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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?

Poseidonium
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  • I think your approach is right , i cannot point the mistake exactly , but their solution seems to be of a different scenario, because i.m.o light will also have a radial velocity in this casr which needs to be resolved – avz2611 May 18 '16 at 02:44

3 Answers3

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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.

Martin Argerami
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  • it is tangential, which is why they needed to resolve it by multiplying with $\cos \theta$ – Poseidonium May 18 '16 at 02:43
  • Yes, but it is the velocity, along a line parallel to the coast, but that intersects the circle at the same point the radius does. So, for every different angle they are using a different "coast". – Martin Argerami May 18 '16 at 02:55
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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.

Robert Israel
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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

avz2611
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