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Derivative of projectile parametric y-component?

  1. Aug 15, 2003 #1
    Hi,

    I'm looking for the derivative of the projectile parametric y-component?

    The y component is:

    y = (vi/k)(1-e^-kt)(sin a) + (g/k^2)(1 - kt - e^-kt)

    I seem to be doing something wrong and my derivative isnt working out, I just want to check it against the final answer to see where I am going wrong.

    Thanks

    Matt
     
  2. jcsd
  3. Aug 15, 2003 #2

    Integral

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    I get
    dy/dt = Vie-ktsin(a)+g(e-kt-1)/k
     
  4. Aug 16, 2003 #3
    Yep, that's it. Thanks, Integral.

    I forgot d/dx [ e^u ] is e^u * du/dx

    Ah, the devil is in the details! Now to step two, how to solve it? I logged both sides, used the power rule and everything else I could think of, but it is not working out. Any other suggestions beside logging?

    [Edited for spelling mistake]
     
  5. Aug 17, 2003 #4
    No ideas? :(
     
  6. Aug 17, 2003 #5

    HallsofIvy

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    What do you mean by "solve it"?? Presumably you don't mean "How did you get that answer" because you said "Yep, that's it". What exactly do you want to do with the derivative?
     
  7. Aug 17, 2003 #6
    Ooops, sorry.

    Im settng the derivative to zero and solving for t.
     
  8. Aug 17, 2003 #7
    The first term contains an exp(-kt). Expand out the second term into two seperate terms, one of which will contain the same exp(-kt) factor. Combine terms containing the exponential, move everything else over to the other side and once the exponential is isolated, then take the natural log of both sides.

    I got something like

    t = (-1/k)*ln[(g/k)*(Vi*sin(a)+g/k)^(-1)]
     
    Last edited: Aug 17, 2003
  9. Aug 17, 2003 #8
    Thanks, Futz. I was combining the exponentials wrong.

    Here it is solved for t:

    [​IMG]
     
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