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, we need to use the formula:
$$
L=\int_a^b \sqrt{1+\left( \frac{dy}{dx} ight)^2} dx
$$
First, we need to find $\frac{dy}{dx}.$ We have:
$$
y=e^x \ln(3x)
$$
$$
\frac{dy}{dx}=e^x \frac{1}{x}+e^x \ln(3)
$$
Next, we need to find the interval of integration. We are given that the interval is $[1,2],$ which means $a=1$ and $b=2.$
Now we can find the integrand, i.e., $\sqrt{1+\left( \frac{dy}{dx} ight)^2}:$
$$
\sqrt{1+\left( \frac{dy}{dx} ight)^2} = \sqrt{1+\left( e^x \frac{1}{x}+e^x \ln(3) ight)^2}
$$
We can simplify this using the identity $\cosh^2 t-\sinh^2t=1:$
\begin{align*}
\sqrt{1+\left( e^x \frac{1}{x}+e^x \ln(3) ight)^2} &= \sqrt{\cosh^2(\ln(3)x)+\sinh^2(\ln(3)x)+2\sinh(\ln(3)x) \cosh(\ln(3)x) \frac{1}{x}}\
&= \sqrt{\left( \frac{1}{2}e^{2\ln(3)x}+\frac{1}{2}e^{-2\ln(3)x} ight) + \left( \frac{1}{2}e^{2\ln(3)x}-\frac{1}{2}e^{-2\ln(3)x} ight) \frac{1}{x}}\
&= \sqrt{\frac{e^{2\ln(3)x}}{2}+\frac{1}{2}\left( \frac{1}{x^2} ight) }
\end{align*}
Now we can write the integral and solve for the arc length:
\begin{align*}
L&=\int_1^2 \sqrt{\frac{e^{2\ln(3)x}}{2}+\frac{1}{2}\left( \frac{1}{x^2} ight) } dx\
&=\frac{1}{\sqrt{2}} \int_1^2 e^{\ln(3)x} \sqrt{1+\left( \frac{1}{(x\sqrt{2})^2} ight)} dx\
&=\frac{1}{\sqrt{2}} \int_1^2 e^{\ln(3)x} \sqrt{1+\left( \frac{1}{2x^2} ight)} dx\
&=\frac{1}{\sqrt{2}} \int_1^2 e^{\ln(3)x} \sqrt{\frac{2x^2+1}{2x^2}} dx\
&=\frac{1}{\sqrt{2}} \int_1^2 \sqrt{2+\frac{1}{2x^2}} e^{\ln(3)x} dx\
&=\frac{\sqrt{2}}{\sqrt{5}} \left[ \sqrt{5}e^{\ln(3)x} \sqrt{2+\frac{1}{2x^2}} -\frac{1}{3}e^{\ln(3)x}\sqrt{\frac{1}{2x^2}+2} ight]_1^2 \
&=\frac{1}{3} \left( 2\sqrt{10} \cdot 3^2 -\sqrt{2} \cdot \frac{1}{3} \cdot 3^2 ight)\
&=\boxed{	extbf{(B) } 11eac44f_ec9f_13d4_bedf_d370e6e587b2_TB7497_11} 
\end{align*}

Find the Arc Length of the Graph of the Function