Math Exam IV - Fall 2011

Transcription

1 Math Exam IV - Fall 2011 December 15, Renato Feres NAME: STUDENT ID NUMBER: General instructions: This exam has 16 questions, each worth the same amount. Check that no pages are missing and notify your proctor if you detect any problems with your copy of the exam. Mark your ID number on the six blank lines on the top of your answer card, using one line for each digit. Print your name on the top of the card. Choose the answer that is closest to the solution and mark your answer card with a PENCIL by shading in the correct box. You may use a 3 5 card with notes and any calculator that does not have graphing functions. GOOD LUCK! 1. Find the point on the plane x + 2y + 3z = 13 closest to the point (1, 1, 1) (A) (3/2, 2, 5/2) (B) (10, 0, 1) (C) (11, 1, 0) (D) (3/2, 5, 1/2) (E) (5, 5/2, 1) (F) (2, 5/2, 2) (G) (6, 2, 1) (H) (3, 2, 2) (I) (2, 1, 3) (J) (4, 3/2, 2) 1

2 Solution: Set f(x, y, z) = (x 1) 2 +(y 1) 2 +(z 1) 2 and g(x, y, z) = x+2y+3z. Then at point of minimum of f under the constraint g = 13 we must have f = λ g. Therefore, 2(x 1), 2(y 1), 2(z 1) = λ 1, 2, 3. Thus It follows that 2(x 1) = y 1, 3(x 1) = z 1. y = 2x 1, z = 3x 2. Using the constraint equation g(x, y, z) = 13 we find 13 = x + 2(2x 1) + 3(3x 2) = 14x 8 or x = 3/2. Therefore, y = 2 and z = 5/2. So the solution is P = (3/2, 2/5/2). 2

3 2. Evaluate the spherical coordinates integral 2π π/ sec φ 3ρ 2 sin φ dρ dφ dθ (A) 0 (B) π (C) 2π (D) 3π (E) 4π (F) 5π (G) 6π (H) 7π (I) 8π (J) 9π 3

4 Solution: Let I stand for the value of the integral. Then, integrate first in θ, we obtain π/3 2 π/3 ( I = 2π 3ρ 2 dρ dφ = 2π 8 cos 3 φ ) sin φ dφ. 0 sec φ 0 Using the u-substitution u = cosφ, so that du = sin φ dφ, we obtain 1 ( I = 2π ) 8 u 3 du = 5π. 1/2 4

5 3. Find the average value of the function f(ρ, φ, θ) = ρ over the solid ball ρ (A) π/2 (B) 3/4 (C) 3/5 (D) π/4 (E) 5/2 (F) 3/2 (G) 3 (H) 4 (I) 2π (J) π 5

6 Solution: Recall that the volume of the ball of radius 1 is V = 4π. Then the 3 average value is I = 1 V 2π π ρ 3 sin φ dρ dφ dθ = 3 2π π dθ sin φ dφ 4π ρ 3 dρ =

7 4. Find the Jacobian (x, y)/ (u, v) of the transformation x = u cosv, y = u sinv (A) 1 (B) u 2 (C) u (D) v (E) cos v (F) sin v (G) u(cosv + sin v) (H) u( sin v + cos v) (I) 1/u (J) 1/u 2 7

8 Solution: The Jacobian determinant is x u x v = cosv u sin v sin v u cosv y u y v = u cos2 v + u sin 2 v = u. 8

9 5. By using the transformation x = u/v, y = uv the integral 2 y ( x 2 + y 2) dxdy + 4 4/y ( x 2 + y 2) dxdy 1 1/y 2 y/4 can be written as b d a c ( ) 2u 3 v + 3 2u3 v du dv. Then the limits of integration a, b, c, d are, respectively, (A) (0, 1, 0, 1) (B) (1, 2, 3, 4) (C) (0, 1, 0, 2) (D) (2, 3, 2, 3) (E) (1, 4, 1, 4) (F) (0, 2, 1, 2) (G) (1, 2, 1, 2) (H) (3, 4, 2, 4) (I) (1, 3, 1, 3) (J) (0, 3, 0, 3) 9

