# Math 023 Spring 2012 Final Same - MATH 23 SPRING 2012 FINAL...

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This preview shows page 1 out of 8 pages. Unformatted text preview: MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS (1) For vectors A = 2i + 3j − k and B = −i + 3k, nd: (a) A − 2B. (5 points per part) Solution: A − 2B = (2i + 3j − k) − 2 (−i + 3k) = 4i + 3j − 7k. (b) A · B. Solution: A · B = (2i + 3j − k) · (−i + 3k) = 2(−1) + 3(0) − 1(3) = −5. (c) The vector projection projB A of A onto B. Solution: projB A = = = = (A · B) B (B · B) ((2i + 3j − k) · (−i + 3k)) (−i + 3k) ((−i + 3k) · (−i + 3k)) −5 (−i + 3k) 10 3 1 i− j 2 2 (d) A × B. Solution: A×B = i j k 2 3 −1 −1 0 3 = (9 − 0) i + (1 − 6) j + (0 + 3) k = 9i − 5j + 3k, which you should check is perpendicular to A and B, which this is. You may think the computation of the j component of the cross-product is the negative of what it should be, but if you look closely, it is correct. (2) Find the line of intersection between the planes P1 dened by 2x + 3y − z = 0 and P2 dened by x + 3y + z = 1. (10 points) 1 MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 2 The direction of the line is the cross-product of the two normal vectors to the planes, Solution: A = i j k 2 3 −1 1 3 1 = (3 + 3) i + (−1 − 2) j + (6 − 3) k = 6i − 3j + 3k, which again you should check is perpendicular to the two normals. Then, you nd a point on both planes. If x = 0, then 3y − z = 0 and 3y + z = 1. Subtract the second equation from the rst, and you get −2z = −1, or z = 1/2. Add the 1 two equations, and you get 6y = 1, or y = 1/6, so 0, 6 , 1 should be on both 2 planes. Check that it satises both equations. Then, the vector equation of the line of intersection, the line containing this point in the direction of A,is 1 1 j + k + t (6i − 3j + 3k) . r(t) = 6 2 Its parametric equations are x = 6t, y = 1 − 3t, z = 1 + 3t. 6 2 (3) Find the arclength of the curve r(t) = et cos(t)i + et sin(t)j, t ∈ [0, 2π]. (10 points) Solution: The arclength is given by (be careful to use the product rule) 2π r dt L = 0 2π (et cos(t) − et sin(t))2 + (et sin(t) + et cos(t))2 dt = 0 2π et = (cos(t) − sin(t))2 + (sin(t) + cos(t))2 dt 0 2π et = cos2 (t) − 2 cos(t) sin(t) + sin2 (t) + cos2 (t) + 2 cos(t) sin(t) + sin2 (t) dt 0 2π = √ et 2dt 0 √ t 2π = 2e 0 √ 2π 2 e −1 . = (4) (5 points per part) (a) If f (x, y) = exy , nd ∂f ∂x . Solution: ∂f = yexy . ∂x (b) If f (x, y) = (2x + 3y)2 , nd ∂f . ∂y Solution: ∂f ∂y = 2 (2x + 3y) 3 = 6 (2x + 3y) . (5) Find the tangent plane to the surface S dened by x2 + y 2 − (1, −1, 2). (10 points) z2 4 = 1 at the point MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 3 Here you use the fact that the surface is a level surface, so the gradient at (1, −1, 2) is perpendicular to the surface there. Solution: N = = x2 + y 2 − z2 4 1 2xi + 2yj − zk 2 (1,−1,2) (1,−1,2) = 2i − 2j − 1k. So the equation of the tangent plane is 2(x − 1) − 2(y + 1) − (z − 2) = 0, or 2x − 2y − z = 2. (6) Find the absolute maximum and the absolute minimum of the function f (x, y) = 2x2 − 4x + y 2 on the disk x2 + y 2 ≤ 9. (15 points) Solution: You have to nd all critical points in the interior, and those on the boundary. Interior: Since f = (4x − 4)i + 2yj, f = 0 only when x = 1 and y = 0. f (1, 0) = 2 − 4 = −2. 