The Shape of a Power Line

A cable hanging between two towers isn't a parabola: it's a catenary, the graph of a hyperbolic cosine. Its slope is the hyperbolic sine, which tells line engineers how steeply the cable meets each tower and how much tension it carries.

How it works

Derivatives of hyperbolic functions

In one sentence

With $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$ , $\frac{d}{dx}\cosh x = \sinh x$ , $\frac{d}{dx}\sinh x = \cosh x$ , and by the chain rule $\frac{d}{dx}a\cosh(x/a) = \sinh(x/a)$ .

General formula

y = {\htmlData{sym=2}{a}}{\htmlData{sym=0}{\cosh}}\frac{x}{\htmlData{sym=2}{{a}}}, \qquad {\htmlData{sym=3}{y'}} = {\htmlData{sym=1}{\sinh}}\frac{x}{\htmlData{sym=2}{{a}}}, \qquad \frac{d}{dx}{\htmlData{sym=0}{\cosh}} x = {\htmlData{sym=1}{\sinh}} x

Each derivative just flips the sign of the $e^{-x}$ term, which turns cosh into sinh and back.

What the symbols mean

$\cosh$
Hyperbolic cosine
(e^x + e^(−x))/2: the hanging-cable curve.
$\sinh$
Hyperbolic sine
(e^x − e^(−x))/2: the cable's slope.
$a$
Catenary parameter
Horizontal tension divided by weight per meter.
$y'$
Slope
How steeply the cable rises at position x.

$\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$ , built from exponentials.
Differentiate the exponentials: $(\cosh x)' = \sinh x$ and $(\sinh x)' = \cosh x$ , no minus signs.
For the cable, the slope at position $x$ is $\sinh(x/a)$ : zero at the lowest point, steepest at the towers.
$a$ is the ratio of the cable's horizontal tension to its weight per meter: pull harder and the sag shrinks.

Learn the math

Derivatives of Hyperbolic FunctionsCalc I · Lesson 21, Derivatives