Chapter 6.4, Problem 19E

To determine

Find the arc length of the curve $x=\sin t\text{ and }y=t^2\text{ for }0\le t\le 2\pi$

Answer to Problem 19E

The arc length of the curve is $40.0562$

Explanation of Solution

Given information: Given curve $x=\sin t\text{ and }y=t^2\text{ for }0\le t\le 2\pi$

Calculation:

This time the case is parametric function so find both $\frac{dx}{dt}\text{ and }\frac{dy}{dt}$

  $\begin{array}{l}\frac{dx}{dy}=\frac{d}{dt}\left(\sin t\right)\\ ={\left(\sin t\right)}^{\prime }\\ =\cos t\\ \frac{dy}{dt}=\frac{d}{dt}\left(t^2\right)\\ ={\left(t^2\right)}^{\prime }\\ =2t\end{array}$

Use the formula:

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dy\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(\cos t\right)}^2+{\left(2t\right)}^2}}dy\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{\cos t^2+4t^2}}dy\\ \text{we have,}n=10\end{array}$

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 $f\left(t\right)=\sqrt{\cos t^2+4t^2},a=0,b=2\pi ,n=10$

  $\begin{array}{l}\Delta y=\frac{2\pi -0}{10}=\frac{2\pi }{10}=\frac{\pi }{5}\\ t_0=0+0\left(\frac{\pi }{5}\right)=0.0\\ t_1=0+1\left(\frac{\pi }{5}\right)=\frac{\pi }{5}\\ t_2=0+2\left(\frac{\pi }{5}\right)=\frac{2\pi }{5}\\ t_3=0+3\left(\frac{\pi }{5}\right)=\frac{3\pi }{5}\\ t_4=0+4\left(\frac{\pi }{5}\right)=\frac{4\pi }{5}\\ t_5=0+5\left(\frac{\pi }{5}\right)=\pi \\ t_6=0+6\left(\frac{\pi }{5}\right)=\frac{6\pi }{5}\\ t_7=0+7\left(\frac{\pi }{5}\right)=\frac{7\pi }{5}\\ t_8=0+8\left(\frac{\pi }{5}\right)=\frac{8\pi }{5}\\ t_9=0+9\left(\frac{\pi }{5}\right)=\frac{9\pi }{5}\\ t_{10}=0+10\left(\frac{\pi }{5}\right)=2\pi \end{array}$

  $\begin{array}{l}f\left(t_0\right)=1.0000\\ f\left(t_1\right)=1.4945\\ f\left(t_2\right)=2.5322\\ f\left(t_3\right)=3.7826\\ f\left(t_4\right)=5.0912\\ f\left(t_5\right)=6.3623\\ f\left(t_6\right)=7.5831\\ f\left(t_7\right)=8.8019\\ f\left(t_8\right)=10.0578\\ f\left(t_9\right)=11.3386\\ f\left(t_{10}\right)=12.6061\end{array}$

Substitute the above Simpson’s rule:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{5}{\int }}f\left(x\right)}dx\approx T_n=\frac{\Delta x}{3}\left[\begin{array}{l}f\left(y_0\right)+4f\left(y_1\right)+2f\left(y_2\right)+4f\left(y_3\right)+2f\left(y_4\right)+4f\left(y_5\right)+\\ 2f\left(y_6\right)+4f\left(y_7\right)+2f\left(y_8\right)+4f\left(y_9\right)+f\left(y_{10}\right)\end{array}\right]\\ =\frac{0.1}{3}\left[\begin{array}{l}f\left(1.0\right)+4f\left(1.1\right)+2f\left(1.2\right)+4f\left(1.3\right)+2f\left(1.4\right)+4f\left(1.5\right)+\\ 2f1.\left(6\right)+4f\left(1.7\right)+2f\left(1.8\right)+4f\left(1.9\right)+f\left(2.0\right)\end{array}\right]\\ =40.0562\end{array}$

Thus, the arc length of the curve is $40.0562$