Chapter 9, Problem 1CRE

To determine

To calculate the arc length of the function $y=\frac{x^5}{10}+\frac{x^{-3}}{6}$ over the interval $\left[1,2\right]$.

Answer to Problem 1CRE

$\text{arc}\text{length}=3.24$

Explanation of Solution

Given:

$y=\frac{x^5}{10}+\frac{x^{-3}}{6}$, $\left[1,2\right]$.

Concept used: The arc length $s$ of $y=f\left(x\right)$ over $\left[a,b\right]$ is,

$s={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left[f^{\prime }\left(x\right)\right]}^2}}dx$

Calculation:

$y=f\left(x\right)=\frac{x^5}{10}+\frac{x^{-3}}{6}$

Differentiating with respect to $x$, we get

$\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{5x^4}{10}-\frac{3x^{-4}}{6}\\ =\frac{x^4}{2}-\frac{x^{-4}}{2}\end{array}$

Now calculate the value of $1+{\left[f^{\prime }\left(x\right)\right]}^2$.

$\begin{array}{c}1+{\left[f^{\prime }\left(x\right)\right]}^2=1+{\left[\frac{x^4}{2}-\frac{x^{-4}}{2}\right]}^2\\ =1+\left[\frac{x^8}{4}+\frac{x^{-8}}{4}-2\cdot \frac{x^4}{2}\cdot \frac{x^{-4}}{2}\right]\\ =1+\left[\frac{x^8}{4}+\frac{x^{-8}}{4}-\frac{1}{2}\right]\\ =\frac{4+x^8+x^{-8}-2}{4}\\ =\frac{x^8+x^{-8}+2}{4}\\ ={\left(\frac{x^4+x^{-4}}{2}\right)}^2\end{array}$

Thus, we get

$\begin{array}{c}\text{arc}\text{length}={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{1+{\left[f^{\prime }\left(x\right)\right]}^2}}dx\\ ={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{{\left(\frac{x^4+x^{-4}}{2}\right)}^2}dx}\\ ={\displaystyle \underset{1}{\overset{2}{\int }}\frac{x^4+x^{-4}}{2}dx}\\ =\frac{1}{2}{\left[\frac{x^5}{5}-\frac{1}{3x^3}\right]}_1^2\\ =\frac{1}{2}\left\{\left[\frac{32}{5}-\frac{1}{24}\right]-\left[\frac{1}{5}-\frac{1}{3}\right]\right\}\\ =\frac{1}{2}\left[6.4-0.0417-0.2+0.33\right]\\ =3.24\end{array}$