Chapter H.1, Problem 70E

a.

To determine

To show: The angle between the tangent line and the radial line is $\psi =\frac{\pi }{4}$

Explanation of Solution

Given: Let point P on the curve $r=f\left(\theta \right)$ , If $\psi$ is the angle between the tangent line at P and the radial line OP .

  

  $\tan \psi =\frac{r}{dr/d\theta }$

Slope of tangent, $m=\tan \varphi$

Slope of tangent at point P on polar curve, $\frac{dy}{dx}=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }$

  $\tan \psi =\frac{r}{dr/d\theta }$

Where, $r=e^{\theta }$ and $\frac{dr}{d\theta }=e^{\theta }$

Substitute into $\tan \psi =\frac{r}{dr/d\theta }$

  $\begin{array}{c}\tan \psi =\frac{e^{\theta }}{e^{\theta }}\\ \tan \psi =1\\ \psi ={\tan }^{-1}1\\ \psi =\frac{\pi }{4}\end{array}$

Hence proved

b.

To determine

To Illustrate: The part (a) by graphing for point $\theta =0\text{ and }\theta =\frac{\pi }{2}$ .

Explanation of Solution

Given: Let point P on the curve $r=f\left(\theta \right)$ , If $\psi$ is the angle between the tangent line at P and the radial line OP .

  

  $\tan \psi =\frac{r}{dr/d\theta }$

Slope of tangent, $m=\tan \varphi$

Slope of tangent at point P on polar curve, $\frac{dy}{dx}=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }$

  $\tan \psi =\frac{r}{dr/d\theta }$

Where, $r=e^{\theta }$ and $\frac{dr}{d\theta }=e^{\theta }$

Now graph for $\theta =0\text{ and }\theta =\frac{\pi }{2}$ of curve and tangent.

  

c.

To determine

To prove: Any polar curve $r=f\left(\theta \right)$ with the property that the angle $\psi$ between the radial line and the tangent line is a constant must be of the form $r=C{e}^{k\theta }$ .

Explanation of Solution

Given: Let point P on the curve $r=f\left(\theta \right)$ , If $\psi$ is the angle between the tangent line at P and the radial line OP .

  

  $\tan \psi =\frac{r}{dr/d\theta }$

Slope of tangent, $m=\tan \varphi$

Slope of tangent at point P on polar curve, $\frac{dy}{dx}=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }$

  $\tan \psi =\frac{r}{dr/d\theta }$

Let $\tan \psi$ is constant k

  $\begin{array}{c}\tan \psi \frac{dr}{d\theta }=r\\ \tan \psi \frac{dr}{r}=d\theta \\ \tan \psi \ln r=\theta +B\left[B\text{ is any constant}\right]\\ \ln r=\frac{\theta +B}{\tan \psi }\\ r=e^{\frac{\theta +B}{\tan \psi }}\\ r=e^{\frac{\theta }{\tan \psi }}{e}^{\frac{B}{\tan \psi }}\\ r=C{e}^{k\theta }\left[\frac{1}{\tan \psi }=k,e^{\frac{B}{\tan \psi }}=C\right]\end{array}$

Hence proved