Unit 3 Test Review, Related Rates and Derivative

SHOW ALL WORK on the worksheet. Due by the end of the period. Find dy/dx of each function using implicit differentiation 1. (x+2)^2 + (y-3)^2 = 37 2(x+2) + 2(y-3)y' = 0 2(y-3)y' = -2(x+2) y' = -(x+2) / (y-3) 2. x^3 + y^3 = 6xy - 1 3x^2 + 3y^2y' = 6y + 6xy' 3y^2y' - 6xy' = 6y - 3x^2 y'(3y^2 - 6x) = 6y - 3x^2 y' = (6y - 3x^2) / (3y^2 - 6x) y' = (2y - x^2) / (y^2 - 2x) 3. Find the second derivative d^2y/dx^2 of x^2 - y^2 = 36 using implicit differentiation 2x - 2yy' = 0 y' = x/y 2nd : y'' = (y * 1 - x * y') / y^2 y'' = (y - x * (x/y)) / y^2 y'' = (y - x^2/y) / y^2 y'' = (y^2 - x^2) / y^3 Since x^2 - y^2 = 36 (from original), then y^2 - x^2 = -36 y'' = -36 / y^3 Find dy/dx of each function using logarithmic differentiation 4. y = (ln(x^2 (x+1))) / (x+2), x > 0 ln y = ln (ln x^2) + ln (x+1) - ln (x+2) 1/y * y' = 1/(ln x^2) * 1/x^2 * 2x + 1/(x+1) - 1/(x+2) 1/y * y' = 2/(x ln x^2) + 1/(x+1) - 1/(x+2) y' = y * [2/(x ln x^2) + 1/(x+1) - 1/(x+2)] y' = (ln(x^2(x+1))) / (x+2) * [2/(x ln x^2) + 1/(x+1) - 1/(x+2)] 5. y = (x ln x) / (sqrt(x+1)), x >= 1 ln y = ln x + ln (ln x) - 1/2 ln (x+1) 1/y * y' = 1/x + 1/(ln x) * 1/x - 1/2 * 1/(x+1) 1/y * y' = 1/x + 1/(x ln x) - 1/(2(x+1)) y' = y * [1/x + 1/(x ln x) - 1/(2(x+1))] y' = (x ln x) / (sqrt(x+1)) * [1/x + 1/(x ln x) - 1/(2(x+1))] Answers 1. y' = -(x+2) / (y-3) 2. y' = (2y - x^2) / (y^2 - 2x) 3. y'' = -36 / y^3 4. y' = (ln(x^2(x+1))) / (x+2) * [2/(x ln x^2) + 1/(x+1) - 1/(x+2)] 5. y' = (x ln x) / (sqrt(x+1)) * [1/x + 1/(x ln x) - 1/(2(x+1))] 6. Dirt is being dumped from a conveyor belt such that it forms a circular cone with base diameter that is twice the height. If dirt is being added at a rate of 8 cubic feet per minute, find the rate of change of the height of the pile when it is 10 feet tall. (V = (1/3) pi r^2 h) Known: dV/dt = 8 Unknown: dh/dt when h = 10 The height is increasing at a rate of 2 / (9 pi) ft/min when the pile is 10 ft tall. V = (1/3) pi r^2 h diameter = 2h, so r = h V = (1/3) pi (h)^2 h V = (1/3) pi h^3 dV/dt = pi h^2 dh/dt 8 = pi (10)^2 dh/dt 8 = 100 pi dh/dt dh/dt = 8 / (100 pi) dh/dt = 2 / (25 pi) Evaluate each limit 7. lim [x->0] (e^(x^2) - 1) / (sin x) lim [x->0] (e^(x^2) - 1) = e^0 - 1 = 0 lim [x->0] (sin x) = 0 lim [x->0] (e^(x^2) - 1) / (sin x) is indeterminate of the form 0/0. L'Hopital's: lim [x->0] (e^(x^2) - 1) / (sin x) = lim [x->0] (2x e^(x^2)) / (cos x) = 0 / 1 = 0 8. lim [x->1] (6x^3 - 2x^2 - 4) / (sin(2x-2)) lim [x->1] (6x^3 - 2x^2 - 4) = 6 - 2 - 4 = 0 lim [x->1] (sin(2x-2)) = sin(0) = 0 lim [x->1] (6x^3 - 2x^2 - 4) / (sin(2x-2)) is indeterminate of the form 0/0. L'Hopital's: lim [x->1] (18x^2 - 4x) / (2 cos(2x-2)) = (18(1) - 4(1)) / (2 cos(0)) = (18 - 4) / (2 * 1) = 14 / 2 = 7 9. y. A particle moves along a straight line. The graph of the particle's velocity, v(t) is shown below for 0 < t < 6. The graph has horizontal tangents at t = 1 and t = 5 a) For what values of t is the position of the particle increasing? Explain your reasoning. [3, 6] because v(t) > 0 b) At what value of t does the particle change direction? Explain your reasoning. t = 3 because v(t) changes sign c) One what intervals, if any, is the acceleration of the particle negative? Explain your reasoning. (0, 1) and (5, 6) because v'(t) is decreasing d) Is the particle speeding up or slowing down at t = 2? Explain your reasoning. at t = 2, v(t) < 0 and v'(t) = a(t) > 0 The particle is slowing down Answers 6. 2 / (25 pi) ft/min 7. 0 8. 7 9. a) [3, 6] 9. b) t = 3 9. c) (0, 1); (5, 6) 9. d) Slowing Down