MATH 124 E MIDTERM 2, v.b Autumn 2018 November 20, 2018 NAME: SIGNATURE: STUDENT ID #: GAB AB AB AB AB AB AB AB AB AB AB AB AB AB QUIZ SECTION: ABB ABB Problem Number of points Points obtained 1 14 2 10 3 10 4 8 5 8 Total 50 Instructions: Any attempt at cheating will be punished. No books or notebooks allowed; you may use an 8.5 11 double-sided, handwritten sheet of notes for personal use (do not share). Place a box around your final answer to each question. No graphing calculators allowed (scientific calculators Ti-30x IIS are the only approved ones). Answers with little or no justification may receive no credit. Answers obtained by guess-and-check work will receive little or no credit, even if correct. Read problems carefully. Raise your hand if you have a question. If you need more space, use additional blank sheets which will be provided by your TA. It is your responsibility to have him/her staple the additional sheets to your exam before you turn it in. Please turn off and put away cell phones. GOOD LUCK! 1
Problem 1. (14pts) Compute the following derivatives. Simplify constants, when possible. a) (3pts) ln(sin(3x)) 1 sin(3x) cos(3x) 3, each term b) (3pts) x cos 1 (ln(x)) (Here cos 1 stands for the inverse cos function, a.k.a. arccos.) ( ) cos 1 1 (ln(x)) + x 1 (ln(x)) 2 1 x ; for Product Rule, 2 for correct terms 2
Problem 1, cont d. c) (4pts) (x+1) 4 (x 2 +1) 2 x 3 sin x One may do this with Product/Quotient Rule. It is not recommended. Use logarithmic differentiation instead to get ln y = ln(x + 1) 4 + ln((x 2 + 1) 2 ) ln x 3 ln sin(x) ln y = 4 ln(x + 1) + 2 ln(x 2 + 1) 3 ln x ln sin(x) () (ln y) = 4 x + 1 + 4x x 2 + 1 3 x 1 sin(x) cos(x) (2pts) y = (x + 1)4 (x 2 + 1) 2 x 3 sin(x) ( 4 x + 1 + 4x x 2 + 1 3 x cos(x) ) sin(x). () d) (4pts) (cos x) x2 This needs logarithmic differentiation. ln y = x 2 ln(cos x). () Product Rule: (ln y) = 2x ln(cos x) + x 2 1 ( sin x) cos x (2pts) ( ) y = (cos x) x2 2x ln(cos x) + x 2 1 ( sin x) cos x () 3
Problem 2. (10pts) A particle is moving along the curve defined by x(t) = sin 3t 2t, y(t) = cos t + 1 2 t. a) (3pts) Calculate the slope of the tangent to the curve at time t. dx = 3 cos 3t 2, dy = sin t + 1 2, Slope is dy/ dx/ = sin t+ 1 2 3 cos 3t 2,.. b) (4pts) For what value of t between 0 and 1 is the tangent line to the curve horizontal? Specify the tangent line at that time. dy = 0, sin t + 1 2 = 0, between 0 and 1 means t = π 6, Check: dx for x = π/6 is 2 0 Tangent line: y = y(π/6) = 3 2 + 1 π 2 6 c) (4pts) At time t = 0, the y-coordinate of the particle s position is y(0) = 1. Use linear approximation to estimate the time t 1 when the y-coordinate of the particle is y(t 1 ) = 1.05. y(0) = 1, y(t 1 ) = 1.05, y = dy t () y 1.05 1 = 0.05 dy/ at t = 0 is 1/2 t y dy = 0.05 1/2 = 0.1, so t 1 0.1 4
Problem 3. (10pts) A curve is implicitly defined by the equation x 2 1 2 xy + y2 = 1. a) (6pts) Find all the points (if any) for which the tangent line to the point is vertical. Differentiate: 2x 1 2 y 1 2 xy + 2yy = 0 Solve: y = 2x y/2 x/2 2y, 2pts Vertical tangent: x/2 2y = 0 (or x = 4y), 2x y/2 0 Substitute x = 4y into the curve equation to get 16y 2 2y 2 + y 2 = 1, or y = ± 1 15 ; get x = 4y = ± 4 15 ( for each pair). b) (4pts) The point (1, 0) is on the curve. Use linear approximation to estimate the value y 0 such that (1.1, y 0 ) is on the curve. y y x, At the point in question x = 1, y = 0, so y = 4 Since x = 1.1 1 = 0.1, y 4 0.1 = 0.4, and so the new y 0 + 0.4 = 0.4 5
Problem 4. (8pts) An oil tanker is leaking oil on the surface of the ocean, producing a cylindrical oil slick of height 0.02m (the height stays constant as the spill increases). The spill is increasing at a rate of 20m 3 /hr. If the volume of a cylinder is given by the formula V = πr 2 h, find the rate at which the radius of the spill is increasing when the radius is 75m. V = πr 2 h with h constant, h = 0.02, so V = 0.02πr 2 dv dr = 2 0.02π r, 4pts dv Hence dr = 0.04π r = 20 0.04π 75 = 20 3π, 2pts 2pts 6
Problem 5. (8pts) Let f be the function defined on [0, 5] as follows: { x + 2, 0 x 2 f(x) = 4 + 6x x 2, 2 < x 5 Note that this function is continuous on [0, 5]. Find the numbers x where the function achieves its absolute maximum and absolute minimum on [0, 5], and state what those maximum and minimum values are. For 2pts, { f (x) = 1, 0 < x < 2 6 2x, 2 < x < 5 Note that at x = 2 we have a critical point ()! And another one at 3. () So for the closed interval method, we must evaluate the function at x = 0, 2, 3, 5, obtaining respectively 2, 4, 5, 1. (3pts) The minimum, 1, is achieved at x = 5, and the maximum, 5, is achieved at 3. () 7