## Precalculus (6th Edition) Blitzer

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 $x=10\text{ inches}$.
(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 “f” 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 $x=10\text{ inches}$.