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:(adsbygoogle = window.adsbygoogle || []).push({});

[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!

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# Trigonometric Identity and Differential Equation question

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