Finding position from velocity (trig function)

[tasveerk]

Homework Statement

S(0)=3, find S(2) position wise.

Homework Equations

V(t)=xsin(x^2)

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]

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

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  \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.

 

[gneill]

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.

 

[SammyS]

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

Yes, I believe you are right about this.

Your earlier suggestion: v(t)=t\sin(t^2) 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.