Without fully opening the exam, check that you have pages 1 through 10.

Transcription

1 MTH 234 Solutions to Exam 2 April 11th 216 Name: Section: Recitation Instructor: INSTRUTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages 1 through 1. Show all your work on the standard response questions. Write your answers clearly! Include enough steps for the grader to be able to follow your work. Don t skip limits or equal signs, etc. Include words to clarify your reasoning. Do first all of the problems you know how to do immediately. Do not spend too much time on any particular problem. Return to difficult problems later. If you have any questions please raise your hand and a proctor will come to you. You will be given exactly 9 minutes for this exam. AADEMI HONESTY Do not open the exam booklet until you are instructed to do so. Do not leave the exam room during the first 2 minutes of the exam. Do not seek or obtain any kind of help from anyone to answer questions on this exam. If you have questions, consult only the proctor(s). Books, notes, calculators, phones, or any other electronic devices are not allowed on the exam. Students should store them in their backpacks. No scratch paper is permitted. If you need more room use the back of a page. Anyone who violates these instructions will have committed an act of academic dishonesty. Penalties for academic dishonesty can be very severe. All cases of academic dishonesty will be reported immediately to the Dean of Undergraduate Studies and added to the student s academic record. I have read and understand the above instructions and statements regarding academic honesty:. SIGNATURE Page 1 of 1

2 MTH 234 Solutions to Exam 2 April 11th 216 Multiple hoice. ircle the best answer. No work needed. No partial credit available. 1. (6 points) Find the rate of change of f(x, y, z) 2x 3 y z at the point (1, 2, 4) and in the direction of v 1, 2, 2. A. 17/3 B. 13/3. 19/3 D. 6 E. 2/3 2. (6 points) The function f(x, y) 2x 3 + 6xy + 3y 2 has: A. one local minimum and one local maximum B. two saddle points. one local minimum and one saddle point D. one local maximum and one saddle point E. two local minima 3. (7 points) Find the absolute maximum of the function f(x, y) x 2 + 4y on the set of points (x, y) that satisfy x 2 + y A. 2 B. 29. D. 25 E. 35 Page 2 of 1

3 MTH 234 Solutions to Exam 2 April 11th (6 points) Let F ye x, x 2, cos z. Then curl(f) is equal to: A. ye x,, sin z B. ye x sin z.,, D., 2x, 2x e x E.,, 2x e x 5. (6 points) The directional derivative of f(x, y) x 3 e 2y in the direction of greatest increase of f at the point (1, ) is: A. 3i B. 3i 2j. 3 D. 5 E (6 points) If we change the order of integration in 6 2 x (x+3y) dz dy dx and integrate first in x, then in y and then in z the integral is given by A. B.. D. E z z 2 3z z 2 z (2z+3y) 6 2z 3y 6 2z y 6 2z y 3 6 z 2 3 y dx dy dz dx dy dz dx dy dz dx dy dz dx dy dz Page 3 of 1

4 MTH 234 Solutions to Exam 2 April 11th (7 points) Evaluate y 2 da where D is the triangle with vertices (, ), (1, 1), and (, 1). A. 1/3 B. 1/6. 1/2 D. 2/3 E. 1/4 D 8. (6 points) Let be the helix 4 cos t, 4 sin t, 3t with t 2π. Evaluate z ds. A. 1π 2 B. 8π 2. 3π 2 D. 15π 2 E. π 2 Page 4 of 1

5 MTH 234 Solutions to Exam 2 April 11th 216 Standard Response Questions. Show all work to receive credit. Please BOX your final answer. 9. (2 points) Use integration in polar coordinates to compute the area of the region in the first quadrant inside the circle (x 1) 2 + y 2 1 and below the line y x. Recall: 2 cos 2 θ 1 + cos(2θ). Solution: onsider the graph of (x 1) 2 + y 2 1 shown below. y y x Putting this into polar coordinates we see it has the equation: (1, ) x (x 1) 2 + y 2 1 x 2 2x y 2 1 r 2 2r cos(θ) r 2 cos(θ) Now we can find the area D da r dr dθ π/4 2 cos(θ) π/4 π/4 π/4 r dr dθ [ r 2 /2 ] 2 cos(θ) dθ [ 2 cos 2 (θ) ] dθ [1 + cos(2θ)] dθ [θ + 12 ] π/4 sin(2θ) π/4 + 1/2 Page 5 of 1

6 MTH 234 Solutions to Exam 2 April 11th (2 points) Evaluate 16z dv E where E is the upper half of the sphere x 2 + y 2 + z 2 1. Solution: onsider switching this into spherical coordinates. 16z dv 16ρ cos φ(ρ 2 sin φ) dρ dφ dθ E 2π π/2 1 π/2 1 π/2 π/2 [ 4π sin 2 φ ] π/2 4π [ 1 2 2] 4π 16ρ 3 sin φ cos φ dρ dφ dθ 32πρ 3 sin φ cos φ dρ dφ [ 8πρ 4 sin φ cos φ ] 1 dφ [8π sin φ cos φ] dφ Page 6 of 1

7 MTH 234 Solutions to Exam 2 April 11th (2 points) Let S be a surface given by r(u, v) uv, v, u, where u 2 + v 2 1. What is the surface area of S? Solution: r u v,, 1 r v u, 1, So therefore: r u r v i j k v 1 u 1 ( 1)i ( u)j + (v )k 1, u, v and so we have r u r v 1 + u 2 + v 2 giving us the surface area of r u r v du dv 1 + u2 + v 2 du dv 1 + r2 (r) dr dθ 2π 1 2π r2 (r) dr dθ 1 + r2 (r) dr π [ 2/3(1 + r 2 ) 3/2] 1 2π [ (2) 3/2 (1) 3/2] 3 2π [ (2) 3/2 1 ] 3 Page 7 of 1

8 MTH 234 Solutions to Exam 2 April 11th (2 points) Evaluate y 3 dx x 3 dy where is the positively oriented circle of radius 2 centered at the origin. Solution: Recall that Green s theorem states that P dx + Q dy D (Q x P y ) da for positively oriented closed curve (like we have). So therefore y 3 dx x 3 dy 3x 2 3y 2 da D 3(x 2 + y 2 ) da D 3(r 2 )r dr dθ D 2π 2 2π 2π 2 [ 3 4 r4 3r 3 dr dθ 3r 3 dr ] 2 3π [16 ] 2 24π Page 8 of 1

9 MTH 234 Solutions to Exam 2 April 11th (2 points) Evaluate z 2 dx + x 2 dy + y 2 dz where is the line segment from (1,, ) to (4, 1, 2). Solution: can be parametrized by r(t) 1 + 3t, t, 2t for t [, 1] so therefore z 2 dx + x 2 dy + y 2 dz t 2 (3dt) + (1 + 3t) 2 (1dt) + t 2 (2dt) (12t 2 + (9t 2 + 6t + 1) + 2t 2 ) dt (23t 2 + 6t + 1) dt [ 23 3 t3 + 3t 2 + t ] 1 Page 9 of 1

10 MTH 234 Solutions to Exam 2 April 11th 216 ongratulations you are now done with the exam! Go back and check your solutions for accuracy and clarity. Make sure your final answers are BOXED. When you are completely happy with your work please bring your exam to the front to be handed in. Please have your MSU student ID ready so that is can be checked. DO NOT WRITE BELOW THIS LINE. Page Points Score Total: 15 Page 1 of 1