Math 13 Solutions to the Final Exam 2010

Math 13, Section 01, Fall 2010 Solutions to the Final Exam 1. Find an equation for the plane that contains the point (0, 3, 0) and the line ~r(t) = 4−t, 1+2t, 3t . Answer. The vector ~a = h−1, 2, 3i is parallel to the line and hence to the plane. Meanwhile, the point ~r(0) = (4, 1, 0) is on the plane, so the vector ~b = h4 − 0, 1 − 3, 0 − 0i = h4, −2, 0i is also parallel to the plane. Thus, we can take the normal vector to be ~a × ~b = h6, 12, −6i; or dividing by 6, let ~n = h1, 2, −1i. Using the point (0, 3, 0) as a point on the plane, then, the equation for the plane is 1(x − 0) + 2(y − 3) − 1(z − 0) = 0, or equivalently, x + 2y − z = 6. 2. Let f (x, y) =  25xy    x2 + y 2    12 if (x, y) 6= (0, 0), if (x, y) = (0, 0). (2a). Compute the directional derivative D~u f (0, 0), where ~u = 3/5, 4/5 . (2b). Prove that f is not continuous at (0, 0).  3h 4h  12h2 , − 12 − f (0, 0) 2 0 5 5 = lim h = lim = 0. Answer. (a). By definition, D~u f (0, 0) = lim h→0 h→0 h h→0 h h (b). If we approach (0, 0) along the x-axis, the value of f approaches 0 lim f (x, 0) = 2 = 0 6= 12 = f (0, 0), and therefore f is not continuous at (0, 0). x→0 x f 3. 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. fx (x, y) = 4x + 8y, and fy (x, y) = 8x + 12y 2 + 4y 3 . Setting both to zero, we see that fx = 0 gives x = −2y, and therefore substituting into fy = 0 gives 4y 3 + 12y 2 − 16y = 0. This last equation factors as 4y(y − 1)(y + 4) = 0, giving y = 0, y = 1, or y = −4. Recalling x = −2y, then, we have three critical points: (0, 0), (−2, 1), and (8, −4). We have fxx = 4, fxy = fyx = 8, and fyy = 12(2y + y 2 ). Thus, the discriminant is D = 48(2y + y 2 ) − 64 = 16(3y 2 + 6y − 4). At (0, 0), we have D = −64 < 0, so there is a saddle point at (0, 0). At (−2, 1), we have D = 80 > 0, and fxx = 4 > 0, so there is a local minimum at (−2, 1). At (8, −4), we have D = 320 > 0, and fxx = 4 > 0, so there is a local minimum at (8, −4). 4. Find the maximum and minimum values of the function f (x, y, z) = (x − 3)2 + (y − 1)2 + (z + 1)2 on the sphere x2 + y 2 + z 2 = 11. Answer. Write g(x, y, z) = x2 + y 2 + z 2 and use Lagrange Multipliers. We have ∇f = h2(x − 3), 2(y − 1), 2(z + 1)i and ∇g = h2x, 2y, 2zi. Thus, ∇f = λ∇g gives x − 3 = λx, y − 1 = λy, z + 1 = λz, along with the original restriction x2 + y 2 + z 2 = 11. The first equation gives (1 − λ)x = 3, which implies that λ 6= 1, and hence x = 3/(1 − λ). Similarly, the next two equations give y = 1/(1 − λ) and z = −1/(1 − λ). Thus, x = 3y and z = −y. Substituting into the fourth equation gives 11y 2 = 11, and hence y = ±1. Thus, there are two points to test: (3, 1, −1) and (−3, −1, 1). We have f (3, 1, −1) = 0 and f (−3, −1, 1) = 36 + 4 + 4 = 44. Thus, the maximum value is 44 and the minimum is 0. 