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Unformatted text preview: 422 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS (d) Show that, once LIN passes LOG, LOG can never catch up. 25. Atmospheric pressure p varies with altitude h according to the equation dp = kp, dh where k is a constant. 31. 32. Given that p is 15 pounds per square inch at sea level and 10 pounds per square inch at 10,000 feet, ﬁnd p at: (a) 5000 feet; (b) 15,000 feet. 26. According to the compound interest formula (7.6.3), if P dollars are deposited in an account now at an interest rate r compounded continuously, then the amount of money in the account t years from now will be Q = Pert . The quantity Q is sometimes called the future value of P (at the interest rate r ). Solving this equation for P , we get P = Qe−rt . In this formulation, the quantity P is called the present value of Q. Find the present value of $20,000 at 6% compounded continuously for 4 years. Find the interest rate r that is needed to have $6000 be the present value of $10,000 over an 8year period. You are 45 years old and are looking forward to an annual pension of $50,000 per year at age 65. What is the presentday purchasing power (present value) of your pension if money can be invested over this period at a continuously compounded interest rate of: (a) 4%? (b) 6%? (c) 8%? The cost of the tuition, fees, room, and board at XYZ College is currently $16,000 per year. What would you expect to pay 3 years from now if the costs at XYZ are rising at the continuously compounded rate of: (a) 5%? (b) 8%? (c) 12%? A boat moving in still water is subject to a retardation proportional to its velocity. Show that the velocity t seconds 33. 27. 28. 34. after the power is shut off is given by the formula v = ce−kt where c is the velocity at the instant the power is shut off. A boat is drifting in still water at 4 miles per hour; 1 minute later, at 2 miles per hour. How far has the boat drifted in that 1 minute? (See Exercise 30.) During the process of inversion, the amount A of raw sugar present decreases at a rate proportional to A. During the ﬁrst 10 hours, 1000 pounds of raw sugar have been reduced to 800 pounds. How many pounds will remain after 10 more hours of inversion? The method of carbon dating makes use of the fact that all living organisms contain two isotopes of carbon, carbon12, denoted 12 C (a stable isotope), and carbon14, denoted 14 C (a radioactive isotope). The ratio of the amount of 14 C to the amount of 12 C is essentially constant (approximately 1/10,000). When an organism dies, the amount of 12 C present remains unchanged, but the 14 C decays at a rate proportional to the amount present with a halflife of approximately 5700 years. This change in the amount of 14 C relative to the amount of 12 C makes it possible to estimate the time at which the organism lived. A fossil found in an archaeological dig was found to contain 25% of the original amount of 14 C. What is the approximate age of the fossil? The Dead Sea Scrolls are approximately 2000 years old. How much of the original 14 C remains in them? In Exercises 35–37, ﬁnd all the functions f that satisfy the equation for all real t . 35. f (t ) = t f (t ). HINT: Write f (t ) − t f (t ) = 0 and multiply 2 the equation by e−t /2 . 36. f (t ) = sin t f (t ). 37. f (t ) = cos t f (t ). 38. Let g be a function everywhere continuous and not identically zero. Show that if f (t ) = g (t ) f (t ) for all real t , then either f is identically zero, or f does not take on the value zero at any t . 29. 30. 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS
Since none of the trigonometric functions are onetoone, none of them have inverses. What, then, are the inverse trigonometric functions? The Inverse Sine
The graph of y = sin x is shown in Figure 7.7.1. Since every horizontal line between −1 and 1 intersects the graph at inﬁnitely many points, the sine function does not have 1 an inverse. However, observe that if we restrict the domain to the interval [− 1 π , 2 π ] 2 (the solid portion of the graph in Figure 7.7.1), then y = sin x is onetoone, and on that interval it takes on as a value every number in [− 1,1]. Thus, if x ∈ [− 1, 1], there is one and only one number in the interval [− 1 π , 1 π ] at which the sine function has 2 2 the value x. This number is called the inverse sine of x, or the angle whose sine is x, and is written sin−1 x. 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS
y 423 y=1 –5π
2 –2 π –3π
2 –π –1π
2 1 π 2 π 3 π 2 2π 5 π 2 x y = –1 y = sin x Figure 7.7.1 Remarks Another common notation for the inverse sine function is arcsin x, read “arc sine of x.” For consistency in the treatment here, we use sin−1 x throughout, but we use both notations in the Exercises. Remember that the “−1” is not an exponent; do not confuse sin−1 x with the reciprocal 1/sin x. The inverse sine function y = sin−1 x, is the inverse of the function y = sin x, domain: [− 1 π , 1 π ], 2 2 range: [− 1, 1]. domain: [− 1, 1], range: [− 1 π , 1 π ] 2 2
Table 7.7.1 x −1π 2 −1π 3 −1π 4 −1π 6 0
1 π 6 1 π 4 1 π 3 1 π 2 sin x −1 √ −1 3 2 √ −1 2 2 −1 2 0
1 2 √ 1 2 2 √ 1 3 2 1 The graphs of these functions are shown in Figure 7.7.2. Each curve is the reﬂection of the other in the line y = x.
