Chapter 11.3, Problem 73E

To determine

To prove: That the substitutions $x=r\cos \theta$ and $y=r\sin \theta$ transform the expression into $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta }$ .

Explanation of Solution

Given information:

The length of the polar curve: $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta }$

The given polar curves are $x=r\cos \theta$ and $y=r\sin \theta$ .

And the length of the polar curve is $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta }$ .

It is known that in a polar coordinate system if $r=f\left(\theta \right)$ is a curve, then the coordinates of the corresponding rectangular coordinate system are,

  $x=r\cos \theta =f\left(\theta \right)\cos \theta$ and $y=r\sin \theta =f\left(\theta \right)\sin \theta$

Now, $\begin{array}{c}{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(\frac{d}{d\theta }\left(f\left(\theta \right)\cos \theta \right)\right)}^2+{\left(\frac{d}{d\theta }\left(f\left(\theta \right)\sin \theta \right)\right)}^2\\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\cos \theta -f\left(\theta \right)\sin \theta \right)}^2+{\left(f^{\prime }\left(\theta \right)\sin \theta +f\left(\theta \right)\cos \theta \right)}^2\\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\cos \theta \right)}^2+{\left(f\left(\theta \right)\sin \theta \right)}^2-2f\left(\theta \right)f^{\prime }\left(\theta \right)\cos \theta \sin \theta +{\left(f^{\prime }\left(\theta \right)\sin \theta \right)}^2+{\left(f\left(\theta \right)\cos \theta \right)}^2+2f\left(\theta \right)f^{\prime }\left(\theta \right)\cos \theta \sin \theta \\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\right)}^2\left({\cos }^2\theta +{\sin }^2\theta \right)+{\left(f\left(\theta \right)\right)}^2\left({\cos }^2\theta +{\sin }^2\theta \right)\end{array}$

Simplify the above expression.

  $\begin{array}{c}{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\right)}^2\left({\cos }^2\theta +{\sin }^2\theta \right)+{\left(f\left(\theta \right)\right)}^2\left({\cos }^2\theta +{\sin }^2\theta \right)\\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\right)}^2\left(1\right)+{\left(f\left(\theta \right)\right)}^2\left(1\right)\\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2\end{array}$

Substitute $f\left(\theta \right)$ for $r$ and $\frac{dr}{d\theta }$ for $f^{\prime }\left(\theta \right)$ in the above expression.

  $\begin{array}{c}{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(f^{\prime }\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2\\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2=r^2+{\left(\frac{dr}{d\theta }\right)}^2\end{array}$

Take square root on each side of the above expression.

  $\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}=\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}$

So, the given length of the polar curve can be written as:

  $\begin{array}{l}L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta }\\ L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta }\end{array}$

Hence, the given statement is proved.