Chapter H.1, Problem 54E

To determine

To find: the points on given curve

Answer to Problem 54E

Vertical: $\left(e^{\frac{\pi }{4}+n\pi },\frac{\pi }{4}+n\pi \right)$ horizontal: $\left(e^{\frac{3\pi }{4}+n\pi },\frac{3\pi }{4}+n\pi \right)$

Explanation of Solution

Given:

  $r=e^{\theta }$

Calculation:

Find $r^{\prime }$ .

Convert $r$ to parametric rectangular form, and find their derivatives.

  $\begin{array}{l}r=e^{\theta }\\ r^{\prime }=e^{\theta }\\ r^{\prime }=\frac{dr}{d\theta }=e^{\theta }\\ x=r\cos \theta \\ =e^{\theta }\cos \theta \\ \frac{dx}{d\theta }=e^{\theta }\left(-\sin \theta \right)+e^{\theta }\cos \theta \\ =e^{\theta }\left(\cos \theta -\sin \theta \right)\\ y=r\sin \theta \\ =e^{\theta }\sin \theta \\ \frac{dy}{d\theta }=e^{\theta }\cos \theta +e^{\theta }\sin \theta \\ =e^{\theta }\left(\cos \theta +\sin \theta \right)\end{array}$

Find $\frac{dy}{dx}$

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ =\frac{e^{\theta }\left(\cos \theta +\sin \theta \right)}{e^{\theta }\left(\cos \theta -\sin \theta \right)}\\ =\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\end{array}$

The slope of the tangent is horizontal when the numerator of $\frac{dy}{dx}$ is $0$ .

( $n$ is an integer)

  $\begin{array}{l}0=\cos \theta +\sin \theta \\ \cos \theta =-\sin \theta \\ \theta =\frac{3\pi }{4}+n\pi , n\in \mathbb{Z}\end{array}$

The slope of the tangent is vertical when the denominator of $\frac{dy}{dx}$ is $0$ .

( $n$ is an integer)

  $\begin{array}{l}0=\cos \theta -\sin \theta \\ \cos \theta =\sin \theta \\ \theta =\frac{\pi }{4}+n\pi , n\in \mathbb{Z}\end{array}$

The vertical tangents are at

  $\theta =\frac{\pi }{4}+n\pi$

Horizontal tangents are at

  $\theta =\frac{3\pi }{4}+n\pi$

  

Plug into $r=e^{\theta }$ to find the $\left(r,\theta \right)$ points.

Vertical tangents at

  $\left(e^{\frac{\pi }{4}+n\pi },\frac{\pi }{4}+n\pi \right)$

Horizontal tangents are at

  $\left(e^{\frac{3\pi }{4}+n\pi },\frac{3\pi }{4}+n\pi \right)$

Conclusion:

Therefore, the vertical point is $\left(e^{\frac{\pi }{4}+n\pi },\frac{\pi }{4}+n\pi \right)$ and the horizontal point is $\left(e^{\frac{3\pi }{4}+n\pi },\frac{3\pi }{4}+n\pi \right)$