Figure 2. The forces acting on a segment of a catenary.
dy =
sds
√
1 + s
2
,
which integrates to y =
√
1 + s
2
. Thus s =
y
2
− 1, which we can substitute into
the original differential equation for the catenary to obtain dy/dx =
y
2
− 1. It is
now straightforward to check that y = (e
x
+ e
−x
)/2 is the solution to this differential
equation that passes through (0, 1).
It remains to verify that the coordinate system assumed in this solution is the same
as that defined by the construction of Figure 1. The key to Leibniz’s verification turns
out to be the intermediate step y =
√
1 + s
2
above. To see this, consider Figure 3,
which is Figure 1(c) with additional notation. We know from above that the catenary
FA L is given by y = (e
x
+ e
−x
)/2 in a certain coordinate system whose origin O
is at a vertical distance OA = 1 below the lowest point of the catenary. Consider the
particular y-value OH = y and the associated arc AL = s, then construct the horizontal
segment AM with the same length s. It follows by the Pythagorean theorem that OM
=
√
1 + s
2
. But above we saw that y =
√
1 + s
2
, which means in terms of this figure
that OH = OM. Thus OHM is an isosceles triangle and so the perpendicular bisector
of its base HM passes through the vertex O. This shows that the construction of Figure
1 does indeed give a way of recovering the coordinate system associated with the
solution y = (e
x
+ e
−x
)/2, as we needed to show. From here it is a simple matter of
algebra to check the final step of Figure 1.
In a 17th-centur y context
Finding logarithms from a catenary may seem like an oddball application of mathe-
matics today, but to Leibniz it was a very serious matter—not because he thought this
method so useful in practice, but because it pertained to the very question of what it
means to solve a mathematical problem. Today we are used to thinking of a formula
such as y = (e
x
+ e
−x
)/2 as the answer to the question of the shape of the catenary,
but this would have been considered a na
¨
ıve view in the 17th century. Leibniz and
his contemporaries discovered this relation between the catenary and the exponential
function in the 1690s, but they never wrote this equation in any form, even though they
understood perfectly well the relation it expresses. Nor was this for lack of familiarity
with exponential expressions, at least in Leibniz’s case, as he had earlier used such
expressions to describe curves with considerable facility [11, No. 6].
Why, indeed, should one express the solution as a formula? What kind of solution
to the catenary problem is y = (e
x
+ e
−x
)/2, anyway? The 17th-century philosopher
VOL. 47, NO. 2, MARCH 2016 THE COLLEGE MATHEMATICS JOURNAL 97