Math 147 Exam II Practice Problems
|
|
- Poppy Waters
- 6 years ago
- Views:
Transcription
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
DIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. However,
More information1 The Derivative and Differrentiability
1 The Derivative and Differrentiability 1.1 Derivatives and rate of change Exercise 1 Find the equation of the tangent line to f (x) = x 2 at the point (1, 1). Exercise 2 Suppose that a ball is dropped
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
3 Differentiation Rules 3.1 The Derivative of Polynomial and Exponential Functions In this section we learn how to differentiate constant functions, power functions, polynomials, and exponential functions.
More informationCalculus I Review Solutions
Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.
More informationMth Review Problems for Test 2 Stewart 8e Chapter 3. For Test #2 study these problems, the examples in your notes, and the homework.
For Test # study these problems, the examples in your notes, and the homework. Derivative Rules D [u n ] = nu n 1 du D [ln u] = du u D [log b u] = du u ln b D [e u ] = e u du D [a u ] = a u ln a du D [sin
More informationGuidelines for implicit differentiation
Guidelines for implicit differentiation Given an equation with x s and y s scattered, to differentiate we use implicit differentiation. Some informal guidelines to differentiate an equation containing
More informationMath Exam 02 Review
Math 10350 Exam 02 Review 1. A differentiable function g(t) is such that g(2) = 2, g (2) = 1, g (2) = 1/2. (a) If p(t) = g(t)e t2 find p (2) and p (2). (Ans: p (2) = 7e 4 ; p (2) = 28.5e 4 ) (b) If f(t)
More information3.4 The Chain Rule. F (x) = f (g(x))g (x) Alternate way of thinking about it: If y = f(u) and u = g(x) where both are differentiable functions, then
3.4 The Chain Rule To find the derivative of a function that is the composition of two functions for which we already know the derivatives, we can use the Chain Rule. The Chain Rule: Suppose F (x) = f(g(x)).
More informationFind the indicated derivative. 1) Find y(4) if y = 3 sin x. A) y(4) = 3 cos x B) y(4) = 3 sin x C) y(4) = - 3 cos x D) y(4) = - 3 sin x
Assignment 5 Name Find the indicated derivative. ) Find y(4) if y = sin x. ) A) y(4) = cos x B) y(4) = sin x y(4) = - cos x y(4) = - sin x ) y = (csc x + cot x)(csc x - cot x) ) A) y = 0 B) y = y = - csc
More informationAPPLICATIONS OF DERIVATIVES UNIT PROBLEM SETS
APPLICATIONS OF DERIVATIVES UNIT PROBLEM SETS PROBLEM SET #1 Related Rates ***Calculators Allowed*** 1. An oil tanker spills oil that spreads in a circular pattern whose radius increases at the rate of
More informationSample Questions Exam II, FS2009 Paulette Saab Calculators are neither needed nor allowed.
Sample Questions Exam II, FS2009 Paulette Saab Calculators are neither needed nor allowed. Part A: (SHORT ANSWER QUESTIONS) Do the following problems. Write the answer in the space provided. Only the answers
More informationSpring 2015 Sample Final Exam
Math 1151 Spring 2015 Sample Final Exam Final Exam on 4/30/14 Name (Print): Time Limit on Final: 105 Minutes Go on carmen.osu.edu to see where your final exam will be. NOTE: This exam is much longer than
More information, find the value(s) of a and b which make f differentiable at bx 2 + x if x 2 x = 2 or explain why no such values exist.
