practice_final_solutions.pdf - Sample Final Exam Solutions...

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Sample Final Exam — Solutions Math 32A/3, Fall 2014 This is a collection of problems that would not be unreasonable for a real midterm exam. WARNING: the inclusion (or exclusion) of a certain topic or type of problem on this sample exam does not guarantee its inclusion (or exclusion) on the actual exam! 1. Recall that the cycloid generated by a circle of radius R is the curve with parametrization r ( t ) = h R ( t - sin t ) , R (1 - cos t ) i . Find the length of this curve when 0 t 2 π . [The identity 1 - cos t = 2 sin 2 ( t/ 2) may be useful.] Solution: We begin by calculating the tangent vector: r 0 ( t ) = h R (1 - cos t ) , - R sin t i . Its length follows from k r 0 ( t ) k 2 = R 2 (1 - cos t ) 2 + R 2 sin 2 t = R 2 (1 - 2 cos t + cos 2 t + sin 2 t ) = 2 R 2 (1 - cos t ) = 4 R 2 sin 2 t 2 , k r 0 ( t ) k = 2 R sin t 2 , where we used the above identity and that sin( t/ 2) 0 when 0 t 2 π . Now, the length of the curve is s = Z 2 π 0 k r 0 ( t ) k dt = 2 R Z 2 π 0 sin t 2 dt = 2 R ( - 2) cos t 2 2 π 0 = - 4 R (cos π - cos 0) = 8 R. 2. Find and classify all critical points of the function f ( x, y ) = 1 4 x 4 + y 4 + 1 2 x 2 - 2 y 2 . [Hint: there are three critical points.] Solution: To find the critical points, we solve f ( x, y ) = h x 3 + x, 4 y 3 - 4 y i ? = h 0 , 0 i . This implies that x ( x 2 + 1) = 0 and y ( y 2 - 1) = 0, so x = 0 and y = 0 , ± 1. Hence, there are three critical points: (0 , - 1) , (0 , 0) , (0 , 1) . The second derivative matrix and its determinant are 2 f ( x, y ) = 3 x 2 + 1 0 0 12 y 2 - 4 , D ( x, y ) = det 2 f ( x, y ) = 4(3 x 3 + 1)(3 y 2 - 1) . Now we apply the Second Derivative Test to each critical point. D (0 , - 1) = 8 > 0, f xx (0 , - 1) = 1 > 0, so (0 , - 1) is a local minimizer; D (0 , 0) = - 4 < 0, so (0 , 0) is a saddle point; D (0 , 1) = 8 > 0, f xx (0 , 1) = 1 > 0, so (1 , 1) is a local minimizer.
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3. (a) Describe the set of all vectors perpendicular to both u = h 1 , 0 , 1 i and v = h 0 , 2 , 3 i . (b) Let f ( x, y ) = 1 2 log( x 2 + y 2 ). At the point P = ( x, y ) 6 = (0 , 0), find a unit vector u such that the directional derivative D u f ( P ) is maximal. What is the maximal value of D u f ( P )? Solution: (a) The vectors u and v are not parallel, so they must span a plane. All vectors perpendicular to u and v are therefore normal to this plane. One normal vector is the cross product, u × v = ( i + k ) × (2 j + 3 k ) = - 2 i - 3 j + 2 k = h- 2 , - 3 , 2 i . Any perpendicular vector is parallel to this, so the set is { t h- 2 , - 3 , 2 i | t R } . (b) The directional derivative D u f ( P ) is maximal when u points in the direction of the gradient f ( P ). Therefore, u = f ( P ) / k∇ f ( P ) k , and the directional derivative at this point is D u f ( P ) = f ( P ) · u = f ( P ) · f ( P ) k∇ f ( P ) k = k∇ f ( P ) k 2 k∇ f ( P ) k = k∇ f ( P ) k .
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