# Homework Help: Arclength math question

1. Feb 28, 2008

### foxjwill

1. The problem statement, all variables and given/known data
Given $$r(t) = x(t)\textbf{i} + y(t)\textbf{j}$$ and

$$\int_0^t \left\|\frac{dr}{dt}\right\| d\tau = t.$$

find, if possible, a closed form expression for x(t) and y(t).

2. Relevant equations

3. The attempt at a solution
I started by applying the fundamental theorem of calculus

$$\left\|\frac{dr}{dt}\right\|=1$$

then evaluated and simplified and got

$$x(t)^2 + y(t)^2 = 1.$$

Is it possible to continue from here?

2. Feb 28, 2008

### e(ho0n3

How are you getting $x(t)^2 + y(t)^2 = 1$?

3. Feb 28, 2008

### Astronuc

Staff Emeritus
Not sure how you determined this from the fundamental theorem.

$$x(t)^2 + y(t)^2 = 1.$$ This applies to a unit circle.

Why t in this equation?
$$\int_0^t \left\|\frac{dr}{dt}\right\| d\tau = t.$$

The integral should provide length, not time.

4. Feb 28, 2008

### foxjwill

Oops. I meant

$$x'(t)^2 + y'(t)^2 = 1.$$

As to how I got there:
$$\frac{d}{dt}\int_0^t \left\|\frac{dr}{dt}\right\| d\tau = \frac{dt}{dt}$$

$$\left\|\frac{dr}{dt}\right\| = 1$$

$$\sqrt{x'(t)^2 + y'(t)^2} = 1$$

$$x'(t)^2 + y'(t)^2 = 1.$$

I'm basically trying to find some vector-valued function r(t) in $$\texttt{r}^2$$ such that $$\forall t, s(t) = t$$, where $$s(t)$$ is arclength.

5. Feb 28, 2008

### e(ho0n3

That's better. You can play around with this by integrating with respect to t. However, you have two unknowns and one equation. I suspect that, without some additional assumptions, you won't be able to determine x(t) and y(t).

Err, why would you do that? Since the integrator is $d\tau$, you can take |dr/dt| out because it doesn't involve $\tau$.

6. Feb 28, 2008

### e(ho0n3

This makes no sense to me. Where did s(t) come from? How are you going from a vector-valued function r(t) to a scalar-valued function s(t)?

7. Feb 28, 2008

### foxjwill

Again, a mistake. >_<
$$\frac{d}{dt}\int_0^t \left\|\frac{dr}{d\tau}\right\| d\tau = \frac{dt}{dt}$$

$$s(t) := \int_0^t \left\|\frac{dr}{d\tau}\right\| d\tau$$

8. Feb 28, 2008

### e(ho0n3

I will assume that the derivative inside the integral is $d/d\tau r(\tau)$ for otherwise the integral would evaluate to 0.

The integral pretty ugly. Taking the derivative of the integral with respect to t doesn't help.

9. Feb 29, 2008

### HallsofIvy

Yes, it does. By the fundamental theorem, mentioned before,
$$\left|\left|\frac{dr}{dt}\right|\right|= 1[tex] just as he said before. That gives, just as he said before, [tex]\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)= 1[/itex]. In fact, that trajectory is the unit circle. 10. Feb 29, 2008 ### e(ho0n3 If you bring the d/dt inside the integral and evaluate the inside of the integral before integrating, then since the integrand is with respect to tau, the derivative with respect to t will be 0. Am I missing something? 11. Feb 29, 2008 ### foxjwill Yes, the fundamental theorem of calculus which says [tex]\frac{d}{dx}\int_a^x f(t) dt = f(x)$$

where a is some constant.

$$\int_a^x f(t) dt$$ is a function of x. In other words, you can't bring the $$\frac{d}{dx}$$ inside the integral because that would be like saying $$\frac{d}{dx}\left [x^2\right] = \left[\frac{d}{dx}(x)\right]^2,$$ which is obviously untrue.

12. Feb 29, 2008

### e(ho0n3

Right. So it does simplify to $x(t)^2 + y(t)^2 = 1$. Thus, x(t) = cos t and y(t) = sin t is a possible solution right?