The length of a telegraph wire is s = 2b (1+

The length of a telegraph wire is $s = 2b\left(1+\dfrac{2f^{2}}{3b^{2}}\right)$, where $2b$ is the distance between the supports and $f$ is the maximum sag. By how much does the sag increase when the length increases by $ds$ because of heating?

Step 1: Simplify the relation

Multiplying out, \[s = 2b+\frac{2b\cdot 2f^{2}}{3b^{2}} = 2b+\frac{4f^{2}}{3b}. \tag{1}\]

The supports do not move, so $b$ is a constant; only $s$ and $f$ change.

Step 2: Differentiate

\[ \frac{ds}{df} = 0+\frac{4}{3b}\cdot 2f = \frac{8f}{3b}. \tag{2} \]

Step 3: Invert

Reading (2) as a relation between differentials, $ds = \dfrac{8f}{3b}\,df$, and solving for $df$: $df = \frac{3b}{8f}\,ds$.

Step 4: Verification by solving explicitly

From (1), $\frac{4f^{2}}{3b} = s-2b \;\Longrightarrow\; f = \frac{\sqrt{3b\bigl(s-2b\bigr)}}{2}$.

\[\frac{df}{ds} = \frac12\cdot\frac{3b}{2\sqrt{3b(s-2b)}} = \frac{3b}{4\sqrt{3b(s-2b)}}\]

\[\text{Since } \sqrt{3b(s-2b)} = 2f, \qquad \frac{df}{ds} = \frac{3b}{4\cdot 2f} = \frac{3b}{8f} \;\;\checkmark\]

\[\text{Numerical check } (b = 25\ \text{m},\ f = 1\ \text{m}):\ s = 50+\frac{4}{75} \approx 50.05333;\ \frac{df}{ds} = \frac{75}{8} = 9.375 \;\;\checkmark\]