Defining and Differentiating Parametric Equations

What Parametric Equations Are

In most algebra and calculus problems, we describe a curve with an equation such as . Here, is written directly as a function of .

A parametric equation describes the same kind of curve differently. Instead of expressing directly in terms of , we define both coordinates using a third variable called the parameter, usually :

As changes, the equations produce different values of and . Together, those values give points on the curve.

The parameter is often interpreted as time, although it does not have to represent time. Thinking of as time is especially useful because it gives the curve a direction: as the parameter increases, the point moves along the curve.

Why Use Parametric Equations?

Parametric equations are particularly useful when a curve is difficult or impossible to describe as a single function .

• They can describe curves that fail the vertical line test.
• They naturally describe the motion of a point along a curve.
• They allow horizontal and vertical position to be modeled independently.
• They make many curves, including circles and cycloids, easier to represent.
• They provide a direct way to calculate slopes without first eliminating the parameter.

For AP Calculus BC, the key idea is that you usually do not need to eliminate the parameter before differentiating. Instead, you can use the derivatives of and directly.

A Simple Parametric Example

Eliminating the Parameter

Sometimes it is useful to eliminate and recover an ordinary Cartesian equation. This can help you recognize the shape of the curve.

Suppose:

Because , we have:

Substitute this into the equation for :

So the parametric equations describe the parabola .

However, there is an important warning: eliminating the parameter can hide information about direction and which portions of the curve are actually traced. For differentiation questions, keeping the parametric form is usually much more efficient.

First Derivatives of Parametric Equations

Suppose a curve is defined parametrically by:

Both and depend on . We want to find the slope of the curve, .

Using the chain rule:

Therefore:

This is one of the most important formulas in the AP Calculus BC parametric-equations unit:

In words: differentiate y with respect to the parameter, differentiate x with respect to the parameter, and divide.

Worked Differentiation Example

Why the Formula Works

The formula for is really an application of the chain rule.

Because depends on , and also depends on , we can think of as depending indirectly on :

Solving for gives:

The cancellation notation is a helpful way to remember the formula, but remember that is not literally an ordinary fraction. The underlying idea is the chain rule.

Horizontal and Vertical Tangents

Parametric equations introduce an important distinction between horizontal and vertical tangents.

Horizontal Tangent

A horizontal tangent has slope zero:

Since , a horizontal tangent usually occurs when:

provided is not zero.

Vertical Tangent

A vertical tangent has an undefined slope. This generally occurs when:

provided is not zero.

This gives you a powerful AP Calculus strategy: when asked to find horizontal or vertical tangents, examine and separately rather than immediately trying to eliminate the parameter.

Worked Example: Horizontal Tangent

Second Derivatives

AP Calculus BC also asks you to find the second derivative of a parametrically defined curve.

Start with:

To differentiate this with respect to , remember that is itself a function of . Therefore:

Using the chain rule:

Since :

Therefore, the standard formula for the second derivative is:

Or, in words: differentiate the first derivative with respect to the parameter, then divide by dx/dt.

Worked Example: Second Derivative

Common Mistakes

AP Calculus Strategy

For parametric-equation questions, a reliable workflow is:

1. Write down x(t) and y(t).
2. Find dx/dt and dy/dt.
3. Use dy/dx = (dy/dt)/(dx/dt) whenever the problem asks for slope.
4. For horizontal tangents, set dy/dt = 0 and verify dx/dt is not zero.
5. For vertical tangents, set dx/dt = 0 and verify dy/dt is not zero.
6. For the second derivative, differentiate dy/dx with respect to t and divide by dx/dt.
7. Substitute the requested parameter value only after obtaining the appropriate derivative expression.

The biggest advantage of the parametric approach is that you can work directly with the equations you are given. There is usually no reason to spend time eliminating the parameter unless the question specifically asks you to identify the Cartesian equation or interpret the curve.

Practice Problems