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### test2primea

Course: MATH 2401, Fall 2008
School: Georgia Tech
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Word Count: 770

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2401 MATH Test 1/2 Fall 05 Hurley 1. (18 points) Compute f x Name: and f y where f (x, y) = (cos x2 )exy . To compute f x we employ the product rule. f = (cos x2 ) (exy ) + (cos x2 )exy x x x = (cos x2 )yexy - 2x(sin x2 )exy = yexy (cos x2 ) - 2xexy (sin x2 ) Because there is only one y, we don't need the product rule for f = x(cos x2 )exy y f . y (I probably won't ask you just to compute the...

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2401 MATH Test 1/2 Fall 05 Hurley 1. (18 points) Compute f x Name: and f y where f (x, y) = (cos x2 )exy . To compute f x we employ the product rule. f = (cos x2 ) (exy ) + (cos x2 )exy x x x = (cos x2 )yexy - 2x(sin x2 )exy = yexy (cos x2 ) - 2xexy (sin x2 ) Because there is only one y, we don't need the product rule for f = x(cos x2 )exy y f . y (I probably won't ask you just to compute the partials, you will need to use them.) 1 2. (16 points) Show that x2 does not exist. (x,y)(0,0) x2 + y 2 lim If we compute the limit along two different lines through (0, 0) and get two different values, the overall limit does not exist. First we try the x-axis, equation y = 0. Plug y = 0 into the given limit. x2 x2 lim = lim 2 x0 x (x,0)(0,0) x2 + 02 = lim 1 x0 =1 Next we try the y-axis, equation x = 0. Plug x = 0 into the given limit. 0 02 = lim 2 lim 2 + y2 y0 y (0,y)(0,0) 0 = lim 0 y0 =0 If the overall limit was c then the limit along all paths approaching (0, 0) would be c. Since we have two different limits along two different paths, the overall limit does not exist. 2 3. (14 points) Let F and G be differentiable functions of 3 variables. Given that the surfaces F (x, y, z) = 0 and G(x, y, z) = 0 intersect at right angles along a curve , what conditions must be satisfied by the partial derivatives of F and G on . The level surfaces F (x, y, z) = 0 and G(x, y, z) = 0 are perpendicular to the gradients of F and G respectively. As F (x, y, z) = 0 and G(x, y, z) = 0 are perpendicluar along , the gradients F and G are perpendicular along . F i+ x G G(x, y, z) = i+ x F (x, y, z) = F j+ y G j+ y F k z G k z Since the gradient vectors are perpendicular along , their dot product is zero. Hence F G F G F G F G 0= + + x x y y z z 0= along . 3 4. (16 points) The temperature in a neighborhood of the origin is given by T (x, y) = T0 + ey sin x. In which direction will a heat seeking particle placed at the origin move? The partical will move in the direction greatest of increase in heat. This is the direction of T (0, 0). T (x, y) = ey cos xi + ey sin xj Plug in (x, y) = (0, 0). T (0, 0) = i So the particle moves right along the x-axis. 4 5. (24 points) Find the maximum volume of a rectangular solid in the first octant (x 0, y 0, z 0) with one vertex at the origin and the other vertex on the plane x+y+ z =1 2 Let (x, y, z) be the vertex of the solid opposite the origin. The volume of the solid is V (x, y, z) = xyz. We maximize V (x, y, z) subject to x + y + z = 1. To keep the solid 2 in the first octant, we also require that x 0, y 0 and z 0. Note that if (x, y, z) is on the boundry of this domain, then once of x, y or z is zero, so V (x, y, z) = 0. We follow the method of Lagrange multipliers. Let g(x, y, z) = x + y + z - 1, so that the constraint x + y + z = 1 can also be written 2 2 g(x, y, z) = 0. We must consider all (x, y, z) so that V (x, y, z) = g(x, y, z), for some constant . So we compute V (x, y, z) = yzi + xzj + xyk 1 g(x, y, z) = i + j + k 2 From V (x, y, z) = g(x, y, z), we get yz = xz = xz = 1 xy = 2 and we don't forget our original constraint 1 x + y + z = 1. 2 Multiplying the first three equati...

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