y
1 2 y π 1 –π –1 1 2 1 2 π x –1 1 x –π y = sin x, x 1 2 [– 1 π, 1 π 2 2 Figure 7.7.2 y = sin–1x, x [–1, 1] Table 7.7.2 x −1 √ −1 3 2 √ −1 2 2 −1 2 0
1 2 √ 1 2 2 √ 1 3 2 1 sin−1 x −1π 2 −1π 3 −1π 4 −1π 6 0 1 π 6
1 π 4 1 π 3 1 π 2 Because these functions are inverses,
(7.7.1) for all x ∈ [− 1, 1], sin (sin−1 x) = x and
(7.7.2) for all x ∈ [− 1 π , 1 π ], 2 2 sin−1 (sin x) = x. Table 7.7.1 gives some representative values of the sine function from x = − 1 π 2 to x = 1 π . Reversing the order of the columns, we have a table for the inverse sine 2 (Table 7.7.2). 424 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS On the basis of Table 7.7.2 one could guess that for all x ∈ [− 1, 1], sin−1 (− x) = −sin−1 x. This is indeed the case. Being the inverse of an odd function (sin (− x) = −sin x for all x ∈ [− 1 π , 1 π ] , the inverse sine is itself an odd function. (Verify this.) 2 2
Example 1 Calculate if deﬁned:
1 π ); 16 9 π ); 5 (a) sin−1 (sin (d) sin−1 (sin
SOLUTION (b) sin−1 (sin 7 π ); 3 (e) sin (sin−1 2). (c) sin (sin−1 1 ); 3 (a) Since 1 π 16 is within the interval [− 1 π , 1 π ], we know by (7.7.2) that 2 2 sin−1 (sin
1 π) 16 = 1 π. 16 (b) Since 7 π is not in the interval [− π/2, π/2], we cannot apply (7.7.2) directly. 3 However, 7 π = 1 π + 2π and sin ( 1 π + 2π ) = sin ( 1 π ) (recall that the sine 3 3 3 3 function is periodic with period 2π ). Thus sin−1 (sin 7 π ) = sin−1 sin 3 (c) By (7.7.1) sin (sin−1 1 ) = 1 . 3 3 (d) Since 9 π is not within the interval [− 1 π , 1 π ], we cannot apply (7.7.2) directly. 5 2 2 However, 9 π = 2π − 1 π . Thus 5 5 sin−1 sin 9 π = sin−1 sin 2π − 1 π 5 5 = sin−1 sin − 1 π 5
by (7.7.2)
1 π 3 + 2π = sin−1 (sin 1 π ) = 1 π . 3 3 ↑ by (7.7.2) = −1π. 5 ↑ (e) The expression sin (sin−1 2) makes no sense since 2 is not in the domain of the inverse sine; there is no angle whose sine is 2. The inverse sine is deﬁned only on [− 1, 1]. If 0 < x < 1, then sin−1 x is the radian measure of the acute angle whose sine is x. We can construct an angle of radian measure sin−1 x by drawing a right triangle with a leg of length x and a hypotenuse of length 1. See Figure 7.7.3. Reading from the ﬁgure we have √ sin (sin−1 x) = x cos (sin−1 x) = 1 − x2 √ x 1 − x2 −1 −1 tan (sin x) = √ cot (sin x) = x 1 − x2 1 1 sec (sin−1 x) = √ csc (sin−1 x) = . 2 x 1−x Since the derivative of the sine function, d (sin x) = cos x, dx 1 x sin–1x √1 – x 2 Figure 7.7.3 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS 425 does not take on the value 0 on the open interval (− 1 π , 1 π ) (recall that cos x > 0 2 2 for − 1 π < x < 1 π ), the inverse sine function is differentiable on the open interval 2 2 (−1,1).† We can ﬁnd the derivative as follows: y = sin−1 x, sin y = x, dy cos y = 1, dx dy 1 1 = =√ dx cos y 1 − x2 In short
(7.7.3)
1 y y 1 √1 – x2 x (–1 < x ≤ 0) x (0 ≤ x < 1) (see the ﬁgure). d 1 (sin−1 x) = √ . dx 1 − x2 If u is a differentiable function of x with u < 1, then, by the chain rule, d 1 du (sin−1 u) = √ ·. 2 dx dx 1−u
