1. **Problem statement:** A woman at point A on the shore of a circular lake with radius 2 miles wants to reach point C, diametrically opposite A, by first rowing to a point B on the shore and then walking from B to C. She rows at 2 mi/h and walks at 4 mi/h. We want to find the angle $\theta$ at which she should row to minimize total travel time. 2. **Setup and variables:** - The lake is a circle with radius $r=2$ miles. - Points A and C lie on the horizontal diameter, with A on the left and C on the right. - Point B lies on the circumference, defined by angle $\theta$ from the horizontal line through A. 3. **Distances:** - Rowing distance $AB$ is the length of the chord from A to B. Since $B$ is on the circle at angle $\theta$ from A, the chord length is: $$AB = 2r \sin\left(\frac{\theta}{2}\right) = 4 \sin\left(\frac{\theta}{2}\right)$$ - Walking distance $BC$ is the arc length along the circle from B to C. Since $B$ is at angle $\theta$ from A, and $C$ is diametrically opposite A (angle $\pi$ from A), the arc length from B to C is: $$BC = r (\pi - \theta) = 2(\pi - \theta)$$ 4. **Travel times:** - Rowing speed $v_r = 2$ mi/h - Walking speed $v_w = 4$ mi/h Time rowing: $$t_r = \frac{AB}{v_r} = \frac{4 \sin(\frac{\theta}{2})}{2} = 2 \sin\left(\frac{\theta}{2}\right)$$ Time walking: $$t_w = \frac{BC}{v_w} = \frac{2(\pi - \theta)}{4} = \frac{\pi - \theta}{2}$$ Total time: $$T(\theta) = t_r + t_w = 2 \sin\left(\frac{\theta}{2}\right) + \frac{\pi - \theta}{2}$$ 5. **Minimizing travel time:** We find $\theta$ that minimizes $T(\theta)$ by setting derivative to zero: $$\frac{dT}{d\theta} = 2 \cdot \cos\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} - \frac{1}{2} = \cos\left(\frac{\theta}{2}\right) - \frac{1}{2} = 0$$ Solve: $$\cos\left(\frac{\theta}{2}\right) = \frac{1}{2}$$ This gives: $$\frac{\theta}{2} = \frac{\pi}{3} \implies \theta = \frac{2\pi}{3}$$ 6. **Check endpoints and confirm minimum:** - At $\theta=0$, $T(0) = 2 \sin(0) + \frac{\pi - 0}{2} = \frac{\pi}{2} \approx 1.5708$ - At $\theta=\frac{2\pi}{3}$, $$T\left(\frac{2\pi}{3}\right) = 2 \sin\left(\frac{\pi}{3}\right) + \frac{\pi - \frac{2\pi}{3}}{2} = 2 \cdot \frac{\sqrt{3}}{2} + \frac{\frac{\pi}{3}}{2} = \sqrt{3} + \frac{\pi}{6} \approx 1.732 + 0.5236 = 2.2556$$ - At $\theta=\pi$, $T(\pi) = 2 \sin\left(\frac{\pi}{2}\right) + 0 = 2 \cdot 1 = 2$ Since $T(0) < T(\frac{2\pi}{3})$ and $T(\pi) < T(\frac{2\pi}{3})$, the critical point is a maximum, not minimum. 7. **Re-examining the problem:** The walking path is along the arc from B to C, so the walking distance is $r(\pi - \theta)$ only if $\theta$ is measured from A to B counterclockwise. The rowing distance is the chord length from A to B. But the problem states the woman rows from A to B and then walks from B to C along the shore. The walking speed is faster, so she wants to minimize rowing distance. 8. **Alternative approach:** The total time function is: $$T(\theta) = \frac{AB}{2} + \frac{BC}{4}$$ Where: $$AB = 2r \sin\left(\frac{\theta}{2}\right) = 4 \sin\left(\frac{\theta}{2}\right)$$ $$BC = r (\pi - \theta) = 2(\pi - \theta)$$ So: $$T(\theta) = \frac{4 \sin(\frac{\theta}{2})}{2} + \frac{2(\pi - \theta)}{4} = 2 \sin\left(\frac{\theta}{2}\right) + \frac{\pi - \theta}{2}$$ Derivative: $$T'(\theta) = 2 \cdot \cos\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} - \frac{1}{2} = \cos\left(\frac{\theta}{2}\right) - \frac{1}{2}$$ Set to zero: $$\cos\left(\frac{\theta}{2}\right) = \frac{1}{2}$$ Solutions: $$\frac{\theta}{2} = \frac{\pi}{3} \implies \theta = \frac{2\pi}{3}$$ Check values at endpoints $\theta=0$ and $\theta=\pi$: - $T(0) = 2 \sin(0) + \frac{\pi}{2} = \frac{\pi}{2} \approx 1.5708$ - $T(\pi) = 2 \sin(\frac{\pi}{2}) + 0 = 2$ - $T(\frac{2\pi}{3}) = 2 \sin(\frac{\pi}{3}) + \frac{\pi - \frac{2\pi}{3}}{2} = 2 \cdot \frac{\sqrt{3}}{2} + \frac{\pi/3}{2} = \sqrt{3} + \frac{\pi}{6} \approx 1.732 + 0.5236 = 2.2556$ Since $T(0) < T(\frac{2\pi}{3})$ and $T(0) < T(\pi)$, the minimum time is at $\theta=0$. 9. **Interpretation:** The woman should row directly across the lake (angle $\theta=0$) and then walk the rest of the way along the shore, minimizing total time. **Final answer:** $$\boxed{\theta = 0}$$ ---