# Key for Exam 3

## Transcription

Key for Exam 3

MATH 2060 April 17, 2015 Exam 3 Name: Side of room: Row: No electronic devices (phones, calculators, computers, etc.) are allowed during the exam. You must have provide clear, logical solutions to each of the problems below to receive full credit. Problem 1 (15 points) 2 Let √ C be the portion of the parabola y = x connecting the point (0, 0) to the point ( 6, 6). Calculate the following line integral: Z x ds C Solution √ Note that C may be parametrized by ~r(t) = ht, t2 i where 1 ≤ t ≤ 6. The line integral is thus Z Z √6 √ x ds = t 1 + 4t2 dt C 0 Performing the substitution u = 1 + 4t2 and du = 8t dt this becomes 25 Z Z 1 2 3/2 1 25 1/2 u du = · u x ds = 8 1 8 3 1 C = 124 1 62 31 3 3 25 /2 − 1 /2 = = = 12 12 6 3 Problem 2 (15 points) Find the flow line ~γ (t) = hx(t), y(t)i of the vector field F~ (x, y) = h3y 2 , 2i which passes through the point (1, 4) at time t = 2. Solution If ~γ (t) is to be a flow line of F~ , then we must have F~ (~γ (t)) = ~γ 0 (t). Equating components gives the system x0 (t) =3y(t)2 y 0 (t) =2 Note that y 0 (t) = 2 implies y(t) = 2t + C. Since this curve is to have y-coordiate 4 at time t = 2 we have 4 = y(2) = 2 · 2 + C, and so C = 0. Plugging this into the first equation we have x0 (t) = 3 (2t)2 = 12t2 and so x(t) = 4t3 + D. At time t = 2 we have 1 = x(2) = 4 · 23 + D = 32 + D and so D = −31. Altogether, we have that ~γ (t) = 4t3 − 31, 2t . MATH 2060 April 17, 2015 Exam 3 Problem 3 (20 points) Let C be D the closed curve by hcos(t), sin(t)i with 0 ≤ t ≤ 2π and let E parametrized H −y x ~ ~ F (x, y) = x2 +y2 , x2 +y2 . Evaluate F · d~r. C Solution Note that at each point of C, x2 + y 2 = 1. I I −y x F~ · d~r = dx + 2 dy 2 2 x +y x + y2 C C Z 2π ((− sin(t)) · (− sin(t)) + cos(t) · cos(t)) dt = 0 Z 2π sin2 (t) + cos2 (t) dt = Z0 2π dt = 0 =2π Problem 4 (15 points) The curve C shown below is parametrized by hcos(t)5 · cos(2t), sin(t)H3 i with 0 ≤ t ≤ 2π. Suppose F~ (x, y) is the vector field hex sin(y), ex cos(y)i. Calculate F~ · d~r. You must C justify your answer: no credit for correct guesses. Solution Notice that F~ (x, y) is defined on all of R2 and ∂ x ∂ x e sin(y) = ex cos(y) = e cos(y), ∂y ∂x ~ ~ so H F is conservative. Since C is a closed curve and F is a conservative vector field, ~ F · d~r = 0. C MATH 2060 April 17, 2015 Exam 3 Problem 5 (15 points) The region below is bounded by a curve C parametrized, with counter-clockwise orientation, by hsin(2t), sin(t)i where 0 ≤ t ≤ π. What is the area of this region? (Hint: Use the trig identity sin(2θ) = 2 sin(θ) cos(θ).) Solution By Green’s theorem, we can calculate the area of the D bounded by C as ZZ Area(D) = 1 dA ID x dy = ZCπ = sin(2t) cos(t) dt 0 Z =2 π sin(t) cos2 (t) dt 0 Now performing the substitution u = cos(t), du = − sin(t)dt we have Z −1 Area(D) = − 2 u2 du Z 1 1 =2 u2 du −1 3 1 2u = 3 −1 2 −2 = − 3 3 4 = 3 MATH 2060 April 17, 2015 Exam 3 Problem 6 (20 points) If a particle with electric charge −4π0 Coloumbs is fixed at the origin in the plane, and a particle with charge q Coloumbs is placed at position (x, y), then by Coloumb’s law we know the force acting on the second particle is given by + * −qy −qx , . F~ (x, y) = 3 3 (x2 + y 2 ) /2 (x2 + y 2 ) /2 Supposing the particle moves from point A to point B along the path given in the figure below, find the work done by the vector field. The point A is on a circle of radius 1 centered at the origin, and the point B is on a circle of radius 3. Solution Since a parametrization of the curve is not given, we can not calculate the work done unless the vector field is conservative. If the vector field were conservative, then its potential function would have to satisfy −3/2 fx (x, y) = − qx x2 + y 2 −3/2 fy (x, y) = − qy x2 + y 2 MATH 2060 April 17, 2015 Exam 3 Integrating the first equation with respect to x gives Z −3/2 dx f (x, y) = −qx x2 + y 2 Z −q 3 = u− /2 du Using u = x2 + y 2 , du = 2x dx 2 −q u−1/2 = + G(y) 2 −1/2 q = 1/2 + G(y) u q =p + G(y). x2 + y 2 Similarly, integrating the second equation above with respect to y gives that f (x, y) = √ q +H(y). Hence the potential function must be of the form f (x, y) = √ q +C 2 2 2 2 x +y x +y for some constant C. This constant does not matter for integrating, so we may assume for this problem that C = 0. As F~ (x, y) is conservative with potential function f (x, y) = √ q , the work done is x2 +y 2 Z F~ · d~r = f (B) − f (A). C Note the denominator in f (x, y) is the distance from the point to the origin. Since B is on a circle of radius 3 centered at the origin and A is on a circle of radius 1 centered at the origin, we have Z q q 2q F~ · d~r = f (B) − f (A) = − = − . 3 1 3 C