Question

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

Answer 1

Tracing the cardioid from pole to pole, in the positive anticlockwise sense, we start in the direction `theta = -pi/4`, from r = 0, and return in the direction `theta=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 (1+cos (theta-alpha))` represents a cardioid through

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

`theta = alpha`.

Assigning a = 2 and `alpha = -pi/2`,

the given equation `r = 2 ( 1 - sin theta )` is obtained.

Here, the graph is periodic, with period `2pi`.

At the pole,`r = 2 ( 1 - sin theta) = 0`, giving

`theta = pi/2 and 5/2pi` in one period `[pi/2, 5/2pi]`

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

the tangent.

Now, the slope of the tangent is `y' = (dy)/(dx)`, in cartesian form.

In polar form, this is

`( r'sin theta + r cos theta )/(r'costheta-r sintheta)`

`=(cos 2theta-sin 2theta+1)/(-cos 2theta-sin2theta-1)`, upon using

`r= 2(1-sintheta)` and simplification.

Starting at `(0, pi/2)`, initial slope is the limit as `theta to pi/2 = -1`,

using L'Hospital rule.

Likewise, at the return to the pole ( now `theta = 5/2pi` ),

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

-pi/4`and theta =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 = -pi/4`, from `O (0, -pi/4)`

and return in the direction `theta=pi/4`, to `O( 0, pi/4)`.

Verification : Practical.