[QUOTE=LaurV;291271]Did not try. Integral calculus was never my stronger point. And you want me to solve a sum of integrals? Are you nuts? :smile:[/QUOTE]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.

[QUOTE=lavalamp;291475]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.[/QUOTE]

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.

FWIW, it's been snowing here for most of the afternoon. Relatively unusual for this time of year.

Lucky. Everywhere in the US except where I am (:rant:) has been plastered in snow.

[QUOTE=Dubslow;291894]Everywhere in the US ... plastered in snow[/QUOTE]

What is "snow"?








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

Well okay, everything north and west of Texas. Even AZ and NM got snow at one point.

Is anyone still working on this or should I post the solution?

[QUOTE=lavalamp;292279]Is anyone still working on this or should I post the solution?[/QUOTE]

I would think it's up to you I get distracted a lot.

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:

[TEX]\frac{dV}{dh} = \frac{\pi}{8}(h^{2} + 4R^{2})[/TEX]

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.

I guess no-one is going to bite then. I have attached my solution.