f05answers2 - Mathematics 2347 Fall 2005 Lecture 2 (Wilson)...

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Unformatted text preview: Mathematics 2347 Fall 2005 Lecture 2 (Wilson) Second Midterm Exam November 17, 2005 ANSWERS Problem 1 (a) Let f (x, y) = x2 + 2y2. Set up7 but do not evaluate, an integral to compute the surface area on the graph of f over the region R which is the triangle with vertices (07 0), (0, 2)7 and (1, 0). ANSWER: The triangle has for its boundaries the lines 36 = 0, y = 0, and y = 2 — 236. One way to set up the integrals would be 1 27290 / / ...dy dm. 0 o For surface area we want the integral of ,/ fl? + f; + 1. In this case f3; = 235 and fy 2 43/. Hence we can use to calculate the surface area 1 27290 / / V4362 + 16y2 dy dx. 0 0 (b) Set up the following integral with the order of integration reversed, i.e. as an integral with dy dm replacing dm dy. Do not evaluate the integral! 7 /12 /jfl(e$ cos(y)) dm dy 18 ANSWER: At the right is a sketch of the region of integration. As a dy dx integral, we see the overall range of x values is from 0 to 1. For any x in that range7 y goes from the lower line y = 1 up to the curved upper boundary. The curve is x = y — 1 or y = $2 + 1. Hence the integral in this form is 18 E 8 GB 1 $2+1 / / (ex cos(y)) dy dx. 0 1 " 0.2 '3" 0.2 034 as BB 1 Problem 2 Set up an iterated integral to compute /// (x + 2y) dV, R where R is the region in space bounded by the cylinder 362 + 22 = 4, the plane 3/ = 0, and the plane 3/ + 2 = 2. You did not have to evaluate this integral, but you were offered 5 extra—credit points for correctly evaluating it. ANSWER: The cylinder has radius 2 and the y—aXis as its center line. The plane y+ z = 2 cuts across it7 slanting down (2 = 2 — y) as you go out in the y direction. The widest part of the (ellipse) where the plane and cylinder meet, in the x direction, is in the x—y plane where z = 0: Picture the circle x2 + 22 = 4 in the plane y = O, and x extends from —2 to 2. If we put dx on the outside, this will give the range for the corresponding integral. If we put dz next, moving inward, the same circle shows that 2 ranges from —\/4 — $2 to x/ 4 — .762. Now for any .76 and 2 corresponding to a point in that circle, the range of y values is from y = 0 at the x—z plane out to the angled plane y = 2 — 2. Thus we can set up the integral as 2 x/4ix2 272 d d d 2 . [Mam/0 (96+ 9) y 2 96 Evaluating this integral is rather messy. Doing the inside integral we get 30(2 — Z) + (2 — 2)? Integrating that in the next stage gives §(4 — x2)3/2 + 435(4 — 352)”2 + 8(4 — $2)1/2. Finally, evaluating the dx integral, we get 207?. Problem 3 (a) For f(x, y) = x2 + yez, find the gradient field €15. ANSWER: grad( f) = V f = 125+ fyj+ f2]? = 2967+ 67+ yezl; (b) Let EQC, y, ,2) be the vector field 2xT+ ezf+ yea/g. —> (i) Find div(F). ANSWER: diU(F) 2 AIM?!— Nyj—l— P2]? where A] = 2x, N = 62, and P = yez. Hence dz’v( F) = 2T+ 07+ gen; (11) Find curl(E). ANSWER: We could conipute this directly. But if we compare F to the answer to part (a), we see that F =gmd(f). Since curl(gmd(f)) is zero for any f, the answer is 6 2 07+ 0j+ 0k. Problem 4 Find all local niaxirna, local niinirna, and saddle points, for f (x, y) = 6x2 — 2x3 + 3y2 + 6mg. Be sure to give both the points at which f takes on the values and the values it takes on there. ANSWER: There is no boundary specified and f as a polynomial is continuous everywhere, so the only candidates are points where the partial derivatives are both zero. Computing, f9; = 1235 — 6x2 + 6y and fy 2 63; + 636. If fy = 0, 6y + 6x = 0, i.e. y 2 —3c. Substituting that in fw =Owe have 12m—6x2—6x20, so 6m—6m2 26$(1—m) =0. Hencem=00r$= 1. If x = O, y = —x = 0, while if x = 1, y = —x = —1. Thus the two candidate points are (0, O) and (1, —1). To check each of these as to whether it gives a saddle point or a local maximum or minimum, we use D = fmfyy — 3y. fm = 12 — 12x, fyy = 6, and fxy = 6. At (0,0), D = 36 > 0 so there is either a local maximum or a local minimum