# 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.