At this point let’s get our first guess as to the limits on \(x\) and \(y\). As noted in previous problems what we’re really finding here is the largest possible ranges for \(x\) and \(y\). In later steps we’ll determine if this the actual set of limits on \(x\) and \(y\) or if we have smaller ranges.

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

\[\begin{array}{ccc} - 1 \le \sin \left( {\frac{1}{4}t} \right) \le 1 & \hspace{1.0in} & - 1 \le \cos \left( {\frac{1}{4}t} \right) \le 1\\ - 4 \le 4\sin \left( {\frac{1}{4}t} \right) \le 4 & \hspace{1.0in} & 0 \le {\cos ^2}\left( {\frac{1}{4}t} \right) \le 1\\ - 4 \le x \le 4 & \hspace{1.0in} & 0 \ge - 2{\cos ^2}\left( {\frac{1}{4}t} \right) \ge - 2\\ & \hspace{1.0in} & 1 \ge 1 - 2{\cos ^2}\left( {\frac{1}{4}t} \right) \ge - 1\\ & \hspace{1.0in} & - 1 \le y \le 1\end{array}\]

Remember that all we need to do is 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. We now have the largest possible set of limits for \(x\) and \(y\).

Now, at this point we need to be a little careful. As noted above what we’ve actually found here are the largest possible ranges 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 fully traced out at least once for the range of \(t\)’s we were given in the problem statement. It is always possible that the parametric 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 \(y\).

Remember that when we talk about the parametric curve getting fully traced out this doesn’t, in general, mean the full parabola we found in Step 1 gets traced out by the parametric equation. All “fully traced out” means, in general, is that whatever portion of the parabola that is described by the set of parametric curves will be completely traced out.

In fact, for this problem, we can see that the parabola from Step 1 will not get fully traced out by the particle regardless of any range of \(t\)’s. The largest possible portion of the parabola that can be traced out by the particle is the portion that lies in the range of \(x\) and \(y\) given above. In a later step we’ll determine if the largest possible portion of the parabola does get traced out or if the particle only traces out part of it.

Show Step 3

Let’s next get the direction of motion for the parametric curve. For this analysis it might be useful to have a quick sketch of the largest possible parametric curve. So, here is a quick sketch of that.

The dashed line is the graph of the full parabola from Step 1 and the solid line is the portion that falls into our largest possible range of \(x\) and \(y\) we found in Step 2. As an aside here note that the two ranges are complimentary. In other words, if we sketch the graph only for the range of \(x\) we automatically get the range for \(y\). Likewise, if we sketch the graph only for the range of y we automatically get the range for \(x\). This is a good check for your graph. The \(x\) and \(y\) ranges should always match up!

When our parametric curve was an ellipse (the previous problem for example) no matter what point we started the analysis at the curve would eventually trace out around the ellipse and end up back at the starting point without ever going back over any portion of itself. The main issue we faced with the ellipse problem was we could rotate around the ellipse in a clockwise or a counter clockwise motion to do this and a careful analysis of the behavior of both the \(x\) and \(y\) parametric equations was required to determine just which direction we were going.

With a parabola for our parametric curve things work a lot differently. Let’s suppose that we “started” at the right end point (this is just randomly picked for no other reason that I’m right handed so don’t think there is anything special about this point!) and it doesn’t matter what \(t\) we use to get to that point.

At this point we know that we are at \(x = 4\) and in order for \(x\) to have that value we must also have \(\sin \left( {\frac{1}{4}t} \right) = 1\). Now, as we increase \(t\) from this point (again it doesn’t matter just what the value of \(t\) is) the only option for sine is for it to decrease until it has the value \(\sin \left( {\frac{1}{4}t} \right) = - 1\). This in turn means that if we start at the right end point we have no option but to proceed along the curve going to the left.

However, we don’t just reach the left end point and then stop! Once we are at \(\sin \left( {\frac{1}{4}t} \right) = - 1\) if we further increase \(t\) we know that sine will also increase until it has the value \(\sin \left( {\frac{1}{4}t} \right) = 1\) and so we must move back along the curve to the right until we are back at the right end point.

Unlike the ellipse however, the only way for this to happen is for the particle to go back over the parabola moving in a rightward direction. Remember that the particle moves to the right or left it must trace out a portion of the parabola that we found in Step 1! Any particle traveling along the path given by the set of parametric equations must follow the graph of the parabola and never leave it.

In other words, if we don’t put any restrictions on \(t\) a particle on this parametric curve will simply oscillate left and right along the portion of the parabola sketched out above. In this case however we do have a range of \(t\)’s so we’ll need to determine a range of \(t\)’s for one trace to fully know the direction of motion information of the particle on this path and we’ll do that in the next step. With a restriction on the range of \(t\)’s it is possible that the particle won’t make a full trace or it might retrace some or all of the curve so we can’t say anything definite about the direction of motion for the particle over the full range of \(t\)’s until the next step when we determine a range of \(t\)’s for one full trace of the curve.

Before we move on to the next step there is a quick topic we should address. We only used the \(x\) equation to do this analysis and never addressed the \(y\)-equation anywhere in the analysis. It doesn’t really matter which one we use as both will give the same information.

Show Step 4

Now we need to determine a range of \(t\)’s for one full trace of the parametric curve. It is important for this step to remember that one full trace of the parametric curve means that no portion of the parametric curve can be retraced.

Note that one full trace does not mean that we get back to the “starting” point. When we dealt with an ellipse in the previous problem that was one trace because we did not need to retrace any portion of the ellipse to get back to the starting point. However, as we saw in the previous step that for our parabola here we would have to retrace the full curve to get back to the starting point.

