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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

$t^3$ depends on $t$ as a direct function. Jacobian ( C, C') vanishes on its independent variables. By a function substitution both can be made identical.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.