2 2 5. Let D be the region in the plane thatZlies Z above the x-axis, inside the circle x + y = 2x, and y dA. outside the circle x2 + y 2 = 1. Compute D Answer. In polar coordinates, the two circles are r = 2 cos θ and r = 1. Their intersection point in the first quadrant occurs when cos θ = 1/2 with 0 ≤ θ ≤ π/2, which means θ = π/3. Thus, Z π/3 Z 2 cos θ ZZ Z Z 2 cos θ 1 π/3 3 1 π/3 2 r sin θ drdθ = y dA = dθ = r sin θ 8 cos3 θ sin θ − sin θ dθ 3 3 r=1 1 0 D 0 0 Z 1 1/2 3 8u − 1 du [u = cos θ, du = − sin θ dθ] = − 3 1 1/2 11 1 1 1 1 11 1 = − − (1 − 2) = + = . = (u − 2u4 ) 3 3 2 8 3 8 3 24 1 p 2 + y 2 + z 2 = 2 and above the cone z = 6. Let E be the solid lyingZinside the sphere x x2 + y 2 in ZZ x dV . the first octant. Compute E Z π/2 Z π/4 Z √2 ZZZ x dV = Answer. In spherical coordinates, ρ3 sin2 φ cos θ dρdφdθ 0 0 E√ 0 Z 2 Z π/4 Z π/2 iπ/4 h ρ4 i√2 h iπ/2 h φ 1 1 1 3 − cos 2φ dφ − sin 2φ cos θ dθ = ρ dρ = sin θ 2 2 2 4 4 0 0 0 0 0 0  π 1 π 1 − − 0 + 0 · (1 − 0) = − . = (1 − 0) · 8 4 8 4 7. Find the volume of the solid bounded by the surface y = x2 and the planes y = z and z = 1. Answer. The planes intersect along the line y = z = 1, and the solid lies below z = 1, above z = y, and to the right of y = x2 . Projecting the solid onto the xy-plane gives the region bounded by the curve y = x2 and the line y = 1. These two curves intersect at (−1, 1) and (1, 1). Thus, the Z 1h Z 1Z 1 Z 1 Z 1Z 1 1 i1 y − y2 (1 − y) dy dx = dz dy dx = volume is dx 2 y=x2 −1 −1 x2 −1 x2 y Z 1 2 1 1 x x3 x5 1 1 1 8 1 =1− + = + = − x2 + x4 dx = − + = 2 2 3 10 −1 3 5 3 5 15 −1 2 2 8. Let C be Z the curve parametrized by ~r(t) = t, 3t, t for 0 ≤ t ≤ 2, and let f (x, y, z) = x + y. f ds. Compute C √ √ Answer. ~r ′ (t) = h1, 3, 2ti, so k~r ′ (t)k = 1 + 9 + 4t2 = 10 + 4t2 . Meanwhile, f (~r(t)) = 4t. Z Z 2 p Z 26 1 26 √ 1 2 2 u du = u2/3 4t 10 + 4t dt [u = 10 + 4t , du = 8t dt] = f ds = Thus, 2 10 3 10 0 C 1 3/2 3/2 = (26 − 10 ). 3 9. Let C be the boundary of the triangle with vertices (0, 0), (2,Z0), and (0, 1), oriented counterclockwise, and let F~ (x, y) = xy + sin(πx3 ), x2 + 5y 2 . Compute F~ · d~r. C Answer. The 2Dcurl is ∂x (x2 + y 2 ) − ∂y (xy + sin(πx3 )) = 2x − x = x, and C bounds a triangle D with hypoteneuse given by 2 − 2y. Thus,Zby Green’s Theorem, Z 1xZ=2−2y Z ZZ 1 1 2 2−2y dy x x dx dy = x dA = F~ · d~r = 0 0 2 0 0 C Z D 1 2 1 2 2 2 − 4y + 2y 2 dy = 2y − 2y 2 + y 3 = 2 − 2 + − 0 = = 3 0 3 3 0 ~ 10. Let G(x, y, z) = x2 − 5yz, xy + z, y 2 − 3xz . ~ (10a). Compute curl G. ~ (10b). Compute div G. ~ equal to the gradient of anything? Why or why not? [If so, you do not need to find (10c). Is G what it is the gradient of.] ~ equal to the curl of anything? Why or why not? [If so, you do not need to find what (10d). Is G it is the curl of.] ~ equal to the divergence of anything? Why or why not? [If so, you do not need to find (10e). Is G what it is the divergence of.] ~ = h2y − 1, −5y + 3z, y + 5zi. Answer. (a). curl G ~ (b). div G = 2x + x − 3x = 0. ~ 6= ~0 [and G ~ is C 1 ]; therefore, G ~ is not the gradient of anything. (c). No, curl G ~ = ~0 and the domain is all of R3 (which has no avocado-like holes); therefore, G ~ is (d). Yes, div G the curl of something. ~ is a vector field. (e). No, divergence is a scalar, and G 11. Let C be the curve in the plane parametrized by ~r(t) = t2 + 1, t3 − 1 for 0 ≤ t ≤ 2. Compute Z y dx + x dy. C Z 2 Z ′ (t3 − 1)(2t) + (t2 + 1)(3t2 ) dt y dx + x dy = Answer. We have ~r (t) = h2t, 3ti, and therefore 0 C Z 2 2 5t4 + 3t2 − 2t dt = t5 + t3 − t2 0 = 32 + 8 − 4 = 36. = 0 12. Let F~ (x, y) = 6x − 3x2 y 2 , 4 − 2x3 y . (12a). Show that F~ is conservative by finding a potential function for F~ . (12b). Let C be the curve parametrized by ~r(t) = t cos(πt/2), t2 