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Old 11-16-2003, 09:26 AM   #1 (permalink)
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Calc question...

A hemispherical bubble is placed on a spherical bubble of radius 1. A smaller hemisperical bubble is placed on the first one. This process continues until n bubbles have been used. Prove that the maximum height of any bubble tower with n bubbles is 1 + n^1/2

I"ve been working on this for a couple hours and have no idea.
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Old 11-16-2003, 12:04 PM   #2 (permalink)
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there seems to be a detail missing here... how does the bubble size change per bubble? maybe i'm misunderstanding what you mean by "hemispherical bubble"... clarify?
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Old 11-16-2003, 07:03 PM   #3 (permalink)
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Yeah I need a picture...
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Old 11-16-2003, 08:29 PM   #4 (permalink)
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Here is the best we can do for a picture...


On those are really hemispheres. The rate of change isn't given, its just assumed to meet be 1+ sqrt(n) in the end.

I know that picture is horrible... My version was worse, someone had to make that for me. I don't do graphic stuff...
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Old 11-16-2003, 08:46 PM   #5 (permalink)
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I dont see how its possible, but *shrug*. If this is from a book, can we get the 'exact' wording of the question?
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Old 11-16-2003, 09:01 PM   #6 (permalink)
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This is the exact wording of the probelm. We really can't do much better. Any thing you are questioning we have also. All we are really looking for i a fresh perspective and possibly new ideas.
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Old 11-16-2003, 10:24 PM   #7 (permalink)
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I'd try usin other theorms to sub in with eachother to try this... thats a damn hard question tho
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Old 11-17-2003, 09:25 AM   #8 (permalink)
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So you're saying 1 + n^(1/2) ?

I get a bit rusty on order of operations...

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Old 11-17-2003, 11:02 AM   #9 (permalink)
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Location: can i use bbcode [i]here[/i]?
hehe, this might be fun, i haven't done calc in forever, but i'll give it a try.

The way I'm starting this out is:

let r(n) be the radii of the (hemi)spheres. So r(1)=1, as defined in the problem. And r(n) < r(n-1). Now we have to show that the summation of all r's has a maximum value of 1+n^(1/2).

Does that sound right?

**edit** Found a stupid error in my logic. We don't just sum up the value of the radii, it's more complicated than that.... *continues scribbling*

ok, so who can figure out this part: if one hemisphere is stacked whose radius is x less than the sphere, how much height is added to the stack?
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Last edited by Moskie; 11-17-2003 at 11:15 AM..
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Old 11-17-2003, 05:03 PM   #10 (permalink)
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This is a really neat problem. The thing that everyone has missed is that it says the *maximum* height is 1+sqrt(n). This is a maximization problem.

I'll get you started on the n=2 case.

Start at the centerline of the big bubble. Then define an angle a up to the bottom of the second bubble. Cos a=R2. the radius of the second bubble, and Sin a=h, the vertical distance from the center of bubble 1 to the bottom of bubble 2.

h=Sin a=sqrt(1-(Cos a)^2)=sqrt(1-R2^2)

The total height of the tower is

H=1+h+R2=1+R2+sqrt(1-R2^2)
(the 1 is from the bottom half of the big bubble)

Maximizing H gives H=1+sqrt(2). Now repeat for arbitrary n.

Sorry that probably sounds like gibberish without a picture to go along.

Last edited by stingc; 11-17-2003 at 05:34 PM..
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Old 11-17-2003, 07:18 PM   #11 (permalink)
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hmm

Edit used periods to help with spacing

Here is the way I did it but I came up with a different solution then the problem. Can anyone find my logic error?

First I make one assumption:

The maximum height of the entire bubble tree can be found by maximizing the height of every touching pair of bubbles.

Now if this is incorrect someone please explain to me why.

Ok the rest will be full of crappy notation that can be written very well in this text editor but I will try.

First we draw a line going horizontal across each radius. This breaks our tree into n+1 intervals (1 interval is the bottom of the base circle radius, 1 interval is the top circle radius, and the rest of the intervals are the distances between consecutive radiuses.

If we denote the distance between the Nth and Nth+1ís circles radius as D(n) and the radius of the Nth circle as R(n) we then have our height

H(n)=R(1)+R(n)+[D(1)+...+D(n-1)]

Ok I hope you are still following (a tablet PC would be nice right about now).

Letís now look at a simple example with 2 circles.

If we set the radius of our larger circle to be R(n) and the smaller circles radius to be R(n+1) then we can say the following.

If we draw a line between the radius of the larger circle and connect it to a point where the two circles touch that distance is R(n). Then if we draw from the bottom circleís radius straight up to the top circles radius and then complete the triangle we have the following triangle.

...... R(n+1)
...... -------
...... |...../
D(n)..|..../ R(n)
...... |../
...... |/
We then have D(n)=sqrt(R(n)^2-R(n+1)^2)

Back to our original height equation for 2 circles we now have

H(n+1)= R(n) + R(n+1) + D(n)

H(n+1)= R(n) + R(n+1) + sqrt(R(n)^2-R(n+1)^2)

We now want to maximize this using R(n+1) as our variable

R(n) is a constant

Hí(n)=1-1*R(n+1)/sqrt(R(n)^2-R(n+1)^2)

Setting this equal to zero and solving gives us

R(n+1)=R(n)/sqrt(2)

This is a FO linear recursion with which gives us a general solution of

R(n)=(1/sqrt(2))^(n-1)*R(1)

So we now know the maximized radius at every height. Our calc work is done and now all we need to do is a simplifying of our original series


H(n)=R(1)+R(n)+[D(1)+...+D(n-1)]

We know R(1)=1
We know R(n)=(1/sqrt(2))^(n-1)

So now letís look at our summation of the Ds

n-1
Sum sqrt[ ((1/sqrt(2))^(2k-2) Ė (1/sqrt(2))^(2k)) ]
k=1


working with what is inside of the outside sqrt we can factor out a (1/sqrt(2))^(2k-2) leaving us with

n-1
Sum sqrt[ ((1/sqrt(2))^(2k-2)*(1 Ė (1/sqrt(2))^2) ]
k=1

OR

n-1
Sum sqrt[ ((1/sqrt(2))^(2k-2)*(1/2) ]
k=1


By putting in the outer square root as a 1/2 power we have

n-1
Sum (1/sqrt(2))^(k-1)*(1/sqrt(2))
k=1

We can then place the last 1/sqrt(2) into our exponent by adding 1 to th power.

n-1
Sum (1/sqrt(2))^k
k=1

Now donít forget about our first 2 terms that we still have. We have the 1 + (1/sqrt(2))^(n-1)

We can add our 1 directly into our sum by lowering k by 1.


.......................n-1
(1/sqrt(2))^(n-1) + Sum (1/sqrt(2))^k
.......................k=0

The summation is now a geometric series with base 1/sqrt(2) giving us the final equation for H(n)

H(n)= (1/sqrt(2))^(n-1) + 1-(1/sqrt(2))^n
..............................--------------------
....................................1-1/sqrt(2)

Placing values into this match the equation 1+sqrt(n) for H(1) and H(2). After that it gets of a minor amount.

My guess is my initial assumption might be wrong.

Last edited by Rekna; 11-17-2003 at 07:25 PM..
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Old 11-25-2003, 06:01 PM   #12 (permalink)
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Sorry, but you lost me at the end of the first line. OK, I admit, I was never really good at math.
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