Re-Parameterizing a Curve: How to Transform from t to tau?

[Bashyboy]

Homework Statement

##C(t) = t =it^2##, where ##-2 \le t \le 2##. Re-parametrize the curve from ##t## to ##\tau## by the following transformation: ##\tau = \frac{t}{2}##.

Homework Equations

The Attempt at a Solution

So, the variable ##\tau## is half of every value ##t## can be. Therefore, I have construct a ##\tau##-interval from the ##t##-interval.

##\frac{-2}{2} \le \tau \le \frac{2}{2} \implies -1 \le \tau \le 1##

So, ##C(\tau) = \tau + i \tau^2##.

One, I am not even certain that this is correct; two, this is absolutely unpalatable. I am having a hard time justifying every step.

At first I thought I would just solve for tau, and substitute the variable in:

##C(2 \tau) = 2 \tau + i (2 \tau)^2##.

But how would I find the interval for ##\tau##? Why would I even want to? Does ##\tau## and ##t## being related by the equation ##\tau = \frac{t}{2}## imply that there intervals are somehow related? How so? Would I think of ##\tau = \frac{t}{2}## as a function, whose domain is ##-2 \le t \le 2##, and the range of this function would be the ##\tau## interval? If so, why?

 

[Orodruin]

You should have gone with your first thought and just solved for tau and substituted. You can also think of ##\tau = t/2## as a function in the way you mention and the range is the interval for ##\tau##. This gives you the same curve ##C##.

 

[Bashyboy]

Okay, but how I find the ##\tau##-interval? What justifies me dividing the endpoints of ##-2 \le t \le 2## by ##2##, and then calling this the ##\tau## interval?

 

[Orodruin]

You are reparametrising with ##\tau = t/2## so any value of ##t## is going to correspond to a value of ##\tau## which is ##t/2##, in particular the endpoints.

 

[Bashyboy]

So, I can regard the codomain (range) of the function ##\tau = \frac{t}{2}## as the ##\tau##-interval?

I have another question which relates to arcs. I was asked to calculate the length of the arc ##C(t) = 1 + it^2##, where ##-2 \le t \le 2##, the length being defined as ##\int_a^b |z'(t)|dt##. I got zero, which was slightly disconcerting. Is this really a good definition of length if it gives an answer of zero? Shouldn't this actually be referred to as displacement of a particle?

 

[Orodruin]

It should not give you zero since ##|z'(t)| \geq 0## - unless ##z## is a constant you should get a positive number.

 

[Bashyboy]

##z(t) = 1 + it^2 \implies z'(t) = 2it##

##\int_{-2}^2 |2it| dt = \int_{-2}^2 2t dt = t^2 \bigg|_{-2}^2 = (2)^2 - (-2)^2 = 0##

 

[Orodruin]

##|2it|## is not equal to ##2t## when ##t## is negative.