Example 2 d [sin−1 (3x2 )] = dx 1 1− (3x2 )2 · d 6x (3x2 ) = √ . dx 1 − 9x 4 NOTE: Although we have not made a point of it, we continue with the convention that if the domain of a function f is not speciﬁed explicitly, then it is understood to be the maximal set of real numbers x for which f (x) is a real number. In this case, the domain is the set of all real numbers x such that −1 ≤ 3x2 ≤ 1, which is the interval √ √ [− 1/ 3, 1/ 3].
Example 3 Show that for a > 0 x dx = sin−1 + C. √ a a2 − x 2 (7.7.4) SOLUTION We change variables so that the a2 becomes 1 and we can use (7.7.3). We set au = x, a du = dx. du √ 1 − u2 x + C. a Then dx = √ a2 − x2 √ a du a2 − a 2 u 2 = a du = √ a 1 − u2 ↑ since a > 0 = sin−1 u + C = sin−1 † See Section 7.1. 426 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS
√ 3 Example 4
SOLUTION Evaluate
0 dx . √ 4 − x2 By (7.7.4) dx x = sin−1 + C. √ 2 2 4−x It follows that
√ 3 √ dx 4− x2 = sin −1 0 x 2 √ 0 3 = sin −1 √ 3 2 − sin−1 0 = π π −0= . 3 3 The Inverse Tangent
Although not onetoone on its full domain, the tangent function is onetoone on the open interval − 1 π , 1 π , and on that interval it takes on as a value every real number 2 2 (see Figure 7.7.4). Thus, for any real number x, there is one and only one number in the open interval − 1 π , 1 π at which the tangent function has the value x. This number 2 2 is called the inverse tangent of x, or the angle whose tangent is x, and is written tan−1 x or arctan x. The inverse tangent function y = tan−1 x, is the inverse of the function y = tan x, domain: (− 1 π , 1 π ), 2 2 range: (− ∞, ∞). domain: (− ∞, ∞), range: (− 1 π , 1 π ) 2 2 The graphs of these two functions are given in Figure 7.7.4. Each curve is a reﬂection
y x = –π
2 x= π
2 y y=
1 2 π
2 π O –
1 π 2 1 π 2 x –
1 2 x π
y=– π
2 y = tan–1x, x real y = tan x, x ( – 1 π, 1 π 2 2 )
Figure 7.7.4 of the other in the line y = x. While the tangent has vertical asymptotes, the inverse tangent has horizontal asymptotes. Both functions are odd functions. Because these functions are inverses,
(7.7.5) for all real numbers x, tan (tan−1 x) = x 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS 427 and
(7.7.6) for all x ∈ (− 1 π , 1 π ), 2 2 tan−1 (tan x) = x. It is hard to make a mistake with the ﬁrst relation since it applies for all real numbers. The second relation requires the usual care: tan−1 (tan 1 π ) = 1 π 4 4 but tan−1 (tan 7 π ) = 7 π . 5 5 We can calculate tan−1 (tan 7 π ) as follows: 7 π = 2 π + π and tan( 2 π + π ) = tan( 2 π ) 5 5 5 5 5 (recall that the tangent function is periodic with period π ). The relation tan−1 ( tan 2 π ) = 5 2 π is valid because 2 π is within the interval (− 1 π , 1 π ). 5 5 2 2 If x > 0, then tan−1 x is the radian measure of the acute angle that has tangent x. We can construct an angle of radian measure tan−1 x by drawing a right triangle with legs of length x and 1 (Figure 7.7.5). Reading from the triangle, we have tan (tan−1 x) = x sin (tan−1 x) = √ sec (tan−1 x) = x 1+ x2 1 cot (tan−1 x) = x
