To this point (in both Calculus I and Calculus II) we’ve
looked almost exclusively at functions in the form or and almost all of the formulas that we’ve
developed require that functions be in one of these two forms. The problem is that not all curves or
equations that we’d like to look at fall easily into this form.
Take, for example, a circle.
It is easy enough to write down the equation of a circle centered at the
origin with radius r.
However, we will never be able to write the equation of a
circle down as a single equation in either of the forms above. Sure we can solve for x or y as the following
two formulas show
but there are in fact two functions in each of these. Each
formula gives a portion of the circle.
Unfortunately we usually are working on the whole circle, or
simply can’t say that we’re going to be working only on one portion of it. Even if we can narrow things down to only one
of these portions the function is still often fairly unpleasant to work with.
There are also a great many curves out there that we can’t
even write down as a single equation in terms of only x and y. So, to deal with some of these problems we
introduce parametric equations. Instead of defining y in terms of x ( ) or x
in terms of y ( ) we define both x and y in terms of a
third variable called a parameter as follows,
This third variable is usually denoted by t (as we did here) but doesn’t have to
be of course. Sometimes we will restrict
the values of t that we’ll use and at
other times we won’t. This will often be
dependent on the problem and just what we are attempting to do.
Each value of t
defines a point that we can plot. The collection of points that we get by
letting t be all possible values is
the graph of the parametric equations and is called the parametric curve.
To help visualize just what a parametric curve is pretend
that we have a big tank of water that is in constant motion and we drop a ping
pong ball into the tank. The point will then represent the location of the ping
pong ball in the tank at time t and
the parametric curve will be a trace of all the locations of the ping pong
ball. Note that this is not always a
correct analogy but it is useful initially to help visualize just what a
parametric curve is.
Sketching a parametric curve is not always an easy thing to
do. Let’s take a look at an example to
see one way of sketching a parametric curve.
This example will also illustrate why this method is usually not the
best.
Example 1 Sketch
the parametric curve for the following set of parametric equations.
Solution
At this point our only option for sketching a parametric
curve is to pick values of t, plug
them into the parametric equations and then plot the points. So, let’s plug in some t’s.
t

x

y

2

2

5

1

0

3



2

0

0

1

1

2

1

The first question that should be asked at this point is,
how did we know to use the values of t
that we did, especially the third choice?
Unfortunately, there is no real answer to this question at this point. We simply pick t’s until we are fairly confident that we’ve got a good idea of
what the curve looks like. It is this
problem with picking “good” values of t
that make this method of sketching parametric curves one of the poorer
choices. Sometimes we have no choice,
but if we do have a choice we should avoid it.
We’ll discuss an alternate graphing method in later
examples that will help to explain how these values of t were chosen.
We have one more idea to discuss before we actually sketch
the curve. Parametric curves have a direction of motion. The direction of motion is given by
increasing t. So, when plotting parametric curves, we
also include arrows that show the direction of motion. We will often give the value of t that gave specific points on the
graph as well to make it clear the value of t that gave that particular point.
Here is the sketch of this parametric curve.
So, it looks like we have a parabola that opens to the
right.
Before we end this example there is a somewhat important
and subtle point that we need to discuss first. Notice that we made sure to include a
portion of the sketch to the right of the points corresponding to and to indicate that there are portions of the
sketch there. Had we simply stopped
the sketch at those points we are indicating that there was no portion of the
curve to the right of those points and there clearly will be. We just didn’t compute any of those points.
This may seem like an unimportant point, but as we’ll see
in the next example it’s more important than we might think.

Before addressing a much easier way to sketch this graph
let’s first address the issue of limits on the parameter. In the previous
example we didn’t have any limits on the parameter. Without limits on the
parameter the graph will continue in both directions as shown in the sketch
above.
We will often have limits on the parameter however and this
will affect the sketch of the parametric equations. To see this effect
let’s look a slight variation of the previous example.
Example 2 Sketch
the parametric curve for the following set of parametric equations.
Solution
Note that the only difference here is the presence of the
limits on t. All these limits do is tell us that we can’t take
any value of t outside of this range. Therefore, the parametric
curve will only be a portion of the curve above. Here is the parametric
curve for this example.
Notice that with this sketch we started and stopped the
sketch right on the points originating from the end points of the range of t’s.
Contrast this with the sketch in the previous example where we had a
portion of the sketch to the right of the “start” and “end” points that we
computed.
In this case the curve starts at and ends at ,
whereas in the previous example the curve didn’t really start at the right
most points that we computed. We need
to be clear in our sketches if the curve starts/ends right at a point, or if
that point was simply the first/last one that we computed.

