Calculating Related Rates: Conical Bin Sand Flow at 10 Minutes

[Jcryan]

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.

 

[Tide]

Are you sure there wasn't any additional information provided - such as the cone angle?

 

[Jcryan]

Well the only thing that was provided that I don't think I mentioned was the height of the bin is 8 m.

 

[Tide]

Well the only thing that was provided that I don't think I mentioned was the height of the bin is 8 m.

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

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

where V is the volume of the deposited sand. Basically, you find the volume of partial (right circular) cone and solve for h.

 

[Aki]

I calculated the rate of change to be 0.0115 m/min.

 

[vladimir69]

related rates problems are cool
i couldn't 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)
\frac{dh}{dt}

where

\frac{dh}{dt} = \frac{dh}{dV}\frac{dV}{dt}

\frac{dV}{dt} = 1.6 and

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

\frac{dh}{dt} = 0.12 m / min

just interested to know if the procedure was right, might have type the equation wrong in maple

 

[Jcryan]

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.

 

[Tide]

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

\frac {\pi}{3} R^2 H

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

\frac {\pi}{3} R^2 H - \frac {\pi}{3} r^2 (H-h)

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

\frac {R}{H} = \frac {r}{H-h}

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

V = \frac {\pi}{3} \frac {R^3 - r^3}{R - r} h

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 R_s then

R_s = r + (R - r) \frac {h}{H}

so the volume of the deposited sand is

V = \frac {\pi}{3} \frac {R_s^3-r^3}{R_s - r} \frac {R_s - r}{R - r} H

or

V = \frac {\pi}{3} \frac {R_s^3 - r^3}{R - r} H

and to complete the derivation replace R_s in V using the equation above and solve for h.

 

[Jcryan]

Thanks Tide