# Finding position from velocity (trig function)

by tasveerk
Tags: function, position, trig, velocity
 P: 24 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.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,235 $$v(t)=\frac{dx}{dt}\quad\to\quad\frac{dx}{x\sin(x^2)}=dt$$ Is the equation on the right what you integrated?
 P: 24 @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.
Mentor
P: 11,200

## Finding position from velocity (trig function)

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?
 Emeritus Sci Advisor HW Helper PF Gold P: 7,235 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.
Mentor
P: 11,200
 Quote by SammyS Now integrate both sides to find t as a function of x.
I think that the LHS is going to prove to be rather difficult to integrate in closed form.
Emeritus
Your earlier suggestion: $$v(t)=t\sin(t^2)$$ is probably correct.