Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.4 The Tangent Problem; The Derivative - 13.4 Assess Your Understanding - Page 917: 44


$16 \pi $ or $\approx 50.265~ft^2/ft$

Work Step by Step

Consider the tangent line that contains the point $(x, c)$. The slope the tangent line to the graph of $f(x)$ at $(x, c)$ can be written as $f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}$ We use the above equation to find the rate of change of surface area with respect to the radius. Consider $S(r)=4\pi r^2$. Then at $r=2$, we have: $S'(2) =\lim\limits_{r \to 2} \dfrac{S(r)-S(2)}{r-2} \\=\lim\limits_{r \to 2} \dfrac{4\pi r^2-(4 \pi)(2^2)}{r-2}\\=\lim\limits_{r \to 2} \dfrac{4\pi (r^2-4) }{r-2}\\=\lim\limits_{r \to 2} \dfrac{ 4\pi (r-2)(r+2) }{r-2}\\=\lim\limits_{r \to 2} 4 \pi (r+2) \\=16 \pi \\ \approx 50.265~ft^2/ft$
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