Problem Set 10

2012 Problem Set 10 Section 17.1: 1) Find the gradient vector field of f (x, y, z) = xcos(y/z) Solution: ∂f ∂f ∂f î + ĵ + k̂ ∂x ∂y ∂z x xy = cos(y/z)î − sin(y/z)ĵ + 2 sin(y/z)k̂ z z ∇f = 2) Give a formula F = M (x, y)î+N (x, y)ĵ for the vector field in the plane that has the properties that F = 0 at (0, 0) and that at any other point (a, b), F is tangent to the circle x2 +y 2 = a2 +b2 and points √ in the clockwise direction with magnitude |F| = a2 + b2 . Solution: F = y î − xk̂ Note that F(0, 0) = 0 p |F| = x2 + y 2 p |F(a, b)| = a2 + b 2 Section 17.2 3) Integrate f (x, y, z) = (x + y + z)/(x2 + y 2 + z 2 ) over the path r(t) = tî + tĵ + tk̂, 0 < a ≤ t ≤ b. Solution: 1 2012 r0 (t) = h1, 1, 1i √ 3 |r0 (t) = t+t+t 1 f (r(t)) = 2 = 2 2 t +t +t t Z √ Z b1 f ds = 3 dt = ln(b/a) C a t 4) Find the mass of a wire that lies along the curve r(t) = (t2 − 1)ĵ + 2tk̂, 0 ≤ t ≤ 1, if the density is ρ = (3/2)t Solution: √ r0 (t) = h0, 2t, 2i, and |r0 (t) = 2 1 + t2 1 Z 0 M= Z ρ|r (t)|dt = 3 0 1 p √ t 1 + t2 dt = 2 2 − 1 0 R 5) Evaluate C xydx + (x + y)dy along the curve y = x2 from (−1, 1) to (2, 4). Solution: Parametrization: r(t) = tî + t2 ĵ −1 ≤ t ≤ 2 x = t, dx = dt; y = t2 , dy = 2tdt. Z Z 2 3  xydx + (x + y)dy = t + (t + t2 ) · 2t dt C −1 Z 2  3  69 = 3 + 2t2 dt = 4 −1 2 Winter 2012 6) Find the circulation and flux of the field F = −y î + xĵ, around and across the closed semicircular path that consists of the semicircular arch r1 (t) = (acost)î + (asint)ĵ, 0 ≤ t ≤ π, followed by the line segment r2 (t) = tî, −a ≤ t ≤ a. Solution: I Z P dx+Qdy = C Z P dx+Qdy+ P dx+Qdy = circulation C1 C2 C1 : x = acost, dx = −asint dt y = asint, dy = acost dt. C2 : x = t, dx = dt, y = 0, dy = 0. P = −y = −asint. Q = x = acost. Z π Z a circulation = [−asint(−asint) + acost(acost)]dt + tdt 0 −a  2 a Z π t 2 = πa2 = a dt + 2 −a 0 I Z Z P dy − Qdx = Flux = C P dy − Qdx + ZCπ1 [−asint(acost) − acost(−asint)]dt = 0Z a + 0dt −a = 0 (7) Find the flow of the velocity field F = −4xy î + 8y ĵ + 2k̂, 3 P dy − Qdx C2 along the curve r(t) = tî + t2 ĵ + k̂, 0 ≤ t ≤ 2. Solution: Z Flow = Circulation = P dx + Qdy + Rdz ZC2 = Z0 2 = [−4t2 (t2 ) + 8(t2 )(2t)]dt 4t3 dt = 16 0 Section 17.3 R 8) Show that the line integral C (1 − ye−x )dx + e−x dy, is path independent and evaluate the integral. C is any path from (0, 1)to(1, 2). Solution: Z (1 − ye−x )dx + e−x dy = C Z P dx + Qdy C −x = ∂Q ∂x = −e , hence the line integral