Solutions to the Spring 2016 Final Exam

Math 211, Sections 01,05, Fall 2018 Solutions to the Spring 2016 Final Exam 1. Let S be the plane that contains the points (1, 0, −2), (3, 1, 0), and (1, 1, 1). Find an equation for the line through the point (3, 2, −1) that is perpendicular to S. Answer. The vector from the second point to the first is h3 − 1, 1 − 0, 0 − (−2)i = h2, 1, 2i. The vector from the third point to the first is h1 − 1, 1 − 0, 1 − (−2)i = h0, 1, 3i. Their cross product ~i ~j ~k ~n = 2 1 2 = h1, −6, 2i is normal to the plane, hence parallel to the desired line. 0 1 3 So we can parametrize the line by ~r(t) = ht + 3, −6t + 2, 2t − 1i. 2. Find and classify (as local minimum, local maximum, or saddle point) every critical point of the function f (x, y) = 2x2 − 8xy + 4y 3 + y 4 . Answer. We compute fx = 4x − 8y and fy = −8x + 12y 2 + 4y 3 . Set fx = 0 gives x = 2y. Setting fy = 0 also, and plugging in x = 2y, gives 4y 3 + 12y 2 − 16y = 0, which factors as 4y(y 2 + 3y − 4) = 0 and hence as 4y(y + 4)(y − 1) = 0. Thus, we get any of y = 0, −4, 1. Recalling x = 2y, this means there are three critical points: (0, 0), (−8, −4), (2, 1). We compute fxx = 4, fxy = −8, and fyy = 24y + 12y 2 = 12y(y + 2). 2 = 0 − 64 < 0, so f has a saddle point at (0, 0). Testing (0, 0), we have D = fxx fyy − fxy Testing (−8, −4), we have D = 4(−48)(−2) − 64 = 4 · 96 − 64 > 0, and fxx = 4 > 0, so f has a local minimum at (−8, −4). Testing (2, 1), we have D = 4(12)(3) − 64 = 4 · 36 − 64 > 0, and fxx = 4 > 0, so f has a local minimum at (2, 1).  3  2x − 3xy if (x, y) 6= (0, 0), 3. Let f (x, y) = x2 + y 2  0 if (x, y) = (0, 0). (a) Compute fx (0, 0) and fy (0, 0). (b) Show that f is not continuous at (0, 0). 2h3 − 0 −0 2 f (h, 0) − f (0, 0) 2h3 Answers. (a): fx (0, 0) = lim = lim h + 0 = lim 3 = lim 2 = 2. h→0 h→0 h→0 h h→0 h h 0 −0 2 f (0, h) − f (0, 0) fy (0, 0) = lim = lim 0 + h = lim 0 = 0. h→0 h→0 h→0 h h 2x3 − 3x2 (b): Approaching along the line y = x, we have lim f (x, x) = lim 2 x→0 x→0 x + x2 2x − 3 3 = lim = − 6= 0 = f (0, 0), so f is not continuous at (0, 0). x→0 2 2 4. Find the maximum and minimum values of the function f (x, y) = x2 y + 3 on the curve x2 + y 4 = 5. Answer. We use Lagrange Multipliers, with g(x, y) = x2 + y 4 . So ∇f = h2xy, x2 i and ∇g = h2x, 4y 3 i. Setting ∇f = λ∇g and g = 5 gives 2xy = 2xλ, x2 = 4y 3 λ, x2 + y 4 = 5. The first equation is 2x(y − λ) = 0, so either x = 0 or λ√= y. √ If x = 0, then the third equation gives y 4 = 5, so y = ± 4 5, giving the two points (0, ± 4 5). If λ = y, the second equation gives x2 = 4y 4 , and hence the third equation gives 5y 4 = 5, so that y 4 = 1 and hence y = ±1. Thus, x2 = 4, so that x = ±2. (Note that these two ± signs are not necessarily in synch.) Thus, we have the four points (±1, ±2). Plugging these six points into f gives √ 4 f (0, ± 5) = 0 + 3 = 3, f (±2, 1) = 4 + 3 = 7, f (±2, −1) = −4 + 3 = −1. Thus, the maximum value is 7, and the minimum value is −1. 