21.3 Applications 701 18. A wheel of radius 5 meters is oriented vertically and spinning counterclockwise at a rate of 7 revolutions per minute. If the origin is placed at the center of the wheel a point on the rim has a horizontal position of (7, 0) at time t = 0. What is the horizontal component of the point’s velocity at t = 2? 19. In this problem you will show that y = C 1 sin kx + C 2 cos kx is a solution to the differential equation y =−k 2 y. Recall that a differential equation is an equation involving a derivative and a function is a solution to the differential equation if it satisfies the differential equation. (a) Show that y 1 = sin kx is a solution to y =−k 2 y.Todothis, first find y 1 . Then write y 1 ? = −k 2 y 1 and verify that the two sides are indeed equal. (b) Show that y 2 = C 1 sin kx is a solution to y =−k 2 y. (c) Show that if y 1 and y 2 are solutions to y =−k 2 y, then y 3 = C 1 y 1 + C 2 y 2 is a solution to y =−k 2 y as well. Conclude that y = C 1 sin kx + C 2 cos kx is a solution to the differential equation y =−k 2 y. 20. For each of the functions below, determine whether the function is a solution to differential equation (i), differential equation (ii), or neither. Differential equations (i) and (ii) are given below. i.y = 16y ii. y =−16y (a) y 1 (t) = sin 16t (b) y 2 (t) = e 4t (c) y 3 (t) = 3 cos 4t (d) y 4 (t) = sin 4t + 1 (e) y 5 (t) = e −16t (f) y 6 (t) =−3e −4t (g) y 7 (t) = e 4t + 3 (h) y 8 (t) =−sin 4t (Hint: Four of the eight functions given are solutions to neither differential equation.) 21. Each of the functions below is a solution to one of the differential equations below. i. y = 9 ii. y = 9y iii. y =−9y Foreach function, determine which of the three differential equations it satisfies. (a) y 1 (t) = 5 sin 3t (b) y 2 (t) = e 3t (c) y 3 (t) = 2 cos 3t (d) y 4 (t) = 4.5t 2 + 3t + 8 (e) y 5 (t) = 4e −3t (f) y 6 (t) = 4.5t 2 − t + 2 702 CHAPTER 21 Differentiation of Trigonometric Functions 22. After having done the previous problem, make up a solution to each of the three differential equations below. i. y = 9 ii. y = 9y iii y =−9y Your answers must be different from the solutions given in the preceding problem, but you can use those answers for inspiration. Check that your answers are right by“plugging them back” into the differential equation. For instance, if you guess that y = e 3t + 1 is a solution to the differential equation y = 9y, test it out as follows. First calculate y . Since y = e 3t · 3, y = 9e 3t . Now see if it satisfies the differential equation. y ? = 9y 9e 3t ? = 9(e 3t + 1) 9e 3t = 9e 3t + 9 So y = e 3t + 1isnot a solution to y = 9y. 23. As you’re riding up an elevator inside the Hyatt Hotel right next to the Charles River, you watch a duck swimming across the Charles, swimming straight toward the base of the elevator. The elevator is rising at a speed of 10 feet per second, and the duck is swimming at 5 feet per second toward the base of the elevator. As you pass the eighth floor, 100 feet up from the level of the river, the duck is 200 feet away from the base of the elevator. (a) At this instant, is the distance between you and the duck increasing or decreasing? At what rate? (b) As you’re watching the duck, you have to look down at more and more of an angle to see it. At what rate is this angle of depression increasing at the instant when you are at a height of 100 feet? Include units in your answer. (Problem written by Andrew Engelward) 24. Approximating the function f(x)=sin x near x = 0 by using polynomials. The point of this problem is