1. For each function, find all of its critical points and then classify each point as a local extremum or saddle point.

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1 Solutions Review for Exam # Math 6. For each function, find all of its critical points and then classify each point as a local extremum or saddle point. a fx, y x + 6xy + y Solution.The gradient of f is f 6x + 6y, 6x + 6y. The critical points occur when 6x + 6y and 6x + 6y. The second equation means that y x. Plugging this into the first equation, we get 6x 6x, which has solutions x and x. Therefore, the critical points are, and,. We compute f xx x, f yy 6, and f xy 6, so At the critical values, we have D f xx f yy f xy 7x 6. D, 6 and D, 6. Therefore, there is a saddle at, and a local minimum because f xx is positive at,. b gx, y xy 4e x Solution.The gradient of g is g y 4x + e x, xye x. Since e x is always positive, the critical points occur when y 4x + and xy. The second equation says that x or y. If x, then y 4, so y ±. If y, then x. So, the critical points are, ± and,. We compute g xx y 4x + e x, g yy xe x, and g xy yx + e x, so D g xx g yy g xy [ xy 4x + 4y x + ] e x. At, ±, we have D, ± 4, so g has a saddle. At,, we have D, 8e, which is positive, and g xx 4, which is negative, so g has a local maximum.

2 . Consider the surface given by z x + y y. a Find the gradient of the surface at the point 4,,. Solution.The gradient is z, y. At the point 4,,, the gradient is z, 4. b Sketch the level curves of the surface corresponding to z,,,,. Then sketch the gradient vector from part a, emanating from the point 4, on the level curve z. Solution.At z k, the level curve is x y y + k y + k, which is a parabola which opens to the right. c Find the equation of the plane tangent to the surface at the point 4,,. Solution.The equation of the tangent plane is z + 4, x 4, y + z + 4x 4 y +.. Let a, b, and P be constants. Use the method of Lagrange multipliers to show that the function fx, y xy P subject to the constraint ax + by P has a maximum value of 4ab. Solution.We want to maximize fx, y xy subject to the constraint function gx, y ax + by P. The gradients are f y, x and g a, b. Setting f λ g, we get y λa and x λb. Plugging these values into the constraint function, we get aλb + bλa P λ P ab, which means x P and y P a b. Therefore, the optimal value of f is f P a, P P b a P b P 4ab. The fact that it is a maximum value instead of a minimum value is not clear, especially because I forgot to specify that a and b are supposed to be positive. For a and b positive, we have y P ax/b, so xp ax fx, y a b b x + P b x, which is a parabola that opens downward, so the critical value is a maximum.

3 4. Let w x y + 4xz. a Find the gradient of w as a function of x, y, z. Solution.The gradient of w is w xy + 4z, x, 4x. b If x s t, y st, and z s + t, use the chain rule to find the gradient of w as a function of s, t. Solution.We compute x x st and s t y s, s t and y t st, so w s w x x s + w y y s + w z z s and xy + 4zst + x t + 4x 4xyst + 8zst + x t + 4x 4s 4 t 4 + 8s t + 6st + s 4 t 4 + 4s t. 5s 4 t 4 + s t + 6st w t w x x t + w y y t + w z z t xy + 4zs + x st + 4x xys + 4zs + x st + 8x s 5 t + 4s + 8s t + s 5 t + 8s t 4s 5 t + 4s + 6s t. The gradient of w with respect to s, t is z z and s t, w 5s 4 t 4 + s t + 6st, 4s 5 t + 4s + 6s t.

4 5. Compute the integral 5 x y da for each of the regions pictured below. S S S S Solution.The region S, as an x-simple set, is so the integral is S { x, y : x, y x }, x 5 x y da 5 x y dy S 5x 4x 5x x4 5 The region S, as a θ-simple set, is so the integral is S 5 x y da S { r, θ : r 4, θ π 4 }, 4 5r r dr dθ 64 dθ 6 θ π/4 5y x y y x π. 5r r4 4 4 dθ The region S, as a y-simple set, is so the integral is S 5 x y da 8y S { x, y : y x 4 y, y }, 4 y y 8y 4y x y dy 5x x xy dy y4 8y 7y + 6 y 4 y y 6. dy

5 6. Consider the triangle T with vertices,,,, and, and with density function δx, y. a Find the mass and center of mass of the triangle directly. Solution. As an x-simple region, the triangle is { x x, y : x, y x} { x, y : x, 4x 9 y x }, so the mass is m x x 9. x/ y x dy + x/ x + The center of mass is computed by + x 4x 9 x y 6x x + 8x dy 4x 9 x [ x x ] x dy + x dy 9 x/ 4x 9 [ x x ] xy 9 + xy x/ 4x 9 [ x + 6x + 8x ] 9 [ x + ] x + 9x 9 5 y [ x x ] y dy + y dy 9 x/ 4x 9 [ x x ] y + y 9 x/ 4x 9 [ x x + 7x 8 ] [ x x + 6x 8x ]

6 b Find the mass and center of mass by performing the change of variables x u + v and y u + v. Solution.The Jacobian of the transformation is [ ] [ ] x/ u x/ v Ju, v det det. y/ u y/ v Solving for u and v, we get u x y and v x + y. The vertices of the triangle in terms of u and v are v,,, 9/, and,. The mass of the original triangle is u m 6 dv du 6v u u The center of mass of the original triangle is computed as 8 du 8u du 8u 9u 9. u x u + 6v dv du 9 u uv + v du 9 9u 8u + 7 du 9 u 9u + 7u 9 5 u y 6u + 6v dv du 9 u 6uv + v du 9 45u 7u + 7 du 9 5u 6u + 7u 9

7 7. Find the volumes of the following solids. a The first octant solid bounded by the coordinate planes and the planes y and x + y + z 6. Solution.This is a y-simple solid. If we do dz dy-integration, then we describe the solid as x 6 y z, z y, y, y/ 6 y z y/ dz dy 6 y z dz dy 6z yz z y/ y dy 4y + 6 dy y 9 y + 6y 5 9. If we do dz, dy-integration, then we describe the solid as z y x, x 6 y, y, b 6 y y/ x/ dz dy y/ y x dy x xy x 6 y y 6 dy 4y + 6 y 9 y + 6y 5 9. The first octant solid bounded by the xz-plane, the xy-plane, the plane y x, and in between the spheres x + y + z and x + y + z 4. dy Solution.Using spherical coordinates, the solid is described by the inequalities π/ ρ, θ π 4, φ π, ρ sin φ dφ dρ dθ ρ dρ dθ 7 dθ 7θ π/4 ρ cos φ ρ dθ 7π. π/ dρ dθ c The solid bounded by the xy-plane, the plane y + z 4, and the cylinder x + y 4. Solution.Using cylindrical coordinates, the solid is described by the inequalities θ < π, r z 4 y 4 r sin θ, π 4 r sin θ r dz dr dθ π π 4r r sin θ dr dθ π 8 8 sin θ dθ 8θ + 8 π cos θ 6π. r r sin θ dθ

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