Question

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

Answer 1

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

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`.