It is now time to take a look at an easier method of
sketching this parametric curve. This method uses the fact that in many,
but not all, cases we can actually eliminate the parameter from the parametric
equations and get a function involving only x
and y. We will sometimes call this the algebraic equation to differentiate it
from the original parametric equations. There will be two small problems
with this method, but it will be easy to address those problems. It is important to note however that we won’t
always be able to do this.
Just how we eliminate the parameter will depend upon the
parametric equations that we’ve got. Let’s see how to eliminate the
parameter for the set of parametric equations that we’ve been working with to
this point.
Example 3 Eliminate
the parameter from the following set of parametric equations.
Solution
One of the easiest ways to eliminate the parameter is to
simply solve one of the equations for the parameter (t, in this case) and substitute that into the other
equation. Note that while this may be the easiest to eliminate the
parameter, it’s usually not the best way as we’ll see soon enough.
In this case we can easily solve y for t.
Plugging this
into the equation for x gives the following algebraic equation,
Sure enough from our Algebra knowledge we can see that
this is a parabola that opens to the right and will have a vertex at .
We won’t bother with a sketch for this one as we’ve
already sketched this once and the point here was more to eliminate the
parameter anyway.
Before we leave this example let’s address one quick
issue.
In the first example we just, seemingly randomly, picked
values of t to use in our table,
especially the third value. There
really was no apparent reason for choosing . It is however probably the most important
choice of t as it is the one that
gives the vertex.
The reality is that when writing this material up we
actually did this problem first then went back and did the first
problem. Plotting points is generally
the way most people first learn how to construct graphs and it does
illustrate some important concepts, such as direction, so it made sense to do
that first in the notes. In practice
however, this example is often done first.
So, how did we get those values of t? Well let’s start off
with the vertex as that is probably the most important point on the
graph. We have the x and y coordinates of the vertex and we also have x and y parametric
equations for those coordinates. So, plug
in the coordinates for the vertex into the parametric equations and solve for
t.
Doing this gives,
So, as we can
see, the value of t that will give
both of these coordinates is . Note that the x parametric equation gave a double root and this will often not
happen. Often we would have gotten two
distinct roots from that equation. In
fact, it won’t be unusual to get multiple values of t from each of the equations.
However, what
we can say is that there will be a value(s) of t that occurs in both sets of solutions and that is the t that we want for that point. We’ll eventually see an example where this
happens in a later section.
Now, from this
work we can see that if we use we will get the vertex and so we included
that value of t in the table in
Example 1. Once we had that value of t we chose two integer values of t on either side to finish out the
table.
As we will see
in later examples in this section determining values of t that will give specific points is something that we’ll need to
do on a fairly regular basis. It is
fairly simple however as this example has shown. All we need to be able to do is solve a
(usually) fairly basic equation which by this point in time shouldn’t be too
difficult.

Getting a sketch of the parametric curve once we’ve
eliminated the parameter seems fairly simple. All we need to do is graph
the equation that we found by eliminating the parameter. As noted already
however, there are two small problems with this method. The first is
direction of motion. The equation involving only x and y
will NOT give the direction of motion of the parametric curve. This is generally
an easy problem to fix however. Let’s take a quick look at the
derivatives of the parametric equations from the last example. They are,
Now, all we need to do is recall our Calculus I
knowledge. The derivative of y with respect to t is clearly always positive.
Recalling that one of the interpretations of the first derivative is
rate of change we now know that as t
increases y must also increase. Therefore, we must be moving up the curve
from bottom to top as t increases as
that is the only direction that will always give an increasing y as t
increases.
Note that the x
derivative isn’t as useful for this analysis as it will be both positive and
negative and hence x will be both
increasing and decreasing depending on the value of t. That doesn’t help with
direction much as following the curve in either direction will exhibit both
increasing and decreasing x.
In some cases, only one of the equations, such as this
example, will give the direction while in other cases either one could be
used. It is also possible that, in some
cases, both derivatives would be needed to determine direction. It will always be dependent on the individual
set of parametric equations.
The second problem with eliminating the parameter is best
illustrated in an example as we’ll be running into this problem in the
remaining examples.
Example 4 Sketch
the parametric curve for the following set of parametric equations.
Clearly indicate direction of motion.
Solution
Before we proceed with eliminating the parameter for this
problem let’s first address again why just picking t’s and plotting points is not really a good idea.
Given the range of t’s
in the problem statement let’s use the following set of t’s.
t

