Math 265H: Calculus III Practice Midterm II: Fall 2014
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1 Name: Section #: Math 65H: alculus III Practice Midterm II: Fall 14 Instructions: This exam has 7 problems. The number of points awarded for each question is indicated in the problem. Answer each question completely. Show all work, and indicate what calculations were done with a calculator. No credit is allowed for mere answers with no work shown. State the reasons that justify any conclusions you make. Please do not write inside of the following box. Question 1: Question : Question 3: Question 4: solutions Question 5: Question 6: Question 7: 1 Total Points:
2 Question 1 (15 points). Let R be the closed region (including the boundary) enclosed by the ellipse x + y 6. Find the maximum and minimum values of f(x, y) x y on R. Solution. We first find the critical points on the interior of the region, by setting the partial derivatives equal to zero: f x xy f y x. This tells us that we have critical points everywhere along the line x, and we also notice that f(x, y) along the line y as well. We find the determinant of the Hessian to try to determine if these are maxima or minima: D f xx f xy y x x 4x. f yx f yy Since D at all points, and it s easy to see that f is negative when y < and positive when y >, we conclude that f has no extreme values on the interior of R. Next, we test along the boundary of R, using Lagrange multipliers. Setting G(x, y) x + y 6, we set up f λg: This leads to the system of equations xy, x λ x, 4y. xy xλ (1) x 4yλ () x + y 6. (3) From (1), if x, then we already know that f and this is not an extreme value. However, if x, then y λ, and substituting into () we find that x 4y. We substitute this into (3) to obtain (4y ) + y 6, which simplifies to y ±1. Putting this back into (3) gives us the points (±, ±1) and (±, 1). Plugging these into f(x, y), we find maximum values of f(x, y) 4 at (±, 1), and minimum values of f(x, y) 4 at (±, 1). Page
3 Question (1 points). Evaluate the integral: 3 9 y y cos(x ) dx dy Solution. hange the order of integration. The new bounds would be written: 9 x y cos(x ) dy dx The integral is then straightforward, and equal to 1 4 sin(81). Page 3
4 Question 3 (1 points). Let E be the region in R 3 enclosed by the planes z and z x + y + 5, and by the cylinders x + y 9 and x + y 4. Evaluate the integral x dv. E Solution. If you consider that in the variables x and y the region E lies in between the two cylinders of radius and 3, respectively, then we only need to know which of the two planes (z and z x+y +5) is the upper bound in z, and which is the lower. Since x+y +5 > everywhere inside x + y 9, we may set our bounds in cylindrical coordinates, with its associated Jacobian: π 3 x+y+5 x dv x dv E π 3 r cos θ+r sin θ+5 (r cos θ)(r) dzdrdθ This integral should be straightforward, but will use the half-angle formula cos θ 1 (1 + cos(θ)). Page 4
5 Question 4 (15 points). Evaluate the integral 4(x + y)e x y dy dx R where R is the triangle in the x, y-plane with vertices (, ), (1, 1), and ( 1, 1). Solution. We will use the change of variables u x + y and v x y, and we first find the Jacobian of this change. Note that we are trying to change a dxdy into a dudv, so the change of variables formula reads: dxdy (x, y) (u, v) dudv That is, we need to invert the system of equations given to us if we wish to take these derivatives. You can check that this inverted system reads: x 1 (u + v) y 1 (u v) (x, y) This means that our Jacobian will be equal to (u, v) 1. Our next task will be to change our bounds into u, v bounds. We can do this by simply translating the three points given to us into the u, v-plane using our transformation. This gives the three points (, ), (, ), and (, ) in the u, v-plane. This immediately shows that our new triangle is bounded by the lines u, v, and v u. Now, we construct our new integral: ( ) 1 4(x + y)e x y dydx 4(u)e (v) dvdu R R u ue v dvdu u ue u du This last integral can be done by parts. Page 5
6 Question 5 (1 points). Let F, G: R 3 R 3 be two 1 vector fields, and k R be any scalar. Use these to show that the curl is linear; that is, it satisfies both curl (k F) k curl (F) curl (F + G) curl F + curl G. Solution. To make these calculations, we name the component functions of both F and G: F(x, y, z) M 1, N 1, P 1 G(x, y, z) M, N, P For the first statement, we multiply F by the scalar k and then take the curl: i j k curl (k F) x y z km 1 kn 1 kp 1 [ y (kp 1) ] [ z (kn 1) i x (kp 1) ] [ z (km 1) j + x (kn 1) ] y (km 1) k [ k P ] [ 1 i k P ] [ 1 j + k N ] 1 k y k N 1 z [ P1 k y N 1 z k curl F. ] i k x k M 1 z [ P1 x M 1 z ] j + k x k M 1 y [ N1 x M ] 1 k y The second statement is similar, with the curl of F + G: i j k curl (F + G) x y z M 1 + M N 1 + N P 1 + P We expand this in the same way that we did for the first statement, and use the additivity of partial derivatives and vectors to simplify. Page 6
