Solutions to the Final Exam, Math 53, Summer 2012

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1 olutions to the Final Exam, Math 5, ummer. (a) ( points) Let be the boundary of the region enclosedby the parabola y = x and the line y = with counterclockwise orientation. alculate (y + e x )dx + xdy. (b) ( points) If the directional derivatives at the point (, ) are given find f x (, ) and f y (, ). (a) Use Green s Theorem. Q x, f(, ) =,, f(, ) =, =, P y (y + e x )dx + xdy = = = 5. = y, so yda = ydydx = y y dx x x x + x dx = x + x5 = (b) The directional derivatives are related to the partial derivatives in the following way, f = f x + f y and, f = f x + f y. Then, evaluating at (, ) we obtain the system of equations f x + f y = f x + f y =, where both partial derivatives are evaluated at (, ). olving the system of equations gives f x (, ) = 6, f y (, ) =.

2 . Let be the surface parametrized by r(u, v) = sin u cos u, sin u, v where the domain of the parameters is = {(u, v) u π, v sin u}. (a) ( points) Find the tangent plane at the point (x, y, z) = (,, ). (b) ( points) alculate (x + )d. (a) We need to calculate r u r v. r u = cos u sin u, sin u cos u,, r v =,,, so r u r v = sin u cos u, sin u cos u,. The point (x, y, z) = (,, ) corresponds to u = π, v =. Then the normal vector to the plane is 6 r u r v ( π 6, ) =,,. The equation of the tangent plane is (x ) (y ) = or simplified (b) (x + )d = x y =. (sin u cos u + ) r u r v dudv. The magnitude of the normal vector is r u r v = ( sin u cos u + (sin u cos u) ) / = (sin u + sin u cos u + cos u) / that simplifies to r u r v = ((sin u + cos u) ) / =. Then (x + )d = = π π sin u = + π (sin u cos u + )dvdu = π sin u cos u + ( cos(u)) du = sin u. sin u cos u + sin u du + u sin(u) π

3 . ( points) efine G = zxe x y, zye x y, e x y + z, H =, x, y and F = G + H. ompute F dr, where is the line segment from (,, ) to (,, ). Hint: alculate the line integrals for G and H separately. Use a different method for each integral. (a) The vector field G is conservative. We look for a potential: f x = zxe x y f = ze x y + g(y, z) f y = zye x y + g y (y, z), Then g y = giving g(y, z) = h(z), so f = ze x y + h(z) f z = e x y + h (z). Then h (z) = z giving h = z + c, where c is a constant. A potential for G is f(x, y, z) = ze x y + z. By the fundamental theorem of line integrals G dr = f(,, ) f(,, ) = e. For H we evaluate the integral directly. A parametrization of is r(t) =,, +t,, = t, t, t, t. Then H dr = = 5., t, + t,, dt = 5 t dt Therefore F dr = 9 e = 9 e.

4 . ( points) Let be the ellipsoid of equation x + y + z = and let (u, v, w) be a point in with u >, v > and w >. The tangent plane to at (u, v, w) has equation ux + vy + wz = and together with the three coordinate planes encloses a (pyramid-like) solid E whose volume equals uvw. Find the point (u, v, w) as in the first paragraph such that E has the minimum possible volume. Write what that volume is. The problem is to minimize Using Lagrange multipliers, uvw subject to the constraint u + v + w u vw = λu, uv w = λv, uvw = λw. =, with u, v, w >. ince u, v, w are nonzero we obtain that λ equals λ = u vw = uv w = uvw. Then, from u vw = uv w we obtain v = u ; and from u vw = uvw we obtain w = u. Using the constraint we see that u =, therefore u =, and then v = and w =. The point is ( ),,, and the minimum volume is =.

5 5. ( points) Let E be the solid enclosed by the paraboloids z = x + y and z = x y and let be the boundary of E with outward pointing normal. alculate F d, where F(x, y, z) = x + y, yz + e z, y z. implify your answer. ince is a closed surface oriented outward we can use the divergence theorem. Now F = x + z z = x, then F d = x dv. To calculate the triple integral we use cylindrical coordinates. The paraboloids are z = r and z = r. The intersection gives r = r so r =. Then E π r x dv = r cos θ rdzdrdθ = r = π(r 6 r6 ) ) = 8π. E π cos θdθ r ( r )dr 5

6 6. Let be the curve consisting of: a line segment from (,, ) to (,, ) followed by the arc of a circle x = cos t, y = sin t, z =, t π, followed by the line segment from (,, ) to (,, ). (a) (5 points) Parametrize the two line segments (with the stated orientations) and verify that lies in the cone of equation z = x + y. (b) (5 points) alculate F dr, where F = yzi + y e y j xyk. (a) For the first line segment from (,, ) to (,, ): r(t) = t,, = t,, t, t. For the second segment from (,, ) to (,, ): r(s) = ( s),, =, s, s, s. To check that the curve lies in the cone, we verify that the parametrizations satisfy the equation of the cone. For the first line segment x + y = t + = t = z, so it satisfies the equation. For the second line segment x + y = + ( s) = s = z, so it satisfies the equation. For the arc of the circle x + y = cos t + sin t = = z, so it satisfies the equation too. (b) We use tokes Theorem where is the part of the cone enclosed by the curve. The curl of F is i j k F = x y z = xi yj + zk. yz y e y xy The cone z = x + y has equation in cylindrical coordinates z = r and the surface can be parametrized in cylindrical coordinates (or cartesian coordinates) as r(r, θ) = r cos θ, r sin θ, r, where θ π and r. Then r r = cos θ, sin θ,, r θ = r sin θ, r cos θ, and the cross product is r r r θ = r cos θ, r sin θ, r which is the upward pointing normal as required by the right hand rule. Then π F d = r cos θ, r sin θ, r r cos θ, r sin θ, r drdθ = π = (π + π ) = π. r cos θ + r sin θ + r drdθ = π + sin θdθ 6

7 7. ( points) Let g be a function of one variable such that the derivatives g, g and g are continuous on R. efine f(x, y) = g ( x + y ), that is, f(x, y) equals the second derivative of g evaluated at x + y. For the disc = {(x, y) x + y 9} calculate xf x + yf y da, in terms of the values of g, g and g at and. The partial derivatives of f are f x = g ( x x + y ) x + y, f y = g ( y x + y ) x + y, so then x xf x + yf y = g ( x + y ) x + y + g ( x + y ) x + y = g ( x + y ) x + y. y Writing the integral in polar coordinates we get xf x + yf y da = π g (r)r rdrdθ = π g (r)r dr. We integrate by parts with u = r, du = rdr, dv = g (r)dr, v = g (r) to get xf x + yf y da = π (g (r)r ) g (r)rdr and a new integration by parts with u = r, du = dr, dv = g (r)dr, v = g (r) gives Evaluating xf x + yf y da = π (g (r)r ( g (r)r )) g (r)dr. xf x + yf y da = π(9g () 6g () + g() g()), where we used the fundamental theorem of calculus to evaluate the integral of g. 7