Frosty's Doom
Frosty the snowman is made from two uniform spherical snowballs, of radii 2R and 3R. The smaller (which is his head) stands on top of 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.
Determine the ratio of Frosty's volume to his initial volume when Frosty is half his initial height, as a common fraction in simplest form a/b. Your answer should be the value of a+b.
(Source: Advanced Problems in Mathematics by Stephen Siklos)
The answer is 261.
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1 solution
Volume of Sphere= 4/3 pi r^3 and Area of Sphere= 4 pi r^2
We are told that the sphere's volume decreases at a rate proportional to its surface area, and the constant proportionality, k, is the same for both small and large balls.
4 pi r^2 dr/dt= k *4 pi*r^2 => dr/dt=k ( all other terms cancel out)
Integrate r with respect to t.
=> r= kt +c
When t=0, r sub 0 = c( constant)
Smaller snowball(s): When t=0, r sub s = 2
r sub s (t) = r sub s (0) +kt => 2 + kt
Bigger snowball(b): When t=0, r sub b =3 r sub b(t) = r sub b (0) + kt => 3 + kt
Taking both snowballs into consideration, at time t,
radius (t)= (2+kt) + ( 3+kt) = 5 + 2kt
Height of snowman(t=0)= diamters of both snowballs= 2 ( 5 +2 kt)= 10 + 4kt=10
After time t,height of snowman is halved. Hence
10+4kt=5
kt= -5/4 = -1.25
This means after time t, smallball of 2 shrunk by 1.25 and bigball of 3 shrunk by 1.25 also, giving us 0.75 and 1.75
The new volumes are 4/3 pi 0.75^3 and 4/3 pi 1.75^3
Old volumes are 4/3 pi 2^3 and 4/3 pi 3^3
S, the new voulmes over old volumes=(4/3 pi 0.75^3+4/3 pi 1.75^3)/( 4/3 pi 2^3+4/3 pi 3^3)= 37/224