Math 147 Exam II Practice Problems

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1 Math 147 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture, all homework problems, all lab assignment problems, and all quiz problems. 1. If f(x) = x 4 2x 3 + 4x 2 10x + 1, find the equation of the tangent line to the graph of y = f(x) at the point (1, 6). Express your answer in slope-intercept form. 2. If f(x) = 2x 3 3x 2 6x + 8, find the equation of the normal line to the graph of y = f(x) at the point (1, 1). Express your answer in slope-intercept form. 3. Differentiate f(x) = (x 4 + 2x + 1)(3x 2 5) and simplify completely. 4. Differentiate f(x) = 3x 7 x 2 + 5x 4 and simplify completely. 5. Suppose that f(5) = 1, f (5) = 6, g(5) = 3, and g (5) = 2. Find the values of: (a) (fg) (5) (b) (f/g) (5) 6. If y = u 3 + u 2 + 1, where u = 2x 2 1, find dy dx 7. Differentiate f(x) = (x 2 + 4x + 6) 5. at x = Differentiate f(x) = x 2 7x. Express your answer using positive exponents. 9. Differentiate f(t) = 3 Express your answer using positive exponents. (2t 2 6t + 1) Differentiate f(t) = (6t 2 + 5) 3 (t 3 7) 4 and simplify completely. ( ) 3 x Differentiate f(x) = and simplify completely. x Differentiate f(x) = x. Express your answer using positive exponents. 13. Suppose that F (x) = f(g(x)), where g(3) = 6, g (3) = 4, f (3) = 2, and f (6) = 7. Find F (3). 14. Differentiate f(x) = x csc x and simplify completely. 15. Differentiate f(x) = sin x 1 + cos x and simplify completely. 16. Differentiate f(x) = sin(x 3 ) + cos 3 x and simplify completely. 17. Differentiate f(x) = sin(cos(tan x)) and simplify completely. 1

2 18. Differentiate f(x) = sec 2 (2x) + cot 1 + x 2 and simplify completely. 19. Differentiate f(x) = e tan x. 20. Differentiate f(x) = e 4x sin(5x) and simplify completely. 21. Differentiate f(x) = e3x and simplify completely. 1 + ex 22. Differentiate f(x) = 10 x Differentiate f(x) = ln x. Express your answer using positive exponents. 24. Differentiate f(x) = ln(sin x) and simplify completely. 25. Differentiate f(x) = x 2 ln(x 3 4) and simplify completely. 26. Differentiate f(x) = 1 ln x 1 + ln x 27. Differentiate f(t) = log 2 (t 4 t 2 + 1). 28. Differentiate f(x) = log(2x + sin x). and simplify completely. 29. Differentiate f(x) = sin 1 (x 2 1) and simplify completely. 30. Differentiate f(x) = x arctan x and simplify completely. 31. Differentiate f(t) = cos 1 2t 1 and simplify completely. 32. If f(x) = 2x + cos x, find 33. If f(x) = x 3 + x 2 + x + 1, find df 1 dx (1) = (f 1 ) (1). 34. Consider the curve defined implicitly by df 1 dx (2) = (f 1 ) (2). x 3 + y 3 = 6xy (a) Find dy dx. (b) Find the equation of the tangent line to the curve at the point (3, 3). Express your answer in slope-intercept form. 35. Consider the curve defined implicitly by x cos y + y cos x = 1 (a) Find dy dx. (b) Find the equation of the tangent line to the curve at the point (0, 1). Express your answer in slope-intercept form. 2

3 36. Consider the curve defined implicitly by (a) Find y = dy/dx. cos(x y) = xe x (b) Find the equation of the tangent line to the curve at the point (0, π/2). Express your answer in slope-intercept form. 37. Consider the curve defined implicitly by (a) Find y = dy/dx. y = ln(x 2 + y 2 ) (b) Find the equation of the tangent line to the curve at the point (1, 0). Express your answer in slope-intercept form. 38. Use logarithmic differentiation to find the derivative of y = (x3 + 1) 4 x (x + 1) 3 (x 2 + 3) Use logarithmic differentiation to find the derivative of f(x) = (sin x) cos x. 40. Find all higher derivatives of f(x) = x 6 2x 5 + 3x 4 4x 3 + 5x 2 6x If f(x) = 1 x 3, find f (2016) (x). 42. If f(x) = sin(2x), find f (50) (x). 43. Find the thousandth derivative of f(x) = xe x. 44. If f(x) = ln(x 1), find f (99) (x). 45. Find y if x 4 + y 4 = The position of a particle is given by the equation s(t) = t 3 6t 2 + 9t where t is measured in seconds and s in feet. (a) Find the velocity and acceleration at time t. (b) What are the velocity and acceleration after 2 s? Include the appropriate units. (c) Find the average velocity from t = 1 to t = 5 s. Include the appropriate unit for velocity. (d) When is the particle at rest? What is the acceleration at these times? Include the appropriate unit for acceleration. 3