So, one full trace of the parametric curve means we move from the right end point to the left end point only or visa-versa and move from the left end point to the right end point. Which direction we move doesn’t really matter here so let’s get a range of \(t\)’s that take us from the left end point to the right end point.

In all the previous problems we’ve used \(t = 0\) as our “starting” point but that won’t work for this problem because that actually corresponds to the vertex of the parabola. We want to start at the left end point so the first part of this process is actually determine a \(t\) that will put us at the left end point.

In order to be at the left end point, \(\left( { - 4,1} \right)\), we need to require that \(\sin \left( {\frac{1}{4}t} \right) = - 1\) which occurs if \(\frac{1}{4}t = \ldots , - \frac{{5\pi }}{2}, - \frac{\pi }{2},\frac{{3\pi }}{2},\frac{{7\pi }}{2}, \ldots \). We also need to require that \(\cos \left( {\frac{1}{4}t} \right) = 0\) which occurs if \(\frac{1}{4}t = \ldots , - \frac{{3\pi }}{2}, - \frac{\pi }{2},\frac{\pi }{2},\frac{{3\pi }}{2}, \ldots \). There are going to be many numbers that are in both lists here so all we need to do is pick one and proceed. From the numbers that we’ve listed here we could use either \(\frac{1}{4}t = - \frac{\pi }{2}\) or \(\frac{1}{4}t = \frac{{3\pi }}{2}\). We’ll use \(\frac{1}{4}t = - \frac{\pi }{2}\), i.e. when \(t = - 2\pi \) , simply because it is the first one that occurs in both lists. Therefore, we will be at the left end point when \(t = - 2\pi \).

Let’s now move to the right end point, \(\left( {4,1} \right)\). In order to get the range of \(t\)’s for one trace this means we’ll need the next \(t\) with \(t > - 2\pi \) (which corresponds to \(\frac{1}{4}t > - \frac{\pi }{2}\)). To do this we need to require that \(\sin \left( {\frac{1}{4}t} \right) = 1\) which occurs if \(\frac{1}{4}t = \ldots , - \frac{{3\pi }}{2},\frac{\pi }{2},\frac{{5\pi }}{2},\frac{{9\pi }}{2}, \ldots \)and we need to that \(\cos \left( {\frac{1}{4}t} \right) = 0\) which occurs if \(\frac{1}{4}t = \ldots , - \frac{{3\pi }}{2}, - \frac{\pi }{2},\frac{\pi }{2},\frac{{3\pi }}{2}, \ldots \). .

The first \(t\) that is in both of these lists with \(\frac{1}{4}t > - \frac{\pi }{2}\) is then \(\frac{1}{4}t = \frac{\pi }{2}\), i.e. when \(t = 2\pi \) . So, the first \(t\) after \(t = - 2\pi \) that puts us at the right end point is \(t = 2\pi \). This means that a range of \(t\)’s for one full trace of the parametric curve is then,

\[ - 2\pi \le t \le 2\pi \]

Note that this is only one possible answer here. Any range of \(t\)’s with a “net” range of \(2\pi - \left( { - 2\pi } \right) = 4\pi \) \(t\)’s, with the endpoints of the \(t\) range corresponding to start/end points of the parametric curve, will work. So, for example, any of the following ranges of \(t\)’s would also work.

\[ - 6\pi \le t \le - 2\pi \hspace{0.25in}\hspace{0.25in}2\pi \le t \le 6\pi \hspace{0.25in}\hspace{0.25in}6\pi \le t \le 10\pi \]

The direction of motion for each may be different range of \(t\)’s of course. Some will trace out the curve moving from left to right while others will trace out the curve moving from right to left. Because the problem did not specify a particular direction any would work.

Note as well that the range \( - 2\pi \le t \le 2\pi \) falls completely inside the given range of \(t\)’s specified in the problem and so we know that the particle will trace out the curve more than once over the full range of \(t\)’s. Determining just how many times it traces over the curve will be determined in the next step.

Show Step 5

Now that we have a range of \(t\)’s for one full trace of the parametric curve we can determine the number of traces the particle makes.

This is a really easy step. We know the total time the particle was traveling and we know how long it takes for a single trace. Therefore,

\[{\mbox{Number Traces = }}\frac{{{\mbox{Total Time Traveled}}}}{{{\mbox{Time for One Trace}}}} = \frac{{34\pi - \left( { - 52\pi } \right)}}{{2\pi - \left( { - 2\pi } \right)}} = \frac{{86\pi }}{{4\pi }} = \frac{{43}}{2} = 21.5{\mbox{ traces}}\] Show Step 6

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

For this sketch we indicated the direction of motion by putting arrow heads going both directions in places on the curve. We also included a set of \(t\)’s for a couple of points to illustrate where the particle is at while tracing out of the curve. The dashed line is the continuation of the parabola from Step 1 to illustrate that our parametric curve is only a part of the parabola.

Here is also the formal answers for all the rest of the information that problem asked for.

\[\begin{array}{lcc} {\mbox{Range of }}x{\mbox{ :}} & \hspace{0.1in} & - 4 \le x \le 4\\ {\mbox{Range of }}y{\mbox{ :}} & \hspace{0.1in} & - 1 \le y \le 1\\ {\mbox{Range of }}t{\mbox{ for one trace : }} &\hspace{0.1in} & - 2\pi \le t \le 2\pi \\ {\mbox{Total number of traces : }} & \hspace{0.1in} & 21.5\end{array}\]