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midterm2solutions-1

# midterm2solutions-1 - Midterm 2 Solutions Math 1A section 1...

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Midterm 2 Solutions, Math 1A, section 1 Thursday, November 6, 2008, 8:10-9:30 am 1. Let p = 0. Show by implicit differentiation that the tangent line to the curve x p + y p = 1 , x > 0 , y > 0 at the point ( x 0 , y 0 ) is given by the equation x p - 1 0 x + y p - 1 0 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 = 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 0 , y 0 ) such that x p 0 + y p 0 = 1 into the point-slope formula for the tangent line: y - y 0 = - ( x 0 /y 0 ) p - 1 ( x - x 0 ) . Multiplying by y p - 1 0 and putting the constants to one side, we obtain y p - 1 0 y + x p - 1 0 x = y p 0 + x p 0 = 1 . Setting x and y = 0, we see that the intercepts are ( a, b ) = ( x 1 - p 0 , y 1 - p 0 ). Then we see that a p/ (1 - p ) + b p/ (1 - p ) = ( x 1 - p 0 ) p/ (1 - p ) + ( y 1 - p 0 ) p/ (1 - p ) = x p 0 + y p 0 = 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 α 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( α ) = x/ 10, so α = arcsin( x/ 10). Then we differentiate using the chain and constant multiple rules: dt = d dt arcsin( x/ 10) = 1 1 - ( x/ 10) 2 dx dt 1 10 = 2 10 2 - x 2 .

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midterm2solutions-1 - Midterm 2 Solutions Math 1A section 1...

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