Question

How do you sketch the graph of the polar equation and find the tangents at the pole of `r=3(1-costheta)`?

Answer 1

The tangents are given by `theta = 0 rarr and larr theta = pi`. See the explanation, for this ticklish problem.

Explanation:

graph{x^2+y^2+3x-3sqrt(x^2+y^2)=0 [-10, 10, -5, 5]}

The cartesian form `x^2+y^2+3x-3sqrt(x^2+y^2=0` is used for this

Socratic graph.

`t = 3( 1 - cos theta )`, with period `2pi`.

`r'= 3 sin theta`

As the point `(r, theta)` is moved on the curve, the tangent vector

rotates. Here, from start at the pole to the finish ( at the return to the

pole), `theta` completes one period `[0, 2pi]`.

With respect to the pole, r = 0 but `theta = 0 and 2pi`, giving the

same direction..

Now, the formula for the slope of the tangent

at `(r, theta ) = ( 3(1-cos theta), theta)`

`slope = (r'sintheta + r cos theta)/(r'cos theta- r sin theta )`.

Here, this is

`(3 sin^2 theta+3(1-cos theta) cos theta)/`
`(3sin theta cos theta-3(1-cos theta)sin theta)`,

At the pole (0, 0), the slope ( in the form `0/0` ) has the limit `0_+`.

Likewise, for the tangent at the finish `(0, 2pi)`, the limit is `0_-`.

The tangents are given by `theta = 0 and theta = pi`.

Upon reading my answer, some eyebrows might be raised.

My approach is practical and real. I have followed the tangent

vector, from start to finish, in the tracing of the cardioid.