Instead of expressing $y$ directly as a function of $x$, parametric equations define both $x$ and $y$ in terms of a third variable called a parameter (usually $t$):

$x = f(t), \quad y = g(t)$

This allows us to describe curves that fail the vertical line test, trace motion over time, and represent complex shapes naturally.

Eliminating the Parameter

To convert from parametric to rectangular form:

1. Solve one equation for $t$
2. Substitute into the other

Example: $x = 2t + 1$, $y = t^2$

$t = \frac{x-1}{2} \implies y = \left(\frac{x-1}{2}\right)^2 = \frac{(x-1)^2}{4}$

Common Parametric Curves

Line

$x = x_0 + at, \quad y = y_0 + bt$

• Direction vector: $\langle a, b \rangle$
• Passes through $(x_0, y_0)$

Circle

$x = h + r\cos t, \quad y = k + r\sin t, \quad 0 \leq t \leq 2\pi$

• Center $(h, k)$, radius $r$

Ellipse

$x = a\cos t, \quad y = b\sin t$

• Eliminates to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Derivatives with Parametric Equations

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad \text{(where } dx/dt \neq 0\text{)}$

Horizontal tangent: $dy/dt = 0$ (and $dx/dt \neq 0$)

Vertical tangent: $dx/dt = 0$ (and $dy/dt \neq 0$)

Second Derivative

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}}$

Arc Length of Parametric Curves

$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt$

Motion Applications

When $t$ represents time:

• Position: $(x(t), y(t))$
• Velocity vector: $\langle dx/dt, dy/dt \rangle$
• Speed: $|\vec{v}| = \sqrt{(dx/dt)^2 + (dy/dt)^2}$

AP Precalculus Tip: Parametric equations are a major topic in Unit 4. Be comfortable eliminating parameters, finding slopes, and interpreting motion.