- #1

- 256

- 0

## Homework Statement

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?

[tex]\left( \frac{1}{4 \pi}\right)[/tex] is the answer

[tex]\frac{dV}{dt} = 1 cm^3 /s[/tex]

[tex]r_{cherry} = 1[/tex] cm because the diameter of the cherry is 2 cm

[tex]r_{hemisphereical glass} = 3[/tex] cm

## Homework Equations

I know that the formula for a sphere is [tex]V= \frac{4}{3} \pi r^3[/tex] so for a hemisphere which is half a sphere would be [tex]V= \frac{2}{3} \pi r^3[/tex]

## The Attempt at a Solution

Here is where the problem arisis... if i use the original Volume of a sphere equation:

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}[/tex]

[tex]1 = 4 \pi (1)^2 \frac{dr}{dt}[/tex]

[tex]1 = 4 \pi \frac{dr}{dt}[/tex]

[tex]\frac{1}{4 \pi} = \frac{dr}{dt}[/tex]

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...

I get the answer alright but if I sub in 1 cm for r then that would mean I am calculating for the cherry NOT the cocktail. So what am i doing wrong? My logic is incorrect but my answer is correct and PLUS i didnt even use the equation for a hemisphere ... only a normal sphere

So what do i do ? is there a better way of solving this problem that actually makes sense and not senselessly putting in values and getting the answer Just like that... Please help I am very confuuuuseeeddddd

All i know is i got the answer but that doesn't explain my understanding. Is this just a lucky 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 don't know how to put it in a quantitative way. I do understand it qualitatively.

Thanks for your help.