# Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1175: 42

The instantaneous rate of change of the volume with respect to the radius is $80\pi \text{ cubic feet}$ per foot when the radius is $x=8\text{ feet}$.

#### Work Step by Step

Consider that the volume of a right circular cylinder of height $5\text{ feet}$ and radius $x\text{ feet}$ is described by the function $f\left( x \right)=5\pi {{x}^{2}}$ Compute the derivative of $f\left( x \right)=5\pi {{x}^{2}}$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ as follows: To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)={{x}^{2}}$. \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{5\pi {{\left( x+h \right)}^{2}}-5\pi {{x}^{2}}}{h} \end{align} Now, simplify $5\pi {{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{5\pi \left( {{x}^{2}}+2xh+{{h}^{2}} \right)-5\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{5\pi {{x}^{2}}+10\pi xh+5\pi {{h}^{2}}-5\pi {{x}^{2}}}{h} \end{align} Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 10\pi x+5\pi h \right)$ Apply the limits, \begin{align} & f'\left( x \right)=10\pi x+5\pi \cdot 0 \\ & =10\pi x \end{align} Now, substitute $x=8$ in $f'\left( x \right)$ to compute the instantaneous change of “f” at $8$. \begin{align} & f'\left( 8 \right)=10\pi \cdot 8 \\ & =80\pi \end{align} Thus, the instantaneous rate of change of the volume with respect to the radius is $80\pi \text{ cubic feet}$ per foot when the radius is $x=8\text{ feet}$.

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