I did a Maths question on mechanics using calculus and physics students disagreed, why are the solutions different?

Question

So, I did the following question and got 2 different answers.

Question: A lighthouse located 300 metres from a straight shoreline sweeps its beam of light around in a circle at a constant rate of 1 revolution every 60s. Let $\theta$ be the angle (in radians) that the beam of light makes with the perpendicular to the shoreline. How fast (in metres per second), is the beam moving along the shoreline when it is at $\frac{\pi}{6}$ radians from the perpendicular?

Solution 1 When the question is done by defining distance along the shore as $x$ and finding $\frac{dx}{dt}$, this is what we get:

$\frac{dx}{dt}=\frac{dx}{d\theta}\frac{d\theta}{dt}=\left(300\sec^2 \frac{\pi}{6}\right)\left(\frac{2\pi}{60}\right)=\frac{40\pi}{3}$

Solution 2 When the question is done by resolving $v=r\omega$, this is what we get:

$\frac{dx}{dt}=v\cos \frac{\pi}{6} = \frac{300}{\cos \frac{\pi}{6}} \cdot \frac{2\pi}{60} \cdot \cos \frac{\pi}{6} =10\pi$

Why would the 2 solutions give different answers?

Answer 1

The second solution is wrong. The equation $v=r\omega$ gives the tangential velocity, which is not what the question asks for: the circle of radius 300 around the lighthouse only touches the short at the perpendicular. Multiplying it by the cosine makes it run along a line parallel to the coast, but the line is moving as the angle changes.

Answer 2

One way to see that solution 2 is wrong is to notice what happens as $\theta \to \pi/2$. The beam goes off to $\infty$, so its speed must increase. But solution 2 wouldn't.

Answer 3

All right here is the answer :your students calculated the tangential velocity , but note that radius of the circle is increasing with time as well which will include a $dr/dt$ term and this needs to be taken into account as well