Parametric Equations
In short: Parametric equations draw a curve with the help of a helper variable t: one formula builds x, another builds y, and each choice of t lands on exactly one point of the path.
PrecalculusFunctions and GraphsF-IF.C.7
What is Parametric Equations?
Parametric equations work like a flight log for a drone: at each moment \(t\), one formula gives the horizontal position and another gives the vertical position.
The pair \(x = f(t)\), \(y = g(t)\) traces a path as \(t\) runs through an interval, and it also shows the direction of travel. Eliminating the parameter, by solving one equation for \(t\) and substituting into the other, gives a single equation in \(x\) and \(y\) whose graph is the same curve (possibly only part of it).
How to spot it
- Parametric equations appear as two formulas sharing one letter, usually t, where one gives x and the other gives y.
- A story follows something moving over time, like a ball, a robot or a point on a wheel.
- A curve is not a function of x, such as a full circle, and still needs an equation.
- The question asks for the position at certain times, or for the path without the time variable.
The method
- For parametric equations, choose several values of \(t\) in the stated interval and compute \(x\) and \(y\) for each.
- Plot the points in order of increasing \(t\) and join them, adding arrows for direction.
- To eliminate the parameter, solve the simpler equation for \(t\) and substitute into the other.
- Check the interval of \(t\) to see which part of the resulting curve is actually drawn.
Parametric Equations: solved example
Problem
A curve is given by \(x = 2t + 1\) and \(y = t^2 – 3\). Find the point for \(t = -2, 0, 2\), then eliminate \(t\).
Solution
At \(t = -2\): \((-3, 1)\). At \(t = 0\): \((1, -3)\). At \(t = 2\): \((5, 1)\).
From the first equation, \(t = \frac{x – 1}{2}\). Substituting gives \(y = \frac{(x – 1)^2}{4} – 3\). The curve is a parabola with vertex \((1, -3)\).
Try it
Question 1
Eliminate \(t\) from \(x = 3\cos t\) and \(y = 3\sin t\).
Show the answer
Square and add: \(x^2 + y^2 = 9\cos^2 t + 9\sin^2 t = 9\). The path is the circle of radius 3 about the origin.
Question 2
Eliminate \(t\) from \(x = t + 4\) and \(y = 2t – 1\).
Show the answer
\(t = x – 4\), so \(y = 2(x – 4) – 1 = 2x – 9\), a line.
Parametric Equations vs. Function
An ordinary function gives y directly from x and passes the vertical line test. Here both coordinates depend on a separate variable, which lets a curve loop or backtrack, as a circle does. The pair as a whole is not a function of x, though each separate formula is a function of t.
Keep going
Lessons
Worksheets and quizzes
- Functions, Domain and Range – Worksheet
- Functions, Domain and Range – Quiz
- Polar Form and De Moivre’s Theorem – Worksheet
- Polar Form and De Moivre’s Theorem – Quiz
Related words
Whole dictionary: A–Z · Precalculus · Functions and Graphs