Question: A curve is given parametrically by `x=frac{9t}{1+t^3}` and `y=frac{9t^2}{1+t^3}`. Sketch the curve and comment on the behaviour as \(t \to -1\).

Solution:

We know that as \(t \to -1\) from below, `1+t^3 to 0` from below (i.e. `1+t^3` tends to zero but is always negative as it does so). This means that as \(t \to -1\) from below:

`x=frac{9t}{1+t^3} to frac{-9}{-0} to infty`  and  `y=frac{9t^2}{1+t^3} to frac{9}{-0} to - infty`

Similarly, as \(t \to -1\) from above, `1+t^3 to 0` from above (i.e. `1+t^3` tends to zero but is always positive as it does so). This means that as \(t \to -1\) from above:

`x=frac{9t}{1+t^3} to frac{-9}{0} to - infty`  and  `y=frac{9t^2}{1+t^3} to frac{9}{0} to infty`

I.e. as \(x \to - \infty\)  ,  \(y \to \infty\) and vice versa. We can start to draw this onto a graph like so:

We can also try to see where the graph crosses the axes. When \(x=0\)  ,    `frac{9t}{1+t^3}=0` and so `t=0`. When `t=0`, `y=frac{9t^2}{1+t^3}=0` and therefore the graph crosses the axes.

We should also consider what happens as \(t \to \pm \infty\). As \(t \to \pm \infty\)  ,   \(x \to \frac{9t}{t^3}=\frac{9}{t^2} \to 0\) and similarly \(y \to \frac{9t^2}{t^3}=\frac{9}{t} \to 0\). Therefore we see that in fact, the graph crosses the origin three times: at `t=0` and \(t=\pm \infty\). We can conjecture that this means that the graph must loop into itself somehow so that it crosses the same point multiple times. At this point, a table of values can shed light on the actual shape of the graph (although this is not very mathematical, it is the method which is often used at A-Level. If you want a more mathematical method, you can differentiate x and y with respect to t to see where the turning points are and that should give you the same result, but I will leave that as an exercise for the reader):

Now we've sketched the graph, but if you sketch it on a computer and zoom out a bit, it seems to tend to being a straight line at infinity:

This makes an answer of "as \(x \to - \infty\)  ,  \(y \to \infty\) and vice versa" somewhat unsatisfying and incomplete. There may be different ways of identifying the straight line that the graph tends to, but my method is as follows:

\[x+y=\frac{9t}{1+t^3}+\frac{9t^2}{1+t^3}\]

\[x+y=\frac{9t+9t^2}{1+t^3}\]

\[x+y=\frac{9t(1+t)}{1+t^3}\]

Now using the sum of two cubes (for a proof of this, see The Difference (and Sum) of Two Powers), \(1+t^3=(1+t)(t^2-t+1)\). Therefore:

\[x+y=\frac{9t(1+t)}{(1+t)(t^2-t+1)}\]

\[x+y=\frac{9t}{t^2-t+1}\]

(Notice that although we are about to set \(t=-1\), we are essentially doing this as a limit, so it is still valid to divide by \((1+t)\) as we have done, since we are considering what happens as \(t\to -1\) and not what happens when \(t=-1\))

Now recall that the graph shoots off to infinity as `t to -1`. Therefore as `t to -1` :

\[x+y \to \frac{-9}{(-1)^2-(-1)+1}\]

\[x+y \to \frac{-9}{3}\]

\[x+y \to -3\]

Therefore as the graph shoots off to infinity, it tends to the line \(x+y=-3\) or \(y=-x-3\). If we draw this line onto our graph we can see that this is indeed correct: