In 1988, Keith Geddes and others involved with the Maple project at the University of Waterloo published a Maple Calculus Workbook of interesting calculus problems and their solutions in Maple. Over the years, I've paged through this book, extracting some of its more unique problems. Recently, I extracted the following problem from this book, and added it to my Clickable Calculus collection, which I use for workshops and web-based presentations.

Of course, we know that a curve has an infinite number of parametrizations, but I had not seen this particular technique for finding one of them. Implementing the calculation in Maple, we easily find

One use of such a parametric representation is in graphing. Figure 1 is a graph of the implicit function.  Note the modifications to the implicitplot command needed to obtain a reasonable plot.

 To determine how the parametrization traces this graph, we can use the animatecurve command, the result of which is shown in Figure 2.

 Click on the graph to obtain the animation toolbar, then run the animation from the toolbar. Alternatively, bring up the Context Menu for Figure 2, and select the "Animation" option.  For , the graph starts at  in the fourth quadrant. As  approaches , the graph approaches , the "transition point" on the portion of the curve lying in the fourth quadrant. This is where the curve has a vertical tangent.

The value of  at the transition point is found by solving

or equivalently, , for .

As , the graph extends to the right, but remains below the -axis. As  increases from zero, the graph traverses the first quadrant from right to left, reaching the origin at . As  increases beyond that, the graph extends downward into the third quadrant. Determining the equivalent information by just taking limits (before seeing the animation) proves to be much more challenging.

Surprisingly, the solve command in Maple has been modified to generate such parametrizations. For example, a parametrization of the unit circle can be obtained with

Notice that in this instance, the starting point at  can be imposed. This is not always possible, as we see from

but

As the help page for solve[parametric]  states, the substitution  is made, and a solution for  as a function of  is sought. However, solutions of the form  are not found.

The parametrization solve[parametric] finds for

is

but there is also a parametrization by rational functions, found with

A standard exercise in calculus is the elimination of the parameter in a given parametric representation of a curve. For example, given , the naive strategy of solving the first equation for  and substituting into the second leads to

At this point, only the student who remains conversant with high school trigonometry knows what to do.  The other students in the calculus class profit greatly with a tool like Maple: the Context Menu option Simplify_Trig gives

from which  soon follows.

(Nearly 40 years ago I had a Dean who characterized mathematics as the discipline where one subject was taught, but another tested. Thus, success in calculus requires a prior mastery of algebra, trigonometry, and even geometry. The calculus exam is often a measure of this mastery. That's why I've always felt so strongly that it really is better with a tool like Maple!)

The first time a student has to determine a parametrization for an explicitly given function  usually occurs in the context of vector calculus, probably first met in the multivariate calculus course. In most such applications, choosing  as the parameter, so  is the parametric representation, takes more than just a casual mention. Students have to think about this for a bit before it becomes "obvious."

From here, it is a short step to the generalization that  can be taken as any function of a parameter such as , and then  completes the parametrization. In fact, as long as  can be solved explicitly for , parametrizations of the form  are always possible.

Setting  and solving for  to obtain a parametrization of  is an implicitization (and not an obvious one) of the explicit scheme in the previous paragraph. (The equation  is solved for , which then gives .)

But suppose it is not possible to solve the equation  explicitly for . The parametrization , is available (at least numerically) by differentiating  with respect to , and solving the differential equation . Applied to the equation , this technique leads to a number of interesting challenges.

First, we have to agree that the curve defined by the given equation does not include the origin. If the equation is written as , then the curve passes through the origin.  Next, in order to implement a numeric solution of the differential equation

initial conditions are needed. These are obtained from the algebraic equation. To obtain the branch that approaches the origin, select an initial point in the third quadrant. Set  and determine the corresponding -coordinate as

Numeric integration of the differential equation is then obtained with

The functions  and  parametrically describe the third-quadrant portion of the curve in Figure 1. (Since  is not defined, the integration becomes meaningless once the origin has been reached.) To obtain the first-quadrant portion of this branch, a new initial point, say,  and  equal to

is selected. (Actually, from Figure 1 we see that at  there are three possible -coordinates. We have computed all three, simplified the expressions, and sorted them from smallest to largest. Clearly, the largest is the one in the first quadrant.)

Numeric integration of the differential equation in the first quadrant is then obtained with

and the functions  and  parametrically describe the first-quadrant portion of the curve in Figure 1.  

Numeric integration of the differential equation along the "lower-left" portion of the fourth-quadrant branch of the curve is obtained with

and along the "upper-right" portion with

Figure 3 provides a graph drawn from the numeric parametrizations.

 
There is an extended discussion of numeric parametrization in the text "Essential Maple 7" by Robert M. Corless, Springer-Verlag, 2002.