Parametric functions only show up on the AP Calculus BC exam. In this article we’ll take a close look at these kinds of functions which turn out to be extremely useful in the sciences.

In fact, this is one case in which the phrase *“It’s not rocket science!”* isn’t really appropriate. Parametric equations play a huge role in rocket guidance systems!

## What are Parametric Functions?

While they may at first seem foreign and confusing, parametric functions are just a more flexible way to track motion in the plane.

Usually you might think of a function as *y* = *f*(*x*). The graph of a function must pass the *Vertical Line Test (VLT)*, and as a result, your options are limited.

Sure you can use typical functions to define parabolas, catenaries and even sinusoidal waves, but what about a the orbit of a planet?

Certainly any circular or elliptical graph fails the VLT. So you can’t write a function of the form *y* = *f*(*x*) for this situation.

Instead, you need to be able to specify both *x* and *y* in terms of some independent **parameter**, *t*. That’s where the term **parametric** comes from.

### Definition

A **parametric function** (or a set of **parametric equations**) is a pair of two functions specifying the *x*– and *y*-coordinates of a point moving through the plane.

Think of each function as a separate control, one for *x* and one for *y*. Perhaps the best physical example of parametric equations is the Etch-A-Sketch.

The Etch-A-Sketch has two knobs, one controlling the vertical, and one controlling the horizontal position of a stylus. When the two knobs work together, almost anything can be drawn!

### Graphing

There is a straightforward way to graph any parametric function.

- Choose a number of sample
*t*-values. If there is a given range of values,*a*≤*t*≤*b*, then you must stick to values within that interval. Otherwise, just pick a wide range of values, both positive and negative. - For each sample
*t*, plug it into*x*=*f*(*t*) and into*y*=*g*(*t*) to find out the corresponding*x*– and*y*-coordinates. - Plot each (
*x*,*y*) pair on the plane. Then connect the dots in the order of increasing*t*.

### Example Graph

Graph the parametric function defined by *x* = *t*^{2} – 2*t* + 1 and *y* = –*t*^{2} + 2.

Because there was no range specified for *t*, let’s just pick a few easy numbers to work with. Remember, use both positive and negative values to get a good sense for how the function behaves.

I like to organize my work in a table.

t | x = t^{2} - 2t + 1 | y = -t^{2} + 2 | (x, y) |
---|---|---|---|

-3 | 16 | -7 | (16, -7) |

-2 | 9 | -2 | (9, -2) |

-1 | 4 | 1 | (4, 1) |

0 | 1 | 2 | (1, 2) |

1 | 0 | 1 | (0, 1) |

2 | 1 | -2 | (1, -2) |

3 | 4 | -7 | (4, -7) |

Next, plot these points on a coordinate plane.

Finally, connect the dots in order of increasing *t*.

## Derivatives of Parametric Functions

Now that you have seen some graphing, let’s talk about slope. As you know, the **derivative measures slope**. But how do you find the derivative of a set parametric equations?

We must be careful, because there are two equations to deal with. Should you take the derivative of *f*(*t*) or *g*(*t*)?

In fact, you’ll have to take the derivative of *both*. Here is the formula for *dy*/*dx*.

### Example — Tangent Line

What is the equation of the tangent line at *t* = π/6 for the parametric function *x* = 3 cos *t*, *y* = 3 sin *t*?

Using the derivative formula, we get:

*dy*/*dx* = (3 cos *t*)/(-3 sin *t*) = -cos *t* / sin *t*.

Plugging in *t* = π/6, the slope is -cos(π/6)/sin(π/6) = -1.732.

Now we also need to know what the *x*– and *y*-coordinates are for the point in question.

*x* = 3 cos(π/6) = 2.598, and *y* = 3 sin(π/6) = 1.5.

Therefore, using the point-slope form, the equation of the tangent line is:

*y* = -1.732(*x* – 2.598) + 1.5

### Derivatives and Velocity

There is another interpretation of the derivative that allows you to compute the **velocity** of an object traveling along a parametric curve.

The idea is to think of a parametric function as a *vector*.

Thinking of *t* as *time*, then both the **velocity vector** and **accelerations** vectors are simply derivatives in each component.

For more information about vector functions, check out this AP Calculus BC Review: Vector-Valued Functions

### Example — Velocity

Find the velocity vector at *t* = 1 for an object traveling according to the parametric function *x* = *t*^{2} – 2*t* + 1, *y* = –*t*^{2} + 2.

First find the derivative of each component function.

**v**(*t*) = (2*t* – 2, -2*t*).

Therefore at time *t* = 1, the velocity vector is:

**v**(1) = (0, -2).

## Length Integrals

Another important formula to memorize is the arc-length integral. If you want to know the length of the curve traced out by the parametric function *x* = *f*(*t*), *y* = *g*(*t*), for *a* ≤ *t* ≤ *b*, then just set up and compute the following integral.

### Example — Length of a Parametric Curve

Find the length of the curve defined by *x* = 3 cos *t*, *y* = 3 sin *t* on the interval [0, π/2].

## Summary

Parametric functions show up on the AP Calculus BC exam. You should know the following.

- How to graph or interpret the graph of a parametric function
- Finding the slope at any given point on a parametric curve
- Computing the velocity vector
- Finding the length of the curve