+ √1
tan–1x 1 x2
x √ 1 + x2 1 cos (tan−1 x) = √ 1 + x2 √ 1 + x2 csc (tan−1 x) = . x Figure 7.7.5 Since the derivative of the tangent function, d 1 (tan x) = sec2 x = , dx cos2 x is never 0 on (− 1 π , 1 π ), the inverse tangent function is everywhere differentiable 2 2 (Section 7.1). We can ﬁnd the derivative as we did for the inverse sine: y = tan−1 x, tan y = x, dy sec2 y = 1, dx dy 1 1 = = cos2 y = 2y dx sec 1 + x2 We have found that 1 d (tan−1 x) = . dx 1 + x2
x2 + √1 x (x ≥ 0) 1 x (x ≤ 0) (see the ﬁgure). y y √1 + x2 (7.7.7) (See Example 3, Section 5.3) If u is a differentiable function of x, then the chain rule gives 1 du d tan−1 u = ·. dx 1 + u2 dx 428 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Example 5 d d 1 tan−1 (ax2 + bx + c) = · (ax2 + bx + c) 2 + bx + c)2 dx dx 1 + (ax 2ax + b = . 1 + (ax2 + bx + c)2
Example 6 Show that, for a = 0, 1 dx x = tan−1 + C. a2 + x2 a a (7.7.8) SOLUTION We change variables so that a2 is replaced by 1 and we can use (7.7.7). au = x, dx = + x2 a du = dx. We set Then a2 1 du a du = 2 u2 a +a 1 + u2 1 1 x = tan−1 u + C = tan−1 + C. a a a ↑ (7.7.7) a2
2 Example 7
SOLUTION Evaluate
0 dx . 4 + x2 1 x dx = tan−1 + C, 2 +x 2 2 By (7.7.8) dx = 4 + x2 22 y so that
2 0 dx 1 x tan−1 = 2 4+x 2 2 2 =
0 1 1 π tan−1 (1) − tan−1 (0) = . 2 2 8 1 –2π –π –1 Inverse Secant
π
2π x y = sec x Figure 7.7.6 The graph of y =sec x is shown in Figure 7.7.6. Note that  sec x ≥ 1 for all x in the domain. If we restrict the domain to the set [0, 1 π ) ∪ ( 1 π , π ]† (the solid portion of the 2 2 graph in Figure 7.7.6), then the secant function is onetoone, and for each number x such that x ≥ 1 there is one and only one number in [0, 1 π ) ∪ ( 1 π , π ] at which the 2 2 secant function has the value x. This number is called the inverse secant of x, or the angle whose secant is x. The inverse secant function y = sec−1 x, domain: (− ∞, −1] ∪ [1, ∞), range: [0, 1 π ) ∪ ( 1 π , π ] 2 2 † Some authors use the set [0, 1 π ) ∪ (π , 3 π ] for the restricted domain of y = sec x. There is no agreedupon 2 2 convention in this matter. Some consequences of this choice are considered in the Exercises. 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS
y y 429 π 1 π 2 π
–1 x –1 1 x y = sec x, x e 0, π 2 )( π , 2 π y = sec–1 x, x e – •, –1 ( 1, • ) Figure 7.7.7 is the inverse of y = sec x, domain: [0, 1 π ) ∪ ( 1 π , π ], 2 2 range: (− ∞, −1] ∪ [1, ∞). The graphs of these two functions are given in Figure 7.7.7. As the graphs indicate, each is an increasing function. The secant function and inverse secant function are related as follows: for all x such that x ≥ 1, for all x ∈ [0, 1 π ) ∪ ( 1 π , π ], 2 2 sec ( sec−1 x) = x, sec−1 (sec x) = x. As you would expect, the second relation requires careful attention: sec−1 (sec 1 π ) = 1 π 3 3 We calculate sec−1 (sec 7 π ) as follows: 4 sec−1 (sec 7 π ) = sec−1 (sec [2π − 1 π ]) = sec−1 (sec 1 π ) = 1 π . 