2 2 Boundary: The boundary is the circle x +y = 9. Here it is best to use Lagrange multipliers. You set f = (4x−4)i+2yj equal to λ g , where g is the constraint equation, x2 + y 2 = 9. So, f = λ g (4x − 4)i + 2yj = λ (2xi + 2yj) , or 4x − 4 = 2λx 2y = 2λy. The second equation is easy to solve; either y = 0 or λ = 1. If y = 0, then to satisfy the constraint, x = ±3, so we get two points (3, 0) and (−3, 0). If λ = 1 we then look at the rst equation, which becomes 4x − 4 = 2x, or 2x = 4, √ x = 2. If x = 2, then to√ satisfy the constraint equation, y = ± 5, which gives √ us two more points, (2, 5) and (2, − 5). We only need to evaluate f on all these points to nd the winner. f (1, 0) = −2 (the interior critical point) f (3, 0) = = = f (−3, 0) = = √ f (2, 5) = = √ f (2, − 5) = 2x2 − 4x + y 2 (3,0) 18 − 12 + 0 6 18 + 12 + 0 30 8−8+5 5 5. So, the maximum value is 30, at (−3, 0) and the minimum value is −2, at (1, 0). MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 1 (7) Evaluate 1 0 1 − y 2 dy dx. 4 (10 points) x Change the order of integration to solve this. Note that the region is the triangle bounded by the lines x = 0, y = x, and y = 1. So Solution: 1 1 1 y 1 − y 2 dy dx = 0 1 − y 2 dx dy x 0 0 1 1 − y2x = 0 1 x=y x=0 dy 1 − y 2 y dy. Set u = 1 − y 2 , = 0 0√ = 1 = = = 1 u − du 2 1√ udu 1 2 0 1 3/2 u 3 1 . 3 1 0 (8) Find the x-coordinate x of the center of mass of the of the region bounded by the lines y = 2 − x, x = 0, and y = 0, if the density ρ is given by ρ = x. (10 points) Solution: x = = = = = = R xdm R dm 2 2−x x (xdydx) 0 0 2 2−x (xdydx) 0 0 2 2−x 2 x dydx 0 0 2 2−x dx 0 yx|0 2 2 2−x dx 0 yx 0 2 0 (2 − x)xdx 2 2 0 (2 − x)x dx 2 x2 − 1 x3 0 3 2 3 1 4 2 3x − 4x 0 4 3 34 43 = 1. = (9) Find the volume of the region bounded by the surfaces z = x2 +y 2 , and z = 2−x2 −y 2 . (10 points) MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 5 I would do this in cylindrical coordinates. Note that the two surfaces intersect when z = 1, along a circle of radius 1, center on the z -axis. Solution: dV V ol = 2π R 1 2−r2 = rdzdrdθ 0 0 1 r2 2r − 2r3 dr = 2π 0 1 r − r4 2 1 2 = 2π 0 = π. (10) (7 points per part) (a) Find the gradient of f (x, y, z) = xy − 2x + 3y + 4yz 2 . Solution: f = (y − 2) i + x + 3 + 4z 2 j + (8yz) k. (b) Find the divergence of F = xyi + y 2 j + 3xzk. Solution: div(F) = ·F = y + 2y + 3x = 3x + 3y. (c) Let F = xyzi + x2 j − xyk. Compute the curl of F, × F. Solution: curl(F) = = ×F i j ∂ ∂x ∂ ∂y x2 k ∂ ∂z xyz −xy = (−x − 0) i + (xy + y) j + (2x − xz) k = −xi + (xy + y) j + (2x − xz) k. (11) Determine whether the vector eld F = (x2 + xy 2 )i + (x2 y + y 2 )j is conservative. If it is conservative, nd a potential function f (x, y) so that f = F. (10 points) Solution: Since this vector eld is dened everywhere, all we have to do is check the mixed partials condition ∂P = ∂Q . But ∂y ∂x ∂(x2 + xy 2 ) ∂y 2y + y2) ∂(x ∂x = 2xy, and = 2xy, so the condition is satised and the vector eld is conservative (you should have expected this, since then you have to nd the potential function, and we want to see whether you can do that). The potential function f (x, y) is then MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 6 obtained by nding P dx (with the "constant of integration" depending on y ) and re-dierentiating to compare to Q f (x, y) = = = P dx (x2 + xy 2 )dx 1 3 1 2 2 x + x y + k(y), 3 2 so, since ∂f ∂y 2 2 (x y + y ) = x2 y + k (y), Q = 1 1 so k (y) = y 2 , k(y) = 1 y 3 , and a potential function f (x, y) = 3 x3 + 1 x2 y 2 + 3 y 3 . 