10 Solution: The union of the regions of integration of the two integrals is the region in the xy-plane bounded by the curves: x = y/4, x = y, x = 1/y, x = 4/y. Substituting x = u/v, y = uv gives, respectively, v 2 = 4, v 2 = 1, u 2 = 1, u 2 = 4. To obtain a one-to-one correspondence between the (u, v) and the (x, y) we take the positive solutions. Thus 1 u 2, 1 v 2. 10

11 6. the parametric equations of a line segment in the xy-plane and a segment of parabola are among the following: (a) r(t) = ti + (1 t)j, 0 t 1 (b) r(t) = (2 cost)i + (2 sint)j, 0 t 2π (c) r(t) = (t 2 1)i + 2tk, 1 t 1 (d) r(t) = ti + tj + tk, 0 t 2 Which parametric equations represent the line segment and the parabola? 15.1, 1, 3, 5, 7 (A) line (c), parabola (d) (B) line (c), parabola (a) (C) line (b), parabola (d) (D) line (a), parabola (d) (E) line (a), parabola (b) (F) line (c), parabola (b) (G) line (b), parabola (a) (H) line (b), parabola (c) (I) line (d), parabola (a) (J) line (a), parabola (c) 11

12 Solution: This is a simple inspection. 12

13 7. Find the line integral F dr of the vector field C F = 3yi + 2xj + 4zk along the straight path C with parametric equation r(t) = ti + tj + tk, 0 t , 7 (A) 3/2 (B) 13/3 (C) 5/2 (D) 1 (E) 7/3 (F) 2 (G) 3 (H) 9/2 (I) 4 (J) 1/2 13

14 Solution: Notice that r (t) = 1, 1, 1. So F dr = 1 3t, 2t, 4t 1, 1, 1 dt = 1 C 0 0 9t dt =

15 8. Find the line integral ( x 2 + y 2) dy C where C is the curve from (0, 0) to (1, 1) shown in the figure. 15.2, 15 (A) 5/2 (B) 3/2 (C) 1/3 (D) 2/3 (E) 4/3 (F) 1 (G) 2 (H) 3 (I) 1/2 (J) 7/3 15

16 Solution: We write C (x2 + y 2 ) dy = C 1 (x 2 + y 2 )dy + C 2 (x 2 + y 2 )dy, where C 1 is the horizontal segment and C 2 the vertical segment of C. On C 1, y is constant so the corresponding integral vanishes. On C 2 the integral is ( x 2 + y 2) 1 ( dy = ) 1 + y 2 dy = C 2 3 =

17 9. Find the work done by the force F = xyi + (y x)j over the straight line from (1, 1) to (2, 3). 15.2, 27 (A) 25/6 (B) 2/3 (C) 13/4 (D) 2/5 (E) 27/6 (F) 4 (G) π (H) 32 (I) 2/7 (J) 0 17

18 Solution: A parametrization of the line is r(t) = 1 + t, 1 + 2t, for 0 t 1. Then the work is given by C F dr = 1 0 (1 + t)(1 + 2t), t 1, 2 dt = 1 0 ( 1 + 5t + 2t 2 ) dt =

19 10. Which of the following vector fields in the plane, if any, are conservative? (a) F = xi + yj (b) F = yi + xj (c) F = yi xj (d) F = yi Indicating by C if conservative and by N if not, these fields are, respectively, 15.3 (A) C, C, C, C (B) C, C, C, N (C) C, C, N, N (D) C, N, N, N (E) N, N, N, N (F) N, N, N, C (G) N, N, C, C (H) N, C, C, C (I) N, C, C, N (J) N, C, N, C 19

20 Solution: In general, if F = Mi+Nj, and M, N are defined over the entire plane (a simply connected region), then the field is conservative if and only if its curl is zero: N x M y = 0. Therefore, (a) N x M y = 0, hence conservative (b) N x M y = 0, hence conservative (c) N x M y = 2, hence not conservative (d) N x M y = 1, hence not conservative 20

21 11. Evaluate the integral (1,2,3) (0,0,0) 2xy dx + ( x 2 z 2) dy 2yz dz where the integrand is an exact differential form. 15.3, 15 (A) 0 (B) 1 (C) 2 (D) 4 (E) 8 (F) 16 (G) 16 (H) 8 (I) 4 (J) 2 21

