8.6 Parametric equations  (Page 4/6)

Finding cartesian equations from curves defined parametrically

When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as $x\left(t\right)=t.$ In this case, $y\left(t\right)$ can be any expression. For example, consider the following pair of equations.

Rewriting this set of parametric equations is a matter of substituting $x$ for $t.$ Thus, the Cartesian equation is $y=x^2-3.$

Finding parametric equations for curves defined by rectangular equations

Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent $x,$ and then substitute it into the $y$ equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for $x$ as the domain of the rectangular equation, then the graphs will be different.

Finding a set of parametric equations for curves defined by rectangular equations

Find a set of equivalent parametric equations for $y={\left(x+3\right)}^2+1.$

An obvious choice would be to let $x\left(t\right)=t.$ Then $y\left(t\right)={\left(t+3\right)}^2+1.$ But let’s try something more interesting. What if we let $x=t+3?$ Then we have

$$\begin{array}{l}y={(x+3)}^2+1\hfill \\ y={\left((t+3)+3\right)}^2+1\hfill \\ y={(t+6)}^2+1\hfill \end{array}$$

The set of parametric equations is

$\begin{array}{l}\hfill \\ x(t)=t+3\hfill \\ y(t)={(t+6)}^2+1\hfill \end{array}$

See [link] .

Got questions? Get instant answers now!

Got questions? Get instant answers now!

Key concepts

• Parameterizing a curve involves translating a rectangular equation in two variables, $x$ and $y,$ into two equations in three variables, x , y , and t . Often, more information is obtained from a set of parametric equations. See [link] , [link] , and [link] .
• Sometimes equations are simpler to graph when written in rectangular form. By eliminating $t,$ an equation in $x$ and $y$ is the result.
• To eliminate $t,$ solve one of the equations for $t,$ and substitute the expression into the second equation. See [link] , [link] , [link] , and [link] .
• Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for $t$ in one of the equations, and substitute the expression into the second equation. See [link] .
• There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.
• Find an expression for $x$ such that the domain of the set of parametric equations remains the same as the original rectangular equation. See [link] .