# Finding position from velocity (trig function)

1. Mar 2, 2011

### tasveerk

1. The problem statement, all variables and given/known data
S(0)=3, find S(2) position wise.

2. Relevant equations

V(t)=xsin(x^2)

3. The attempt at a solution
I tried to integrate with u-substitution and I got -t^4/4cos(t^2). I tested it by taking the derivative and it didn't work out.

Last edited: Mar 2, 2011
2. Mar 2, 2011

### SammyS

Staff Emeritus
$$v(t)=\frac{dx}{dt}\quad\to\quad\frac{dx}{x\sin(x^2)}=dt$$

Is the equation on the right what you integrated?

3. Mar 2, 2011

### tasveerk

@SammyS,
Thanks for the reply, but I have never seen the method you used before. I understand that you manipulated the first equation to get the second, but I do not know why. If you could explain it a bit or give me a link to a website that explains it I would appreciate it.

4. Mar 2, 2011

### Staff: Mentor

Are you sure that your velocity function is v(t) = x*sin(x^2), where x is a distance? Is it possible that it's v(t) = t*sin(t^2) instead?

5. Mar 2, 2011

### SammyS

Staff Emeritus
If  $$\frac{dx}{dt}=x\sin(x^2)\,,$$

then  $$\frac{1}{x\sin(x^2)}\ \frac{dx}{dt}\,dt=dt\,.$$

But,  $$\frac{dx}{dt}\,dt=dx\,.$$

Therefore,  $$\frac{dx}{x\sin(x^2 )}=dt$$

Now integrate both sides to find t as a function of x.

6. Mar 2, 2011

### Staff: Mentor

I think that the LHS is going to prove to be rather difficult to integrate in closed form.

7. Mar 2, 2011

### SammyS

Staff Emeritus
Your earlier suggestion: $$v(t)=t\sin(t^2)$$ is probably correct.