# Related Rates questions

1. Jan 22, 2005

### 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.

2. Jan 23, 2005

### Tide

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

3. Jan 23, 2005

### Jcryan

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

4. Jan 25, 2005

### Tide

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.

5. Jan 25, 2005

### Aki

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

6. Jan 25, 2005

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)
$$\frac{dh}{dt}$$

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

Last edited: Jan 26, 2005
7. Jan 25, 2005

### 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.

8. Jan 26, 2005

### 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.

9. Jan 26, 2005

Thanks Tide