The parametric equations for a curve in the plane consists of a pair of equations
Each value of the parameter t gives values for
x and y; the point is the corresponding point on
the curve.
For example, consider the parametric equations
Here are some points which result from plugging in
some values for t:
The graph of the curve looks like this:
These are parametric equations for the circle :
You can sometimes recover the x-y equation of a parametric curve by eliminating t from the parametric equations. In this case,
Notice that the graph of a circle is not the graph of a function. Parametric equations can represent more general curves than function graphs can, which is one of their advantages.
These parametric equations represent a spiral:
This is also not the graph of a function .
Example. Find the x-y equation for
Solve the x-equation for t:
Plug this expression for t into the y-equation:
Example. Find the x-y equation for
Notice that
So
This is the standard form for the equation of an ellipse.
In some cases, recovering an x-y equation would be difficult or impossible. For example, these are the parametric equations for a hypocycloid of four cusps:
In this case, it would be difficult to eliminate t to obtain an x-y equation.
What about going the other way? If you have a curve (or an x-y equation), how do you obtain parametric equations?
Note first that a given curve can be represent by infinitely many
sets of parametric equations. For example, all of these sets of
parametric equations represent the unit circle :
Even so, it can be difficult to find parametrizations for curves.
Let's start with an easy case. If you have x-y equations in which x
or y is solved for, it's easy. For example, to parametrize , set
. Then
. A parametrization is given by
To parametrize , set
. Then
, so
This is a parametrization of . (This is
how you can graph x-in-terms-of-y equations on your calculator.)
Here's another important case. If and
are points, the line through
and
may be parametrized by
It is easiest to remember this in the vector form
Notice that when , I have
, and when
,
. Thus, if you let
, you get the
segment from
to
.
Example. Find parametric equations for the
line through and
.
Thus,
An analogous result holds for lines in 3 dimensions (or in any number of dimensions).
As an example of a more general method of parametrizing curves, I'll
consider parametrizing by slope. The idea is
to think of a point on the curve as the
intersection point of the curve and the line
:
The slope t will be the parameter for the curve.
Example. Find parametric equations for the Folium of Descartes:
Set . Then
Therefore, . A
parametrization is given by
The first and second derivatives give information about the shape of a curve. Here's how to find the derivatives for a parametric curve.
First, by the Chain Rule,
Solving for gives
Example. Find the equation of the tangent line to the curve
When ,
and
. The tangent line is
Example. At what points on the following curve is the tangent line horizontal?
Find the derivative:
The tangent line is horizontal when , and
for
and for
.
When ,
and
. When
,
and
. There are horizontal
tangents are
and at
.
To find the second derivative, I differentiate the first derivative.
Since will come out in terms of t, I want to be
sure to differentiate
with respect to
t. Use the Chain Rule again:
That is,
Example. Find and
at
for
First,
So
When , I have
.
Next,
So
When , I have
.
Copyright 2020 by Bruce Ikenaga