How do you find the equation of the tangent lines to the polar curve #r=sin(2theta)# at #theta=2pi# ?

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Answer 1 · 2014-09-25T07:39:16

The equation of the tangent line is #y=0# (in red), which looks like this.

Graph of a tangent line

Let us look at some details.

Let us find the coordinates #(x_1,y_1)# of the point when #theta=2pi#.

Since

#{(x(theta)=sin(2 theta)cos theta),(y(theta)=sin(2theta)sin theta):}#,

we have #(x_1,y_1)=(x(2pi),y(2pi))=(0,0)#.

Let us now find the slope #m# of the tangent line.

By differentiating with respect to #theta#,

#{(x'(theta)=2cos(2theta)cos theta-sin(2 theta)sin theta),(y'(theta)=2cos(2theta)sin theta+sin(2theta)cos theta):}#.

So, we have #m={y'(2pi)}/{x'(2pi)}=0/2=0#

Hence, the equation of the tangent line is #y=0#.