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

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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.