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Question about two-variable function

  1. Nov 27, 2011 #1
    If I have X(a,t) = .5at2 and a and t were both functions of F
    a(F) = F-g
    t(F) = gT/F

    and I wanted to know how much X changes with respect to F, how would I do that?
  2. jcsd
  3. Nov 27, 2011 #2
    Are g and T constant?
  4. Nov 27, 2011 #3
    yes, they are
  5. Nov 27, 2011 #4
    First step substitute (F - g) for a and gT/F for t to get x in terms of F and the constants.
  6. Nov 27, 2011 #5
    Oh, so is this right?

    x(F)= g2T2/2F - g3T2/2F2
    x(F) = g2T2/2(F-1-gF-2)
    x'(F) = g2T2/2(-F-2+2gF-3)
  7. Nov 27, 2011 #6
  8. Nov 27, 2011 #7
    Okay, so
    x(F) = g2T2/2F - g3T2/2F2
    x'(F) = -g2T2/2F2 + g3T2/F3

    That is correct?
  9. Nov 27, 2011 #8
    I would have a line showing

    [itex]\frac{g^{2}T^{2}F^{-1}}{2}[/itex] - [itex]\frac{g^{3}T^{2}F^{-2}}{2}[/itex]
  10. Nov 27, 2011 #9
    Your differentiation is correct.
  11. Nov 27, 2011 #10
    Right, that is before taking the derivative?

    I am not sure if I need to take the derivative now, so I guess let me lay it on ya

    gT = Ft
    gX = Fx

    a = F-g
    t = gT/F

    x = 1/2at2
    x = g2T2F-1/2 - g3T2F-2/2

    If I am trying to find F for a given g, X, and T

    I could just put x = gX/F in for x and get

    gF-1X = g2T2F-1/2 - g3T2F-2/2
    divide by gF-1
    gives me
    X=gT2/2 - g2T2F-1/2
    multiply by F and rearrange a bit to get
    gT2F/2 - FX = g2T2/2

    pull out F for F((gT2-2X)/2) = g2T2/2

    divide for

    F = g2T2/(gT2-2X)

    Any of that look right?
  12. Nov 27, 2011 #11
    I do not know what you are doing now.

    Your final line x'(F) = -g2T2/2F2 + g3T2/F3

    was ok. That is you were asked to find dX/dF and you found it because your differentiation was ok.
  13. Nov 27, 2011 #12
    Right, the last post was not using the derivative, I might not need to.

    If the non-differentiated version is correct, and i know the equation
    gX = Fx
    can I not solve for x = gX/F
    substitute that back into the equation I had gotten for x(F), and then I have F in terms of only g, X, and T, getting rid of x and t?

    The resultant equation should be correct to give me F for a given g, X and T, right?
  14. Nov 27, 2011 #13
    From where did you get gX = Fx?
  15. Nov 27, 2011 #14
    Right, so what I am looking at here is g = gravity, X is the total displacement, F is a force that is greater than and opposite to gravity, and x is a distance less than X, at s=0 the velocity is 0 and s=X the velocity is 0, so the potential energy given by gX must equal the energy put into the system Fx

    Or to keep this as he math sub-forum, it is a given from post#10
    Last edited: Nov 27, 2011
  16. Nov 27, 2011 #15
    Energy put in system will be given by Fx ONLY if F is constant and if F is in the same direction of the displacement x.
  17. Nov 27, 2011 #16
    Which they both indeed are
  18. Nov 27, 2011 #17
    A constant force F whose magnitude is greater than that of gravity, is working against gravity to give a displacement of x, it gains (F-g)x kinetic energy, and gx potential energy is put into the system. At that point F drops to 0, so the system's energy remains constant, it's kinetic energy is put into potential energy by gravity until KE reaches 0, and (F-g)x+gx = gX which reduces to Fx = gX
  19. Nov 27, 2011 #18
    It is getting rather late over here...Sorry.
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