20120227, 22:55  #1 
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 20120227 at 23:01 
20120227, 23:20  #2 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
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 20120227 at 23:40 
20120227, 23:30  #3 
Oct 2007
Manchester, UK
1362_{10} Posts 

20120227, 23:35  #4 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
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 20120227 at 23:42 
20120227, 23:57  #5 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 

20120228, 01:31  #6  
Oct 2007
Manchester, UK
2·3·227 Posts 
Quote:


20120228, 03:46  #7  
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}×613 Posts 
Quote:
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 20120228 at 03:59 Reason: table format fkd up, repaired by using plain text 

20120228, 04:53  #8 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
Forget it, I found my mistake, missed a division by 4pi/3. This is the correct "simulation":

20120228, 13:49  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
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. 

20120228, 21:37  #10 
Oct 2007
Manchester, UK
552_{16} 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? 
20120229, 09:20  #11  
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
Quote:
Quote:
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 20120229 at 09:25 
