Math 265 Final Exam Full Name: Fall 2012 Instructor/TA: 1. Equation for plane AB = h4, 1, −2i, AC = h3, 2, −1i so a normal vector is h3, −2, 5i. An equation is then 3x − 2y + 5z = 4 (or equivalent forms). b) Plane intersects y-axis at (0, −2, 0). 2. Motion in the plane v = ht cos t, t sin ti, a = hcos t − t sin t, sin t + t cos ti. b) components of acceleration aT = a.T = 1, aN = a.N = a.h− sin t, cos ti = t. 3. Tangent plane Normal to the given surface is n = h2x − 3y − 2, −3x + 5y + 3, −1i. This needs to be parallel to h2, −4, 1i. So there has to be a number c such that one vector equals c times the other, 2x − 3y − 2 = 2c −3x + 5y + 3 = −4c −1 = 1c So c = −1, 2x − 3y = 0 and −3x + 5y = 1. This is true only for the point (x, y) = (3, 2). 4. Local Extrema ∇f = h4x3 − 4y − 14x + 4, −4x + 8y − 8i equals h0, 0i exactly for y = x/2 + 1 and (substitute for y into the other equation) 4x3 − 2x − 4 − 14x + 4 = 4x(x2 − 4) = 0, so x = 0, ±2. This gives the points (0, 1), (2, 2), (−2, 0). Classification. We compute fxx fyy fxy D = = = = 12x2 − 14 8 −4 8(12x2 − 14) − 16 = 32(x2 − 4) and get for the points listed above D = −128, 0, 0 (in this order). So (0, 0) is a saddle, the SPT is inconclusive for the two other points. Math 265 Final Page 2 5. Order of integration The inequalities y 2 ≤ x ≤ 2 − y mean the region is bounded on the left by the parabola y 2 = x, on the right by the line 2 − y = x. The inequalities 0 ≤ y ≤ 1 say that the region is bounded on top and bottom by the lines y = 0, 1, but a sketch shows that the line y = 1 is not actually needed. We find the smallest and biggest x-value √ as x = 0, 2 (both happen for y = 0). Then we solve all inequalities for y. Both y ≤ x and y ≤ 2 − x need to be true, so √ 0 ≤ y ≤ min( x, 2 − x). So the integral equals 2 Z Z √ min( x,2−x) f (x, y) dy dx. 0 0 6. Spherical / Cylindrical coordinates Describe solid with inequalities, translate to spherical ρ ≤ 4 cos φ, cos phi ≤ sin φ. This gives limits π/4 ≤ φ ≤ π/2 and Z π/2 Z 4 cos φ Z V = π/4 0 2π ρ4 sin3 φ dθ dρ dφ. 0 The integral over θ could be in any other position, too. 7. FTC for line integrals / independent of patch / conservative a) We try to find a potential function f (x, y) such that fx = y − so f = xy + 1 x 1 , x2 + C(y). Then fy = x + C 0 (y) which means f (x, y) = xy + 1 1 − + D. x y Since F = ∇f , F is conservative. b) We can use the FTC for line integrals, Z 1 9 F · dr = f (4, 1) − f (1, 1) = 4 + − 1 − 1 = . 4 4 C 8. Gauss’ Divergence Theorem says that the flux integral in question equals ZZ ZZZ F · n dS = div F dV. ∂S S It turns out that div F is really simple, div F = 2z. Math 265 Final Page 3 For the volume integral, we best use cylindrical coordinates. Note that r2 ≤ z ≤ 2 − r gives an upper bound for r (sketch the functions r2 and 2 − r! the biggest possible value for r happens where their graphs meet, at r = 1). So 0 ≤ r ≤ 1, and ZZ Z 1 Z 2−r Z 2π F · n dS = 2rz dθ dz dr ∂S 0 r2 0 Z 1 r5 − r(2 − r)2 dr = 2π Z0 1 3π r5 − r3 + 4r2 − 4r dr = 2π(1/6 − 1/4 + 4/3 − 2) = − . = 2π 2 0