Answer
a) The average rate of change in the area is $4.1\pi \text{ square inches}$ per inch as radius changes from $2$ to $2.1\text{ inches}$ and $4.01\pi \text{ square inches}$ per inch as radius changes from $2$ to $2.01\text{ inches}$.
b) The instantaneous rate of change of the area with respect to the radius is $4\pi \text{ square inches}$ per inch when $x=2\text{ inches}$.
Work Step by Step
(a)
Consider that the area of a circle is described by the function $f\left( x \right)=\pi {{x}^{2}}$, radius changes from $\text{2 inches}$ to $2.1\text{ inches}$
Here, as the radius changes from $2$ to $2.1$, this implies $a=2$ 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:
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 2+0.1 \right)-f\left( 2 \right)}{0.1} \\
& =\frac{f\left( 2.1 \right)-f\left( 2 \right)}{0.1}
\end{align}$
Now, substitute $x=2.1$ and $x=2$ respectively in the function $f\left( x \right)=\pi {{x}^{2}}$ and simplify.
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\pi {{\left( 2.1 \right)}^{2}}-\pi {{\left( 2 \right)}^{2}}}{0.1} \\
& =\frac{4.41\pi -4\pi }{0.1} \\
& =4.1\pi
\end{align}$
Thus, the average rate of change in the area is $4.1\pi $ square inches per inch as the radius changes from $2$ to $2.1\text{ inches}$.
Now, consider that the radius changes from $2\text{ inches}$ to $2.01\text{ inches}$.
Here, as radius changes from $2$ to $2.01$, this implies $a=2$ 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:
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 2+0.01 \right)-f\left( 2 \right)}{0.01} \\
& =\frac{f\left( 2.01 \right)-f\left( 2 \right)}{0.01}
\end{align}$
Now, substitute $x=2.01$ and $x=2$ respectively in the function $f\left( x \right)=\pi {{x}^{2}}$ and simplify.
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\pi {{\left( 2.01 \right)}^{2}}-\pi {{\left( 2 \right)}^{2}}}{0.01} \\
& =\frac{4.0401\pi -4\pi }{0.01} \\
& =4.01\pi
\end{align}$
Thus, the average rate of change in the area is $4.01\pi $ square inches per inch as the radius changes from $2$ to $2.01\text{ inches}$.
(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 $x=2$ in $f'\left( x \right)$ to compute the instantaneous change of âfâ at $2$.
$\begin{align}
& f'\left( 2 \right)=2\pi \cdot 2 \\
& =4\pi
\end{align}$
Thus, the instantaneous rate of change of the area with respect to the radius is $4\pi \text{ square inches}$ per inch when $x=2\text{ inches}$.