4 4 4 4 If x > 1, then sec−1 x is the radian measure of the acute angle that has secant x. We can construct an angle of radian measure sec−1 x by drawing a right triangle with hypotenuse of length x and a side of length 1 (Figure 7.7.8). The values sec (sec−1 x) = x √ x2 − 1 −1 sin (sec x) = x √ tan (sec−1 x) = x2 − 1 can all be read from the triangle. The derivative of the secant function, d ( sec x) = sec x tan x, dx x csc (sec−1 x) = √ 2−1 x cos (sec−1 x) = 1 x 1
sec 1x 1
_ but sec−1 (sec 7 π ) = 7 π . 4 4 x √x 2 – 1 cot (sec−1 x) = √ x2 − 1 Figure 7.7.8 430 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS is nonzero (in fact, positive) on (0, 1 π ) ∪ ( 1 , π ). Therefore, the inverse secant function 2 2 is differentiable for x > 1. We can ﬁnd the derivative as we did for the inverse sine and inverse tangent functions: y = sec−1 x sec y = x dy sec y tan y =1 dx dy 1 = . dx sec y tan y To complete the calculation, we need to express the product sec y tan y in terms of x. We know that sec y = x. Since √ tan2 y = sec2 y − 1 = x2 − 1, x2 − 1, which, with sec y = x, gives sec y tan y = ± x x2 − 1. If 0 < y < 1 π , then sec y and tan y are both positive; if 1 π < y < π , then sec y and 2 2 tan y are both negative. In each case, the product sec y tan y is positive. It follows that √ sec y tan y = x x2 − 1 and therefore
(7.7.9) We have tan y = ± d 1 sec−1 x = √ . dx x x2 − 1 Note that the derivative is positive, conﬁrming the fact that the inverse secant function is increasing. If u is a differentiable function of x, with u > 1, then, by the chain rule, d 1 du sec−1 u = √ ·. dx u u2 − 1 dx Example 8 d 1 d sec−1 (2x3 ) = · (2x3 ) dx 2x3  (2x3 )2 − 1 dx = 6x2 6x2 3 = =√ . √ √ 3  4x 6 − 1 2 x  4x 6 − 1 2x 2x x 4x6 − 1 Example 9 Show that x x2 − 1 √ dx = sec−1 x + C . SOLUTION If x > 1, sec−1 x = sec−1 x and d 1 1 (sec−1 x) = √ =√ . 2−1 dx x x x x2 − 1 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS 431 If x < −1, sec−1 x = sec−1 (− x) and d 1 −1 1 [sec−1 (− x)] = (− 1) = =√ . √ √ dx  − x x2 − 1 −x x2 − 1 x x2 − 1 1 Thus, it follows that f (x) = sec−1 x is an antiderivative for f (x) = √ . x x2 − 1 We leave it as an exercise to show that if a > 0 is a constant, then 1 1 dx = sec−1 √ 2 − a2 a xx x a + C. (7.7.10) The Other Trigonometric Inverses
There are three other trigonometric inverses: the inverse cosine, y = cos−1 x, is the inverse of y = cos x, x ∈ [0, π ]; the inverse cotangent, y = cot−1 x, is the inverse of y = cot x, x ∈ (0, π ); the inverse cosecant, y = csc−1 x, is the inverse of y = csc x, x ∈ [− 1 π , 0] ∪ (0, 1 π ]. 2 2 Figure 7.7.9 illustrates each of these inverses between 0 and 1 π in terms of right 2 triangles. The differentiation formulas for these functions are as follows: −1 d d (cos−1 x) = √ = − (sin−1 x) dx dx 1 − x2 −1 d d (cot−1 x) = = − (tan−1 x) 2 dx 1+x dx d −1 d (csc−1 x) = √ = − (sec−1 x). 2−1 dx dx x x θ
1 1– x2
_ _ √1 +x 2 θ
1 x θ
