# Show two parametrizations to be equal

Given the two curves \begin{align*}&\mathcal{C}\left\{\begin{matrix}u = t\\v = t\end{matrix}\right., & t\in [0,1]\\ \\ &\mathcal{C'}\left\{\begin{matrix}u = t^3\\v = t^3\end{matrix}\right., & t\in [0,1]\end{align*} Show these to be equivalent.

Intuitively I see that they must represent the same line, but I fail to see how I can show this mathematically.

• Hint: if we let $u=t^2, v=t^2$ be $C''$ on the same interval $[0,1]$, would $C$ and $C''$ also be the same line? – daOnlyBG Jan 27 '16 at 19:35
• @daOnlyBG I intuitively understand that $\mathcal{C}^{(n)}: u = t^n, v = t^n, n > 0, t \in [0,1]$ would make the same line, but I don't see how I can prove this... – Frank Vel Jan 27 '16 at 19:47
• lol, I think you're making this harder than it needs to be. If $a=c$ and $b=c$, how are $a,b$ related? – daOnlyBG Jan 27 '16 at 19:48
• @daOnlyBG Maybe I am, I just think that it would be too simple and that there is another way... – Frank Vel Jan 27 '16 at 19:51
• I don't know of anyone who succeeded by over-complicating matters :) Go ahead and use direct subsitution – daOnlyBG Jan 27 '16 at 19:51

$t^3$ depends on $t$ as a direct function. Jacobian ( C, C') vanishes on its independent variables. By a function substitution both can be made identical.
• :) by inspection, as there is a $single$ parameter here. It happens naturally for function of a single independent variable. – Narasimham Jan 27 '16 at 20:01