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\begin{document}

\newpage

% You only need to change the NAME OF SCENARIO in the next line and delete the footnote. See above for instructions on toggling from Teacher Version to Student Version.
\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\ICE MELT}


\markboth{Ice Melt}{Ice Melt}






\author{Brian Winkel,
Director SIMIODE\\
Chardon OH 44024USA\\
BrianWinkel@simiode.org
}
\makeStitlePDFLaTex


\begin{abstract}
We offer up the claim of a store catalog  that   its ice ball mold allows users to  ``\ldots make ice balls that outlast cubes and won't water drinks down.'' We ask students to build a mathematical model to defend or contradict this claim.
\end{abstract}


\section*{STATEMENT}

\begin{figure}[h]
\centering
\includegraphics[width=14cm]{1-20-IceMelt-BrookstoneAd-Paint.eps}

\vspace*{-13 true cm}

\parbox{9.8 true cm}{\caption{\label{fig:1-20-IceMelt-BrookstoneAd-Paint} Copy of notice from {\it Brookstone Holiday Gift Guide\/} \cite{Brookstone2013} in which it is claimed that Brookstone's spherical ice ball mold ``Lets you make ice balls that outlast cubes and won't water drinks down.''}}

\end{figure}


In  a Brookstone catalog \cite{Brookstone2013} (see Figure~\ref{fig:1-20-IceMelt-BrookstoneAd-Paint}), received by the author on 21 November 2013, there is pictured a ``Rock Glass with Ice Ball Maker,'' a device for making spherical ice balls to put in drinks. The claim is that this device, ``Lets you make ice balls that outlast cubes and won't water drinks down.''

\begin{enumerate}
\item Use mathematical modeling to defend or contradict this claim.
\item Design an experiment to test this claim.
\item Carry out the experiment and use any data in (2) to validate your model in (1).
\end{enumerate} 

\teacher{
\section*{NOTES FOR TEACHERS}


Suppose we consider the same fixed volume of ice in two shapes: (a) a cube and (b) a sphere. What would the relative dimensions of these shapes be?  Could we determine and compare their volumes over time? Until all the ice is gone?

Given a cube of volume $V$ and side length, $s= V^{1/3}$, then the sphere of the same volume would have to have a radius of  $r = \left(\frac{3 V}{4 \pi }\right)^{1/3}$. This is true because if we set the volume of a sphere of radius $r$ equal to a fixed volume $V$, i.e. $\frac{4}{3} \pi r^3 = V$,   then $r = \left(\frac{3 V}{4 \pi }\right)^{1/3}$.

Now we will make the reasonable and key assumption  that  the melting of the ice volume is only proportional to the surface area of the volume's shape.  

Indeed, as was suggested by one reviewer, Terrance Quinn of Middle Tennessee State University, this problem really revolves around the principle (based on the assumption that the melting of the ice volume is  proportional to the surface area of the volume's shape) that to determine the amount of ice left we simply integrate the surface area  times the linear rate of melting per unit time over a fixed time interval. Then we could do comparisons. While this is most certainly true we note that solving a differential equation is comparable to performing an integral and so we believe a differential equation model is quite acceptable for attacking this problem.

Let $V_C(t)$ be the volume of the cube of ice at time $t$ and $V_S(t)$ be the volume of the sphere of ice at time $t$.  Thus, for our models we seek to determine formulae for $dV_C/dt$ (rate of melting of ice in the cube) and $dV_S/dt$ (rate of melting of ice in the sphere) in terms of  respective surface areas for a fixed volume, say, $V$. 

We will then solve these two differential equations and determine the time at which both volumes are zero and  track the volumes over time, comparing them as time goes on.