Math 171 Exam II Summary Sheet and Sample Stuff (NOTE: The questions posed here are not necessarily a guarantee of the type of questions which will be on Exam II. This is a sampling of questions I have
More informationMATH 10550, EXAM 2 SOLUTIONS. 1. Find an equation for the tangent line to. f(x) = sin x cos x. 2 which is the slope of the tangent line at
MATH 100, EXAM SOLUTIONS 1. Find an equation for the tangent line to at the point ( π 4, 0). f(x) = sin x cos x f (x) = cos(x) + sin(x) Thus, f ( π 4 ) = which is the slope of the tangent line at ( π 4,
More informationFind the slope of the curve at the given point P and an equation of the tangent line at P. 1) y = x2 + 11x - 15, P(1, -3)
Final Exam Review AP Calculus AB Find the slope of the curve at the given point P and an equation of the tangent line at P. 1) y = x2 + 11x - 15, P(1, -3) Use the graph to evaluate the limit. 2) lim x
More information1 + x 2 d dx (sec 1 x) =
Page This exam has: 8 multiple choice questions worth 4 points each. hand graded questions worth 4 points each. Important: No graphing calculators! Any non-graphing, non-differentiating, non-integrating
More informationMATH 135 Calculus 1 Solutions/Answers for Exam 3 Practice Problems November 18, 2016
MATH 35 Calculus Solutions/Answers for Exam 3 Practice Problems November 8, 206 I. Find the indicated derivative(s) and simplify. (A) ( y = ln(x) x 7 4 ) x Solution: By the product rule and the derivative
More information1. Compute the derivatives of the following functions, by any means necessary. f (x) = (1 x3 )(1/2)(x 2 1) 1/2 (2x) x 2 1( 3x 2 ) (1 x 3 ) 2
Math 51 Exam Nov. 4, 009 SOLUTIONS Directions 1. SHOW YOUR WORK and be thorough in your solutions. Partial credit will only be given for work shown.. Any numerical answers should be left in exact form,
More informationm(x) = f(x) + g(x) m (x) = f (x) + g (x) (The Sum Rule) n(x) = f(x) g(x) n (x) = f (x) g (x) (The Difference Rule)
Chapter 3 Differentiation Rules 3.1 Derivatives of Polynomials and Exponential Functions Aka The Short Cuts! Yay! f(x) = c f (x) = 0 g(x) = x g (x) = 1 h(x) = x n h (x) = n x n-1 (The Power Rule) k(x)
More informationMATH 151, SPRING 2018
MATH 151, SPRING 2018 COMMON EXAM II - VERSIONBKEY LAST NAME(print): FIRST NAME(print): INSTRUCTOR: SECTION NUMBER: DIRECTIONS: 1. The use of a calculator, laptop or computer is prohibited. 2. TURN OFF
More informationUNIT 3: DERIVATIVES STUDY GUIDE
Calculus I UNIT 3: Derivatives REVIEW Name: Date: UNIT 3: DERIVATIVES STUDY GUIDE Section 1: Section 2: Limit Definition (Derivative as the Slope of the Tangent Line) Calculating Rates of Change (Average
More informationMATH 162. Midterm Exam 1 - Solutions February 22, 2007
MATH 62 Midterm Exam - Solutions February 22, 27. (8 points) Evaluate the following integrals: (a) x sin(x 4 + 7) dx Solution: Let u = x 4 + 7, then du = 4x dx and x sin(x 4 + 7) dx = 4 sin(u) du = 4 [
More informationSolutions to Math 41 Final Exam December 10, 2012
Solutions to Math 4 Final Exam December,. ( points) Find each of the following limits, with justification. If there is an infinite limit, then explain whether it is or. x ln(t + ) dt (a) lim x x (5 points)
More informationMA 113 Calculus I Fall 2016 Exam Final Wednesday, December 14, True/False 1 T F 2 T F 3 T F 4 T F 5 T F. Name: Section:
MA 113 Calculus I Fall 2016 Exam Final Wednesday, December 14, 2016 Name: Section: Last 4 digits of student ID #: This exam has five true/false questions (two points each), ten multiple choice questions
More informationAP Calculus BC Chapter 4 AP Exam Problems. Answers
AP Calculus BC Chapter 4 AP Exam Problems Answers. A 988 AB # 48%. D 998 AB #4 5%. E 998 BC # % 5. C 99 AB # % 6. B 998 AB #80 48% 7. C 99 AB #7 65% 8. C 998 AB # 69% 9. B 99 BC # 75% 0. C 998 BC # 80%.
More informationMath 2413 General Review for Calculus Last Updated 02/23/2016
Math 243 General Review for Calculus Last Updated 02/23/206 Find the average velocity of the function over the given interval.. y = 6x 3-5x 2-8, [-8, ] Find the slope of the curve for the given value of
More informationa x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).
