Fall 14 MTH 34 FINAL EXAM December 8, 14 Name: PID: Section: Instructor: DO NOT WRITE BELOW THIS LINE. Go to the next page. Page Problem Score Max Score 1 5 5 1 3 5 4 5 5 5 6 5 7 5 8 5 9 5 1 5 11 1 3 1a 8 1b 4 4 13 16 14a 4 5 14b 1 14c 6 6 16a 6 7 16b 6 17 1 8 9 15a 18 8 1 1 15b 19 1 8 1 16 Total Score
Fall 14 MTH 34 FINAL EXAM December 8, 14 Name: PID: Section: Instructor: READ THE FOLLOWING INSTRUCTIONS. Do not open your exam until told to do so. No calculators, cell phones or any other electronic devices can be used on this exam. Clear your desk of everything excepts pens, pencils and erasers. If you need scratch paper, use the back of the previous page. Without fully opening the exam, check that you have pages 1 through 1. Fill in your name, etc. on the first page and on this page. Show all your work. 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. You will be given exactly 1 minutes for this exam. I have read and understand the above instructions:. SIGNATURE SCORE:
Multiple Choice. Circle the best answer. No work needed. No partial credit available. 1. A parametric equation for the line through (1,, 3) and perpendicular to the plane z x y 5 is given by: (a) r(t) t 1, t 1, 3t + 1 (b) r(t) 1 t, t, t + 3 (c) r(t) t 1, t 1, 3t + 1 (d) r(t) 1 t, t, 3 + t Answer (e) None of the above. Suppose x + 3xy 7. Which of the following is true? (a) dy dx 3x x + 3y (b) dy x + 3y dx 3x (c) dy dx 3y x + 3y (d) dy + 3y x Answer dx 3x (e) None of the above 3. Let f x + 3xy where x v + sin u and y u + sin v. Which of the following is true? (a) f (3u v + 5 sin v)(cos u) + ( 3v + 3 sin u) u (b) f (3u v + sin u + 3 sin v)(cos u) + ( 3v + 3 sin u) Answer u (c) f ( 3u + v sin u 3 sin v) + ( 3v + 3 sin u)(cos v) u (d) f ( v + sin u) + 3(u + sin v) u (e) None of the above 4. The directional derivative of f(x, y) x (y 1) y at the point (, ) in the direction of the vector i j is given by: (a) x(y 1), x 4y 1, 5 (b) x(y 1), x 4y, 4 1, (c) 4, 4 Answer 5, (d) 6, 9 4 (e) None of the above 1
Fill in the Blanks. No work needed. Only possible scores given are, 3, and 5. 4x y 5. Evaluate the limit if it exists. Write DNE if the limit does not exist. lim (x,y) (1,) x y 4 4x + y 6. Evaluate the limit if it exists. Write DNE if the limit does not exist. lim (x,y) (,) x y DNE 7. If dr dt (3t + 1)i + (4t 3 + 1)j k and r(1) 3i + k then r(t) t 3 + t 5, t 4 + t, t + 3 8. The integral 1 4t + e t + (π cos(πt)) dt expresses the length of the curve r(t) 1 + t, e t, sin(πt) between the points (1, 1, ) and (, e, ). (Do not evaluate) 9. 4(x ) + 4(y ) 1(z + 4) is an equation of the tangent plane to the surface z x (y 1) y at the point (,, 4). 1. If 4 x 3 f(x, y, z) dz dy dx 3 a b f(x, y, z) dx dy dz then a and b 4 y. Extra Work Space.
Standard Response Questions. Show all work to receive credit. Please put your final answer in the BOX. 11. (1 points) Find where the lines L 1 (t) 4 + t, 5 + t, 6 + t and L (s) 3 s, 6 s, 1 + s intersect Solve the system of equations: 5 + t 6 s 4 + t 3 s To give us s and t 3. Plugging t 3 into L 1 (t) we get L 1 ( 3) 4 3, 5 3, 6 3 1,, 3 Giving us x 1, y, and z 3. 1. (8+41 points) Given the points A(1, 1, 1), B(, 1, ), and C(,, 3). (a) Find an equation of a plane that through the points A, B, and C. Consider the vectors AB 1,, 1 and AC 1, 1,. Then n AB AC i j k 1 1 1 1 ( + 1)i ( 1)j + (1 )k 1, 1, 1 So we get (x 1) (y 1) + (z 1) z 1 x + y (b) Find the area of triangle ABC. Using the work we did in (a) we know Area of triangle ABC 1 1, 1, 1 1 3 3 3
