Expert: Josh - 7/20/2005

Question
My 16 year old daughter is struggling with these types of problems on her summer school homework. If you give me the answer to the following problem i can be sure that i am not about to show her the wrong example. I think i know what the answer is though but i am not sure. Here is the problem:
The day before

Princess Amira demands only the most luxurious of accommodations.
So for her bath, she had glacial ice from Terror Mountain shipped
to her palace in the Lost Desert. There, the ice was carved into
a perfect sphere, exactly 2 meters in diameter.
The ice sphere melted into her hexagonal bath measuring 1.5 meters
on a side, and 0.8 meters deep.

If the desert heat uniformly melts the surface of the ice at a rate
of 1.5mm of depth per minute, how many minutes will it take for her
bath to be completely full? Please round up to the nearest minute,
and assume the volume does not change between solid and liquid states.


Princess Amira was furious. "All I ask is to have a bath full of
freshly sun-melted glacial ice, and my servants can't even get that
right!" She ordered them to empty out her bath and start over with
new glacial ice.

This time, they returned with three ice spheres: one was 0.9 meters
in diameter; one was 1.5 meters in diameter; and the last was 1.8
meters in diameter. They all started melting at the same time.
Also, the water in the bath evaporated at a rate of 0.1mm of depth per
minute.

Otherwise assuming the same conditions as the information above provides,
how many minutes will elapse before the bath is completely full?
Please round up to the nearest minute.


Answer
Hi Kirstin,

I will attempt to solve this problem using two methods.
The first method uses Calculus. It is more powerful, but might be a little hard to understand. The second method uses geometric arguments; this might be more intuitive.
I won't actually find the numerical answer for you, but I'll show you how to set up and solve this type of problems.

Let r= radius of ice sphere, s=side of hexagon, h=depth of bath.

Data: (all dimensions measured in meters unless otherwise stated)
r=1, s=1.5, h=0.8

Relevant facts:
[#1] Volume of sphere: V = (4/3)*pi*r^3.
[#2] Ice melting rate (think of this as reduction in diameter, or radius, with time): dr/dt = -1.5x10^(-3) [meter/minute]
Note 1: 1 millimeter equals (1/1000) meter. Alternatively, can write this as 1x10^(-3) meter.
Note 2: Even though the radius of the ice changes uniformly, the increase in the volume of water happens at a non-uniform rate. It slows progressively with time.

Method 1: Calculus approach

Let the change in the volume of water (resulting from the melting ice sphere) during period "dt" be denoted by "dW".

Now, "dW", is given by the instantaneous surface area of the sphere, multiplied by the thickness of the shell (visualize peeling layers of an onion, each slice actually contributes some volume, no matter how thin). Now, surface area of sphere is given by 4*pi*r^2. So, the rate of change
dW/dt = (4*pi*r^2) dr/dt ..........[#3]

From geometry (draw some diagrams), the hexagon consists of 6 identical equilateral triangles, like this
A
.
D...............C
.
B
Note: The edge AC actually has the same length as the side of the hexagon.
So, the length, |AC|=|BC|=s. |AD|=|DB|=s/2.
Using Pythagoras theorem, |DC|=sqrt(|AC|^2-|AD|^2)=(sqrt(3)/2)*s
or from trigonometry, |DC|= |AC|*cos(30)=(sqrt(3)/2)*s gives the same answer.

Therefore, each equilateral triangle has area, A_equitri= |DC|*|AD|=(sqrt(3)/2)*s *s/2 = s^2 * sqrt(3)/4
Total area of hexagon, A_hex = 6 * A_equitri = [3*sqrt(3)/2] * s^2
Total volume of hexagon bath, V_hex = A_hex * h = [3*sqrt(3)/2] * s^2 * h ..........[#4]
Substituting the values for "s" and "h", V_hex = 2.7 * square_root_of(3). It has a numerical value of approximately 4.676 [cubic meters].

To fill the bath completely, set the volume of water, W, equals to V_hex (from [#4]).

From [#3], we have dW/dt = (4*pi*r^2) "dr/dt".
To obtain W, we need to integrate dW/dt with respect to time, t.
Problem is that the expression dW/dt is a function of the radius "r".
Our goal is to rewrite "r" as a function of "t".
r(t) = ro - (dr/dt)*t, because the sphere shrinks at a uniform rate from its initial radius, ro=1, dr/dt is the rate at which the radius is reduced, t is the elapsed time....[#5]

Putting this into dW/dt, we have
dW/dt = [4*pi*(ro-(dr/dt)*t)^2] * (dr/dt) = 4 pi [ro^2-2 ro(dr/dt) t+(dr/dt)^2 t^2] (dr/dt).
To avoid clutter, let K=dr/dt [=1.5*10^(-3)]
W = Integral (dW/dt) dt = Integral 4 pi K (ro^2-2ro*Kt+K^2*t^2) dt
=4 pi K [ro^2 t - ro K t^2 + K^2*(t^3/3)] evaluate at limits t=0 and t=T.
Simplifying, the volume of water melted from the ice after T minutes is given by
W=4 pi K [ro^2 T - ro K T^2 + K^2*(T^3/3)] ....[#6]
Setting [#6] equal to [#4], we solve for "T"
4 pi K [T - K T^2 + K^2*(T^3/3)] = [3*sqrt(3)/2] * s^2 * h
...note: this equation contains only one unknown, "T", the values of "ro","s","h" and K=dr/dt are known.
===
Method 2: Geometric approach (you probably would have done it this way)
View volume of water, W, as the difference between two concentric spheres.
Outer one has volume given by (4/3) pi ro^3.
Inner one changes as a function of time, (4/3) pi r^3 = (4/3) pi [ro-(dr/dt)t]^3.
Therefore, W = (4/3) pi [ro^3 - [ro-(dr/dt)t]^3 ]
= (4/3) pi [ro^3 - [ro^3 - 3 ro^2 (dr/dt)t + 3 ro (dr/dt)^2 t^2 - (dr/dt)^3 t^3] ] ...binomial expansion has been used here
= (4/3) pi [3 ro^2 (dr/dt)t - 3 ro (dr/dt)^2 t^2 + (dr/dt)^3 t^3]
= 4 pi [ro^2 (dr/dt)t - ro (dr/dt)^2 t^2 + (1/3) (dr/dt)^3 t^3] ....this is identical to [#6]
===
Second part.
Essentially the same set up. Substitute different values for "ro" in expression [#5] and proceed from there.
ro={0.45, 0.75, 0.9} etc.

The rates at which the ice melts from sphere 1, 2 and 3 are given by,
dW1/dt = [4*pi*(0.45-(dr/dt)*t)^2] * (dr/dt), with time constraint 0<= t <= ro/(dr/dt), where ro=0.45
dW2/dt = [4*pi*(0.75-(dr/dt)*t)^2] * (dr/dt), with time constraint 0<= t <= ro/(dr/dt), where ro=0.75
dW3/dt = [4*pi*(0.9-(dr/dt)*t)^2] * (dr/dt), with time constraint 0<= t <= ro/(dr/dt), where ro=0.9

Total positive contribution, dW/dt = dW1/dt + dW2/dt + dW3/dt
but it's possible that one has totally melted, while the remaining spheres are still melting.

Take into account evaporation (negative contribution) E= -A_hex*1x10(-3) * t.