You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and
More informationSOLUTIONS FOR PRACTICE FINAL EXAM
SOLUTIONS FOR PRACTICE FINAL EXAM ANDREW J. BLUMBERG. Solutions () Short answer questions: (a) State the mean value theorem. Proof. The mean value theorem says that if f is continuous on (a, b) and differentiable
More information1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
Math120 - Precalculus. Final Review. Fall, 2011 Prepared by Dr. P. Babaali 1 Algebra 1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
More informationMath 222 Spring 2013 Exam 3 Review Problem Answers
. (a) By the Chain ule, Math Spring 3 Exam 3 eview Problem Answers w s w x x s + w y y s (y xy)() + (xy x )( ) (( s + 4t) (s 3t)( s + 4t)) ((s 3t)( s + 4t) (s 3t) ) 8s 94st + 3t (b) By the Chain ule, w
More informationMath 180, Exam 2, Practice Fall 2009 Problem 1 Solution. f(x) = arcsin(2x + 1) = sin 1 (3x + 1), lnx
Math 80, Exam, Practice Fall 009 Problem Solution. Differentiate the functions: (do not simplify) f(x) = x ln(x + ), f(x) = xe x f(x) = arcsin(x + ) = sin (3x + ), f(x) = e3x lnx Solution: For the first
More informationFree Response Questions Compiled by Kaye Autrey for face-to-face student instruction in the AP Calculus classroom
Free Response Questions 1969-010 Compiled by Kaye Autrey for face-to-face student instruction in the AP Calculus classroom 1 AP Calculus Free-Response Questions 1969 AB 1 Consider the following functions
More informationMath 180, Final Exam, Fall 2012 Problem 1 Solution
Math 80, Final Exam, Fall 0 Problem Solution. Find the derivatives of the following functions: (a) ln(ln(x)) (b) x 6 + sin(x) e x (c) tan(x ) + cot(x ) (a) We evaluate the derivative using the Chain Rule.
More informationAP Calculus Free-Response Questions 1969-present AB
AP Calculus Free-Response Questions 1969-present AB 1969 1. Consider the following functions defined for all x: f 1 (x) = x, f (x) = xcos x, f 3 (x) = 3e x, f 4 (x) = x - x. Answer the following questions
More informationDuVal High School Summer Review Packet AP Calculus
DuVal High School Summer Review Packet AP Calculus Welcome to AP Calculus AB. This packet contains background skills you need to know for your AP Calculus. My suggestion is, you read the information and
More informationSolution: It could be discontinuous, or have a vertical tangent like y = x 1/3, or have a corner like y = x.
1. Name three different reasons that a function can fail to be differentiable at a point. Give an example for each reason, and explain why your examples are valid. It could be discontinuous, or have a
More informationWORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I.
WORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I. DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE Contributors: U. N. Iyer and P. Laul. (Many problems have been directly taken from Single Variable Calculus,
More informationChapter 8: Radical Functions
Chapter 8: Radical Functions Chapter 8 Overview: Types and Traits of Radical Functions Vocabulary:. Radical (Irrational) Function an epression whose general equation contains a root of a variable and possibly
More informationTangent Lines Sec. 2.1, 2.7, & 2.8 (continued)
Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued) Prove this Result How Can a Derivative Not Exist? Remember that the derivative at a point (or slope of a tangent line) is a LIMIT, so it doesn t exist whenever
More informationFinal Exam Review Exercise Set A, Math 1551, Fall 2017
Final Exam Review Exercise Set A, Math 1551, Fall 2017 This review set gives a list of topics that we explored throughout this course, as well as a few practice problems at the end of the document. A complete
More information5 t + t2 4. (ii) f(x) = ln(x 2 1). (iii) f(x) = e 2x 2e x + 3 4
Study Guide for Final Exam 1. You are supposed to be able to determine the domain of a function, looking at the conditions for its expression to be well-defined. Some examples of the conditions are: What
More informationMath 1431 Final Exam Review
Math 1431 Final Exam Review Comprehensive exam. I recommend you study all past reviews and practice exams as well. Know all rules/formulas. Make a reservation for the final exam. If you miss it, go back
More informationOld Math 220 Exams. David M. McClendon. Department of Mathematics Ferris State University
Old Math 0 Exams David M. McClendon Department of Mathematics Ferris State University Last updated to include exams from Spring 05 Contents Contents General information about these exams 4 Exams from 0
More information1. Determine the limit (if it exists). + lim A) B) C) D) E) Determine the limit (if it exists).
