1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about two-variable function
Loading...