3
    The curve $C$ is given by the Cartesian equation: $y^2=x^3+x^2$. Show that the function $\overrightarrow{r}:\mathbb{R^2} \rightarrow \mathbb{R}$ defined by $\overrightarrow{r}(t)=(t^2-1,t^3-t)$, represents the parameterization of the curve $C$.

I solved this question (and correct me if I'm wrong) by the following:

let $x=t^2-1$, then $y^2=(t^2-1)^3-(t^2-1)^2$

$y^2=(t^6-3t^3+3t^2-1)+(t^3-2t^2+1)$

$y^2=t^6-2t^4+t^2=t^2(t^4-2t^2+1)=t^2(t^2-1)^2$

then we get:

$y=t(t^2-1)=t^3-t$

My question is, how do I find the parameterization of $C$ if the question hadn't already defined $\overrightarrow{r}(t)=(t^2-1,t^3-t)$?

Knowing that $x=t^2-1$ made the question trivial, is there a way to figure what $x$ is equal to?

Bshara Zahran
  • 181
  • 3
    First, to prove rigorously that the given parameterization is correct, it should be proved that: (1) For any $(x,y)$ on curve $C$, there exists a $t$ such that $(x,y)=(t^2-1,t^3-t)$, (2) For any $t$, the point $(t^2-1,t^3-t)$ is on curve $C$. Here you've only verified (2). Second, this parameterization might be a by-product in the study of elliptic curves, but I'm not sure if there's any other natural way to derive this. – Ѕᴀᴀᴅ Jun 02 '18 at 07:43
  • Not sure the name strophoid is the right name for this curve. The common point is that they have an $\alpha$ shape... – Jean Marie Feb 04 '23 at 14:46

1 Answers1

6

As Alex Francisco commented, usually the plane curve defined by a cubic polynomial equation is an elliptic curve. Those cannot be parametrized by rational functions. The exceptions are the cubic curves with a singular point. Here you see that both partial derivatives vanish at the origin, a point on your curve, so we have a singularity at the origin.

The presence of a singularity changes everything (in the language of algebraic geometry the genus of the curve changes from one to zero). A recurring trick in algebraic geometry is to "blow up the singularity". We use the ratio $t=y/x$ as a parameter $t$ (if the singularity were at the point $(x_0,y_0)$ instead of the origin we would use $t=(y-y_0)/(x-x_0)$ instead. The idea is that using $t$ instead of $y$ simplifies the equation on such occasions.

Let's see. If $t=y/x$ then $y=tx$. Plugging this into your equation gives $$ t^2x^2=y^2=x^3+x^2. $$ Due to the singularity only terms quadratic (or higher) in $x$ and $y$ appeared in the original equation. Therefore this substitution made both sides divisible by $x^2$. If we assume (temporarily) that $x\neq0$ we can cancel that factor $x^2$ and end up with $$ t^2=x+1. $$ Lo and behold! We just solved that $x=t^2-1$! Consequently $$ y=tx=t(t^2-1). $$ Observe that if $x=0$ then $y=0$ is the only possibility. We do get $x=0$ from two different choices of $t$, namely $t=\pm1$. Both of those values, of course, give us the origin as the point $(x(t),y(t))$. If you plotted this curve you will see that it has two branches throught the origin. When $t\approx1$ you get points on the ascending branch (with slope close to $+1$) but with $t\approx-1$ you get points on the descending branch. Given that $t$ is the slope of the line segment from the origin to the point $(x,y)$ this was to be expected. Blowing up separates those two branches - in the $(x,t)$ coordinate system the origin is no longer "a double point", and the singularity is gone.

As an exercise I invite you to use the same trick and derive a rational parametrization for the points of the Folium of Descartes $$ x^3+y^3=3xy. $$ It also has a singularity at the origin.

Jyrki Lahtonen
  • 140,891
  • ♦ Interesting, can you provide a reference or a book to further study such subjects? (curves in general). And great answer! – Bshara Zahran Jun 02 '18 at 09:01
  • 1
    Many recommend Fulton's book, but it requires tools from abstract algebra. Quite a bit of algebra is necessary if you dig deeper, but IIRC there also was a version starting out with more examples. – Jyrki Lahtonen Jun 02 '18 at 09:40