When plotting graphs in polar coordinates, how does one know when to

1. Aug 13, 2012

hale2bopp

When plotting graphs in polar coordinates, how does one know when to make the graph sharp (at θ=0) (as in for the graph for r=1-cosθ) as opposed to a dimple (r=3/2 + cos θ) ?

2. Aug 13, 2012

micromass

Re: Limacons

A graph in polar coordinates is given by $r=f(\theta)$.
Now, we can express such a graph in cartesian coordinates. So, if $r=f(\theta)$, then we can use the formulas

$$x=r\cos(\theta),~y=r\sin(\theta)$$

to come up with the following form of the graph in cartesian coordinates:

$$(f(\theta)\cos(\theta),f(\theta)\sin(\theta)$$

For example, given $r=a-\cos(\theta)$ (with a constant), we can write this in cartesian coordinates as

$$((a-\cos(\theta))\cos(\theta),(a-\cos(\theta))\sin(\theta))$$

Now, the use of this is simpy that we can now investigate our curve using analysis. So, we can find the "velocity vector" at a point by taking derivatives. The derivative of our above function now becomes

$$(\sin(\theta)(2\cos(\theta)-a),a\cos(\theta)-\cos(2\theta))$$

Now, if a=3/2, then our derivative in 0 is $(0,1/2)$

So, we can deduce that in 0, our function is going up with a speed of 1/2.

What if a=1? Then our derivative in 0 is (0,0). This is a weird result. It means that at 0, our velocity vector is zero and thus the curve just stands still. This is the explanation of why you get a sharp point (= a cusp) when a=0, but just a smooth line when a=3/2.