For our cube of side $s$ we have the rate of change of the volume ($V_C$) of the cube $s^3$ is proportional to the surface area $6 s^2$ we obtain:
\begin{equation}\label{eq:CubeRate}
dV_C/dt =  - k (6 s^2) = -6 k  V_C^{2/3}\, . 
\end{equation}

For our sphere of radius $r$ we have the rate of change of the volume ($V_S$) of the sphere $\frac{4}{3}\pi r^3$ is proportional to the surface area $4 \pi r^2$ and using $r = \left(\frac{3 V}{4 \pi }\right)^{1/3}$ we obtain:
\begin{equation}\label{eq:SphereRate}
dV_S/dt =  - k (4 \pi r^2) = -6^{2/3} \pi^{1/3} k  V_S^{2/3}\, . 
\end{equation}


\subsection*{Analyzing  melting with specific numerical initial value of $V(0)$ and rate constant $k$}

To get a feel for this situation let us use an arbitrary common volume of 10 units for our sphere and our cube with a rate constant of, say, $k = 0.3$.   the units of $k$ are length/time.   Thus upon using  $V_C(0) = 10$ and $k = 0.3$ in (\ref{eq:CubeRate}) we solve for the real branch of (\ref{eq:CubeRate}) to find

\begin{equation}
V_C(t) = -0.216 t^3+2.32679 t^2-8.35486 t+10 \, ,
\end{equation}
from which we can determine that $V_C(3.59077) = 0$ and so the cube is entirely melted  at time $tc = 3.59077$. 

Upon using  $V_S(0) = 10$ and $k = 0.3$ in (\ref{eq:SphereRate}) we solve for the real branch of (\ref{eq:SphereRate}) to find

\begin{equation}
V_S(t) = \frac{1}{3} \left(-0.339292 t^3+4.53465 t^2-20.202 t+30\right)\, ,
\end{equation}
from which we can determine that $V_S(4.45506) = 0$ and so the sphere  is entirely melted at time $ts = 4.45506$. Indeed, if we plot the volumes over time (see Figure~\ref{fig:1-20-Plots}) we see that the ice ball sphere does last longer. Thus it would appear that a sphere of ice with the same volume as a cube of ice will last longer and so the claims of Brookstone's marketing are true, at least for this parameter set.


\begin{figure}[h]
\centering
\includegraphics[width=12.4cm]{1-20-Plots.eps}

\parbox{11.1 true cm}{\caption{\label{fig:1-20-Plots} Plots of the volumes of our cube (thin and orange) and sphere (thick and purple) show  that  the sphere's volume  has longer life than the cube's volume. Here we are using initial volume of 10 units and a value of $k = 0.3$.}}


\end{figure}


\subsection*{Analyzing  melting  of volumes with general  (non-numerical) initial value of $V(0)$ and rate constant $k$}

We denote $V_I$ as the initial volume of ice. In (\ref{eq:CubeRate}) we solve for the real branch of (\ref{eq:CubeRate}) to find the volume (${\rm VolC}(t)$) of the cube at time $t$ to be:




$${\rm VolC}(t) = -8 k^3 t^3+12 k^2 t^2 \sqrt[3]{V_I}-6 k t
   V_I^{2/3}+V_I \, ,$$

\noindent from which we can find (a triple multiplicity root) the time at which the ice of the cube is all gone to be

$$tC = \frac{\sqrt[3]{V_I}}{2 k}\, . $$


While in (\ref{eq:SphereRate}) we solve for the real branch of (\ref{eq:SphereRate}) to find the volume (${\rm VolS}(t)$) of the sphere at time $t$ to be:

$${\rm VolS}(t) = \frac{1}{3} \left(-4 \pi  k^3 t^3+6 \sqrt[3]{6} \pi ^{2/3} k^2
   t^2 \sqrt[3]{V_I}-3\ 6^{2/3} \sqrt[3]{\pi } k t
   V_I^{2/3}+3 V_I\right)\, ,$$
from which we can find (also a triple multiplicity root) the time at which the ice of the sphere is all gone to be

$$tS = \frac{\sqrt[3]{\frac{3}{\pi }} \sqrt[3]{V_I}}{2^{2/3} k} \, . $$



Now if we examine the ratio ($tS/tC$) of the time it takes the ice of the sphere to be zero to the time it takes the ice in the comparable (same initial volume) cube to be zero  we find that
$$tS/tC = \sqrt[3]{\frac{6}{\pi }} = 1.2407\, .$$
This means   the time it takes the ice of the sphere to be zero is ALWAYS 1.2407 times the time it takes the ice in the comparable (same initial volume) cube to be zero. Thus we see that it takes almost 25\% longer for the ice in the sphere to completely melt than it takes for the ice in the cube to do so.  

