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(Two parameterizations): Image

$$\vec r_1(t) = (t^3, t + 1), t ∈ [0, 1] $$

$$\vec r_2(t) = (t^6, t^2 + 1), t ∈ [0, 1]$$

How can I show that these two parameterizations represent the same line in plane?

Thought that maybe line integral would be the way, however I am not given any $f(x,y,z)$ function to integrate over. When using $f(x,y,z) = 1$ the integral becomes different.

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  • $\begingroup$ First it would be helpful to show that $r_1(s)$ and $r_2(t)$ are linearly dependent for all $s,t$. Then showing that they have the same start and endpoints, which should be obvious. Lastly, here we can use monotonicity to show that the lines are actually the same. $\endgroup$ – Pink Panther Apr 1 at 13:06
  • $\begingroup$ Write y in terms of t. Now write x in terms of y. $\endgroup$ – Paul Apr 1 at 13:11
  • $\begingroup$ @Paul So in this case r1(r2) = r2(r1) ? This makes r1(r2) = (t^18, t^2 + 2) , but r2(r1) = (t^18, t^2 + 2t + 2) $\endgroup$ – dondeman Apr 1 at 13:34
  • $\begingroup$ @Panther If they share the same endpoints would that mean that they are the same line? And what do you mean by your first comment "r1(s) and r2(t)", and what do you mean by monotonicity to show that the lines are the same? Thank you and sorry for the stupidity $\endgroup$ – dondeman Apr 1 at 13:50
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    $\begingroup$ These are the same lines because you obtain the second from the first by reparametrizing. Just define $t=s^2$. When $t\in[0,1]$, $s\in[0,1]$ and the reparametrization is smooth and monotonic. $\endgroup$ – GReyes Apr 1 at 13:56

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