91Ó°ÊÓ

Suppose a uniform flexible cable is suspended between two points \((\pm L, H)\) at equal heights located symmetrically on either side of the \(x\) -axis (Fig. \(1.4 .12\) ). Principles of physics can be used to show that the shape \(y=y(x)\) of the hanging cable satisfies the differential equation $$ a \frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}} $$ where the constant \(a=T / \rho\) is the ratio of the cable's tension \(T\) at its lowest point \(x=0\) (where \(y^{\prime}(0)=0\) and its (constant) linear density \(\rho\). If we substitute \(v=d y m y s \operatorname{lash} d x, d v / d x=d^{2} y / d x^{2}\) in this second- order differential equation, we get the first-order equation $$ a \frac{d v}{d x}=\sqrt{1+v^{2}} $$ Solve this differential equation for \(y^{\prime}(x)=v(x)=\) \(\sinh (x / a)\). Then integrate to get the shape function $$ y(x)=a \cosh \left(\frac{x}{a}\right)+C $$ of the hanging cable. This curve is called a catenary, from the Latin word for chain.

A river \(100 \mathrm{ft}\) wide is flowing north at \(w\) feet per second. A dog starts at \((100,0)\) and swims at \(v_{0}=4 \mathrm{ft} / \mathrm{s}\), always heading toward a tree at \((0,0)\) on the west bank directly across from the dog's starting point. (a) If \(w=2 \mathrm{ft} / \mathrm{s}\), show that the dog reaches the tree. (b) If \(w=4 \mathrm{ft} / \mathrm{s}\) show that the dog reaches instead the point on the west bank \(50 \mathrm{ft}\) north of the tree. (c) If \(w=6 \mathrm{ft} / \mathrm{s}\), show that the dog never reaches the west bank.

In the calculus of plane curves, one learns that the curvature \(\kappa\) of the curve \(y=y(x)\) at the point \((x, y)\) is given by $$ \kappa=\frac{\left|y^{\prime \prime}(x)\right|}{\left[1+y^{\prime}(x)^{2}\right]^{3 / 2}}, $$ and that the curvature of a circle of radius \(r\) is \(\kappa=1 / r\). [See Example 3 in Section \(11.6\) of Edwards and Penney, Calculus: Early Transcendentals, 7 th edition (Upper Saddle River, NJ: Prentice Hall, 2008).] Conversely, substitute \(\rho=y^{\prime}\) to derive a general solution of the second-order differential equation $$ r y^{\prime \prime}=\left[1+\left(y^{\prime}\right)^{2}\right]^{3 / 2} $$ (with \(r\) constant) in the form $$ (x-a)^{2}+(y-b)^{2}=r^{2}. $$ Thus a circle of radius \(r\) (or a part thereof) is the only plane curve with constant curvature \(1 / r\).