math1a - quiz 4 - solns - The most-missed point on this...

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Math 1A Quiz 4 Solution February 29th, 2008 Name SID 1. Show that the sum of the x - and y -intercepts of any tangent line to the curve x + y = c is equal to c . Solution: We use implicit differentiation to find y 0 : differentiating x + y = c gives 1 2 x - 1 / 2 + 1 2 y - 1 / 2 y 0 = 0. Solving for y 0 , we get y 0 = - y x . Now, let ( a, b ) be a point on this curve, so a + b = c . At this point, we get y 0 = - b a , so the equation to the tangent line at this point (using point-slope form) is ( y - b ) = - b a ( x - a ). To find the x -intercept, we plug in y = 0 and solve for x , which gives x = a + a b b = a + a b . Similarly, to solve for the y -intercept, we set x = 0 and solve for y , which gives y = b + a b . Thus thus sum of the x - and y -intercepts is a + 2 a b + b = ( a + b ) 2 . Since a + b = c , the sum of the x - and y -intercepts is c 2 = c . Note:
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Unformatted text preview: The most-missed point on this problem was that you must plug in a and b into the formula for y . The reason is that the formula y =- y x tells you the slope of the tangent at a given point ( x, y ), but you have to actually pick a point and plug in the values to get the slope at that point. Otherwise the tangent line equation which youll come up with wont even be the equation for a line. 2. Let f ( x ) = (1 + sin(cos(sin(2 x )))) 100 . Find f ( x ). Solution: Many applications of chain rule give: 100 (1+sin(cos(sin(2 x )))) 99 cos(cos(sin(2 x ))) (-sin(sin(2 x ))) cos(2 x ) 2 . 1...
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This note was uploaded on 09/17/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at University of California, Berkeley.

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