Math 241, Final Exam. 12/8/11.
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1 Math 1, Final Exam. 1/8/11. Name: No notes, calculator, or text. There are points total. Partial credit may be given. Write your full name in the upper right corner of page 1. Do problem 1 on p.1, problem on p.,... (or at least present your solutions in numerical order. Circle or otherwise clearly identify your final answer. 1. (5 points: Give the cosine of the angle between the vectors u i j + k, v i j + k. cos θ u v u v (15 points: Lines. Find parametric equations for the line L through the point (1,, 3 that is parallel to the plane Q : x + y + z and perpendicular to the line L 1 : x 1 + t; y 1 t; z t. Solution: The vector n 1, 1, 1 is normal to Q; the vector v 1,, 1 is the direction vector of L 1. It follows that L has direction vector i j k n v i(1 + j(1 + k( 3 i + j k. It follows that L has parametric equations x 1 + 3t; y + t; z 3 t, t R. 3. (15 points: Planes. Find the equation of the plane that passes through the point (1,, 3 and contains the line L : x 3t; y 1 + t; z t, t R. Solution: The line has direction vector v 3, 1, 1. We set t in the line to obtain a second point in the plane, (, 1, ; therefore, a second vector in the plane is u (1,, 3 (, 1, 1, 1, 1. A normal vector to the plane is i j k v u i(1 + 1 j( k(3 1 i j + k.
2 Hence, an equation for the plane is (x 1 (y + (z 3 x y z 6 x y + z x y + z.. (15 points: Tangent plane. Find the equation of the plane tangent to z x y + xy 3 at the point (1, 1, 3. Solution: With z f(x, y x y + xy 3, we compute It follows that f x xy + y 3 f x (1, 1 ; f y x + 6xy f y (1, 1 7. (x 1 + 7(y 1 z 3 x + 7y 7 z 3 x + 8y z (15 points: Chain rule. (a (1 points: Let z e xy, x t cos t, y t sin t. Compute dz/dt. You may leave your answer in the variables x, y, and t. Solution: The chain rule gives dz dt z dx x dt + z dy y dt (y e xy ( t sin t + cos t + (xye xy (t cos t + sin t. (b (5 points: Suppose that z F (x, y, x G(u, v, u H(t, v I(t. Write down the form of the chain rule that you would use to compute dz/dt. (Use a tree diagram to show variable dependencies. dz dt z x x u du dt + z x x dv v dt. 6. (15 points: Gradient and directional derivative. Let F (x, y, z x 3 y 3y z. (a (1 points: Compute the directional derivative of F at (1, 1, 1 in the direction of the vector u, 3, 6. Is F increasing or decreasing at this point? F 6x y i+(x 3 6yz j 3y k; it follows that F (1, 1, 1 6 i + 8 j 3 k. Furthermore, we have u, 3, 6, 3, 6. We compute u D u F (1, 1, 1 6, 8, 3, 3, 6 7 Since D u (1, 1, 1 <, the function is decreasing
3 (b (5 points: In what direction from (1, 1, 1 is the directional derivative of F a maximum? Your answer need not be a unit vector. Solution: The direction of maximum increase is F (1, 1, 1 6, 8, (15 points: Local extrema. Let f(x, y x 3 + y 3 3x 1y +. Find the point(s (x, y at which f(x, y has a local maximum, minimum, or saddle. f x 3x 3 x ±1; f y 3y 1 y ±. The critical points are {(1,, (1,, ( 1,, ( 1, }. In order to apply the second derivative test, we require f xx 6x, f yy 6y, f xy, D 6x 6y 36xy. We test the points as follows: (1, : f xx > and D > (1, is a local minimum. (1, : D < (1, is a saddle point. ( 1, : D < ( 1, is a saddle point. ( 1, : f xx < and D > ( 1, is a local maximum. 8. (15 points: Absolute extrema. Find the absolute maximum of f(x, y x + 3y + 1 on the triangular region D in the first quadrant bounded by: Clearly explain your answer. L 1 : x (y-axis; L : y (x-axis; L 3 : x + y. Solution: Since f x and f y 3, there are no critical points inside D. Hence, the extrema of f(x, y lie on the boundary of D. L 1 : We have u 1 (y f(, y 3y + 1. Since u 1(y 3, the extrema on L 1 occur at the endpoints: f(, 1 and f(, 7. L : We have u (x f(x, x + 1. Since u (x, the extrema on L occur at the endpoints: f(, 1 and f(, 5. L 3 : We have u 3 (x f(x, x x + 3( x x. Since u 3(x 1, the extrema on L 3 occur at the endpoints: f(, 7 and f(, 5. It follows that the absolute maximum of f(x, y on D is f(, 7. 3
4 9. (15 points: Lagrange multipliers. Use Lagrange multipliers to find the point(s (x, y, z with x, y, z which maximize f(x, y, z xyz subject to the constraint x+y+z 6. Solution: Let g(x, y, z x + y + z. We have Equating components gives f yz i + xz j + xyz k λ( i + j + k λ g. yz λ, xz λ xz λ, xyz λ xyz λ. Now, yz xz implies that y x. Similarly, yz xyz implies that x z. We substitute in the constraint equation to obtain ( x x + + x x 6 x 3, y 3, z (15 points: Polar coordinates. Set up an integral in polar coordinates for the volume of the solid E that lies below the plane z 6 and above the paraboloid z 3x + 3y in the first octant. (How do the plane and paraboloid intersect? Solution; We have V (6 3x + 3y da D (6 3r r dr dθ 3 ( r dr dθ. 11. (15 points: Rectangular coordinates. Suppose that E is the tetrahedron (a polyhedron with four vertices and four triangular faces, three of which meet at each vertex bounded by the planes x y, x + y, y z, z (xy-plane. Express the integral dv as an iterated integral with dv dy dx dz. Do not evaluate. E (Draw a quick sketch. What is the shadow (projection of E on the xz-plane? E dv z x z x dy dx dz. 1. (15 points: Cylindrical coordinates. Convert the integral 3 9 x 9 x y 3 y x + y dz dy dx
5 from rectangular to cylindrical coordinates. Do not evaluate. 3 9 x 9 x y r y x + y dz dy dx r 3 sin θ dz dr dθ. 3 9 r r sin θ r r dz dr dθ 13. (15 points: Spherical coordinates. Use spherical coordinates to evaluate the triple integral xz dv E where E is the solid region that lies within the sphere x + y + z and below the cone z x + y in the first octant. π (ρ sin φ cos θ(ρ cos φ(ρ sin φ dρ dφ dθ ( ( ( cos θ dθ sin φ cos φ dφ ( ] 1 u 3 3 π (3 5 ( 3 15 (1 ( / 3 ρ dρ ( ( ( π sin θ ] π ρ sin φ cos φ cos θ dρ dφ dθ ( 1 3 u du u 15 8 ( ρ 5 5 ] (. 1. (15 points: Change of variable in a double integral. Compute the double integral x y x + y da R over the square R with vertices {(,, (1, 1, (,, (1, 3}. Use a suitable change of variable. Solution: Let u x y and v x + y. We have (u, v (x, y It follows that (x, y (u, v
6 Therefore, we have R x y x + y da 1 S u du dv. v It remains to determine S and to do the evaluation. Since the transformation is linear, it will map the square to a quadrilateral whose vertices are: (, (, ; (1, 1 (, ; (, (, ; (1, 3 (,. The quadrilateral S is in fact, a square. We obtain R x y x + y da 1 u v du dv 1 ( u v (ln ln ( ln ln ln. ] dv dv v 6
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