Beam on the Rocks

There is no wrong diagram here as long as it shows a fixed lighthouse, a straight shoreline, and the beam at several angles. Most groups will guess "no, it speeds up near the ends" from intuition about how a rotating line sweeps across a wall; a smaller number will guess "yes, constant" by over-generalising from "constant angular speed."

At \(\theta = 0\): \(x = d\tan(0) = 0\), the spot sits directly opposite the lighthouse. Matches the picture.

As \(\theta \to 90°\): \(\tan\theta \to \infty\), so \(x \to \infty\). Physically, the beam would have to be pointing almost exactly along the shore itself, and the spot would be arbitrarily far away. Nothing breaks mathematically, but it's a strong hint that the spot's speed can't stay constant as \(\theta\) grows, an infinite \(x\) at a finite \(\theta\) of \(90°\) is only possible if the spot is moving faster and faster to get there.

This is the identical chain-rule move students already use for implicit differentiation in \(x\) and \(y\); only the label on the "hidden" variable has changed from \(y\) to \(x(t)\). If a group writes \(2x\) with no \(dx/dt\), ask them what \(\dfrac{d}{dx}(y^2)\) is and get them to substitute \(x \to t\), \(y \to x\) themselves.

Differentiating \(x = d\tan\theta\) with respect to \(t\), with \(d\) constant:

Using \(\dfrac{dx}{dt} = d\sec^2\theta\,\dfrac{d\theta}{dt} = 200(0.05)\sec^2\theta = 10\sec^2\theta\):

• \(\theta = 0°\): \(\sec^2 0° = 1 \Rightarrow \dfrac{dx}{dt} = 10\) m/s
• \(\theta = 30°\): \(\sec^2 30° = \dfrac{4}{3} \Rightarrow \dfrac{dx}{dt} = \dfrac{40}{3} \approx 13.33\) m/s
• \(\theta = 60°\): \(\sec^2 60° = 4 \Rightarrow \dfrac{dx}{dt} = 40\) m/s

Model answer: the beam's angular speed \(\dfrac{d\theta}{dt}\) really is constant, but the spot's linear speed \(\dfrac{dx}{dt}\) is that constant multiplied by \(\sec^2\theta\), and \(\sec^2\theta\) itself grows as \(\theta\) grows. Near \(\theta = 0\) the beam is nearly perpendicular to the shore and a small turn moves the spot only a little; near \(\theta = 90°\) the same small turn sweeps the spot a huge distance, because the shore is almost parallel to the beam. So "constant turning" and "constant sliding" are simply different quantities, related by a factor that itself changes.

Nothing physical is travelling at this speed. The lighthouse itself only rotates at \(0.05\) rad/s, and no material object moves along the shore at all; the "spot" is just wherever a straight beam happens to intersect a straight shoreline at each instant, a geometric intersection point, not a thing with mass. Two independent geometric constructions can cross paths arbitrarily fast without transporting any matter or energy along the shore, so no law of physics (including the speed limits that apply to matter and information) is violated.

At \(\cos^2\theta = \dfrac23\), \(\sin^2\theta = \dfrac13\), so \(\tan\theta = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}\).

For \(\theta < 35.3°\), \(10\sec^2\theta < 15\), the spot is slower than the researcher: she can walk ahead of it comfortably. For \(\theta > 35.3°\), the spot is faster than her constant \(15\) m/s and only gets faster still.

Because \(\sec^2\theta\) is strictly increasing on \([0°, 90°)\) and diverges to infinity as \(\theta \to 90°\), once the beam passes the critical angle it never slows back down relative to her; a fixed human walking speed can delay being overtaken but not prevent it, since the spot's speed has no upper bound while hers does. Yes, guaranteed, eventually, if she keeps walking that same stretch for long enough.