Question

Find the points on the given curve where the tangent line is horizontal or vertical
r=e^(theta)

Answer 1

ur soo pretty

Answer 2

$$
r=e^\theta
$$
using polar coordinates
$$
x=r\cos(\theta)=e^\theta\cos\theta\\
y=r\sin(\theta) =e^\theta\sin\theta
$$
Horizontal tangents occur where \(\cfrac{dy}{dx}=0\) and vertical tangents occur where \(\cfrac{dy}{dx} \to \infty \implies dx \to 0\).
$$
\cfrac{dy}{dx}=\cfrac{\cfrac{dy}{d\theta}}{\cfrac{dx}{d\theta}}\\
\cfrac{dy}{dx}=\cfrac{e^\theta\cos\theta +e^\theta\sin\theta}{-e^\theta\sin\theta+e^\theta\cos\theta}
$$
So horizontal tangents occur where
$$
e^\theta\cos\theta +e^\theta\sin\theta=0\\
\implies \theta=\cfrac{-\pi}{4}
$$
Vertical tangents occur where
$$
-e^\theta\sin\theta+e^\theta\cos\theta=0\\
\implies \theta=\cfrac{\pi}{4}
$$
So, there is a horizontal tangent at
$$
<-\cfrac{\pi}{4},e^{\left(-\cfrac{\pi}{4}\right )}>
$$

and a vertical tangent at
$$
<\cfrac{\pi}{4},e^{\left(\cfrac{\pi}{4}\right )}>
$$
Which is the answer to your question.

Note that in rectangular coordinates, the horizontal an vertical tangents occur at
$$
<e^{-\left(\cfrac{\pi}{4}\right )}\cos\cfrac{\pi}{4},e^{-\left(\cfrac{\pi}{4}\right )}\sin\cfrac{\pi}{4}>=<0.32,-0.32>
$$
and
$$
<e^{\left(\cfrac{\pi}{4}\right )}\cos\cfrac{\pi}{4},e^{\left(\cfrac{\pi}{4}\right )}\sin\cfrac{\pi}{4}>=<1.55,1.55>
$$
respectively.

Here is a plot of your equation:

http://www.wolframalpha.com/input/?i=r%3De^theta

If you enable interactivity, you can confirm roughly our work here.

Does all this make sense?

Answer 3

Yes, thank you!