1. geometric Setup

Let the bridge consist of two segments resting in the semi-cylindrical groove of diameter $D = 40 \text{ cm}$. Since the angle in a semicircle is a right angle ($90^\circ$), the lengths of the segments, $l$ and $L_2$, satisfy the Pythagorean theorem:

$$ l^2 + L_2^2 = D^2 \implies L_2 = \sqrt{D^2 – l^2} $$

The ant travels down the segment of length $l$ with speed $2v$ and up the segment $L_2$ with speed $v$.

2. Function Analysis

The total time $t$ as a function of the length $l$ (the “down” segment) is:

$$ t(l) = \frac{l}{2v} + \frac{\sqrt{40^2 – l^2}}{v} $$

To investigate the behavior of $t$, we differentiate with respect to $l$:

$$ \frac{dt}{dl} = \frac{1}{2v} + \frac{1}{v} \left( \frac{1}{2\sqrt{40^2 – l^2}} \cdot (-2l) \right) $$ $$ \frac{dt}{dl} = \frac{1}{2v} \left[ 1 – \frac{2l}{\sqrt{40^2 – l^2}} \right] $$

3. Finding the Critical Point

We analyze the sign of the derivative to determine where the time increases or decreases.

$\frac{dt}{dl} > 0$ (Time is increasing) when:

$$ 1 > \frac{2l}{\sqrt{40^2 – l^2}} \implies \sqrt{40^2 – l^2} > 2l $$ $$ 40^2 – l^2 > 4l^2 \implies 40^2 > 5l^2 $$ $$ l < \frac{40}{\sqrt{5}} \approx 17.9 \text{ cm} $$

Conclusion from Derivative:

• For $l < 17.9 \text{ cm}$, $\frac{dt}{dl} > 0$, so time increases as $l$ increases.
• At $l \approx 17.9 \text{ cm}$, time reaches a local maximum.
• For $l > 17.9 \text{ cm}$, $\frac{dt}{dl} < 0$, so time decreases as $l$ increases.

4. Optimization Conclusion

We want to minimize $t$. The graph shows that for $l > 17.9$, the time decreases continuously as $l$ increases. Therefore, to achieve the smallest time, we must choose the largest possible value for $l$.

Given the constraint that the maximum wire length available is $38 \text{ cm}$, we set:

$$ l = 38 \text{ cm} $$

Substituting this back into the time equation with $v = 0.5 \text{ cm/s}$:

$$ t = \frac{38}{1.0} + \frac{\sqrt{40^2 – 38^2}}{0.5} $$ $$ t = 38 + 2 \times \sqrt{156} $$ $$ t = 38 + 2(12.49) \approx 38 + 25 = 63 \text{ s} $$