Trigonometric Identity and Differential Equation question

[bitrex]

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:

$$\frac{dy}{dx} = tan(\phi) \frac{ws}{T_0}$$

$$\frac{d^2y}{dx^2} = \frac{w}{T_0}\frac{ds}{dx}$$

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

ds/dx is the secant of phi, or something...any pointers would be appreciated!

 

[HallsofIvy]

So what you are saying, then, is that y= y(x) is the equation of some graph and $\phi(x)$ is the angle that graph makes, at each point, with a horizontal line. In other words, $tan(\phi)$ is the slope of the tangent line: $tan(\phi(x))= dy/dx$ and s(x) is the arclength from some point to (x, y(x)).

 

[bitrex]

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.