# Frosty is Melting!

## Problem

As each snowball melts, the radius of each snowball decreases by the same amount each minute. During melting each snowball remains spherical and uniform.

- When Frosty is half his initial height, show that the ratio of his volume to his initial volume is 37 : 224 .

- What is this ratio when Frosty is one-tenth of his initial height?

*Adapted from STEP Mathematics I, 1991, Q2. Question reproduced by kind permission of Cambridge Assessment Group Archives. The question remains Copyright University of Cambridge Local Examinations Syndicate ("UCLES"), All rights reserved.*

## Getting Started

The volume of a sphere is given by $V=\frac 4 3 \pi r^3$

**Can you use this to find the initial volume of Frosty in terms of $R?$**

Remember that Frosty is made of two snowballs!

**What is the initial height of Frosty?What is the height of Frosty when it is half its height?**

Given that the radius of each snowball decreases by the same amount each minute, can you find the radii of each sphere when Frosty is half his height?

You can now find the new volume of Frosty in terms of $R$.

The last part of the question, "What is this ratio when Frosty is one-tenth of his initial height?" looks very similar to the first part, but there is a crucial difference! To answer this part, think about what happens to a snowman as it melts.

## Student Solutions

**When Frosty is half his initial height, show that the ratio of his volume to his initial volume is 37 : 224 **

Ziyao from Garden International School in Malaysia, Jaeuk from Brighton College Abu Dhabi in the UAE and Callum from Ilkley Grammar school in the UK used a similar method to solve this problem. This is Ziyao's work:

Note that the formula for the volume of a sphere will come in very handy: $\frac43\pi r^3$

In the first question, Frosty has melted to half of his original height. It is given that the radius of the snowballs decreases by the same amount, so to get the radius of the spheres at half height we do the following below:

The height is $2R\times2+3R\times2=10R.$ $10R\div2=5R$ and so the height has decreased by $5R.$ Therefore each sphere’s diameter has decreased by $5R\div2=2.5R$ according to the information in the quotation. The radii of the spheres after the height has decreased by $\frac12$ will be $\frac{4R - 2.5R}2=0.75R,$ $\frac{6R - 2.5R}2=1.75R.$

__Calculating the volumes__

The original volume [of the smaller snowball] is $8R^3\times\frac43\times\pi=32R^3\frac\pi3$ (applying the formula for the volume of a sphere)

[The original volume of the larger snowball is] $27R^3\times\frac43\times\pi=36R^3\pi$

$\dfrac{32R^3\pi}3+\dfrac{108R^3\pi}3=\dfrac{140R^3\pi}3$

The volume when the height has decreased by $\frac12$:

$\left(\frac34R\right)^3\times\frac43\times\pi=1\times\left(\frac34\right)^2R^3\times\pi=\dfrac{9R^3\pi}{16},$ (again, applying the formula for the volume of a sphere)

$\left(\frac74R\right)^3\times\frac43\times\pi=\dfrac{343R^3\pi}{64}\times\frac43\times\pi=\dfrac{343R^3\pi}{48},$

and $\dfrac{9R^3\pi}{16}+\dfrac{343R^3\pi}{48}=\dfrac{27R^3\pi}{48}+\dfrac{343R^3\pi}{48}=\dfrac{370R^3\pi}{48}$

__The final ratio__

The ratio is $\dfrac{370R^3\pi}{48}:\dfrac{140R^3\pi}3$

multiply both sides by $48$ to get $370R^3\pi:2240R^3\pi$

divide both sides by $R^3\pi$ to get $370:2240$

and finally, divide by $10$ on both sides to get $37:224$

J from EIGCDN in Switzerland and Yonwoo solved the problem by writing a formula for Frosty's height. This is J's work:

The initial height of the snowman is $2\times2R+2\times3R$ (sum of diameters) which is equal to $10R.$

To find the initial volume, we simply substitute the values $2R$ and $3R$ into the general formula:

$\frac43\pi(2R)^3+\frac43\pi(3R)^3=\frac43\pi(8R^3+27R^3)=\frac43\pi(35R^3)=\frac{140}3\pi R^3$

$5R$ is half of the initial snowman's height. Let $D$ be the amount each radius decreases by every minute. After $x$ minutes, the top snowball has a radius of $2R-xD$ and the bottom one has a radius of $3R-xD.$

Using these facts, we can create a formula for the height of the snowman after $x$ minutes:

$2\times(2R-xD)+2\times(3R-xD)=4R-2xD+6R-2xD=10R-4xD$

At the point where the total height is $5R,$ we get $10R-4xD=5R\Rightarrow4xD=5R\Rightarrow xD=\frac54R.$ By plugging this into the expressions for the top and bottom radii, we get:

Top radius $=2R-\frac54R=\frac34R$

Bottom radius $=3R-\frac54R=\frac74R$

Next, we can find the volumes of the snowballs.

From here, J's solution is the same as Ziyao's solution above.

**What is this ratio when Frosty is one-tenth of his initial height?**

J continued using the same formula, but it led to some very dangerous maths:

So far, we have built up quite a few formulae; now, we can finally apply them.

The snowman's height is $R$ when the height reaches one-tenth of the initial height, $10R.$ Remember that the formula for the height after $x$ minutes is $10R-4xD$ so $10R-4xD=R\Rightarrow4xD=9R\Rightarrow xD=\frac94R.$

We can now substitute into the formulae for top / bottom radii:

Top radius $=2R-\frac94R=-\frac14R$

Bottom radius $=3R-\frac94R=\frac34R$

This part seems a little weird. Negative magnitudes are nonexistent. However, we can proceed because the volume of the bottom snowball is greater than the top radius and so the total volume is still positive. Also note that in this scenario, the top snowball has already melted.

The total volume is:

$\frac43\pi\left(-\frac1{64}R^3+\frac{27}{64}R^3\right)=\frac43\pi\left(\frac{13}{32}R^3\right)=\frac{52}{96}\pi R^3=\frac{13}{24}\pi R^3$

As usual, we ignore $\pi R^3$ and focus on the other coefficient. We compare $\frac{13}{24}$ and $\frac{140}{3}$; by scaling them up to the same denominator, we get $\frac{13}{24}$ and $\frac{1120}{24}.$ The ratio therefore is $13 : 1120$ which cannot be simplified further.

Jaeuk and Yonwoo also realised that the smaller snowball would have melted completely, and they adjusted their calculations to get correct answers. This is Jaeuk's work (click here to see a larger version):

## Teachers' Resources

This problem requires students to use the information in the question to find formulae and then apply these to solve the problem. Students are also required to find the volume of spheres and simplify ratios, and the question needs them to apply their algebraic manipulation skills.

Some possible starting points are described in the getting started section.

Usually when the dimensions of a solid are reduced to one half, the volume is reduced to one eighth of the original. Why is this is not the case with Frosty?

This problem has been adapted from the original one so that students do not need any knowledge of calculus to solve it. Find the original problem here.

This problem is one of a collection designed to develop students' carbon numeracy; we hope it will encourage students to think about the issues surrounding climate change. You can find the complete collection here.