Please do not write on. Calc AB Semester 1 Exam Review 1. Determine the limit (if it exists). 1 1 + lim x 3 6 x 3 x + 3 A).1 B).8 C).157778 D).7778 E).137778. Determine the limit (if it exists). 1 1cos
More informationCalculus I 5. Applications of differentiation
2301107 Calculus I 5. Applications of differentiation Chapter 5:Applications of differentiation C05-2 Outline 5.1. Extreme values 5.2. Curvature and Inflection point 5.3. Curve sketching 5.4. Related rate
More informationFinal Exam Solutions
Final Exam Solutions Laurence Field Math, Section March, Name: Solutions Instructions: This exam has 8 questions for a total of points. The value of each part of each question is stated. The time allowed
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More informationSection 3.8 Related Rates
Section 3.8 Related Rates Read and re-read the problem until you understand it. Draw and label a picture which gives the relevant information (if possible). Introduce notation. Assign a symbol to every
More informationSANDERSON HIGH SCHOOL AP CALCULUS AB/BC SUMMER REVIEW PACKET
SANDERSON HIGH SCHOOL AP CALCULUS AB/BC SUMMER REVIEW PACKET 017-018 Name: 1. This packet is to be handed in on Monday August 8, 017.. All work must be shown on separate paper attached to the packet. 3.
More informationDEPARTMENT OF MATHEMATICS
DEPARTMENT OF MATHEMATICS A2 level Mathematics Core 3 course workbook 2015-2016 Name: Welcome to Core 3 (C3) Mathematics. We hope that you will use this workbook to give you an organised set of notes for
More informationMath 611b Assignment #6 Name. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Math 611b Assignment #6 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find a formula for the function graphed. 1) 1) A) f(x) = 5 + x, x < -
More informationf(x 0 + h) f(x 0 ) h slope of secant line = m sec
Derivatives Using limits, we can define the slope of a tangent line to a function. When given a function f(x), and given a point P (x 0, f(x 0 )) on f, if we want to find the slope of the tangent line
More information2. Which of the following is an equation of the line tangent to the graph of f(x) = x 4 + 2x 2 at the point where
AP Review Chapter Name: Date: Per: 1. The radius of a circle is decreasing at a constant rate of 0.1 centimeter per second. In terms of the circumference C, what is the rate of change of the area of the
More informationPractice problems from old exams for math 132 William H. Meeks III
Practice problems from old exams for math 32 William H. Meeks III Disclaimer: Your instructor covers far more materials that we can possibly fit into a four/five questions exams. These practice tests are
More informationM408 C Fall 2011 Dr. Jeffrey Danciger Exam 2 November 3, Section time (circle one): 11:00am 1:00pm 2:00pm
M408 C Fall 2011 Dr. Jeffrey Danciger Exam 2 November 3, 2011 NAME EID Section time (circle one): 11:00am 1:00pm 2:00pm No books, notes, or calculators. Show all your work. Do NOT open this exam booklet
More informationMath 229 Mock Final Exam Solution
Name: Math 229 Mock Final Exam Solution Disclaimer: This mock exam is for practice purposes only. No graphing calulators TI-89 is allowed on this test. Be sure that all of your work is shown and that it
More informationChapter 4. Section Derivatives of Exponential and Logarithmic Functions
Chapter 4 Section 4.2 - Derivatives of Exponential and Logarithmic Functions Objectives: The student will be able to calculate the derivative of e x and of lnx. The student will be able to compute the
More informationMLC Practice Final Exam
Name: Section: Recitation/Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through. Show all your work on the standard response
More informationName Date Period. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
AB Fall Final Exam Review 200-20 Name Date Period MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. ) The position of a particle
More information2. Find the midpoint of the segment that joins the points (5, 1) and (3, 5). 6. Find an equation of the line with slope 7 that passes through (4, 1).
