Math 11 Fall 2018 Practice Final Exam

Transcription

1 Math 11 Fall 218 Practice Final Exam Disclaimer: This practice exam should give you an idea of the sort of questions we may ask on the actual exam. Since the practice exam (like the real exam) is not long enough to cover everything we studied, there may be topics on the real exam that are not on the practice exam, and vice versa. Anything covered in assigned reading, class, WeBWorK, or written homework is fair game. Advice: A good way to use the practice exam is to first study and prepare for the exam. Then take a couple of hours, sit in a quiet place, and take the practice exam as if it were the real exam. That should tell you which areas you should study further. About the real exam: There may be short answer questions that will be graded only on the answer, and there will definitely be questions on which we grade on your work, your explanations, as well as the answer. 1. TRUE or FALSE? (No partial credit; you need not show your work.) (a) If S is the unit sphere centered at the origin, oriented outward and the flux integral S F d A is zero, then F =. Solution. False. Two other ways for the flux integral to be zero are that the vector field is always tangent to the surface so that F N = always, or that part of the surface contributes positive flux, and the remaining part of the surface contributes exactly the negative of that flux. In either case the flux integral would be zero. (b) If S 1 is a rectangle with area 1 and S 2 is a rectangle with area 2, then 2 S 1 F d A = S 2 F d A. Solution. False. The flux does not only depend on the area, but also on the vector field and the orientation of the surface with respect to the field. We have an easy counterexample in a nonzero constant vector field, with one rectangle oriented so that its normal vector is parallel to the field vectors, and the other rectangle oriented so that its normal vector is perpendicular to the field vectors. Then One integral would be nonzero, and the other zero. (c) If S is the unit sphere centered at the origin, oriented outward and F = x i+y j+z k, then the flux integral S F d A is positive. Solution. True. We observe that the normal vectors of S and the field vectors are always pointing in the same direction, so we are integrating a positive function, yielding a positive integral value. 1

2 (d) If S is an open-ended circular cylinder centered about the z-axis, oriented away from the z-axis, and F = 3, 2, 6, then the flux of F through S is zero. Solution. True. Since the surface is symmetric and the vector field is constant, we expect some nice behavior, possibly an integral value of zero. To be sure of this, we need to check whether there is cancellation. We can consider pairs of points on the cylinder on opposite sides of the z-axis. The normal vectors at these points have opposite direction, so F N at the two points cancel. Since each point on the cylinder has a corresponding point on the other side of the z-axis, we end up with perfect cancellation. This shows that the flux is zero. (e) If u and ( v are ) non-zero and not parallel, the angle θ from u to v is equal to cot 1 u v. u v Solution. True. The conditions on u and v eliminate the possibility that u v =, so the formula is well-defined. Using the geometric formulas for dot product and magnitude of cross product, we find that Thus, θ = cot 1 ( u v u v u v u v ). = u v cos θ u v sin θ = cot θ. (f) If r(t) is twice differentiable and r(t) = c > is a constant then r (t) is orthogonal to r(t) for all t and r (t) r(t) < for all t. Solution. Not necessarily true. The condition r(t) = c > means that the curve travels on a sphere of radius c, and since r(t) is differentiable, the tangent vector r (t) always exists. This vector must point in the direction of travel, which on the surface of a sphere is perpendicular to the radius. Thus, r (t) is orthogonal to r(t) for all t. Algebraically, we can use that c 2 = r(t) 2 = r(t) r(t), and by differentiating the ends, we have = 2r (t) r(t), so r (t) is orthogonal to r(t) for all t. We can also argue geometrically that since r (t) is the rate of change of r (t) as t increases, and as a curve on the sphere travels, the tangent vector always dips down towards the center of the sphere, that r (t) tips toward the center. That means that we expect the angle between r (t) and r(t) to be acute, so that the dot product r (t) r(t) is positive. To make this intuition precise, we can compute (r (t) r(t)) = r (t) r(t) + r (t) r (t) = r (t) r(t) + r (t) 2. Note that since r (t) r(t) =, its derivative must be zero as well. Thus, r (t) r(t) = r (t) 2. 2

3 The right side must be negative or zero. Since we cannot rule out the case r (t) = (when the parametrization stops moving), it is not necessarily true that r (t) r(t) <. (g) The function u(x, y) = e 3x cos(3y) satisfies Laplace s equation 2 u x u y 2 =. Solution. True. We will compute the partial derivatives and plug them into Laplace s equation. For the first partials, we have u x = 3e 3x cos(3y), For the second partials, we find 2 u x 2 = 9e 3x cos(3y), u x = 3e 3x sin(3y). 2 u y 2 = 9e 3x cos(3y). Thus, 2 u x + 2 u 2 y = 2 9e 3x cos(3y) 9e 3x cos(3y) =. 3