1 cos 1x x (a) cot 1x x (b) csc 1x √x 2 –1 (c) _ Figure 7.7.9 You are asked to verify these formulas in the Exercises. Since the derivatives of these functions differ from the derivatives of their corresponding cofunctions by a constant factor, these last three trigonometric inverses are not needed for ﬁnding antiderivatives. Remark In Figure 7.7.9a, we let θ denote the other acute angle of the right triangle. Since the two acute angles in a right triangle are complementary, we have θ + cos−1 x = π . 2 432 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS But note that sin θ = x, which means that θ = sin−1 x. Therefore, sin−1 x + cos−1 x = π 2 or cos−1 x = π − sin−1 x. 2 You can verify that the corresponding relations hold for the pairs {tan x, cot x} and {sec x, csc x}: cot−1 x = π − tan−1 x 2 and csc−1 x = π − sec−1 x. 2 This is another reason why it is sufﬁcient to restrict our attention to sin−1 x, tan−1 x, and sec−1 x. EXERCISES 7.7
Determine the exact value. 1. (a) tan 0; 2. (a) arc sec 2; 3. (a) cos−1 − 1 ; 2
−1 √ 4. (a) sec (sec−1 [− 2/ 3]); 5. 6. 7. 8. (a) cos (sec−1 2); (a) arcsin (sin [11π/6]); (a) cos−1 (sec [7π/6]); (a) cos sin−1 [ 3 ] ; 5 2 cos [ 1 ] 2 sin−1 [ 1 ] + 2
−1 √ (b) sin (− 3/2). √ (b) tan−1 ( 3). √ (b) sec−1 (− 2).
−1 33. Show that for a > 0 dx a2 − (x + b)2 = sin−1 x+b a + C. (7.7.11) (b) sec (arccos[− 1 ]). 2 (b) (b) (b) (b) arctan (sec 0). arctan (tan [11π/4]). sec−1 (sin [13π/6]). sec tan−1 [ 4 ] . 3
−1 34. Show that for a = 0 1 dx = tan−1 a2 + (x + b)2 a x+b a (7.7.12) + C. 9. (a) sin 10. (a) cos ;
−1 (b) cos 2 sin [4] 5 . 35. (a) Verify (7.7.10). (b) Show that for a > 0 dx (x + b) (x + b)2 − a2 = 1 −1 sec a x + b a + C. sin [− 1] ; √ (b) tan sin−1 [ 2/2] + cos−1 [ 1 ] . 2 √ 12. y = tan−1 x. 14. f (x) = e x sin−1 x. 16. f (x) = etan x . 18. v = tan−1 (e x ). √ 20. y = sec−1 x2 + 2. 22. f (x) = ln (tan−1 x).
−1 Differentiate the function. 11. y = tan−1 (x + 1). 13. f (x) = sec−1 (2x2 ). 15. f (x) = x sin−1 2x. 17. u = (sin−1 x)2 . tan−1 x 19. y = . x √ 21. f (x) = tan−1 2x. 23. y = tan (ln x). 24. g (x) = sec−1 (cos x + 2). √ 25. θ = sin−1 ( 1 − r 2 ). 27. g (x) = x2 sec−1 1 . x
−1 Give the domain and range of each function, and verify the differentiation formula. d −1 36. f (x) = cos−1 x; (cos−1 x) = √ . dx 1 − x2 37. f (x) = cot−1 x; 38. f (x) = csc−1 x; d −1 . (cot−1 x) = dx 1 + x2 d −1 (csc−1 x) = √ . dx x x2 − 1 26. θ = sin−1 28. θ = tan−1 30. f (x) = esec r . r+1 1 1 + r2
−1 x Evaluate the integral. .
1 39.
0 dx . 1 + x2 √ dx 1 − x2 . 1 40.
−1 1 dx . 1 + x2 √ dx 4 − x2 dx . 29. y = sin [sec−1 (ln x)]. . 41.
0 √ 1/ 2 √ x 31. f (x) = c2 − x2 + c sin−1 , c > 0. c x x − sin−1 , c > 0, 32. y = √ 2 − x2 c c 42.
0 8 5 43.
0 dx . 25 + x2 44.
5 x x2 − 16 √ . 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS
3/2 433 45.
0 3 dx . 9 + 4x 2 5 46.
2 dx . 9 + (x − 2)2 what point on the road is the angle θ subtended by the billboard a maximum (see the ﬁgure)? 47.
3/2 6 dx . √ x 16x2 − 9 dx . √ (x − 3) x2 − 6x + 8 dx 4 − (x + 3)2 ex dx. 1 + e2x
ln 3 s k 48.
