Find the arc length of the graph of the function over the interval .

Question

Find the arc length of the graph of the function  over the interval  . ​

Quizplus · Accepted Answer

To find the arc length of the graph of the function $f(x)=\sqrt{1+16x^2}$ over the interval $[-3,2],$ we use the formula:

$$L=\int_a^b\sqrt{1+\left[f'(x)ight]^2}\ dx$$

where $f'(x)$ is the derivative of $f(x).$

For $f(x)=\sqrt{1+16x^2},$ we have $f'(x)=\frac{8x}{\sqrt{1+16x^2}},$ so

\begin{align*}
L&=\int_{-3}^2\sqrt{1+\left[\frac{8x}{\sqrt{1+16x^2}}ight]^2}\ dx \
&=\int_{-3}^2\sqrt{1+\frac{64x^2}{1+16x^2}}\ dx \
&=\int_{-3}^2\sqrt{\frac{17x^2+1}{1+16x^2}}\ dx \
&=\int_{-3}^2\frac{\sqrt{17x^2+1}}{\sqrt{1+16x^2}}\ dx.
\end{align*}

We can't immediately integrate this, so we need to use a trigonometric substitution. Let $x=\frac{1}{4}	an	heta,$ so $dx=\frac{1}{4}\sec^2	heta\ d	heta,$ and we have:

\begin{align*}
L&=\int_{-3}^2\frac{\sqrt{17x^2+1}}{\sqrt{1+16x^2}}\ dx \
&=\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\frac{\sqrt{17\cdot\frac{1}{16}	an^2	heta+1}}{\sqrt{1+\frac{1}{4}	an^2	heta}}\cdot\frac{1}{4}\sec^2	heta\ d	heta \
&=\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\frac{\sqrt{17	an^2	heta+16}}{4\sqrt{	an^2	heta+4}}\sec^2	heta\ d	heta \
&=\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\frac{\sqrt{17	an^2	heta+16}}{4\sqrt{	an^2	heta+4}}\cdot\frac{	an^2	heta+4}{	an^2	heta+4}\sec^2	heta\ d	heta \
&=\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\frac{1}{4}\sqrt{17	an^2	heta+68}\ d	heta \
&=\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\frac{1}{4}\sqrt{17	an^2	heta+17\cdot 4}\ d	heta \
&=\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\frac{1}{4}\sqrt{17\sec^2	heta}\cdot\sqrt{1+4\cdot\frac{1}{17}}\ d	heta \
&=\frac{1}{4}\sqrt{17}\cdot\frac{\sqrt{21}}{\sqrt{17}}\int_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})}\sec	heta\ d	heta \
&=\frac{\sqrt{21}}{4}\ln\left|\sec	heta+	an	hetaight|\bigg|_{\arctan(-\frac{3}{4})}^{\arctan(\frac{2}{4})} \
&=\frac{\sqrt{21}}{4}\ln\left|\frac{\sqrt{65}+3}{5}\cdot\frac{5}{\sqrt{5}}ight| \
&=\frac{\sqrt{21}}{4}\ln(\sqrt{65}+3).
\end{align*}

Therefore, the answer is $\boxed{	extbf{(C) } \frac{\sqrt{21}}{4}\ln(\sqrt{65}+3)}.$

Find the Arc Length of the Graph of the Function