How do you find the slope of the polar curve $r = cos (2 θ)$ $r=cos(2theta)$ at $θ = \frac{π}{2}$ $theta=pi/2$ ?

Question

71288 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-16T01:35:01

By converting into parametric equations,

${(x(theta)=r(theta)cos theta=cos2theta cos theta), (y(theta)=r(theta)sin theta=cos2theta sin theta):}$

$x'(theta)=-sin2theta cos theta-cos2theta sin theta$

$x'(pi/2)=-sin(pi)cos(pi/2)-cos(pi)sin(pi/2)=1$

$y'(theta)=-sin2thetasin theta+cos2theta cos theta$

$y'(pi/2)=-sin(pi)sin(pi/2)+cos(pi)cos(pi/2)=0$

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.

How do you find the slope of the polar curve r=cos(2theta)r=cos(2θ)r=cos(2theta) at theta=pi/2θ=π2theta=pi/2 ?