Thus with our general argument it would appear that a sphere of ice of the same volume as a cube of ice will last longer and so the claims of Brookstone's marketing are true.

\subsection*{Tasks (2) and (3) Considered}


Tasks (2) and (3) in the Statement of the problem could be very difficult to carry out, for one would have to actually purchase or build an ice sphere mold and then a cube mold with the same volume or using two shapes, perhaps of different volumes determine experimentally the  $k$ value and use it in a model of same volume shapes and then compare results. 


\subsection*{Other issues}

Why do we consider the shapes to the ice sphere and ice cube to remain a sphere and cube, respectively, through time?

Maybe, if we consider the liquid in the glass with the ice as a closed system that the changing temperature of the liquid causes our $k$ to be non-constant, perhaps causing us to build a more complicated model, say using Newton's Law of Cooling as well. For when the cube melts faster the surrounding water cools faster, thus slowing the rate of melting. How do these two phenomena work together? 

Is there a different shape that would provide a ``better'' experience?  Maybe there is some shape whose volume could even go to zero before the sphere of cube, but have most of the melting happen toward the very end, when the drink is already finished? Or is such a shape impossible?


Other factors could include how much of the cube sticks out of the water when floating on top, and the temperature of the surrounding air, in that case.

\section*{Another Approach From Scratch}

One could consider this problem without differential equations by comparing the surface area to volume ratio for two shapes of the same volume, a sphere and a cube. For the shape which has a larger surface area to volume ratio would melt faster since the melting (or loss in volume) takes place at the surface.

If we have the two equal volume expressions of the two shapes - sphere and cube, where the volume of the  sphere is in terms of its radius, $r$,and the volume of the cube is in terms of its lateral edge, $s$,
\begin{equation}\label{eq:compare}
V_{\rm cube} = \frac{4 \pi r^3}{3} = s^3 = V_{\rm cube}
\end{equation}
then we can solve for $s$ in terms of $r$
\begin{equation}\label{eq:solve}
s = \left(\frac{4 \pi}{3}\right)^{1/3} r
\end{equation}
and write both volumes and surface areas in terms of the same variable, namely, $r$. 

For the sphere, $SA_{\rm sphere}$ is the surface are of the sphere and $V_{\rm sphere}$ is the volume of the sphere and our surface area to volume ratio is,
\begin{equation}\label{eq:SphereRatio}
\frac{SA_{\rm sphere}}{V_{\rm sphere}} = \frac{4 \pi r^2}{\frac{4 \pi r^3}{3}} = \frac{3}{r}
\end{equation}


For the cube, $SA_{\rm cube}$ is the surface are of the cube and $V_{\rm cube}$ is the volume of the cube and our surface area to volume ratio is,
\begin{equation}\label{eq:CubeRatio}
\frac{SA_{\rm cube}}{V_{\rm cube}} = \frac{6 s^2}{s^3} = \frac{6}{s} = \frac{6}{\left(\frac{4 \pi}{3}\right)^{1/3} r} =  6 \left(\frac{3}{4 \pi}\right)^{1/3} r \approx \frac{3.7221}{r}\, .
\end{equation}

Thus the surface area to volume ratio for the sphere of the same volume as a cube is less and it will therefore melt slower.

While we recognize and can show this argument to students (if they do not produce it themselves) we also want to use the tools of our interest, namely differential equations, to solve the problem as well. We can get more specific information through  the differential equations approach  beyond answering the immediate question as to which shape of ice lasts longer. We can find out just how much longer the ice sphere will last compared to the time the ice cube will last.
}
\begin{thebibliography}{99}

\bibitem{Brookstone2013} Brookstone. 2013. {\it Brookstone Holiday Gift Guide\/}.  November 2013. p. 43.  


\end{thebibliography}




\end{document}
