At this point we can get a good idea on what the limits on \(x\) and \(y\) are going to be so let’s do that. Note that often we won’t get the actual limits on \(x\) and \(y\) in this step. All we are really finding here is the largest possible range of limits for \(x\) and \(y\). Having these can sometimes be useful for later steps and so we’ll get them here.

We can use our knowledge of sine and cosine to determine the limits on \(x\) and \(y\) as follows,

Note that to find these limits in general we just start with the appropriate trig function and then build up the equation for \(x\) and \(y\) by first multiplying the trig function by any coefficient, if present, and then adding/subtracting any numbers that might be present (not needed in this case). This, in turn, gives us the largest possible set of limits for \(x\) and \(y\). Just remember to be careful when multiplying an inequality by a negative number. Don’t forget to flip the direction of the inequalities when doing this.

Now, at this point we need to be a little careful. What we’ve actually found here are the largest possible inequalities for the limits on \(x\) and \(y\). This set of inequalities for the limits on \(x\) and \(y\) assume that the parametric curve will be completely traced out at least once for the range of \(t\)’s we were given in the problem statement. It is always possible that the curve will not trace out a full trace in the given range of \(t\)’s. In a later step we’ll determine if the parametric curve does trace out a full trace and hence determine the actual limits on \(x\) and \(y\).

Before we move onto the next step there are a couple of issues we should quickly discuss.

First, remember that when we talk about the parametric curve tracing out once we are not necessarily talking about the ellipse itself being fully traced out. The parametric curve will be at most the full ellipse and we haven’t determined just yet how much of the ellipse the parametric curve will trace out. So, one trace of the parametric curve refers to the largest portion of the ellipse that the parametric curve can possibly trace out given no restrictions on \(t\).

Second, if we can’t completely determine the actual limits on \(x\) and \(y\) at this point why did we do them here? In part we did them here because we can and the answer to this step often does end up being the limits on \(x\) and \(y\). Also, there are times where knowing the largest possible limits on \(x\) and/or \(y\) will be convenient for some of the later steps.

Finally, we can sometimes get these limits from the sketch of the parametric curve. However, there are some parametric equations that we can’t easily get the sketch without doing this step. We’ll eventually do some problems like that.

Before we sketch the graph of the parametric curve recall that all parametric curves have a direction of motion, i.e. the direction indicating increasing values of the parameter, \(t\) in this case.

In previous problems one method we looked at was to build a table of values for a sampling of \(t\)’s in the range provided. However, as we discussed in Problem 4 of this section tables of values for parametric equations involving trig functions they can be deceptive and so we aren’t going to use them to determine the direction of motion for this problem.

Also, as noted in the discussion in Problem 4 it also might help to have the graph of sine and cosine handy to look at since we’ll be talking a lot about the behavior of sine/cosine as we increase the argument.

So for this problem we’ll just do the analysis of the behavior of sine and cosine in the range of \(t\)’s we were provided to determine the direction of motion. We’ll be doing a quicker version of the analysis here than we did in Problem 4 so you might want to go back and check that problem out if you have trouble following everything we’re going here.

Let’s start at \(t = 0\) since that is the first value of \(t\) in the range of \(t\)’s we were given in the problem. This means we’ll be starting the parametric curve at the point \(\left( {0, - 4} \right)\).

Now, what happens if we start to increase \(t\)? First, if we increase \(t\) then we also increase 2\(t\), the argument of the trig functions in the parametric equations. So, what does this mean for \(\sin \left( {2t} \right)\) and \(\cos \left( {2t} \right)\)? Well initially, we know that \(\sin \left( {2t} \right)\)will increase from zero to one and at the same time \(\cos \left( {2t} \right)\) will also have to decrease from one to zero.

So, this means that \(x\) (given by \(x = 3\sin \left( {2t} \right)\)) will have to increase from 0 to 3. Likewise, it means that \(y\) (given by \(y = - 4\cos \left( {2t} \right)\)) will have to increase from -4 to 0. For the \(y\) equation note that while the cosine is decreasing the minus sign on the coefficient means that \(y\) itself will actually be increasing.

Because this behavior for \(x\) and \(y\) must be happening at simultaneously we can see that the only possibility is for the parametric curve to start at \(\left( {0, - 4} \right)\)and as we increase the value of \(t\) we must move to the right in the counter clockwise direction until we reach the point \(\left( {3,0} \right)\).

Okay, we’re now at the point \(\left( {3,0} \right)\), so \(\sin \left( {2t} \right) = 1\) and \(\cos \left( {2t} \right) = 0\). Let’s continue to increase \(t\). A further increase of \(t\) will force \(\sin \left( {2t} \right)\) to decrease from 1 to 0 and at the same time \(\cos \left( {2t} \right)\) will decrease from 0 to -1.

In terms of \(x\) and \(y\) this means that, at the same time, \(x\) will now decrease from 3 to 0 while \(y\) will continue to increase from 0 to 4 (again the minus sign on the \(y\) equation means \(y\) must increase as the cosine decreases from 0 to -1). So, we must be continuing to move in a counter clockwise direction until we reach the point \(\left( {0,4} \right)\).

For the remainder we’ll go a little quicker in the analysis and just discuss the behavior of \(x\) and \(y\) and skip the discussion of the behavior of the sine and cosine.

