How do you find the slope of the polar curve #r=1+sin(theta)# at #theta=pi/4# ?

1 Answer
Oct 10, 2014

By converting into parametric equations,

#{(x(theta)=r(theta)cos theta=(1+sin theta)cos theta),(y(theta)=r(theta)sin theta=(1+sin theta)sin theta):}#

By finding the derivatives using Product Rule,

#x'(theta)=cos theta cdot cos theta + (1+sin theta)cdot(-sin theta)#

#=(cos^2theta-sin^2theta)-sin theta#

#=cos2theta-sin theta#

#Rightarrow x'(pi/4)=cos(pi/2)-sin(pi/4)=-1/sqrt{2}#

#y'(theta)=cos theta cdot sin theta+(1+sin theta)cdot cos theta#

#=2sin theta cos theta+cos theta#

#=sin2theta+cos theta#

#Rightarrow y'(pi/4)=sin(pi/2)+cos(pi/4)=1+1/sqrt{2}#

The slope #m# we are looking for is:

#m={dy}/{dx}|_{theta=pi/4}={y'(pi/4)}/{x'(pi/4)}={1+1/sqrt{2}}/{-1/sqrt{2}} =-(sqrt{2}+1)#

I hope that this was helpful.