practice21answers - Answer Key 4 MATH 20550: Calculus III...

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Unformatted text preview: Answer Key 4 MATH 20550: Calculus III Exam II Name: October 26, 2010 Instructor: As a member of the Notre Dame community, I will not participate in or tolerate academic dishonesty. Please sign Record your answers to the multiple choice problems by placing an × through one letter for each problem on this page. There are 10 multiple choice questions worth 6 points each and 3 partial credits problems worth 14 points each. On the partial credit problems try to simplify your answer and indicate your final answer clearly. You must show your work and all important steps to receive credit. You may use a calculator if you wish. Score DO NOT WRITE IN THIS COLUMN Multichoice 11 12 13 Total 1. a • c d e 6. a b c • e 2. • b c d e 7. a b • d e 3. a • c d e 8. a b • d e 4. a b c • e 9. a b c d • 5. • b c d e 10. a • c d e 1 1 Practice Exam 1 ￿￿ Problem (#11). Evaluate D x dA where D is the region bounded by the curves y = x and y = x2 − x. Solution. The region D looks like this: and the intersections of the to curves occur at x = 0 and x = 2. (See this by solving the equation x = x2 − x.) Thus the integral we need to compute is ￿￿ 2 x x dy dx 0 So, let’s do it: ￿2￿x 0 x2 −x x dy dx = = = = = = x2 − x ￿ ￿ 2 ￿￿ x 0 x2 −x x dy ￿ dx ￿ 2 ￿￿ x x x2 −x dy dx ￿02 x[x − (x2 − x)]dx ￿02 2 3 0 x − x dx ￿ 2 x3 x4 2 3 −4 0 because x is a const. with resp. to dy 4 3 Problem (#12). Find the absolute maximum value of the function f (x, y ) = x2 y 3 on the region D = {(x, y ) | x2 + y 3 ≤ 1} Solution. 1. We first find the critical points of f (x, y ). We compute ∇f = ￿2xy 3 , 3x2 y 2 ￿ This is defined for all x, y , so the only critical points are where ∇f (x, y ) = ￿ 0. Now, ∇f = ￿ when at least one of x or y is 0. Noting that f (x, 0) = f (0, y ) = 0 0 for all x, y , we know that none of these points (x, 0) or (0, y ) yields the 1 absolute maximum we seek; f (x, y ) does attain positive values on the region D. 2. We find the the maxima of f (x, y ) on the boundary of D, ￿ ￿ ∂ D = (x, y ) : x2 + y 3 = 1 This means we need to solve the following constrained optimization problem: Maximize f (x, y ), Subject to g (x, y ) = x2 + y 3 = 1. To solve this, we will use Lagrange multipliers. Setting ∇f = λ∇g , find ourselves with the following system of equations: 2xy 3 = 2λx 3x2 y 2 = 3λy 2 x2 + y 3 = 1 We know that points (x, y ) with x = 0 or y = 0 are irrelevant (they cannot yield the maximum), so we can assume from now on that x ￿= 0 and y ￿= 0; in particular, we can divide by x or y whenever we want. With this assumption, the system yields that, 2y 3 = 2λ, 3x2 = 3λ and then that x2 = λ = y 3 Substituting these into the equation x2 + y 3 = 1, we get 2x2 = 1, 2y 3 = 1 2+ 3 So, the extrema of f (x, y ) on ￿ boundary {(x, y ) | x￿ y￿= 1} occur at the ￿ ￿￿ the points ( 1 , 3 1 ) and (− 1 , 3 1 ). Moreover, f ( 1 , 3 1 ) = 1/4 and 2 2 2 2 2 ￿ ￿2 f ( − 1 , 3 1 ) = 1/4, so both points are absolute maxima of f on D, and 2 2 the maximum value is 1/4. Problem (#13). Find the tangent plane to the surface x2 + y 3 − xy = z 2 + x at the point (1, 1, 0). 2 Solution. Let f (x, y, z ) = x2 + y 3 − xy − z 2 − x. The surface we are interested in is really the level surface f (x, y, z ) = 0, and f (1, 1, 0) = 0 – that is, the point (1, 1, 0) is on this surface. It follows that the gradient vector ∇f (1, 1, 0) is orthogonal to the tangent plane of the surface f (x, y, z ) = 0 at (1, 1, 0). Moreover, this tangent plane is determined by the base point r0 = ￿1, 1, 0￿ and normal vector n = ∇f (1, 1, 0). We compute: ∇f = ￿2x − y − 1, 3y 2 − x, −2z ￿ ∇f (1, 1, 0) = ￿0, 2, 0￿ A vector equation of the tangent plane, then, is ￿0, 2, 0￿ · ￿x − 1, y − 1, z ￿ = 0 which can be seen to be equivalent to the equation y = 1 once you compute the dot product. 3 ...
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