Math 241, Final Exam. 5/4/13.

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1 Math 41, Final Exam. 5/4/13. No notes, calculator, or text. There are points total. Partial credit may be given. ircle or otherwise clearly identify your final answer. Name: 1. (15 points): Find the equation of the plane that passes through the point (1,, 3) and contains the line given by parametric equations x = 3t, y = 1 + t, z = t. State your equation in the form ax + by + cz = d. Solution: The line lies in the plane and has direction vector v = 3, 1, 1. To get a second vector parallel to the plane, we use the given point and the point (, 1, ) to get u = 1, 1, 1. A normal vector to the plane, therefore, is i j k u v = = i( 1 1) j( 1 3)+ k(1 3) = i+4 j k = ( i j + k). With normal vector n = i j + k and the point (, 1, ), the plane equation is 1(x ) (y 1) + 1(z ) = x y + + z = x y + z =.. (15 points): Suppose that w = f(x, y, z), y = g(s, t), and z = h(t). Write down the form of the chain rule that you would use to compute w/ s and w/ t. Solution: We have w s = w y y s, w t = w y y t + w dz z dt. 3. (15 points): Find parametric equations for the line normal to the surface sin(xyz) = x + y + 3z at the point (, 1, ). Solution: Let F (x, y, z) = sin(xyz) x y 3z. Then the normal line has direction vector at (, 1, ) given by F (, 1, ) = yz cos(xyz) 1, xz cos(xyz), xy cos(xyz) 3 = 1,, 3 = 1,, 5 = 1,, 5. (,1,) For convenience, we use the parallel vector v = 1,, 5. The parametric equations are x = + t, y = 1 + t, z = 5t; t R.

2 4. (15 points): Let F (x, y, z) = xy + xz y + z. (a) (1 points): Find the directional derivative of F (x, y, z) at the point (1,, 1) in the direction of the vector v = 1, 1,. Solution: We first normalize v to get u = 1,1, = (1/ 6) 1, 1,. Next, we compute F (1,, 1) = y + z, x y, x + z =, 5, 4 1,1, (1,,1) We now have D u F (1,, 1) = F (1,, 1) u =, 5, 4 (1/ 6)(1, 1, ) = 13/ 6. (b) (5 points): Find the maximum rate of change of F (x, y, z) at the point (1,, 1). Solution: From (a), we have F (1,, 1) =, 5, 4. The maximum rate of change is, 5, 4 = = (15 points): The function f(x, y) = x y 8xy + y + 5 has three critical points. Find the critical points and determine whether each is a local maximum, local minimum, saddle point, or none of these. Solution: We find critical points by solving the system f x = 4xy 8y = 4(x )y =, f y = x 8x+y = (x 4x+y) =. The first equation gives y = or x =. When y =, the second equation becomes x 4x = x(x 4) =, which gives the two points (, ) and (4, ). When x =, the second equation gives y = 4, and hence the point (, 4). For the Second Derivative Test, we compute f xx = 4y, f yy =, and f xy = 4x 8. We now test the three points. (a) (, ). We compute D = 8 8 = 64 <. It follows that (, ) is a saddle point. (b) (4, ). We compute D = 8 8 = 64 <. It follows that (4, ) is a saddle point. (c) (, 4). We compute D = 16 = 3 >. Since f xx(, 4) = 16 >, we conclude that (, 4) is a local minimum.

3 6. (15 points): Use Lagrange multipliers to find the point(s) on the surface z = x 3y 1 closest to (,, ). (You are asked to minimize the distance from a point (x, y, z) on the surface to the point (,, ).) Solution: It suffices to minimize the square of the distance, f(x, y, z) = x +y +z subject to the constraint g(x, y, z) = x 3y z = 1. We compute f = x, y, z = x, y, z, g = x, 6y, 1. It follows that we must solve the system x = λx, y = 6λy, z = λ, x 3y z = 1. We first suppose that x. The first equation gives λ = 1/. Hence, we get z = 1/ from the third equation. The second equation then becomes y = 6(1/)y, so y =. To get x, we use the fourth equation: 1 = x 3y z = x ( 1/) = x + 1/. It follows that x = ±1/. We obtain the points (±1,, 1/) when x. When x =, we consider the possibilities y = and y. If y, the second equation implies that λ = 1/6. Therefore, the third equation gives z = 1/6. To get y, we substitute in the fourth equation: 1 = x 3y z = 3y 1/6 <. We conclude that it is not possible for y, so we must have y =. When x = and y =, the fourth equation gives z = 1. We obtain a third point, (,, 1). It remains to evaluate f(x, y, z) at the three points: f(±1/,, 1/) = = 3 ; f(,, 1) = 1. 4 We conclude that the points (±1/,, 1/) are closest to (,, ) on the surface g(x, y, z) = 1, at a distance of 3/4 = 3/. 3

