 x 2 3 cos 3 6 y At x cos 1.155x 2(x 2 1) or y 3 (a) y 1 y 1) or y 1 (x 2 1) or 2 4 2(x 3.464 2. 1 1 and 2 0.866x 2 (a) y 2, and at that instant 12(2) 9 3 m/sec . 3 tan 2 3 2 1) or (x sec 4 (x 3 5 2 sec , and 3 2 x (b) y 0 at t 3(2)2 2 3 (c) Right: v 0 for 0 t 1, 3 left: v 0 for 1 t 3 (d) a v x 33 2 2 9 0 for t: 3(t 3 2 2 (b) y 3 2 3 cos ( /3) 2 sin ( /3) y 1 3 1, y 33 2 y dx dt dv dt 31. (a) v :x dy/dt dx/dt (a) y sec x tan 2 x) 2y sec x tan x (2y 2)2 2 3 2) (sec x sec x tan 2 x) 2 sec 2 x tan 2 x (2y 2)3 v)u uv x0 (1 v)2 3 dy dx sec x tan x, (2y (1 35. At t sec x tan x , 2y 2 (2y 2)(sec 3 x y 30. d (sec x) dx 2y) 443 1.443 111 1.5 74 mph. 444 Cumulative Review 40. (a) Increasing in [ 0.7, 2] (where f 0), decreasing in [ 2, 0.7] (where f 0), and has a local minimum at x 0.7. 2x 2 (b) y 3x (e) Local (and absolute) maximum of approximately 3.079 at 3 2 x 6 3 2 and x 6 3 ; local (and absolute) minimum of 0 at x at x (f) 0 and 2 ( 1.042, 1.853) [ 3, 3] by [ 15, 10] 23 x 3 (c) f (x) f (0) 41. f (x) f (x) x2 x2 1: f (x) 3x 3x 43. (a) f has an absolute maximum at x minimum at x 3. 32 x 3x C; choose C so that 2 23 32 x x 3x 1. 3 2 cos x cos x C; choose C so that f (0) 1. (b) f has a point of inflection at x 1 and an absolute 2. 1 (x 1)2 2 7 x2 2 2: (c) The function f (x) 3, 1 x 2 3 , 2 2 x 3 42. is one example of a function with the given properties. [ 2.35, 2.35] by [ 0.5, 3.5] f (x) is defined on [ 2, 2]. f (x) 2x 4 x x2 4 f (x) 0 for x to find x The graph of y 0, x [ 3.7, 5.7] by [ 3, 5] 3x 3 8x x3 2 4 x2 ; solve 44. y = 2 1 26 . 3 A(x) 4x f (x) is shown. x2 , and the area of the rectangle for x 16 x2 1 x 16 x 2. 16 x2 2(8 x 2) 2 16 A (x) x x2 16 A (x) (b) 2) 2 by 4 [ 2.35, 2.35] by [ 10, 10] 2, 2 possible area is A(2 (a) 0 when x = 26 26 , 0, 3 3 45. f 26 26 ,0 , ,2 3 3 equation is y Use NDER to plot f (x) and find that f (x) x 1.042. 0 for y 2x 2 and y , and so 2. The maximum 8, with dimensions 2. 2 and f 4 x2 16 0 is sec 4 2 4 2x 2 4 4 1.414x tan 4 2. The or 0.303 46. V s 3 dV 3s 2 ds Since ds 0.01s, the error of the volume calculation is approximately dV 3s 2(0.01s) 0.03s 3 0.03V, or 3%. [ 2.35, 2.35] by [ 15, 5] (c) Approximately ( 1.042, 1.042) (d) Approximately ( 2, 1.042), (1.042, 2) Cumulative Review 47. Let s be the rope length remaining and x be the horizontal distance from the dock. s2 (a) x 52, ds dt 1.5, and which means that for s dx dt speed s 8 s2 s 1.9 ft/sec ds , which for 25 dt ( 1.5) 64 1 dx x(ln x)2 2e Therefore, 0.15 rad/sec. e 1 e (3 0.179 in./min. t 2)i 2 V dV h , so dt 12 4 ( 9) 25 1 1 1 ln 2 h 4 36 25 ex 57. Let u x 2 e cot (e (b) (1)(1.8 1 … 6.4 6.4 0 50. x dx x dx 2 1 2 165 in. 12 x 2 x dx 2 2 0) 1 0 0 12 x 2 x x 2 x dx ( 2 1 0 ) 4 x 2 2 sin 1 x 2 2, 2 2 2. Alternately, observe that the region under the curve and above the x-axis is a semicircle of radius 2, so the area is 1 (2)2 2 3 52. 1 x2 ln 3 /4 53. 0 1. 2 1) dx cot u du (csc2 u Since 1) du u 1 C (e x 1) 1) 2. 3 1 13 dx x x 3 26 9.765 3 sec2 x dx ln x 9 1 /4 tan x 1 0 ln 3 C C is an arbitrary constant, we may redefine C cot (e x ex 1) ds s , so du . 2 2 ds 1 ds ds 2 2[(s/2)2 s2 4 4(s/2)2 4 1 1 s tan 1 u C tan 1 C 2 2 2 59. Let u 2 4 2 (ln 3)j C. 1 2 1] 2.5 51. Using Number 29 in the Table of Integrals, with a 2 2i 58. Let u 165 in. … 12.6 0.409 2i csc2 u and write the solution as 16.2) ln 2 ln 2 1 cot (e x the rate of about 0.458 in./min. 1.8 C. e x dx. 1, so du cot u dh is negative, the level in the cone is falling at dt 49. (a) (1)(0 1 ln x C (ln 3)j 0.458 in./min. Since 1 3 5 3 (ln t)j dh dh and dt dt 2 1 u du Use the identity cot2 u V be the volume of the coffee in the cone. 3 12 1 1 (b) Now let h be the level of the coffee in the cone, and let 1 h2 h 32 dV 4 h 2 dt x 1 j dt t 2t)i (3t 56. 9 16 x dx x(ln x)2 2e 1 ln x 25 dV/dt 16 4 1 dx. x u the volume of the coffee in the pot. dh V , so dt 16 1 dx x ln x, so du 48. (a) Let h be the level of the coffee in the pot, and let V be h 4 2 dx x Then 5 4 x 7 25 5 8 ds , 25 dt 2 1 55. Let u ( 1.5) 64 8 ft becomes s s2 8 ft, d s sec 1 , so dt 5 (b) dx dt 4 54. 445 sin (x cos3 (x cos (x 3) dx 3) 1 2 cos2 (x 3) 3) , so du (u C 3 ) du sin (x 12 u 2 3) dx. C du u2 1 446 Cumulative Review 60. Use integration by parts. u e x du dv e x e 65. Use integration by parts. x dx 1 sin 2x 2 v du 1x e sin 2x 2 cos 2x dx 1x e sin 2x dx 2 e 1 sin 2x dx 2 1 cos 2x 4 dv x u cos x x 2 cos x x dx v du dv x 2x cos x dx 2 cos x dx dx v x 2 sin x dx e Then 2 sin x x 2 cos x 2x sin x 2 x cos x cos 2x dx 1x e sin 2x 2 1x e cos 2x 4 1 e x cos 2x dx 4 2x sin x x 2) cos x (2 so 2 sin x dx 2 cos x 2x sin x C C The graph of the slope field of the differential equation x e 61. v 2x dx sin x dx Now let x du e dv x 2 sin x dx Now let u x2 u cos 2x dx x x2 x cos 2x dx 2 5x 6 2 Solving A x2 1)(x 6) 6) B (2 sin 2x 5 (x A (x x e B (x 1, B cos 2x) A x 1) 6A dy dx C 1 x x 2) cos x (2 2x sin x is shown below. B (A x 2 sin x and the antiderivative y 6 B) x (B 6A) 1 ,B 7 2 yields A 2 8 1 . Then 5x 6 7(x 6) 7(x 1) x2 8 1 dx dx x 2 5x 6 7(x 6) 7(x 1) 8 1 1 (x 6)8 ln x 6 ln x 1 C ln 7 7 7 x1 8 so 7 x x2 5 [3 2 62. Area Volume 2(8.3) 25 359 2(9.9) … 2(8.3) [ 5, 5] by [ 10, 10] 66. Use integration by parts. C 3] u 359; x du dv v ex xe x ex dx x e x dx 8975 ft 3 e x dx xe x e x dx C e x(x 1) C Confirm by differentiation: 63. y C (t 1) 7 and y 2 1 1 2t e C; y (0) 2 1 2t 7 1 e . 2 2 t1 1 2 1 C 2, so dx [e (x dx 67. (a) y 64. y y y ⇒y 1 cos 2 sin 2 1 cos 2 sin 2 1 sin 2 cos 4 1 sin 2 cos 4 C1, and y 1 . 2 1 2 1) ex C] (x Ce kt, with 6,000 1)e x xe x Ce k(2) and 10,000 Ce k(5). 0⇒ 2 Then 10,000 6,000 e k(5 2) , so 5 3 e 3k and therefore 5 C2, and y 1 2 4 2 0 k ln 3 3 Furthermore, C 0.170. 6,000 e 2k of bacteria is given by y (b) About 4268 4268. The approximate number 4268e 0.170t. Cumulative Review 68. Let t be the time in minutes where t 0 represents right 72. The region has four congruent portions, so /2 Area now, and let T(t) be the number of degrees above room 4 sin 2x dx 1 cos 2x 2 4 0 temperature. Then we may write T(t) T(0) 50 and T( 15) 13 1 ln 10 15 k kt where 50 and kt gives t ln 0.1 k 2 74. Solve y dy y 0.08y 1 dx 500 500 dy 0.08 dx y(500 y) (500 y) y dy 0.08 dx y(500 y) 1 1 dy 0.08 dx y 500 y ln y 9 2 0 y 500 y 0.08x y (1 y 70. C2e dy dx dy y4 dy y4 ) 500C2e 2 cos 1 2 cos 3 2 cos 4)(x 3) (x 3) dx 77. Solve 4x x x3 2 dx 2 x2 0 or x ln y 4 x2 2 y 4 e C1e(x y Ce(x 3x 2 2 /2) 3x /2) 3x 0 17 x 47 1 14 1 0.224 4 14 x 4 4 2 0 (4x 2 x 3) dx 128 3 0 134.041. 4 4 2 x 2) dx 2 x(4x 43 x 3 2 d 0 to find the limit of integration: 0 C1 d 4. By the cylindrical shell method, Volume 3) dx cos 2 2 42.412 4 (x 16.039. cos )2 d 9(1 1 sin 2 4 0) 1 0 1 27 2 1 2 1 cos 2 2 2 sin 76. Volume (y 3) dy cos2 ) d (1 9 (3 2 0.08x 500 Ce 0.08x 1 0 93 22 C2e 0.08x 1 2 r2 d 2 2 . Then (y 2 2) 2 Integrate both sides. C1 2 21)/2 2 1 2 0 21 21)/2 [(y 9 2 0.08x 1 0⇒y (1 75. Area 2x 2) dx (8 2 to find the integration limits: Area 0 y y 5 y 9 2 ln 500 3 (1 y2 3)] dx 64 3 2 131.6 minutes, or about 2 hours and 12 minutes from now. 2 (x 2 2 2 23 x 3 2 4 0 3 to find the integration limits: x 2) [(5 8x /2 2. Then 2 6.13°C above room temperature. 