1 problem found
Frosty the snowman is made from two uniform spherical snowballs, of initial 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 both snowballs. During melting each snowball remains spherical and uniform. When Frosty is half his initial height, find the ratio of his volume to his initial volume. If \(V\) and \(S\) denote his total volume and surface area respectively, find the maximum value of \(\dfrac{\mathrm{d}V}{\mathrm{d}S}\) up to the moment when his head disappears.
Solution: \(V_h = \frac43 \pi r_h^3, S_h = 4 \pi r_h^2\) \(\frac{\d V_h}{\d t} = -k4\pi r_h^2 \Rightarrow 4\pi r_h^2 \frac{\d r_h}{\d t} = -k 4\pi r_h^2 \Rightarrow \frac{\d r_h}{\d t} = -k\) Therefore \(r_h = 2R - kt, r_b = 3R - kt\). The height will halve when \(2kt = \frac{5}{2}R \Rightarrow kt = \frac{5}{4}R\) and the two sections will have radii \(\frac{3}{4}R\) and \(\frac{7}{4}R\) and the ratio of the volumes will be: \begin{align*} \frac{\frac{3^3}{4^3}+\frac{7^3}{4^3}}{2^3+3^3} = \frac{37}{224} \end{align*} \begin{align*} && \frac{\d V}{\d t} &= -4\pi k(r_h^2+r_b^2) \\ && \frac{\d S}{\d t} &= -8\pi k (r_h+r_b) \\ \Rightarrow && \frac{\d V}{\d S} &= \frac{r_h^2 + r_b^2}{2(r_h+r_b)} \\ &&&= \frac{(2R-kt)^2+(3R-kt)^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rkt+2k^2t^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rs + 2s^2}{2(5R-2s)} \end{align*} Where \(s = kt\) and \(0 \leq s \leq 2R\). We can maximise this but differentiating wrt to \(s\). \begin{align*} \Rightarrow && &= \frac{(-10R+4s)(10R-4s)+4(13R^2-10Rs+2s^2)}{4(5R-2s)^2} \\ &&&= \frac{-48R^2+40Rs-8s^2}{4(5R-2s)^2} \\ &&&= \frac{-8(s-2R)(s-3R)}{4(5R-2s)^2} \\ &&&<0 \end{align*} Therefore it is largest when \(s = 0\), ie \(\frac{13R^2}{10R} = \frac{13}{10}R\)