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!

Related Rates questions

  1. Jan 22, 2005 #1
    Ok. This is the question I got from a friend of the family. There is a conically shaped bin in which sand is poured in at a rate of 2.4 m3 per minute. The sand escapes out the bottom of the conically shaped bin at 0.8 m3 per minute. The opening at the top has a radius of 5 m and the opening in which the sand is escaping is 0.5 m. The question asks for the rate of depth change after 10 minutes. I took Calculus but absolutely hate word questions. Anyway the most I can figure is that they are asking for the rate of change of the height of sand in the container.
     
  2. jcsd
  3. Jan 23, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Are you sure there wasn't any additional information provided - such as the cone angle?
     
  4. Jan 23, 2005 #3
    Well the only thing that was provided that I don't think I mentioned was the height of the bin is 8 m.
     
  5. Jan 25, 2005 #4

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    That is essential information!

    If R is the radius at the top, r the radius at the bottom and H is the height of the bin then the height of the deposited sand relative to the bottom is

    [tex]h = \left[ \left( \frac {3V}{\pi} \frac {R-r}{H} + r^3 \right)^{1/3} - r\right] \frac {H}{R-r}[/tex]

    where V is the volume of the deposited sand. Basically, you find the volume of partial (right circular) cone and solve for h.
     
  6. Jan 25, 2005 #5

    Aki

    User Avatar

    I calculated the rate of change to be 0.0115 m/min.
     
  7. Jan 25, 2005 #6
    related rates problems are cool
    i couldnt figure out Tide's part but can see how he did it.
    anyway i worked through the question and got a different answer to aki, here is what i did
    we want to find (where h is defined in Tide's post)
    [tex]\frac{dh}{dt}[/tex]

    where

    [tex]\frac{dh}{dt} = \frac{dh}{dV}\frac{dV}{dt}[/tex]

    [tex]\frac{dV}{dt} = 1.6 [/tex] and

    [tex]\frac{dh}{dV} = 0.07513809118[/tex] after 10 mins ( i took V=16 after 10 mins)
    so

    [tex]\frac{dh}{dt} = 0.12 m / min [/tex]

    just interested to know if the procedure was right, might have type the equation wrong in maple
     
    Last edited: Jan 26, 2005
  8. Jan 25, 2005 #7
    Ok. Sorry about missing the height thing. Had to reread the post just to realize I forgot it. Sorry to bother you again while I get where vladimir69 got his equation to solve the problem I was wondering if Tide could expound on how he developed the formula he created.
     
  9. Jan 26, 2005 #8

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Jcryan,

    I can get you started.

    To calculate the volume of a partial (right circular) cone suppose the top radius is R and the bottom radius is r and the separation between the top and bottom is h. If it were a complete cone with the bottom extending downward to a vertex then its volume would simply be

    [tex]\frac {\pi}{3} R^2 H[/tex]

    where H is the height of the complete cone. For our partial cone, the volume will be the difference

    [tex]\frac {\pi}{3} R^2 H - \frac {\pi}{3} r^2 (H-h)[/tex]

    We don't know what H is but we do know that the large and small cones are similar so that

    [tex]\frac {R}{H} = \frac {r}{H-h}[/tex]

    which you can use in the equation for the volume of the partial cone to find

    [tex]V = \frac {\pi}{3} \frac {R^3 - r^3}{R - r} h[/tex]

    To address your particular problem I have to change notation just a little. I'm going to represent the height of the bin by H and the height of the sand's upper surface to be h. If the radius of the upper surface of the sand is [itex]R_s[/itex] then

    [tex]R_s = r + (R - r) \frac {h}{H}[/tex]

    so the volume of the deposited sand is

    [tex]V = \frac {\pi}{3} \frac {R_s^3-r^3}{R_s - r} \frac {R_s - r}{R - r} H[/tex]

    or

    [tex]V = \frac {\pi}{3} \frac {R_s^3 - r^3}{R - r} H[/tex]

    and to complete the derivation replace [itex]R_s[/itex] in V using the equation above and solve for h.
     
  10. Jan 26, 2005 #9
    Thanks Tide
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Related Rates questions
  1. Related Rates Question (Replies: 3)

Loading...