## Tag Archives: application of basic calculus

### What is a catenary or a chain curve? A nice little application of basic calculus

Let us continue our exploration of basic calculus and its application. As you will discover, the invention of calculus is a triumph of the human intellect, it is a fountain head of many ideas in pure mathematics as well as mine of applications to sciences and engineering and even economics and humanities!

Hanging cables

Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable’s weight per unit length is w and horizontal tension at its lowest part is a vector of length H. If we choose a coordinate system for the plane of the cable in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cable lies at the point $y=H/w$ on the y-axis. (Fig 1), then it can be shown that the cable lies along the graph of the hyperbolic cosine

$y=(H/w)\cosh (wx/H)$.

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena meaning ‘chain’.

a) Let $P(x,y)$ denote an arbitrary point on the cable. Fig 2 displays the tension at P as a vector of  length (magnitude) T, as well as the tension at H at the lowest point A. Show that the cable’s slope at P is

$\tan \phi =dy/dx=\sinh (wx/H)$

b) Using the result from part (a) and the fact that the tension at P must equal H (the cable is not moving or swinging), show that $T=wy$. This means that the magnitude of the tension at $P(x,y)$ is exactly equal to the weight of y units of cable.

2) (Continuation of above problem). The length of arc AP in Fig 2 is

$s=(1/a)\sinh (ax)$ where $a=w/H$. Show that the coordinates of P may  be expressed in terms of s as

$x=(1/a) (\sinh)^{-1} (as)$ and $y=\sqrt (s^{2}+1/a^{2})$.

3) The sag and horizontal tension in a cable. The ends of a cable 32 feet long and weighing 2 pounds per foot are fastened at the same level in posts 30 feet apart.

i) Model the cable with the equation

$y=(1/a)\cosh (ax)$ for $-15 \leq x \leq 15$

Use information from problem 2  above to show that a satisfies the equation

$16a=\sinh 15a$

ii) Estimate the horizontal tension in the cable at the cable’s lowest point.

I hope you enjoy a lot 🙂

More later…

Nalin Pithwa