What's the relationship between the height of something in a hollow hemisphere...

[Howard]

I'm trying to find the rate at which the height of a liquid decreases in a hemispherical bowl (r = 30cm) when the height of the liquid is 15 cm...

I get dr/dt = -250cm^3/(pi*r^2 min), but I don't see how that has anything to do with the question. 🙁

 

[Howard]

Maybe what I'm looking for is how to find the volume of the liquid in a sphere at any given height, h <= diameter...

 

[BigPoppa]

Assuming the bowl is in the shape of a U:

Volume of a sphere = 1/3(pi)(r)^3.
Volume of a hemisphere = 1/6(pi)(r)^3.
Radius liquid = height of liquid, ergo r = h.

And my mind just went blank. Related rates used to be my friend.

 

[Howard]

Nah, the radius of the liquid does not = the height of the liquid.

Besides, the formula for the volume of a sphere is 4/3, not 1/3. 😀

 

[BigPoppa]

Ahh crap, thats right. The derivative of the area = the surface area. 4/3(pi)r^3 -> 4pi(r)^2. Is the bowl not a perfect sphere? If it is, the radius of the liquid has to be the same as the height of the liquid. If not, we kind of need an equation for the shape of the bowl.

 

[Howard]

Well, imagine a bowl with water in it. The body of water is not a sphere, or a hemisphere, or a cone or cylinder or whatever. It's a section of a sphere. Now how can I use the height of that water to find its volume?

 

[BigPoppa]

Originally posted by: Howard
Well, imagine a bowl with water in it. The body of water is not a sphere, or a hemisphere, or a cone or cylinder or whatever. It's a section of a sphere. Now how can I use the height of that water to find its volume?

How can the body of water be anything BUT a perfect hemisphere? In relation the bowl it is not, however. I believe I just contradicted myself. Lemme think on this some more.

 

[Howard]

Originally posted by: BigPoppa

Originally posted by: Howard
Well, imagine a bowl with water in it. The body of water is not a sphere, or a hemisphere, or a cone or cylinder or whatever. It's a section of a sphere. Now how can I use the height of that water to find its volume?

How can the body of water be anything BUT a perfect hemisphere? In relation the bowl it is not, however. I believe I just contradicted myself. Lemme think on this some more.

So are you saying a body of water in a sphere, no matter th eheight, is a perfect hemisphere?

 

[BigPoppa]

It is a perfect hemisphere no matter how much water has been removed from the bowl. But, it is not a perfect hemisphere in relation to the original bowl, which is where I believe the problem of my interpretation is.

 

[Crypticburn]

Originally posted by: BigPoppa
It is a perfect hemisphere no matter how much water has been removed from the bowl. But, it is not a perfect hemisphere in relation to the original bowl, which is where I believe the problem of my interpretation is.

Wrong. If the height of the water, h, is not equal to the radius of the sphere, r, containing the hemisphere, then the water does not make a hemisphere. You can think of this many ways, but here are a few simple examples:

When h = 2*r, then the water makes an entire sphere
When h = r/2, the it could be described as half a pointy tear drop


To solve this problem, the most simple approach is the method of washers.:

Step 1: Find the radius of the circle describing the top of the water (rd) at an arbitrary height (h)
Draw a circle and the vertical diameter.
Draw a horizontal line lower than the center of the circle, and define it as height h from the bottom of the circle
Draw a a radius from the center of the shpere to where the horizontal line intersects the circle.

This makes a right triangle, by the sides rd, r-h, and the hypotenuse r.

Therefore rd^2 = r^2 - (r-h)^2 = 2*r*h - h^2

Step 2: Find the area of the circle describing the top of the water (A)

A = pi * rd^2 = 2*r*h*pi - h^2*pi

Step 3: Find the volume of the water (V)

V = integral ( A*dh ) from 0 to h
V = pi*r*h^2-1/3*h^3*pi

Step 4: Verify some known solutions:

V(h) = pi*r*h^2-1/3*h^3*pi
V(2r) = pi*r*(2r)^2-1/3*(2r)^3*pi = 4/3*pi*r^3 (correct)
V(r) = pi*r*(r)^2-1/3*(r)^3*pi = 2/3*pi*r^3 (correct)
V(0) = pi*r*(0)^2-1/3*(0)^3*pi = 0 (correct)

Crypticburn

 

[Howard]

Argh! I haven't learned integration yet!

Here's the problem in full:

Fruit punch is being ladled out of a hemispherical bowl of radius 30cm at a rate of 500cm^3/min. How fast is the level of the fruit punch falling at the time the depth (from the bottom of the sphere to the surface of the fruit punch) is 15cm?

The answer is 0.24cm/min...

 

[Howard]

^

 

[Howard]

^

 

[Howard]

^

 

[RossGr]

A little research (CRC Stantard Math Tables) gives the volume of interest as

V= pi h^2 (3R-h)/3


You know that

dV/dt = dV/dh * dh/dt


take the derivatives and plug in your numbers.

 

[Howard]

I assume that R is for radius? And is the /3 for everything, or just (3R-h)?

 

[RossGr]

Yes R is the radius of the sphere,

Think about it... does it make any difference if the /3 is just the one factor or all of the factors?