Finding position from velocity (trig function)

  • Thread starter tasveerk
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  • #1
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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.
 
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Answers and Replies

  • #2
SammyS
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[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?
 
  • #3
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@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
gneill
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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
SammyS
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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.
 
  • #6
gneill
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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.
 
  • #7
SammyS
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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: [tex]v(t)=t\sin(t^2)[/tex] is probably correct.
 
  • #8
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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.
 

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