find the arc length of the graph of the equation y = x^(3/2) from (0,0) to (4,8). you may use a calculator…

Explanation:

Step1: Find the derivative of y

Given $y = x^{\frac{3}{2}}$, by the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=\frac{3}{2}x^{\frac{1}{2}}$.

Step2: Use the arc - length formula

The arc - length formula for a function $y = f(x)$ from $x=a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Here, $a = 0$, $b = 4$, and $(y')^{2}=\frac{9}{4}x$. So, $L=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}dx$. Let $u = 1+\frac{9}{4}x$, then $du=\frac{9}{4}dx$ and $dx=\frac{4}{9}du$. When $x = 0$, $u = 1$; when $x = 4$, $u=1 + 9=10$. The integral becomes $L=\frac{4}{9}\int_{1}^{10}\sqrt{u}du$.

Step3: Integrate $\sqrt{u}$

Since $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=\frac{1}{2}$, $\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}+C$. So, $L=\frac{4}{9}\times\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{1}^{10}=\frac{8}{27}(10^{\frac{3}{2}}-1^{\frac{3}{2}})$.

Step4: Calculate the value

$L=\frac{8}{27}(10\sqrt{10}-1)\approx\frac{8}{27}(10\times3.1622777 - 1)=\frac{8}{27}(31.622777-1)=\frac{8}{27}\times30.622777\approx9.073$.

Answer:

$9.073$