4 2. A vector field F = f(x), g(y), is plotted on the xy-plane below. Determine the sign of the flux across the parametrized surface r(u, v) = 2 cos v, 2 sin v, u, u 1, π/2 v π/2. Solution. We first determine the shape of the surface and the direction of the normals for the surface. We note that the parametrization describes a cylinder centered at the z-axis, and the bounds truncate the cylinder between the planes z = and z = 1, and on the positive side of the xz-plane. We calculate the normal vectors as i j k r u r v = 1 = 2 cos v, 2 sin v,. 2 sin v 2 cos v By testing v =, we see that the normals are pointing into the cylinder. We now compare the direction of the field vectors with the direction of the normal vectors which are pointing towards the origin. For most of the semicircular crosssection, we note that the field vectors are pointing roughly in the opposite direction to the normal vectors. In particular, the vectors in the first quadrant and surrounding the x-axis do this. The remaining vectors in the fourth quadrant appear to be generally pointing perpendicularly to the normal vectors, so we do not expect these to contribute much to the flux. Thus, our observations suggest that the flux should be negative. 4

5 3. The vector field F is plotted below. (a) Is the divergence of F positive, negative, or zero at (, 1.5) and ( 2, 1)? (b) Is the curl of F positive, negative, or zero at (1, 1)? Solution. Intuitively, the divergence of a vector field is positive where the vectors spread, and negative where the vectors converge. At (, 1.5) the vectors are spreading, so the divergence appears to be positive, and at ( 2, 1) the vectors are converging, so the divergence appears to be positive. We can also reason by studying the partial derivative expression div F = P x + Q y. Traveling through the point (, 1.5) in the positive x-direction, the x-component of the vectors are increasing, so P x is positive. Through the point (, 1.5) in the positive y-direction, the y-component of the vectors also are increasing, so P y is positive. Thus, their sum, div F, must be positive. Likewise, we conclude that div F at ( 2, 1) must be negative. The curl of a vector field is positive where vectors around a point would cause a tiny windmill to spin counterclockwise, and negative if the spin is clockwise. Since the vector nortwest of the point (1, 1) appears to be shorter than the vector southeast of that point, a tiny windmill would be coerced to spin counterclockwise. Thus, the curl is positive. Using the partial derivative expression curl F = Q x P y, we could argue that Q x looks very small, possibly zero, while P y appears to be negative, so curl F would be positive. 5

6 4. Use change of variables to evaluate D (x 2y) dx dy where D is the square with vertices (, ), (3, 1), (4, 2), and (1, 3). Solution 1. We first determine the boundary of the region D. This is a square region bounded by the parallel lines y = 3x, y = 3x 1 and parallel lines y = (1/3)x, y = (1/3)x + (1/3). We use this to determine a transformation from the uv plane to the xy plane. We may rearrange the first pair of equations as y 3x =, y 3x = 1 and the second pair of equations as y + (1/3)x =, y + (1/3)x = 1/3 so that by choosing u = y 3x, v = y + (1/3)x, the bounds 1 u, v 1/3 describe the region D. We next express x and y in terms of u and v: x = (3/1)(v u), y = (9v + u)/1. We write the transformation as T (u, v) = ((3/1)(v u), (9v + u)/1). To execute the change of variables, we need to calculate the Jacobian J(u, v): J(u, v) = 3/1 1/1 3/1 9/1 = 3 1. Since the Jacobian is constant, we know that the transformation scales areas linearly. Thus, we can check our calculations by comparing the areas of the two regions. on the xy plane, the square region has area 1, while on the uv plane, the rectangular region has area 1/3. Reassuringly, the areas scale by a factor of 3/1 from the uv plane to the xy plane. Now we are prepared to set up and evaluate the integral: D (x 2y) dx dy = 1/3 1 = 3 2 = 3 2 = 1 2 = ( 3 1 (v u) 2 ) 3 (9v + u) dv du 1 1 1/3 [ 3v 2 (3v + u) dv du 2 + uv (5 + u) du [5u + u2 2 ] 1/3 Solution 2. Alternatively, we could pick the rectangle S in the uv-plane with vertices (, ), (1, ), (, 1), (1, 1) and find a transformation that sends it to D. The two sides of S can be described by the vectors 1, and, 1. So we can pick T to be a linear map that sends i = 1, to 3, 1 and j =, 1 to 1, 3. So T (ui) = 3u, u and 6 ] 1 =. du