x

y

0

5

0


0

2


5

0


0

2


5

0

The question that we need to ask now is do we have enough
points to accurately sketch the graph of this set of parametric equations? Below are some sketches of some possible
graphs of the parametric equation based only on these five points.
Given the nature of sine/cosine you might be able to
eliminate the diamond and the square but there is no denying that they are
graphs that go through the given points.
The last graph is also a little silly but it does show a graph going
through the given points.
Again, given the nature of sine/cosine you can probably
guess that the correct graph is the ellipse.
However, that is all that would be at this point. A guess.
Nothing actually says unequivocally that the parametric curve is an
ellipse just from those five points.
That is the danger of sketching parametric curves based on a handful
of points. Unless we know what the
graph will be ahead of time we are really just making a guess.
So, in general, we should avoid plotting points to sketch
parametric curves. The best method,
provided it can be done, is to eliminate the parameter. As noted just prior to starting this
example there is still a potential problem with eliminating the parameter
that we’ll need to deal with. We will eventually discuss this issue. For now, let’s just proceed with
eliminating the parameter.
We’ll start by eliminating the parameter as we did in the
previous section. We’ll solve one of
the of the equations for t and plug this into the other equation.
For example, we could do the following,
Can you see the problem with doing this? This is
definitely easy to do but we have a greater chance of correctly graphing the
original parametric equations by plotting points than we do graphing this!
There are many ways to eliminate the parameter from the
parametric equations and solving for t is usually not the best way to
do it. While it is often easy to do we will, in most cases, end up with
an equation that is almost impossible to deal with.
So, how can we eliminate the parameter here? In this case all we need to do is recall a
very nice trig identity and the equation of an ellipse. Let’s notice that we could do the following
here.
Eliminating the
middle steps gives the following algebraic equation,
and so it looks like we’ve got an ellipse.
Before proceeding with this example it should be noted
that what we did was probably not all that obvious to most. However, once it’s been done it does
clearly work and so it’s a nice idea that we can use to eliminate the
parameter from some parametric equations involving sines and cosines. It won’t always work and sometimes it will
take a lot more manipulation of things than we did here.
An alternate method that we could have used here was to
solve the two parametric equations for sine and cosine as follows,
Then, recall
the trig identity we used above and these new equations we get,
So, the same
answer as the other method. Which
method you use will probably depend on which you find easier to use. Both are perfectly valid and will get the
same result.
Now, let’s continue on with the example. We’ve identified that the parametric
equations describe an ellipse, but we can’t just sketch an ellipse and be
done with it.
First, just because the algebraic equation was an ellipse
doesn’t actually mean that the parametric curve is the full ellipse. It is always possible that the parametric
curve is only a portion of the ellipse.
In order to identify just how much of the ellipse the parametric curve
will cover let’s go back to the parametric equations and see what they tell
us about any limits on x and y.
Based on our knowledge of sine and cosine we have the following,
So, by starting with sine/cosine and “building up” the
equation for x and y using basic algebraic manipulations
we get that the parametric equations enforce the above limits on x and y. In this case, these
also happen to be the full limits on x
and y we get by graphing the full
ellipse.
This is the second potential issue alluded to above. The parametric curve may not always trace
out the full graph of the algebraic curve.
We should always find limits on x
and y enforced upon us by the
parametric curve to determine just how much of the algebraic curve is
