Parametric Functions

What Are Parametric Functions?

A parametric function expresses both x and y in terms of a third variable called the parameter, often $t$.

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

As t changes, the pair (x,y) traces out a curve in the plane.

Why use parametric equations?

• They model motion (position of particle at time t)

• They describe curves that cannot be written as a simple function y=f(x

• They allow direction and speed interpretation in physics

Example of a Parametric Curve

$$x(t)=\cos(t) , y(t)=\sin(t)$$

As t goes from 0 to $$2\pi$$, the point (x,y) moves around the unit circle.

Eliminating the Parameter

Sometimes we convert parametric equations to a Cartesian equation (remove $t$).

Example:

$$x=2t+1 , y=t^{2}$$

Solve for t from the first equation:

$$t=\frac{x-1}{2}$$​

Plug into $y$:

$$y=\left(\frac{x-2}{2}\right)^{2}$$

This gives a normal Cartesian function.

Derivatives of Parametric Functions

For parametric equations:

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

The derivative $$\frac{dy}{dx}$$​ is:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$​

This is extremely important for AP Calculus.

Example:

$$x=t^{2}+1 , y=t^{3}-3t$$

Derivative each x(t) and y(t) respect by t:

$$\frac{dx}{dt}=2t , \frac{dy}{dt}=3t^{2}-3$$

Thus:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^{2}-3}{2t}$$​

Second Derivative for Parametric Functions

Sometimes we need curvature or concavity.

$$\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}$$​

AP BC exams often test this.

Tangent Lines for Parametric Curves

Horizontal tangent:

$$\frac{dx}{dt}\neq0 , \frac{dy}{dt}=0$$

Vertical tangent:

$$\frac{dx}{dt}=0 , \frac{dy}{dt}\neq0$$

Arc Length of a Parametric Curve (BC Topic)

For a curve:

$$x=f(t) , y=g(t) , t=a to b$$

Arc length:

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

This is required for AP Calculus BC.

Example:

$$x(t)=t , y(t)=t^{2} , t=0 to 1$$

$$\frac{dx}{dt}=1 , \frac{dy}{dx}=2t$$

$$L=\int_{0}^{1}\sqrt{1^{2}+(2t)^{2}}=\int_{0}^{1}\sqrt{1+4t^{2}}$$

(This integral requires trig substitution or numeric approximation.)

Motion Along a Curve (BC)

If a particle’s position is given by parametric equations:

$$x(t) , y(t)$$

Then:

Velocity vector:

$$<x^\prime(t) , y^\prime(t)>$$

Speed:

$$\sqrt{(x^\prime(t))^{2}+(y^\prime(t))^{2}}$$​

Acceleration:

$$<x^{\prime\prime}(t) , y^{\prime\prime}(t)>$$