Math 13, Section 01, Solutions to the Final Exam

Math 13, Section 01, Spring 2009 Solutions to the Final Exam 1. (15 points) Let S be the plane that contains the points (0, 0, 1), (4, 2, 0), and (1, −3, 2). Find an equation for the line through the point (1, 0, −2) that is perpendicular to S. Answer. Subtracting the first point from the second and third, we see that the vectors ~a = h4, 2, −1i and ~b = h1, −3, 1i lie in the plane. Their cross product is ~n = ~a × ~b = h−1, −5, −14i. Thus, ~n is orthogonal to the plane and hence parallel to the desired line. Replacing ~n by −~n for convenience, then, the line is given by ~r(t) = ht + 1, 5t, 14t − 2i.  3 2  2x + 3xy − 3y if (x, y) 6= (0, 0), 2. (20 points) Let f (x, y) = x2 + 2y 2  0 if (x, y) = (0, 0). (a). Prove that f is not continuous at (0, 0). √ √ (b). Compute the directional derivative D~u f (0, 0), where ~u = h1/ 2, 1/ 2i. 3 −3y 2 = − 6= 0 = f (0, 0). Thus, f is Answer. (a). Along the y-axis, we have lim f (0, y) = lim 2 y→0 y→0 2y 2 discontinuous at (0, 0). 2h3 3h2 3h2 √ √ √ 3+√ 2−√ 2 f ((0, 0) + h~u) − f (0, 0) 2 h3 / 2 2 2 2 (b). D~u f (0, 0) = lim = lim = . = lim  h2 h→0 h→0 h→0 h(3h2 /2) h 3 2h2  h √ 2+√ 2 2 2 3. (25 points) Find and classify (as local minimum, local maximum, or saddle point) every critical point of the function f (x, y) = xy 2 − 6x2 − 3y 2 + 7. Answer. We have fx = y 2 − 12x and fy = 2xy − 6y, both of which are always defined. Setting them both to zero, we get 12x = y 2 and (x − 3)y = 0. By the second equation, either x = 3 or y = 0. If x = 3, then the first equation gives y 2 = 36, and hence y = ±6. If y = 0, then the first equation gives x = 0. Thus, there are three critical points: (0, 0), and (3, ±6). The second derivatives are fxx = −12, fxy = fyx = 2y, and fyy = 2x − 6. Thus, the discriminant 2 = 24(3 − x) − 4y 2 . At (3, ±6), we have D = −4 · 36 < 0, so that both of those is D = fxx fyy − fxy points are saddle points. At (0, 0), we get D = 3 · 24 > 0 and fxx = −12 < 0. so that there is a local maximum at (0, 0). 4. (25 points) Find the maximum and minimum values of the function f (x, y) = x2 y subject to the constraint x2 + y 2 = 9. Answer. Let g(x, y) = x2 + y 2 and use Lagrange Multipliers. We have fx = 2xy, fy = x2 , gx = 2x, and gy = 2y. So we have to solve the system of equations 2xy = 2λx, x2 = 2λy, and x2 + y 2 = 9. The first equation gives x(y −λ) = 0, so either x = 0 or λ = y. If x = 0, then the last equation gives y 2 = 9, and so y = ±3. If x 6= 0, then λ = y, and so√the second equation √ gives x2 = 2y 2 , turning 2 the third equation into 3y = 9. That gives y = ± 3, and hence x = ± 6, with no correlation between the two ± signs. √ √ Thus, we have six points to check: √ two √ of the √ form (0, ±3), √ and√four of the √ form (± 6, ± 3). We compute f√ (0, ±3) = 0, f (± 6, 3) = √ 6 3, and f (± 6, − 3) = −6 3. Thus, the maximum value of f is 6 3, and the minimum is −6 3. 5. (30 points) Let C be the path in the xy-plane that begins at (0, 3), runs (counterclockwise) through the second quadrant along the arc of the circle of radius 3 centered at the origin to the point (−3, 0), then moves right along the x-axis to the origin, and finally moves up the y-axis to return to the starting point (0, 3). Compute Z C 6x2 y dx + (2x3 − xy) dy. q(0,3) ' q- (-3,0) 6 q(0,0) Answer. Let D denote the quarter-disk bounded by C; note with RR that C is oriented positively 2 respect to D. Thus, by Green’s Theorem, the integral is D Qx − Py dA, where P = 6x y and Q = 2x3 − xy. We compute Qx − Py = 6x2 − y − 6x2 = −y. Thus, converting to polar coordinates, the Z Z integral is Z π Z 3    Z 3 Z π π  r 3 3  r2 dr = cos θ sin θdθ (−r sin θ)r dr dθ = − −y dA = 3 0 π/2 0 π/2 π/2 0 D = (−1 − 0)(9 − 0) = −9. 