is path independent. Then Z (1 − ye−x )dx + e−x dy = f (1, 2) − f (0, 1), ∂P ∂y C where fx = (1 − ye−x ) and fy = e−x fy = e−x =⇒ =⇒ =⇒ =⇒ f (x, y) = ye−x + g(x) fx = −ye−x + g 0 (x) = 1 − e−x g 0 (x) = 1. Then g(x) = x + C f (x, y) = ye−x + x + C 4 Then Z (1 − ye−x )dx + e−x dy = f (1, 2) − f (0, 1) = 2e−1 . C 9) Show that the differential form Z (1,2,3) 2xydx + (x2 − z 2 )dy − 2yzdz, (0,0,0) is exact. Then evaluate the integral Solution: ∂Q ∂P ∂R ∂R = ∂Q ∂x = 2x, ∂z = ∂x = 0, ∂z = ∂y = −2z, hence the differential form is exact. Then Z (1,2,3) 2xydx+(x2 −z 2 )dy−2yzdz = f (1, 2, 3)−f (0, 0, 0), ∂P ∂y (0,0,0) where fx = 2xy, fy = x2 − z 2 , and fz = −2yz. fx = 2xy Then, Z (1,2,3) =⇒ =⇒ =⇒ =⇒ Then, =⇒ =⇒ f (x, y, z) = x2 y + g(y, z) fy = x2 + gy = x2 − z 2 gy = −z 2 . Then, g(y, z) = −yz 2 + h(z) f (x, y, z) = x2 y − yz 2 + h(z) fz = −2yz + h0 (z) = −2yz h0 (z) = 0 =⇒ h(z) = C f (x, y, z) = x2 y − yz 2 + C 2xydx+(x2 −z 2 )dy−2yzdz = f (1, 2, 3)−f (0, 0, 0) = −16, (0,0,0) 10) Find the work done by F = (x2 +y)î+(y 2 +x)ĵ+zez k̂ over the following paths from (1, 0, 0) to (1, 0, 1). 5 2012 (a) The line segment x = 1, y = 0, 0 ≤ z ≤ 1. (b) The helix r(t) = (cost)î + (sint)ĵ + (t/2π)k̂, 0 ≤ t ≤ 2π. (c) The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = x2 , y = 0 from (0, 0, 0) to (1, 0, 1). Solution: ∂Q ∂Q ∂P ∂P ∂R ∂R ∂y = ∂x = 1, ∂z = ∂x = 0, ∂z = ∂y = 0, hence F is conservative and the work done is independent of the path. Thus, Z Z Work = F·Tds = ∇f ·Tds = f (1, 0, 1)−f (1, 0, 0), C C for all the paths in (a), (b), and (c). 2 fx = x + y =⇒ x3 + xy + g(y, z) f (x, y, z) = 3 fy = x + gy = y 2 + x y3 2 + h(z) gy = y . Then, g(y, z) = 3 y3 x3 f (x, y, z) = + xy + + h(z) 3 3 fz = h0 (z) = zez Z h(z) = zez dz = zez − ez + C =⇒ f (x, y, z) = =⇒ =⇒ =⇒ =⇒ Then, x3 y3 + xy + + zez − ez + C. 3 3 Then, Z Z Work = ∇f ·Tds = f (1, 0, 1)−f (1, 0, 0) = 1. F·Tds = C C 6 11) Suppose that F = ∇f is a conservative vector field and Z (x,y,z) g(x, y, z) = F · dr. (0,0,0) Show that ∇g = F. Solution: Since F = ∇f , Z (x,y,z) g(x, y, z) = F · dr = f (x, y, z) − f (0, 0, 0). (0,0,0) Then ∇g = ∇{f (x, y, z) = f (0, 0, 0)} = ∇f = F Thus, ∇g = F. Section 17.4 12) Evaluate the line integral below along the given positively oriented curve using Green’s Theorem. Z xe−2x dx + (x4 + 2x2 y2)dy, C where C is the boundary of the region between the circles x2 + y 2 = 1 and x2 + y 2 = 4. 