4 47. The position of a particle is given by the equation s(t) = t 2 + t + 4 where s is measured in meters and t in seconds. (a) Find the average velocity of the particle from t = 0 to t = 3 s. appropriate unit for velocity. Include the (b) Find the instantaneous velocity of the particle after 3 s. Include the appropriate unit for velocity. (c) State the name of a theorem which guarantees that there is some time T (0, 3) such that the instantaneous velocity of the particle at time T seconds is equal to the average velocity found in part (a)? Find the time T guaranteed by this theorem. 48. Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cubic centimeters per second. How fast is the radius of the balloon increasing when the diameter is 50 centimeters? 49. A ladder 10 ft long rests against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 2 ft/s. (a) How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 4 ft from the wall? (b) How fast is the angle between the top of the ladder and the wall changing when the angle is π/4 radians? 50. A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m 3 /min, find the rate at which the water level is rising when the water is 3 m deep. 51. Car A is traveling west at 50 mi/h and car B is traveling north at 60 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection? 52. Use a linear approximation to find an approximate value for Find the absolute extrema of f(x) = x 3 3x on [ 1, 3]. 54. Find the value of c which satisfies the Mean Value Theorem for f(x) = x x + 5 interval [1, 10]. on the 55. Suppose that f(x) is continuous on the interval [2, 5] and differentiable on the interval (2, 5). Show that if 1 f (x) 4 for all x [2, 5], then 3 f(5) f(2) 12. 4

5 Solutions Solutions may contain errors or typos. If you find an error or typo, please notify me at 1. If f(x) = x 4 2x 3 + 4x 2 10x + 1, find the equation of the tangent line to the graph of y = f(x) at the point (1, 6). Express your answer in slope-intercept form. The slope is m = f (1), which we calculate as follows: f (x) = 4x 3 6x 2 + 8x 10 f (1) = = 4 Therefore, the equation of the tangent line is y + 6 = 4(x 1) y = 4x 2 2. If f(x) = 2x 3 3x 2 6x + 8, find the equation of the normal line to the graph of y = f(x) at the point (1, 1). Express your answer in slope-intercept form. The slope of the tangent line is m 1 = f (1) which we calculate as follows: f (x) = 6x 2 6x 6 f (1) = 6 Thus, the slope of the normal line is m 2 = 1/m 1 = 1/6. Therefore, the equation of the tangent line is y 1 = 1 (x 1) 6 y 1 = 1 6 x 1 6 y = 1 6 x

6 3. Differentiate f(x) = (x 4 + 2x + 1)(3x 2 5) and simplify completely. By the Product Rule, we have f (x) = (x 4 + 2x + 1) d dx (3x2 5) + (3x 2 5) d dx (x4 + 2x + 1) = (x 4 + 2x + 1)(6x) + (3x 2 5)(4x 3 + 2) = 6x x 2 + 6x + 12x 5 + 6x 2 20x 3 10 = 18x 5 20x x 2 + 6x 10 Notice that we could first multiply the factors: f(x) = (x 4 + 2x + 1)(3x 2 5) = 3x 6 + 6x 3 + 3x 2 5x 4 10x 5 = 3x 6 5x 4 + 6x 3 + 3x 2 10x 5 and then differentiate to obtain f (x) = 18x 5 20x x 2 + 6x Differentiate f(x) = 3x 7 x 2 + 5x 4 By the Quotient Rule, we have and simplify completely. f (x) = (x2 + 5x 4)D(3x 7) (3x 7)D(x 2 + 5x 4) (x 2 + 5x 4) 2 = (x2 + 5x 4)(3) (3x 7)(2x + 5) (x 2 + 5x 4) 2 = 3x2 + 15x 12 (6x 2 + x 35) (x 2 + 5x 4) 2 = 3x2 + 14x + 23 (x 2 + 5x 4) 2 6

