Chapter 6.4, Problem 6E

To determine

Find the length of the curves $x=t\cos t,y=t\sin t0\le t\le 2\pi$

Answer to Problem 6E

The length of the curve is $21.2563$

Explanation of Solution

Given information: Given curves $x=t\cos t,y=t\sin t0\le t\le 2\pi$

Calculation:

To find the length of the curve $x=t\cos t,y=t\sin t0\le t\le 2\pi$ ,

Use the formula:

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}\text{ dt}$ .

Know that:

  $\begin{array}{l}\frac{dx}{dt}={\left(t\cos t\right)}^{\prime }{}_t=\cos t-t\sin t,\\ \frac{dy}{dt}={\left(t\sin t\right)}^{\prime }{}_t=\sin t+t\cos t.\end{array}$

So,

  $\begin{array}{l}L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(\cos t-t\sin t\right)}^2+{\left(\sin t+t\cos t\right)}^2}}dt\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(\cos t\right)}^2-2t\sin t\cos t+t^2{\left(\sin t\right)}^2+{\left(\sin t\right)}^2+2t\sin t\cos t+t^2{\left(\cos t\right)}^2}}dt\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{1+t^2}}dt\left({\left(\sin x\right)}^2+{\left(\cos x\right)}^2=1\right)\\ \approx 21.2563\end{array}$

Thus the length of the curve is $21.2563$