5. Let D be the quarter of the disk x2 + y2 ZZ ≤ 1 that lies in the first quadrant. Compute y 3 dA. D Z π/2 Z 1 (r3 sin3 θ)r dr dθ Answer. In polar coordinates, the integral becomes 0 0   r5 1  Z π/2  Z π/2 Z 1  3 4 sin θ dθ = r dr = (1 − cos2 θ) sin θ dθ [u = cos θ, du = − sin θ dθ] 5 0 0 0 0 Z 1  1  u3   0 0 1 1 2 = (1 − u2 ) du = −0 − −u = 0− +1 = . 5 5 3 5 3 15 1 1 6. Let C be the path that runs Z from (0, 0) to (2, 4) along the line y = 2x, and then back to (0, 0) along
cos(πx2 ) + 5xy dx + x2 dy. the curve y = x2 . Compute C Answer. Let D be the thin, banana-shaped region bounded by C. (So D has the curve y = x2 as its lower boundary, and the line y = 2x as its upper boundary.) Note that C is oriented negatively with respect to D. Let P = cos(πx2 ) + 5xy and Q = x2 . By Green’s Theorem, we have Z ZZ ZZ Z 2 Z 2x Z 2 2x P dx + Q dy = − Qx − Py dA = − 2x − 5x dA = 3x dy dx = 3xy dx y=x2 C Z D D 0 x2 0 2 3 2 6x2 − 3x3 dx = 2x3 − x4 = 16 − 12 − (0 − 0) = 4. = 4 0 0 Z xy ds. 7. Let C be the straight line segment in the plane from (1, 5) to (2, 2). Compute C Answer. (I’ll give the solution anyhow.) Parametrize √ segment by ~r(t) = h1 + t, 5 − 3ti for √ the line 0 ≤ t ≤ 1. So ~r 0 (t) = h1, −3i, and hence k~r 0 (t)k = 1 + 9 = 10. Z Z 1 √ √ Z 1 Thus, xy ds = (1 + t)(5 − 3t) 10 dt = 10 5 + 2t − 3t2 dt C 0 0 √  √ √ 1 = 10 5t + t2 − t3 = 10(5 + 1 − 1 − 0) = 5 10. 0 8. Let F~ (x, y, z) = h2x − 3y, 2yz − 3x, y 2 − 5i. (a) Show that F~ is conservative by finding a potential function f (x, y, z) for F~ . (b) Let C be the curve parametrized by ~r(t) = ht2 − 3, sin(πt), t2 + 1i Z for −1 ≤ t ≤ 2. Compute F~ · d~r. C Answer. (a): A potential function f must have ∇f = F~ . In particular, fx = 2x − 3y, and hence f = x2 − 3xy + g(y, z), for some function g(y, z). Differentiating with respect to y, we have fy = −3x + gy . However, ∇f = F~ gives fy = 2yz − 3x, and hence gy = 2yz, so that g(y, z) = y 2 z + h(z), for some function h(z). That is, f = x2 − 3xy + y 2 z + h(z). Differentiating with respect to z, we have fz = y 2 + h0 (z); but since fz = y 2 − 5, we have h0 (z) = −5. Thus, we may choose h(z) = −5z. Letting f (x, y, z) = x2 − 3xy + y 2 z − 5z, then, we have ∇f = F~ . (b): The curve C starts at ~r(−1) = (−2, 0, 2) and ends at ~r(2) = (1, 0, 5). Thus, by the Fundamental Theorem of Line Integrals, Z F~ · d~r = f (1, 0, 5) − f (−2, 0, 2) = (1 − 0 + 0 − 25) − (4 − 0 + 0 − 10) = −24 + 6 = −18. C 9. Let E be the solid bounded by the surfaces x = y2, ZZZ x = 1, z = y, and z = 1. Compute 2x dV . E Answer. The projection of the solid onto the xy-plane is the region between x = y 2 (the sideways parabola opening towards the positive x-axis) and the line x = 1 (parallel to the y-axis). On this region, we have y ≤ 1, and hence over this region, the plane z = y lies below the plane z = 1. Thus, the Z Z Z Z Z Z Zintegral is Z Z Z 1 1 1 1 1 1 1 1 2x(1 − y) dx dy dx dy = 2xz 2x dz dx dy = z=y −1 y 2 −1 y 2 −1 y 2 y E Z 1 Z 1 Z 1 1 1 − y 4 − y + y 5 dy (1 − y 4 )(1 − y) dy = dy = x2 (1 − y) = 2 x=y −1 −1 −1  y5 y2 y6 1 1 1 1  1 1 1 2 8 =y− = 1− − + − + − −1+ − + =2− = . 5 2 6 −1 5 2 6 5 2 6 5 5 2x dV = 10. Let S be the portion of the surface z = 3x2 + 3y 2 that lies in the first octant and below the plane z = 3. Let C be the boundary of S, oriented counterclockwise when viewed from above. That is, C consists of three arcs: • from (0, 0, 0) to (1, 0, 3) along the curve z = 3x2 in the xz-plane; • from (1, 0, 3) to (0, 1, 3) along a quarter-circle in the plane z = 3; and • from (0, 1, 3) back down to (0, 0, 0) along the curve z = 3y 2 in the yz-plane. Z Compute xz dx − yz dy + sin4 z dz. C Z xz dx − yz dy + sin4 z dz = Answer. Write F~ = hxz, −yz, sin4 zi. By Stokes’ Theorem, we have C ZZ ~ ~ (curl F ) · dS, where (by the right hand rule) S is oriented upward. S Parametrize S by ~r(r, θ) = hr cos θ, r sin θ, 3r2 i for 0 ≤ θ ≤ π/2 and 0 ≤ r ≤ 1. ~k ~i ~j Then ~rr ×~rθ = cos θ sin θ 6r = h−6r2 cos θ, −6r2 sin θ, ri, which has positive z-coordinate and −r sin θ r cos θ 0 hence is oriented upward, i.e., correctly for our purposes. ~k ~i ~j ~ Meanwhile, curl F = ∂x ∂y ∂z = hy, x, 0i. Thus, the integral is xz −yz sin4 z Z π/2 Z 1 Z π/2 Z 1 hr sin θ, r cos θ, 0i · h−6r2 cos θ, −6r2 sin θ, ri dr dθ = −12r3 sin θ cos θ dr dθ 0 0 0 0  Z π/2  Z 1  3 = −3 sin θ cos θ dθ 4r dr [u = sin θ, du = cos θ dθ] 0 0 Z 1  1   3u2 1   3  3 = − (1 − 0) = − + 0 = − = −3u du r4 2 0 2 2 0 0 11. (25 points) Let E be the solid in the first octant bounded by the sphere x2 + y 2 + z 2 = 4 and the coordinate planes. Let S be the surface ofZ ZE, oriented outward. ~ Let F~ (x, y, z) = hxy, y 2 − 2yz, z 2 i. Compute F~ · dS. S ZZ Answer. By the Divergence Theorem, ~= F~ · dS S ZZZ div(F~ ) dV . We compute E div(F~ ) = y + (2y − 2z) + 2z = 3y. Using spherical coordinates, then, the integral is Z π/2 Z π/2 Z 2  Z π/2  Z π/2  Z 2 2 (3ρ sin φ sin θ)(ρ sin φ) dρ dφ dθ = sin θ dθ sin φ dφ 0 0 0 0 0 0 2  3ρ3 dρ  3ρ4 2  φ 1 π/2 1 − cos θ (1 − cos 2φ) dφ = (−0 + 1) − sin 2φ 2 4 0 2 4 0  π 0  = 12 − 0 − (0 − 0) = 3π 4 =  π/2  Z π/2  0 (12 − 0)