Parametric Equations

Skill 1: Parametric Equations

You need to understand what parametric equations mean and how to plot graphs from them.

Example: Sketch the graph of $x=t-1$ and $y=t^{2}$.

Create a table of points:

Now plot these points on a graph.

You should also be able to answer questions such as “what is the value of $y$ when $t=4$” and “what value of $t$ corresponds to $y=1$”. All questions of this form can be solved by substituting in the right values into the parametric equation.

Example: For the graph $x=t-1$ and $y=t^{2}$, find the value of $y$ when $x=5$.

$x=t-1$

$5=t-1$

$t=6$

$begin{aligned}y&=t^{2}[1.2em]&=6^{2}[1.2em]&=36end{aligned}$

Skill 3: Intersections of Graphs and Parametric Equations

To find the intersection of a parametric equation and Cartesian equation on a graph, substitute in the values for $t$ for $x$ and $y$ into the Cartesian equation, then solve for $t$, then find the $x$ and $y$ values corresponding to that value of $t$.

Example: Find where the graph $x=3t+4$ and $y=2t^{2}$ intersects the line $y=x+1$

$y=x+1$

Substitute in expressions in $t$

$2t^{2}=3t+4+1$

$2t^{2}=3t+5$

$2t^{2}-3t-5=0$

$(2t-5)(t+1)=0$

$t=dfrac{5}{2}$ or $t=-1$

$x=3t+4$

$x=3timesdfrac{5}{2}+4$ or $x=3times(-1)+4$

$x=dfrac{15}{2}+4$ or $x=-3+4$

$x=dfrac{23}{2}$ or $x=1$

$y=x+1$

When $x = dfrac{23}{2}$, $y=dfrac{23}{2}+1 = dfrac{25}{2}$

When $x = 1$, $y=1+1 = 2$

Skill 4: Finding the Cartesian Equation

It is possible to get back to the cartesian equation of a graph from its parametric form. The way to do this is to rearrange the equation of one of $x$ or $y$ into the format $t=f(x)$ or $t=f(y)$, then substitute in this expression for $t$ into the other equation.

Example: Find the Cartesian equation of $y=t^{2}$, $x=t+5$

$x=t+5$

$t=x-5$

$begin{aligned}y&=t^{2}[1.2em]&=(x-5)^{2}end{aligned}$

Skill 5: Using Trig Identities to find the Cartesian Equation

Note: For more about trig identities, go to the Trigonometric Identities page.

For parametric equations involving $theta$ and trigonometric functions, it is often necessary to use trig identities to recover the cartesian form.

Example: Find the Cartesian equation of $x=1+cos(theta)$, $y=cos(2theta)$

$x=1+cos(theta)$

$cos(theta)=x-1$

$begin{aligned}y&=cos(2theta)[1.2em]&=2cos^{2}(theta)-1[1.2em]&=2(x-1)^{2}-1[1.2em]&=2(x^{2}-2x+1)-1[1.2em]&=2x^{2}-4x+2-1[1.2em]&=2x^{2}-4x+1end{aligned}$

Skill 6: Modelling with Parametric Equations

Parametric equations can be used to model real life situations. However, this can place restrictions on the parameter because some scenarios are not possible in real life.

• The equations could model for a parameter where only some $x$ and $y$ values make sense, e.g. positive values.
• Restrictions could exist to avoid repeating values of $x$ and $y$, such as limiting $theta$ to $0<theta<2pi$.
• Avoiding dividing by zero is another reason for restrictions, for example if we had $x=dfrac{1}{t-1}$ we must say $tneq 1$.

The values of the parameter that are allowed are collectively called the domain of the parameter.

Example: A tennis ball is thrown into the air. Its horizontal distance from the point of throwing is $x=25t$, and its vertical distance from the point of throwing is $y=9t-t^{2}$. Find its vertical distance when it has travelled $50$m horizontally.

We can approach this problem in the same way we would approach a problem that is not modelling a real thing.

$begin{aligned}x&=25t[1.2em]&=50end{aligned}$

$25t=50$

$t=2$

$begin{aligned}y&=9t-t^{2}[1.2em]&=9times2-2^{2}[1.2em]&=18-4[1.2em]&=14text{ m}end{aligned}$