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I've a doubt concerning parametrization of $f(x) = x^2$. I'm studying curves can be parametrized in different ways, changing the speed with which they travel their path. I'm pretty sure that if I'll write, $$ f(t)=(t,t^2)$$ and $$ f(t)=(2t,4t^2)$$ they represents the same curve travelled with different speed.

I'm not sure about it if we write something like, $$ f(t)=(t^2,t^4)$$
Does the fact that the components of the latter aren't anymore multiples of the previous ones change something?
I know the question can seem a bit ambiguous and wide, but any explanation or advice on the topic is welcomed with enthusiasm!

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    $f(2t, 4t^2)$ would be the correct parameterization. But the speed at which you move along the curve is irrelevant. All that matters is that for any $t$, you map to a unique (and the correct) $(x,y)$. There is a problem with $f(t^2,t^4)$ if the domain of $x$ includes the negative real numbers. – 2017-02-02
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    Second is $(2t,4t^2)$. – 2017-02-02
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    I suggest writing explicitly the coordinates $x$, $y$ and using different letters for different parameters. For instance, at the beginning you have that $y=x^2$ can be represented by $x=t$, then obviously $y=t^2$. If you say $x=2 u$, then $y=x^2\neq 2 u^2$; then $x=v^2$, $y=x^2=v^4$. The parameter is arbitrary. – 2017-02-02
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    Yeah sorry guys, Thank you for pointing it out! – 2017-02-02

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Your notation is a bit muddy but your question is a good one.

To clear the notation, your first parameterization is $$\left\{ \matrix{ x=t\\f = t^2} \right. $$ and your second is $$ \left\{ \matrix{ x=2t\\f = 2t^2} \right. $$ and these are not equivalent curves; but if you meant $$ \left\{ \matrix{ x=2t\\f = (2t)^2} \right. $$ then those would be equivalent only with different "speeds" along the two parameterizations.

Your third curve $$ \left\{ \matrix{ x=t^2\\f = (t^{2})^{2} } \right. $$ would be the same except for one important fact: The range of $x$ is no longer $(-\infty,\infty)$, it is instead $(0,\infty)$. So this is a parameterization of only the left half of the original curve.

That is, you have the right idea but you need to be careful that the range of $x$ is fully covered.

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    Thank you very much! I know it can be very boring answering this kind of questions, so I appreciate it even more! thank you also for pointing out the mistake and answering anyway! – 2017-02-02