find the arclength of $y = 3x^{3/2}$ on $2leq xleq3$

Explanation:

Step1: Find the derivative of $y$

The derivative of $y = 3x^{3/2}$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=\frac{d}{dx}(3x^{3/2})=3\times\frac{3}{2}x^{1/2}=\frac{9}{2}\sqrt{x}$.

Step2: Use the arc - length formula

The arc - length formula for a function $y = f(x)$ on the interval $[a,b]$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Here, $a = 2$, $b = 3$, and $(y')^{2}=\frac{81}{4}x$. So, $L=\int_{2}^{3}\sqrt{1+\frac{81}{4}x}dx$. Let $u = 1+\frac{81}{4}x$, then $du=\frac{81}{4}dx$ and $dx=\frac{4}{81}du$. When $x = 2$, $u=1+\frac{81\times2}{4}=1+\frac{162}{4}=\frac{4 + 162}{4}=\frac{166}{4}=\frac{83}{2}$. When $x = 3$, $u=1+\frac{81\times3}{4}=1+\frac{243}{4}=\frac{4+243}{4}=\frac{247}{4}$.

Step3: Evaluate the integral

$L=\frac{4}{81}\int_{\frac{83}{2}}^{\frac{247}{4}}\sqrt{u}du=\frac{4}{81}\times\frac{2}{3}u^{3/2}\big|{\frac{83}{2}}^{\frac{247}{4}}=\frac{8}{243}\left[u^{3/2}\right]{\frac{83}{2}}^{\frac{247}{4}}$. $=\frac{8}{243}\left(\left(\frac{247}{4}\right)^{3/2}-\left(\frac{83}{2}\right)^{3/2}\right)=\frac{8}{243}\left(\frac{247^{3/2}}{8}-\frac{83^{3/2}}{2\sqrt{2}}\right)=\frac{1}{243}\left(\frac{247^{3/2}}{1}-\frac{4\times83^{3/2}}{\sqrt{2}}\right)$. $=\frac{1}{243}\left(247\sqrt{247}- \frac{4\times83\sqrt{83}}{\sqrt{2}}\right)=\frac{1}{243}\left(247\sqrt{247}-2\sqrt{2}\times83\sqrt{83}\right)$. $=\frac{1}{243}\left(247\sqrt{247}-166\sqrt{166}\right)$.

Answer:

$\frac{1}{243}\left(247\sqrt{247}-166\sqrt{166}\right)$