to show you how the values of sin x can be approxi- mated numerically with a very high degree of accuracy. It is an introduction to Taylor polynomials. (a) Find the equation of the line tangent to f(x)=sin x at x = 0. (b) Find the equation of a quadratic Q(x) = a + bx + cx 2 such that the function Q(x) and its nonzero derivatives match those of sin x at x = 0. In other words, Q(0) = f(0), Q (0)=f (0), and Q (0) = f (0). The quadratic that you found is the quadratic that best “fits” the sine curve near x = 0. In fact, the “quadratic” turns out not to really be a quadratic at all. Sine is an odd function, so there is no parabola that “fits” the sine curve well at x = 0. (c) Find the equation of a cubic C(x) = a + bx + cx 2 + dx 3 such that the function C(x) and its nonzero derivatives match those of sin x at x = 0. In other words, C(0) = f(0),C (0)=f (0),C (0) = f (0), and C (0) = f (0). The cubic that you found is the cubic that best “fits” the sine curve near x = 0. Using a calculator, on the same set of axes graph sin x, the tangent line to sin x at x = 0, and C(x), the cubic you found. Now “zoom in” around x = 0. Can 21.4 Derivatives of Inverse Trigonometric Functions 703 you see that near x = 0 the line is a good fit to the sine curve but the cubic is an even better fit for small x and the cubic hugs the sine curve for longer? The next set of questions asks you to investigate how good the fit is. (d) Use C(x) from part (c) to estimate the following, and then compare with the actual value using a calculator. sin(0.01) sin(0.1) sin(0.5) sin(1) sin(3) (e) (Challenge) Find the “best” fifth degree polynomial approximation of sin(x) for x near 0 by making sure that the first five derivatives of the polynomial match those of sin(x) when evaluated at x = 0. Graph sin x along with its first, third, and fifth degree polynomial approximations on your graphing calculator. The higher the degree of the polynomial, the better the fit to sin x near x = 0, right? (f) Using a calculator, on the same set of axes graph sin x and the polynomial given below. x − x 3 6 + x 5 120 − x 7 5040 + x 9 362880 This polynomial is an even better fit than the last one, right? Now graph the difference between sin x and this polynomial; in other words, graph y = sin x − x − x 3 6 + x 5 120 − x 7 5040 + x 9 362880 . On approximately what interval is the difference between sin x and this polynomial less than 0.005? 21.4 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS We can find the derivatives of inverse trigonometric functions using implicit differentiation. The Derivative of arcsin x Let y = arcsin x. Our goal is to find dy dx . y = arcsin xyis the angle between − π 2 and π 2 whose sine is x. sin y = x Therefore, the sine of y is x. (sine and arcsine are inverse functions.) d dx sin y = d dx x Differentiate each side with respect to x. (cos y) dy dx = 1 Remember the Chain Rule on the left-hand side, since y is a function of x. dy dx = 1 cos y Solve for dy dx . 704 CHAPTER 21 Differentiation of Trigonometric Functions We’d like to express dy dx in terms of x, not y. We know that x = sin y and that sin y and cos y are related by sin 2 y + cos 2 y = 1. Solving for cos y gives us cos y =± 1 − sin 2 y =± 1 − x 2 . Do we want the positive square root, or the negative square root? y is an angle between − π 2 and π 2 ,socosyis nonnegative; we choose √ 1 − x 2 . Therefore, dy dx = 1 √ 1−x 2 . d dx arcsin x = 1 √ 1 − x 2 Alternatively, once we know dy dx = 1 cos y , we can replace y by arcsin x obtaining dy dx = 1 cos(arcsin x) . cos(arcsin x) is the cosine of the angle between −π/2 and π/2 whose sine is x.Wecan draw a triangle with an acute angle θ whose sine is x, use the Pythagorean Theorem to solve for the length of the third side, and find cos(arcsin x) = √ 1 − x 2 , as illustrated in Figure 21.12. 