7 5. Suppose that f(5) = 1, f (5) = 6, g(5) = 3, and g (5) = 2. Find the values of: (a) (fg) (5) By the Product Rule, we have (fg) (5) = f (5)g(5) + f(5)g (5) = 6( 3) + 1(2) = 16 (b) (f/g) (5) By the Quotient Rule, we have (f/g) (5) = f (5)g(5) f(5)g (5) [g(5)] 2 = 6( 3) 1(2) = 20 ( 3) If y = u 3 + u 2 + 1, where u = 2x 2 1, find dy dx By the Chain Rule, we have At x = 2, u = 2(2) 2 1 = 7, and so at x = 2. dy dx = dy du du dx = (3u2 + 2u)(4x) dy dx = ( )(4 2) = = Differentiate f(x) = (x 2 + 4x + 6) 5. By the Chain Rule, we have f (x) = 5(x 2 + 4x + 6) 4 d dx (x2 + 4x + 6) = 5(x 2 + 4x + 6) 4 (2x + 4) = (10x + 20)(x 2 + 4x + 6) 4 7

8 8. Differentiate f(x) = x 2 7x. Express your answer using positive exponents. We first rewrite f as By the Chain Rule, we have f(x) = (x 2 7x) 1/2 f (x) = 1 2 (x2 7x) 1/2 d dx (x2 7x) 1 = 2 (2x 7) x 2 7x 2x 7 = 2 x 2 7x 9. Differentiate f(t) = 3 Express your answer using positive exponents. (2t 2 6t + 1) 8 We first rewrite f as By the Chain Rule, we have f(t) = 3(2t 2 6t + 1) 8 f (t) = 24(2t 2 6t + 1) 9 d dt (2t2 6t + 1) = 24(2t 2 6t + 1) 9 (4t 6) = 24(4t 6) (2t 2 6t + 1) Differentiate f(t) = (6t 2 + 5) 3 (t 3 7) 4 and simplify completely. By the Product and Chain Rules, we have f (t) = d dt [(6t2 + 5) 3 ](t 3 7) 4 + (6t 2 + 5) 3 d dt [(t3 7) 4 ] = 3(6t 2 + 5) 2 (12t)(t 3 7) 4 + (6t 2 + 5) 3 4(t 3 7) 3 (3t 2 ) = 36t(6t 2 + 5) 2 (t 3 7) t 2 (6t 2 + 5) 3 (t 3 7) 3 By using common factors, we can simplify the answer as f (t) = 12t(6t 2 + 5) 2 (t 3 7) 3 [3(t 3 7) + t(6t 2 + 5)] = 12t(6t 2 + 5) 2 (t 3 7) 3 (9t 3 + 5t 21) 8

9 11. Differentiate f(x) = ( ) 3 x 6 and simplify completely. x + 7 By the Chain and Quotient Rules, we have ( ) 2 ( ) x 6 f d x 6 (x) = 3 x + 7 dx x + 7 ( ) 2 x 6 (1)(x + 7) (x 6)(1) = 3 x + 7 (x + 7) 2 3(x 6)2 13 = (x + 7) 2 (x + 7) 2 = 39(x 6)2 (x + 7) Differentiate f(x) = x. Express your answer using positive exponents. We first rewrite f as By the Chain Rule, we have f(x) = (1 + x) 1/3 f (x) = 1 3 (1 + x) 2/3 d dx (1 + x) ( ) 1 1 = 3(1 + x) 2/3 2 x 1 = 6 x(1 + x) 2/3 Notice that the Chain Rule has been used twice. 13. Suppose that F (x) = f(g(x)), where g(3) = 6, g (3) = 4, f (3) = 2, and f (6) = 7. Find F (3). By the Chain Rule, we have F (3) = f (g(3))g (3) = f (6)(4) = 7(4) = 28 9