4 7. (15 points): Evaluate the triple integral π 1 zx sin(xyz) dx dy dz. (The easiest way to do this integral is to arrange so that you do not have to use integration by parts.) Solution: We compute π 1 π 1 = = = π zx sin(xyz) dx dy dz = ( ) ] zx cos(xyz) dx dz = xz ( x sin(xz) ( = sin(πx) ( cos(π) = + π ) 4π π 1 π 1 1 π ( x cos(xz) + x) dx dz = ) ] π 1 + xz dx = x ) + πx dx = cos(πx) ( cos 4π + 4π ) = π. zx sin(xyz) dy dx dz ( x cos(xyz)) ] dx dz ( x cos(xz) + x) dz dx ) ] π + xz dx ( sin(xz) + πx ] 1 8. (15 points): onvert the double integral 6 6y y 6y y 1 + (x + y ) 1/ dx dy to polar coordinates. (onsider drawing a quick sketch of the region in the xy-plane.) Do not evaluate. Solution: In order to sketch the region, we study the x-limits. We have x = ± 6y y x = 6y y x + y 6y + 9 = 9 x + (y 3) = 9. Hence, the region is a disk of radius 3 centered at (, 3); it is tangent to the x-axis at the point (, ). It follows that θ π. For bounds on the polar variable r, we convert x + y = 6y to r = 6r sin θ to obtain r = 6 sin θ. To conclude, we have 6 6y y 6y y π 6 sin θ 1 + (x + y ) 1/ dx dy = r 1 + r dr dθ. 4

5 9. ( points): onsider the triple integral 4 x 4 y f(x, y, z) dz dy dx. (Draw a quick sketch of the three-dimensional region of integration.) (a) (1 points): onvert the triple integral to a triple integral in rectangular coordinates with dv = dx dz dy. Solution: The region has four sides; three are planes, and the fourth is curved. The planes are z = (xy-plane), x = (yz-plane), and y + z = 4. The fourth side has an edge on the z-axis ( z 4), an edge in the xy-plane (y = x), and an edge where the plane y + z = 4 meets the parabolic cylinder y = x. We have 4 y y dx dz dy. (b) (1 points): onvert the triple integral to a triple integral in rectangular coordinates with dv = dy dz dx. (What is the equation for the boundary of the projection on the xz-plane?) Solution: We note that the equation for the projection on the xz-plane arises from the intersection of y+z = 4 with y = x, which gives z = 4 x or x = (1/4)(z 4). This is a parabola in xz-plane with vertex at x =, z = 4 which opens up in the positive x-direction. We have 4 4 x 4 z x dy dz dx. 1. (15 points): Set up but do not evaluate a triple integral in spherical coordinates for the volume of the solid bounded above by the sphere x + y + z = 4 and below by the cone 3z = x + y with x, y, z. (Draw a quick sketch of the solid.) Solution: The solid is a cone (above the xy-plane) with a spherical cap. To compute bounds on the spherical variable φ, we see that 3z = x + y gives ±z 3 = x + y, so we have x + y tan φ = = ±z 3 = ± 3. Since the solid lies above the xy-plane, we have z z tan φ = 3 with θ π. Hence, we get φ = π/3. The integral giving the volume in spherical coordinates is then Volume = π/ π/3 ρ sin φ dρ dφ dθ. 5