7 Solution: Z Z −2x 4 2 2 xe dx + (x + 2x y )dy = P dx + Qdy C C   ZZ ∂Q ∂P − = dA ∂x ∂y R ZZ = (4x3 + 4xy 2 )dA R Z 2π Z 2 (4r3 cos3 θ + 4r3 cosθsin2 θ)rdrdθ = 0 Z 2π 1Z 2 (cos3 θ + cosθsin2 θ)r4 drdθ = 4 Z0 2π 1 Z 2 = 4 cosθdθ r4 dr 0 1 = 0 13) Use Green’s Theorem to find the counterclockwise circulation and outward flux of the field F = xy î+y 2 ĵ around and over the boundary of the region enclosed by the curves y = x2 and y = x in the first quadrant. Solution: ZZ  I circulation = P dx + Qdy = C Z ZR ∂Q ∂P − ∂x ∂y (0 − x)dA Z 1Z x = − xdydx = R 0 = − 8 1 12 x2  dA ZZ  Flux = Z ZR  ∂P ∂Q + dA ∂x ∂y = (y + 2y)dA ZR1 Z x ydydx = 3 x2 0 = 1 5 14) Use a Green’s Theorem area formula to find the area of the region enclosed by√ the curve r(t) = t2 î + √ ((t3 /3) − t)ĵ, − 3 ≤ t ≤ 3. Solution: 3 x = t2 , dx = 2t dt; y = t3 − t, dy = (t2 − 1) dt. ZZ Area = dA R I 1 xdy − ydx = 2  Z √3  3 1 t = t2 (t2 − 1) − ( − t)(2t) dt √ 2 − 3 3  Z √3  4 1 t = + t2 dt √ 2 − 3 3 √ 8 3 = 5 Section 17.5 15) Which of the fields are conservative and which are not? Find the curl and divergence of each field. 9 (i) F = −y î + xĵ (ii) F = yz î + xz ĵ + xy k̂ (iii) F = (ysinz)î + (xsinz)ĵ + (xycosz)k̂ (iv) F = y î + (x + z)ĵ − y k̂ Solution: Recall that ∇ = (i) ∂P ∂y = −1 6= servative. ∂ ∂x î ∂Q ∂x + ∂ ∂y ĵ + ∂ ∂z k̂ = 1. Thus, the field is not con- Divergence: ∇ · F = 0 Curl: ∇ × F = 2k̂ (ii) ∂P ∂R = z = ∂Q ∂x ; ∂z = y = ∂x ; the field is conservative. ∂P ∂y Divergence: ∇ · F = 0 Curl: ∇ × F = 0. (iii) ∂P ∂y = sinz ∂R ∂y , Thus, ∂Q ∂z =x= ∂R ∂y ,. Thus, Since F is conservative ∂P ∂R ∂Q = ∂Q ∂x ; ∂z = ycosz = ∂x ; ∂z = xcosz = the field is conservative. Divergence: ∇ · F = −xysinz Curl: ∇ × F = 0. Since F is conservative (iv) ∂P ∂y ∂R ∂y ∂Q ∂P ∂R = 1 = ∂Q ∂x ; ∂z = 0 = ∂x ; but ∂z = 1 6== = −1, Thus, the field is not conservative. Divergence: ∇ · F = 0 Curl: ∇ × F = −2î. 10 16) Show that (a) any vector field of the form F = f (x)î + g(y)ĵ + h(z)k̂ where f , g, h are differentiable functions, is irrotational. (b) any vector field of the form F = f (y, z)î+g(x, z)ĵ+ h(x, y)k̂ is incompressible. Solution: (i) ∇×F = î ĵ k̂ ∂ ∂x ∂ ∂y ∂ ∂z f (x) g(y) h(z)   ∂ ∂ h(z) − g(y) î = ∂y ∂z   ∂ ∂ − h(z) − f (x) ĵ ∂x ∂z   ∂ ∂ + g(y) − f (x) k̂ ∂x ∂y = 0 Hence any vector field F = f (x)î + g(y)ĵ + h(z)k̂ is irrotational. (ii) ∇·F= ∂ ∂ ∂ f (y, z) + g(x, z) + h(x, y) = 0. ∂x ∂y ∂z Hence any vector field of the form F = f (y, z)î + g(x, z)ĵ + h(x, y)k̂ is incompressible. 11