10 14. Differentiate f(x) = x csc x and simplify completely. By the Product Rule, we have f (x) = (1) csc x + x( csc x cot x) = csc x x csc x cot x 15. Differentiate f(x) = sin x 1 + cos x By the Quotient Rule, we have f (x) = and simplify completely. cos x(1 + cos x) sin x( sin x) (1 + cos x) 2 = cos x + cos2 x + sin 2 x (1 + cos x) 2 = cos x + 1 (1 + cos x) 2 1 = 1 + cos x In simplifying the answer, we have used the identity sin 2 x + cos 2 x = Differentiate f(x) = sin(x 3 ) + cos 3 x and simplify completely. By the Chain Rule, we have f (x) = cos(x 3 )(3x 2 ) + 3 cos 2 x( sin x) = 3x 2 cos(x 3 ) 3 cos 2 x sin x 17. Differentiate f(x) = sin(cos(tan x)) and simplify completely. By the Chain Rule, we have f (x) = cos(cos(tan x)) d [cos(tan x)] dx = cos(cos(tan x))[ sin(tan x)] d (tan x) dx = cos(cos(tan x)) sin(tan x) sec 2 x Notice that the Chain Rule has been used twice. 10

11 18. Differentiate f(x) = sec 2 (2x) + cot 1 + x 2 and simplify completely. By the Chain Rule, we have f (x) = 2 sec(2x) d dx [sec(2x)] csc2 1 + x d 2 dx ( 1 + x 2 ) = 2 sec(2x)[2 sec(2x) tan(2x)] csc [ ] x 2 2 d 1 + x 2 dx (1 + x2 ) = 4 sec 2 (2x) tan(2x) csc [ ] x x (2x) 2 = 4 sec 2 (2x) tan(2x) csc ( ) 2 x 1 + x x Differentiate f(x) = e tan x. By the Chain Rule, we have f (x) = e tan x d (tan x) dx = e tan x sec 2 x 20. Differentiate f(x) = e 4x sin 5x and simplify completely. By the Product and Chain Rules, we have f (x) = d dx (e 4x ) sin 5x + e 4x d (sin 5x) dx = 4e 4x sin 5x + e 4x (5 cos 5x) = e 4x ( 4 sin 5x + 5 cos 5x) 21. Differentiate f(x) = e3x and simplify completely. 1 + ex By the Quotient and Chain Rules, we have f (x) = 3e3x (1 + e x ) e 3x (e x ) (1 + e x ) 2 = 3e3x + 3e 4x e 4x (1 + e x ) 2 = 3e3x + 2e 4x (1 + e x ) 2 11

12 22. Differentiate f(x) = 10 x2. By the Chain Rule, we have f (x) = (ln 10)10 x2 d dx (x2 ) = (ln 10)10 x2 (2x) 23. Differentiate f(x) = ln x. Express your answer using positive exponents. By the Chain Rule, we have f (x) = = = 1 2 d (ln x) ln x dx ln x x 1 2x ln x 24. Differentiate f(x) = ln(sin x) and simplify completely. By the Chain Rule, we have f (x) = cos x sin x = cot x 25. Differentiate f(x) = x 2 ln(x 3 4) and simplify completely. By the Product and Chain Rules, we have f (x) = d dx (x2 ) ln(x 3 4) + x 2 d dx [ln(x3 4)] = 2x ln(x 3 4) + x 2 3x 2 x 3 4 = 2x ln(x 3 4) + 3x4 x

13 26. Differentiate f(x) = 1 ln x 1 + ln x By the Quotient Rule, we have and simplify completely. f (x) = 1 x (1 + ln x) (1 ln x) 1 x (1 + ln x) 2 2 x = (1 + ln x) 2 2 = x(1 + ln x) Differentiate f(t) = log 2 (t 4 t 2 + 1). By the Chain Rule, we have f (t) = 4t 3 2t (t 4 t 2 + 1) ln Differentiate f(x) = log(2x + sin x). By the Chain Rule, we have f (x) = 2 + cos x (2x + sin x) ln Differentiate f(x) = sin 1 (x 2 1). By the Chain Rule, we have f (x) = = = 1 d 1 (x2 1) 2 dx (x2 1) 1 1 (x4 2x 2 + 1) 2x 2x 2x2 x 4 13

14 30. Differentiate f(x) = x arctan x and simplify completely. By the Product and Chain Rules, we have f (x) = d dx (x) arctan x + x d dx (arctan x) = (1) arctan 1 x + x ( d x x) dx = arctan x + x 1 x x = arctan x x + 2(x + 1) 31. Differentiate f(t) = cos 1 2t 1 and simplify completely. By the Chain Rule, we have f 1 d (t) = 1 ( 2t 1 2t 1) 2 dt 1 2 = 1 (2t 1) 2 2t 1 1 = 2 2t 2t If f(x) = 2x + cos x, find df 1 dx (1) = (f 1 ) (1). To use the theorem for the derivative of an inverse function, we find f (x) = 2 sin x. Additionally, we need to know f 1 (1) and we can find it by inspection. Since f(0) = 1, f 1 (1) = 0. Therefore, (f 1 ) (1) = 1 f [f 1 (1)] = 1 f (0) = 1 2 sin 0 =

