Question

You are given the polar curve r = e^θ

Answer 1

The tangent to $r(\theta)=e^\theta$ has slope $\frac{\mathrm dy}{\mathrm dx}$, where

$\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}$

By the chain rule, we have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

and by the product rule,

$\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta$

$\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta$

so that with $\frac{\mathrm dr}{\mathrm d\theta}=e^\theta$, we get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}$

The tangent line is horizontal when the slope is 0; this happens for

$-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi$

where $n$ is any integer. In the interval $0\le\theta\le2\pi$, this happens for $n=1,2$, or

$\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4$

i.e at the points

$(r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)$

and

$(r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)$