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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.

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  • $\begingroup$ 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? $\endgroup$ – daOnlyBG Jan 27 '16 at 19:35
  • $\begingroup$ @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... $\endgroup$ – Frank Vel Jan 27 '16 at 19:47
  • $\begingroup$ lol, I think you're making this harder than it needs to be. If $a=c$ and $b=c$, how are $a,b$ related? $\endgroup$ – daOnlyBG Jan 27 '16 at 19:48
  • $\begingroup$ @daOnlyBG Maybe I am, I just think that it would be too simple and that there is another way... $\endgroup$ – Frank Vel Jan 27 '16 at 19:51
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    $\begingroup$ I don't know of anyone who succeeded by over-complicating matters :) Go ahead and use direct subsitution $\endgroup$ – daOnlyBG Jan 27 '16 at 19:51
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$t^3$ depends on $t$ as a direct function. Jacobian ( C, C') vanishes on its independent variables. By a function substitution both can be made identical.

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  • $\begingroup$ Is it possible to show this without the Jacobian? $\endgroup$ – Frank Vel Jan 27 '16 at 19:50
  • $\begingroup$ :) by inspection, as there is a $single $ parameter here. It happens naturally for function of a single independent variable. $\endgroup$ – Narasimham Jan 27 '16 at 20:01

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