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How do I derive this equation? HELP

  1. Sep 13, 2008 #1
    How do I derive this equation?!? HELP!!!

    I started with the equation:
    y=yinitial + (vinitialsintheta)t -1/2gt^2

    I plugged in t=x/(vinitialcostheta)

    to get: y=yinitial + x (tantheta) - [(gx^2)/(2(vinitial^2)(cos^2theta))]

    Using derivatives, how do I get from this equation:

    y=yinitial + x (tantheta) - [(gx^2)/(2(vinitial^2)(cos^2theta))]

    to this equation:

    (dy)/(dtheta)=
    x(sec^2theta)-[(gx^2(tantheta)(sec^2theta))/(vinitial^2)]

    I am completely and totally and utterly confused, bewildered, perplexed, and a list of other things.
    Plese help!!!!!
     
  2. jcsd
  3. Sep 13, 2008 #2
    Re: How do I derive this equation?!? HELP!!!

    To get from y=yinitial + x (tantheta) - [(gx^2)/(2(vinitial^2)(cos^2theta))]

    to this equation:

    (dy)/(dtheta)=
    x(sec^2theta)-[(gx^2(tantheta)(sec^2theta))/(vinitial^2)] ---

    Taking y and x not to vary with theta.. then they are taken as constants.
    differentiating yinitial wrt theta gets 0,
    differentiating x tantheta wrt theta gets x sec^2 theta (differentiate tangent to get sec^2)

    You could use quotient rule for the last term to get
    [gx^2 . (2 cos theta sin theta)] / [2 vinitial (cos^4 theta)]

    before simplifying to get the eventual equation..
     
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