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Math 10C (Overholser) Quiz 2 Oct. 15, 2010 Name: TA: Section: Instructions: You may use a calculator but must show all steps needed to derive your answer. Answers with no work shown will receive no credit. No simplification of solutions is required for full credit: you may leave expressions such as sin(4) or 1 2 as they are. 1. (5 points) Find the Taylor polynomial of degree 2, P 2 ( x ), for the function f ( x ) = 4 1 + x for x near a = 0. Solution: f ( x ) = 4 1 + x = (1 + x ) 1 4 so f (0) = 1 f 0 ( x ) = 1 4 (1 + x ) - 3 4 so f 0 (0) = 1 4 f 00 ( x ) = 1 4 ( - 3 4 )(1 + x ) - 7 4 so f 00 (0) = - 3 16 The formula for P 2 ( x ) with a = 0 is f (0) + f 0 (0) x + f 00 (0) 2 x 2 so plugging in we get P 2 ( x ) = 1 + 1 4 x - 3 32 x 2

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2. (5 points) The volume V of material needed to construct a 3 foot long section of pipe can be expressed in terms of the outer radius x of the pipe and the inner radius y of the pipe by the formula V ( x, y ) = 3 π ( x 2 - y 2 ). (a) Graph the single variable function
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Unformatted text preview: V (2 ,y ). Please label your axes and mark several points on each axis. Figure 1: V (2 ,y ) = 3 π (4-y 2 ) (b) Graph the single variable function V ( x, 1). Please label your axes and mark several points on each axis. Figure 2: V ( x, 1) = 3 π ( x 2-1) (c) Explain in the context of the problem what the graphs in (a) and (b) tell you about the volume of material needed to construct a pipe as a function of the two radi. How does the volume of material needed change as a function of x , the outer radius? As a function of the inner radius y ? Solution: The volume of material needed to make the pipe is a decreasing function of the inner radius and an increasing function of the outer radius. Page 2...
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