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April 7, 2021

Eliminating the Parameter

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Textbook Reference: Section 8.6 Parametric Equations

We now know three ways to describe a curve with an equation:

  • We can describe curves explicitly, showing directly how one variable is a function of another. For example, a line can be described as \(y=mx+b\), or a parabola might be described as \(x=y^2-1\).
  • We can describe curves implicitly, showing how \(x\) and \(y\) are related but without having one variable neatly solved for. For example, a line can be described in general form as \(Ax+By=C\), or the unit circle is described as \(x^2+y^2=1\). (The general form is nice because it even handles the case of vertical lines, which can't be written as \(y=mx+b\).
  • We can describe curves parametrically, showing how \(x\) and \(y\) are each separate functions of a parameter \(t\). For example, a circle can be parametrized as \(x=\cos(t)\), \(y=\sin(t)\).

Going between these different representations is sometimes easy, sometimes hard, depending on the curve

For example, the explicit line \(y=-2x+1\) can easily be made implicit by rearranging it into \(2x+y=1\), or it can be parametrized as \(x=t,y=-2t+1\). (You should verify that all three of these describe the same line.)

On the other hand, it may be hard (or even impossible!) to turn an implicit or parametric curve into an explicit one.

What we're going to focus is how to start with a parametric curve and "eliminate the parameter" to turn it into an explicit or an implicit curve.

Eliminating the parameter

Suppose we have the parametric equations \[\begin{cases}x=t^2+2\\ y=2t-1\end{cases}\] Can we figure out what kind of graph this will be without plotting a bunch of points?

Well, notice that we can take our \(y\) equation and solve for \(t\): \[ \begin{align*} y &= 2t-1\\ y+1 &= 2t\\ \dfrac{y+1}{2} &= t \end{align*} \] Then we can replace \(t\) with this expression in our other equation: \[ \begin{align*} x & = t^2 + 2\\ x & = \left(\dfrac{y+1}{2}\right)^2+2 \end{align*} \] And there we go — we now have a relationship between \(x\) and \(y\) directly! It appears that the graph is a parabola. We can clean things up just a little bit by rearranging some terms: \[x-2=\dfrac14(y+1)^2\] This tells us everything we need to graph the parabola. We can even imagine a particle "traveling" along the parabola as \(t\) increases.

This technique is useful for finding the explicit representation of many curves. For other curves, we'll need to use other techniques we'll go over in class ... which will just happen to tie a bunch of things you've learned about together!

Preview Activity 24

Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)

  1. Consider the following parametric equations from the last Preview Activity: \[ \begin{cases} x = 1+2t\\ y = 3-t \end{cases} \] Eliminate the parameter to get an equation for \(y\) explicitly in terms of \(x\).

  2. Let's tackle the same problem but in more generality: \[ \begin{cases} x = h+at\\ y = k+bt \end{cases} \] Eliminate the parameter to get an equation for \(y\) explicitly in terms of \(x\). (Assume \(a\neq 0\).)
    What's the slope? Why does this make sense with how you learned slope in middle/high school?

  3. Consider the parabola defined by the explicit equation \(y=(x+3)^2+1\).

    1. Explain why the parametric equations \[\begin{cases}x=t\\ y=(t+3)^2+1\end{cases}\] parametrize this curve. (How does this suggest a way to parametrize the graph of any function \(y=f(x)\)?)
    2. Come up with a different parametrization of the same parabola: \[\begin{cases}x=t-3\\ y=\text{???}\end{cases}\] Check your answer on Desmos to see if you're getting it right. (Hint: Solve for \(t\) in the first equation.)
  4. Answer AT LEAST one of the following questions:
    1. What was something you found interesting about this activity?
    2. What was an "a-ha" moment you had while doing this activity?
    3. What was the muddiest point of this activity for you?
    4. What question(s) do you have about anything you've done?

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