Parametrizing A Curve

Illustrated · 2 min

+2 contributors

Contributors

Introduction

In this next lesson in the topic of parametric equations, we will talk about parametrization.

### $\underline{\textbf{Parametrization}}$

$\textbf{Parametrization}$ is the process of finding the pair of parametric equations that describe a function.

Okay, let's say we have a function, $y = x^{2} - 1$ To determine the two parametric equations that define this function, we first have to make our independent variable, in this case $x$ , a $\text{function}$ of another independent variable, conventionally $t$ . This essentially makes $x$ equal to a function. So doing that we have, $x = t$ or more accurately, $x(t) = t$ Next, we also have to alter how we define our dependent variable $y$ . Since, $y = x^{2} - 1$ and $x$ is now a function of $t$ , $y$ , therefore, is now: $y = [x(t)]^{2} - 1$ In this case, since $x$ is equal to $t$ for any given point, we can just say $x$ = $t$ and from this conclusion we can also say that $y = t^{2} - 1$

Let's look at more difficult example below:

Here we have a quadratic function which we are trying to parametrize. So we first start by setting our independent variable, in this case again $x$ , equal to a function of the variable $t$ . $f(t)$ is not just equal to $t$ as it was in our first example, in this case $f(t)$ is equal to $t + 2$ . So in this case when we make $x$ a function of $t$ we get $x(t) = t +2$

Since we have adjusted out dependent variable $x$ , e now have to adjust our dependent variable $y$ . We do this by replacing all the $x$ variables we see with what $x$ is now equal to: $x(t) = t + 2$ After doing this, we can expand out the equation to simplify.

At the end, this is what our function, originally $y = 3x^{2} + 4x + 1$ , becomes when it is represented by parametric equations!

(We could also write these equations as: $y(t) = 3t^{2} + 10t + 11$ $x(t) = t + 2$ since $x$ and $y$ are now functions of $t$ ).

You can test if this answer is correct by checking with a point. If the point is on the curve, then it is correct!

Looking Further

We can take this concept a step further when dealing in three dimensions (or other dimensional spaces)! All you have to do is additionally consider the $z$ variable.