6. (30 points) Let S denote the sphere in R3 of Z Zradius 2 centered at the origin, oriented outward, 2 2 2 y ~ and let F~ (x, y, z) = hy z, yz , x e i. Compute F~ · dS. S Answer. We can use the Divergence Theorem, so we compute div F~ = 0 + z 2 + 0 = z 2 . Denoting the spherical coordinates, we have Z Z solid inside Z Zthe Z sphere asZE and Z using Z 2π π 2 ~= F~ · dS z 2 dV = (ρ cos φ)2 ρ2 sin φ dρ dφ dθ S E 0 0 0   Z 2  Z 2π  Z π 2 ρ4 dρ [u = cos φ, du = − sin φ dφ] cos φ sin φ dφ dθ = 0 0 0 Z  ρ5 2    u3 1  32  −1 2 32 128π u2 du = 2π − = 2π − 0 = 2π · · = . 5 0 3 −1 5 3 5 15 1 7. (25 points) For each of the following vector fields, either find a potential function (i.e., a function that it is the gradient of) or prove that the vector field is not conservative. (a). F~ (x, y) = hx2 − cos(2y), y 3 + 2x sin(2y)i. ~ (b). G(x, y, z) = h2xy − x2 , z 3 , 3yz 2 i. ∂Q ∂P Answer. (a). The 2D curl of F~ is − = 2 sin(2y) − 2 sin(2y) = 0, on all of R2 . Since R2 is ∂x ∂y simply-connected, F~ is conservative, so there is some f with ∇f = F~ . x3 Then fx = x2 − cos(2y), and so f (x, y) = − x cos(2y) + g(y) for some function g. Therefore 3 y4 . Thus, fy = −2x sin(2y) + g ′ (y), which means g ′ (y) = y 3 , and hence we can choose g(y) = 4 x3 y 4 f (x, y) = + − x cos(2y) is a potential function for F~ . 3 4 ~ = hRy − Qz , Pz − Rx , Qx − Py i = h3z 2 − 3z 2 , 0 − 0, 0 − 2xi, which is (b). We compute curl G ~ is not conservative. h0, 0, −2xi = 6 ~0. Thus, G 8. (30 points) Let C be the curve that lies in the surface Zz = x3 − xy + 2 directly above the sin(x2 ) dx + yz dy − y 2 dz, if C is boundary of the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1. Compute oriented clockwise when viewed from above. C Answer. Let S be the portion of the surface z = x3 − xy + 2 bounded by C, so that S is parametrized by ~r(x, y) = hx, y, x3 −xy+2i for (x, y) ∈ [0, 2]×[0, 1]. We have ~rx = h1, 0, 3x2 −yi, and ~ry = h0, 1, −xi, so that ~rx ×~ry = hy − 3x2 , x, 1i. However, that vector is pointing upward, which (by the right hand rule) is the wrong way for the orientation on C; so we use ~ry ×~rx = h3x2 − y, −x, −1i instead. Meanwhile, the vector field F~ = hsin(x2 ), yz, −y 2 i has curl F~ = h−2y − y, 0, 0i = −3y~i. Thus, by Stokes’ Theorem, Z Z Z Z Z ZZ 2 2 1 1 ~= 3y 2 − 9x2 y dy dx −3y(3x2 − y) + 0 + 0 dy dx = (curl F~ ) · dS F~ · d~r = 0 0 C Z S Z 20 0 2 9 2 21 9 2 3 32 3 y − x y = 1 − x dx = x − x = 2 − 12 − (0 + 0) = −10. dx = 2 2 2 0 y=0 0 0 BONUS A. (2 points) Let R be region in the first quadrant of the xy-plane bounded to the upper right by y = 4/x, to the lower right by√y = x, to the lower left by yZ Z = 1/x, and to the upper √ u left by y = 9x. Use the transformation x = cos(πxy) dy dx. , y = v u to compute v R √ √ Answer. The curve y = 4/x becomes v u = 4v/ u, and hence u = 4; similarly, y = 1/x becomes √ √ u = 1. Meanwhile, y = x becomes v u = u/v, and hence v 2 = 1, meaning either v = 1 or v = −1; however, the region R is in the first quadrant and so has x > 0, so that it must be v = 1. Similarly, y = 9x