Chapter 8.4, Problem 13E

To determine

To Find: The length of the curve.

Answer to Problem 13E

The length of the curve $=\frac{53}{6}$ .

Explanation of Solution

Given information: curve $\text{x=}\left(\frac{y^3}{3}\right)\text{+}\frac{1}{4y}\text{ from y}=1\text{ to y}=3$ .

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 $\text{x=}\left(\frac{y^3}{3}\right)\text{+}\frac{1}{4y}\text{ from y}=1\text{ to y}=3$

  $\frac{dx}{dy}=y^2-\frac{1}{4y^2}$

Which is continuous on $\left[1,3\right]$ . Therefore

  $\begin{array}{l}\frac{dx}{dy}=y^2-\frac{1}{4y^2}\\ L={\displaystyle \underset{1}{\overset{3}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy\\ L={\displaystyle \underset{1}{\overset{3}{\int }}\sqrt{1+{\left(y^2-\frac{1}{4y^2}\right)}^2}}dy\\ L={\displaystyle \underset{1}{\overset{3}{\int }}\sqrt{\frac{16y^8+8y^4+1}{16y^4}}}dy\\ L={\displaystyle \underset{1}{\overset{3}{\int }}\frac{4y^4+1}{4y^2}}dy\\ L={\displaystyle \underset{1}{\overset{3}{\int }}{y}^2+\frac{1}{4y^2}}dy\\ L=\frac{53}{6}\end{array}$