Help me with this Calculus problem

Discussion in 'Physics & Math' started by HonorAndStrength, Oct 8, 2006.

  1. HonorAndStrength I know nothing Registered Senior Member

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    friend needed answer, its been solved now, ty
     
    Last edited: Oct 9, 2006
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  3. James R Just this guy, you know? Staff Member

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    What have you done so far?
     
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  5. Facial Valued Senior Member

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    Will you need to know the mass?
     
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  7. Zephyr Humans are ONE Registered Senior Member

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    Which of Newton's 'laws' do you think might be useful in part 2?
     
  8. Roman Banned Banned

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    What's tanh(x+C)? Is that supposed to be tan(x+C), or the tangent of [h(x+C)]?
     
  9. Zephyr Humans are ONE Registered Senior Member

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  10. alyosha Registered Senior Member

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    For the second part, it looks like a differential equation that I either don't know how to solve or have forgotten how.

    ma = F + kv^2

    Where a, the acceleration, is the second derivative of the position function with respect to time, and v, the velocity, is the first derivative.
     
  11. Roman Banned Banned

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  12. D H Some other guy Valued Senior Member

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    I suspect you mean ma = F - kv^2. Air drag makes things slow down, not speed up.

    Look at the mathworld web site that Zephyr pointed out. It has the answer to your question.
     
  13. Pete It's not rocket surgery Registered Senior Member

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    It could be correct, if it's an object being slowed by reverse thrust + parachute, or something.
     
  14. D H Some other guy Valued Senior Member

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    If that were the case he wouldn't have got the tanh solution. That solution arises from m*dv/dt = F - kv^2.
     
  15. alyosha Registered Senior Member

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    Hey, I didn't come up with the formula! heh.
     
  16. D H Some other guy Valued Senior Member

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    Nonsense results arise if both F and k are positive. In particular, the speed will reach infinity in a finite time.

    Equations of motion:
    dx/dt = v
    m dv/dt = F + k*v^2
    with initial states at t=0 given by x=x0, v=v0

    Suppose F,k are both positive. Define
    v_c = sqrt(F/k)
    t_c = m/
    sqrt(F*k)
    u = v/v_c
    u0 = v0/v_c
    tau = t/t_c
    Dividing both sides of the velocity equation of motion by v_c yields
    du/dt = sqrt(k/F)*F/m + sqrt(k/F)*k/m*F/k*u^2
    or
    du/dt = 1/t_c*(1+u^2)
    du/dtau = du/dt*dt/tau = 1+u^2
    which has solution
    u(tau) = tan(tau+atan(u0))
    or
    v(t) = v_c*tan(t/t_c+atan(v0/v_c))

    The velocity goes to infinity as t -> t_c*(pi/2-atan(v0/v_c))

    If k is negative, the trigonometric functions become hyperbolic functions by Osborne's Rule -- or just rewrite as m dv/dt = F - k*v^2
     
  17. HonorAndStrength I know nothing Registered Senior Member

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    edited : fixed now
     
    Last edited: Oct 9, 2006
  18. Pete It's not rocket surgery Registered Senior Member

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    Hi HonorAndStrength,
    Replacing the whole original post is kind of bad form. It ruins the historical record of the discussion.
    Adding a note after the original question would be better. Something like this:

    Short question: find the integral of tanh(x + C) with respect to x.

    Long question: given that the force on an object is F + kv^2, where F is some arbitrary (but constant) force, k is a constant coefficient, and v is the velocity of the object, find the distance the object has travelled after t seconds.

    Edit: friend needed answer, its been solved now, ty
     

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