Math 129: Pre-Calculus Spring 2018 Practice Problems for Final Exam Name (Print): 1. Find the distance between the points (6, 2) and ( 4, 5). 2. Find the midpoint of the segment that joins the points (5,
More informationMAC 2311 Calculus I Spring 2004
MAC 2 Calculus I Spring 2004 Homework # Some Solutions.#. Since f (x) = d dx (ln x) =, the linearization at a = is x L(x) = f() + f ()(x ) = ln + (x ) = x. The answer is L(x) = x..#4. Since e 0 =, and
More informationCALCULUS I: FIU FINAL EXAM PROBLEM COLLECTION: VERSION WITHOUT ANSWERS
CALCULUS I: FIU FINAL EXAM PROBLEM COLLECTION: VERSION WITHOUT ANSWERS FIU MATHEMATICS FACULTY NOVEMBER 2017 Contents 1. Limits and Continuity 1 2. Derivatives 4 3. Local Linear Approximation and differentials
More informationHello Future Calculus Level One Student,
Hello Future Calculus Level One Student, This assignment must be completed and handed in on the first day of class. This assignment will serve as the main review for a test on this material. The test will
More information10550 PRACTICE FINAL EXAM SOLUTIONS. x 2 4. x 2 x 2 5x +6 = lim x +2. x 2 x 3 = 4 1 = 4.
55 PRACTICE FINAL EXAM SOLUTIONS. First notice that x 2 4 x 2x + 2 x 2 5x +6 x 2x. This function is undefined at x 2. Since, in the it as x 2, we only care about what happens near x 2 an for x less than
More informationMultiple Choice. Circle the best answer. No work needed. No partial credit available. is continuous.
Multiple Choice. Circle the best answer. No work needed. No partial credit available. + +. Evaluate lim + (a (b (c (d 0 (e None of the above.. Evaluate lim (a (b (c (d 0 (e + + None of the above.. Find
More informationAB CALCULUS SEMESTER A REVIEW Show all work on separate paper. (b) lim. lim. (f) x a. for each of the following functions: (b) y = 3x 4 x + 2
AB CALCULUS Page 1 of 6 NAME DATE 1. Evaluate each it: AB CALCULUS Show all work on separate paper. x 3 x 9 x 5x + 6 x 0 5x 3sin x x 7 x 3 x 3 5x (d) 5x 3 x +1 x x 4 (e) x x 9 3x 4 6x (f) h 0 sin( π 6
More informationMath 121: Calculus 1 - Fall 2012/2013 Review of Precalculus Concepts
Introduction Math 11: Calculus 1 - Fall 01/01 Review of Precalculus Concepts Welcome to Math 11 - Calculus 1, Fall 01/01! This problems in this packet are designed to help you review the topics from Algebra
More informationx+1 e 2t dt. h(x) := Find the equation of the tangent line to y = h(x) at x = 0.
Math Sample final problems Here are some problems that appeared on past Math exams. Note that you will be given a table of Z-scores for the standard normal distribution on the test. Don t forget to have
More informationFinal Examination 201-NYA-05 May 18, 2018
. ( points) Evaluate each of the following limits. 3x x + (a) lim x x 3 8 x + sin(5x) (b) lim x sin(x) (c) lim x π/3 + sec x ( (d) x x + 5x ) (e) lim x 5 x lim x 5 + x 6. (3 points) What value of c makes
More informationCHAPTER 3: DERIVATIVES
(Exercises for Section 3.1: Derivatives, Tangent Lines, and Rates of Change) E.3.1 CHAPTER 3: DERIVATIVES SECTION 3.1: DERIVATIVES, TANGENT LINES, and RATES OF CHANGE In these Exercises, use a version
More informationPractice Questions From Calculus II. 0. State the following calculus rules (these are many of the key rules from Test 1 topics).
