## Precalculus (6th Edition) Blitzer

a) The average rate of change in the area is $8.1\pi \text{ square inches}$ per inch as radius changes from $4$ to $4.1\text{ inches}$ and $8.01\pi \text{ square inches}$ per inch as radius changes from $4$ to $4.01\text{ inches}$. b) The instantaneous rate of change of the area with respect to the radius is $8\pi \text{ square inches}$ per inch when $x=4\text{ inches}$.
(a) Consider that the area of a circle is described by the function $f\left( x \right)=\pi {{x}^{2}}$, radius changes from $\text{4 inches}$ to $4.1\text{ inches}$ Here, as the radius changes from $4$ to $4.1$, this implies $a=4$ and $h=0.1$ Now, compute the average rate of change of the area of a circle by substituting the above values in the difference quotient as follows: \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f4\left( 2+0.1 \right)-f\left( 4 \right)}{0.1} \\ & =\frac{f\left( 4.1 \right)-f\left( 4 \right)}{0.1} \end{align} Now, substitute $x=2.1$ and $x=2$ respectively in the function $f\left( x \right)=\pi {{x}^{2}}$ and simplify. \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\pi {{\left( 4.1 \right)}^{2}}-\pi {{\left( 4 \right)}^{2}}}{0.1} \\ & =\frac{16.81\pi -16\pi }{0.1} \\ & =8.1\pi \end{align} Thus, the average rate of change in the area is $8.1\pi$ square inches per inch as the radius changes from $4$ to $4.1\text{ inches}$. Now, consider that the radius changes from $4\text{ inches}$ to $4.01\text{ inches}$. Here, as radius changes from $4$ to $4.01$, this implies $a=4$ and $h=0.01$ Now, compute the average rate of change of the area of a circle by substituting the above values in the difference quotient as follows: \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 4+0.01 \right)-f\left( 4 \right)}{0.01} \\ & =\frac{f\left( 4.01 \right)-f\left( 4 \right)}{0.01} \end{align} Now, substitute $x=4.01$ and $x=4$ respectively in the function $f\left( x \right)=\pi {{x}^{2}}$ and simplify. \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\pi {{\left( 4.01 \right)}^{2}}-\pi {{\left( 4 \right)}^{2}}}{0.01} \\ & =\frac{16.0801\pi -16\pi }{0.01} \\ & =8.01\pi \end{align} Thus, the average rate of change in the area is $8.01\pi$ square inches per inch as the radius changes from $4$ to $4.01\text{ inches}$. (b) Consider that the area of a circle is described by the function $f\left( x \right)=\pi {{x}^{2}}$ Compute the derivative of $f\left( x \right)=\pi {{x}^{2}}$ 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{\pi {{\left( x+h \right)}^{2}}-\pi {{x}^{2}}}{h} \end{align} Now, simplify $\pi {{\left( x+h \right)}^{2}}$ 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{\pi \left( {{x}^{2}}+2xh+{{h}^{2}} \right)-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{x}^{2}}+2\pi xh+\pi {{h}^{2}}-\pi {{x}^{2}}}{h} \\ \end{align} Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2\pi x+\pi h \right)$ Apply the limits, \begin{align} & f'\left( x \right)=2\pi x+\pi \cdot 0 \\ & =2\pi x \end{align} Now, substitute $x=4$ in $f'\left( x \right)$ to compute the instantaneous change of “f” at $4$. \begin{align} & f'\left( 4 \right)=2\pi \cdot 4 \\ & =8\pi \end{align} Thus, the instantaneous rate of change of the area with respect to the radius is $8\pi \text{ square inches}$ per inch when $x=4\text{ inches}$.