Challenge Level

You might find the following useful!

- Surface area of a sphere: $A=4 \pi r^2$
- Volume of a sphere: $V=\frac 4 3 \pi r^3$

**Can you write down a differential equation connecting $V$ and $A$, using what you have been told about the rate at which the volume decreases?**

We are told that for each snowball the volume decreases at a rate which is proportional to the surface area, so we have: $$\dfrac{\mathrm{d} V}{\mathrm{d} t}=-kA \quad \text{where} \quad k>0$$

**Can you rewrite the differential equation in terms of $r$?**

Substituting for $V$ and $A$ gives: $$ \quad \; \; \; \dfrac{\mathrm{d}}{\mathrm{d} t}(\frac 4 3 \pi r^3)=-k\times 4 \pi r^2$$ $$ \; \; \implies 4 \pi r^2 \dfrac{\mathrm{d} r}{\mathrm{d} t} =-k\times 4 \pi r^2 $$ $$ \implies \dfrac{\mathrm{d} r}{\mathrm{d}
t} =-k$$

It might be helpful to use notation like $r_1=$ radius of head, $r_2=$ radius of body.

**Can you solve the differential equation to find $r_1$ and $r_2$ in terms of $t$?**

Integrating with respect to $t$ gives: $$r=-kt + c$$ Using the initial values of $r_1$ and $r_2$ gives: $$r_1=-kt+2R$$ $$r_2=-kt+3R$$

**Can you write the height of Frosty ($h$) as a function of $r_1$ and $r_2$? (it might help to draw a sketch of Frosty!) Can you write $h$ in terms of $t$?**

If you draw a picture of two circles, one on top of the other, you should find that $h=2r_1+2r_2$. Substituting the expressions for $r_1$ and $r_2$ gives $h=2(-kt+2R)+2(-kt+3R) = -4kt+10R$

**What is the initial value of $h$? Can you find an expression for $t$ when the height is half the initial height? **

**Can you use this value of $t$ to find out the radii of the two snowballs at this time? Use this to find the volumes of each one.**

**What is the total volume of Frosty at the start? What is the total volume when the height is half the initial height? Can you find a simplify a ratio between these two total volumes?**

**At what point does Frosty's head disappear?**

**In terms of $R$, how tall is Frosty when his height is one tenth of the initial height? What does that tell you?**