Lesson 10.6 : Parametric Equations

New Concept

Parametric equations describe a curve using a parameter, like time $t$. Instead of $y=f(x)$, we use $x(t)$ and $y(t)$ to define coordinates, revealing not just the path's shape but also its direction of movement.

What’s next

Next, you’ll practice parameterizing curves and converting them to rectangular form through a series of interactive examples and challenge problems.

Property

Suppose $t$ is a number on an interval, $I$. The set of ordered pairs, $(x(t), y(t))$, where $x = f(t)$ and $y = g(t)$, forms a plane curve based on the parameter $t$.
The equations $x = f(t)$ and $y = g(t)$ are the parametric equations. When we parameterize 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, $x$, $y$, and $t$.

Examples

• To parameterize the curve $y = x^2 + 5$ with $x(t) = t$, we substitute $t$ for $x$ to get the parametric pair: $x(t) = t$ and $y(t) = t^2 + 5$.

• To model an object moving from $(1, 2)$ to $(9, 5)$ in 3 seconds, the x-position changes by 8 units, so $x(t) = \frac{8}{3}t + 1$. The y-position changes by 3 units, so $y(t) = \frac{3}{3}t + 2 = t + 2$.

Property

For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for $t$. We substitute the resulting expression for $t$ into the second equation. This gives one equation in $x$ and $y$.

Examples

• Given $x(t) = t^2 + 3$ and $y(t) = t - 1$, solve for $t$ in the $y$ equation to get $t = y + 1$. Substitute this into the $x$ equation: $x = (y + 1)^2 + 3$, which simplifies to $x = y^2 + 2y + 4$.

• To eliminate the parameter in $x(t) = e^{t}$ and $y(t) = 2e^{-t}$, rewrite the second equation as $y = \frac{2}{e^t}$. Since $x = e^t$, substitute $x$ to get the Cartesian equation $y = \frac{2}{x}$.

Property

To eliminate the parameter from trigonometric equations, we use trigonometric identities and the Pythagorean Theorem. For parametric equations $x(t) = a \cos t$ and $y(t) = b \sin t$, we first solve for $\cos t$ and $\sin t$: $\frac{x}{a} = \cos t$ and $\frac{y}{b} = \sin t$. Substituting these into the Pythagorean identity $\cos^2 t + \sin^2 t = 1$ gives the Cartesian equation:

Examples

• Given $x(t) = 5 \cos t$ and $y(t) = 2 \sin t$, we get $\frac{x}{5} = \cos t$ and $\frac{y}{2} = \sin t$. The Cartesian equation is $(\frac{x}{5})^2 + (\frac{y}{2})^2 = 1$, or $\frac{x^2}{25} + \frac{y^2}{4} = 1$.

• If $x(t) = 3 \cos t$ and $y(t) = 3 \sin t$, then $\frac{x}{3} = \cos t$ and $\frac{y}{3} = \sin t$. This leads to $(\frac{x}{3})^2 + (\frac{y}{3})^2 = 1$, which simplifies to the circle equation $x^2 + y^2 = 9$.

Property

When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” The simplest method is to solve one equation for the parameter $t$ and substitute the resulting expression into the second equation.

Examples

• For $x(t) = 2t + 5$ and $y(t) = t - 3$, solve the $y$ equation for $t$ to get $t = y + 3$. Substitute this into the $x$ equation: $x = 2(y + 3) + 5$, which simplifies to $x = 2y + 11$.

• Given $x(t) = t^3$ and $y(t) = t^6 + 1$, notice that $y = (t^3)^2 + 1$. Since $x = t^3$, we can substitute $x$ directly to get the Cartesian equation $y = x^2 + 1$.

Property

There are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy is valid if it produces an equivalent graph. The most straightforward method is to let $x(t) = t$. Then, substitute $t$ for every $x$ in the original equation to find the corresponding equation for $y(t)$.

Examples

• To find parametric equations for $y = (x - 4)^2 + 3$, we can set $x(t) = t$. Substituting $t$ for $x$ gives $y(t) = (t - 4)^2 + 3$. The parametric equations are $x(t) = t$ and $y(t) = (t - 4)^2 + 3$.

• For the same equation, $y = (x - 4)^2 + 3$, we could choose a different substitution. Let $x(t) = t + 4$. Then $y = ((t + 4) - 4)^2 + 3$, which simplifies to $y(t) = t^2 + 3$. The parametric pair is $x(t) = t + 4$ and $y(t) = t^2 + 3$.