Hyperbolic Functions

Definitions
\[ \newcommand{\ex}{\mathrm{e}} \sinh(x) = \frac{\ex^x-\ex^{-x}}{2} \qquad \mathrm{csch}(x) = \frac{1}{\sinh(x)} \] \[ \newcommand{\ex}{\mathrm{e}} \cosh(x) = \frac{\ex^x+\ex^{-x}}{2} \qquad \mathrm{sech}(x) = \frac{1}{\cosh(x)} \] \[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \qquad \coth(x) = \frac{1}{\tanh(x)} \]
Basic Identities
\[ \sinh(-x) = -\sinh(x) \qquad \cosh(-x) = \cosh(x) \] \[ \sinh^2(x) - \cosh^2(x) = 1 \qquad 1 - \tanh^2(x) = \mathrm{sech}^2(x) \] \[ \sinh(x+y) = \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \] \[ \cosh(x+y) = \cosh(x)\cosh(y) + \sinh(x)\sinh(y) \]
Derivative Formulas
\[ \frac{\mathrm{d}}{\mathrm{d}x} \sinh(x) = \cosh(x) \qquad \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{csch}(x) = -\mathrm{csch}(x)\coth(x) \] \[ \frac{\mathrm{d}}{\mathrm{d}x} \cosh(x) = \sinh(x) \qquad \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{sech}(x) = -\mathrm{sech}(x)\tanh(x) \] \[ \frac{\mathrm{d}}{\mathrm{d}x} \tanh(x) = \mathrm{sech}^2(x) \qquad \frac{\mathrm{d}}{\mathrm{d}x} \coth(x) = -\mathrm{csch}^2(x) \]
Logarithmic Formulas for Inverse Hyperbolic Functions
\[ \mathrm{arsinh}(x) = \ln\left(x+\sqrt{x^2+1}\right) \] \[ \mathrm{arcosh}(x) = \ln\left(x+\sqrt{x^2-1}\right),\, x\geq 1 \] \[ \mathrm{artanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right),\, |x| \lt 1 \]
Derivatives of Inverse Hyperbolic Functions
\[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arsinh}(x) = \frac{1}{\sqrt{1+x^2}} \qquad \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arcsch}(x) = -\frac{1}{|x|\sqrt{x^2+1}} \] \[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arcosh}(x) = \frac{1}{\sqrt{x^2-1}} \qquad \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arsech}(x) = -\frac{1}{x\sqrt{1-x^2}} \] \[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{artanh}(x) = \frac{1}{1-x^2},\, |x| \lt 1 \qquad \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arcoth}(x) = \frac{1}{1-x^2},\, |x| \gt 1 \]
When a cable of uniform linear density is hung between two poles (in a uniform gravitational field) it’ll sag in a “u”-shaped curve under its own weight. This curve is called a catenary Considered as a curve \(y = f(x)\) in the \((x,y)\)-plane, symmetric about the \(y\)-axis, and with its lowest point on the \(y\)-axis, it can be shown that the catenary curve must satisfy the differential equation \[ \ddot{y} = \frac{\rho g}{T}\sqrt{1 + \dot{y}^2} \] where \(\rho\) is the cable’s linear density, \(g\) is acceleration due to gravity, \(T\) is the tension on the cable at its lowest point, \(\dot y\) is the first derivative of \(f(x)\), and \(\ddot y\) is the second derivative of \(f(x)\).
  1. Verify that this function \(f\) is a solution to that differential equation. Note that \(C\) is just some vertical offset. \[ y = f(x) = \frac{T}{\rho g}\cosh\left(\frac{\rho g x}{T}\right) + C \]
  2. Investigate the units of every parameter involved in the formula \(f(x)\). In particular verify that the argument of \(\cosh\) and the output of \(\cosh\) are both scalar, unit-less quantities.

  3. Desmos

    Suppose you have two utility poles embedded in the ground 100 meters apart with a power line of linear density 2kg/m running between them fastened 10 meters up each pole. Use \(g = 9.80\)m/s² for gravity in Grand Junction due to elevation.
    1. If the power line is rated for a maximum tension of 20,000N. How much will the line sag at its lowest point if we hang it such that it experiences this tension at its lowest point?
    2. If we hang the power line at this maximum tension, at what angle \(\theta\) will the line be descending where it initially hangs from the pole?

    3. We don’t want to push the line near its limits, maxing out its tension, though. We only need to allow for a few meters of clearance under the wire at the lowest point. Hanging the line such that it allows for 5 meters of clearance, compute the force of tension the cable is experiencing at its lowest point as precisely as possible. 

      g = 9.8
define f(x) { return ( x*cosh(100*g/x)-x-10*g ) }
define ff(x) { return  ( -100*g*sinh(100*g/x)/x + cosh(100*g/x) -1 ) }