15 33. If f(x) = x 3 + x 2 + x + 1, find df 1 dx (2) = (f 1 ) (2). To use the theorem for the derivative of an inverse function, we find f (x) = 3x 2 + 2x x 3 + x 2 + x + 1 Additionally, we need to know f 1 (2) and we can find it by inspection. Since f(1) = 2, f 1 (2) = 1. Therefore, (f 1 ) 1 (2) = f [f 1 (2)] = 1 f (1) = = = Consider the curve defined implicitly by (a) Find dy dx. x 3 + y 3 = 6xy Differentiating both sides with respect to x, regarding y as a function of x, and using the Chain Rule on the y 3 term and Product Rule on the 6xy term, we have Solving for y, we obtain 3x 2 + 3y 2 y = 6y + 6xy x 2 + y 2 y = 2y + 2xy (y 2 2x)y = 2y x 2 y = 2y x2 y 2 2x (b) Find the equation of the tangent line to the curve at the point (3, 3). Express your answer in slope-intercept form. When x = y = 3, the slope of the tangent line is y = So the equation of the tangent line is 2(3) (3) = 1 y 3 = 1(x 3) y = x

16 35. Consider the curve defined implicitly by (a) Find dy dx. x cos y + y cos x = 1 Differentiating both sides with respect to x, regarding y as a function of x, and using the Product and Chain Rules, we have Solving for y, we obtain (1) cos y x sin y(y ) + y cos x y sin x = 0 cos y xy sin y + y cos x y sin x = 0 xy sin y + y cos x = cos y + y sin x y ( x sin y + cos x) = cos y + y sin x y = cos y + y sin x x sin y + cos x (b) Find the equation of the tangent line to the curve at the point (0, 1). Express your answer in slope-intercept form. When x = 0 and y = 1, the slope of the tangent line is y = So the equation of the line is cos sin 0 0 sin 1 + cos 0 = cos 1 y = 1 (cos 1)x 16

17 36. Consider the curve defined implicitly by (a) Find dy dx. cos(x y) = xe x Differentiating both sides with respect to x, regarding y as a function of x, and using the Chain Rule on the cos(x y) term and Product Rule on the xe x term, we have Solving for y, we obtain sin(x y)(1 y ) = (1)e x + x(e x ) sin(x y) + y sin(x y) = e x (1 + x) y sin(x y) = sin(x y) + e x (1 + x) y = 1 + ex (1 + x) sin(x y) (b) Find the equation of the tangent line to the curve at the point (0, π/2). Express your answer in slope-intercept form. When x = 0 and y = π/2, the slope of the tangent line is y = 1 + e0 (1 + 0) sin(0 π/2) = = 0 So the tangent line is the horizontal line y = π/2. 17

18 37. Consider the curve defined implicitly by (a) Find dy dx. y = ln(x 2 + y 2 ) Differentiating both sides with respect to x, regarding y as a function of x, and using the Chain Rule on the ln(x 2 + y 2 ) term, we have y = 2x + 2yy x 2 + y 2 Solving for y, we obtain y (x 2 + y 2 ) = 2x + 2yy y (x 2 + y 2 ) 2yy = 2x y (x 2 + y 2 2y) = 2x y = 2x x 2 + y 2 2y (b) Find the equation of the tangent line to the curve at the point (1, 0). Express your answer in slope-intercept form. When x = 1 and y = 0, the slope of the tangent line is So the equation of the line is y = 2(1) (0) = 2 y 0 = 2(x 1) y = 2x 2 18