actually sketched out by the parametric equations.
Therefore, in this case, we now know that we get a full
ellipse from the parametric equations.
Before we proceed with the rest of the example be careful to not
always just assume we will get the full graph of the algebraic equation. There are definitely times when we will not
get the full graph and we’ll need to do a similar analysis to determine just
how much of the graph we actually get.
We’ll see an example of this later.
Note as well that any limits on t given in the problem statement can also affect how much of the graph
of the algebraic equation we get. In
this case however, based on the table of values we computed at the start of
the problem we can see that we do indeed get the full ellipse in the range . That won’t always be the case however, so
pay attention to any restrictions on t
that might exist!
Next, we need to determine a direction of motion for the
parametric curve. Recall that all
parametric curves have a direction of motion and the equation of the ellipse
simply tells us nothing about the direction of motion.
To get the direction of motion it is tempting to just use the
table of values we computed above to get the direction of motion. In this case, we would guess (and yes that
is all it is a guess) that the curve traces out in a
counterclockwise direction. We’d be
correct. In this case, we’d be
correct! The problem is that tables of
values can be misleading when determining a direction of motion as well see
in the next example.
Therefore, it is best to not use a table of values to
determine the direction of motion. To
correctly determine the direction of motion we’ll use the same method of
determining the direction that we discussed after Example 3. In other words, we’ll take the derivative
of the parametric equations and use our knowledge of Calculus I and trig to
determine the direction of motion.
The derivatives of the parametric equations are,
Now, at we are at the point and let’s see what happens if we start
increasing t. Let’s increase t from to . In this range of t’s we know that sine is always positive and so from the
derivative of the x equation we can
see that x must be decreasing in
this range of t’s.
This, however, doesn’t really help us determine a
direction for the parametric curve.
Starting at no matter if we move in a clockwise or
counterclockwise direction x will
have to decrease so we haven’t really learned anything from the x derivative.
The derivative from the y parametric equation on the other hand will help us. Again, as we increase t from to we know that cosine will be positive and so y must be increasing in this
range. That however, can only happen
if we are moving in a counterclockwise direction. If we were moving in a clockwise direction
from the point we can see that y would have to decrease!
Therefore, in the first quadrant we must be moving in a
counterclockwise direction. Let’s
move on to the second quadrant.
So, we are now at the point and we will increase t from to . In this range of t we know that cosine will be negative and sine will be
positive. Therefore, from the
derivatives of the parametric equations we can see that x is still decreasing and y
will now be decreasing as well.
In this quadrant the y
derivative tells us nothing as y
simply must decrease to move from . However, in order for x to decrease, as we know it does in this quadrant, the direction
must still be moving a counterclockwise rotation.
We are now at and we will increase t from to . In this range of t we know that cosine is negative (and hence y will be decreasing) and sine is also negative (and hence x will be increasing). Therefore, we will continue to move in a
counterclockwise motion.
For the 4^{th} quadrant we will start at and increase t from to . In this range of t we know that cosine is positive (and hence y will be increasing) and sine is negative (and hence x will be increasing). So, as in the previous three quadrants, we
continue to move in a counterclockwise motion.
At this point we covered the range of t’s we were given in the problem statement and during the full
range the motion was in a counterclockwise direction.
We can now fully sketch the parametric curve so, here is the
sketch.

