f10exam2solutions - Math 113 Exam #2 Solutions 1. The...

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1. The equation (4 - x ) y 2 = x 3 , determines a curve called a cissoid , pictured below. What is the equation of the tangent line to the cissoid at the point (2 , 2)? 0 1 2 3 4 -3 -2 -1 1 2 3 Answer: Differentiating the above equation on both sides yields - 1 · y 2 + (4 - x ) · 2 yy 0 = 3 x 2 or, equivalently, (4 - x ) · 2 yy 0 = 3 x 2 + y 2 . Solving for y 0 yields y 0 = 3 x 2 + y 2 2 y (4 - x ) . Therefore, at the point (2 , 2), the slope of the tangent line is given by plugging in ( x,y ) = (2 , 2) in the above equation: Slope = 3(2) 2 + 2 2 2(2)(2) = 16 8 = 2 . Therefore, the slope of the tangent line is 2, and so, by the point-slope formula, the tangent line is given by y - 2 = 2( x - 2) or, equivalently, y = 2 x - 2 . 2. Consider the function f ( x ) = 5 sin x. At which values of x does the graph of f have a vertical tangent line? Answer: A line is vertical when its slope is infinite (either + or -∞ ). Since the slope of the tangent line to the graph is given by the derivative, the tangent line will be vertical when the derivative approaches ±∞ . Now, we compute the derivative using the Chain Rule: f 0 ( x ) = 1 5 (sin x ) - 4 / 5 · cos x = cos x 5(sin x ) 4 / 5 . 1
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.

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f10exam2solutions - Math 113 Exam #2 Solutions 1. The...

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