50e x2 8⇒x 2x 2 Area (b) Solving 5 69. x2 73. Solve 5 0.0175. k(120) (a) 50e 65, giving T0 T0e 78. The average value is the integral divided by the interval length. Using NINT, 71. Use EULERT. x y 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 0.1 0.2095 0.3285 0.4568 0.5946 0.7418 0.8986 1.0649 1.2411 1.4273 447 1 average value sin x dx 0.763 0 79. y sec 2 x, so we may use NINT to obtain /4 Length (sec 2 x)2 dx 1 2.556. /4 80. dx dt cos t and dy dt 1 sin t, so we may use NINT to obtain /2 Length cos 2 t (1 sin t)2 dt /2 /2 2 /2 2 sin t dt 4. 448 Cumulative Review 81. Using NINT, 88. (a) The work required to raise a thin disk at height y from dr 2 2 Length d 0 dy 82. dx 2 83. dx dt 2 1 d 6.110. the bottom is 0 e x/2 4 cos t and /2 Area x e dy dt 15 15 y 2(12 1 (y 22,500 sin t, so 3 y) dy 2 12y ) dy 14 y 4 15 2 cos t 2 sin t 2 2 (1 4y 3 10 0 70,686 ft-lb. 22,500 275 (b) y). 0 10 8.423. 257 sec 4 min, 17 sec sin t) dt /2 0 Total work 0 dx 2 sin t 0 10 y2 dy (12 2 60 (weight)(distance) 1 x/2 e , so we may use NINT to obtain 2 2 e x/2 2 2 e x/2 1 dx 2 0 Area 0 d 2 sin t dt 3.470. 89. The sideways force exerted by a thin disk at depth y is its edge area times the pressure, or dr 84. d 1, so we may use NINT to obtain Area 2 2 sin +1d (2 dy)(849y) 1698 y dy. H 32.683. Total force /2 849 H 2, where H is depth. 1698 y dy 0 1 x2 2 x 85. Volume 2 0 1 0 4 12 x 42 9 280 dx 2x 5/2 (x 849 H 2 Solve: 40,000 40,000 and V 849 ⇒H 12.166 ft 3. H x 4) dx 4 7/2 x 7 15 x 5 ln x 90. Use l’Hôpital’s Rule: lim 1 x→ 0 f (x) 0.101. 1 x lim x 1 t2 (2 tan x) dx 2 0 lim /4 8 tan x dx 0 2 (sec x 0 tan x 1 b x lim b→ 8 4 2 2 93. 2 3 dt t2 3 5.394. dx x x e Since lim (ln b 0 e dx b→ F kx ⇒ 200 300 N, x k (0.8) ⇒ k 300 250 1.2 (b) Work 250x dx 0 125x 2 x dx 2 e x dx 0 x e dx 0 lim e b→ x b 1, the 0 original integral converges. 250 N/m, so for 1.2 m. . Both integrals diverge. e lim 1 , and x 1 ln x 0 b x 2, ln 2) e x dx 0 87. (a) F converges. 4 b→ dx ), g(t) dt converges, we conclude that 92. Use the comparison test: for x 0 8 1, and 1) dx /4 8 f (t) g(t) f (t) dt 3 /4 8 t→ 2 and 4 1 . Since f and g are both continuous on [3, t2 g(t) 2 0. x x. 91. Use the limit comparison test with f (t) Volume x→ 2 ln x grows slower than g (x) 86. Use the region’s symmetry: /4 2 lim 1/2 x→ x 1 0 1.2 180 J b 4r dr 94. 1 r b→1 0 1 b2 lim ( 4 b→1 0 10 95. 0 1 dx 1 x 0 1 Since 0 1 4) 1 x r lim b→1 2 4 1 x dx 1x b dx x 01 1 lim b→1 b lim b→1 diverges. ln (1 x) , the original integral 0 r2 b 0 4. The integral converges. 10 dx dx 1 4r dr lim 2 449 Cumulative Review 2 1 dx 96. 3 0 x 0 1 b 3 0 b→1 1 1 lim x a→1 1 1) x x b 2/3 1 1 3 (x 2 lim a→1 0 x6 . By the Alternating 6! x7 1 Series Estimation Theorem, error 0.001. 7! 7! 1 dx 3 a 101. The first six terms of the Maclaurin series are dx 3 2 3 (x 2 lim x dx lim b→1 2 dx 3 2 1)2/3 102. f (0) 1 x for 1 … ( 1)nx n 2x 1 1 2 1 , 3 1)2/3 3(0 2 , 9 2 5 … (3n ( 1)n 1 3n 2 1)2/3 9(0 … f (n)(0) 2x 4x 2 … 8x 3 ( 1)n2nx n 4) , so the Taylor series is … 1 x 3 2584 253 x x 4! 34 3! 33 2 5 … (3n 4) n … ( 1)n 1 x . n! 3n 1 1, so the interval of convergence is 1 . 2 x 98. (a) 2 x2 2! 32 … Since by the Ratio Test cos t 2 (t 2)2 (t 2)4 (t 2)6 … (t 2)2n ( 1)n 2! 4! 6! (2n)! 4n t 12 t4 t8 nt … ( 1) 6! 2! 4! (2n)! 1 1 … lim n→ lim 9 x 5(2!) 13 x 9(4!) … x 13(6!) ( 1)n x 4n 1 (4n 2 5 … (3n 4)(3n (n 1)!3n 1 (3n 1)x (n 1)(3) n→ yields 5 an 1 an lim Integrating each term with respect to t from 0 to x x x5 5! 1. Substituting 2x for x yields 1 2x for … x3 x 1 1 x2 x x4 4! 1 1, f (0) f (0) 97. We know that 1 x3 3! 0. a The whole integral converges. 1 x2 2! x …. 1)(2n)! n→ x 103. Using the Ratio Test, lim ; Since the cosine series converges for all 4)x n an 1 an 2 lim n1 n→ 3 3n 2 1 , so 3 the series converges. real numbers, so does the integrated series, by, the 104. Note that an term-by-term integration theorem (Section 9.1, n!3n 2 5 … (3n 1 x , the radius of convergence is 1. n→ (b) 1)x n Test, since n1 Theorem 2). 1 for every n. By the Direct Comparison n 1 2 diverges, so does . n n1 n 105. Use the alternating series test. 99. ln (2 2x) ln [2(x 2 ln 2 x 2 x 1)] 3 x 3 ln 2 4 … x 4 an 1 Since by the Ratio Test lim an n→ ln (x ( 1)n lim n→ 1) 1x n n n n … 1 x x , the Note that ∑ n0 ( 1)n n1 ∑( n0 1)nun, where un Since each un is positive, un lim un n→ un 1 1 n 1 . for all n, and 0, the original series converges. series converges for n1 1 x ( 1)n converges, but n n 100. Let f (x) sin x. Then f (x) cos x, f (4)(x) f (x) 106. Using the Ratio Test, 1. 1 does not. n 1 cos x, f (x) sin x, and so on. At x lim n→ sin x, 2 the sine terms are zero and the cosine terms alternate between 1 and (x 1, so the Taylor series is 2 )3 (x 2 )5 3! 5! (x 2 )2n 1 …. ( 1)n (2n 1)! 2) (x … an 1 an lim n→ series converges. 3n 1 (n 1)! n! 3n lim n→ 3 n 1 0, and the 450 Cumulative Review 107. (a) Using the Ratio Test, a lim n→ n1 n1 (x lim n→ an n 2) 1 n 3 x x 2)n (x means that the series converges for or dr dt dv dt 112. (a) v(t) 1 x 1. Furthermore, at x a(t) 2 , which 2 1, 3 x 3 x (c) At x 2 x n→ lim n→ x (n (n nx ln 2 (n) 1) ln 2 (n n→ n 2t. 2t and 50 16t 2 100(sin 45 )t 13 130, we have t 2 13 and so 52 13 y n ln2 n xn to easily clear the 35-ft tree. 1) 1 50 2 2 75.92 ft, high enough 52 114. Since r cos x, r sin y, the Cartesian equation is x y 2. The graph is a line with slope 1 and y-intercept 2. 1) that the series converges for 16 5 lim 1 50 4. an 1 an 2 ln n 1) n→ ln (n n lim 16t 2 When x xn 1 1)ln2 (n 2 sin t dt 100(cos 45 )t y 1 n→ sin t)2 dt 113. Yes. The path of the ball is given by 1 and the radius of 108. (a) Using the Ratio Test, lim (1 /2 1 lim ( cos t)2 /2 3 /2 convergence is 1. (b) 3 /2 /2 convergence is sin t) j (cos t)j v(t) dt n1 ∑ (sin t)i 3 /2 3, the series 1 , which diverges, and at x 1, the series is n ( 1)n , which converges. The interval of n (1 (b) Using NINT, the distance traveled is is ∑ n1 ( cos t)i x , which means x 1. At x 115. 1, the series converges by the Integral Test: 2 1 dx x(ln x)2 b lim b→ 1 ln b lim b→ 2 1 ln 2 1 dx x(ln x)2 1 ln 2 convergence interval is 1 ln x [ 3, 3] by [ 0.5, 3.5] b The shortest possible -interval has length 2 . 2 116. x . So the 1 x 1 and the radius of convergence is 1. (b) 1 x r cos y r sin dy dx 1 cos 2 , cos sin cos sin dy/d dx/d sin cos sin 2 cos 2 2 sin cos (c) Nowhere 109. 1 22 110. 1 cos 2, ( 3)2 3 2 3 13 , 1 sin 3 Zeros of cos 3 13 1 3 , 22 dy : d sin 2 cos , 1 (2 cos 111. dy dx t 3 /4 dy/dt dx/dt t 3 /4 tangent vectors are 4 , 5 3 sin t 4 cos t t 3 /4 1 32 42 4, 3 3 3 . The normal vectors are , 5 5 3 . The 4 4 and 5 sin 34 ,. 55 sin 0 1) 2 , 3 0, 0 4 , or 3 2 dx : d 2 sin cos , 0 1 2 0 or cos 0, 0 2 cos2 1)(cos Zeros of 43 , and 55 cos 2 3 or 5 , 3 2 451 Appendix A2 dx 0 at d 2 4 dx dy and at , and vertical tangents 0, 0 3 3 d d 5 at , at , and at . 3 3 dy 0 For 0 (or 2 ), becomes and l’Hôpital’s Rule dx 0 There are horizontal tangents dy d 0, leads to dy dx s Appendix A2 (pp. 581–584) 1. Step 1: The formula holds for n …x x Step 2: Suppose x 1 2 x2 x1 x1 k … x2 Then, x1 1, because x1 xk … x2 … x1 . xk . 