1. polar coordinates

if $r = e^{\theta}$ find points on curve where tangent is horizontal and vertical.

I got $(e^{n \pi - \frac{\pi}{4}}, \pi n - \frac{\pi}{4})$ for horizontal tangents.

Is this correct?

2. Hello, shilz222!

If $r = e^{\theta}$ find points on curve where tangent is horizontal and vertical.

I got $\left(e^{n \pi - \frac{\pi}{4}},\,\pi n - \frac{\pi}{4}\right)$ for horizontal tangents.

Is this correct? . . . . Yes!

First, we need $\frac{dy}{dx}$

There is a formula for this, but we can derive it.

We know the conversion formulas: . $\begin{array}{c}x \:=\:r\cos\theta \\ y \:=\:r\sin\theta\end{array}$

Then: . $\begin{array}{ccc}\frac{dx}{d\theta} & = & \text{-}r\sin\theta + r'\cos\theta \\ \frac{dy}{d\theta} & = & r\cos\theta + r'\sin\theta\end{array}$

. . Hence: . $\frac{dy}{dx} \;=\;\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \;=\;\frac{4\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta}$

Since $r = e^{\theta}$ and $r' = e^{\theta}$, we have: . $\frac{dy}{dx} \;=\;\frac{e^{\theta}\cos\theta + e^{\theta}\sin\theta}{\text{-}e^{\theta}\sin\theta + e^{\theta}\cos\theta}$

Divide top and bottom by $e^{\theta}\!:\;\;\frac{dy}{dx}\;=\;\frac{\cos\thet a + \sin\theta}{\text{-}\sin\theta + \cos\theta}$

We have horizontal tangents when the numerator equals zero.

. . $\cos\theta + \sin\theta \:=\:0\quad\Rightarrow\quad \sin\theta \:=\:-\cos\theta\quad\Rightarrow\quad\frac{\sin\theta}{\ cos\theta} \:=\:-1$

. . Therefore: . $\tan\theta\:=\:-1\quad\Rightarrow\quad\boxed{\theta \:=\:-\frac{\pi}{4} + \pi n}$

We have vertical tangents when the denominator equals zero.

. . $-\sin\theta + \cos\theta\:=\:0\quad\Rightarrow\quad\sin\theta \:=\:\cos\theta\quad\Rightarrow\quad\frac{\sin\the ta}{\cos\theta} \:=\:1$

. . Therefore: . $\tan\theta \:=\:1\quad\Rightarrow\quad\boxed{\theta \:=\:\frac{\pi}{4} + \pi n}$

3. But $x = e^{\theta} \cos \theta$ and $y = e^{\theta} \sin \theta$. Where the the $e^{\theta}$ go?