# Reparameterizing curves

1. Jun 16, 2004

### mathrocks

I have a question about reparameterizing curves. My specific question is "Reparameterize the curve with respect to the arc length measured from the point where t=0 in the direction of increasing t. r(t)=e^t*sint i +e^t*cost j"

I understand the whole process about finding the answer, but I have one important problem, when they give you a point "t=0" does that point plug into the derivative of r(t) and you solve for |r(t)| from there? And if I replace that t=0 with a point (1,1) or something, how do I go about dealing with that?

Thanks!

2. Jun 16, 2004

### arildno

First of all:
This has nothing to do with (re)-parametrizing a curve; rather, it is to identify the point in the plane on the curve that the parameter value t=0 determines.

Note that we have the arclength-formula from the beginning of a curve at t=0:
$$s(t)=\int_{0}^{t}||\frac{d\vec{r}}{d\tau}||d\tau$$

Hence, the first thing you have to do in order to reparametrize, is to calculate the integral!

The second thing to, is to invert the relationship between t and s; i.e., rather than having the function s(t), you must find the correct function t(s).

the third thing you must do, is to substitute t(s) for t in your parametric equation.

3. Jun 16, 2004

### mathrocks

Right, I understand I have to find the integral first, but to make the integral easier, my teacher told use to first determine |r'(t)|, which gives you a number and when finding the integral you simply add a 't' to the end of the number which gives you the function s(t). And then you you find t(s) and plug it into the original equation.

My question was when you are given a point (1,1) rather than t=0 how do you find the |r'(t)| since the point (1,1) corresponds to (x,y)??

4. Jun 16, 2004

### master_coda

There's no reason to make the integral easier. This isn't a complicated integral that needs some "trick" to help you see the solution. And since the trick just seems to be confusing you (and I'm pretty sure it's not even correct, unless I'm misunderstanding your description of it) there really isn't any reason to use it. You're better off just solving things the proper way. As was already said, the proper thing to do to get s(t) is to solve:

$$s(t)=\int_0^t\lVert \mathbf{r}^{\prime}(\tau) \rVert\;d\tau$$

Once you solve that you just find t(s) and then plug it into r(t).

Edited to add: if you really insist, you can find the t value associated with some (x,y) by solving e^t*sin(t) = x and e^t*cos(t) = y for t in terms of x and y. You can then use that t value to find ||r'(t)|| at (x,y). But that's very messy.

Last edited: Jun 16, 2004
5. Jun 17, 2004

### arildno

Hint:
$$\frac{d\vec{r}}{dt}=e^{t}((\cos(t)-\sin(t))\vec{i}+(\cos(t)+\sin(t))\vec{j})$$
$$||\frac{d\vec{r}}{dt}||=\sqrt{2}e^{t}$$