1. ## Filling Spherical Tank

I have a related rate question. I am an amateur mathematician...and I emphasize "amateur" here...and I have a question. I solved a related rate problem regarding the filling of a hemispherical tank. It is, I think, a standard problem in most calc 1 texts. Then I begin to wonder about how a full spherical tank fills. The first half of the filling of a spherical tank is the filling of the lower half and the second half is the filling of the upper hemisphere. I don't believe that the same method for calculating the rate of increase of the height per time is the same for filling the upper half is the same as for the lower half.
Then I began to think what happens if you just split the tank in two spherical halves and then put the upper half on the ground flat side down and began to fill it, assuming the flat side is down. I don't believe the same approach would work to finding dh/dt..would it? Or is dh/dt constant because the bottom is flat? I know that unlike when the bottom side fills, the height isn't always equal to the radius.
thanks...

2. ## Re: Filling Spherical Tank

Originally Posted by scottshannon
I have a related rate question. I am an amateur mathematician...and I emphasize "amateur" here...and I have a question. I solved a related rate problem regarding the filling of a hemispherical tank. It is, I think, a standard problem in most calc 1 texts. Then I begin to wonder about how a full spherical tank fills. The first half of the filling of a spherical tank is the filling of the lower half and the second half is the filling of the upper hemisphere. I don't believe that the same method for calculating the rate of increase of the height per time is the same for filling the upper half is the same as for the lower half.
Then I began to think what happens if you just split the tank in two spherical halves and then put the upper half on the ground flat side down and began to fill it, assuming the flat side is down. I don't believe the same approach would work to finding dh/dt..would it? Or is dh/dt constant because the bottom is flat? I know that unlike when the bottom side fills, the height isn't always equal to the radius.
thanks...
if the rate of water supply is constant then dh/dt varies with time. this is true even if you have the hemisphere inverted on the ground.

3. ## Re: Filling Spherical Tank

Well...lets say the tank is spherical and is 20 ft in diameter. As the tank begins filling from the bottom, the formula dV/dt =(4/3) pi r^2 (dh/dt) applies. One can find dh/dt at any height h as the lower half is being filled..ok..I get that. Now suppose that the level passes the 10 ft mark. When the level gets to the 15 Ft mark can the same formula be applied? That's my question. I understand that dh/dt is not constant.

4. ## Re: Filling Spherical Tank

Originally Posted by scottshannon
Well...lets say the tank is spherical and is 20 ft in diameter. As the tank begins filling from the bottom, the formula dV/dt =(4/3) pi r^2 (dh/dt) applies. One can find dh/dt at any height h as the lower half is being filled..ok..I get that. Now suppose that the level passes the 10 ft mark. When the level gets to the 15 Ft mark can the same formula be applied? That's my question. I understand that dh/dt is not constant.
i assume that you have correctly deduced $\frac{dV}{dt}=(4/3) \pi r^2 \cdot \frac{dh}{dt}$.
this same formula works for any height but you have to keep in mind that when you are calculating the $\frac{dh}{dt}$ at 10 ft mark then you have to first find the value of $r$ when the height is 10 ft and then use THIS value of $r$ in the formula. Similarly for 15 ft mark.
I am sorry for misinterpreting your question.

5. ## Re: Filling Spherical Tank

That's quite alright. Here is what bothers me. When the lower half is being filled, the radius is always equal to the hight..right? But when the upper half is being filled then this is no longer the case. It seems to me that when one calculates the volume via integration, the fact that r = h is an integral part of the derivation. SO when the top half is being filled whether it is sitting on the ground or not then you are saying the formula still applies?

Can you tell me how to use latex on this site so that I can write integrals and other mathematical symbols.

6. ## Re: Filling Spherical Tank

In my own mind I am not convinced that what applies to the solution in the lower half applies to the upper half.

7. ## Re: Filling Spherical Tank

forget my previous posts. I am representing $r$ as the radius of the sphere so $r$ is of course constant. I derived the formula myself and got $\pi (2rh-h^2) \frac{dh}{dt}=\frac{dV}{dt}$. This is true for both lower and upper half. Now this formula doesn't match with yours so first we should clarify that. Please post the derivation of the formula you posted in post #3.

To use latex symbols:
for example if you want to write $\alpha$ you should write ["tex"]\alpha["/tex"] WITHOUT THE INVERTED COMMAS. for more help there is a separate thread on the forum for latex help.

8. ## Re: Filling Spherical Tank

would you mind showing me how you derived your equation? It should be the same as the one earlier for the lower half.

9. ## Re: Filling Spherical Tank

In the lower half h=r so the formula you have is the same as the one earlier. In the upper half r is not equal to h so that's why the earlier formula doesnt work. can you show me your derivation?

10. ## Re: Filling Spherical Tank

Originally Posted by scottshannon
In the lower half h=r so the formula you have is the same as the one earlier. In the upper half r is not equal to h so that's why the earlier formula doesnt work. can you show me your derivation?
why h=r in the lower half? i am taking r to be the radius of the sphere.

11. ## Re: Filling Spherical Tank

yes I realize that but in the top half of the sphere the height is no longer equal to the radius r of the sphere. As you are moving upward from the bottom, at the halfway point it is the last point where r=h. After that r is no longer h. h>r. Isn't that correct? It is for this reason that dh/dt isn't governed in the top half by the same formula as in the bottom half. Now that's what I am thinking but I am not sure. I do believe that your formula is correct. Can you show me how you derived it?

12. ## Re: Filling Spherical Tank

Originally Posted by scottshannon
yes I realize that but in the top half of the sphere the height is no longer equal to the radius r of the sphere. As you are moving upward from the bottom, at the halfway point it is the last point where r=h. After that r is no longer h. h>r. Isn't that correct? It is for this reason that dh/dt isn't governed in the top half by the same formula as in the bottom half. Now that's what I am thinking but I am not sure. I do believe that your formula is correct. Can you show me how you derived it?
my point is "r>h in the lower half and r<h in the upper half."
i will post the derivation once you agree with this. I think your point is that "r=h in the lower half and h>r in the upper half" which is wrong according to me.

13. ## Re: Filling Spherical Tank

yes I agree with that.
Here is the approach I took to deriving the volume:

I know that the area of a circle is pi*r^2. I took an infinite disk with radius x so that x = (r^2 - y^2)^1/2.

The disk has thickness dy so I integrated dV = pi*(r^2 - y^2)dy and I integrated it from -r to h which would be the height of the tank above the x axis. I wanted to use calc 1 methods. The circle is centered on the origin. If you would. derive it that way I would appreciate it because I am not getting it to come out.

14. ## Re: Filling Spherical Tank

So I integrated from -r to h and got pi*r^2 - [pi*y^3]/3 evaluated from -r to some height h where h>r. My plan was then to take the derivative of both sides with respect to t with r being a constant but I am not getting what you got which I know is probably the correct formula.

15. ## Re: Filling Spherical Tank

let the sphere have radius $R$ and be centered at $(0,R)$.

$0 \le h \le 2R$

$V = \pi \int_0^h [R^2 - (y-R)^2] \, dy = \pi \int_0^h (2Ry - y^2) \, dy$

$\frac{dV}{dt} = \pi (2Rh - h^2) \cdot \frac{dh}{dt}$

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