12 1 θ x √1–x 2 Figure 21.12 The Derivative of arctan x We can use the same method to compute d dx arctan x. y = arctan xyis the angle between − π 2 and π 2 whose tangent is x. tan y = x d dx tan y = d dx x Use the Chain Rule on the left-hand side. (sec 2 y) dy dx =1 dy dx =cos 2 y 12 Knowing that θ ∈ [−π/2, π/2] assures us that cos θ is nonnegative. 21.4 Derivatives of Inverse Trigonometric Functions 705 To express dy dx in terms of x we can replace y with arctan x. cos(arctan x) is the cosine of the angle between −π/2 and π/2 whose tangent is x. Notice that cosine is positive on this interval. We can draw a triangle with an acute angle θ whose tangent is x, use the Pythagorean Theorem to solve for the length of the third side, and find cos(arctan x) = 1 √ 1+x 2 , as illustrated in Figure 21.13. 1 θ x √1+x 2 Figure 21.13 cos y = 1 √ 1+x 2 ,so dy dx =cos 2 y = 1 1+x 2 . d dx arctan x = 1 1 + x 2 EXERCISE 21.4 Using the same method demonstrated in the previous two examples, show the following. d dx arccos x = −1 √ 1 − x 2 COMMENT Surprisingly, the derivatives of the inverse trigonometric functions involve neither trigonometric functions nor inverse trigonometric functions. We will find it handy to keep inverse trigonometric functions and their derivatives in mind later on, when searching for a function whose derivative is 1 1+x 2 or 1 √ 1−x 2 . We mention this here because the inverse trigonometric functions are probably not the first things you might think of when looking for such functions! ◆ EXAMPLE 21.4 Differentiate. (a) f(x)=x arcsin(x 2 ) (b) j(x)=5tan −1 x 2 (c) g(x) =tan √ x · tan −1 √ x SOLUTIONS (a) Use the Product Rule. The Chain Rule tells us that d dx arcsin(mess) = 1 √ 1−(mess) 2 · (mess) . Here mess = x 2 . f (x) = x · d dx [arcsin(x 2 )] + arcsin(x 2 ) = x · 1 1 − (x 2 ) 2 · 2x + arcsin(x 2 ) = 2x 2 √ 1 − x 4 + arcsin(x 2 ) 706 CHAPTER 21 Differentiation of Trigonometric Functions (b) The Chain Rule tells us that d dx tan −1 (mess) = 1 1+(mess) 2 · (mess) , where mess = x 2 . j (x) = 5 1 + (x 2 ) 2 · 2x = 10x 1 + x 4 CAUTION (mess) is the derivative of x 2 , not of x 4 . (c) g(x) =tan √ x · tan −1 √ x. Again we need the Product Rule together with the Chain Rule. g (x) = sec 2 √ x · 1 2 1 √ x · tan −1 √ x + tan √ x · 1 1 + ( √ x) 2 · 1 2 1 √ x = 1 2 √ x sec 2 √ x · tan −1 √ x + tan √ x 1 + x ◆ PROBLEMS FOR SECTION 21.4 For Problems 1 through 6, differentiate the function given. 1. y = 3 tan x − 4 tan −1 x 2. f(x)=3arctan 2 √ x 3. y = sin x · arcsin x 4. y = √ tan −1 x 5. y = x tan −1 x 6. y = arctan(e x ) e 7. (a) Show that the derivative of arccos x is −1 √ 1−x 2 . (b) What is the domain of arccos x? (c) What is the range of arccos x? (d) Where is the graph of arccos x decreasing? (e) Where is the graph of arccos x concave up? Concave down? If there is a point of inflection, where is it? (f) Graph f(x)=arccos x. 8. Differentiate f(x)=3cos 1 x 2 +1 + x arctan 1 x . 9. Compute d dx sin −1 x cos −1 x . Is it the same as d dx tan −1 x? 21.5 Brief Trigonometry Summary 707 21.5 BRIEF TRIGONOMETRY SUMMARY Unit Circle Definitions of Trigonometric Functions sin x = v cos x = u tan x = sin x cos x = v u 2 1 10– 0.42 v u (cos 2, sin 2) = P(2) u 2 + v 2 = 1 Right-Triangle Definitions of Trigonometric Functions sin x = opp hyp csc x = 1 sin x cos x = adj hyp sec x = 1 cos(x) tan x = opp adj cot x = 1 tan(x) x hyp opp adj Radians and Degrees. π radians = 180 ◦ Special triangles 1 2 1 2 π 4 √2 2 1 2 2 2 π 3 π 6 √3 2 Note: The shortest side is opposite the smallest angle and the largest side is opposite the largest angle. 