Finding position from velocity (trig function)

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.

SammyS

[tex]v(t)=\frac{dx}{dt}\quad\to\quad\frac{dx}{x\sin(x^2)}=dt[/tex]

Is the equation on the right what you integrated?

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.

gneill

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?

SammyS

If  [tex]\frac{dx}{dt}=x\sin(x^2)\,,[/tex]

then  [tex]\frac{1}{x\sin(x^2)}\ \frac{dx}{dt}\,dt=dt\,.[/tex]

But,  [tex]\frac{dx}{dt}\,dt=dx\,.[/tex]

Therefore,  [tex]\frac{dx}{x\sin(x^2 )}=dt[/tex]  

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

gneill

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

SammyS

Yes, I believe you are right about this.

Your earlier suggestion: [tex]v(t)=t\sin(t^2)[/tex] is probably correct.

tasveerk

I switched t with x in the equation. Now that I think about it I'm even unsure of why I did this. Anyway, I solved the problem by taking t out of the function before integrating. Thank you all for the quick replies.