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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.
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  3. Jan 23, 2005 #2


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    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


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    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


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    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} = \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)

    [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


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    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]


    [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
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