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Melting a Cube


Date: 12/20/2001 at 13:51:46
From: Asma
Subject: Differentiation (calculus)

Question: How long does it take for an ice cube to melt?  

Assume it retains its cubical shape as it melts: V = s^3, s is edge 
length. 

Assume the cube volume decreases at a rate proportional to its 
surface area: dv/dt = -k(6s^2)  K>0 a constant.

We need one more piece of information: how long will it take for a 
specific percentage of the ice cube to melt? This depends on many 
factors that are determined from the situation in which the ice cube 
sits.  Let us set the conditions here that the cube loses 1/4 of its 
volume in the first hour.

Thus given V = s^3 and dV/dt = -k(6s^2)
V = Vo when t=0 (initially)
V = (3/4).Vo  when t = 1 hour
Find the volume of t when V = 0
Note/hint: side length s also changes; so S(initial) is side length 
at t = 0rs.


Date: 12/21/2001 at 03:32:26
From: Doctor Jeremiah
Subject: Re: Differentiation (calculus)

Hi Asma,

We need to know s in terms of t (how fast s is changing).
If we differentiate V = s^3, we get:

               V = s^3
       d/dt( V ) = d/dt( s^3 )
           dV/dt = 3s^2 ds/dt

But dV/dt = -k(6s^2) so:

        -k(6s^2) = dV/dt = 3s^2 ds/dt
        -k(6s^2) = 3s^2 ds/dt
             -2k = ds/dt

So ds/dt is a constant value, a straight line. The advantage of a 
straight line is that it doesn't matter how far apart two points are, 
the slope is always accurate (not true of dV/dt = -k(6s^2) because 
it's a parabola).

The slope of a straight line is:

       m = ds/dt = (s1 - s0)/(t1 - t0)

And that means:

       m = ds/dt = (s1 - s0)/(t1 - t0)
             -2k = (s1 - s0)/(1 - 0)
             -2k = (s1 - s0)

We have two points on this graph:

  s1 is the side length at t = 1 and
  s0 is the side length at t = 0

              V0 = s0^3
        V0^(1/3) = s0

              V1 = s1^3
           3V0/4 = s1^3
   (3V0/4)^(1/3) = s1

So then:

       m = ds/dt = (s1 - s0)/(t1 - t0)
             -2k = (s1 - s0)/(1 - 0)
             -2k = (s1 - s0)
             -2k = ((3V0/4)^(1/3) - V0^(1/3))

Now we know what k is.

Since it is a straight line slope, the straight line would have a form 
of y = m x + b or, with our variables, s = -2k x + C. We can also get
that (even though we don't need to) if we integrate the slope like 
this:

           ds/dt = -2k
              ds = -2k dt

            /      /
            | ds = | -2k dt
            /      /

               s = -2k t + C

Which, as I said, is a straight line where -2k is the slope (m) and 
C is the y-intercept.

So now we have a straight line and two points on that line, and the 
slope:

               s = -2k t + C

       m = ds/dt = (s1 - s0)/(t1 - t0)

point at t = 0: s0 = V0^(1/3)
point at t = 1: s1 = (3V0/4)^(1/3)

At this point it's easy to solve because we just have to substitute in 
these two points and solve for the answers.

So now we have a straight line and we know the slope is 
-2k = (3V0/4)^(1/3) - V0^(1/3), so we can substitute in a known value 
to get C:

               s = -2k t + C
              s0 = -2k t0 + C
        V0^(1/3) = -2k 0 + C
        V0^(1/3) = C

Now we can put these back into our line equation:

               s = -2k t + C
               s = ((3V0/4)^(1/3) - V0^(1/3)) t + V0^(1/3)

And solve for when s = 0 (we don't need to solve for when the volume 
gets to 0 because the side length will get to 0 at the same time)

               s = ((3V0/4)^(1/3) - V0^(1/3)) t + V0^(1/3)
               0 = ((3V0/4)^(1/3) - V0^(1/3)) t + V0^(1/3)
       -V0^(1/3) = ((3V0/4)^(1/3) - V0^(1/3)) t

Which gives us:

               t = -V0^(1/3) / ((3V0/4)^(1/3) - V0^(1/3))

So if we know the original volume V0, we can calculate how big t is 
when s = 0.

What saved us is that the slope ds/dt (how fast s is changing) was a 
straight line. I don't see any other way to solve this.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/   


Date: 12/21/2001 at 07:37:00
From: Asma
Subject: Differentiation (calculus)

Thank you Dr. Jeremiah - you have been a lot of help. I appreciate 
your time and thinking spend on my problem. Once again thank you.
    
Associated Topics:
High School Calculus

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