Math 132. Practice Questions From Calculus II I. Topics Covered in Test I 0. State the following calculus rules (these are many of the key rules from Test 1 topics). (Trapezoidal Rule) b a f(x) dx (Fundamental
More informationTest one Review Cal 2
Name: Class: Date: ID: A Test one Review Cal 2 Short Answer. Write the following expression as a logarithm of a single quantity. lnx 2ln x 2 ˆ 6 2. Write the following expression as a logarithm of a single
More informationAmherst College, DEPARTMENT OF MATHEMATICS Math 11, Final Examination, May 14, Answer Key. x 1 x 1 = 8. x 7 = lim. 5(x + 4) x x(x + 4) = lim
Amherst College, DEPARTMENT OF MATHEMATICS Math, Final Eamination, May 4, Answer Key. [ Points] Evaluate each of the following limits. Please justify your answers. Be clear if the limit equals a value,
More information3. Go over old quizzes (there are blank copies on my website try timing yourself!)
final exam review General Information The time and location of the final exam are as follows: Date: Tuesday, June 12th Time: 10:15am-12:15pm Location: Straub 254 The exam will be cumulative; that is, it
More informationMath 152 Take Home Test 1
Math 5 Take Home Test Due Monday 5 th October (5 points) The following test will be done at home in order to ensure that it is a fair and representative reflection of your own ability in mathematics I
More informationMA1021 Calculus I B Term, Sign:
MA1021 Calculus I B Term, 2014 Final Exam Print Name: Sign: Write up your solutions neatly and show all your work. 1. (28 pts) Compute each of the following derivatives: You do not have to simplify your
More informationAP Calculus Related Rates Worksheet
AP Calculus Related Rates Worksheet 1. A small balloon is released at a point 150 feet from an observer, who is on level ground. If the balloon goes straight up at a rate of 8 feet per second, how fast
More informationV = π 3 r2 h. dv dt = π [ r 2dh dt r2. dv 3 dt +2rhdr dt
9 Related Rates Related rates is the phrase used to describe the situation when two or more related variables are changing with respect to time. The rate of change, as mentioned earlier, is another expression
More informationAP Calculus AB Chapter 4 Packet Implicit Differentiation. 4.5: Implicit Functions
4.5: Implicit Functions We can employ implicit differentiation when an equation that defines a function is so complicated that we cannot use an explicit rule to find the derivative. EXAMPLE 1: Find dy
More informationMath 121: Calculus 1 - Fall 2013/2014 Review of Precalculus Concepts
Introduction Math 121: Calculus 1 - Fall 201/2014 Review of Precalculus Concepts Welcome to Math 121 - Calculus 1, Fall 201/2014! This problems in this packet are designed to help you review the topics
More informationMath 250 Skills Assessment Test
Math 5 Skills Assessment Test Page Math 5 Skills Assessment Test The purpose of this test is purely diagnostic (before beginning your review, it will be helpful to assess both strengths and weaknesses).
More informationMath 2413 t2rsu14. Name: 06/06/ Find the derivative of the following function using the limiting process.
