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Unformatted text preview: Midterm 2 Solutions, Math 1A, section 1 Thursday, November 6, 2008, 8:109:30 am 1. Let p 6 = 0. Show by implicit differentiation that the tangent line to the curve x p + y p = 1 , x > , y > at the point ( x , y ) is given by the equation x p 1 x + y p 1 y = 1. Show that the xintercept a and yintercept b of the tangent line satisfy a p/ (1 p ) + b p/ (1 p ) = 1 if p 6 = 1. Solution: Implicitly differentiate, using the power, sum, and chain rules: d dx ( x p + y p ) = px p 1 + py p 1 y = d dx (1) = 0 . Solving for y , we get y = ( x/y ) p 1 . We plug in the point ( x , y ) such that x p + y p = 1 into the pointslope formula for the tangent line: y y = ( x /y ) p 1 ( x x ) . Multiplying by y p 1 and putting the constants to one side, we obtain y p 1 y + x p 1 x = y p + x p = 1 . Setting x and y = 0, we see that the intercepts are ( a, b ) = ( x 1 p , y 1 p ). Then we see that a p/ (1 p ) + b p/ (1 p ) = ( x 1 p ) p/ (1 p ) + ( y 1 p ) p/ (1 p ) = x p + y p = 1 . 2. A ladder 10 ft. long leans against a vertical wall. If the bottom of the ladder slides away from the base of the wall at a speed of 2 ft./s. , how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft. from the base of the wall? Solution: Let be the angle between the ladder and the wall, let x ft. be the distance from the base of the ladder and the wall, and let t s. be the time. The given information implies that dx dt = 2. Then sin( ) = x/ 10, so = arcsin( x/ 10). Then we differentiate using the chain and constant multiple rules:...
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This note was uploaded on 10/31/2010 for the course MATH 53603 taught by Professor Christ during the Fall '08 term at University of California, Berkeley.
 Fall '08
 CHRIST

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