Chapter 7.4, Problem 14E

To determine

To find: The length L of a curve between two points.

Answer to Problem 14E

The length L of the curve $x=\frac{y^4}{4}+\frac{1}{8y^2}$ between points $y=1$ to $y=2$ is $\approx 3.9$ units.

Explanation of Solution

Given data: The given curve is $x=\frac{y^4}{4}+\frac{1}{8y^2}$ and L extends from $y=1$ to $y=2$ .

Method/Formula used:

The length L of the curve $x=g\left(y\right)$ between points $y=c$ to $y=d$ is given by

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

Calculation:

The given curve is

  $x=\frac{y^4}{4}+\frac{1}{8y^2}$ … (2)

Differentiate (2) with respect to y ,

  $\begin{array}{c}\frac{dx}{dy}=\frac{d}{dy}\left(\frac{y^4}{4}\right)+\frac{d}{dy}\left(\frac{1}{8y^2}\right)\\ =\frac{4}{4}{y}^{4-1}+\left(-2\right)\frac{1}{8}{y}^{-2-1}\\ =y^3-\frac{1}{4}{y}^{-3}\\ =y^3-\frac{1}{4y^3}\end{array}$

Therefore,

  $\begin{array}{c}1+{\left(\frac{dx}{dy}\right)}^2=1+{\left(y^3-\frac{1}{4y^3}\right)}^2\\ =1+y^6-2\cdot y^3\cdot \frac{1}{4y^3}+\frac{1}{16y^6}\\ =1+y^6-\frac{1}{2}+\frac{1}{16y^6}\\ =y^6+\frac{1}{2}+\frac{1}{16y^6}\end{array}$

Simplify,

  $\begin{array}{c}1+{\left(\frac{dx}{dy}\right)}^2=\left(y^3\right)+2\cdot y^3\cdot \frac{1}{4y^3}+\frac{1}{16y^6}\\ =\left(y^3\right)+2\cdot y^3\cdot \frac{1}{4y^3}+\frac{1}{16y^6}\\ ={\left(y^3+\frac{1}{4y^3}\right)}^2\end{array}$

Therefore, the arc length L of the curve between points $y=1$ to $y=2$ is

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

Substitute, ${\left(y^3+\frac{1}{4y^3}\right)}^2$ for $1+{\left(\frac{dx}{dy}\right)}^2$

  $\begin{array}{c}L={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{{\left(y^3+\frac{1}{4y^3}\right)}^2}}dy\\ ={\displaystyle \underset{1}{\overset{2}{\int }}\left(y^3+\frac{1}{4y^3}\right)}dy\\ ={\left|\frac{y^{3+1}}{3+1}\right|}_1^2+\frac{1}{4}{\left|\frac{y^{-3+1}}{-3+1}\right|}_1^2\\ ={\left|\frac{y^4}{4}\right|}_1^2+\frac{1}{4}{\left|\frac{y^{-2}}{-2}\right|}_1^2\end{array}$

Simplify,

  $\begin{array}{c}L=\frac{1}{4}\left[{\left(2\right)}^4-{\left(1\right)}^4\right]-\frac{1}{8}\left[{\left(2\right)}^{-2}-{\left(1\right)}^{-2}\right]\\ =\frac{15}{4}+\frac{5}{32}\\ =\frac{125}{32}\\ \approx 3.9\end{array}$

The length L of the curve $x=\frac{y^4}{4}+\frac{1}{8y^2}$ between points $y=1$ to $y=2$ is $\approx 3.9$ units.