In traditional Cartesian equations, a relationship is expressed between two variables—typically x and y. However, in many real-world scenarios, both variables depend on a third variable, often time. This is where parametric equations come in. Instead of one equation, we use two or more to define the coordinates as functions of an independent parameter.
A parametric equation defines a group of quantities as functions of one or more independent variables, called parameters. In two dimensions, parametric equations typically take the form:
Here, t is the parameter, and f, g are parametric functions describing how x and y change over time.
Example 1:
Let:
This defines a unit circle. As t varies from 0 to 2π, the coordinates (x, y) trace a circular path.
Example 2:
Projectile motion in physics is described by:
This pair of parametric equations models the position of a projectile over time.
When you graph the ordered pairs (x(t), y(t)), the resulting plot is called a parametric curve. These curves are useful for:
Standard Parametric Form (2D): x = f(t), y = g(t)
Standard Parametric Equation of a Circle:
Parametric Equation of a Line:
If a line passes through point () and has direction vector ⟨a, b⟩, then the parametric form is:
This is extremely useful in 2D and 3D geometry for describing the trajectory of points, particles, or lines in space.
The purpose of a parametric equation is to:
In short, parametric equations extend the power of mathematics to describe dynamic systems with elegance and precision.
Sometimes, it's useful to convert parametric equations back to Cartesian form by eliminating the parameter.
Example:
Given: x = 2t + 1, y = 3t - 4
Solve for t from the first equation:
Substitute into the second:
This gives the Cartesian form:
Example 1: Given the parametric equations:
Eliminate the parameter t to find the Cartesian equation.
Solution:
From y = 2t, solve for t:
Substitute into :
Answer:
Example 2: Find the parametric equations of a line passing through point (2, -1) and having direction vector ⟨3, 4⟩.
Solution:
The general form is:
Substitute:
x = 2 + 3t, y = –1 + 4t
Answer: x = 2 + 3t, y = -1 + 4t
Example 3: Given , find .
Solution:
First, compute derivatives:
Then use:
Answer:
Example 4: Given: for 0 ≤ t ≤ 2π, identify the curve.
Solution:
Use the identity:
Answer: The curve is a circle of radius 1 centered at the origin.
Example 5: Write the parametric equations for a circle of radius 4 centered at (2, -3).
Solution:
The general form:
Substitute h = 2, k = –3, r = 4:
Answer:
(Session 2025 - 26)