Rearranging to eliminate the R variable in volume equation

[FiveAlive]

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?

 

[murmillo]

What are R, r, and h?

 

[FiveAlive]

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

 

[Dick]

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?

 

[FiveAlive]

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]

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.

 

[FiveAlive]

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 can't 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]

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 can't 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

R=5h/20+3 is ok, for h=0. But at h=20 you want R=5, it's NOT ok there. Fix the slope. What should be multiplying h instead of 5/20?