How do you find the slope of the polar curve #r=3sec(2theta)# at #theta=pi/6# ?

1 Answer
Sep 21, 2014

The slope of the curve #r=3sec(2theta)# at #theta=pi/6# is #{3sqrt{3}}/5#.

Let us look at some details.

We can write

#{(x=rcos theta=3sec(2theta)cos theta),(y=rsin theta=3sec(2theta)sin theta):}#

#{dx}/{d theta}=3[2sec(2theta)tan(2theta)cdot cos theta+sec(2 theta)cdot(-sin theta)]#

#=3sec(2theta)[2tan(2theta)cos theta-sin theta]#

#{dy}/{d theta}=3[2sec(2theta)tan(2theta)cdot sin theta+sec(2theta)cdot cos theta]#

#=3sec(2theta)[2tan(2theta)sin theta+cos theta]#

So,

#{dy}/{dx}={{dy}/{d theta}}/{{dx}/{d theta}}={3sec(2theta)[2tan(2theta)sin theta+cos theta]}/{3sec(2theta)[2tan(2theta)cos theta-sin theta]}#

by cancelling out #3sec(2theta)#,

#={2tan(2theta)sin theta+cos theta}/{2tan(2theta)cos theta-sin theta}#

Now, we can find the slope #m# by pluggin in #theta=pi/6#.

#m={dy}/{dx}|_{theta=pi/6}={2(sqrt{3})(1/2)+sqrt{3}/2}/{2(sqrt{3})(sqrt{3})/2-1/2}={3sqrt{3}}/{5}#