sin(πt/2) for 1 ≤ t ≤ 2. Compute Z C F~ · d~r. Answer. (a). [Note: as suggested by the problem, I’m just going to go ahead and find the potential function, rather than checking the curl first. The only reason to check the curl is to not waste time looking for a potential function that isn’t there. But this problem is saying the potential function is there, and we are just supposed to find it.] We need a function f (x, y) such that fx = 6x − 3x2 y 2 and fy = 4 − 2x3 y. Antidifferentiating the first, we get f = 3x2 − x3 y 2 + g(y) for some function g(y). Thus, fy = −2x3 y + g ′ (y), so it suffices to find a function g(y) such that g ′ (y) = 4; the choice g(y) = 4y works. Thus, the function f (x, y) = 3x2 + 4y − x3 y 2 has ∇f = F~ , and thus it is a potential function for F~ . Z F~ · d~r = f (~r(2)) − f (~r(1)) (b). By the Fundamental Theorem of Line Integrals, = f (−2, 0) − f (0, 1) = 12 − 4 = 8. C OPTIONAL BONUS A. Recall that on the homework, you verified that the vector field F~ = Z D −y x E ∂Q ∂P P, Q = , − = 0, but you computed has F~ · d~r and worked out that x2 + y 2 x2 + y 2 ∂x ∂y C1 it was notZzero, where C1 is the circle of radius 1 centered at the origin, oriented counterclockwise. Compute F~ · d~r, where C2 is the limaçon r = 4 + sin θ, oriented counterclockwise. C2 ZZZ Qx − Py dA = Answer. Let D be the region inside C2 and outside C1 . By Green’s Theorem, D Z Z F~ · d~r − F~ · d~r. (That’s because C2 is the outer boundary, oriented counterclockwise; but C2 C1 C1 isZ the inner boundary and so needs to be oriented backwards.) However, since Qx − Py = 0, we Z F~ · d~r. F~ · d~r = get C1 C2 Parametrizing C1 by ~r(t) = hcos t, sin ti for 0 ≤ t ≤ 2π, we get Z Z 2π Z Z − sin t(− sin t) + cos t(cos t) dt = F~ · d~r = F~ · d~r = C2 C1 0 2π dt = 2π. 0 OPTIONAL BONUS B. Find a vector field F~ (x, y, z) such that curl(F~ ) = x − yz, y − xz, xy − 2z . Answer. There are many correct answers. [As soon as you find one solution F~ , just add ∇f , for your favorite function f (x, y, z), and you have another solution.] So let’s start by trying to rig a choice of F~ = hP, Q, Ri by looking at the first coordinate and (fairly arbitrarily) trying to get Ry = x and Qz = yz, so that Ry − Qz = x − yz. So we want R = xy + g(x, z) for some g. Looking at the second coordinate, then, we need Pz −Rx = y −xz, which means Pz = y −xz +Rx = 2y − xz + gx . That is, P = 2yz − xz 2 /2 + h(x, y, z), where hz = gx . Looking at the third coordinate, we need Qx −Py = xy−2z, which means Qx = x−2z+Py = xy+hy . However, recall Qz = yz, so that Q = yz 2 /2 + k(x, y). Thus, Qx = kx ; so kx = xy + hy , meaning that k = x2 y/2 + a(x, y, z), where ax = hy . Note, then, Q = yz 2 /2 + x2 y/2 + a. At this point, any choice of the function g, h, a works, provided they fit the only restrictions we have listed: that hz = gx and ax = hy . The easiest way to do that is to pick all three functions to be zero. Thus, we can choose F~ = h2yz − xz 2 /2, z 2 + x2 y/2, xyi. A quick check shows that curl F~ = hx − yz, y − xz, xy − 2zi, as desired. OPTIONAL BONUS C. A massive oil spill earlier in 2010 occurred when an offshore drilling rig exploded, leaving the oil well open on the bottom of the ocean. What was the name of the drilling rig itself? Answer. Deepwater Horizon.