## Question

###### A flexible chain of length $L$ is suspended between two poles of equal height separated by a distance $2 M$ (Figure I3). By Newton's laws, the chain describes a catenary $y=a \cosh \left(\frac{x}{a}\right)$, where a is the number such that $L=2 a \sinh \left(\frac{M}{a}\right) .$ The sag $s$ is the vertical distance from the highest to the lowest point on the chain. Suppose that $L=160$ and $M=50$. (a) Use Newton's Method to find a value of $a$ (to two decimal places) satisfying $L=2 a

A flexible chain of length $L$ is suspended between two poles of equal height separated by a distance $2 M$ (Figure I3). By Newton's laws, the chain describes a catenary $y=a \cosh \left(\frac{x}{a}\right)$, where a is the number such that $L=2 a \sinh \left(\frac{M}{a}\right) .$ The sag $s$ is the vertical distance from the highest to the lowest point on the chain. Suppose that $L=160$ and $M=50$. (a) Use Newton's Method to find a value of $a$ (to two decimal places) satisfying $L=2 a \sinh (M / a)$. (b) Use Eq. (3) and the Linear Approximation to estimate the increase in sag $\Delta s$ for changes in length $\Delta L=1$ and $\Delta L=5$. (c) $\quad$ Compute $s(161)-s(160)$ and $s(165)-s(160)$ directly and compare with your estimates in (b).