## More Difficult Parametric Sketching

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):

 t -5 -3 -2 -1 -0.5 0 0.5 1 2 3 5 x 0.3629 1.0334 2.5714 $$\to \pm \infty$$ -5.142 0 4 4.5 2 0.9642 0.3571 y -1.814 -3.115 -5.142 $$\to \pm \infty$$ 2.5714 0 2 4.5 4 2.8928 1.7857

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: