Skip to main content

Section 4.3 The one-parameter family of catenaries through a point

Our next step is to determine the number and the character of the catenaries (4.2.4) which pass through the two given points 1 and 2. The plan is to find the equation of the one-parameter family of these catenaries passing through the point 1, and then to determine how many of them pass through the second point 2. The equation \(y_1 = b \cosh \frac{x_1-a}{b}\) is the condition that the catenary (4.2.4) shall pass through the point 1. It is satisfied, as one readily verifies, when \(a\) and \(b\) are expressed in terms of a new parameter \(\alpha\) in the form

\begin{equation} a=x_{1}-y_{1} \frac{\alpha}{\cosh \alpha}, \quad b=\frac{y_{1}}{\cosh \alpha},\label{eqn-4-6}\tag{4.3.1} \end{equation}

and the family of catenaries through the point 1 is therefore

\begin{equation} y=\frac{y_{1}}{\cosh\alpha} \cosh\left(\alpha+\frac{x-x_{1}}{y_{1}} \cosh \alpha \right)=y(x, \alpha),\label{eqn-4-7}\tag{4.3.2} \end{equation}

where \(y(x, \alpha)\) is simply a convenient symbol for the more complicated expression preceding it.

In deducing the properties of the one-parameter family of catenaries through the point 1, we shall need the first and second derivatives with respect to \(x\) and \(\alpha\) of the function \(y(x, \alpha)\) defined in (4.3.2). Derivatives with respect to \(x\) will be denoted by primes and with respect to \(\alpha\) by subscripts, while the subscript 1 will be used to designate values of \(x, y, y'\) at the point 1. If we remember the formulas

\begin{equation*} \frac{d}{du} \cosh u = \sinh u, \quad \frac{d}{du} \sinh u = \cosh u \end{equation*}

mentioned above, we find readily the values

\begin{align} y \amp= \frac{y_1}{\cosh \alpha} \cosh\left(\alpha + \frac{x - x_1}{y_1} \cosh \alpha \right) = y (x, \alpha) \tag{4.3.3}\\ y' \amp= \sinh \left(\alpha + \frac{x-x_1}{y_1} \cosh\alpha\right), \quad y_1' = \sinh \alpha \tag{4.3.4}\\ y_\alpha \amp= \frac{y' \cdot y_1'}{\cosh \alpha} \left(x - \frac{y}{y'} - x_1 + \frac{y_1}{y_1'}\right), \tag{4.3.5} \end{align}

where in calculating the last derivative hyperbolic sines and cosines have been replaced except in one instance by their values in terms of \(y, y', y_1 , y_1'\) from the first three equations.