Another increase in \(t\) will force \(x\) to decrease from 0 to -3 and at the same time \(y\) will have to also decrease from 4 to 0. The only way for this to happen simultaneously is to move along the ellipse starting that \(\left( {0,4} \right)\) in a counter clockwise motion until we reach \(\left( { - 3,0} \right)\).

Continuing to increase \(t\) and we can see that, at the same time, \(x\) will increase from -3 to 0 and \(y\) will decrease from 0 to -4. Or, in other words we’re moving along the ellipse in a counter clockwise motion from \(\left( { - 3,0} \right)\) to \(\left( {0, - 4} \right)\).

At this point we’ve gotten back to the starting point and we got back to that point by always going in a counter clockwise direction and did not retrace any portion of the graph and so we can now safely say that the direction of motion for this curve will always counter clockwise.

We have to be very careful here to continue the analysis until we get back to the starting point and see just how we got back there. It is possible, as we’ll see in later problems, for us to get back there by retracing back over the curve. This will have an effect on the direction of motion for the curve (i.e. the direction will change!). In this case however since we got back to the starting point without retracing any portion of the curve we know the direction will remain counter clockwise.

Let’s now think about how much of the ellipse is actually traced out or if the ellipse is traced out more than once for the range of \(t\)’s we were given in the problem. We’ll also be able to verify if the ranges of \(x\) and \(y\) we found in Step 2 are the correct ones or if we need to modify them (and we’ll also determine just how to modify them if we need to).

Be careful to not draw any conclusions about how much of the ellipse is traced out from the analysis in the previous step. If we follow that analysis we see a full single trace of the ellipse. However, we didn’t ever really mention any values of \(t\) with the exception of the starting value. Because of that we can’t really use the analysis in the previous step to determine anything about how much of the ellipse we trace out or how many times we trace the ellipse out.

Let’s go ahead and start this portion out at the same value of \(t\) we started with in the previous step. So, at \(t = 0\) we are at the point \(\left( {0, - 4} \right)\). Now, when do we get back to this point? Or, in other words, what is the next value of \(t\) after \(t = 0\) (since that is the point we choose to start off with) are we at the point \(\left( {0, - 4} \right)\)?

In order to be at this point we know we must have \(\sin \left( {2t} \right) = 0\) (only way to get \(x = 0\)!) and we must have \(\cos \left( {2t} \right) = 1\) (only way to get \(y = - 4\)!). Note the arguments of the sine and cosine! That is very important for this step.

Now, for \(t > 0\) we know that \(\sin \left( {2t} \right) = 0\) at \(2t = \pi ,2\pi ,3\pi , \ldots \) and likewise we know that \(\cos \left( {2t} \right) = 1\) at \(2t = 2\pi ,4\pi ,6\pi , \ldots \). Again, note the arguments of sine and cosine here! Because we want \(\sin \left( {2t} \right)\) and \(\cos \left( {2t} \right)\) to have certain values we need to determine the values of 2\(t\) we need to achieve the values of sine and cosine that we are looking for.

The first value of 2\(t\) that is in both lists is \(2t = 2\pi \). This now tells us the value of \(t\) we need to get back to the starting point. We just need to solve this for \(t\)!

So, we will get back to the starting point, without retracing any portion of the ellipse, important in some later problems, when we reach \(t = \pi \).

But this is in the middle of the range of \(t\)’s we were given! So, just what does this mean for us? Well first of all, provided the argument of the sine/cosine is only in terms of \(t\), as opposed to \({t^2}\) or \(\sqrt t \) for example, the “net” range of \(t\)’s for one trace will always be the same. So, we got one trace in the range of \(0 \le t \le \pi \) and so the “net” range of \(t\)’s here is \(\pi - 0 = \pi \) and so any range of \[t\]’s that span \(\pi \) will trace out the ellipse exactly once.

This means that the ellipse will also trace out exactly once in the range \(\pi \le t \le 2\pi \). So, in this case, it looks like the ellipse will be traced out twice in the range \(0 \le t \le 2\pi \).

This analysis also has shown us that the parametric curve traces out the full ellipse in the range of \(t\)’s given in the problem statement (more than once in fact!) and so we know now that the limits of \(x\) and \(y\) we found in Step 2 are in fact the actual limits on \(x\) and \(y\) for this curve.

Before we leave this step we should note that once you get pretty good at the direction analysis we did in Step 3 you can combine the analysis Steps 3 and 4 into a single step to get both the direction and portion of the curve that is traced out. Initially however you might find them a little easier to do them separately.

Finally, here is a sketch of the parametric curve for this set of parametric equations.

For this sketch we included a set of \(t\)’s to illustrate a handful of points and their corresponding values of \(t\)’s. For some practice you might want to follow the analysis from Step 4 to see if you can verify the values of \(t\) for the other three points on the graph. It would, of course, be easier to just plug them in to verify, but the practice of the process of the Step 4 analysis might be useful to you.

Also, because the problem asked for it here are the formal limits on \(x\) and \(y\) for this parametric curve.

You should also take a look at problems 4 and 6 in this section and contrast the number of traces of the curve with this problem. The only difference in the set of parametric equations in problems 4, 5 and 6 is the argument of the trig functions. After going through these three problems can you reach any conclusions on how the argument of the trig functions will affect the parametric curves for this type of parametric equations?