Chapter 8.4, Problem 11E

To determine

To Find: The length of the curve.

Answer to Problem 11E

The length of the curve $=12$ .

Explanation of Solution

Given information: curve $y=\frac{{\left(x^2+2\right)}^{\frac{3}{2}}}{3}\text{ from }x=0\text{ to }x=3$ .

Formula used: if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$

  $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx$

If y is a smooth function of x on $\left[a,b\right]$

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If x is a smooth function of y on $\left[c,d\right]$

Calculation:

curve $y=\frac{{\left(x^2+2\right)}^{\frac{3}{2}}}{3}\text{ from }x=0\text{ to }x=3$

  $\frac{dy}{dx}=x\sqrt{x^2+2}$

Which is continuous on $\left[0,3\right]$ . Therefore

  $\begin{array}{l}\frac{dy}{dx}=x\sqrt{x^2+2}\\ L={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx\\ L={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{1+{\left(x\sqrt{x^2+2}\right)}^2}}dx\\ L={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{x^4+2x^2+1}}dx\\ L={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{{\left(x^2+1\right)}^2}}dx\\ L={\displaystyle \underset{0}{\overset{3}{\int }}\left(x^2+1\right)}dx\\ L={\left[\frac{x^3}{3}+x\right]}_0^3\\ L=12\end{array}$