22 Solution: Let f(x, y, z) denote the potential. Then f x = 2xy, f y = x 2 z 2, and f z = 2yz. The antiderivative in y of f y is f(x, y, z) = x 2 y yz 2 + g(x, z). Differentiating this expression in x gives: 2xy = f x = 2xy + g x (x, z). So g x (x, z) = 0, hence g(x, z) = h(z), a function of z only. Now, differentiating f(x, y, z) = x 2 y yz 2 + h(z) in z gives: Therefore, h(z) is a constant, and Therefore, the value of the integral is 2yz = f z = 2yz + h z (z). f(x, y, z) = x 2 y yz 2 + C. f(1, 2, 3) f(0, 0, 0) =

23 12. Evaluate the integral 4x 3 y dx + x 4 dy C where C is a closed path in the xy-plane. 15.4, 31 (A) 4π (B) 3π (C) 2π (D) π (E) 0 (F) π (G) 2π (H) 3π (I) 4π (J) 1 23

24 Solution: Let M = 4x 3 y, N = x 4. These are smooth functions over the entire coordinate plane. Notice that N x M y = 0, so this is the integral of an exact differential. Therefore, 4x 3 y dx + x 4 dy = 0 for an arbitrary closed curve C. C 24

25 13. Suppose that a simple closed curve C in the plane encloses a region R (so that C is the boundary of R). If xdy y dx = 8 what is the area of R? (A) 1/16 (B) 1/8 (C) 1/4 (D) 1/2 (E) 1 (F) 2 (G) 4 (H) 8 (I) 16 (J) 32 C 25

26 Solution: It is an immediate consequence of Green s theorem that 1 xdy y dx = Area of R. 2 Therefore, the area of R is 4. C 26

27 14. Find the area of the surface cut from the paraboloid x 2 + y 2 z = 0 by the plane z = , 37 (A) 13π (B) 13π/3 (C) 13π/6 (D) 26π/3 (E) π/3 (F) 26π (G) 3π/2 (H) 6π (I) 5π (J) π 27

28 Solution: We wish to find the area of the surface z = x 2 +y 2 that lies above the disc in the xy-plane of radius 2 centered at the origin. A possible choice of parametrization is r(x, y) = x, y, x 2 + y 2. Note that so the surface area element is i j k r x r y = 1 0 2x = 2x, 2y, y dσ = r x r y dxdy = (x 2 + y 2 ) dxdy. Therefore, the area we want is the area integral over the disc of radius 2 of 1 + 4r2. In polar coordinates Area = 2π 2 A u-substitution u = 1 + 4r 2 gives r2 r dr dθ = 2π 1 + 4r2 r dr. Area = π udu = 13π 3. 28

29 15. Find the area of the ellipse cut from the plane z = 2x by the cylinder x 2 +y 2 = , 43 (A) 2π (B) 3π (C) 2π (D) 5π (E) 6π (F) 7π (G) 2 2π (H) 3π (I) 10π (J) 11π 29

30 Solution: We use the parametrization r(x, y) = x, y, 2x. Then i j k r x r y = = 2, 0, and dσ = r x r y dxdy = 5dA. Integrated over the disc of radius 1 gives 5π for the area of the ellipse. 30

31 16. Find the flux F n dσ of the vector field F = zk across the portion of the S sphere x 2 +y 2 +z 2 = a 2 in the first octant in the direction away from the origin. 15.6, 21 (A) a 3 (B) 6a 3 π (C) 2a 2 π (D) aπ/6 (E) 2π (F) aπ/2 (G) a 2 π/3 (H) a 3 π/6 (I) a 2 (J) 2a 31

32 Solution: The normal vector field is n = a 1 x, y, z, so F k = z 2 /a = a cos 2 φ. Recall that the area element on the sphere of radius a takes the form Therefore, the flux is S F n dσ = dσ = a 2 sin φ dφ dθ. π/2 π/2 0 0 a 3 cos 2 φ sin φ dφ dθ = a3 π 6. 32