becomes v = 3. Thus, R is the image of the rectangle S = [1, 4] × [1, 3] in the uv-plane under the given transformation. √ √ The absolute value of the determinant of the Jacobian is |xu yv − xv yu | = |(1/(2v u)) · u − √ √ (− u/v 2 )(v/(2 u))| = |1/(2v) + 1/(2v)| = 1/v. Meanwhile, xy = u. Thus, the integral is Z 4Z 3  3  1 4 cos(πu) = (ln 3 − 0)(0 − 0) = 0. dv du = ln |v| sin(πu) v π 1 1 1 1 BONUS B. (2 points) Let C be the curve in R3 that starts at the point (0, 4, 2), goes to the point (2, 0, 0) via ~r(t) = ht, 4 − t2 , 2 − ti for 0 ≤ t ≤ 2; and then goes from (2, 0, 0) to the point 2 (0, 4, 0) along Z the arc of the parabola y = 4 − x in the xy-plane. cos(ln x) dx + z dy + y 2 dz. Compute C Answer. [Note: due to a last-minute change I made, the integral in this problem is actually improper, because cos(ln x) is undefined on the plane x = 0, which includes two of the endpoints of C. Technically that can be changed by using (ε, 4 − ε2 , 2 − ε) for the starting point, and (ε, 4 − ε2 , 0) for the ending point, and then taking the limit as ε ց 0. [And the vertical line segment C ′ that appears below would go from (ε, 4 − ε2 , 0) to (ε, 4 − ε2 , 2 − ε).] But, as is common with improper integrals that converge, it all works if you just compute the thing blindly. So, since I did not intend for the problem to be that complicated when I wrote it, I awarded full credit for the simplified method in grading, and I will only do that simplified method here.] Let C ′ be the line segment from (0, 4, 0) to (0, 4, 2). So C and C ′ together form the boundary 2 of a surface S which is sort of a triangular slice ofZthe parabolic Z cylinder yZ Z= 4 − x . Writing ~ We F~ = hcos(ln x), z, y 2 i, Stokes’ Theorem tells us that F~ · d~r + (curl F~ ) · dS. F~ · d~r = C C′ S will compute the second and third integrals. For the second, parametrize C ′ by ~r(t) = h0, 4, ti for t ∈ [0, 2]. Then ~r ′ (t) = ~k, and y = 4 on C ′ ; so Z Z 2 ~ 16 dt = 32. F · d~r = C′ 0 Meanwhile, we can parametrize S by ~r(x, z) = hx, 4 − x2 , zi, for 0 ≤ x ≤ 2 and 0 ≤ z ≤ 2 − x. We have ~rx = h1, −2x, 0i and ~rz = h0, 0, 1i, and therefore ~rx × ~rz = h−2x, −1, 0i. However, by the right hand rule, this is pointing the wrong way, so instead we use ~rz × ~rx = h2x, 1, 0i. We have curl F~ = h2y − 1, 0, 0i. Thus, Z 2 Z 2 Z 2−x ZZ ~= (14x − 4x3 )(2 − x) dx (2(4 − x2 ) − 1)(2x) dz dx = (curl F~ ) · dS 0 0 0 S Z  2 2 14 16  8 · 23 184 2 3 4 2 14 3 4 4 5 28x−14x −8x +4x dx = 14x − x −2x + x = 8 7− −4+ = = = . 3 5 0 3 5 15 15 0 Finally, then, the original integral is the difference of the two we computed; that is, Z 296 184 − 32 = − . F~ · d~r = 15 15 C BONUS C. (1 point) Name the largest battle (in terms of numbers of participants and casualties) in the history of human warfare. Answer. The Battle (or Siege) of Stalingrad, fought between the Nazis and the Soviets from July 1942 through February 1943. Casualty estimates on both sides totalled about 2 million. (By contrast, the Battle of the Bulge, the largest battle involving US forces, had fewer than 200,000 total casualties.) BONUS D. (1 point) About two weeks ago, a US senator changed political parties. What is that senator’s name? Answer. Arlen Specter. (Senator from Pennsylvania. Was a Republican until recently, but now is a Democrat. Interestingly, he was originally a Democrat long ago but switched to the Republican party back in 1966.)