Chapter 3.1, Problem 27E

To determine

To find : $dr/d\theta$ .

Answer to Problem 27E

  $\frac{dr}{d\theta }=\frac{\left(\theta \cos \theta +\sin \theta \right)}{2\sqrt{\theta \sin \theta }}$

Explanation of Solution

Given information :

  $r=\sqrt{\theta \sin \theta }$ .

Formula needed :

Product rule: - $\frac{d}{dx}\left(uv\right)=v\frac{d}{dx}\left(u\right)+u\frac{d}{dx}\left(v\right)$ .

Chain rule of derivative

If $y=f\left(u\right),u=g\left(x\right)$ , then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ .

Assume $u=\theta \sin \theta$ then $r=\sqrt{u}$ .

Differentiate $r=\sqrt{u}$ with respect to $\theta$ .

  $\begin{array}{c}\frac{dr}{d\theta }=\frac{d}{d\theta }\left(\sqrt{u}\right)\\ =\frac{1}{2\sqrt{u}}\cdot \frac{du}{d\theta }\end{array}$

Differentiate $u=\theta \sin \theta$ with respect to $\theta$ and apply product rule.

  $\begin{array}{c}\frac{du}{d\theta }=\frac{d}{d\theta }\left(\theta \sin \theta \right)\\ =\theta \cdot \frac{d}{d\theta }\left(\sin \theta \right)+\sin \theta \cdot \frac{d}{d\theta }\left(\theta \right)\\ =\theta \cdot \cos \theta +\sin \theta \cdot 1\\ =\theta \cos \theta +\sin \theta \end{array}$

Substitute the value of $u$ and $\frac{du}{d\theta }$ in $\frac{dr}{d\theta }=\frac{1}{2\sqrt{u}}\cdot \frac{du}{d\theta }$ .

  $\begin{array}{c}\frac{dr}{d\theta }=\frac{1}{2\sqrt{\theta \sin \theta }}\cdot \left(\theta \cos \theta +\sin \theta \right)\\ =\frac{\left(\theta \cos \theta +\sin \theta \right)}{2\sqrt{\theta \sin \theta }}\end{array}$