6 ( ) x y 11. (15 points): Evaluate the double integral cos da, where R is the trapezoidal R x + y region in the first quadrant with vertices (1, ), (3, ), (, 1), and (, 3). (Use a suitable change of variables from (x, y) to (u, v) to evaluate a double integral over a region S in the uv-plane. To get limits on the new variables u and v, it may be useful to draw S; you can do this by plotting a few well-chosen points. What are the equations for the boundary of S?) Solution: We use the change of variables u(x, y) = x y, v(x, y) = x + y. Then we have J 1 = =, so J = 1/. We observe that points (x, y) in R map to points (u(x, y), v(x, y)) in S. Since R is a trapezoid and u(x, y), v(x, y) are linear, the region S must be a quadrilateral (4 sides, 4 vertices). The vertices of S are (u(, 1), v(, 1)) = ( 1, 1), (u(, 3), v(, 3)) = ( 3, 3), (u(1, ), v(, 1)) = (1, 1), (u(3, ), v(3, )) = (3, 3). Hence, S is a trapezoid with sides v = 1, v = 3 (parallel), and u = v, u = v. We now compute cos R 3 ( ) x y da = 1 x + y 3 v 1 v = 1 (v sin 1 (v sin( 1))) dv = 1 1 ( 3 = (sin 1) 1 ) = 4 sin 1. ( u ) cos du dv = 1 v v sin 1 dv = sin 1 1 ( ( u )) ] v v sin v 3 1 v dv v dv 6

7 1. Some exams had problem (a); some had problem (b). (a) 16.3, #15: Let F (x, y, z) = yz i + xz j + (xy + z) k, and let be the line segment from (1,, ) to (4, 6, 3). (b) 16.3, #18: Let F (x, y, z) = e y i + xe y j + (z + 1)e z k, and let be the curve given by r(t) = t i + t j + t 3 k, t 1. i. (1 points) Find a function f such that F = f. Solution: For (a), we seek f(x, y, z) with f/ x = yz, f/ y = xz, and f/ x = xy + z. First, we compute f(x, y, z) = f dx + g(y, z) = yz dx + x g(y, z) = xyz + g(y, z). Hence, we have f/ y = xz + g y (y, z) = xz. It follows that g y (y, z) =. Therefore, we obtain g(y, z) = g y (y, z) dy + h(z) = h(z), which gives f(x, y, z) = xyz + h(z). To conclude, we compute f/ z = xy + h (z) = xy + z. We obtain h(z) = h (z) dz = z dz = z + K We now have f(x, y, z) = xyz + z + K. For (b), we seek f(x, y, z) with f/ x = e y, f/ y = xe y, and f/ z = (z + 1)e z. First, we compute f(x, y, z) = f dx + g(y, z) = e y dx + g(y, z) = x xe y + g(y, z). Hence, we have f/ y = xe y + g y (y, z) = xe y. It follows that g y (y, z) =. Therefore, we obtain g(y, z) = g y (y, z) dy + h(z) = h(z), which gives f(x, y, z) = xe y + h(z). To conclude, we compute f/ z = h (z) = (z + 1)e z. Using integration by parts with u = z + 1 and dv = e z dz, we obtain h(z) = h (z) dz = (z+1)e z dz = (z+1)e z e z dz = (z+1)e z e z = ze z +K. We now have f(x, y, z) = xe y + ze z + K. ii. (5 points) Use part (a) to evaluate F d r along the given curve. Solution: For (a), the Fundamental Theorem of alculus for line integrals gives F d r = f d r = f(4, 6, 3) f(1,, ) = 81 4 = 77. For (b), the curve has initial point r() =,, and terminal point r(t) = 1, 1, 1. The Fundamental Theorem of alculus for line integrals gives F d r = f d r = f(1, 1, 1) f(,, ) = e. 7

8 , #1 (15 points): Let F (x, y) = y cos x, x + y sin x, and let be the triangle from (, ) to (, 6) to (, ) to (, ). Use Green s Theorem to evaluate F d r. (heck the orientation of the curve before applying the theorem.) Solution: We have P (x, y) = y cos x, Q(x, y) = x + y sin x. We note that is a simple closed curve enclosing the triangle D in the first quadrant with sides given by y =, x =, and y = 3x. Since is oriented negatively (in the clockwise direction), we use Green s Theorem as follows: F d r = P (x, y) dx + Q(x, y) dy = P (x, y) dx + Q(x, y) dy ( + Q = x P ) 3x da = ((x + y cos x) y cos x)) dy dx y = D 3x x dy dx = 3x dx = = 16. 8