1 xk xk 1 by the triangle inequality. sin (0) 4 sin (0) cos (0) cos (0) 2 cos 2 (0) 2 sin 2 (0) 0 0, so this is So, by the transitivity of , x1 x1 x2 … xk 1 … x2 xk 1 . another horizontal tangent line. The mathematical induction principle now guarantees the Horizontal tangents: original formula for all n. At 0 or 2 , we have r coordinates are (0, 0), so the tangent is y 2 , we have r 3 At are 33 3 , , so the tangent is 44 3 4 3 , 4 3 3 , so 4 1 rk 1 rk 1 rk rk 1 . Then 1r rk 1 rk 1 1r r) 1 rk 2 . 1r 1 1 r k 1(1 r 3. Step 1: The formula holds for n x kx k 1 1, because d (x) dx 1. kx k 1. Then x k) x dk (x ) dx xk kx k xk xk d (x) dx 1)x k. (k The mathematical induction principle now guarantees the original formula for any positive integer n. 1 1 cos , sin 2 32 3 At 1 3 , , so the tangent is x 44 , we have r (2 cos , 2 sin ) ( 2, 0), so the tangent is x 1 51 5 cos , sin 2 32 3 x 1 . In summary, the horizontal tangents are y 4 3 4 , and y 2 and x 1 . 4 2. 1 and the Cartesian coordinates 2 are 3 1 . 4 2 and the Cartesian coordinates are 5 , we have r 3 At x … r2 r dk (x ) dx d k1 d (x ) (x dx dx are y 1 r2 . r 1 1 … r2 r Step 2: Suppose . 1 and the Cartesian coordinates 2 , we have r 3 Step 2: Suppose 1 r) original formula for every positive integer n. Vertical tangents: At r)(1 1r 1, because The mathematical induction principle now guarantees the 3 and the Cartesian 2 3 43 4 cos , sin 2 32 3 3 (1 r 1 33 . y 4 4 , we again have r At 3 the tangent is y 1 0. 3 and the Cartesian coordinates 2 3 23 2 cos , sin 2 32 3 coordinates are 2. Step 1: The formula holds for n 0 and the Cartesian 1 , 4 3 3 4 3 4 , so the tangent is 0, , and the vertical tangents are 4. Step 1: The formula holds for n 1, because f (x1) f (x1). Step 2: Suppose f (x1x2 … xk) f (x1) f (x2) … f (xk). Then by the given property, f (x1x2 … xk 1) f (x1x2 … xk) f (xk 1) f (x1) f (x2) … f (xk 1). The mathematical induction principle now guarantees the original formula for every positive integer n. 5. Step 1: The formula holds for n Step 2: Suppose 2 31 2 32 2 32 … 1 2 31 32 3k 1 1, because … 2 3k 1 1 2 3k 1 3k 1 1 3k 1 2 3 1 1 . 3 1 . Then 3k 2 3k 1 1. The mathematical induction principle now guarantees the original formula for all positive integers n. 452 Appendix A2 6. Experiment: 9. Step 1: The formula holds for n 5 6 1(1 12 n 1 2 3 4 7 n! 1 2 6 24 120 720 5040 n3 1 8 27 64 125 216 1/2)(1 3 … k(k k 3. Then (k Step 2: Suppose k! 4, k 1), and so k (k 1)k of 3 , (k 2 k 1 3 1)! 2 1 (since k k2 (9/2)k 2 k 2 2k 1 2 and 1)2 (13/2)k 1)(k 3/2)(k 3 (k The mathematical induction principle now guarantees the original inequality for all n . 1)2 3(k (k 1)3. (k (k 1) 3 1) . So 1) , and thus by the transitivity 1)! 1) 3 3 3 (k 1) 2 (k 1/2)(k 3 1)2 (k 1)k 3. (k k(k k2 1/2)(k 3 1/2)(k k3 6, because 216. 1 k2 … k(k Step 1: The inequality holds for n For k 22 1. 22 Then 12 720 1) Step 2: Suppose 12 343 1, because 1)[(k 1) 2) 1/2][(k 3 1) 1] . The mathematical induction principle now 6. guarantees the original formula for all positive 7. Experiment: integers n. n 1 2 3 4 5 6 2n 2 4 8 16 32 64 n2 1 4 9 16 25 36 10. Step 1: The formula holds for n Step 1: The inequality holds for n Step 2: Suppose 2k since 1 k k 2. For k 1 3, k 1 , and so k2 2k 2 2k 25. Step 2: Suppose 13 1 k ,2 2k k Then 13 2k 1 (k by the transitivity of 1 . Then 2k 1 8 3, because 2 2 2k and the fact that 1 4 3 2 8 3. 1 . 8 1 , and 4 1 k1 1 ,2 . 8 8 The mathematical induction principle now guarantees the original inequality for n 1)2 k2 4 (k k 2 1)3 1)(k 1 (k 2) 2 1)2 1)2 (k 1)(k 2 . 1)3 (k 4 1) 2 k (k k3 (k 1) 2 k 2 (k 1)2. 5. 8. Step 1: The inequality holds for n Step 2: Suppose 2k … 23 2 The mathematical induction principle now guarantees the original inequality for all n … 23 1. Then by the 2k + 1. And thus k2 1. 2 k(k transitivity of 2k 5, because 32 1) 2 1(1 13 1, because (k 2)2 4 (k 1)2 . The mathematical induction principle now guarantees the original formula for all positive integers n. 11. (a) Step 1: The formula holds for n 1 ∑ ( ak k1 1 ak) 1 k1 k1 i ∑ ak ∑ bk i Step 2: Suppose ∑ (ak bk ) k1 i1 ∑ ( ak k1 1, because a1 b1 . i ∑ ak ∑ bk. Then k1 k1 i bk) i ∑ ak ∑ ( ak k1 i ∑ bk k1 i1 k1 ai 1 (ai bi 1 bi 1) 1 i1 k1 bk) k1 ∑ ak ∑ bk. The mathematical induction principle now guarantees the original formula for every positive integer n. Appendix A3 (b) Step 1: The formula holds for n 1 1 ∑ ( ak k1 k1 (pp. 584–592) k1 b1 . 1. k1 a c b 4 9 i 1 2 4 7 x ∑ ak ∑ bk. Then bk ) i1 a1 k1 i i Step 2: Suppose ∑ (ak s Appendix A3 1, because 1 ∑ ak ∑ bk bk) 453 k1 i ∑ ( ak ∑ ( ak bk) k1 bk) k1 i i ∑ ak ∑ bk k1 i1 ai k1 1 bi 1) 1 bi 1 2 Step 1: x ⇒ k1 ∑ ak ∑ bk. 1 2 Step 2: The value of The mathematical induction principle now guarantees ⇒ 4 9 1 2 x 1 x 2 4 ⇒ 9 ⇒ 1 i1 k1 (ai 1 2 1 2 1 , or 18 4 ⇒ 7 1 2 which assures x 4 is the smaller value, 7 x 1 . 14 1 . 18 the original formula for every positive integer n. 2. a c b 2.7591 3 3.2391 3 ⇒ x (c) Step 1: The formula holds for n 1 1, because 1 ∑ cak ∑ ak c k1 Step 1: x ca1. k1 i i Step 2: Suppose ∑ cak k1 k1 k1 i 1 i ∑ cak cak 1 1 ∑ ak c ak k1 c ∑ ak k1 cak k1 i1 i c ⇒ i1 ∑ ak. Then ∑ cak c ∑ ak k1 1 3 x 3 2.7591 ⇒ 0.2409, 3 3.2391 ⇒ 0.2391. or The value of . 3 3 Step 2: 1 x ⇒ 2.7591 which assures x x 3 3.2391 is the smaller value, The mathematical induction principle how guarantees 0.2391. the original formula for every positive integer n. 3. Step 1: x n (d) Step 1: The formula ∑ c n c holds for n 1, ⇒ 3 ⇒ 3 x 3 Step 2: From the graph, 3 k1 3 x because 1 ∑c 1c k1 c. or i Step 2: Suppose ∑ c ∑c ic c (i 4. Step 1: x 1) c. k1 the original formula for every positive integer n. 12. Step 1: The formula holds for n ⇒ ( 1) ⇒ The mathematical induction principle now guarantees 1 x 0.39. 0.77, or 1 (and every real 9 25 x 1 1 Step 2: From the graph, number x), because x1 0.41; thus 0.39, i c. Then k1 i1 x1 3.41 ⇒ 3 2.61 ⇒ 16 7 ⇒ 9 9 16 9 ⇒ 0.36; thus 25 25 1 1 0.36. The mathematical induction principle now guarantees the 5. Step 1: (2x 2) ( 6) 0.02 ⇒ 2x 4 0.02 ⇒ 0.02 2x 4 0.02 ⇒ 4.02 2x 3.98 ⇒ 2.01 x 1.99 Step 2: x ( 2) ⇒ x2 ⇒ 2x 2⇒ 0.01. original formula for every positive integer n (and every real 6. Step 1: Step 2: Suppose x xk number x). 1 x. k xk x x k k x . Then xk x xk x x k 1. x11 0.1 ⇒ 0.1 ⇒ 0.9 x 1 1.1 ⇒ 0.81 ⇒ 0.19 x 0.21 Step 2: x 0 ⇒ x ⇒ x 1 1 0.1 x 1 1.21 0.19. 454 Appendix A3 19 x 3 1⇒ 1 19 x 3 1 19 x 4 ⇒ 4 19 x 16 ⇒2 ⇒ 4 x 19 16 ⇒ 15 x 3 or 3 x 15 Step 2: x 10 ⇒ x 10 ⇒ 10 x 10. Then 10 3 ⇒ 7, or 10 15 ⇒ 5; thus 5. 7. Step 1: 8. Step 1: x 2 4 0.5 ⇒ 0.5 x 2 4 0.5 2 ⇒ 3.5 x 4.5 ⇒ 3.5 x 4.5 ⇒ 4.5 x 3.5, for x near 2. Step 2: x ( 2) ⇒ x2 ⇒ 2x 2. Then 2 4.5 ⇒ 4.5 2 0.1213, or 2 3.5 2 3.5 0.1292; thus 4.5 2 0.121. 9. (a) lim x 2 x→ 5 6x 5 x5 4, x lim x→ 5 lim (x x→ 5 4.05 x 1 5.05 Step 2: x x 5 lim f (x) lim (6x 4) x→1 1: (4 ⇒ sin 1 ⇒1 1 ⇒ sin 1 sin ⇒ sin 1 Choose 0.5 ⇒ 2 2 0.5 4) 0.5 ⇒ 6x 0.5 ⇒ 3 ⇒ 4 1 . Choose 12 0.05. x 1 2. 2x 0.5 6 0.5 13 . 12 6 1 . 1 , or 1 4 1 . 12 1 1 (sin 1 (sin 1 (sin 1 (sin 1 0.01) 0.01) 0.0182, or 0.01) 0.01) 1 0.0188. 0.018. sin x, y2 sin 1 1, and 1. The curve intersects the lines at 0.98175 1 0.01825 and at 1.01878 1 0.01878. We may choose [0.9, 1.1] by [0.78, 0.88] 3 ; 4 2 6 x 0.05, or x→1 2 ⇒x sin 1 0.01) . 1 sin 1 1 (sin 1 1 0.01 x x x Then 1 sin 1 0.01) ⇒ 1 0.01 sin x (sin 1 sin Step 2: x sin 1 0.01 5 2, so lim f (x) 2 1: (6x ⇒1 ⇒ sin 1 0.018. 