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Unformatted text preview: Midterm 2 Solutions, Math 1A, section 1 Thursday, November 6, 2008, 8:10-9:30 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 x-intercept a and y-intercept 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 point-slope 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