# Paramaterizing a curve

1. Apr 16, 2012

### fleazo

Hello. I had always been annoyed when dealing with parametric equations but it finally dawned on me how awesome the concept it. I want to make sure I understand things properly.

So say I have a regular straight line given by rectangular equation y=2x+1

Now I can graph this. Now say I want to consider a particle moving in motion along the path of that line. But say I want that particle to move with exponential speed, say e^t, so shouldn't I then be able to paramaterize the curve to represent that? Is there any simple algorithmic process to do so?

Thanks!

2. Apr 16, 2012

### victor.raum

Sure, it's a simple procedure for straight lines. Your particle's velocity vector will be in line with the vector $(1, 2)$, and will have a magnitude of $e^t$. The normalization of $(1, 2)$ is $(\frac{1}{\sqrt 5}, \frac{2}{\sqrt 5})$. The equation for your velocity vector will thus be \begin{align}v_x(t) &= \frac{e^t}{\sqrt 5}\\ \\ \\ \\ v_y(t) &= \frac{2e^t}{\sqrt 5}\end{align}

You integrate to get your curve (or in this case "line") parameterization, which in this case comes out identical to the velocity because of the nature of the exponential function: \begin{align}r_x(t) &= \frac{e^t}{\sqrt 5}\\ \\ \\ \\ r_y(t) &= \frac{2e^t}{\sqrt 5}\end{align}

Then you throw in a constant of integration to represent the fact that your particle doesn't go through the origin. We'll say that at $t=0$ the particle is at location $(0, 1)$, so the equations become \begin{align}r_x(t) &= \frac{e^t}{\sqrt 5}\\ \\ \\ \\ r_y(t) &= 1 + \frac{2e^t}{\sqrt 5}\end{align}

If you differentiate those $r_x$ and $r_y$ functions you get the velocity functions. And if you take those and pop them into $\sqrt{v_x^2 + v_y^2}$ you ought to get $e^t$; that is, unless I made a mistake somewhere, because actually I rushed while writing this. Gota run!