19 38. Use logarithmic differentiation to find the derivative of y = (x3 + 1) 4 x (x + 1) 3 (x 2 + 3) 2 We take logarithms of both sides of the equation: ln y = 4 ln(x 3 + 1) ln(x5 + 2) 3 ln(x + 1) 2 ln(x 2 + 3) Differentiating implicitly with respect to x gives y y = 12x2 x x4 2x x + 1 Solving for y, we obtain ( 12x y 2 = y x x4 2x x + 1 = (x3 + 1) 4 ( x x 2 (x + 1) 3 (x 2 + 3) 2 x x ) x x x x4 2x x + 1 4x ) x Use logarithmic differentiation to find the derivative of f(x) = (sin x) cos x. Let y = (sin x) cos x. We take logarithms of both sides of the equation: ln y = ln[(sin x) cos x ] = (cos x) ln(sin x) Differentiating implicitly with respect to x gives y y y y = ( sin x) ln(sin x) + (cos x) cos x sin x = (sin x) ln(sin x) + cos x cot x Solving for y, we obtain y = y [ (sin x) ln(sin x) + cos x cot x] = (sin x) cos x [ (sin x) ln(sin x) + cos x cot x] 19

20 40. Find all higher derivatives of f(x) = x 6 2x 5 + 3x 4 4x 3 + 5x 2 6x + 7. Differentiating repeatedly, we have f (x) = 6x 5 10x x 3 12x x 6 f (x) = 30x 4 40x x 2 24x + 10 f (x) = 120x 3 120x x 24 f (4) (x) = 360x 2 240x + 72 f (5) (x) = 720x 240 f (6) (x) = 720 f (7) (x) = 0 In fact, f (n) (x) = 0 for all n If f(x) = 1 x 3, find f (2016) (x). Differentiating repeatedly, we have f(x) = x 3 = 1 x 3 f (x) = 3x 4 = 3 x 4 f (x) = ( 4)( 3)x 5 = 12 x 5 f (x) = x 6 f (4) (x) = x 7 f (5) (x) = x 8 = 7! 2x 8. f (n) (x) = ( 1) n (n + 2)(n + 1)(n) 4 3 x (n+3) f (n) (x) = ( 1)n (n + 2)! 2x n+3 Therefore, the 2016th derivative is f (2016) (x) = 2018! 2x

21 42. If f(x) = sin(2x), find f (50) (x). The first few derivatives of f(x) = sin(2x) are f (x) = 2 cos(2x) f (x) = 4 sin(2x) f (x) = 8 cos(2x) f (4) (x) = 16 sin(2x) f (5) (x) = 32 cos(2x) We see that the successive derivatives occur in a cycle of length 4, the coefficients are powers of 2 and, in particular, whenever n is a multiple of 4. Therefore, f (n) (x) = 2 n sin(2x) f (48) (x) = 2 48 sin(2x) and, differentiating two more times, we have f (50) (x) = 2 50 sin(2x) 43. Find the thousandth derivative of f(x) = xe x. The first few derivatives of f(x) = xe x are f (x) = e x xe x = e x (1 x) f (x) = e x (1 x) e x = e x (x 2) f (x) = e x (x 2) + e x = e x (3 x) f (4) (x) = e x (3 x) e x = e x (x 4). f (n) (x) = ( 1) n e x (x n) Therefore, the 1000th derivative is f (1000) (x) = e x (x 1000) 21

22 44. If f(x) = ln(x 1), find f (99) (x). The first few derivatives of f(x) = ln(x 1) are f 1 (x) = = (x 1) 1 x 1 f (x) = (x 1) 2 1 = (x 1) 2 f (x) = 2(x 1) 3 = 2 (x 1) 3 f (4) (x) = 3 2(x 1) 4 = 3! (x 1) 4 f (5) (x) = 4 3 2(x 1) 5 =. 4! (x 1) 5 f (n) (x) = ( 1) n+1 (n 1)!(x 1) n = ( 1)n+1 (n 1)! (x 1) n Therefore, the 99th derivative is f (99) (x) = 98! (x 1) 99 22

23 45. Find y if x 4 + y 4 = 16. Differentiating the equation implicitly with respect to x, we obtain Solving for y gives 4x 3 + 4y 3 y = 0 4y 3 y = 4x 3 y = x3 y 3 Using the Quotient Rule and remembering that y is a function of x, we have y = 3x2 (y 3 ) x 3 (3y 2 y ) y 6 = 3x2 y 3 3x 3 y 2 y y 6 Substituting the expression we obtained for y into this expression, we obtain ( ) x 3 3x 2 y 3 3x 3 y 2 y y = 3 y 6 = 3x2 y 4 + 3x 6 y 7 = 3x2 (y 4 + x 4 ) y 7 But the values of x and y must satisfy the original equation x 4 + y 4 = 16. So the answer simplifies to y = 3x2 (16) y 7 = 48x2 y 7 23

