Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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




Similar Discussions: Trigonometric Identity and Differential Equation question
Loading...