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# must1 - Preliminary Exam II Solutions 1 Find dy by implicit...

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Preliminary Exam II – Solutions 1 . Find dy dx by implicit differentiation given that cos( πy ) - 3 sin( πx ) = 1 . Solution: Differentiation yields - π sin( πy ) dy dx - 3 π cos( πx ) = 0 hence π sin( πy ) dy dx = - 3 cos( πx ) sin( πy ) . 2 . An object is moving along the circle x 2 + y 2 = 16. If dy dt = 2 when it reaches the point with coordinates x = 3 and y = 1, find dx dt at that point. Solution: Differentiating with respect to the variable t yields: 2 x dx dt + 2 y dy dt = 0 . Substituting the given information yields: 3 dx dt + 2 = 0 . Thus dx dt = - 2 3 . 3 . Find the absolute extrema of the function f ( x ) = x 3 - 3 2 x 2 + 5 on the closed interval [ - 1 , 2]. Solution: Differentiation yields f 0 ( x ) = 3 x 2 - 3 x = 3 x ( x - 1) hence the critical numbers are x = 0 , 1. The candidates are - 2 , 0 , 1 , 2 and 4 f ( - 1) = ( - 1) 3 - 3 2 ( - 1) 2 + 5 = - 1 - 3 2 + 5 = 5 2 f (0) = 5 f (1) = 1 - 3 2 + 5 = 9 2 f (2) = 2 3 - 3 2 2 2 + 5 = 8 - 6 + 5 = 7 . Thus (a). max = 7 min = 5 / 2 1

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2 4 . Find all of the critical numbers of the function f ( x ) = x 2 + 1 x . Solution: Differentiation yields f 0 ( x ) = x 2 - 1 x 2 . The critical numbers are obtained by setting the numerator equals zero as well as setting the denominator equals. Thus the critical numbers are x = 0 , 1 , - 1. 5 . Compute d 2 y dx 2 for the curve x 2 - y 2 = 25 . First we compute dy/dx : 2 x - 2 y dy dx = 0 hence dy dx = x y .
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must1 - Preliminary Exam II Solutions 1 Find dy by implicit...

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