4 −2 49.
−3 . 50.
ln 2 1/ 2 e−x dx. √ 1 − e−2x √ 1 3 − 4x 2 θ ln 2 51.
0 52.
0 dx. 72. A person walking along a straight path at the rate of 6 feet per second is followed by a spotlight that is located 30 feet from the path. How fast is the spotlight turning at the instant the person is 50 feet past the point on the path that is closest to the spotlight? 73. (a) Show that x2 a2 x a − x2 + sin−1 , a>0 2 2 a √ is an antiderivative of f (x) = a2 − x2 . F (x ) =
a Calculate the indeﬁnite integral. 53. √ x 1 − x4 dx. 54. sec2 x dx. √ 9 − tan2 x √ dx 4x − x 2 . 55. x dx. 1 + x4 sec2 x dx. 9 + tan2 x sin−1 x dx. √ 1 − x2 dx x 1 − (ln x )2 . 56. 57. 58. cos x dx. 3 + sin2 x tan−1 x dx. 1 + x2 dx . x[1 + (ln x)2 ] 1 1 − x2 √ 1 dx = sin−1 3 is NOT 59. 61. 60. 62.
3 (b) Use the result in part (a) to calculate and interpret your result as an area. 74. Let f (x) = tan−1 (a) Show that f (x) = a+x , x = 1/a. 1 − ax
−a a2 − x2 dx 63. Explain why the formula correct. 64. Explain why the formula
0 √
1 65. 66. 67. 68. 69. 70. 71. dx = sec−1 (1) − x x2 − 1 sec−1 (− 1) = − π is NOT correct. Find the area of the region bounded by the graph of y = √ 1/ 4 − x2 and the xaxis between x = −1 and x = 1. Find the area of the region bounded by the graph of y = 3/(9 + x2 ) and the xaxis between x = −3 and x = 3. Sketch the region bounded by the graphs of 4y = x2 and y = 8/(x2 + 4) and ﬁnd the area. √ The region bounded by the graph of y = 1/ 4 + x2 between x = 0 and x = 2 is revolved about the xaxis. Find the volume of the resulting solid. The region in Exercise 68 is revolved about the yaxis. Find the volume of the resulting solid. √ The region bounded by the√ graph of y = 1/(x2 x2 − 9) and the xaxis between x = 2 3 and x = 6 is revolved about the yaxis. Find the volume of the resulting solid. A billboard k feet wide is perpendicular to a straight road and is s feet from the road. At what point on the road would a motorist have the best view of the billboard; that is, at
−1 1 , x = 1/a. 1 + x2 (b) Show that there is no constant C such that f (x) = tan−1 x + C for all x = 1/a. (c) Find constants C1 and C2 such that f (x) = tan−1 x + C1 f (x) = tan−1 x + C2 for x < 1/a for x > 1/a. c 75. Let f (x) = tan−1 (1/x), x = 0. (a) Use a graphing utility to graph f and g (x) = tan−1 x. What do you notice about these two graphs? (b) Use your graph of f to estimate the limits
x →0 + lim f (x) and x →0 − lim f (x) −1 , x = 0. 1 + x2 (d) Show that there does not exist a constant C such that f (x) + tan−1 x = C for all x. (e) Find constants C1 and C2 such that (c) Show that f (x) = f (x) + tan−1 x = C1 f (x) + tan−1 x = C2 for x > 0 for x < 0. 434 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS c 76. Evaluate sin−1 x x →0 x lim numerically. Justify your answer by other means. c 77. Estimate the integral
0.5 0 1 dx √ 1 − x2 by using the partition {0, 0.1, 0.2, 0.3, 0.4, 0.5} and the intermediate points
∗ x1 = 0. 05, ∗ x4 ∗ x2 = 0. 15, ∗ x3 = 0. 25, c 79. Use a CAS to ﬁnd the exact values of the following expressions: (a) tan ( arccos [3/5] − arctan [12/13]). (b) cos ( sin−1 [4/5] − cos−1 [4/5]). (c) sin ( arctan [− 12/5] − arccos [5/13]). Verify your answers analytically. c 80. Use a CAS to ﬁnd a oneterm expression for f : 3 (a) f (x) dx = x2 − 9 + 3 tan−1 √ + C. 2−9 x
