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Trigonometric Identity and Differential Equation question

  1. Oct 8, 2009 #1
    I'm looking over the differential equation describing a hanging cable in a textbook, and I probably need to review my trigonometric derivatives and integrals again because I'm not seeing how they got the following:

    [tex]\frac{dy}{dx} = tan(\phi) \frac{ws}{T_0}[/tex]

    [tex]\frac{d^2y}{dx^2} = \frac{w}{T_0}\frac{ds}{dx}[/tex]

    [tex]\frac{ds}{dx} = [1 + (\frac{dy}{dx})^2]^\frac{1}{2} [/tex]

    ds/dx is the secant of phi, or something...any pointers would be appreciated!
  2. jcsd
  3. Oct 9, 2009 #2


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    So what you are saying, then, is that y= y(x) is the equation of some graph and [itex]\phi(x)[/itex] is the angle that graph makes, at each point, with a horizontal line. In other words, [itex]tan(\phi)[/itex] is the slope of the tangent line: [itex]tan(\phi(x))= dy/dx[/itex] and s(x) is the arclength from some point to (x, y(x)).
    Last edited by a moderator: Oct 10, 2009
  4. Oct 9, 2009 #3
    Thanks Halls - I see now how they got there - ds/dx is the derivative of the formula for arc length which I apparently temporarily forgot. :wink:
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