# Homework Help: Re-Parameterizing a Curve

1. Nov 30, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
$C(t) = t =it^2$, where $-2 \le t \le 2$. Re-parametrize the curve from $t$ to $\tau$ by the following transformation: $\tau = \frac{t}{2}$.

2. Relevant equations

3. 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?

2. Nov 30, 2014

### Orodruin

Staff Emeritus
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 $C$.

3. Nov 30, 2014

### 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?

4. Nov 30, 2014

### Orodruin

Staff Emeritus
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.

5. Nov 30, 2014

### 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?

6. Nov 30, 2014

### Orodruin

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

7. Nov 30, 2014

### Bashyboy

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

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

8. Nov 30, 2014

### Orodruin

Staff Emeritus
$|2it|$ is not equal to $2t$ when $t$ is negative.