45 ◦ -45 ◦ -90 ◦ 30 ◦ -60 ◦ -90 ◦ 708 CHAPTER 21 Differentiation of Trigonometric Functions Graphs of the Trigonometric Functions π –π –2π 2π Graph of f(x) = tan(x)Graph of f(x) = cos(x) x Domain: x ≠ + πn Range: (– ∞, ∞)Domain: (– ∞, ∞) Range: [–1, 1] π 2 x 1 –1 π –π 2π –2π Graph of f(x) = sin(x) Domain: (– ∞, ∞) Range: [–1, 1] x 1 –1 π –π 2π –2π The Relationship Between Graphs and Equations y = A sin(Bx) + K A cos(Bx) + K amplitude =|A|, period = 2π |B| , balance value =K y = A sin(Bx + C) should be rewritten as y = A sin B x + C B ; to read off amplitude =|A|,period = 2π |B| , shift the graph of y = A sin(Bx) to the left C B units. Inverse Trigonometric Functions We must restrict the domain of sine, cosine, and tangent in order to construct inverse functions, because a function must be one-to-one to have an inverse function. arcsin(x) is the angle between −π/2 and π/2 whose sine is x. domain of arcsin x:[−1, 1] arctan(x) is the angle between −π/2 and π/2 whose tangent is x. domain of arctan x: (−∞, ∞) arccos(x) is the angle between 0 and π whose cosine is x. domain of arccos x:[−1, 1] NOTE If A is positive then arcsin(A), arccos(A), and arctan(A) are all angles between 0 and π/2 (provided A is in the function’s domain). If A is positive and in the function’s domain, then arcsin(−A) =−arcsin(A), arccos(−A) = π − arccos(A), and arctan(−A) =−arctan(A). Arc Length. The arc length of a circle of radius R subtended by an angle of x radians is Rx. Derivatives Function Derivative Function Derivative f(x)=sin(x) f (x) = cos(x) f (x) = arcsin(x) f (x) = 1 √ 1−x 2 f(x)=cos(x) f (x) =−sin(x) f (x) = arccos(x) f (x) =− 1 √ 1−x 2 f(x)=tan(x) f (x) = sec 2 (x) f (x) = arctan(x) f (x) = 1 1+x 2 21.5 Brief Trigonometry Summary 709 Remember the Chain Rule d dx sin[g(x)] = cos[g(x)] · dg dx d dx arcsin[g(x)] = 1 1 − [g(x)] 2 · dg dx d dx cos[g(x)] =−sin[g(x)] · dg dx d dx arccos[g(x)] = −1 1 − [g(x)] 2 · dg dx d dx tan[g(x)] = sec 2 [g(x)] · dg dx d dx arctan[g(x)] = 1 1 + [g(x)] 2 · dg dx Trigonometric Identities Pythagorean Identity sin 2 A + cos 2 A = 1 It follows that sin 2 A cos 2 A + cos 2 A cos 2 A = 1 cos 2 A so tan 2 A + 1 = sec 2 A sin 2 A sin 2 A + cos 2 A sin 2 A = 1 sin 2 A so 1 + cot 2 A = csc 2 A. Addition Formulas and the Double-Angle Formulas sin(A + B) = sin A cos B + sin B cos A cos(A + B) = cos A cos B − sin A sin B It follows that sin(2A) = 2 sin A cos A and cos(2A) = cos 2 A − sin 2 A. General Triangles A B C c a b Law of Cosines c 2 = a 2 + b 2 − 2ab cos C You can think of −2ab cos C as a “correction term” for the Pythagorean Theorem; it is positive when C>π/2and negative when C<π/2. Law of Sines a sin A = b sin B = c sin C . have to look down at more and more of an angle to see it. At what rate is this angle of depression increasing at the instant when you are at a height of 100 feet? Include units in your answer. (Problem. Derivatives of Inverse Trigonometric Functions 705 To express dy dx in terms of x we can replace y with arctan x. cos(arctan x) is the cosine of the angle between −π/2 and π/2 whose tangent is. must be one -to- one to have an inverse function. arcsin(x) is the angle between −π/2 and π/2 whose sine is x. domain of arcsin x:[−1, 1] arctan(x) is the angle between −π/2 and π/2 whose tangent is