# Cardiod parametric equation problem

1. Sep 28, 2013

### mahler1

The problem statement, all variables and given/known data.
Let $C_a$ a cardioid given in polar coordinates by $r_a= a+cos(\theta)$ with a being a parameter, and $\theta$ $\in [0,2\Pi]$

a)Prove that, for a>1, $C_a$ is a smooth curve.
b)Calculate the tangent line to the curve $C_a$ in the cartesian coordinates point $(x,y)=(0,a)$

My problem is I don't understand how this curve is, I mean, I googled and saw how it looks like, but I don't know the parameterization.

By the way, I accidently pressed enter, so I opened an empty thread and I couldn't check the spell in the title, it says "cardiod"", when it should be "cardioid"", does anyone know how can I correct the title?

Last edited: Sep 28, 2013
2. Sep 28, 2013

### LCKurtz

I wouldn't worry to much about the misspelling in the title but you can ask a moderator to do it for you if it bothers you. Not sure what is bothering you about the parameterization. It is just a polar equation in the usual form $r = f(\theta)$. Surely you have studied polar coordinates and their graphs??

3. Sep 28, 2013

### Zondrina

For part (a), does the tangent vector vanish at all or is it continuous?

4. Sep 29, 2013

### mahler1

I got confused, the parameterization in polar coordinates is $ψ(t)=(f(\theta)cos(\theta),f(\theta)sin(\theta))$

5. Sep 30, 2013

### mahler1

Sorry, I don't understant the question. The tangent vector doesn't vanish if the curve is smooth, I'm not sure if that what you've asked.

6. Sep 30, 2013

### Zondrina

Exactly, if the tangent vector doesn't vanish anywhere, then it is continuous. That is the exact condition you need for the curve to be smooth.

7. Sep 30, 2013

### LCKurtz

Use either $\theta$ or $t$ for the parameter, not both. But yes, for a curve given as $r = f(\theta)$, that is the x-y parameterization$$\Psi(\theta) = \langle f(\theta)\cos(\theta), f(\theta)\sin(\theta)\rangle$$You can get the slope by$$\frac{dy}{dx}= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$