2 and 2x) ⇒x 1 sin x 5. 0.05; thus 2x) Step 2: x 0.01 0.01 x 5 5.05 ⇒ lim (4 x sin 1 5. 4.95 ⇒ x→1 (b) Step 1: x Then 1 x x 5 5 x→1 0.841 x 0.05 3.95, x ⇒ Then x→1 ⇒ y3 4 4.95, x ( 5) ⇒ (b) Step 1: sin x 1) 0.05 5)(x 1) x5 ⇒ 10. (a) lim f (x) ( 4) (x 0.05 ⇒ sin 1 x→1 Alternately, graph y1 6x 5 x5 ⇒ ⇒ 5)(x 1) x5 5. x2 (b) Step 1: (x 11. (a) lim (sin x) 13 12 455 Appendix A3 1 ( 1)2 4 x 12. (a) lim 2 x→ 1 x (b) Step 1: 4 x2 4 ⇒ 0.1 x2 x 4 13 7 ⇒ 30 30 x2 4 28 13 2 72 ⇒x x x 30 30 30 28 7 Then x 2 x 0 30 30 15 421 15 x 421 15 1.1551 13 3.4628. Thus Step 2: x 1 or ⇒ 1 x x 1 3 421 7 ⇒ . ⇒ 1 901 4 , y2 1 3 0.78833 1 1 3 1 1 , 1. , 1 1}, 3 0.15513 and at 1 3 3 1 1 for x near 3 1 ⇒ 3 3. 3 3 1 3 ⇒ 3 3 . 3 1 3 1 3 3 3 3 x2 3 , x x 3 3 ⇒ 3 ⇒ 3 Then 3 x 3 1 3 3 x 3 ⇒ 0.1, and 1 3 1 3 1 Step 2: x 1 3 , 3. 3 Choose min 15. (a) (5 ⇒I 3 3 1 3 , 3 1 3. 3 ⇒ 2 ⇒ 2 2 x ⇒ ⇒ )5 (5, 5 0 ) 0.21167. We may choose (b) lim x→5 0.155. 16. (a) 4 ⇒I 5 (4 (4 (b) lim ) 2 4 x→4 [ 1.5, 0] by [ 0.15, 0.55] 1 1 1 x2 1 x2 1 x2 3 0.1. The curve intersects the lines at 1.15513 1 ⇒ min {1 3 or x ⇒ 1 ⇒ 1 3 1 or 13 x2 1. 1 1 3 ⇒ 0.155. Alternately, graph y1 1. 1 x 1 1 x2 14. Step 1: 1. 15 near x x 1 or 901 13 15 901 28 0.1551, 13 15 421 1 7 22 421 0.2117. 7 Choose y3 x x x Then ⇒ 901 1 that is, the smaller of the two distances. 15 13 1 1 1 ⇒ Choose x ⇒ ( 1) ⇒ 15 x Then 5.0740, and also 7 901 1 ⇒ 0.7883 or 7 x 1 x2 1 1 Step 2: x 52 for x near 1. 30 13 2 52 x x , 30 30 which using the quadratic formula implies x x2 ⇒ 0.1 ⇒ 1 ⇒ 1 3 x 0.1 1, x 2 13. Step 1: For x ⇒1 1 3 x 1 3 , 4) x 0 17. If L, c, and k are real numbers and lim f (x) x→c for any 0, there is a ⇒ k f (x) kL Proof: For any is a . Therefore, 0 x c . 0, let 0 such that 0 k 0 such that 0 L, show that k x x c c . Since lim f (x) x→c ⇒ f (x) ⇒ k f (x) L, there L kL . 456 Appendix A5.1 18. If L, M, and c are real numbers and lim f (x) L and x→c lim g(x) M, show that for any x→c x Proof: f (x) g(x) (L f (x) M g(x) L 0, there is a ⇒ f (x) such that 0 c g(x) M) f (x) f (x) there exist 1 , x c 1 ⇒ f (x) 0 x c 2 ⇒ g(x) min { 1, 2 2 2 M, for f (x) f (x) . g(x) (L 2 M) 2 g( f (c)) x x c f (c) f2(x) x→c L1 L2 f3(x)] lim [ f1(x) x→c ⇒ g( f (x)) c L3 L3, by two applications of the Sum Rule. To generalize: Step 1 (n 1): lim f1(x) that g( f (x)) g f is continuous at c. Step 2: Suppose lim [ f1(x) L1 … L2 lim [ f1(x) … h)2 (y 0)2 (y (y 2)2 y (0, 4) fk(x)] (0, 2) … … fk(x)] Lk 1 –3 (0, 0) 1): lim f1(x) 2. (x [x (x L1, as given. x→c x→c y (0, 8) lim (f1(x)) f2(x) … fk(x)) Lk 1 n factors lim (anxn an 1xn lim anx x→c n lim an 1x x→c an lim x n aa 1 cn 1 an c an 1 lim x n x→c cc…c 1 n factors … ax 1 … n1 x→c x→c n , by the Product Rule. lim (x x … x) x→c 1 (–1, 5) …L k x→c 22. lim f (x) 1 …a c 1 … a0 lim a1x x→c a1 lim x x→c (0, 2) cn –5 a0) lim a0 x→c lim a0 x→c f (c), where in addition to the items given in the problem, the Constant Multiple Rule was used (to move the coefficients out of the scope of the limit signs). 10)2 10 lim ( f1(x) f2(x) … fk 1(x)) L1 L2 x h)2 (y k)2 a 2 ( 1)]2 (y 5)2 ( 1)2 (y 5)2 10 x→c x→c 3 Lk 1, by the Sum Rule. L1 L2 … Lk. Then x→c a2 22 5 Step 2: Suppose lim ( f1(x) f2(x) … fk(x)) 21. lim x n k)2 2)2 4 fk 1(x)] f2(x) L2 20. Step 1 (n 1. (x (x x2 Lk. Then lim[ f1(x) L1 … f2(x) x→c f2(x) x→c (pp. 593–606) L1 as given. x→c g( f (c)) . f (c), so . But from the . So for any g( f (c)) . f2(x)] 0 0 0 such that s Appendix A5.1 19. lim [ f1(x) 0 there is a 0 such that ⇒ f (x) there is a M x→c ⇒ g( f (x)) f (c) ⇒ g( f (x)) f (c) 0 g(x) lim g(x) The second inequality also holds when f (x) 0 L ⇒ f (x) such that 0 and }. Then ⇒ f (x) c M by the x→c M x g(x) g(x) f (c) . g(c) x→ c 24. From the continuity of g, for any continuity of f, there is a L 0 M . such that 2 0 Choose L 0 M) L, lim g(x) x→c any (L L triangle inequality. Since lim f (x) lim f (x) f (x) x→c g(x) 23. By the Quotient Rule, lim 5 x , which means Appendix A5.1 3. Complete the squares. x 2 y 2 4x 4y 4 0 x 2 4x 4 y 2 4y 4 4 (x 2)2 (y 2)2 22 Center ( 2, 2); radius 2 13. x2 2 1⇒c y2 vertices are ( 14. y y2 4 2 1 ⇒ foci are ( 1, 0); 4 1 5 ⇒ foci are (0, 2, 0) 1⇒c x2 1 457 vertices are (0, 2); asymptotes are y 5); 2x 4 12x ⇒ 4p 15. y 2 2 –4 x –2 2 12 ⇒ p 3; focus is (3, 0), directrix is 3 y x 4 4. Complete the squares. x 2 y 2 4x 4y 0 x 2 4x 4 y 2 4y 4 8 (x 2)2 (y 2)2 (2 2)2 Center (2, 2); radius 2 2 x = –3 y 2 = 12x 2 F –4 –2 2 x 4 –2 –4 y 5 5 1 y ⇒ 4p 4 4x 2 ⇒ x 2 16. y x focus is 0, 1 ⇒p 4 1 , directrix is y 16 1 16 y 5. The circle with center at (1, 0) and radius 2 plus its interior. 6. The region exterior to the unit circle and interior to the circle with center at (0, 0) and radius 2. 1 4 1 8 –1 4 7. y 2 8x ⇒ 4p 8⇒p directrix is x 8. y 2 directrix is x 9. x 2 directrix is y 10. x 2 17. 16x 2 y2 9 y2 9 2⇒p 1 25 16 3 foci are ( 3, 0) 1 1 ; focus is 0, , 2 2 1 2 1⇒c 2 F1 4 9 –4 13 F2 –2 2 –2 13, 0); vertices are ( 2, 0); 3 x 2 1⇒c vertices are (0, y2 16 b2 a2 x2 25 y asymptotes are y x2 4 400 ⇒ 25y 2 ⇒c 3 , 2 3 2 ⇒ foci are ( 12. 3 ; focus is 0, 2 6⇒p 2y ⇒ 4p x –1 8 –1 4 1; focus is ( 1, 0), 1 directrix is y x2 11. 4 y=– 1 16 4⇒p 6y ⇒ 4p 1 4 F 2; focus is (2, 0), 2 4x ⇒ 4p y = 4 x2 9 3) 4 5 ⇒ foci are (0, 5); 4 x 1 ; 16 458 Appendix A5.1 18. 3x 2 6⇒ 2y 2 ⇒c a2 x2 2 b2 foci are (0, y2 3 3 2 b and a 1 4 ⇒b 3 2 4 (3) 3 x 4⇒ 9 y y F2 5 –1 1 x 2 (y + 2)2 = 8(x – 1) x = –1 F1 –5 V 2, 0), Vertices: ( 2, 0) ⇒ a ⇒ b2 a2 20. Foci: (0, ⇒ b2 c2 ( 2)2 4 5) ⇒ a 4), Vertices: (0, 25 9⇒ 16 21. x 2 y 2 1 ⇒ c asymptotes are y foci are ( 2, 0) 2 2, c x2 2⇒ 4 5, c y 25 a2 x, 1 26. (a) 1 x2 16 y2 9 1 ⇒ center is (0, 0), vertices are ( 4, 0) and ( 1 a2 and (4, 0), c 4 1 b2 F 2 y2 2 2 x 9 x 5 –2 19. Foci: ( 3 25. (a) y 2 8x ⇒ 4p 8 ⇒ p 2 ⇒ directrix is x 2, focus is (2, 0), and vertex is (0, 0); therefore the new directrix is x 1, the new focus is (3, 2), and the new vertex is (1, 2). 1) 2 –2 4 x⇒a 3 y2 1 16 24. Vertices: ( 3, 0), Asymptotes: y 1 7 ⇒ foci are ( 7, 0) b2 7, 0); therefore the new center is (4, 3), the new vertices are (0, 3) and (8, 3), and the new foci are 2; (4 7, 3). (b) y y 8 –x x y= y= 4 2 F1 –4 F2 –2 2 V1 x 4 (4, 3) F1 F2 V2 –2 8 –4 22. 8y 2 16 ⇒ 2x 2 ⇒c a2 b2 asymptotes are y y2 2 x2 8 1 27. (a) 2 8 10; 10) c a2 b2 x 5 x (b) y ⇒a x⇒c 2), Asymptotes: y 1⇒a 1⇒b 4 (x – 2)2 y2 – =1 16 9 F1 b a y = 3(x – 2) 10 F1 and 5 ⇒ foci are ( 5, 0) and 25 and (7, 0), and the new asymptotes are y y= 2 23. Foci: (0, 3x , 4 vertices are ( 2, 0) and (6, 0), the new foci are ( 3, 0) F2 2 1 ⇒ center is (0, 0), vertices are ( 4, 0) (5, 0); therefore the new center is (2, 0), the new 5 –x y2 9 and (4, 0), and the asymptotes are y x , foci are (0, 2 y y= x2 16 x b ⇒ c2 1 ⇒ y2 a2 x2 b2 1 –10 2 2a 2 ⇒ 2 F2 C V1 10 V2 2a 2 –10 y = – 3(x – 2) 4 x 3(x 2) 4 . 459 Appendix A5.1 28. Original parabola: y 2 4x; vertex is (0, 0); y 2 4x ⇒ 4p 4 ⇒ p 1, so focus is (1, 0) and directrix is x 1. New parabola: (y 3)2 4(x 2); vertex is ( 2, 3), focus is ( 1, 3), directrix is x 3. 