1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rearranging to eliminate the R variable in volume equation

  1. Nov 10, 2011 #1
    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?
     
  2. jcsd
  3. Nov 10, 2011 #2
    What are R, r, and h?
     
  4. Nov 10, 2011 #3
    R=5 cm r= 3 cm h= 20 cm
     
  5. Nov 10, 2011 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  6. Nov 11, 2011 #5
    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?
     
  7. Nov 11, 2011 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  8. Nov 11, 2011 #7
    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
     
  9. Nov 11, 2011 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
    Last edited: Nov 11, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rearranging to eliminate the R variable in volume equation
  1. Rearranging Equations (Replies: 3)

  2. Eliminating variables (Replies: 1)

  3. Rearranging Equations (Replies: 6)

  4. Rearranging Equation (Replies: 1)

Loading...