Okay, that was a really long example. Most of these types of problems aren’t as
long. We just had a lot to discuss in
this one so we could get a couple of important ideas out of the way. The rest of the examples in this section
shouldn’t take as long to go through.
Now, let’s take a look at another example that will illustrate
an important idea about parametric equations.
Example 5 Sketch
the parametric curve for the following set of parametric equations.
Clearly indicate direction of motion.
Solution
Note that the only difference in between these parametric
equations and those in Exampel 4 is that we replaced the t with 3t.
We can eliminate the parameter here using either of the methods we discussed
in the previous example. In this case
we’ll do the following,
So, we get the
same ellipse that we did in the previous example. Also note that we can do the same analysis
on the parametric equations to determine that we have exactly the same limits
on x and y. Namely,
It’s starting
to look like changing the t into a
3t in the trig equations will not
change the parametric curve in any way.
That is not correct however.
The curve does change in a small but important way which we will be
discussing shortly.
Before
discussing that small change the 3t
brings to the curve let’s discuss the direction of motion for this curve. Despite the fact that we said in the last
example that picking values of t
and plugging in to the equations to find points to plot is a bad idea let’s
do it any way.
Given the range
of t’s from the problem statement
the following set looks like a good choice of t’s to use.
t

x

y

0

5

0


0

2


5

0


0

2


5

0

So, the only
change to this table of values/points from the last example is all the nonzero
y values changed sign. From a quick glance at the values in this
table it would look like the curve, in this case, is moving in a clockwise
direction. But is that correct? Recall we said that these tables of values
can be misleading when used to determine direction and that’s why we don’t
use them.
Let’s see if
our first impression is correct. We
can check our first impression by doing the derivative work to get the
correct direction. Let’s work with
just the y parametric equation as
the x will have the same issue that
it had in the previous example. The
derivative of the y parametric
equation is,
Now, if we
start at as we did in the previous example and start
increasing t. At the derivative is clearly positive and so
increasing t (at least initially)
will force y to also be
increasing. The only way for this to
happen is if the curve is in fact tracing out in a counterclockwise
direction initially.
Now, we could
continue to look at what happens as we further increase t, but when dealing with a parametric curve that is a full
ellipse (as this one is) and the argument of the trig functions is of the
form nt for any constant n the direction will not change so
once we know the initial direction we know that it will always move in that
direction. Note that this is only true
for parametric equations in the form that we have here. We’ll see in later examples that for
different kinds of parametric equations this may no longer be true.
Okay, from this
analysis we can see that the curve must be traced out in a counterclockwise
direction. This is directly counter to
our guess from the tables of values above and so we can see that, in this
case, the table would probably have led us to the wrong direction. So, once again, tables are generally not
very reliable for getting pretty much any real information about a parametric
curve other than a few points that must be on the curve. Outside of that the tables are rarely
useful and will generally not be dealt with in further examples.
So, why did our
table give an incorrect impression about the direction? Well recall that we mentioned earlier that
the 3t will lead to a small but
important change to the curve versus just a t? Let’s take a look at
just what that change is as it will also answer what “went wrong” with our
table of values.
Let’s start by
look at . At we are at the point and let’s ask ourselves what values of t put us back at this point. We saw in Example 3 how to determine
value(s) of t that put us at
certain points and the same process will work here with a minor modification.
Instead of
looking at both the x and y equations as we did in that example
let’s just look at the x
equation. The reason for this is that
we’ll note that there are two points on the ellipse that will have a y coordinate of zero, and . If we set the y coordinate equal to zero we’ll find all the t’s that are at both of these points
when we only want the values of t
that are at .
So, because the
x coordinate of five will only
occur at this point we can simply use the x
parametric equation to determine the values of t that will put us at this point.
Doing this gives the following equation and solution,
Don’t forget
that when solving a trig equation we need to add on the “ ” where n
represents the number of full revolutions in the counterclockwise direction
(positive n) and clockwise
direction (negative n) that we
rotate from the first solution to get all possible solutions to the
equation.
Now, let’s plug
in a few values of n starting at . We don’t need negative n in this case since all of those would result in negative t and those fall outside of the range
of t’s we were given in the problem
statement. The first few values of t are then,
We can stop
here as all further values of t
will be outside the range of t’s
given in this problem.
So, what is
this telling us? Well back in Example
4 when the argument was just t the
ellipse was traced out exactly once in the range . However, when we change the argument to 3t (and recalling that the curve will
always be traced out in a counterclockwise direction for this problem) we
are going through the “starting” point of two more times than we did in the previous
example.
In fact, this
curve is tracing out three separate times.
The first trace is completed in the range . The second trace is completed in the range and the third and final trace is completed
in the range . In other words, changing the argument from t to 3t increase the speed of the trace and the curve will now trace
out three times in the range !
This is why the
table gives the wrong impression. The
speed of the tracing has increased leading to an incorrect impression from
the points in the table. The table seems
to suggest that between each pair of values of t a quarter of the ellipse is traced out in the clockwise
direction when in reality it is tracing out three quarters of the ellipse in
the counterclockwise direction.
Here’s a final sketch of the curve and note that it really
isn’t all that different from the previous sketch. The only differences are the values of t and the various points we included. We did include a few more values of t at various points just to illustrate
where the curve is at for various values of t but in general these really aren’t needed.

