Answer
a) The average rate of change in the area is $20.1\text{ square inches}$ per inch as x changes from $10$ to $10.1\text{ inches}$ and $20.01\text{ square inches}$ per inch as x changes from $10$ to $10.01\text{ inches}$.
b) The instantaneous rate of change of the area with respect to x is $20\text{ square inches}$ per inch when $x=10\text{ inches}$.
Work Step by Step
(a)
Consider that the area of the square is described by the function $f\left( x \right)={{x}^{2}}$, x changes from $10\text{ inches}$ to $10.1\text{ inches}$
Here, as x changes from $10$ to $10.1$, this implies $a=10$ and $h=0.1$
Now, compute the average rate of change of the area of a square by substituting the above values in the difference quotient as follows:
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 10+0.1 \right)-f\left( 10 \right)}{0.1} \\
& =\frac{f\left( 10.1 \right)-f\left( 10 \right)}{0.1}
\end{align}$
Now, substitute $x=10.1$ and $x=10$ respectively in the function $f\left( x \right)={{x}^{2}}$ and simplify.
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{{{10.1}^{2}}-{{10}^{2}}}{0.1} \\
& =\frac{102.01-100}{0.1} \\
& =20.1
\end{align}$
Thus, the average rate of change in the area is $20.1$ square inches per inch as x changes from $10$ to $10.1\text{ inches}$.
Now, consider that x changes from $10\text{ inches}$ to $10.01\text{ inches}$.
Here, as x changes from $10$ to $10.01$, this implies $a=10$ and $h=0.01$
Now, compute the average rate of change of the area of a square by substituting the above values in the difference quotient as follows:
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 10+0.01 \right)-f\left( 10 \right)}{0.01} \\
& =\frac{f\left( 10.01 \right)-f\left( 10 \right)}{0.01}
\end{align}$
Now, substitute $x=10.01$ and $x=10$ respectively in the function $f\left( x \right)={{x}^{2}}$ and simplify.
$\begin{align}
& \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{{{10.01}^{2}}-{{10}^{2}}}{0.01} \\
& =\frac{100.2001-100}{0.01} \\
& =20.01
\end{align}$
Thus, the average rate of change in the area is $20.01$ square inches per inch as x changes from $10$ to $10.01\text{ inches}$.
(b)
Consider that the area of the square is described by the function $f\left( x \right)={{x}^{2}}$
Compute the derivative of $f\left( x \right)={{x}^{2}}$ 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 $f\left( x \right)={{x}^{2}}$.
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h}
\end{align}$
Now, simplify ${{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}+2xh+{{h}^{2}}-{{x}^{2}}}{h}$
Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives,
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2x+h \right)$
Apply the limits,
$\begin{align}
& f'\left( x \right)=2x+0 \\
& =2x
\end{align}$
Now, substitute $x=10$ in $f'\left( x \right)$ to compute the instantaneous change of âfâ at $10$.
$\begin{align}
& f'\left( 6 \right)=2\cdot 10 \\
& =20
\end{align}$
Thus, the instantaneous rate of change of the area with respect to x is $20\text{ square inches}$ per inch when $x=10\text{ inches}$.
