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2012-03-01, 23:59   #12
lavalamp

Oct 2007
Manchester, UK

2×3×227 Posts

Quote:
 Originally Posted by LaurV Did not try. Integral calculus was never my stronger point. And you want me to solve a sum of integrals? Are you nuts?
The question asks for dV/dh, so if you find a formula for V in terms of h, then differentiate, you'd have an answer.

2012-03-02, 01:21   #13
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by lavalamp The question asks for dV/dh, so if you find a formula for V in terms of h, then differentiate, you'd have an answer.
well V = V1+V2 = ((4/3) *Pi) *(R1^3 + R2^3) ; h=D1+D2 = 2*(R1+R2)

h^3 = 8*R1^3 + 24*R2*R1^2 + 24*R2^2*R1 + 8*R2^3 = 8*(R1^3 + R2^3) + 24*(R2*R1^2 + R2^2*R1)

V-h^3 = -(8-((4/3)*Pi))*(R1^3+R2^3) + 24*(R2*R1^2 + R2^2*R1) that's about how far I got so far with Pari's help.

Last fiddled with by science_man_88 on 2012-03-02 at 01:22

 2012-03-04, 16:17 #14 xilman Bamboozled!     "𒉺𒌌𒇷𒆷𒀭" May 2003 Down not across 2ACC16 Posts FWIW, it's been snowing here for most of the afternoon. Relatively unusual for this time of year.
 2012-03-04, 20:13 #15 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2012-03-04, 20:53   #16
Wacky

Jun 2003
The Texas Hill Country

21018 Posts

Quote:
 Originally Posted by Dubslow Everywhere in the US ... plastered in snow
What is "snow"?

(Last year, we were even asking "What is rain?")

 2012-03-04, 22:34 #17 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
 2012-03-08, 01:03 #18 lavalamp     Oct 2007 Manchester, UK 2·3·227 Posts Is anyone still working on this or should I post the solution?
2012-03-08, 01:37   #19
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts

Quote:
 Originally Posted by lavalamp Is anyone still working on this or should I post the solution?
I would think it's up to you I get distracted a lot.

 2012-03-08, 20:23 #20 lavalamp     Oct 2007 Manchester, UK 2×3×227 Posts Well I posted a slightly harder version of this problem than the one that was given to me, so I'll post the original now which gives more of a hint to the answer. If there are still no takers after a while I'll post my solution. 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. When frosty is half his initial height, show that the ratio of his volume to his initial volume is 37 : 224. Let V and h denote Frosty's total volume and height, respectively, at time t. Show that, for 2R < h <= 10R: $\frac{dV}{dh} = \frac{\pi}{8}(h^{2} + 4R^{2})$ And derive the corresponding expression for 0 <= h < 2R. Sketch dV/dh as a function of h, for 4R >= h >= 0, hence give a rough sketch of V as a function of h.
2012-03-13, 03:11   #21
lavalamp

Oct 2007
Manchester, UK

2×3×227 Posts

I guess no-one is going to bite then. I have attached my solution.
Attached Files
 frosty.pdf (146.9 KB, 582 views)

Last fiddled with by lavalamp on 2012-03-13 at 03:17

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