(b) (c) f (x) dx = 2 arcsin f (x) dx = x2 sec x − x + 2 x x3 − 4 2 4 − x2 + C . = 0. 35, ∗ x5 = 0. 45. Note that the sine of your estimate is close to 0.5. Explain the reason for this. 1 c 78. Use a graphing utility to draw the graph of f (x) = 1 + x2 on [0, 10]. n (a) Calculate 0 f (x) dx for n =250, 500, 750, 1000. (b) What are the numbers in part (a) getting “close” to? (c) Determine the value of n 1 lim dx. n→∞ 0 1 + x 2 x2 − 1 + C . (d) Verify your results by differentiating the righthand sides. c 81. Use a graphing utility to graph f (x) = tan−1 x and g (x) = cos−1 x together. (a) Use a CAS to ﬁnd the point of intersection of the graphs. (b) Use a CAS to ﬁnd the area of the region bounded by graphs of f , g , and the xaxis. c 82. Repeat Exercise 81 with the functions f (x) = sin−1 x and g (x) = cos−1 x. PROJECT 7.7 Refraction
Dip a straight stick in a pool of water and it appears to bend. Only in a vacuum does light travel at speed c (the famous c of E = mc2 ). Light does not travel as fast through a material medium. The index of refraction n of a medium relates the speed of light in that medium to c: n= c . speed of light in the medium
Index of refraction ni n1 θ1 Like the Law of Reﬂection (see Example 5, Section 4.5), Snell’s Law of Refraction can be derived from Fermat’s Principle of Least Time. Problem 1. A light beam passes from a medium with index of refraction n1 through a plane sheet of material whose top and bottom faces are parallel and then out into some other medium with index of refraction n2 . Show that Snell’s law implies that n1 sin θ1 = n2 sin θ2 regardless of the thickness of the sheet or its index of refraction. θi θr Index of refraction nr θ2 n2 When light travels from one medium to another, it changes direction; we say that light is refracted. Experiment shows that the angle of refraction θr is related to the angle of incidence θi by Snell’s law. ni sin θi = nr sin θr . A star is not where it appears to be. Consider a beam of light traveling through an atmosphere whose index of refraction varies with height, n = n(y). The light follows some curved path y = y(x). Think of the atmosphere as a succession of parallel slabs. When a light ray strikes a slab at height y, it is traveling at some angle θ to the vertical; when it emerges at height y + y, it is traveling at a slightly different angle, θ + θ . 7.8 THE HYPERBOLIC SINE AND COSINE 435 Problem 2. a. Use the result in Problem 1 to show that dθ d y/dx 1 dn . = −cot θ = n dy dy 1 + (dy/dx)2
2 2 b. Verify that the slope of the light path must vary in such a way that 1 + (dy/dx)2 = (constant) [n(y)]2 . c. How must n vary with height y for the light to travel along a circular arc? 7.8 THE HYPERBOLIC SINE AND COSINE
Certain combinations of the exponential functions e x and e−x occur so frequently in mathematical applications that they are given special names. The hyperbolic sine (sinh) and hyperbolic cosine (cosh) are the functions deﬁned by
(7.8.1) sinh x = 1 (e x − e−x ), 2 cosh x = 1 (e x + e−x ). 2 The reasons for these names will become apparent as we go on. Since d d ( sinh x) = dx dx
1 (e x 2 − e −x ) = 1 ( e x + e −x ) 2 and we have d d ( cosh x) = [ 1 (e x + e−x )] = 1 (e x − e−x ), 2 dx dx 2 (7.8.2) d ( sinh x) = cosh x, dx d ( cosh x) = sinh x. dx In short, each of these functions is the derivative of the other. The Graphs
We begin with the hyperbolic sine. Since sinh (− x) = 1 (e −x − ex ) = − 1 (e x − e−x ) = − sinh x, 2 2 the hyperbolic sine is an odd function. The graph is therefore symmetric about the origin. Since d ( sinh x) = cosh x = 1 (e x + e−x ) > 0 2 dx the hyperbolic sine increases everywhere. Since d d2 ( sinh x) = ( cosh x) = sinh x = 1 (e x − e−x ), 2 dx2 dx you can see that for all real x, ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston  Downtown.
 Spring '10
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