Chapter 6.4, Problem 18E

To determine

Find the arc length of the curve $x=y+\text{}\sqrt{y}\text{ for }0\le x\le \text{5}$

Answer to Problem 18E

The arc length of the curve is $1.0420$

Explanation of Solution

Given information: Given curve $x=y+\text{}\sqrt{y}\text{ for }0\le x\le \text{5}$

Calculation:

Want to find the length of the line $x=y+\text{}\sqrt{y}\text{ for }0\le x\le \text{5}$

First we want to find $\frac{dx}{dy}$

  $\begin{array}{l}\frac{dx}{dy}=\frac{d}{dy}\left(y+\sqrt{y}\right)\\ =y^{\prime }+{\left(\sqrt{y}\right)}^{\prime }\\ =1+\frac{1}{2\sqrt{y}}\end{array}$

Using the formula:

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left(\frac{dx}{dy}\right)}^2+1}}dy\\ ={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left[1+\frac{1}{2\sqrt{y}}\right]}^2+1}}dy\\ \text{we have}\\ n=10\end{array}$

Recall Simpson’s rule:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\approx S_n=\Delta x\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+4f\left(x_3\right)+\mathrm{...}+f\left(x_{n-2}\right)+4f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

Where $\Delta x=\frac{b-a}{n},$

  $n$ must be even

So in our case

  $f\left(y\right)=\sqrt{{\left[1+\frac{1}{2\sqrt{y}}\right]}^2+1},a=1,b=2,n=10$

  $\begin{array}{l}\Delta y=\frac{2-1}{10}=\frac{1}{10}=0.1\\ y_0=1+0\left(0.1\right)=1.0\\ y_1=1+1\left(0.1\right)=1.1\\ y_2=1+2\left(0.1\right)=1.2\\ y_3=1+3\left(0.1\right)=1.3\\ y_4=1+4\left(0.1\right)=1.4\\ y_5=1+5\left(0.1\right)=1.5\\ y_6=1+6\left(0.1\right)=1.6\\ y_7=1+7\left(0.1\right)=1.7\\ y_8=1+8\left(0.1\right)=1.8\\ y_9=1+9\left(0.1\right)=1.9\\ y_{10}=1+10\left(0.1\right)=2.0\end{array}$

  $\begin{array}{l}f\left(1.0\right)=1.0541\\ f\left(1.1\right)=1.0508\\ f\left(1.2\right)=1.0480\\ f\left(1.3\right)=1.0454\\ f\left(1.4\right)=1.0432\\ f\left(1.5\right)=1.0412\\ f\left(1.6\right)=1.0394\\ f\left(1.7\right)=1.0377\\ f\left(1.8\right)=1.0362\\ f\left(1.9\right)=1.0348\end{array}$

Substitute the above Simpson’s rule:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{5}{\int }}f\left(x\right)}dx\approx T_n=\frac{\Delta x}{3}\left[\begin{array}{l}f\left(y_0\right)+4f\left(y_1\right)+2f\left(y_2\right)+\\ 4f\left(y_3\right)+2f\left(y_4\right)+4f\left(y_5\right)+2f\left(y_6\right)+4f\left(y_7\right)+2f\left(y_8\right)+4f\left(y_9\right)+f\left(y_{10}\right)\end{array}\right]\\ =\frac{0.1}{3}\left[\begin{array}{l}f\left(1.0\right)+4f\left(1.1\right)+2f\left(1.2\right)+4f\left(1.3\right)+2f\left(1.4\right)+4f\left(1.5\right)+2f1.\left(6\right)+\\ 4f\left(1.7\right)+2f\left(1.8\right)+4f\left(1.9\right)+f\left(2.0\right)\end{array}\right]\\ =1.0420\end{array}$

Thus, the arc length of the curve is $1.0420$