2 x 6 29. Original ellipse: (0, 3); c 2 2 y 9 3 ⇒ foci are (0, 3); center is (0, 0). New ellipse: 2)2 (x ( 2, 2h x2 2 b/2 x4 b2 2h 4x 3 dx b2 x 0 hb 2 ; 8 0 b2 h 2 1 3 Volume of the Cone: V2 b2 h 4 1 3 hb 2 ; 12 3 V 22 1; vertices are 9 4); foci are ( 2, 1 3); center is 38. (a) y 2 y2 ; the volume of the solid formed by k kx ⇒ x revolving A about the y-axis is 1). kx x2 30. Original hyperbola: 4 ( 2, 0); c 2 b/2 4h 2 x dx b2 2 xh 0 therefore V1 1)2 (y 6 ( 2, 2) and ( 2, b/2 V1 1; vertices are (0, 3) and 6⇒c 9 37. Volume of the Parabolic Solid: y2 5 5⇒c 4 1; vertices are (2, 0) and 3 ⇒ foci are (3, 0) and 5 ( 3, 0); center is (0, 0); asymptotes are y New hyperbola: 2)2 (x 2)2 (y 4 2 kx y2 2 dy k 0 k2 0 x 2 kx ; the 5 y 4 dy volume of the right circular cylinder formed by x. revolving the rectangle about the y-axis is 1; vertices are (4, 2) 5 V1 V2 x2 kx ⇒ the volume of the solid formed by and (0, 2); foci are (5, 2) and ( 1, 2); center is (2, 2); 5 asymptotes are y 2 (x 2) revolving B about the y-axis is V3 2. 31. Original hyperbola: y 2 x 2 1; vertices are (0, 1) and (0, 1); c 2 1 1 ⇒ c 2 ⇒ foci are (0, 2); center is (0, 0); asymptotes are y x. New hyperbola: (y 1)2 (x 1)2 1; vertices are ( 1, 2) and ( 1, 0); foci are ( 1, 1 2); center is ( 1, 1); asymptotes are y (x 1) 1. 32. x 2 4x y 2 12 ⇒ x 2 4x 4 y 2 12 4 ⇒ (x 2)2 y 2 16; this is a circle: center at C( 2, 0), a4 33. 2x 2 2y 2 28x 12y 114 0 ⇒ x 2 14x 49 y 2 6y 9 57 49 9 ⇒ (x 7)2 (y 3)2 1; this is a circle: center at C(7, 3), a 1 4y 34. x 2 2x 4y 3 0 ⇒ x 2 2x 1 ⇒ (x 1)2 4(y 1); this is a parabola: V( 1, 1), F( 1, 0) 3 1 ⇒ x2 1 35. x 2 ⇒ (x 5y 2 4x 2)2 5y 2 5⇒ 4x 2)2 (x 5y 2 4 5 y2 4 ellipse: the center is ( 2, 0), the vertices are (2 5, 0); c a2 b2 5 1 4:1. (b) The volume of the solid formed by revolving B about x the x-axis is V1 2 ⇒ the foci are ( 4, 0) and (0, 0) 36. x 2 y 2 2x 4y 4 ⇒ x 2 2x 1 (y 2 4y 4) 1 ⇒ (x 1)2 (y 2)2 1; this is a hyperbola: the center is (1, 2), the vertices are (2, 2) and (0, 2); c a2 b2 11 2 ⇒ the foci are (1 2, 2); the asymptotes are y 2 (x 1) ( kt)2 dt 0 x k 0 t dt kx 2 . 2 The volume of the right circular cylinder formed by revolving the rectangle about the x-axis is ( kx)2x kx 2 ⇒ the volume of the solid formed by revolving A about the x-axis is V3 1; this is an V1 4 x 2 kx . Therefore we can see the ratio of V3 to V1 is 5 V2 1 V2 V2 V1 kx 2 ratio of V3 to V1 is 1:1. kx 2 2 kx 2 . Therefore the 2 460 Appendix A5.2 39. Let P1( p, y1) be any point on x p, and let P(x, y) be a point where a tangent intersects y 2 4px ⇒ 2y y2 dy dx 4p ⇒ tangent line from P1 is ⇒ y2 y2 yy1 yy1 1 2 ⇒ y2 y2 4p dy dx ( p) 2 y2 , we have 4p 12 y 2 yy1 42. 9x 2 2p 2 y12 y1 2 2p y12 y1 4p 2 4p (y12 y1 2 4 2 9 4 4p 2. Therefore y12 y1 4p 2 43. x 2 24) 1 ⇒ the lines are 4p 2) x2 on the interval 0 4 x2 4 4x 1 and the height is 2y) ⇒ A (x) x 2. The area of x2 (since the length is 2x 4 x2 x2 41 x2 . 4 1 2 x 4 ⇒4 1 x2 x x2 4 0⇒4 1 0 ⇒ x2 2 0⇒0 2⇒x 3 2 2 0 3 ( y3 30 y C⇒C 0 we have that A( 2) C; y drA drB dt dt ⇒ rA 0; therefore y ⇒ rB d (r dt A rB) 0 a constant s Appendix A5.2 4 is the maximum ⇒c ⇒V 2 9 2 9x 0 36 ⇒ y 2 92 x 4 92 x and we use the positive root 4 2 9 22 92 9 x dx 2 9 x dx 0 4 4 3 32 x 40 x2 25 y2 16 1 b2 a2 c a 2. 9 400 ⇒ 25 16 3 25y 2 x 2 wx 2 is the 2H 46. PF will always equal PB because the string has constant length AB FP PA AP PB. ⇒e ⇒y 0 when equation of the cable’s curve. 45. 0 3 ; F( 3, 0); directrices are 5 5 a 25 3= e 3 5 2. 2x 2 ⇒c 1 b2 2 1 ; F(0, a2 1); directrices are 1 c a ⇒e y2 2 2 ⇒ x2 y2 2 24 y 0 a e 2 2 1 2 3 y 2)2 dy 0 area, when the length is 2 2 and the height is 4y 2 y 24 wx 2 2H C 1 3 2 (only the x2 4 1 w x2 H2 w(0)2 2H w x dx H 1. 16x 2 41. (a) Around the x-axis: 9x 2 dx 8 (pp. 606–611) A(2) 4 y 2 on the interval y 2)2 dy y 2) dy (1 positive square root lies in the interval). Since A(0) 2 x2 9 56 43 1 (1 3 0 44. y x 4 Thus A (x) 4 on 4 2 4x 8 1⇒x ⇒V the inscribed rectangle is given by 2x 2 1 3 2 2 x2 24 3 3 A(x) 8 3 16 y2 2 1 4 3 2 ⇒y 9 x3 43 4) dx 64 3 36 4 4⇒V x (x 2 3 (56 4 perpendicular. 40. Let y 9x 2 36 ⇒ y 2 4y 2 2p and m2 2 ⇒ m1m2 0 the interval 2 the slopes of the two tangents from P1 are m1 4y 9 4 16p 2 3 2 2p y 0 4y12 2y1 ⇒y x y1 2p 2 ⇒ y 2 2p 2 yy1 y 42 y 9 4 4 ⇒V 2p ; then the slope of a y 36 ⇒ x 2 4y 2 42 y and we use the positive root 9 3 4 22 42 4 y dy 2 4 y dy 0 9 9 4 33 y 16 27 0 ⇒x 4px. Now 2p 2. Since x 2px 2p dy dx (b) Around the y-axis: 9x 2 1 Appendix A5.2 3. 3x 2 ⇒c a2 F(0, x2 2 6⇒ 2y 2 y2 3 b2 3 ⇒ 16 ; 2 therefore 3 1 4. 6x 9y x2 54 ⇒ 9 2 ⇒c a 1 2 b ; F( y 9 x 25 1 9 c a 3⇒e 3 –4 –2 36 ⇒b 1600 64 ⇒ ae x2 100 y2 94.24 and c ⇒ 100 0.1 ⇒ a 7 ⇒ b2 y 4900 9 ⇒ 5 Then PF 5)2 ⇒ (x 5)2 2 4 9 ⇒ x2 y2 4⇒ ae e2 ae x 16 y 15 1 is a model of Pluto’s orbit. 5 94.24 x 4851 2 4)2 y2 ⇒ x2 8x 16 3 y2 Therefore, a 5 2 52 x 9 1)2 (x 3, b 2 and the major axis is vertical. The ⇒c 18 x 5 81 5 y 4 e2 16 ⇒ e 2 (y 0)2 ae 1; c2 9 a2 4 c a 5 3 4 4 5 3 y 8 16)2 32x 256) 1 –5 5 –2 32 22 x 95 . 5 5 5), the , and the directrices are 3 a e 4 and 1 1 ⇒e . Then 4 2 1 x 16 2 b2 5; therefore the foci are F(1, 4 eccentricity is e 1 16 ⇒ c 4)2 (y 4 5 12 (x 4 2 x y2 48 ⇒ 64 48 y2 3 other axis is from (3, 4) to ( 1, 4) and is 4 units long. . center is the point C(1, 4) and the ellipse is given by y 4 1 (x 4 5 9 x 4)2 (x 5 and 5 ⇒e 9 ⇒ e2 2 x 9 13. One axis is from (1, 1) to (1, 7) and is 6 units long; the 5 9 y2 16 ⇒ ⇒ (x 3 4 15; therefore, 5 ⇒c 5 5 x 9 5 1 PD ⇒ 2 ⇒ x2 5.76 49 5 9 0)2 (y y2 5x 16 ⇒ PF 15 ⇒ b b2 16 2 y 4900 9 10. Focus: ( 4, 0), Directrix: x a e b2 ⇒ 1 2 70 ae 9 5 ⇒2 e2 e 5 5 PD 3 (x ⇒ x2 a2 1 9. Focus: ( 5, 0), Directrix: x ⇒ ⇒ b2 4; c2 1 and a 1 4 2 x 4851 a e take c 40 0.25 1 70(0.1) ae 2 c a 12. The eccentricity e for Pluto is 0.25, ⇒ e 10 and 2.4 ⇒ b 2 70), e 8. Vertices: (0, 6 c 8 e 0.2 y2 1 1536 8 and a 0.24 ⇒ a 10(0.24) 3 0.5 1 x2 1536 ⇒ 1600 7. Vertices: ( 10, 0), e c y 36 0.2 ⇒ c 6. Foci: ( 8, 0), e 2 2 x 27 ⇒ 27 9 c e 3 and a 2 ⇒ b2 x 4 –4 0.5 ⇒ c 3), e 2 –2 3 5. Foci: (0, 1 2 3 33 1 3; 4 6 3 a e 0 2 b2 y 3, 0); directrices are 3 x y2 6 9⇒b a2 2 3 2 5; c 2 4 and a b2 ⇒ b2 25 3 3 a e 0 1 c a 1⇒e 2 1); directrices are y 4 ⇒ take c 5 11. e 1 461 462 Appendix A5.2 14. Using PF e PD, we have ⇒ (x 4) ⇒ x2 8x 5 9 ⇒ x2 15. 9x 2 2 y 20 ⇒ 5x 2 a 2 b 2 a e 0 16. y 2 a2 x2 8 b2 a e 8 16 ⇒ ⇒c x2 2 a2 2 2 x2 8 2⇒ 2 ⇒ (x 4) ⇒ x2 8x x2 2 y2 5(y 2 12y x a2 y b2 1) 45 4 4 180 c2 e2a 2 1⇒ c ⇒c a a 2; e a2 2 x a2 y a 2(e2 a 2(e2 ea 1); 2 1; the asymptotes 1) e2 1x. As e increases, the slopes of the asymptotes increase and the hyperbola 5 10 approaches a single straight line. 25. The ellipse must pass through (0, 0) ⇒ c 8 10 ( 1, 2) lies on the ellipse ⇒ x ; 2 2 0 c2 c a 1 and e a2 9 1 2 5 a e 10 3⇒c c2 a2 9 1 ⇒ x2 8 2 ⇒ e2 4)2 2) y2 2(x 2 x 8 y 8 ae (y 4x 2 y 8 1 3a 2 1 a e 2 2x 2 4 and 2. 2 2(x 8⇒ 8 ⇒ x2 2⇒e (x 2 3⇒c 2⇒c 4) 0)2 2b 4 ⇒ the equation is 4x 2 and b 4x 2 3 c a 3 and e a 0; the point 8. The ellipse is tangent to the x-axis ⇒ its center is on the y-axis, so a ⇒c 16 36) 1 y2 4y (y 2) 4 ⇒ 4x 2 4 2 4 1⇒a 3⇒e c a y2 a2 b2 3 2 3 2 . 4y 2)2 (y 2 and b standard symbols) ⇒ c2 2 0 2 1 4 e2 4) 2 a e 0 3⇒a 1) and e y 4 4y 2 e2a 2 ⇒ b2 2 92 (y 4 9 2x; 5; asymptotes are y 2PD ⇒ Then PF (x 1 60y b2 ⇒ b2 a2 10 1 21. Focus (4, 0) and directrix x ⇒ 6) 36 thus, 8 x 8 b2 1 ⇒ b2 ae e2 1) 1 2 2x 2 y 2 20. Foci ( 3, 0) and e ⇒ 2x 0)2 2 6y 8x 2 y 3 of this hyperbola are y 3 ⇒ b2 3a ⇒a 5y 2 2. 