Differentiating Parametrically Defined Functions (Preview)

What a Parametric Function Is

Most of the functions you've differentiated so far have been written explicitly as . In that setup, x is the independent variable and y is directly expressed in terms of x.

A parametrically defined curve works differently. Instead of writing y directly as a function of x, we define both coordinates using a third variable called the parameter, usually :

As changes, both and change. Together, the ordered pair traces out a curve in the coordinate plane.

Think of as a clock. At one value of , the particle is at one point. As increases, the particle moves through a sequence of points, creating the curve.

Why Parametric Equations Are Useful

Parametric equations are especially useful when describing motion. If an object moves through the plane, it is often more natural to describe its horizontal and vertical positions separately as functions of time.

For example, if an object has position

then tells us its horizontal position and tells us its vertical position at time .

The important AP Calculus question is then: How steep is the curve at a particular value of the parameter?

In other words, how can we find when the curve is given in terms of rather than directly in terms of ?

The Key Formula

Here is the central result you need to know:

This formula comes from the chain rule. Since both and depend on , we can think of the derivative of y with respect to x as a ratio of their rates of change with respect to the parameter.

In words:

This is one of the most important formulas in the parametric portion of AP Calculus BC.

Where the Formula Comes From

Suppose the parameter is , and both coordinates depend on it:

By the chain rule,

Solving for gives:

So the parametric derivative is really a chain-rule relationship. You are comparing how quickly y changes with the parameter to how quickly x changes with the parameter.

A Step-by-Step Method

1. Differentiate the equation for x with respect to the parameter. Find .
2. Differentiate the equation for y with respect to the parameter. Find .
3. Divide. Use .
4. Substitute the requested parameter value, if the problem asks for the slope at a particular value of .

The key habit is simple: differentiate both equations with respect to the parameter first, then form the ratio.

Worked Examples

The Most Important Condition

Notice that the formula

requires the denominator to be nonzero. If

then the usual formula for is undefined.

Geometrically, this is important. If , the curve has no horizontal movement at that instant. The tangent line may be vertical, depending on what is happening to .

This is one reason parametric curves can behave differently from ordinary functions of .

Parametric Equations Can Trace Curves That Aren't Functions of x

A major advantage of parametric equations is that they can describe curves that fail the vertical line test.

Consider the familiar circle:

A circle cannot be written globally as a single function , because many values of correspond to two different values of .

But the circle has a natural parametric representation:

As moves through an interval, the point traces the circle. This makes parametric equations especially useful for curves involving motion, loops, and other shapes that are difficult or impossible to represent as one ordinary function of .

A Preview of Higher-Level Parametric Calculus

Once you know how to find , several more AP Calculus BC ideas become possible.

• Finding horizontal and vertical tangent lines.
• Finding the second derivative of a parametric curve.
• Determining whether a parametric curve is concave up or concave down.
• Finding the area under or between parametric curves.
• Analyzing the motion of a particle using position, velocity, and acceleration.

For now, the essential idea is the first one: to find the slope of a parametrically defined curve, divide the rate of change of y by the rate of change of x.

Common Mistakes

Practice Problems