here. Since f$(0, 0) = 12 > 0, it is a local minimum. The value of f at this point is f(0, 0) = 0. At (1, —1), D = —36 < 0, so there is a saddle point. If we look at points along the x—axis, where y = 0, f(x, 0) = 6x2 — 2x3. (For very large positive values of .76 this becomes arbitrarily large and negative, while for large negative values of x it becomes large and positive. Hence there is no global maximum or minimum for the whole plane.) Problem 5 Evaluate the line integral / (1 — m)d8 where C is a portion of the circle of radius 2 and center 0 (0,0), traversed from (2,0) to (0,2). ANSWER: The curve can be parametrized as m = 2cost, y = 2sint, 0 S t S The function 1 — m becomes 1 — 2 cost. y’ = 2 cost and m’ = —2 sint. We can evaluate the integral as /§(1 — 2cost)\/4sin2t + 4costht = 2/§(1 — 2cost) dt = 2 [t — 2sint]0g 2 7T — 4. 0 0 Problem 6 What are the largest and smallest values that f (m, y) = my takes on, for (m, y) on the ellipse $2 y2 — — = 1? 8 + 2 At what points does f achieve those values? ANSWER: We can view this as a constrained maximum / minimum (Lagrange multiplier) prob— lem with the objective function f(m, y) = my and constraint g(m, y) = “38—2 + g — 1 = 0. Then $1” = yT+ mfi and $9 = fi—l— yj To find where these are parallel we solve 6f 2 A($g) for some value A. We need to find m and y (and A perhaps as a tool for finding m and y) that satisfyy=A(§),m=Ay,and%+%—1=O. From the first equation we get % = 2, while the second yields % = (We divided by m and by A: If A = 0 then both m and y are 0, which does not fit g(m, y) = 0, so we can assume A 7A 0. If m = 0 and y = A with A 7E 0, then y = 0, so again we have the point (0,0) which is not on the ellipse.) Setting these equal we have 2 = i or A2 = 4. Hence A = :2. If A = 2, m = Ay gives m 2 2y. Using g(m,y) = 0 we have % + g = 1, y2 = 1, so y = ::1. If y = 1, m 2 2y 2 2, so we have the point (2,1). If y = —1 we get the point (—2, —1). If A = —2, so m 2 —2y, we get in the same way the points (—2, 1) and (2, —1). Now we have four points to consider, ( 2, 1). We evaluate the function f(m, y) = my at each of these. At (2,1) we get f(m,y) = 2. A (2,—1) we get f(m,y) = —2. At (—2,1) we get f(m,y) = —2. At (—2, —1) we get f(m,y) = 2. Hence the maximum value, f(m,y) = 2, occurs at the points (2, 1) and (—2, —1), and the minimum value, f(m, y) = —2, at (2, —1) and (—2, 1). Problem 7 A thin plate covers the region in the plane bounded by the m—aXis7 the line x = 17 and the curve 3/ = fl, Where y 2 0. The density of this plate is given by the function 6(36, y) = x + y. (a) Use an integral to evaluate the mass of this plate. ANSWER: The plate is shown at the right. We compute the mass by integrating the density function through the region, 1 1 1 1 y4 13 M=// dd=/— ___3d:_ 0 y2($+y)$y 0<2+y 2 y 96 20 (b) Find the moment My; of the plate about the m—axis. ANSWER: M 1 1 d d 3 m—/0 AZWHW) x y—l—O- (c) Find the moment Ag of the plate about the y—aXis. ANSWER: A] 1 1 d d 19 311—1142347544» 95 y—E- (d) Find the coordinates (E, y) of the center of mass of this plate. ANSWER: - — _ _ m The .76 coordinate x — M — 273. The y coordinate y = = %. Problem 8 Find the volume of the region that is above the cone 9b = g and inside the sphere p = 1. At the right is a figure cut away so that you can see inside the sphere. ANSWER: We set up the integral in spherical coordinates. The radius p goes from 0 at the center out to 1 on the sphere. The horizontal direction 9 goes all the way around from 0 to 27?. And the vertical angle p goes from 0 at the top, along the z—aXis, down to Hence the volume integral is 271’ 1 g 2 . 71— / ps1ngz§dgz5dpd6=—. 0 0 0 3 ...
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This note was uploaded on 05/15/2008 for the course MATH 234 taught by Professor Dickey during the Fall '08 term at University of Wisconsin.

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f05answers2 - Mathematics 2347 Fall 2005 Lecture 2 (Wilson)...

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