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Old 2012-02-27, 22:55   #1
lavalamp
 
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Default 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
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Old 2012-02-27, 23:20   #2
Dubslow
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Hmm... I think I've missed something.

Let r be the time varying radius, such that initially r(0) = R. Then h=(2*2r+2*3r)=10r (and h(0) = 10R). Therefore dV/dh = 1/10*dV/dr. dV/dr = dv[SUB]2[/SUB]/dr+dv[SUB]3[/SUB]/dr; but dv[SUB]x[/SUB]/dr is precisely the surface area; so dV/dr=A[SUB]2[/SUB]+A[SUB]3[/SUB] = 4*pi*( (2r)[SUP]2[/SUP] + (3r)[SUP]2[/SUP] ) = 4*13*pi*r[SUP]2[/SUP], and so dV/dh = 2/5*13*pi*r[SUP]2[/SUP], independent of dV/dt. Equivalently, substituting h=10r gives dV/dh = 2/5*13*pi*(h/10)[SUP]2[/SUP] = 2/500*13*pi*(h)[SUP]2[/SUP] (However, dV/dt, is necessary (along with dV/dr) to solve for dr/dt.) (In the three sphere case, dV/dh = 2/5*29*pi*r[SUP]2[/SUP], where 29 = 13+4*4.)

Last fiddled with by Dubslow on 2012-02-27 at 23:40
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Old 2012-02-27, 23:30   #3
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Quote:
Originally Posted by Dubslow View Post
dV/dh = 1/10*dV/dR
No, R is a constant.
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Old 2012-02-27, 23:35   #4
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Quote:
Originally Posted by lavalamp View Post
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
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Old 2012-02-27, 23:57   #5
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Quote:
Originally Posted by lavalamp View Post
Extra difficulty: What if frosty is made of three spheres of radii 2R, 3R and 4R?
I.e., one of the classical snowmen.
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Old 2012-02-28, 01:31   #6
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Quote:
Originally Posted by Dubslow View Post
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.
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Old 2012-02-28, 03:46   #7
LaurV
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Quote:
Originally Posted by lavalamp View Post
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
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Old 2012-02-28, 04:53   #8
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Forget it, I found my mistake, missed a division by 4pi/3. This is the correct "simulation":
Attached Thumbnails
Click image for larger version

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Old 2012-02-28, 13:49   #9
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Quote:
Originally Posted by lavalamp View Post
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.
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Old 2012-02-28, 21:37   #10
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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?
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Old 2012-02-29, 09:20   #11
LaurV
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Quote:
Originally Posted by lavalamp View Post
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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