Question

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

Answer 1

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.