So, we saw in the last two examples two sets of parametric
equations that in some way gave the same graph. Yet, because they traced
out the graph a different number of times we really do need to think of them as
different parametric curves at least in some manner. This may seem like a
difference that we don’t need to worry about, but as we will see in later
sections this can be a very important difference. In some of the later
sections we are going to need a curve that is traced out exactly once.
Before we move on to other problems let’s briefly acknowledge
what happens by changing the t to an nt in these kinds of parametric
equations. When we are dealing with
parametric equations involving only sines and cosines and they both have the
same argument if we change the argument from t to nt we simply change
the speed with which the curve is traced out.
If we will increase the speed and if we will decrease the speed.
Let’s take a look at a couple more examples.
Example 6 Sketch
the parametric curve for the following set of parametric equations. Clearly identify the direction of
motion. If the curve is traced out
more than once give a range of the parameter for which the curve will trace
out exactly once.
Solution
We can eliminate the parameter much as we did in the
previous two examples. However, we’ll
need to note that the x already
contains a and so we won’t need to square the x.
We will however, need to square the y as we need in the previous two examples.
In this case the algebraic equation is a parabola that
opens to the left.
We will need to be very, very careful however in sketching
this parametric curve. We will NOT get
the whole parabola. A sketch of the
algebraic form parabola will exist for all possible values of y.
However, the parametric equations have defined both x and y in terms of sine and cosine and we know that the ranges of
these are limited and so we won’t get all possible values of x and y here. So, first let’s
get limits on x and y as we did in previous examples. Doing this gives,
So, it is clear from this that we will only get a portion
of the parabola that is defined by the algebraic equation. Below is a quick sketch of the portion of
the parabola that the parametric curve will cover.
To finish the sketch of the parametric curve we also need
the direction of motion for the curve.
Before we get to that however, let’s jump forward and determine the
range of t’s for one trace. To do this we’ll need to know the t’s that put us at each end point and
we can follow the same procedure we used in the previous example. The only difference is this time let’s use
the y parametric equation instead
of the x because the y coordinates of the two end points of
the curve are different whereas the x
coordinates are the same.
So, for the top point we have,
For, plugging
in some values of n we get that the
curve will be at the top point at,
Similarly, for
the bottom point we have,
So, we see that
we will be at the bottom point at,
So, if we start
at say, ,
we are at the top point and we increase t
we have to move along the curve downwards until we reach at which point we are now at the bottom
point. This means that we will trace
out the curve exactly once in the range .
This is not the
only range that will trace out the curve however. Note that if we further increase t from we will now have to travel back up the curve
until we reach and we are now back at the top point. Increasing t again until we reach will take us back down the curve until we
reach the bottom point again, etc. From this analysis we can get two more
ranges of t for one trace,
As you can
probably see there are an infinite number of ranges of t we could use for one trace of the curve. Any of them would be acceptable answers for
this problem.
Note that in the process of determining a range of t’s for one trace we also managed to
determine the direction of motion for this curve. In the range we had to travel downwards along the curve
to get from the top point at to the bottom point at . However, at we are back at the top point on the curve
and to get there we must travel along the path. We can’t just jump back up to the top point
or take a different path to get there.
All travel must be done on the path sketched out. This means that we had to go back up the
path. Further increasing t takes us back down the path, then up
the path again etc.
In other words, this path is sketched out in both
directions because we are not putting any restrictions on the t’s and so we have to assume we are
using all possible values of t. If we had put restrictions on which t’s to use we might really have ended
up only moving in one direction. That
however would be a result only of the range of t’s we are using and not the parametric equations themselves.
Note that we didn’t really need to do the above work to
determine that the curve traces out in both directions.in this case. Both the x and y parametric
equations involve sine or cosine and we know both of those functions
oscillate. This, in turn means that
both x and y will oscillate as well.
The only way for that to happen on this particular this curve will be for the
curve to be traced out in both directions.
Be careful with the above reasoning that the oscillatory
nature of sine/cosine forces the curve to be traced out in both
directions. It can only be used in
this example because the “starting” point and “ending” point of the curves
are in different places. The only way
to get from one of the “end” points on the curve to the other is to travel
back along the curve in the opposite direction.
Contrast this with the ellipse in Example 4. In that case we had sine/cosine in the
parametric equations as well. However,
the curve only traced out in one direction, not in both directions. In Example 4 we were graphing the full
ellipse and so no matter where we start sketching the graph we will
eventually get back to the “starting” point without ever retracing any
portion of the graph. In Example 4 as
we trace out the full ellipse both x
and y do in fact oscillate between
their two “endpoints” but the curve itself does not trace out in both
directions for this to happen.
Basically, we can only use the oscillatory nature of
sine/cosine to determine that the curve traces out in both directions if the
curve starts and ends at different points.
If the starting/ending point is the same then we generally need to go
through the full derivative argument to determine the actual direction of
motion.
So, to finish this problem out, below is a sketch of the
parametric curve. Note that we put
direction arrows in both directions to clearly indicate that it would be
traced out in both directions. We also
put in a few values of t just to
help illustrate the direction of motion.

