# A Bike and a Catenary

**Figure 1.**The catenary and the tractrix.

*fixed length*, which we take to be \(1\), with the velocity of \(R\) constrained to the line \(RF\).

Figure 2 summarizes the connection: all normals to the tractrix are tangent to the catenary (so that the tractrix is the *involute* of the catenary). Equivalently, if we let a string \(ATR_0\) hug the catenary and—keeping the end \(A\) fixed—unwind the end \(R\) while holding the string taut, \(R\) will sweep a tractrix.

**Figure 2.**As the string unwraps from the catenary, its end

*R*describes the tractrix.

To prove this connection, consider an arbitrary position of the “bike” \(RF\) in Figure 2, and let \(T\) be the point of intersection of the normal through \(R\) and the normal to \(MN\) at \(F\). The point \(T\) is automatically the center of curvature of the tractrix at \(F\); leaving out the proof of this fact (which I will address in next month’s column), we show that \(T\) traces out a catenary, i.e., that \(y= FT= \cosh x\), where \(x\) is the coordinate of \(F\) on the line. From Figure 3,

\[y= \frac{1}{\sin \theta}. \tag1 \]

According to Figure 4,

\[\theta ^\prime = \frac{d\theta }{dx} = - \sin \theta.\tag2 \]

**Figure 3.**Explanation of (1).

\[y ^\prime = \frac{dy}{dx} = \cot \theta.\]

One more differentiation and one more use of \((2)\) gives

\[y ^{\prime\prime} = -\frac{1}{\sin ^2 \theta } (- \sin \theta ) =\frac{1}{\sin \theta } = y,\]

so that \(y\) is a combination of \(\cosh x\) and \(\sinh x\). And since \(y(0)= 1\) and \(y ^\prime (0) = 0\), we determine that \(y= \cosh x\), the equation of a catenary, as claimed.

**Figure 4.**Explanation of (2).

The figures in this article were provided by the author.