1 4 x 2 3 y 2 3)2 y2 2 and 12 2 3 ⇒ x2 1 ae 4⇒e (y 4 x2 (y 2 2 19. Vertices (0, ⇒ y2 ⇒ y2 y2 ⇒ 4(x 2 2)2 4x 2x (y 1 ⇒ e2 2 (x 1)2 (x ⇒ c2 1 10); directrices are y ⇒c 8 5; asymptotes are y 10 c a F(0, 3x 2 24. c 2 y2 8 b2 16 ⇒ 2x 2 ⇒e 2; 4); directrices are 2 ⇒c 4 c a 4⇒e 10, 0); directrices are x 18. 8y 2 4 23. 2 10 c a F( 4x 2 a2 ⇒e y2 ⇒ 5 ; 4 c a 2)2 ⇒ 4x 2 x; F(0, 2PD ⇒ ⇒ x2 5⇒e 8 1 2 ⇒2 2 e ⇒ (x 1. 1 8 2y 2 17. 8x 2 y2 20 ⇒ x2 y2 8 8⇒ 0 x2 36 1 9 1 ae ⇒2 2 e Then PF 3 x; F( 5, 0); directrices are 4 asymptotes are y y 81) 180 or 1 ⇒c 2 22. Focus ( 2, 0) and directrix x 16 . 5 x2 ⇒c 18x y2 9 16 asymptotes are y x 9 a e 9y 2 x2 16 2 x 3 y2 2 42 (x 9 144 ⇒ 16y 2 9) y2 16 y2 ⇒c 4 (x 9 2 4)2 (x 0. Next, 4 1 (now using the 4 1 3 0 Appendix A5.2 26. We first prove a result which we will use: let m1 and m2 be b2 . Since tan cy0 Similarly, tan the slopes of two nonparallel, nonperpendicular lines. Let are both less than 90 , we have tan , and 463 and . be the acute angle between the lines. Then tan y m2 m1 m1m2 1 let 1 and . (To see this result for positive-slope lines, α P(x0, y0) β be the angle of inclination of the line with slope m1, 2 F1(–c, 0) be the angle of inclination of the line with slope m2. Assume m1 m2. Then 1 Then tan tan ( 2 tan since m1 1 1 ) and m2 2 and we have tan 1 tan tan tan 1 1 2 tan 2 1 2 m1 . m2 1 m1 m2 , 27. To prove the reflective property for hyperbolas: b 2x 2 .) 2 Now we prove the reflective property of ellipses (see the x F2(c, 0) a 2y 2 2b 2x a 2b 2 2a 2yy 0 2 y accompanying figure): 2 2b x 2 2a yy Let P(x0, y0) be a point of tangency (see the accompanying b 2x . a2y ⇒y 0 figure). The slope from P to F( c, 0) is and from P to F2(c, 0) it is Let P(x0, y0) be any point on the ellipse a a b x0 2 x02 a 2y0 F2( c, 0) be the foci. Then mPF mPF 2 y0 x0 c . Let and . Let F1(c, 0) and y0 1 x0 c and F1PA will show that tan tan . be the angles between the y0 a 2y0 tan x0 y F1(–c, 0) c 2 b x0y0 1 a 2y0(x0 b 2x02 2 c) b 2x0c a 2y02 a y0x0 a y0c b 2x0y0 b 2x0c (b 2x02 a 2y02) 2 2 2 a y0 c (a b 2x0c a 2b 2 2 a y0c 2 c x0y0 b 2)x0y0 b2 . cy0 x0 and define the angles tangent line and PF1 and PF2 respectively. Then b 2x0 y0 c y0 x0 c . Let the tangent through P meet the x-axis in point A, 2 bx0 ⇒ y (x0) bx a 2y P(x0, y0) αβ A x F2(c, 0) and F2 PA . We 464 Appendix A5.3 10. 3x 2 6xy 3y 2 4x 5y 12 ⇒ B 2 4AC 62 4(3)(3) 0 ⇒ Parabola 27. continued From the preliminary result in Exercise 26, x0b 2 x0 c x0b 2 1 y0 y0a 2 x0 x0 b 2 c y02a 2 x0y0a 2 y0 a 2 c x0y0b 2 2 2 21xy 14. 25x ⇒ B 2 4AC 4y 212 2 x0b 2c x0y0c 2 x0 y0 1 and y0a 2 c tan x0 x0b c y0a 2 are acute angles, we have tangent to the hyperbola at P bisects A 17. cot 2 C therefore x APB. Since APC 2 2 ⇒ APC, and the m APB 2 12 2 A m LAPB 2 2 ⇒x 2 2 ⇒ 1. x 3xy y ⇒ B 2 4AC 2 2 2 x 2 2 2 2 2 2 2 2 y 2 2 2 y 4 ⇒ Hyperbola 2 y sin, y y ,y y cos 2 0⇒2 2 x y 2 y 1 2 2 ; ⇒ 4 x sin 2 x 2 2 x 2 y cos y 2 y ; 2 x 2 2 y 2 y 1 12 12 12 12 12 xy y x y x xy 2 2 2 2 2 12 3 12 y 1⇒ x2 y 1 ⇒ 3x 2 y 2 2 2 2 2 ⇒x x0 ( 3)2 4(1)(1) 5 0 ⇒ Hyperbola 2 18xy 27y 5x 7y 4 2. 3x ⇒ B 2 4AC ( 18)2 4(3)(27) 0 ⇒ Parabola 3. 3x 2 7xy ⇒ B 2 4AC ⇒ Ellipse 17y 2 1 ( 7)2 4(3) 17 4. 2x 2 ⇒ B2 2y 2 x y 0 ( 15)2 4(2)(2) 15xy 4AC 1 2 s Appendix A5.3 (pp. 612–618) 2⇒x 2 4 x sin x 2 x x cos x 2 2 1 therefore x hyperbola are perpendicular. 2 B 90 , the tangents to the ellipse and ⇒ 2 2 y ,y 2 C 0 ⇒ Ellipse y sin , y y 12 y 2 ⇒x 180 and m APC 2 2 2 18. cot 2 m APC 2 x x 2 0⇒2 x cos 2 ⇒x APB are a linear pair, so that 2 0 ⇒ Hyperbola 41 435x 9y 72 0 4(3)(12) 0 ⇒ Parabola 0 1 B tan , and . 0 0 ⇒ Ellipse 3 350x 0 4(25)(4) 16. 3x 2 12xy 12y 2 ⇒ B 2 4AC 122 b2 . Since tan y0c 2 28. The tangent to the ellipse of P bisects and 0.01 2 x0b 2 y0 0 ⇒ Hyperbola 3xy 2y 17y 2 0 15. 6x ⇒ B 2 4AC 32 4(6)(2) 39 b2 . In a similar manner, y0c y0a 2c 7 4(2)(3) 13. x 2 3xy 3y 2 6y 7 ⇒ B 2 4AC ( 3)2 4(1)(3) c x02b 2 a 2b 2 14y 1 4(3)(2) 1 12. 2x 2 4.9xy 3y 2 4x ⇒ B 2 4AC ( 4.9)2 ⇒ Hyperbola y0 y0a 2 tan 11. 3x 2 5xy 2y 2 7x ⇒ B 2 4AC ( 5)2 0.477 ⇒ Ellipse A 19. cot 2 0 C 5. x 2 2xy y 2 2x y 2 ⇒ B 2 4AC 22 4(1)(1) 0 ⇒ Ellipse 0 0 ⇒ Parabola ⇒3 6. 2x 2 y 2 4xy 2x 3y 6 ⇒ B 2 4AC 42 4(2)( 1) 24 0 ⇒ Hyperbola 7. x 2 4xy 4y 2 3x 6 ⇒ B 2 4AC 42 4(1)(4) 8. x 2 y 2 3x ⇒ B 2 4AC 2y 02 9. xy y 2 3x ⇒ Hyperbola 5 ⇒ B2 10 4(1)(1) 4AC 3 ⇒x 2 3 2 4 12 3 3 2 1 0 3 1 2 ⇒ 4x 3x 2 16y ⇒ 3 x sin 1 x 2 1 y 2 x 2 ⇒2 3 3 2 y 2 1 y 2 8 0 ⇒ Ellipse (circle) 4(0)(1) 1 y ,y 2 x 1 x 2 0 ⇒ Parabola 1 y sin , y x cos x 2 1 23 therefore x 1 3 B 1 x 2 3 2 2 y 8 3 2 y 3 2 x 0 0 ⇒ Parabola y 1 y 2 6 ; y cos 465 Appendix A5.3 A 20. cot 2 C therefore x 3 3 12 2 52 y 2 A 21. cot 2 2 2 2 x 2 2 A 22. cot 2 y 3 3 x 2 2 x y A 23. cot 2 2 1 y 3 2 2 2 2 x sin 3 3 y 2 2 2 2 2x 2 2 2 2 2 2 x 1 y 2 2 8 2 2 2 ⇒ 4x 2 + 2y 2 2 y 2 y 2 2 x 2 1 2 2 2 2 2 2 y 2 y 2 0 ⇒ hyperbola 0⇒2 ⇒ 2 4 y sin , 2 ⇒x 2 2 x 2 y, 2 2 2 2 x 2 2 y 2 2 x 2 y 2 y 19 19 ⇒ Ellipse 4 C 3 B 3 2 12 y 2 2 3 2 3y ⇒2 3 3 y sin , y 1 y ,y 2 x x 1 43 1 x 2 ⇒ 5x ( 1) x cos 3 2 1 x 2 4 3 3 2 3 2 x sin 6 y cos y 1 y 2 x ⇒ 1 x 2 3 2 2 y 7 7 ⇒ Hyperbola 2 y cos 27. cot 2 A C 14 2 16 B 3 ⇒ cos 2 4 3 (if we 5 choose 2 in Quadrant I); 2 2 y 8 2 2 x y 2 2 y cos or sin 2 4y 1 thus sin 0 0⇒x ; y x 2 A ; y 2 x y 0 y x 2y ⇒ x sin x 2 2 2 2 y cos y 2 x therefore x 2 x x 2 2 2 ; ⇒3 0⇒2 2 y ,y 2 2 8 2 y sin , y 2 2 2 2 ⇒x 22 2 3 2 x x sin ; y cos 1 2 x 2 2 26. cot 2 3 3x 12 y 2 x 2 3 y cos 3 1 y 2 x 1 2 2 x cos 2 2 2x C 4 1 x cos 2 2 ⇒ 3 3 3 y ,y B 2 y 2x B x sin y ⇒2 y sin , y C therefore x 2 2 ⇒ 2 x 2 x 2 therefore x 1 ⇒ Parallel horizontal lines 2 2 2 2 12 y 2 A 25. cot 2 2 y ,y 2 2 y ⇒3 1 y x 2 2 y cos 2 1 ⇒ Parallel horizontal lines 2 3 2 x sin 2 3 2 1 2 ⇒2 2 x cos ⇒3 x 2 2 y 2 1 x 2 ⇒x 2 4 ⇒x ; 0⇒2 y sin , y y x 12 2 2 2 B ⇒x ⇒ 4y 2 2 2 C therefore x 2 ⇒ 2 y sin , y x 2⇒y 2 2 2 0⇒ x 2 ⇒ ellipse 2 0⇒2 y ,y 2 x 2 y x 2 2 2 x 2 y 2 2 2 2 2 ⇒ y 0 1 2 2 x 3 2 0 x cos 2 5y 1 x cos 2 ⇒ 2y 2 2 2 therefore x ⇒x 1 x 2 C B 1 1 B 2 y cos y 1 y 2 x A 24. cot 2 ; 2 y 1⇒x C therefore x ⇒ 2 3 2 6 x sin 2 3 3 1 2 ⇒x 1 x 2 ⇒ 3 y sin a, y 12 y 2 2x ⇒2 3 1 y ,y 2 x ⇒x 1 3 x 2 2 ⇒ 2 x cos 3 ⇒x ⇒ 1 B 0 ⇒ Parabola 1 cos 2 2 cos 2 2 2 and cos 5 1 1 (3/5) 2 (3/5) 2 1 5 1 and 5 2 5 y ; 466 Appendix A5.3 A 28. cot 2 C 4 1 B 3 ⇒ cos 2 4 4 3 (if we 5 1 1 cos 2 2 1 (3/5) 2 cos 2 2 1 or sin 1 ⇒ sin C 1 ⇒2 2 0.23, cos ⇒ 0.88x E 3.12y 13.28 0.88, B 1.20, and F 0.74x 1.20y 0.00, C ⇒ sin B 1 ⇒2 5 0.10, cos 0.00, C 12 2 0, an F 11.31 ⇒ 1 ,C 2 2.05, 2 3.05y 2 2.99x 0.30, and 0.30y 7 0, a hyperbola 4 31. tan 2 1 ⇒ sin C 4 ⇒2 3 0.45, cos 5.00, D ⇒ 5.00y 2 32. tan 2 0, and F 0 or y 0.32, cos 20.00, D ⇒ 20.00y 26.57 0.00, B 0.00, 5 1.00, parallel lines 3 ⇒2 4 12 18 2 53.13 ⇒ 0.89; then A 0, E 5 ⇒ sin C 4 36.87 ⇒ 0.95; then A 0, E 49 18.48, 18.48x 7.65y 86 0, 0, and F 18.43 0.00, B 2 B 2 y 2 1⇒x 1 ,B 2 2 2 2 1 ,F 2 2 y 0.00, 0, 1 2 2 1 (see part (a) above), 2 1 0, F a⇒ x 2 2 12 y 2 a 2a 0 we have 37. The one curve that meets all three of the stated criteria is the ellipse x 2 4xy 5y 2 1 0. The reasoning: The symmetry about the origin means that ( x, y) lies on the graph whenever (x, y) does. Adding Ax 2 Bxy Cy 2 Dx Ey F 0 and A( x)2 B( x)( y) C( y)2 D( x) E( y) F 0 and dividing by the result by 2produces the equivalent equation Ax 2 Bxy Cy 2 F 0. Substituting x 1, y 0 (because the point (1, 0) lies on the curve) shows further that A F. Then Fx 2 Bxy Cy 2 F 0. By implicit differentiation, 2Fx By Bxy 2Cyy 0, so substituting x 2, y 1, and y 0 (from Property 3) gives 4F B 0 ⇒B 4F ⇒ the conic is Fx 2 4Fxy Cy 2 F 0. Now substituting x 2 and y 1 again gives 4F 8F C F 0 ⇒ C 5F ⇒ the equation is now Fx 2 4Fxy 5Fy2 F 0. Finally, dividing through by F gives the equation x 2 4xy 5y 2 1 0. 