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

Jan 3, 2017

The tangents are given by $\theta = 0 \rightarrow \mathmr{and} \leftarrow \theta = \pi$. 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.

$t = 3 \left(1 - \cos \theta\right)$, with period $2 \pi$.

$r ' = 3 \sin \theta$

As the point $\left(r , \theta\right)$ 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 $\left[0 , 2 \pi\right]$.

With respect to the pole, r = 0 but $\theta = 0 \mathmr{and} 2 \pi$, giving the

same direction..

Now, the formula for the slope of the tangent

at $\left(r , \theta\right) = \left(3 \left(1 - \cos \theta\right) , \theta\right)$

$s l o p e = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$.

Here, this is

(3 sin^2 theta+3(1-cos theta) cos theta)/
$\left(3 \sin \theta \cos \theta - 3 \left(1 - \cos \theta\right) \sin \theta\right)$,

At the pole (0, 0), the slope ( in the form $\frac{0}{0}$ ) has the limit ${0}_{+}$.

Likewise, for the tangent at the finish $\left(0 , 2 \pi\right)$, the limit is ${0}_{-}$.

The tangents are given by $\theta = 0 \mathmr{and} \theta = \pi$.