# Thread: Cocktail and cherry related rate

1. ## Cocktail and cherry related rate

A cocktail is being poured into a hemispherical glass, which contains a cherry, at a uniform rate of 1 cm3/s. The cherry has a diameter of 2 cm and the glass has a radius of 3 cm. How fast is the level of the cocktail rising at the moment when half the cherry is submerged? (Answer is $\frac{1}{4 \pi}$

$V = \frac{4}{3} \pi r^3$ for a full sphere
Now, for a hemisphere which is half a sphere $\frac{\frac{4}{3}}{2} = \frac{2}{3}$

$V = \frac{2}{3} \pi r^3$ Volume of a hemisphere

Given

$\frac{dV}{dt} = 1 cm^3/s$

$r_{cherry} = 1$ cm because the diameter of the cherry is 2 cm
$r_{hemisphereical glass} = 3$ cm

$V = \frac{4}{3} \pi r^3$
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

$1 = 4 \pi (1)^2 \frac{dr}{dt}$

$1 = 4 \pi \frac{dr}{dt}$

$\frac{1}{4 \pi} = \frac{dr}{dt}$

By the way, notice that I didn't use the formula for the hemispherical glass but the formula for the sphere which would mean the cherry but the question asks how fast is the level of the cocktail rising.... which now I become confused...........

All i know is i got the answer but that doesn't explain my understanding. Is this just a lukcy shot or am i actually correct ???? is there a better way of solving the problem that makes much more sense and still arrives at the correct answer? Please help.

The only thing i understanding is with no cocktail the cherry sits at the bottom by itself. When a little bit of cocktail comes in a little surface at the bottom of the cherry is covered with cocktail and as the cocktail level rises more of the surface is covered starting from the bottom and moving to center of the cherry which is the radius of the cherry. I just dont know how to put it in a quantitative way. I do understand it qualitatively.