24 46. The position of a particle is given by the equation s(t) = t 3 6t 2 + 9t where t is measured in seconds and s in feet. (a) Find the velocity and acceleration at time t. The velocity function is the derivative of the position function: v(t) = s (t) = 3t 2 12t + 9 The acceleration function is the derivative of the velocity function: a(t) = v (t) = s (t) = 6t 12 (b) What are the velocity and acceleration after 2 s? Include the appropriate units. The velocity after 2 s means the instantaneous velocity when t = 2. That is, v(2) = 3(2) 2 12(2) + 9 = 3 ft/s Similarly, the acceleration after 2 s is a(2) = 6(2) 12 = 0 ft/s 2 (c) Find the average velocity from t = 1 to t = 5 s. Include the appropriate unit for velocity. The average velocity from t = 1 to t = 5 is Change in position Change in time = s(5) s(1) 5 1 = = 4 f/s (d) When is the particle at rest? What is the acceleration at these times? Include the appropriate unit for acceleration. The particle is at rest when v(t) = 0. That is, 3t 2 12t + 9 = 3(t 2 4t + 3) = 3(t 1)(t 3) = 0 and this is true when t = 1 or t = 3. Thus, the particle is at rest after 1 s and after 3 s. The acceleration at these times is a(1) = 6(1) 12 = 6 ft/s 2 a(3) = 6(3) 12 = 6 ft/s 2 24

25 47. The position of a particle is given by the equation s(t) = t 2 + t + 4 where s is measured in meters and t in seconds. (a) Find the average velocity of the particle from t = 0 to t = 3 s. Include the appropriate unit for velocity. The average velocity from t = 0 to t = 3 is Change in position Change in time = s(3) s(0) 3 0 = = 2 3 m/s (b) Find the instantaneous velocity of the particle after 3 s. Include the appropriate unit for velocity. The velocity function is the derivative of the position function. By the Chain Rule, we have v(t) = s 2t + 1 (t) = 2 t 2 + t + 4 After t = 3 seconds, the instantaneous velocity is v(3) = 2(3) (3) = 7 8 m/s 25

26 (c) State the name of a theorem which guarantees that there is some time T (0, 3) such that the instantaneous velocity of the particle at time T seconds is equal to the average velocity found in part (a)? Find the time T guaranteed by this theorem. Since s(t) is continuous and differentiable on the interval [0, 3], the Mean Value Theorem guarantees that there is some time T (0, 3) such that the instantaneous velocity of the particle at time T seconds is equal to the average velocity from t = 0 to t = 3 s. To calculate this time T, let Therefore, we have v(t ) = 2 3 2T T 2 + T + 4 = 2 3 3(2T + 1) = 4 T 2 + T + 4 6T + 3 = 4 T 2 + T + 4 (6T + 3) 2 = 16(T 2 + T + 4) 36T T + 9 = 16T T T T 55 = 0 Using the Quadratic Formula, we obtain T = 20 ± (20) 2 4(20)( 55) 2(20) = 1 ± Since the value of T must lie in the interval (0, 3), we reject the negative value of T and obtain T = s 26

27 48. Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cubic centimeters per second. How fast is the radius of the balloon increasing when the diameter is 50 centimeters? The volume V of a sphere with radius r is V = 4 3 πr3 We differentiate both sides of this equation with respect to t. Using the Chain Rule, we have dv dt = dv dr dr = 4πr2 dr dt dt Now we solve for the unknown quantity: dr dt = 1 dv 4πr 2 dt Substituting r = 25 and dv/dt = 100 in this equation, we obtain dr dt = 1 4π(25) = π The radius of the balloon is increasing at the rate of 1/(25π) cm/s. 27

28 49. A ladder 10 ft long rests against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 2 ft/s. (a) How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 4 ft from the wall? We first draw a diagram and label it as in the figure below. θ y 10 x By the Pythagorean Theorem, x 2 + y 2 = 100 Differentiating each side with respect to t using the Chain Rule, we have 2x dx dt + 2y dy dt = 0 and solving this equation for the desired rate, we obtain dy dt = x dx y dt When x = 4, the Pythagorean Theorem gives y = 84 and so, substituting these values and dx/dt = 2, we have dy dt = 4 84 (2) = 4 21 The top of the ladder slides down the wall at the rate of 4/ 21 ft/s. 28