13. (16 points) Find and classify (max, min or saddle point) the critical points of f(x, y) 1 + 6x + 6y 3x y y 3. (Hint: There are four.) Taking partial derivatives we get f x 1x 6xy 6x( y) and f y 1y 3x 3y 3(y 4y+x ). Since these are polynomials they are never undefined so we get critical points when 6x( y) and y 4y+x giving us the critical points (, ), (, 4), (, ), (, ). Now we calculate second derivatives to classify them: So we get: f x x y or y 4 or y x or x f xx 1 6y f yy 1 6y f xy 6x D(, ) (1)(1) () 144 > f xx (, ) 1 (, ) is a local min D(, 4) ( 1)( 1) () 144 > f xx (, 4) 1 (, 4) is a local max D(, ) ()() ( 1) 144 < (, ) is a saddle pt D(, ) ()() (1) 144 < (, ) is a saddle pt 4
14. (4+1+6 points) Let F x + y + yz, x + xz, xy + 1. (a) Show that curl(f). i j k x y z x + y + yz x + xz xy + 1 (x x)i (y y)j + ((1 + z) (1 + z))k,, (b) Find a potential function f such that f F. f x dx x + y + yz dx f x + xy + xyz + g(y, z) (integrate ( )) f y x + xz + g y (y, z) (take partial derivative) x + xz x + xz + g y (y, z) (Plug in Q for f y ) g y (y, z) (algebra) g(z) g(y, z) (integrate) f x + xy + xyz + g(z) (substitute into ( )) f z xy + g z (z) (take partial derivative) xy + 1 xy + g z (z) (Plug in R for f z ) 1 g z (z) (Algebra) z + K g(z) (integrate) f x + xy + xyz + z + K (substitute into ( )) (c) Evaluate F dr where C is a curve from the point (,, ) to the point (, 1, 1). C F dr f(, 1, 1) f(,, ) (4 + + + 1) () 9. C 5
15. (8+816 points) Evaluate the integrals (a) 1 1 x 4e (x/y) dy dx. This can not be so easily integrated. Lets switch the limits of integration. Sketch a mini picture (, ) (1, 1) 1 1 x 4e (x/y) dy dx 1 y 4e (x/y) dx dy 1 [4ye (x/y)] y 1 1 dy [ 4ye 1 4y ] dy 4y(e 1) dy [ y (e 1) ] 1 (e 1) (b) 3 9 x cos(x + y ) dy dx. Polar coordinates for the win. Sketch a mini picture (, ) (3, ) 3 9 x cos(x + y ) dy dx 3 π π 3 cos(r ) r dr dθ r cos(r ) dr [ sin(r ) ] 3 π sin(9) 6
16. (6+61 points) (a) Express (x ) + y 4 in terms of polar coordinates. Solve for r. Simplify as much as possible. (r cos θ ) + (r sin θ) 4 r cos θ 4r cos θ + 4 + r sin θ 4 r 4r cos θ r(r 4 cos θ) r 4 cos θ (r is only a dot, not a circle. Extraneous.) (b) Express (DO NOT EVALUATE) a triple integral in cylindrical coordinates for the volume of the portion of the sphere x + y + z 16 contained within the cylinder (x ) + y 4. The surface we enter our region of integration through is z 16 x y 16 r and we exit through z 16 x y [ 16 r. Finally we should note that a possible domain of r 4 cos θ to cover the circle once is π, π ]. So we get. E 1 dv r dz dr dθ 4 cos θ 16 r π/ 16 r r dz dr dθ 17. (1 points) Evaluate the integral Sketch a mini picture in D then 3D (3, ) (, ) (,, ) x z (3,, ) These should suffice to give us: y 3 3 9 y 9 x y (4z) dz dx dy in spherical coordinates. 9 y 9 x y (4z) dz dx dy (4ρ cos φ)ρ sin φ dρ dφ dθ π π/ 3 4ρ 3 sin φ cos φ dρ dφ dθ [ ρ 4 sin φ cos φ ] 3 dφ π [81 sin φ cos φ] dφ π [ 81 sin φ ] π/ 4 π 4 [81] 7
18. (1 points) Find the work done by the field F yi xj + z k over the path r(t) cos ti + sin tj + tk, t π. y, x, z dr y, x, z sin t, cos t, 1 dt [ t 3 3 t sin t, cos t, t sin t, cos t, 1 dt sin t cos t + t dt t 1 dt ] π/ π3 4 π 19. (1 points) Use Green s Theorem to evaluate the line integral and C is the positively oriented triangle shown below. C F dr where F 6y + sin x 1+x, 3y 9 cos(y ) 4x 4 y (, ) C > > > (, ) x C F dr 6y + sin x C 1 + x 4, 3y9 cos(y ) 4x dr ( 4 6) da D 1 da D [ ] 1 1 ()() 8
. (1 points) Find the surface area of the paraboloid z 5 x y that lies above the plane z 13. These words give us f(x, y) 5 x y and 5 x y 13 9 x + y Giving us D {(x, y) 9 x + y }. Using the surface area formula we get: 1 + [f x ] + [f y ] da 1 + 16x + 16y da D D π 3 π 4π 96 3 1 + 16r r dr dθ 3r 3 1 + 16r dr [ π 3(3) (1 + 16r ) 3/ [ ] (1 + 16(9)) 3/ 1 ] 3 π 4 [ (145) 3/ 1 ] 1. (16 points). Let F (xy z)i + (1x + yz )j + (zx sin x)k and let S be the sphere x + y + z 4. Use the Divergence Theorem to find evaluate F ds. S S F ds E E div F dv (y + z + x ) dv (ρ )ρ sin φ dρ dφ dθ E π π π π [ ρ 5 (ρ )ρ sin φ dρ dφ dθ 5 sin φ ] dφ 64π 5 [ cos φ]π 18π 5 9