Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1176: 77

Answer

The rate of change of the area of the circle is equal to the circumference of the circle.

Work Step by Step

The area of a circle is given by the formula $f\left( x \right)=\pi {{x}^{2}}$, where x is the radius of the circle. The rate of change is given by the formula, ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ Find the rate of change of the area of the circle by the above formula, $\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{\left( x+h \right)}^{2}}-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi \left( {{x}^{2}}+{{h}^{2}}+2xh \right)-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{x}^{2}}+\pi {{h}^{2}}+2\pi xh-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{h}^{2}}+2\pi xh}{h} \end{align}$ On further simplification $\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \pi h+2\pi x \right) \\ & =\pi \left( 0 \right)+2\pi x \\ & =2\pi x \end{align}$ It is known that the circumference of the circle is $2\pi x$, where x is the radius of the circle. Therefore, the rate of change of the area of the circle is equal to the circumference of the circle.
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