3 2 A +C is optional. (12) Find the line integral C x2 sin(x) + y 2 dx + (x + ey ) dy , where C is the circle of radius 1, centered at the origin, traversed in the positive direction (counterclockwise). (10 points) This has to be a Green's theorem problem. Recall the trick that ydA = yArea, where y is the y -coordinate of the center of gravity, which R in this case is at the origin, so y = 0. Solution: x2 sin(x) + y 2 dx + (x + ey ) dy = C (1 − 2y) dA R = Area − 2yArea = π. 2 2 (13) Find the surface integral S f dS of the function f (x, y, z) = x +y over the sphere x2 + y 2 + z 2 = 1. (10 points) Solution: This is not a ux problem, so you can't use any of the theorems. You have to do that directly. Parametrize the sphere by r(u, v) = cos(u) sin(v)i + sin(u) sin(v)j + cos(v)k. This comes from spherical coordinates, with ρ = 1, u = θ and v = φ. From that you can see that you cover the whole sphere taking u ∈ [0, 2π] and, for each u, v ∈ [0, π]. Then, it will take a while to compute ∂r ∂r ∂r ∂r ∂u × ∂v directly, but in the end it comes out that ∂u × ∂v = sin(v). You can see this by again considering spherical coordinates and remembering how we got dV , but you should probably work out that cross product. Then, 2π π x2 + y 2 dS = S cos2 (u) sin2 (v) + sin2 (u) sin2 (v) sin(v)dvdu 0 0 2π π sin3 (v)dvdu = 0 0 π 1 − cos2 (v) sin(v)dv = 2π 0 You can do the u integral with no problem since what you get out of the v integral will have no u-dependence. Then set t = cos(v), so dt = − sin(t)dt. MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 7 Also, when v = 0, t = 1 and when v = π , t = −1, so −1 x2 + y 2 dS = 2π S 1 − t2 (−dt) 1 = 2π 1 1 t − t3 3 −1 8π = . 3 Since this is an integral of a positive function, it denitely has to have a positive answer. (14) Find the ux integral S (xi + yj + zk) · n dS , where the surface S is that portion of the paraboloid z = 4 − x2 − y 2 above the xy -plane, and n is the upward-pointing normal. (15 points) Solution: No, you cannot use the divergence theorem here on that surface. The standard method will be to do this directly. Since, for the ux of a vector eld F = P i + Qj + Rk over a graph z = f (x, y) with the upward normal can be evaluated as ∂f ∂f Q + R dA, F · ndS = − P− ∂x ∂y S A where A is the region in the xy -plane which is the domain of f (x, y), and here f (x, y) = 4 − x2 − y 2 (xi + yj + zk) · n dS = (−(−2x)x − (−2y)y + z) dA x2 +y 2 ≤4 S 2x2 + 2y 2 + 4 − x2 − y 2 = dA x2 +y 2 ≤4 2π 2 4 + r2 rdrdθ = 0 = 2π 0 1 2r2 + r4 4 2 0 = 24π. (15) Find the ux of the vector eld F = (x2 + 2x) i + x2 + yz j + cos(x)k outward across the sphere x2 + y 2 + z 2 = 1. (15 points) Solution: This one uses the theorems. It is a ux outward, so it has to use the divergence theorem. F · ndS = S · FdV V = ((2x + 2) + z + 0) dV V = 2xV ol + 2V ol + zV ol 8π = , 3 using that fact again that xdV = xV ol (etc., for y and z ) relating the integrals to the center of gravity, which in this case is the origin, along with the volume of the sphere being 4π (radius)3 . 3 MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 8 (16) Find the line integral C F · dr, if the vector eld is F = (sin(x) + y 2 ) i + ey j + x k, and the curve C is the intersection of the paraboloid z = x2 + y 2 with the cylinder (x − 1)2 + y 2 = 1 in space. The curve C is oriented positively (counter-clockwise) when viewed from above. (14 points) Solution: Only one left, so it has to use Stokes' theorem. Well, actually, it uses Stokes' theorem since it's integrating a vector eld over a closed curve in space which bounds a surface S  the part of the paraboloid z = x2 + y 2 which is inside the cylinder (x − 1)2 + y 2 = 1. F · dr = × F · ndS, C S where the curl of F, × F, is i ×F = j k ∂ ∂x ∂ ∂y ey ∂ ∂z (sin(x) + y 2 ) x = (0 − 0) i + (0 − 1) j + (0 − 2y) k, so F · dr = C × F · ndS S (−j − 2yk) · ndS = S (−(0)(2x) − (−1)(2y) − 2y) dA = (x−1)2 +y 2 ≤1 = 0. ...
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