# How do you sketch the graph of the polar equation and find the tangents at the pole of r=2(1-sintheta)?

Jan 7, 2017

Tracing the cardioid from pole to pole, in the positive anticlockwise sense, we start in the direction $\theta = - \frac{\pi}{4}$, from r = 0, and return in the direction $\theta = \frac{\pi}{4}$.

#### Explanation:

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

I think that it is my duty to answer in the way that is right to me.

In polar curves through the pole, some do have a node at the pole.

This node has two tangents ( including the pair, in the opposite

directions ).

In (2-D or 3-D) polar coordinates, the pole is a point that has to be

assigned an ad hoc direction. for the chosen application. Here it is

graphing..

In general, $r = a \left(1 + \cos \left(\theta - \alpha\right)\right)$ represents a cardioid through

the pole r = 0 ( theta?), with axis of symmetry along

$\theta = \alpha$.

Assigning a = 2 and $\alpha = - \frac{\pi}{2}$,

the given equation $r = 2 \left(1 - \sin \theta\right)$ is obtained.

Here, the graph is periodic, with period $2 \pi$.

At the pole,$r = 2 \left(1 - \sin \theta\right) = 0$, giving

$\theta = \frac{\pi}{2} \mathmr{and} \frac{5}{2} \pi$ in one period $\left[\frac{\pi}{2} , \frac{5}{2} \pi\right]$

Tracing any curve is done by moving the marker in the direction of

the tangent.

Now, the slope of the tangent is $y ' = \frac{\mathrm{dy}}{\mathrm{dx}}$, in cartesian form.

In polar form, this is

$\frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$

$= \frac{\cos 2 \theta - \sin 2 \theta + 1}{- \cos 2 \theta - \sin 2 \theta - 1}$, upon using

$r = 2 \left(1 - \sin \theta\right)$ and simplification.

Starting at $\left(0 , \frac{\pi}{2}\right)$, initial slope is the limit as $\theta \to \frac{\pi}{2} = - 1$,

using L'Hospital rule.

Likewise, at the return to the pole ( now $\theta = \frac{5}{2} \pi$ ),

the slope is 1. We can say that these directions are given by #theta =

-pi/4$\mathmr{and} \theta = \frac{\pi}{4}$, and make it easy to realize.

Tracing the cardioid from pole to pole, in the positive anticlockwise

sense, we start in the direction $\theta = - \frac{\pi}{4}$, from $O \left(0 , - \frac{\pi}{4}\right)$

and return in the direction $\theta = \frac{\pi}{4}$, to $O \left(0 , \frac{\pi}{4}\right)$.

Verification : Practical.