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Finding position from velocity (trig function)

  1. Mar 2, 2011 #1
    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. jcsd
  3. Mar 2, 2011 #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?
     
  4. Mar 2, 2011 #3
    @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.
     
  5. Mar 2, 2011 #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?
     
  6. Mar 2, 2011 #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.
     
  7. Mar 2, 2011 #6

    gneill

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    I think that the LHS is going to prove to be rather difficult to integrate in closed form.
     
  8. Mar 2, 2011 #7

    SammyS

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    Yes, I believe you are right about this.

    Your earlier suggestion: [tex]v(t)=t\sin(t^2)[/tex] is probably correct.
     
  9. Mar 3, 2011 #8
    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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