MATH 317 Fall 2016 Assignment 5

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1 MATH 37 Fall 26 Assignment 5 6.3, 6.4. ( 6.3) etermine whether F(x, y) e x sin y îı + e x cos y ĵj is a conservative vector field. If it is, find a function f such that F f. enote F (P, Q). We have Q x e x cos y and P y e x cos y. Since Q x P y in R 2, F is conservative. Since f x e x sin y, we have f(x, y) e x sin ydx e x sin y + h(y) f y e x cos y e x cos y + h (y) Thus h (y) and h(y) c. We conclude f(x, y) e x sin y + c. 2. etermine whether F(x, y) e x cos y îı + e x sin y ĵj is a conservative vector field. If it is, find a function f such that F f. enote F (P, Q). We have Q x e x sin y and P y e x sin y. Since Q x P y, F is not conservative. 3. (#7) etermine whether F(x, y) (ye x + sin y)îı + (e x + x cos y)ĵj is a conservative vector field. If it is, find a function f such that F f. enote F (P, Q). We have Q x e x + cos y and P y e x + cos y. Since Q x P y in R 2, which is simply connected, we conclude that F is conservative. Since f x ye x + sin y, we have f(x, y) (ye x + sin y)dx ye x + x sin y + h(y) f y e x + cos y e x + cos y + h (y) Thus h (y) and h(y) c. We conclude f(x, y) ye x + x sin y + c. 4. (#5) Let F(x, y, z) yz îı + xz ĵj + (xy + 2z) k. (a) Find a function f such that F f and (b) use part (a) to evaluate F d r along the line segment from (,, 2) to (4, 6, 3). (a) Since f x yz, we have f(x, y, z) yzdx xyz + h(y, z)

2 Thus f y xz + h y xz Thus h y and h h(z). We also have f z xy + h (z) xy + 2z Thus h (z) 2z and h(z) z 2 + c. We conclude f(x, y) xyz + z 2 + c. (b) We may take c. F d r f(4, 6, 3) f(,, 2) ( + 4) (#9) Show that the line integral I 2xe y dx + (2y x 2 e y )dy, with any path from (, ) to (2, ), is independent of path and evaluate the integral. enote P 2xe y and Q 2y x 2 e y. Note Q x 2xe y and P y 2xe y. Since Q x P y in R 2, which is simply connected, F (P, Q) is conservative and hence the line integral is independent of path. We now look for f(x, y) so that F f: We have f(x, y) 2xe y dx x 2 e y + h(y) f y 2y x 2 e y x 2 e y + h (y) Thus h (y) 2y and h(y) y 2 + c. We can choose c and get f(x, y) x 2 e y + y 2. Thus I f(2, ) f(, ) (4/e + ) ( + ) 4/e. Alternative solution. Once we have checked that F is conservative, we can choose any path to evaluate I. For example we can choose to be the union of the line segment from (, ) to (2, ), and 2 the line segment from (2, ) to (2, ). We have dy on and dx on 2. Thus I 2 2xdx + (2y 4e y )dy [ x 2] 2 + [ y 2 + 4e y] [4 ] + [ + 4/e ( + 4)] 4/e. 6. (#2) Suppose you re asked to determine the curve that requires the least work for a force field F to move a particle from one point to another point. You decide to check first whether F is conservative, and indeed it turns out that it is. How would you reply to the request? If F is conservative, then F d r is independent of path. This means that the work done along all piecewise-smooth curves that have the described initial and terminal points is the same. Your reply: It doesn t matter which curve is chosen. 2

3 7. etermine whether or not the given set is (a) open, (b) connected, and (c) simply connected. (i) {(x, y) < x < 3} (ii) 2 {(x, y) < x < 3} (iii) 3 {(x, y) < x 2 + y 2 4} (iv) 4 {(x, y) < x 2 + y 2 4, y > } (a) and 2 are open. 3 and 4 are not open. (b), 3 and 4 are connected. 2 is not connected. (c) and 4 are simply connected. 2 and 3 are not simply connected. Remark. The set (iii) is *not* simply connected as it is missing the origin. Also, a set can be neither open nor closed (as is the case for (iii), (iv) ). 8. ( 6.4#7) Use Green s Theorem to evaluate the line integral I (y + e x )dx + (2x + cos y 2 )dy where is the positively oriented boundary of the region enclosed by the parabolas y x 2 and x y 2. enote P y + e x and Q 2x + cos y 2. By Green Theorem, I Q x P y da (2 )da Area() x x 2 dydx x /2 x 2 dx 9. (#9) Use Green s Theorem to evaluate the line integral I is the positively oriented circle x 2 + y 2 4. [ 2 3 x3/2 ] 3 x3 3. y 3 dx x 3 dy where enote P y 3 and Q x 3. enote the disk x 2 + y 2 4 by. By Green Theorem, I Q x P y da ( 3x 2 3y 2 )da 3 (x 2 + y 2 )da Using polar coordinates r, θ, I 3 2π 2 r 2 rdrdθ 3 2π [r 4 /4 ] 2 24π.. (#) Use Green s Theorem to evaluate the line integral I F d r where is the triangle from (, ) to (, 4) to (2, ) to (, ) and F (y cos x xy sin x, xy +x cos x). heck the orientation of the curve before applying the theorem. enote the region inside the triangle by. Note (xy + x cos x) x (y cos x xy sin x) y y + cos x x sin x cos x + x sin x y. 3

4 By Green Theorem, and noting that the triangle is negatively oriented, 2 4 2x 2 I y da y dy dx 2 (4 2x)2 dx 2 2 (x 2) 2 dx 2 3 [ (x 2) 3 ] (#2) Use Green s Theorem to evaluate the line integral I F d r where F (e x + y 2, e y + x 2 ), and consists of the arc of the curve y cos x from (, ) to (π/2, ), and the line segment from (π/2, ) to (, ). heck the orientation of the curve before applying the theorem. enote the enclosed region by. Note (e y + x 2 ) x (e x + y 2 ) y 2x 2y By Green Theorem, and noting that the curve is negatively oriented, π/2 cos x I (2x 2y) da ( 2x + 2y)dy dx π/2 [ 2xy + y 2 ] cos x dx π/2 ( 2x cos x + cos 2 x ) dx Since the first integrand is odd, its integral is zero. By symmetry, we can replace the second integrand cos 2 x by 2 (cos2 x + sin 2 x) 2. Thus I π/2 2 dx π (#9) Find the area under one arch of the cycloid x t sin t, y cos t, t 2π. For cycloid see enote by the cycloid from (, ) to (2π, ), by 2 the line segment from (, ) to (2π, ), and by the region they enclose. Let F (P, Q) (, x). Since Q x P y, by Green Theorem, and noting has negative orientation, Area() da F d r + F d r. 2 4

5 On 2, F d r (, x) (dx, ). Thus, Area() F d r 2π t sin t dt + 2π 2π (, t sin t) ( cos t, sin t)dt sin 2 t dt [t cos t sin t] 2π + π 3π. Alternative solution. Since dx/dt cos t > for < t < 2π, we can solve t t(x) for x 2π and y y(t(x)). Thus Area() 2π 2π y(t(x))dx 2π 2π dt 2 cos tdt + 2π + π 3π. y(t) dx dt dt where we have used symmetry for the last integral. 2π 2π (cos t) 2 dt ( cos t) 2 dt 5

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