49 38. If A 2 2 36. Yes, the graph is a hyperbola: with AC 4AC 0 and B 2 4AC 0. 7 ⇒ 2.05x E ⇒x 5.65 2.99, E 12 y 2 ⇒x 3 3 0.995; then A 3.05, D 10.45y sin 45 cos 45 D 1 ( 3) 10.45, D –86 cos 45 sin 45 (b) A 2 2 0.92; 0.00, C 7.65, and F 35. (a) A ellipse 30. tan 2 0.55, B 22.5 an ellipse 0.97; then A 2 45 ⇒ 0.38, cos ⇒ 0.55x 26.57 ⇒ 0.74, E 2 1⇒2 9 then A and cos 5 5 3 3.12, D 2 5 2 and cos 1 1 (3/5) 2 1 5 29. tan 2 2 ⇒ sin choose 2 in Quadrant II); thus sin 7 34. tan 2 C, then B B cos 2 (C A) sin 2 0, parallel lines B cos 2 . 5 33. tan 2 3 ⇒ sin C 2 5⇒2 0.63, cos 0.05, D ⇒ 5.05x 2 a hyperbola 0.05y 78.69 ⇒ 0.77; then A 5.07, E 2 5.07x 39.35 5.05, B 6.19, and F 6.19y 1 Then 0.00, 1 0, 4 ⇒2 2 ⇒B B cos 2 0 so the xy-term is eliminated. 39. 90 ⇒ x x cos 90 y sin 90 and y x sin 90 y cos 90 x x2 b2 y2 (b) 2 a (a) (c) x 2 y2 a2 x2 b2 y y 1 1 2 a2 (d) y mx ⇒ x (e) y mx m( y ) ⇒ y b⇒x m( y ) 1 x m b⇒y 1 x m b m 467 Appendix A6 180 ⇒ x 40. and y x cos 180 x sin 180 2 y sin 180 y cos 180 43. (a) B 2 4AC 42 4(1)(4) 0, so the discriminant indicates that this conic is a parabola. x y (b) The left-hand side of x 2 4xy 4y 2 6x 12y 9 0 factors as a perfect square: (x 2y 3)2 0 ⇒ x 2y 3 0 ⇒ 2y x 3; thus the curve is a degenerate parabola (i.e., a straight line). 2 (a) x a2 y b2 1 (b) x2 a2 y2 b2 1 2 (c) x 2 y (d) y mx ⇒ (e) y mx 41. (a) B 2 b⇒ 2x x y 1 1 ⇒ b⇒y mx b 1 ⇒ hyperbola 4(0)(0) dy dx mx m( x ) 0 ⇒ y(x y 2x (c) y m( x ) ⇒ y y 4AC (b) xy 44. (a) B 2 4AC 62 4(9)(1) 0, so the discriminant indicates that this conic is a parabola. a2 2 2 (x 1) 2x 2x ⇒ y 1) x 1 2, the slope of y ⇒ 1)2 (x 2 2 ⇒ (x 1 A 1)2 Since B2 4 3 or x 1; x 3⇒y 3 ⇒ (3, 2 AC 4AC B 2 Cy 2 1 1 ⇒ the area is C 2 1 C (because B ⇒x 1 and A ⇒ the semi-axes are 2x 0 45. Assume the ellipse has been rotated to eliminate the xy-term ⇒ the new equation is A x and we want 1 dy dx (b) The left-hand side of 9x 2 6xy y 2 12x 4y 4 0 factors as a perfect square: (3x y 2)2 0 ⇒ 3x y 2 ⇒y 3x 2; thus the curve is a degenerate parabola (i.e., a straight line). . 4A C 4A C 4A C 2 0) we find that the area is 4AC 3) is a B2 as claimed. point on the hyperbola where the line with 2 is normal ⇒ the line is slope m y x 3 46. (a) A C (A cos2 B cos sin C sin2 ) 2 2 (A sin B cos sin C cos ) A(cos2 sin2 ) C(sin2 cos2 ) A C 2(x 3) or y 1⇒y 2x 1 ⇒ ( 1, 3; 1) is a point on the hyperbola where the line with slope m 2x E 2 is normal ⇒ the line is y E sin )2 ( D sin (b) D 2 E 2 (D cos 2 cos ) D 2 cos2 2DE cos sin E2 sin2 D 2 sin2 2DE sin cos E 2 cos2 2 2 2 2 2 D (cos sin ) E (sin cos2 ) 2 2 D E s Appendix A6 1 2(x 1) or y 3 (pp. 618–627) 1. sinh x 1 [ 9.4, 9.4] by [ 6.1, 6.1] 42. (a) False: let A ⇒ parabola C 1, B 2 ⇒ B2 4AC 0 (b) False: see part (a) above (c) True: AC 0 ⇒ ⇒ hyperbola 4AC 0 ⇒ B2 4AC 0 3 4 5 4 3 ⇒ cosh x 1 sinh2 x 4 9 25 5 sinh x , tanh x 16 4 16 cosh x 3 , coth x 5 4 , and csch x 5 1 sin x 1 tanh x 4 3 1 5 , sech x 3 32 4 1 cosh x 468 Appendix A6 4 ⇒ cosh x 3 2. sinh x 16 9 1 25 9 tanh x 17 ,x 15 3. cosh x 17 2 15 289 225 17 , sech x 8 169 25 13 , sech x 12 2 dy dt 16. y x 1 (4e 0) 4 1 (4) 4 dy dx tanh t 1 t t 2 tanh t 2 tanh t 1 2 )]( t 1 t 1 e 2t tanh ln x x2 1 sinh x 5 12 e ln x 2 ln x e 2 ln x 1 e ln x ln x 2 lnx e e 7. cosh 5x sinh 5x 1 t cosh z sinh z coth z ln (cosh z) ⇒ dy dz sinh z cosh z tanh z (sech )(1 ln sech ) sech tanh sech ( sech sech (sech ) tanh )(1 tanh (sech ln sech ) tanh )(1 (sech 1 x x 20. y 8. cosh 3x sinh 3x 9. (sinh x ⇒ cosh x)4 e3x sinh x) ln (cosh2 x e 5x e 5x e 3x 2 e 3x e 3x 2 ex e x ex ln (cosh x ln 1 x) (b) cosh 2x cosh (x x) cosh2 x sinh2 x e x4 2 (csch )(1 dy d ln sech )] csch coth csch csch ln csch )( csch coth (1 coth ) ln csch )(csch coth ) 3x (e x)4 (csch 21. y ⇒ cosh x sinh x coth )(1 (csch e 4x 0 coth )(ln csch ) sinh x sinh x 1 ln csch ) 1 tanh2 x 2 1 sinh x (2 tanh x)(sech2 x) 2 cosh x ln cosh x dy dx tanh x (tanh x)(sech2 x) (tanh x)(tanh2 x) cosh x cosh x (1 ln csch ) (csch ) (1 sinh x) sinh x cosh x tanh )(ln sech ) e 5x e 2 2 sinh2 x) 11. (a) sinh 2x sinh (x 2 sinh x cosh x 5x e 2 tanh )[1 ln sech ) 2 2 e 5x ) ) dy dz dy ⇒ d x4 1 2x 2 2 10. ln (cosh x e 1/2 1 (2t)(tanh t 19. y 2 1)](2) dy dt 18. y e ln x x x 3 (tanh t 1/2)(t ln (sinh z) ⇒ 1 tanh x 2 1 x2 ⇒ )(t 2) (sech 6. sinh (2 ln x) 2 cosh 1 [cosh (2x 2 1 1/2 t (2t 1/2) 2 t [sech2(t 1/2)] sech2 In Exercises 5–10, graphical support may consist of showing that the graph of the original expression minus the simplified one is the line y 0. 2 e 2t 1/2 tanh t 1/2 t 1 5 , and csch x 13 (e x ) 1 1 3 x 3 6 cosh x e 17. y 12 , coth x 13 1 cosh x 5. 2 cosh (ln x) ) (e x 1) [sech2 (t 15 8 12 , 5 12 5 13 5 x2 e t 1 sinh x cosh2 x ex 2 e 1) ⇒ t tanh 2 1 tanh x 15 , and csch x 17 144 25 1 ⇒ 8 , 15 8 , coth x 17 0 ⇒ sinh x sinh x cosh x tanh x 64 225 15. y 1 sech 1 cosh x 13 ,x 5 4. cosh x cosh2 x 1 8 15 17 15 (e x ) 1 sinh (2x 2 cosh (2x x2 e 2 dy x ⇒ dx 3 6 sinh 14. y 3 , 5 1 cosh x 0 ⇒ sinh x sinh x cosh x tanh x x e 13. y 3 4 1 ex sinh2 x 1 (2e x )(2e x ) 4 4 , 5 1 sinh x and csch x 12. cosh2 x 1x (e 4 5 , sech x 4 1 tanh x coth x 42 3 1 5 , 3 4 3 5 3 sinh x cosh x sinh2 x 1 tanh3 x (tanh x)(1 sech2 x) ) 469 Appendix A6 22. y ⇒ 1 coth2 x 2 cosh x 1 (2 coth x)( csch2 x) sinh x 2 dy dv 2 coth x (coth x)(csch x) 2 23. y (x (x 2 (4x 2 24. y (4x 1) dy ⇒ dx (x 2 (2x) 2x csch x) 1) (4x 2 (4x 1 ln x e e ln 2x ⇒ e ln 2x 4x 1) 2 4x 2 4x 1 dy dx 1 x 33. y 4 1 1 25. y ⇒ sinh x 1 sinh (x 1/2 1 x 1/2 2 dy dx ⇒ 1 (x 1/2 )2 1 ) 2 x(1 x 1 cosh (2 (2) dy ⇒ dx x 1) 1 (x 2 1) 1 x 1 4x 4x 27. y 7x (1 dy ⇒ d 1 1 (1 ) 1 sinh 36. y (sec x)(tan x) ⇒ 2 ( dy d 2 ( 2 2 2 2 ( 1) 37. (a) If y 1 2) 1 (2 1)2 ( 1 t) coth 1 1) 1 2) tanh 1 ( 1 ( 1) 1 2) tanh (2 2) tanh (2 1 2 ) tanh ( 1) tan dy dx (sec x)(tan x) (sinh x) (1 t (1 t) coth x 2 dy dx sech x, which verifies the formula. 