Hanging Out to Dry

Let $l$ be the length of the cable, $b$ be the separation between the poles, $h$ be the height of the poles.

Shape of a hanging cable (known as Catenary ) is given by $f(x)=a\cosh\left(\dfrac{x}{a}\right)$

Equation of a hanging cable that is tangent to the $x$ -axis (ground) is $f(x)=a\cosh\left(\dfrac{x}{a}\right)-a$

Since the graph is symmetric about $y$ -axis

Arc length of the curve from $x=0$ to $x=\frac{b}{2}$ is $\displaystyle \int_0^{\frac{b}{2}}\sqrt{1+\left(\dfrac{d}{dx}\left(a\cosh\left(\dfrac{x}{a}\right)-a\right)\right)^2}\,dx=\frac{l}{2}$

Solving the integral and rearranging terms gives $\implies \color{#3D99F6}\sinh\left(\dfrac{b}{2a}\right)=\dfrac{l}{2a}$

Also, $h=f\left(\dfrac{b}{2}\right)=a\cosh\left(\dfrac{b}{2a}\right)-a$

Rearranging the terms gives $\implies \color{#D61F06}\cosh\left(\dfrac{b}{2a}\right)=\dfrac{h+a}{a}$

Solving the red and blue simultaneous equations $\implies \boxed{b=\dfrac{l^2-4h^2}{4h}\ln\left(\dfrac{l+2h}{l-2h}\right)}$

Putting $l=16$ and $h=4$ $\implies\boxed{b=12\ln{3}}$

A suspended cable hangs in a catenary , and if we let the ground be the $x$ -axis, then a catenary with the minimum separation between the poles will be one that passes through the origin at its axis of symmetry, which follows the equation $y = a \cosh (\frac{x}{a}) - a$ , with an arc length of $s = 2a \sinh (\frac{x}{a})$ .

The first equation can be rearranged to $\cosh (\frac{x}{a}) = \frac{y + a}{a}$ and the second to $\sinh (\frac{x}{a}) = \frac{s}{2a}$ . Then using the identity $\cosh^2(t) - \sinh^2(t) = 1$ , we have $(\frac{y + a}{a})^2 - (\frac{s}{2a})^2 = 1$ , and since $y = 4$ when $s = 16$ , we have $(\frac{4 + a}{a})^2 - (\frac{16}{2a})^2 = 1$ , and solving this gives us $a = 6$ . So the equation of the catenary is $y = 6 \cosh (\frac{x}{6}) - 6$ .

We want to find the separation along the ground, which is $2x$ when $y = 4$ . So $4 = 6 \cosh (\frac{x}{6}) - 6$ , and since $\cosh t = \frac{1}{2}(e^t + e^{-t})$ , our equation becomes $4 = 6 \cdot \frac{1}{2}(e^{\frac{x}{6}} + e^{-\frac{x}{6}}) - 6$ , which solves to $x = 6 \ln 3$ . This means the separation along the ground is $2x = 2 \cdot 6 \ln 3 = 12 \ln 3$ , and so $\alpha = 12$ and $\beta = 3$ , and $\alpha + \beta = 12 + 3 = \boxed{15}$ .

The cable hangs in a catenary $y \; = \; a \cosh \tfrac{x}{a} \hspace{2cm} -X \le x \le X$ for some $a,X > 0$ . Thus $\frac{dy}{dx} \; = \; \sinh \tfrac{x}{a} \hspace{2cm} \frac{ds}{dx} \; = \; \cosh \tfrac{x}{a}$ and so the length of the cable is $16 \; = \; \int_{-X}^X \cosh \tfrac{x}{a}\,dx \; = \; 2a \sinh \tfrac{X}{a} \; = \; 4a \sinh \tfrac{X}{2a} \cosh \tfrac{X}{2a}$ If the centre of the cable just brushes the ground, then $4 \; = \; a\big(\cosh\tfrac{X}{a} - 1\big) \; = \; 2a\sinh^2\tfrac{X}{2a}$ These two equations tell us that $\tanh \tfrac{X}{2a} = \tfrac12$ , so that $X = a\ln3$ . Thus $\sinh\tfrac{X}{a} = \tfrac43$ , and hence $a=8\,\mathrm{cosech}\tfrac{X}{a} = 6$ . Thus we deduce that $X = 6\ln3$ , and hence the separation of the poles is $12\ln3$ .

If the poles were further apart, the cable would hang higher. Thus the answer is $12 + 3 = \boxed{15}$ .