Math 209 Solutions to Assignment 7. x + 2y. 1 x + 2y i + 2. f x = cos(y/z)), f y = x z sin(y/z), f z = xy z 2 sin(y/z).

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1 Math 29 Solutions to Assignment 7. Find the gradient vector field of the following functions: a fx, y lnx + 2y; b fx, y, z x cosy/z. Solution. a f x x + 2y, f 2 y x + 2y. Thus, the gradient vector field is f x + 2y, 2 x + 2y b Thus, the gradient vector field is Due: 2 Noon on Thursday, November, 25 x + 2y i + 2 x + 2y j. f x cosy/z, f y x z siny/z, f z xy z 2 siny/z. f cosy/z, x xy siny/z, z z 2 siny/z cosy/z i x xy siny/z j + siny/z k. z z2 2. Suppose fx, y x 2 y 2. Find f ds where a is formed from the edges of a triangle with vertices at,, 2, and, 2. b is a circle of radius 2 centered at the origin. Solution. a For convenience, we denote by, 2 and 3 the line segments from, to 2,, from 2, to, 2 and from, 2 to,, respectively. Then and the parametric equations of, 2 and 3 as given as follows: : x t + 2t 2t t, y t + t t, 2 : x 2 t + t 2 t t, y t + 2t t +, 3 : x t + t t y 2 t + t 2 2t, t, Thus, f ds f ds + f ds + f ds 2 3 4t 2 t 2 5 dt + 2 t 2 t dt + t 2 4 t 2 5 dt 3 5. t 2 dt t dt t 3t 2 dt

2 b The paramentric equation of is x 2 cos t y 2 sin t, t 2π. Thus 3. Evaluate f ds 8 2π 2π 4 cos 2 t 4 sin 2 t 4 sin 2 t + 4 cos 2 t dt cos2t 4 sin2t 2π. x + yz dx + 2x dy + xyz dz, where consists of line segments from,, to 2, 3, and from 2, 3, to 2, 5, 2. Solution. We denote by and 2 the line segments from,, to 2, 3, and from 2, 3, to 2, 5, 2, respectively. Then Thus, Therefore : xt, yt, zt,, t + t2, 3, + t, 3t,, t, 2 : xt, yt, zt 2, 3, t + t2, 5, 2 2, 3 + 2t, + t, t. x + yz dx + 2x dy + xyz dz 2 x + yz dx + 2x dy + xyz dz + + t + 3t t 3 + 3t + t dt 7 + t dt 2, t + t t + t dt 4 + t + 4t 2 dt The formula for a cycloid is given parametrically by t sint, cost. Find the length of the curve over one cycle t 2π. Solution. The length is 2π x t 2 + y t 2 dt 2π 2π cos t 2 + sin 2 t dt 2π 2 2 cos t dt 2 sin t dt Determine whether or not F is a conservative vector field, if it is, find a function f such that F f. a b Fx, y 2x cos y y cos xi + x 2 sin y sin xj. Fx, y ye x + sin yi + e x + x cos yj. 2

3 Solution. a Let Since P x, y 2x cos y y cos x, Qx, y x 2 sin y sin x. P y 2x sin y cos x Q x throughout the open, simply connected domain R 2, it follows that F is conservative. Now assume f F. Then f x P 2x cos y y cos x, and hence On the the hand, since f y Q, that is f P x, y dx + gy x 2 cos y y sin x + gy. x 2 cos y y sin x + gy x 2 sin y sin x, y x 2 sin y sin x + g y x 2 sin y sin x. Thus, g y and gy K, where K is a constant. Therefore b Let Then fx, y x 2 cos y y sin x + K. P x, y ye x + sin y, Qx, y e x + x cos y. P y e x + cos y Q x throughout R 2. Thus, F is conservative. Now assume f F. Then fx, y P x, y dx + gy ye x + x sin y + gy. Since f y Q, ye x + x sin y + gy e x + x cos y. y It follows that g y and hence gy K. Therefore 6. Evaluate a Fx, y F dr along the given curve : fx, y ye x + x sin y + K. y 2 + x 2 i + 2y arctan x j, : rt t 2 i + 2t j, t. b Fx, y, z y 2 cos zi + 2xy cos zj xy 2 sin zk, : rt t 2 i + sin t j + t k, t π. Solution. a Let P x, y y2 x 2, Qx, y 2y arctan x. + 3

4 Then it s easy to verify that f F P, Q, where fx, y y 2 arctan x. It follows by the fundamental theorem that F dr f dr fr fr f, 2 f, π. b Let Then it s easy to verify that fx, y, z xy 2 cos z. f y 2 cos z, 2xy cos z, xy 2 sin z F. Therefore, F dr f dr frπ fr fπ 2,, π f,,. 7. Show that the line integral is independent of path and evaluate the integral: ye x dx + e x dy, where is any path from, to, 2. Solution. Let F ye x, e x P, Q. Suppose the equation of is given by Then Let r rt. ye x dx + e x dy fx, y x + ye x. P dx + Q dy F dr. then it is easy to verify that f F. Therefore F is conservative and the integral F dr is independent of path. Moreover, by the fundamental theorem, F dr f dr f, 2 f, 2e. 8. Find the work done by the force field Fx, y y 2 /x 2 i 2y/x j in moving an object from P, to Q4, 2. Solution. Let fx, y y2 x. Then f y2 x 2, 2y x F. 4

5 Therefore the work done by the force field Fx, y y 2 /x 2 i 2y/x j in moving an object from P, to Q4, 2 is F dr f dr f4, 2 f,. 9. Show that if the vector field F P i + Q j + R k is conservative and P, Q, R have continuous first-order partial derivatives, then y z R z R y. Solution. Suppose F P i + Q j + R k is conservative. Then there exists a function fx, y, z such that f x P, f y Q, f z R. Therefore y f xy, z f xz, z f yz, x f yx, R x f zx, R y f zy. It then follows by lairaut s theorem that y z R z R y.. Let Fx, y y i + x j x 2 + y 2. a Show that y x. b Show that F dr is not independent of path. [Hint: onsider the upper and lower halves of the circle x 2 + y 2 from, to, ] Proof. a Since P x, y y x 2 + y 2, Qx, y x x 2 + y 2, y x 2 + y 2 + y 2y x 2 + y 2 2 y2 x 2 x 2 + y 2 2, x x 2 + y 2 x 2x x 2 + y 2 2 y2 x 2 x 2 + y 2 2. Thus y x. 5

6 b We denote by and 2 the upper and lower halves of the circle x 2 + y 2 from, to,, respectively. Then Thus, Thus 2 F dr : rt cos t, sin t, t π, 2 : rt cos t, sin t, t π. π F dr 2 and 2 π y x 2 + y 2 dx + x x 2 + y 2 dy sin t sin t + cos t cos t dt π, y x 2 + y 2 dx + x x 2 + y 2 dy sin t sin t + cos t cos t dt π. F dr is dependant on path. Note: This happens because the domain of definition of F is R 2 \, }, which is not simply connected. 6