Parametric Equations

Parametric Equations are equations that help define a curve using a new variable, called a parameter. Instead of expressing $y$ directly in terms of $x$ (or vice versa), both $x$ and $y$ are expressed in terms of a third variable, usually denoted as $t$. This method is particularly useful for simplifying complex shapes, modeling objects in motion and representing relationships where the variables depend on a hidden factor like time.

Parameterizing a Curve

When an object moves along a curve in a given direction and in a given amount of time, we can describe its position given by $x$ and $y$ coordinates. However, both $x$ and $y$ vary over time making them functions of time. For this reason, we can add another variable $t$ which is the parameter upon which both $x$ and $y$ depend on. So, when we are parameterizing a curve, we are translating a single equation in two variables, such as $x$ and $y$, into an equivalent pair of equations in three variables, such as $x$, $y$, and $t$.

In mathematical terms, suppose $t$ is a number on an interval $I$ and the set of ordered pairs, $(x(t), y(t))$, where $t \in I$, $x = f(t)$, and $y = g(t)$ forms a curve on the parameter $t$. Then, $x = f(t)$ and $y = g(t)$ are called the parametric equations of the curve.

$\underline{Example}$

Parameterize the curve given by $x = y^3 - 2y$.

There can be multiple ways to parameterize a curve as long as the resulting equations satisfy the original equation. We could always let $x = t$ and solve for $y$ in terms of $t$ but this equation is already solved for $x$. This means if we let $y = t$, it would be much easier to express $x$ in terms of $t$.

So, letting $y = t$, we can substitute this into the original equation to get $x = t^3 - 2t$.

Therefore, the parametric equations of the curve are $x(t) = t^3 - 2t$ and $y(t) = t$.

note

Sometimes, when we are parameterizing a curve, the curve can not be a function because it fails the vertical line test. However, parametric equations can still represent these curves because they do not require $y$ to be expressed as a function of $x$ or vice versa. This means parametric equations allow us to describe more complex curves that would otherwise be impossible to represent with standard functions.

Eliminating the Parameter

Sometimes, we may want to eliminate the parameter from parametric equations to express the relationship between $x$ and $y$ directly. This process involves solving one of the parametric equations for the parameter and substituting it into the other equation.

$\underline{Example}$

Eliminate the parameter for the equations $x(t) = t^2$ and $y(t) = \ln (t)$ where $t > 0$. Write the final equation as a rectangular equation.

Lets first write $x(t)$ in terms of $t$ so that we can substitute it into $y(t)$. From $x(t) = t^2$, we can take the square root of both sides to get $t = \sqrt{x}$.

Now, substituting $t = \sqrt{x}$ into $y(t) = \ln (t)$ gives us $y = \ln(\sqrt{x})$.

So, the rectangular equation that relates $x$ and $y$ is $y = \ln(\sqrt{x})$ where $x > 0$.

Trigonometric Equations

We can eliminate the parameter from trigonometric equations using straightforward substituion as well. However, we may need to use some familiar trigonometric identities to help us simplify the final equation.

The identities we may need to use are...

1. $x(t) = a \cos(t)$ and $y(t) = b \sin(t)$ where $a$ and $b$ are constants.
2. $\dfrac{x}{a} = \cos(t)$ and $\dfrac{y}{b} = \sin(t)$ which we get by solving for $\cos(t)$ and $\sin(t)$ respectively.
3. $\cos^2(t) + \sin^2(t) = 1$ which is the Pythagorean identity.
4. $\left(\dfrac{x}{a}\right)^2 + \left(\dfrac{y}{b}\right)^2 = 1$ which we get by substituting into the Pythagorean identity.

$\underline{Example}$

Eliminate the parameter from the equations $x(t) = 2 \cos(t)$ and $y(t) = 3 \sin(t)$. Write the final equation as a Cartesian equation.

Firstly, let's solve in terms of $\cos (t)$ and $\sin (t)$ so that we can substitute them into the Pythagorean identity to eliminate them. From $x(t) = 2 \cos(t)$, we can solve for $\cos (t)$ to get $\cos (t) = \dfrac{x}{2}$ and from $y(t) = 3 \sin(t)$, we can solve for $\sin (t)$ to get $\sin (t) = \dfrac{y}{3}$.

We can now substitute $\cos (t)$ and $\sin (t)$ into the Pythagorean identity $\cos^2(t) + \sin^2(t) = 1$ to get...

$$\begin{array}{cccccccccccccc} \cos^2(t) + \sin^2(t) &=& 1 \\[1em] \left(\dfrac{x}{2}\right)^2 + \left(\dfrac{y}{3}\right)^2 &=& 1 \\[1em] \dfrac{x^2}{4} + \dfrac{y^2}{9} &=& 1 \end{array}$$

So, the Cartesian equation that relates $x$ and $y$ is $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.