1 sin (tanh x) sech2 x 29. y 1 sec x, 0 C, then cosh x cosh2 x cosh x sinh2 x (b) If y 1 sec2 x (sec x)(tan x) tan x 1 22 1 1 (tan x)2 sec x sec x sec x sec x (sec x) ⇒ 1 cosh (2 ) sec2 x 1 tanh x ln 2 2 dy dx 1 ( 1) tanh 2 1 2 sec2 x sec x tan2 x 28. y (ln 2)2 dy d (tan x) ⇒ sec2 x 1 1 sech 12 2 1 2⇒ 1 sec2 x 1 x2 ln 2 12 2 3 1 x 2 1 2 1 3 ) tanh x x 1 2 ln (2) csch 35. y 1 1 1 2 1 2 1 1 2 1 ( 2x) sech 2 dy d 34. y 1/2 sech x) 1 x 11 csch 1)1/2 ) (2(x x) x2 1/2 1)1/2]2 [2(x 1 cosh 2 1 ln (1) 26. y 1 x2 1 x 1 (1 2 x 1 x1 2 sech 1 x 2)1/2 ln 1 x x x (1 1 x 1 1 x 1 (1 1 x x 2 sech x 2)1/2 sech ln x (1) sech x2 sech x2 1 ln x dy 2x ⇒ dx 1 x1 1 x2 2 1) 2 32. y x x x2 1 2 e ln x 2x 1) 2 1 x 2 (x 1 1 1 2 1 x sech 1 2 (coth x)(1 1) csch (ln 2x) 2 dy dx coth x 1) sech (ln x) 2 1) xx ⇒ 3 (coth x)(coth x) 2 cos 1x 31. y ln sinh x (t 1/2 sech2 x sech x 2 tanh x 1 ) C, then dy dx sech x, which verifies the formula. ⇒ 1/2 dy dt (1 1 (1/2)t 1 (t 1/2 )2 t) coth 1 ( 1) coth 1 38. If y dy dx t t 2 (t 1/2 ) x2 sech 1 x 2 x sech x sech 30. y ⇒ (1 dy dt 1 t 2) coth (1 2t coth 1 t 1 x x2 1 C, then x 1 2 x 1 x2 2x 41 x2 x, which verifies the formula. t 1 t 2) 1 1 1 2 1 2 t2 ( 2t) coth 1 t 39. If y dy dx x2 1 2 x coth coth 1 x 1 x 2 1 x x 2 2 which verifies the formula. C, then 1 1 x2 1 2 x coth 1 x, 470 Appendix A6 40. If y 1 x tanh dy then dx 1 ln (1 2 x 1 tanh x x2) 1 2x 2 1 x2 1 x x2 1 1 tanh sech x, 2x and du 2 dx. x dx 5 sinh x 2 x 6 cosh 2 ln 3 dx C x 5 12 sinh u 2 sech 51. Let u sinh x, du cosh x dx, the lower limit is e ln 2 e ln 4 4 cosh (3x ln 2) dx 4 sinh (3x 3 ln 2) 4 sinh u 3 C 52. Let u 1 4 4 ln 4 15/8 cosh x dx sinh x 15 3 ln ln 8 4 cosh 2x, du 15 . 8 2 2 1 du u 15 4 ln 8 3 ln 2 15/8 3/4 ln u C 1 x and du dx. 7 7 x sinh u tanh dx 7 du 7 cosh u x 7 ln cosh C1 7 x/7 x/7 e e 7 ln C1 2 e ln 4 coth x dx 3 and the upper 4 2 2 ln 4 ln 2 1 2 2 ln 2 e C 3 dx. 4 cosh u du 3 C C C ln 2 and du C csch u coth u du csch (ln t) limit is sinh (ln 4) 3x t dt . t C sinh (ln 2) 12 cosh u du ln 3 5 cosh C csch (ln t) coth (ln t) dt t C 1 dx. 2 ln 3 and du x 12 sinh 2 44. Let u 5 cosh u 2 sech u tanh u du ln t and du csch u 5 sinh u du 43. Let u cosh 2x 2 C 1 dx. 5 x and du 5 42. Let u 50. Let u cosh u 2 dt . t 2 t 2( sech u) 1 sinh u du 2 sinh 2x dx t tanh t which verifies the formula. 41. Let u dt t 1/2 and du t 49. Let u C, 3/4 ln 5 2 0.916 2 sinh (2x) dx, the lower limit is 45. Let u 7 ln e x/7 e 7 ln e x/7 e x/7 x/7 3 coth d 3 ln sinh C1 3 3 ln e / 3 e 3 ln e / 3 e /3 2 sech x tanh x 48. Let u 1 and du 2 x 1 dx 2 1 C 2 (5 csch2 (5 coth u coth (5 C C e /3 e ln 4 ln 2 C1 /3 ln 4 (e 2 e ln 2 ln 2 1 32 3 32 ln 2 54. tanh u 0 sinh 2 0 (1 2 ln 2 d C 2 ln 2 2 e e ln 4 3 32 ln 4 1 8 0 )d 1 2 2 ln 2 e e 2 2 2 ln 2 d 2 2 0 ln 2 e 4e 2 ln 2 2 dx. ( coth u) 2 ln 4 ln2 4e ln 2 C d 0.787 ln 2 dx. 0.377 ln 2 ln 4 2 1 8 C1 e 2 e 2 1) d 1 du u 1 17 ln 2 8 1 e 2e 2 2 C 2 d 2 ln 2 e C1 2 3 ln 2 csch2 u du x) dx ln 2 2e cosh ln 4 sech u du x) and du x) 3 ln sinh u 3 ln /3 47. Let u 53. 17/8 1 sinh 2x dx 0 2 cosh 2x 1 17 ln ln 1 2 8 ln 2 . 2 ln 2 tanh 2x dx 1 17/8 ln u 2 1 3 cosh u 3 du sinh u 3 ln 2 C ln 4 e 17 . 8 2 C1 e ln 4 cosh (ln 4) 1 4 4 0 d 1 and the upper limit is C1 cosh (2 ln 2) 7 ln 2 and du 46. Let u cosh 0 7 ln cosh u ln 2 0 0 e 2 1 4 1 ln 4 3 4 0.636 Appendix A6 /4 55. 1 sinh u e1 1 1 e e e d sinh (1) 1 e ln 10 e 1 e 1 e 4 sinh2 ln 10 2 (sinh 0 0)] sec d , the 9.9 e e 1 d 2 2(cosh 1 2 2 1 1.086, where u 2 2 ln 10 2 ln 10 ln e 2 1 62. 1 2 1 sin , du cos ln d, 2 1, so (cosh2 x 0 sinh2 x) dx 2 ln 1 8 cosh ln 2 ln 2 sinh u 0 cosh u du 0 e ln 2 sinh (0) x dx 16 16(sinh 2 e2 1 dt, the lower limit is t 2 e cosh u du e 2 2 e e 3 e e ln ln 3 e ln 3 ln 3 ) x 1 3 3 ln 63. 199 16 sinh u 199 (1 ln 1 2 tanh x tanh2 x) dx (2 0 2 tanh x)2 dx (1 0 2 tanh x sech2 x) dx 199 ln 199 2 ln (cosh x) tanh x 0 39.227, where dx 2 ln 2 cosh x dx 2 0 1 (cosh x 2 ln 2 1 [(sinh 0 0) 2 1 e (0 0) 2 1 2 1 1 2 (1/2) 2 1 4 2 ln 2 , the lower limit is 2 ln 199 2 ln [cosh (ln 199 2 ln 199)] tanh (ln 199) x 2 1 and the upper limit is 0 3 1 2 2 ln u 3 1 3 ln 3 e 3 e e ln 1 1 2x 1 1/2 x dx 2 x 1/2, du ln 3 1 1 e ln 3 e 0 2 8(e 2 3 3 ln sinh 1) e e 0 ln t, du 1 3 e ln 3 2 tanh x ln 3 e ln ln 2 e 2 2 x 16 3 e ln 3 0 and the upper limit is ln 2 1 ln 3 sech2 x dx e ln 0 and the upper limit is 3 , where u 4 2 59. 2 ln 10 5.295 sinh2 x 1 dx 10 3 2 cosh (ln t) dt t 2 58. 2 e 1 sinh (ln 2) 4 sinh u du 0 cosh 0) the lower limit is sin 0 57. ln 10 1 10 1 1 2 cosh u 0 sin e 2 2 sinh (sin ) cos 0 ln 10 0 ln 10) 0 56. dx x 61. cosh2 x /2 1 2 sinh x 1 and the upper limit is 4 cosh x 2 1) dx 1. 4 4 0 (cosh x 0 e ln 10 2 tan , du ln 10 x dx 2 2[(sinh (ln 10) e 2 2.350, where u lower limit is tan 0 sinh ( 1) 2 1 60. cosh u du 1 e1 2 tan 1 cosh (tan ) sec2 /4 471 4 2. e ln 199 e ln 199 2 0 cosh x 1 dx ln 2 2 1 0 1) dx sinh x x 2 ln 2 e ln 2 2 e ln 199 e ln 199 ( 199 ln 199 ln 2 1 ln 2 2 3 8 ln 2 0.722 ln ln 199 ln 2)] ln 2 3 8 199 e ln 199 ln ln 199 (sinh ( ln 2) ln 2 e ln ln 1.214 199 10,000 199 1/ 199)2 4 2 1/199 4 99 100 199 199 199 199 1 1 2 ln 199 100 1/ 199 1/ 199 99 100 472 Appendix A6 1 cosh 2x ⇒ y 2 64. (a) y ln 5 ln (sinh 2x)2 dx 1 0 sinh ax b a 0 ln 5 0 ak sin kt bk cos kt ak 2 cos kt k 2(a cos kt 1 sinh2 ax bk 2 sin kt k 2 s(t) ⇒ acceleration is sin kt) k 2 implies cosh2 ax that the acceleration is directed toward the origin. b 0 cosh ax dx sinh ab a f (x) (b) s(t) f ( x) ds dt d 2s ⇒2 dt f (x) and O(x) a cosh kt ⇒ f ( x) . Then 2 2 f (x) f ( x) f (x) f ( x) 2 f (x) E(x) O(x) 2 2 2 f ( x) f ( ( x)) f (x) f ( x) f (x). Also, E( x) 2 2 f ( x) f ( ( x)) E(x) ⇒ E(x) is even, and O( x) 2 f (x) f ( x) O(x) ⇒ O(x) is odd. 2 65. (a) Let E(x) b sin kt proportional to s. The negative constant ( y )2 cosh2 ax dx 0 ds dt d 2s ⇒2 dt cosh 2x dx 2x a cos kt ⇒ 5 0 1 e 2x e 2 2 1 ln 5 sinh 2x 2 0 1 1 6 5 4 5 5 1 (b) y cosh ax ⇒ 1 a b ⇒L 68. (a) s(t) sinh 2x ⇒ L b sinh kt ak sinh kt bk cosh kt ak 2 cosh kt k 2 (a cosh kt bk 2 sinh kt k 2 s(t) ⇒ acceleration is sinh kt) proportional to s. The positive constant k 2 implies that the acceleration is directed away from the origin. Consequently, f (x) can be written as a sum of an even 69. and an odd function. ex (b) Even part: odd part: e x e 2 e x 2 f (x) 2 x 1 ⇒y f (x) f (x) f (x) 2 0⇒C 70. y f ( x) 2 4 cosh f (x) f (x) mg sech2 k g sech2 Then m dv dt mg kv 2 mg tanh2 mg sech2 4 4 sech C; x 1 dy 2 dx 1 and (x) sinh2 1 x 2 sinh ln 81 4 4 gk t. m kv 2 are equal to the same quantity, so the differential equation is satisfied. Furthermore, the initial condition is satisfied because v(0) mg tanh 0 k 0. 2 ln 36 4 ln 6 4 4 ln 6 cosh2 ln 81 4 x ; 4 dy 2 dx dx ln 16 ln 16 1 cosh x dx 2 ln 81 2 2 sinh ln 16 2 2 sinh 4 [ln (9 4)2 2 gk t m x 4 x ln 81 2 ln 16 4 [ln (81 16) 2 gk t m x2 1 2y 1 x dx 4 cosh2 ln 16 gk t and m mg tanh k k dv and mg dt gk t m f (x) gk m 0 ln 16 gk t. m mg sech2 mg 1 Thus, m f (x) 2 dv dt 8 f ( x) 2 f (x) 2 67. Note that 1 dx ln 81 2 f (x) x2 (x) the surface area is S ln 81 f ( x) x2 1 x ⇒1 4 0 (b) If f is odd, then f (x) x dx 0⇒y sech y f (x) 1 x2 1 x2 1 x sinh x 2 x 2 1 x x 2 f ( x) 1 ⇒y cosh x 66. (a) If f is even, then f (x) dy dx 2 sinh (ln 9) (e ln9 1 9 9 80 9 e 2 sinh (ln 4)] ln 9 ) 4 (e ln 4 e ln 4 )] 1 4 15 4 320 135 36 16 ln 6 455 9 248.889 Appendix A6 a cosh (x/a) ⇒ y 71. y ⇒y sinh (x/a) (1/a) cosh (x/a) (1/a) sinh2 (x/a) 1 Also, y (0) sinh (0) cosh2 (x/a) (1/a) (y )2. (1/a) 1 0 and y(0) a cosh (0) a. 72. (a) Let the point located at (cosh x, 0) be called T. Then A(u) area of the triangle the curve y 1 from A to T cosh u 1 cosh u sinh u 2 ⇒ A(u) (b) A(u) x2 OTP minus the area under 1 cosh u sinh u 2 x2 1 cosh u 1 x2 1 dx. 1 dx ⇒ A (u) 1 (cosh2 u 2 1 cosh2 u 2 1 (cosh2 u 2 (c) A (u) A(0) sinh2 u) cosh2 u 1 sinh2 u 2 sinh2 u) 1 (1) 2 1)(sinh u) sinh u 1 ⇒ A(u) 2 0⇒C ( u 2 2 1 2 C, and from part (a) we have 0 ⇒ A(u) u ⇒u 2 2A 473 ...
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