# Thread: Particle moving on a line.

1. ## Vector calculus on particle movement

Hello,
I'm currently working on a question which is about a particle moving across along a line. The particle is moving along the line log (x) at a constant speed of one. When t = 0, the particle is at point (1,0). I have to find a) the velocity vector and b) the acceleration vector at (1,0).
I've start off by making $r(t) = (x(t), \log (x(t))$. To find the velocity vector, I must differentiate $r(t)$. So: $v(t) = r'(t) = (x'(t), \frac{x'(t)}{x(t)})$. This is where I get stuck. I know that the speed of the particle is 1 at any time t. So $\mid v(t)\mid = x'(t)\sqrt{1+\frac{1}{(x(t))^2}}=1$. I feel I need to find the value x'(t). Is this right? If so, how do I find it?
Thanks for your help.

2. This looks to be a differential equations problem. You've got $\frac{dx}{dt} \sqrt{1 + 1/x^2} = 1$ and you want to find $x$. This is a separable equation which can be solved by taking $\sqrt{1+1/x^2} dx = dt$ and integrating both sides. The fact that when t = 0, x = 1 gives you an initial condition. The problem is that integral on the left hand side is pretty nasty. I'm not sure that you'll be able to solve for x after all is said and done. Maybe see if someone on the differential equations board can figure this out.

3. Thanks, mate. I'll do just that.