31aPracticeFinal-solutions

Solution 15 ft 6 ft x s 15 x s 6 s s 2 3 x ds dt 2 3

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Solution: 15 ft 6 ft X S 15 x + s = 6 s s = 2 3 x ds dt = 2 3 dx dt We are given dx dt = 3 So ds dt = 2 ft per second. 9
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6. (10 points) Let f ( x ) = x 2 - 5 x - 6 and F ( X ) = R x 0 f ( t ) dt . a ) Find the critical points of F ( x ) and determine whether they are local minima, maxima, or neither. Solution: By the Fundamental Theorem of Calculus F 0 ( x ) = f ( x ) = x 2 - 5 x - 6 = ( x - 6)( x + 1). So F ( x ) has critical points at x = - 1 , 6. The second derivative F 00 ( x ) of F ( x ) is just f 0 ( x ) = 2 x - 5. We have F 00 ( - 1) = - 7 F 00 (6) = 7 so by the second derivative test, F has a local maximum at x = - 1 and a local minimum at x = 6. b ) Find the points of inflection of F ( x ) and determine the change in concavity (up to down or down to up) of these points. Solution: From the above, F 00 ( x ) = 2 x - 5, which has a zero at x = 5 2 . To the left of this point, F 00 ( x ) is negative, and to the right of this point, F 00 ( x ) is positive. So at x = 5 2 , F ( x ) changes from concave down to concave up. 10
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7. (10 points) Find the points on the ellipse 4 x 2 + y 2 = 4 that are farthest away from the point (1 , 0). Solution: The squared distance D between points (1 , 0) and ( s, t ) is given by the distance formula D = ( s - 1) 2 + ( t - 0) 2 = ( s - 1) 2 + t 2 if ( s, t ) also lies on this ellipse, then 4 s 2 + t 2 = 4, so t 2 = 4 - 4 s 2 , so D = ( s - 1) 2 + t 2 = ( s - 1) 2 + (4 - 4 s 2 ) = - 3 s 2 - 2 s + 5 This ellipse has center (0 , 0), with a vertical major axis of length 4 ( a = 2) and a horizontal minor axis of length 2 ( b = 1). Therefore, the possible values for s lie in [ - 1 , 1], and we wish to maximize D , as a function of s , on this interval (which means we also need to test the endpoints s = - 1 , 1). We compute dD/ds = - 6 s - 2, which has a zero at s = - 1 / 3. Testing the points we need to test, we get D ( - 1) = 4 D ( - 1 / 3) = 16 / 3 D (1) = 0 so the points on the ellipse that are furthest away from the point (1 , 0) have x -coordinate equal to - 1 / 3. Plugging this into the equation of the ellipse, we find that there are two such points: 11
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t 2 = 4 - 4 - 1 3 2 = 4 - 4 9 = 32 9 so t = ± 4 2 3 , and the coordinates of the two points are ( - 1 3 , 4 2 3 ) and ( - 1 3 , - 4 2 3 ).
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