A friend from work found his old Maths exam from uni in the 1970's and I am giving it a bash! Im a bit stuck on a question so thought id ask the hive mind for some help!
A bowl is in the shape of a hemisphere of inside radiues $r$ cm. Initially the bowl is full of water which leaks through a small hole at its lowest point. Show that, when the water surface is at a height $h$ cm above the hole, the volume of water in the bowl is
$$\pi\left(rh^2-\frac{h^3}{3}\right) \text{cm}^3.$$ If the rate of leakage at that time is $\pi k{h^\frac{1}{2}}\text{cm}^3 / \text{sec}$, where $k$ is constant, show that the rate of fall of the surface is then $$\frac{k}{2rh^{\frac{1}{2}}-h^{\frac{3}{2}}} \text{cm\sec}.$$ Find the value of $h$ for which the rate of fall is least.
Any help will be appreciated!
Thanks in advance everyone!
Best Answer
The volume of the water can be integrated with the disk method in cylindrical coordinates for the sphere $\rho^2+z^2= r^2 $,
$$V(h)= \int_{r-h}^r \pi \rho^2 (z)dz = \pi\int_{r-h}^r (r^2-z^2) dz =\pi\left(rh^2-\frac{h^3}{3}\right) $$
Then, evaluate
$$\frac{dV}{dt} = \pi k{h^\frac{1}{2}}= \pi (2rh-h^2) \frac{dh}{dt} $$
which leads to the rate of surface fall
$$\frac{dh}{dt} = \frac{k}{2rh^{\frac{1}{2}}-h^{\frac{3}{2}}} \>\text{cm\sec}$$
Set $\frac{d^2h}{dt^2}=0$ to obtain $h=\frac 23r$ for the least rate of fall.