7 Question 6 (5 points each). Evaluate the following line integrals. You may use any theorems or definitions that you like to help you, but you should be clear in your explanations and reasoning. (a) F ds, where F(x, y) (y + x)i + (3x + y)j, and is the ellipse x 9 + y 3 1, oriented counterclockwise in R. Solution. This vector field is not conservative, so we cannot use the fundamental theorem. Instead, we first parameterize our curve and take its derivative: Therefore: F d π π π π π c(t) 3 cos t, 3 sin t, t [, π] c (t) 3 sin t, 3 cos t (y + x), 3x + y 3 sin t, 3 cos t dt ( 3 sin t + 3 cos t), 9 cos t + 3 sin t 3 sin t, 3 cos t dt 6 3 sin t cos t + 4 sin t cos t dt 6 3 sin t cos t cos t + 4 sin t cos t dt cos t + 4 sin t cos t dt Now one of these integrals will use a half-angle formula, and the other is a u-substitution. Page 7
8 (b) 8xyz ds, where is parameterized by: c(t) 3i + 1tj + 5tk, t [, ]. Solution. This integral is scalar-valued, so we have no theorems to aid us. Instead, we simply calculate: c (t), 1, 3 c (t) Therefore: 8xyz d xyz c (t) dt xyz(13) dt (3)(1t)(5t)(13) dt 64t t dt Page 8
9 (c) sin x cos y dx + (xy + cos x sin y) dy, where is the closed curve traveling in a straight line from (, ) to (1, 1), and then back to the origin along the curve y x. Solution. Here, our piecewise smooth path is closed, but again this vector field is not conservative so we cannot use the fundamental theorem to say that the integral is equal to zero. However, we can use Green s Theorem instead. We note: N y sin x sin y x M sin x sin y y Therefore: F d R R 1 y ( N x M ) da y y da y y dxdy. The rest of this integral is straightforward. Page 9
10 (d) F ds, where F(x, y, z) ln y, x y + 1y3 z, 3y 4, and is the unit circle centered on the y-axis and lying in the plane y. Solution. This vector field is conservative, with potential function f(x, y, z) x ln y + 3y 4 z, and the path is closed, so by the Fundamental Theorem of alculus, the integral is equal to zero. Page 1
11 (e) (cos x cos y sin x sin z) dx sin x sin y dy + cos x cos z dz, where is given by: c(t) t, sin t, t + π, t [, π]. Solution. This vector field is also conservative, with potential function f(x, y, z) sin x cos y + cos x sin z, but here the path is not closed. However, we may still use the fundamental theorem, by evaluating the potential function at the endpoints of the curve. We first find those endpoints: ( c(π) π,, 5π ) ( c(),, π ) So our integral is given by: F d f(c(π)) f(c()) ( f π,, 5π ) ( f,, π ) 1 1 We note here that despite the fact that this integral turned out to equal zero, it was not necessarily zero, as in the last example. Page 11
12 Question 7 (15 points). Imagine rotating the curve y sin x, x π around the x-axis to form a closed surface S in R 3. (a) Find an appropriate parametrization Φ for S. Solution. Since S is a surface of rotation, it is best to parameterize it using cylindrical coordinates. The rotation is happening in y and z, with symmetry along the x-axis, so we parameterize with: Here u [, π] and v [, π] φ(u, v) u, (sin u) cos v, (sin u) sin v (b) Use your parametrization Φ to find the equation of the tangent plane to S at the point ( π, 3, 1 ). Solution. We find our normal vector: φ(u, v) u, sin u cos v, sin u sin v φ u 1, cos u cos v, cos u sin v φ v, sin u sin v, sin u cos v φ u φ v sin u cos u, sin u cos v, sin u sin v Now we would like to evaluate this normal function at our point, but we must figure out which value of (u, v) corresponds to the given point. Since it is clear that u π, we look next at the y and z components of our parametrization; v must satisfy both: sin π cos v cos v 3 sin π sin v sin v 1 A quick look at the unit circle tells us that this occurs at v 5π. So we evaluate our 6 normal: ( π [φ u φ v ], 5π ) 3, 6, 1 We now construct our plane: ( x π ) 3 + ( ) 3 y + 1 ( z 1 ) Page 1
13 (c) Set up an integral that would give the surface area of this closed surface. You need not finish the integral, but you should simplify it as much as possible. Solution. Most of the work has been done in the above parts; we are only left to find the magnitude of the element φ u φ v. So: φ u φ v u + sin u cos v + sin u sin v u + sin u So our integral would look like: S φ u φ v ds π π u + sin u dudv Page 13
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