Here is a related rates you may like to try (Perito?). At least, it is different than the same old cliché ones we have seen for years.

A hemi-spherical swimming pool has a radius of 10 meters and is completely full of water. A stone is dropped into the pool, at the center of the surface of the water, and falls vertically at a constant rate of one meter every five seconds. A light at the edge of the pool casts a shadow of the stone on the opposite side of the pool, as shown in the accompanying diagram. Find the rate at which the shadow is moving along the side of the pool at the instant when the stone lies at a depth of 5 meters. [HINT: You may use the fact that, in the accompanying diagram,θ = 2α .]

Think about s=r{\theta} and {\alpha}=\frac{\theta}{2}

I tried it... it's of a too high level for me. Nevertheless, I pretty sure there're is sine involved... or another trig...

0.32 m/s.

Sorry no time to post calculations in Math type format.

Will show my work once I get round to typing in whatever format is used to post math equations...

I tried it... it's of a too high level for me. Nevertheless, I pretty sure there're is sine involved...

I used tan instead of sine. tan({\alpha})=\frac{D}{10}

S=2r{\alpha}

\frac{dS}{dt}=20\frac{d{\alpha}}{dt}

Now, find \frac{d{\alpha}}{dt} and it's done.

Thanks Chris! Just didn't how to proceed! I just realised that that question involved implicit differentiation...

EDIT: Oops, you just posted before mine appeared... I got it, thanks to Chris... :)

Here is the solution I came up with:

Remember the old circle geometry theorem which says that {\theta}=2{\alpha}, as in the hint?

We know that s=r{\theta}. Therefore, s=2r{\alpha}

We are told that r=10. So, it is a constant.

\frac{ds}{dt}=20\frac{d{\alpha}}{dt}... [1]

From the problem statement, ds/dt is what we want. So, we have to find \frac{d{\alpha}}{dt}, plug it in and we have it.

tan({\alpha})=\frac{D}{10}

sec^{2}({\alpha})\frac{d{\alpha}}{dt}=\frac{1}{10} \frac{dD}{dt}

we are told that dD/dt=1/5 meter per second.

When D=5, then sec^{2}(tan^{-1}(\frac{1}{2}))\frac{d{\alpha}}{dt}=\frac{1}{10} \;\ \cdot \;\ \frac{1}{5}

Solving, we find that \frac{d{\alpha}}{dt}=\frac{2}{125}

Plug into the [1] and we get \frac{ds}{dt}=\frac{8}{25} \;\ \frac{m}{sec}=.32.

Which agrees with Chris's solution.