asst6-f10-solns

asst6-f10-solns - MATH 137 Assignment 6 - Solutions 1. (a)...

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MATH 137 Assignment 6 - Solutions 1. (a) lim x π e sin x - 1 x - π To express this as a derivative, notice that since sin π = 0, we can express the limit as lim x π e sin x - e sin π x - π = lim x π f ( x ) - f ( π ) x - π = f 0 ( π ) where f ( x ) = e sin x The derivative is calculated using the chain rule: f 0 ( x ) = e sin x cos x = f 0 ( π ) = e sin π cos π = e 0 ( - 1) = - 1 Therefore, lim x π e sin x - 1 x - π = - 1 (b) lim x 0 2 - 5 32 - x x If we let f ( x ) = 2 - 5 32 - x , we note that f (0) = 0 and the limit is of the form lim x 0 2 - 5 32 - x x = lim x 0 f ( x ) - f (0) x - 0 = f 0 (0) Differentiating: f 0 ( x ) = - 1 5 (32 - x ) - 4 / 5 ( - 1) = 1 5 (32 - x ) - 4 / 5 = f 0 (0) = 1 5 (32) - 4 / 5 = 1 5 ± 1 32 4 / 5 ² = 1 80 lim x 0 2 - 5 32 - x x = 1 80 2. To find the equation of the tangent line to the curve y 3 + yx 2 + x 2 - 3 y 2 = 0 at the point (0 , 3), we differentiate implicitly to find the slope of the tangent: 3 y 2 dy dx + x 2 dy dx + 2 xy + 2 x - 6 y dy dx = 0 At (0 , 3): (27 - 18) dy dx
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This note was uploaded on 01/18/2012 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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asst6-f10-solns - MATH 137 Assignment 6 - Solutions 1. (a)...

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