Here is the problem: A rotating light L is situated at sea 180 metres from the nearest point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.
I'm having some problems with this rate of change question. I'm not too sure where to really start. I thought to use the chain rule to solve this question. So far, I have dd/dt = dd/dr * dr/dt, where dr/dt is 0.1rev/sec. I'm not sure how to find dd/dr though. By using the circumference of a circle as d=2pir, differentiating this with respect to r, removes the r to make it d^1=2pi. This doesn't allow for the 180m or 300m information given to be included and gives the wrong answer. I also found this similar question at Lighthouse problem: my answer does not match the key , but I am struggling to apply the ideas here in this question to mine.
Any help on whether I am on the right track or how I should solve this question would be really useful. Thanks.
Best Answer
Let $x$ be the distance of the light ray along the shore from $P$ and $\theta$ be the corresponding acute angle between the light ray and $LP.$
We have that $$\frac{\mathrm d\theta}{\mathrm d t}=\frac{2\pi}{10}=\frac\pi5;\\\tan\theta=\frac x{180};\\x=300\implies\theta=\arctan\frac{500}{300}=\arctan\frac53.$$
Thus, $$\frac{\mathrm d x}{\mathrm d t}=\frac{\mathrm d x}{\mathrm d \theta}\times\frac{\mathrm d \theta}{\mathrm d t}\\ =180\sec^2\theta\times\frac\pi5.$$ At $x=300,$ $$\frac{\mathrm d x}{\mathrm d t}=36\pi\sec^2\left(\arctan\frac53\right)\\ =36\pi\left(\frac{\sqrt{25+9}}3\right)^2\\ =136\pi.$$