MAIN CONCEPTS
The shape of a catenary is defined by the equation
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We set our coordinate system such that the y-axis passes through the lowest point of the chain. Let s be the arc length from this point to a variable point ( x, y ), and let  be the linear density of the chain, that is the weight per unit length. Here is where some physics is used. The part of the chain from the lowest point to the variable point (x, y) is in static equilibrium under the action of three forces: the tension  at the lowest point directed horizontally left; the variable tension  at ( x, y ) directed in the tangential direction of the curve; and the downward force  from the weight of the chain. A diagram illustrates these forces.

We’ll now do what physics students call, “resolving forces”. Since this part of the chain is in static equilibrium, we can equate the forces in the horizontal and vertical direction. This gives us,

 and

We divide the equations to eliminate and get

Since  describes the gradient of the curve at a certain point, we can write

replacing the constants with a convenient a.

Here is where some differentiation intuition is needed. We want to eliminate the s term in the equation but it will be clumsy to substitute the explicit formula for the arc length. How about substituting the differential form ds which is much neater. To do this, we need to differentiate w.r.t x.

This is the differential equation of the catenary. Solving this gives us the required shape. We can do that be two successive integrations. First, we need to introduce the auxiliary variable , so we now have

By separating the variables and integrating we get

 

Remember that nice little integral formula we got from our hyperbolic sine. Well, we’ll use it now. The equation above follows that form so we immediately write

We see from the graph that when , where the curve intersects the y-axis, the graph is zero or  as we earlier define . It follows that =0 and so the equation reduces to

Replacing p we then solve the differential equation:

We would like to make the equation a little neater so what we’ll do is that we’ll move the coordinate axis such that the origin is at just the right level of  whe . This gives us  and finally the shape of the catenary takes the form

Who would have thought an innocent chain hanging on its two ends is defined by something as usual as the hyperbolic cosine function. Now you can tell your friends that a catenary takes this form, and not the common  everyone thought of.