Download Exam with Answers - Calculus and Analytic Geometry I | MATH 124 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Math 124 Exam 2 answers 20 November 2001 1. (20 points) Compute these derivatives: (a) d dx (tan(3x)) = Solution. 3 sec2(3x) (chain rule). (b) If y = x √ 2x, then y′ = Solution. Use logarithmic differentiation: take logs of both sides: ln y = ln(x √ 2x), so ln y = √ 2x ln(x). Now differentiate: y′ y = ( √ 2x)′ ln(x) + √ 2x(ln(x))′ = 1 √ 2x ln x + √ 2x x . Solve for y′, replacing y by x √ 2x: y′ = x √ 2x ( 1 √ 2x ln x + √ 2x x ) . (There are various equivalent ways of writing the answer.) (c) ( x2 + 3 e−x − x )′ = Solution¿ Quotient rule: (e−x − x)(2x) − (x2 + 3)(−e−x − 1) (e−x − x)2 . 2. (20 points) Use implicit differentiation to find the x-coordinates of all points on the curve 2yey = x4 − 2x3 + x2 where the tangent line is horizontal. Solution. Differentiate both sides (left side by the product rule, right side by the power rule): 2y′ey + 2yy′ey = 4x3 − 6x2 + 2x. Solve for y′: y′ = 4x3 − 6x2 + 2x 2ey + 2yey = 2x3 − 3x2 + x ey + yey . Mathematics 124 EA Name: The tangent line is horizontal when y′ = 0. This quantity is zero when the numerator is zero—that is, when 2x3 − 3x2 + x = 0. Factor out an x: x(2x2 − 3x + 1) = 0. Factor the quadratic using the quadratic formula (or by eyeballing it): x(2x − 1)(x − 1) = 0. This is zero when x = 0 , x = 1/2 , or x = 1 . 3. (20 points) At 3:00, two cars start moving from the same point; one (“car A”) heads due north, the other (“car B”) due west. Car A maintains a constant speed of 40 kilometers per hour. Car B’s speed varies, though. At 3:45 (that is, 3/4 of an hour after they started), car B is stuck at a traffic light (not moving) 40 kilometers west of the starting point. How fast is the distance between the cars increasing at 3:45? Solution. Let z(t) be the distance between the two cars at t hour after 3:00. We want to know dz dt . Let x be the distance from B to its starting point; let y be the distance from A to its starting point. Here’s a picture. B x A yz u u So z2 = x2 + y2. Differentiate with respect to t: 2z dz dt = 2x dx dt + 2y dy dt . At 3:45, car A is 30 km north of its starting point: y = 30. Also, it is moving at 40 kph: dy dt = 40. Car B is 40 km from its starting point: x = 40. It is stationary, so dx/dt = 0. Finally, by the Pythagorean theorem, z = 50. Plug all these numbers in: 2 · 50 dz dt = 2 · 40 · 0 + 2 · 30 · 40. So dz dt = 30 · 40/50 = 24. The answer is 24 kph. 4. (20 points) A point moves along the y-axis in such a way that its position at time t is given by y(t) = 5te−t 2 . Assume t ≥ 0. (a) Find all times t ≥ 0 when the acceleration of the particle is zero. Solution. The acceleration is the second derivative of position, y ′′. The velocity is y′ = 5(e−t 2 − 2t2e−t2) = 5e−t2(1 − 2t2), so y′′ = 5(e−t2(−4t) − 2t(1 − 2t)e−t2) =