Name: 06/06/014 Math 413 trsu14 1. Find the derivative of the following function using the limiting process. f( x) = 4x + 5x. Find the derivative of the following function using the limiting process. f(
More informationNote: Final Exam is at 10:45 on Tuesday, 5/3/11 (This is the Final Exam time reserved for our labs). From Practice Test I
MA Practice Final Answers in Red 4/8/ and 4/9/ Name Note: Final Exam is at :45 on Tuesday, 5// (This is the Final Exam time reserved for our labs). From Practice Test I Consider the integral 5 x dx. Sketch
More informationMathematics 111 (Calculus II) Laboratory Manual
Mathematics (Calculus II) Laboratory Manual Department of Mathematics & Statistics University of Regina nd edition prepared by Patrick Maidorn, Fotini Labropulu, and Robert Petry University of Regina Department
More informationHave a Safe and Happy Break
Math 121 Final EF: December 10, 2013 Name Directions: 1 /15 2 /15 3 /15 4 /15 5 /10 6 /10 7 /20 8 /15 9 /15 10 /10 11 /15 12 /20 13 /15 14 /10 Total /200 1. No book, notes, or ouiji boards. You may use
More informationCalculus II - Fall 2013
Calculus II - Fall Midterm Exam II, November, In the following problems you are required to show all your work and provide the necessary explanations everywhere to get full credit.. Find the area between
More informationMath 121: Calculus 1 - Winter 2012/2013 Review of Precalculus Concepts
Introduction Math 11: Calculus 1 - Winter 01/01 Review of Precalculus Concepts Welcome to Math 11 - Calculus 1, Winter 01/01! This problems in this packet are designed to help you review the topics from
More information2.1 The derivative. Rates of change. m sec = y f (a + h) f (a)
2.1 The derivative Rates of change 1 The slope of a secant line is m sec = y f (b) f (a) = x b a and represents the average rate of change over [a, b]. Letting b = a + h, we can express the slope of the
More informationPractice Exam 1 Solutions
Practice Exam 1 Solutions 1a. Let S be the region bounded by y = x 3, y = 1, and x. Find the area of S. What is the volume of the solid obtained by rotating S about the line y = 1? Area A = Volume 1 1
More informationMath 121: Final Exam Review Sheet
Exam Information Math 11: Final Exam Review Sheet The Final Exam will be given on Thursday, March 1 from 10:30 am 1:30 pm. The exam is cumulative and will cover chapters 1.1-1.3, 1.5, 1.6,.1-.6, 3.1-3.6,
More informationMATH 3A FINAL REVIEW
MATH 3A FINAL REVIEW Guidelines to taking the nal exam You must show your work very clearly You will receive no credit if we do not understand what you are doing 2 You must cross out any incorrect work
More informationMATH 18.01, FALL PROBLEM SET # 6 SOLUTIONS
MATH 181, FALL 17 - PROBLEM SET # 6 SOLUTIONS Part II (5 points) 1 (Thurs, Oct 6; Second Fundamental Theorem; + + + + + = 16 points) Let sinc(x) denote the sinc function { 1 if x =, sinc(x) = sin x if
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Determine algebraically whether the function is even, odd, or neither even nor odd. ) f(x)
More informationm(x) = f(x) + g(x) m (x) = f (x) + g (x) (The Sum Rule) n(x) = f(x) g(x) n (x) = f (x) g (x) (The Difference Rule) thing CFAIHIHD fkthf.
. Chapter 3 Differentiation Rules 3.1 Derivatives of Polynomials and Exponential Functions Aka The Short Cuts! Yay! f(x) c f (x) 0 g(x) x g (x) 1 h(x) x n h (x) n x n-1 (The Power Rule) k(x) c f(x) k (x)
More informationMath 2250 Final Exam Practice Problem Solutions. f(x) = ln x x. 1 x. lim. lim. x x = lim. = lim 2
Math 5 Final Eam Practice Problem Solutions. What are the domain and range of the function f() = ln? Answer: is only defined for, and ln is only defined for >. Hence, the domain of the function is >. Notice
More informationAP Calculus AB Semester 1 Practice Final
Class: Date: AP Calculus AB Semester 1 Practice Final Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Find the limit (if it exists). lim x x + 4 x a. 6
More informationCalculus I Practice Exam 2
Calculus I Practice Exam 2 Instructions: The exam is closed book, closed notes, although you may use a note sheet as in the previous exam. A calculator is allowed, but you must show all of your work. Your
More informationFinal Exam Review Problems
Final Exam Review Problems Name: Date: June 23, 2013 P 1.4. 33. Determine whether the line x = 4 represens y as a function of x. P 1.5. 37. Graph f(x) = 3x 1 x 6. Find the x and y-intercepts and asymptotes
More informationChapter 2 Derivatives
Contents Chapter 2 Derivatives Motivation to Chapter 2 2 1 Derivatives and Rates of Change 3 1.1 VIDEO - Definitions................................................... 3 1.2 VIDEO - Examples and Applications
More informationImplicit Differentiation and Related Rates
Math 3A Discussion Notes Week 5 October 7 and October 9, 05 Because of the mierm, we re a little behind lecture, but this week s topics will help prepare you for the quiz. Implicit Differentiation and
More information