# Parametric Equation Problem

1. Oct 29, 2007

### Feldoh

1. The problem statement, all variables and given/known data
A particle moves in the xy-plane so that its position at any time t, 0 =< t =< pi, is given by:
$$x(t) = \frac{t^2}{2}-ln(1+t)$$
$$y(t) = 3sint$$

-- At that time is the particle on the y-axis on the interval? Find the speed and acceleration vector of the particle at this time.

2. Relevant equations
None?

3. The attempt at a solution
I'm actually lost -- I believe that when x(t) = 0 it will be on the y-axis but I must be doing the algebra wrong I can't seem to get an exact solution...

2. Oct 29, 2007

### Dick

You can't solve it exactly algebraically. A well reasoned guess for a value of t such that x(t)=0 will give you the answer though.

3. Oct 29, 2007

### Feldoh

So my method is right up to the point I was at? I should mention I needed to find a t value such that 0 < t < pi, so I don't think 0 is the answer, I should have added that, sorry. Eye-balling the graph it's something like t=1.25 but my teacher claims that it is possible to get an exact solution, but not that you've confirmed it I really don't think there is...

Oh and, velocity is <x'(t), y'(t)> or (x'(t))i+(y'(t))j and acceleration is <x''(t), y''(t)> or (x''(t))i+(y''(t))j

4. Oct 29, 2007

### Dick

What's wrong with t=0?

5. Oct 29, 2007

### Feldoh

Nothing I just forgot to add that the problem asks for a value of t > 0. I know it works but the problem isn't asking for it I don't think. My mistake I should have clarified:(

Last edited: Oct 29, 2007
6. Oct 29, 2007

### Dick

Your post does say t>=0. You've eyeballed the t>0 root correctly. But there's no way you can solve for that one in terms of elementary functions.