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 2012-02-27, 22:55 #1 lavalamp     Oct 2007 Manchester, UK 2·3·227 Posts Melting Snowman Frosty the snowman is made from two uniform spherical snowballs, of radii 2R and 3R. The smaller (which is his head) stands on top the larger. As each snowball melts, its volume decreases at a rate which is directly proportional to its surface area. The constant of proportionality being the same for each snowball. During melting, the snowballs remain spherical and uniform. Let V and h denote Frosty's total volume and height, respectively, at time t. Find dV/dh in terms of h and R. Hint 1: Two expressions for dV/dh will be needed, one for 0 <= h < 2R and another for 2R < h <= 10R Hint 2: When frosty is half his initial height, the ratio of his volume to his initial volume is 37 : 224. Extra difficulty: What if frosty is made of three spheres of radii 2R, 3R and 4R? Last fiddled with by lavalamp on 2012-02-27 at 23:01
 2012-02-27, 23:20 #2 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2012-02-27, 23:30   #3
lavalamp

Oct 2007
Manchester, UK

136210 Posts

Quote:
 Originally Posted by Dubslow dV/dh = 1/10*dV/dR
No, R is a constant.

2012-02-27, 23:35   #4
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts

Quote:
 Originally Posted by lavalamp No, R is a constant.
Well then substitute r for R, where dr/dt =/= 0. dV/dh cannot depend on R, only on r. To see this, consider we have initial radius Q[SUB]1[/SUB]=4R and Q[SUB]2[/SUB]=6R. After such time has passed that the snowman has melted to spheres of size R and 1.5R, the rate dV/dh cannot possibly be different regardless of whether or not we started at 2R or Q=4R.

Last fiddled with by Dubslow on 2012-02-27 at 23:42

2012-02-27, 23:57   #5

"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·641 Posts

Quote:
 Originally Posted by lavalamp Extra difficulty: What if frosty is made of three spheres of radii 2R, 3R and 4R?
I.e., one of the classical snowmen.

2012-02-28, 01:31   #6
lavalamp

Oct 2007
Manchester, UK

2·3·227 Posts

Quote:
 Originally Posted by Dubslow Well then substitute r for R, where dr/dt =/= 0. dV/dh cannot depend on R, only on r. To see this, consider we have initial radius Q[SUB]1[/SUB]=4R and Q[SUB]2[/SUB]=6R. After such time has passed that the snowman has melted to spheres of size R and 1.5R, the rate dV/dh cannot possibly be different regardless of whether or not we started at 2R or Q=4R.
Just because their initial radii start in the ratio 2:3 does not mean that they will always be in this ratio.

2012-02-28, 03:46   #7
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

24×613 Posts

Quote:
 Originally Posted by lavalamp Hint 2: When frosty is half his initial height, the ratio of his volume to his initial volume is 37 : 224.
Are you sure? for me the ratio goes to 25.8 no matter how I compute it, I even had a numerical example emulated in excel (can see him melting). No matter how fast/slow I melt him, I always get the same thing. I can melt it in an instant, or in a billion years. For example, melting it in a single step:
Code:
r1       r2            s1            s2             h              v        dr1          dr2
230.35    345.53    666802.87    1500306.45    575.88    224000020.66    88.5979    199.3453
141.76    146.18    252514.86    268540.86      287.94    25017223.21     33.5516    35.6809
.
(I arranged the initial radii to have the volume 224 to be easy to see, but the initial radii does not mater, and melt it to half in a single step, there is a "melting speed" parameter which can't be seen on the table, which takes positive real values up to infty)

In this case the report is 25 to 224. Its limit (when the melting speed limits to 0) is somewhere at 25.79 or so.

Last fiddled with by LaurV on 2012-02-28 at 03:59 Reason: table format fkd up, repaired by using plain text

 2012-02-28, 04:53 #8 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 24·613 Posts Forget it, I found my mistake, missed a division by 4pi/3. This is the correct "simulation": Attached Thumbnails
2012-02-28, 13:49   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by lavalamp Frosty the snowman is made from two uniform spherical snowballs, of radii 2R and 3R. The smaller (which is his head) stands on top the larger. As each snowball melts, its volume decreases at a rate which is directly proportional to its surface area. The constant of proportionality being the same for each snowball. During melting, the snowballs remain spherical and uniform. Let V and h denote Frosty's total volume and height, respectively, at time t. Find dV/dh in terms of h and R. Hint 1: Two expressions for dV/dh will be needed, one for 0 <= h < 2R and another for 2R < h <= 10R Hint 2: When frosty is half his initial height, the ratio of his volume to his initial volume is 37 : 224. Extra difficulty: What if frosty is made of three spheres of radii 2R, 3R and 4R?

here's my 2 cents (volume of a sphere) / (surface area of a sphere) = 1/3*r which is constant algebraically for both. dV for a sphere is given on a Wikipedia page, and the one we are figuring out has 2 spheres so its sum( dV1,dV2). as radius changes so does diameter, sum(diameter 1,diameter 2) = height.

 2012-02-28, 21:37 #10 lavalamp     Oct 2007 Manchester, UK 55216 Posts So you got the ratio right LaurV, but it is a precise mathematical thing, so can you get it analytically rather than numerically? Nice snowman graphic btw. Does it help the simulation?
2012-02-29, 09:20   #11
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

24·613 Posts

Quote:
 Originally Posted by lavalamp So you got the ratio right LaurV, but it is a precise mathematical thing, so can you get it analytically rather than numerically?
Did not try. Integral calculus was never my stronger point. And you want me to solve a sum of integrals? Are you nuts?

Quote:
 Nice snowman graphic btw. Does it help the simulation?
Yes, it shows the ratio of the spheres. I made a macro (5 VBA lines in a loop for all lines in the sheet) to show the spheres, but they became small and difficult to see, then I modify it to keep the snowman the same size. In this case you can see an animation with the spheres changing the size (the ratio between them, but they occupy the same space) when I run the macro.

Hint:
Code:

ActiveSheet.Shapes.Range(Array("Oval 1")).ShapeRange.ScaleWidth blablabla, msoFalse, msoScaleFromTop
....
Selection.ShapeRange.ScaleHeight blablabla, msoFalse, msoScaleFromBottom

Last fiddled with by LaurV on 2012-02-29 at 09:25

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