7 T (vj) = v, 3v. In other words T (u, v) = 3u + v, u + 3v. Now we can set up an integral as above, with u 1 and 1 v 1. Note that this strategy works in an even more general setting. If D 2 is D shifted a units to the right and b units up, i.e. D 2 has vertices (a, b), (3 + a, b 1), (4 + a, 2 + b), and (1 + a, 3 + b), then we just need to add a and b to each of the components of T, i.e. T (u, v) = 3u + v + a, u + 3v + b. (You could double check, if you like, that the vertices of S are mapped to the vertices of D 2.) 5. Let F be the vector field F = sin x + y 2, x 2 y and let C be the triangle with vertices (, ), (2, ), and (, 3), oriented counterclockwise. Compute the line integral C F dr. Solution. We first apply Green s theorem to reduce the integration work: C F dr = = = x/ x/2 2 x (x2 y) y ( sin x + y 2 ) dy dx 2(x 1)y dy dx (x 1)(x 2) 2 dx = An object moves according to parametric equations x = cos 2 t, y = 2t 2, z = sin 2 t. Find the following, as functions of time. (a) The position vector r(t). (b) the velocity vector v(t). (c) The acceleration vector a(t). (d) The speed of the object. (e) The tangential component of acceleration. (f) The normal component of acceleration. (g) The point on the curve which is half the distance along the curve between the points r() and r ( π 2 ). Solution. (a) r(t) = cos 2 t, 2t 2, sin 2 t. (b) v(t) = 2 cos t sin t, 4t, 2 sin t cos t = sin(2t), 4t, sin(2t). (c) a(t) = 2 cos(2t), 4, 2 cos(2t). (d) v(t) = 2 sin 2 (2t) + 16t 2. (e) comp v(t) a(t) = v(t) a(t) = 2 sin(4t)+16t v(t) 2 sin 2 (2t)+16t 2. 7

8 (f) v(t) a(t) v(t) = 128t2 cos 2 (2t) 32t sin(4t)+64 sin 2 (2t) 2 sin 2 (2t)+16t 2. (g) The arc length function is s(t) = t v(θ) dθ. This is hard to integrate, so we only describe the idea one needs to find t such that s(t) = s(π/2)/2. 7. Let f(x, y) = 1 2 x y2 ln(xy) where the domain of f is {(x, y) x >, y > }. (a) Find all critical points of f. (b) Classify each critical point of f as either a local maximum, local minimum, or saddle point. Solution. To find critical points, we identify points such that f x = and f y =. Since f x (x, y) = x 1/x and f y (x, y) = y 1/y, the only point in the domain satisfying the condition is (1, 1). To classify this point, we calculate the discriminant: D(x, y) = f xx f yy (f xy ) 2 = (1 + 1x ) (1 + 1y ), 2 2 which is always positive. We also note that f xx >. Thus, (1, 1) is a local minimum. 8. Let S be the surface z = sin x cos y, x π, π/2 y π/2, oriented upward. (a) Parametrize the surface as r(u, v). (b) Find a parametrization for S (the surface with opposite orientation). (c) Set up the surface area integral for S. (d) Set up the surface integral of f(x, y, z) = xyz over S. (e) Set up the surface integral of F (x, y, z) = x, y, z over S. Solution. (a) r(u, v) = u, v, sin u cos v. (b) r 1 (u, v) = v, u, sin v cos u. (c) (d) (e) π π/2 π/2 π π/2 π/2 π π/2 π/2 r u r v dv du = π π/2 π/2 xyz cos u cos 2 v + sin 2 u sin 2 v + 1 dv du. x, y, z (r u r v ) dv du = π π/2 π/2 cos u cos 2 v + sin 2 u sin 2 v + 1 dv du. ( u cos u sin v + v sin u sin v + sin u cos v) dv du. 8

9 9. Let S be the torus parametrized by r(u, v) = (a + b cos v) cos u, (a + b cos v) sin u, b sin v where u 2π, v 2π and < b < a. Use the parametrization to find the surface area of S. Solution. We calculate r u r v = b(a + b cos v), so the surface area is 2π 2π b(a + b cos v) dv du = (2πa)(2πb). 1. Let C be the positively oriented boundary of the surface S defined by z = 2 x 2 y 2, where x 1, y, and z. Let F be the vector field F = xyz,, xyz. (a) Directly compute the line integral C F dr. (b) Use Stokes Theorem to find curlf ds, where the surface is oriented so that S the normals have a positive dot product with k. (c) Using your answer from part (b), show that F is not conservative. Solution. We assume the surface has outward pointing normals. The boundary is composed of four piecewise curves, which we parametrize so that the orientation of the whole curve is positive: C 1 : r 1 (t) = t,, 2 t 2, t 1, C 2 : r 2 (t) = 1, t, 1 t 2, t 1, C 3 : r 3 (t) = 2 cos t, 2 sin t,, π/4 t π/2, C 4 : r 4 (t) =, 2 t, 2 ( 2 t) 2 t 2. We now calculate each line integral. Since F = on C 1, C 3, and C 4, it remains to calculate the line integral 1 F dr = t(1 t 2 ),, t(1 t 2 ), 1, 2t dt C 2 = 2 1 [ t 3 = 2 3 t5 5 = 4/15. t 2 (1 t 2 ) dt ] 1 By Stokes Theorem, curlf ds = F dr = 4/15. If F were conservative, then S C it would be the gradient of a potential field. However, the curl of a gradient is always zero, so the flux would be zero. However, we have found it to be nonzero, so F cannot be conservative. 9

10 11. Use the Divergence Theorem to compute the surface integral F ds, where F = S x, y 2, z 3 and S is the boundary of the pyramid with vertices (,, ), (1,, ), (, 1, ), and (,, 1). Solution. By the Divergence Theorem, the surface integral is equal to F dv. We can set up this integral as Have fun integrating this. 1 1 y 1 x y E (1 + 2y + 3z 2 ) dz dy dx. 1