How does $(t^3, t^2)$ represent $x^{2/3}$?

Question

I just learnt about this parameterization and somehow I am not being able to wrap my head around it. How on earth is the curve $\left<x, x^{2/3}\right> = \left<t^3, t^2\right>$.

I got this from Claudio Arezzo's class on differential geometry and I have verified it manually but I still can't believe it / comprehend it.

Answer 1

If you make the variable substitution $x=t^3$ in the formula $\langle x, x^{2/3} \rangle$ you get $$\langle t^3, (t^3)^{2/3} \rangle = \langle t^3,t^2 \rangle $$ You can also do this in the reverse direction, substitute $x=t^{1/3}$ into the formula $\langle t^3,t^2\rangle$ and you get $$\langle (x^{1/3})^3, (x^{1/3})^2 \rangle = \langle x,x^{2/3} \rangle $$ This tells you that, as subsets of the coordinate plane, we have an equation $$\{(x,x^{2/3}) \mid x \in \mathbb R \} = \{(t^3,t^2) \mid t \in \mathbb R\} $$ which is what is really meant by the not-really-very-rigorous equation $ \langle x,x^{2/3}\rangle = \langle t^3,t^2 \rangle$.

And by the way, it should not really be a big surprise that one can parameterize geometric shapes using different parameters. Just as an example, the line $x+y=1$ can be parameterized as $\langle x,1-x\rangle$ or as $\langle 1-y,y \rangle$ as well as in multiple other ways.

Answer 2

Start with the parametric graph $$ \begin{cases}x=t^3\\y=t^2\end{cases} $$

We want to eliminate the parameter $t$ and just get $y=f(x)$. To do this:

Solve $x=t^3$ for $t$ to get $t = x^{1/3}$. Then substitute this into $y=t^2$ to get $y = (x^{1/3})^2$.