To this point we’ve seen examples that would trace out the
complete graph that we got by eliminating the parameter if we took a large
enough range of t’s. However, in the previous example we’ve now
seen that this will not always be the case.
It is more than possible to have a set of parametric equations which
will continuously trace out just a portion of the curve. We can usually determine if this will happen
by looking for limits on x and y that are imposed up us by the
parametric equation.
We will often use parametric equations to describe the path
of an object or particle. Let’s take a
look at an example of that.
Example 7 The
path of a particle is given by the following set of parametric equations.
Completely describe the path of this particle. Do this by sketching the path, determining
limits on x and y and giving a range of t’s for which the path will be traced
out exactly once (provide it traces out more than once of course).
Solution
Eliminating the parameter this time will be a little
different. We only have cosines this
time and we’ll use that to our advantage.
We can solve the x equation
for cosine and plug that into the equation for y. This gives,
This time the algebraic equation is a parabola that opens
upward. We also have the following
limits on x and y.
So, again we only trace out a portion of the curve. Here is a quick sketch of the portion of
the parabola that the parametric curve will cover.
Now, as we discussed in the previous example because both
the x and y parametric equations involve cosine we know that both x and y must oscillate and because the “start” and “end” points of the
curve are not the same the only way x
and y can oscillate is for the
curve to trace out in both directions.
To finish the problem then all we need to do is determine
a range of t’s for one trace. Because the “end” points on the curve have
the same y value and different x values we can use the x parametric equation to determine
these values. Here is that work.
So, we will be at the right end point at and
we’ll be at the left end point at . So,
in this case there are an infinite number of ranges of t’s for one trace. Here
are a few of them.
Here is a final sketch of the particle’s path with a few
value of t on it.

We should give a small warning at this point. Because of the ideas involved in them we
concentrated on parametric curves that retraced portions of the curve more than
once. Do not, however, get too locked
into the idea that this will always happen.
Many, if not most parametric curves will only trace out once. The first one we looked at is a good example
of this. That parametric curve will
never repeat any portion of itself.
There is one final topic to be discussed in this section
before moving on. So far we’ve started
with parametric equations and eliminated the parameter to determine the
parametric curve.
However, there are times in which we want to go the other
way. Given a function or equation we
might want to write down a set of parametric equations for it. In these cases we say that we parameterize the function.
If we take Examples 4 and 5 as examples we can do this for
ellipses (and hence circles). Given the
ellipse
a set of parametric equations for it would be,
This set of parametric equations will trace out the ellipse
starting at the point and will trace in a counterclockwise
direction and will trace out exactly once in the range . This is a fairly important set of parametric
equations as it used continually in some subjects with dealing with ellipses
and/or circles.
Every curve can be parameterized in more than one way. Any of the following will also parameterize
the same ellipse.
The presence of the will change the speed that the ellipse rotates
as we saw in Example 5. Note as well
that the last two will trace out ellipses with a clockwise direction of motion
(you might want to verify this). Also
note that they won’t all start at the same place (if we think of as the starting point that is).
There are many more parameterizations of an ellipse of
course, but you get the idea. It is
important to remember that each parameterization will trace out the curve once
with a potentially different range of t’s. Each parameterization may rotate with
different directions of motion and may start at different points.
You may find that you need a parameterization of an ellipse
that starts at a particular place and has a particular direction of motion and
so you now know that with some work you can write down a set of parametric
equations that will give you the behavior that you’re after.
Now, let’s write down a couple of other important
parameterizations and all the comments about direction of motion, starting
point, and range of t’s for one trace
(if applicable) are still true.
First, because a circle is nothing more than a special case
of an ellipse we can use the parameterization of an ellipse to get the
parametric equations for a circle centered at the origin of radius r as well. One possible way to parameterize a circle is,
Finally, even though there may not seem to be any reason to,
we can also parameterize functions in the form or . In these cases we parameterize them in the
following way,
At this point it may not seem all that useful to do a
parameterization of a function like this, but there are many instances where it
will actually be easier, or it may even be required, to work with the
parameterization instead of the function itself. Unfortunately, almost all of these instances
occur in a Calculus III course.