# Rearranging to eliminate the R variable in volume equation

A water glass (10 cm diameter at the top, 6 cm diameter at the bottom, 20 cm in height) is being filled at a rate of 50 cm^3/min. Find the rate of change of the height of the water after 5 seconds.

V=(∏* (h/3)) * (R^2 + r^2 +(R*r))

I've been messing around with this for a bit and made a bit of leeway but I'm unsure of how to eliminate R and r from my equation. If it was a cone I would use the r/h=5/20 and rearrange to 5h/20 and replace my 'r' then differentiate. However that ratio doesn't seem to work on a truncated cylinder and I have two radius-es. Any suggestions?

What are R, r, and h?

R=5 cm r= 3 cm h= 20 cm

Dick
Homework Helper
R=5 cm r= 3 cm h= 20 cm

Ok. So r is your lower radius. That's a constant. As the glass fills R and h will change. But there is a linear relation between them since the edge of the glass is a straight line. When h=0 cm then R=3 cm. When h=20 cm R=5 cm. What's the linear relation?

OK cool I think I get what you're going at. If I put those point on a graph the slope would be a slope of R/H=10/1. I can sub that in for R and put 3 in for r because it's a constant. Simplified it might look a little something like this

V= 100H^3(1/3)∏ + 90H^2(1/3)∏ + 9H(1/3)∏

I'm still a little uncertain what to do with the issue of time. Could I take the derivative of the height (20 cm) ,when the glass is full, get the rate of change and then attempt to find the rate algebraically?

Dick
Homework Helper
OK cool I think I get what you're going at. If I put those point on a graph the slope would be a slope of R/H=10/1. I can sub that in for R and put 3 in for r because it's a constant. Simplified it might look a little something like this

V= 100H^3(1/3)∏ + 90H^2(1/3)∏ + 9H(1/3)∏

I'm still a little uncertain what to do with the issue of time. Could I take the derivative of the height (20 cm) ,when the glass is full, get the rate of change and then attempt to find the rate algebraically?

But R=10h doesn't work. Remember, when h=0 cm then R=3 cm. When h=20 cm R=5 cm. If I put h=0 into R=10h I get 0, not 3. Can you fix that? For the other part of your question how much water is in the glass after 5s? Use that to find R and h at the time you want to find the rates.

Could I use R=5H/20 and plug 3 in as a constant? If you plug in 0, the answer will still be 0 however I am a little unsure what else to try since I cant think of a number that i can multiply times 0 and not get 0, unless I just add 3 on to the end like

R= 5h/20 + 3

Dick