How do you find the slope of the polar curve #r=cos(2theta)# at #theta=pi/2# ?

Question

Luke Alvarez · Accepted Answer

By converting into parametric equations, #{(x(theta)=r(theta)cos theta=cos2theta cos theta),
(y(theta)=r(theta)sin theta=cos2theta sin theta):}# By Product Rule, #x'(theta)=-sin2theta cos theta-cos2theta sin theta# #x'(pi/2)=-sin(pi)cos(pi/2)-cos(pi)sin(pi/2)=1# #y'(theta)=-sin2thetasin theta+cos2theta cos theta# #y'(pi/2)=-sin(pi)sin(pi/2)+cos(pi)cos(pi/2)=0# So, the slope #m# of the curve can be found by #m={dy}/{dx}|_{theta=pi/2}={y'(pi/2)}/{x'(pi/2)}=0/1=0# I hope that this was helpful.

Evelyn Agard · Answer

undefined To find the slope of the polar curve $ r = \cos(2\theta) $ at $ \theta = \frac{\pi}{2} $, first, differentiate $ r $ with respect to $ \theta $ to find $ \frac{dr}{d\theta} $. Then, plug in $ \theta = \frac{\pi}{2} $ into $ \frac{dr}{d\theta} $ to find the slope at that point.

Charles Anctil · Answer

undefined To find the slope of the polar curve $r = \cos(2\theta)$ at $\theta = \frac{\pi}{2}$, differentiate the equation $r = \cos(2\theta)$ with respect to $\theta$ and then evaluate it at $\theta = \frac{\pi}{2}$.

$\frac{dr}{d\theta} = -2\sin(2\theta)$

Evaluate at $\theta = \frac{\pi}{2}$:

$\frac{dr}{d\theta}\Bigg|_{\theta = \frac{\pi}{2}} = -2\sin(\pi) = 0$

So, the slope of the polar curve $r = \cos(2\theta)$ at $\theta = \frac{\pi}{2}$ is $0$.