29 (b) How fast is the angle between the top of the ladder and the wall changing when the angle is π/4 radians? Using the diagram, we have sin θ = x 10 Differentiating each side with respect to t using the Chain Rule, we have cos θ dθ dt = 1 dx 10 dt and solving this equation for the desired rate, we obtain dθ dt = sec θ dx 10 dt Substituting θ = π/4 and dx/dt = 2, we have dθ dt = sec(π/4) 2 (2) = 10 5 The angle between the top of the ladder and the wall is increasing at the rate of 2/5 rad/s. 29

30 50. A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m 3 /min, find the rate at which the water level is rising when the water is 3 m deep. We first sketch the cone and label it as in the figure below. 2 r 4 h The volume V of the cone of water with height h and radius r is V = π 3 r2 h but it is very useful to express V as a function of h alone. In order to eliminate r we use the similar triangles to write and the expression for V becomes r h = 2 4 r = h 2 V = π 3 ( ) 2 h h = π 2 12 h3 Differentiating each side with respect to t using the Chain Rule, we have dv dt = π dh h2 4 dt and solving this equation for the desired rate, we obtain dh dt = 4 πh 2 dv dt Substituting h = 3 and dv/dt = 2, we have dh dt = 4 π(3) (2) = 8 2 9π The water level is rising at the rate of 8/(9π) m/min. 30

31 51. Car A is traveling west at 50 mi/h and car B is traveling north at 60 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection? We first draw a diagram and label it as in the figure below. x A y z B By the Pythagorean Theorem, z 2 = x 2 + y 2 Differentiating each side with respect to t using the Chain Rule, we have 2z dz dt dz dt = 2x dx dt = 1 z + 2y dy dt ( x dx dt + y dy dt ) When x = 0.3 and y = 0.4, the Pythagorean Theorem gives z = 0.5, so dz dt = 1 [0.3( 50) + 0.4( 60)] = The cars are approaching each other at a rate of 78 mi/h. 31

32 52. Use a linear approximation to find an approximate value for Let f(x) = 3 x = x 1/3 and a = 64. It follows that f (x) = 1 3x 2/3 Evaluating at a = 64, we obtain f(64) = 4 and f (64) = 1 3(64) 2/3 = 1 3(4) 2 = 1 48 Therefore, the linearization of f(x) at a = 64 is L(x) = f(64) + f (64)(x 64) = (x 64) 48 Thus, a linear approximation of is L(64.1) = ( ) 48 = ( ) = = Using a calculator, the actual value of is accurate to three decimal places is Thus, the approximation 32

33 53. Find the absolute extrema of f(x) = x 3 3x on [ 1, 3]. To find the critical values of f, we set f (x) = 3x 2 6x = 3x(x 2) = 0. Thus, the critical values are x = 0 and x = 2 which both lie in the interval [ 1, 3]. Evaluating the function at the critical values and the endpoints of the interval, we obtain f( 1) = 3 f(0) = 1 f(2) = 3 f(3) = 1 Therefore, f has an absolute maximum value of 1 and and absolute minimum value of 3. The graph of y = f(x) is given below. 33

34 54. Find the value of c which satisfies the Mean Value Theorem for f(x) = x x + 5 interval [1, 10]. on the First, note that f is continuous on the interval [1, 10] and differentiable on (1, 10). By the Mean Value Theorem, there exists a number c (1, 10) such that By the Quotient Rule, Moreover, f (c) = f(10) f(1) 10 1 f (x) = x + 5 x (x + 5) 2 = f(10) f(1) (x + 5) 2 = = 1 18 Therefore, we want to calculate the value of c (2, 5) which satisfies Solving for c, we obtain 5 (c + 5) 2 = 1 18 (c + 5) 2 = 90 c + 5 = ± 90 c + 5 = ±3 10 c = ± Since we seek the value of c which lies in the interval (1, 10), we take c = Suppose that f(x) is continuous on the interval [2, 5] and differentiable on the interval (2, 5). Show that if 1 f (x) 4 for all x [2, 5], then 3 f(5) f(2) 12. By the Mean Value Theorem, there exists a number c (2, 5) such that f (c) = Since 1 f (x) 4 on this interval, we have f(5) f(2) f(5) f(2) 3 Multiplying this inequality by 3, we obtain 4 3 f(5) f(2) 12 34