Expert: Paul Klarreich - 3/11/2011

Question
I really need help on this problem. I think I have figured it out to number 3, but I don't understand what to do with the numbers for the chain rule and how to finish the rest. So far I have:
V=s cubed, A= 6s squared,
dV/dt= -k * A


How long will it take an ice cube to melt completely? You left out an ice cube ant 1/4 of it melts in 1 hour. Use the following steps to find out how long it will take before it is melted completely.

1-let s represent the side length of the ice cube, and observe that s is a function of time t. Write expressions which describe the volume V and surface area A of the cube as functions of s (which you should observe, in turn makes them functions of t).

2-melting takes place at the surface of the cube, so you decided that it is reasonable to assume that the cubes volumn V decreases at a rate that is proportional to its surface area A. saying that a is proportional to b means that there is constant k so that a=kb.) write an equation that describes dV/dt in terms of s. Is your constant k a positive or negative number?

3-use the chain rule and your answer from #1 to write an expression which relates dV/dt to ds/dt

4-now you have two expressions (your answers from 2 and 3 for dV/dt).set them equal to find ds/dt in terms of the constant k.

5-use your answer from #4 to write an equation which relates s0=s(0) to s1=s(1), where t is measured in hours, and use it to find tmelt (the melting time) in terms of the quantity s1/s0.

6-taking into account that 1/4 of the cube melted in 1 hour, find an approximation to s1/s0 using the function you wrote in #1 for V in terms of s.

7-find tmelt. how much longer will you have to wait for the ice cube to melt completely

Answer
Questioner: Emily
Country: United States
Category: Calculus
Private: No
Subject: calc problem
Question: I really need help on this problem. I think I have figured it out to number 3, but I don't understand what to do with the numbers for the chain rule and how to finish the rest. So far I have:
V=s cubed, A= 6s squared,
dV/dt= -k * A

How long will it take an ice cube to melt completely? You left out an ice cube ant 1/4 of it melts in 1 hour. Use the following steps to find out how long it will take before it is melted completely.

1-let s represent the side length of the ice cube, and observe that s is a function of time t. Write expressions which describe the volume V and surface area A of the cube as functions of s (which you should observe, in turn makes them functions of t).

2-melting takes place at the surface of the cube, so you decided that it is reasonable to assume that the cubes volumn V decreases at a rate that is proportional to its surface area A. saying that a is proportional to b means that there is constant k so that a=kb.) write an equation that describes dV/dt in terms of s. Is your constant k a positive or negative number?

3-use the chain rule and your answer from #1 to write an expression which relates dV/dt to ds/dt

4-now you have two expressions (your answers from 2 and 3 for dV/dt).set them equal to find ds/dt in terms of the constant k.

5-use your answer from #4 to write an equation which relates s0=s(0) to s1=s(1), where t is measured in hours, and use it to find tmelt (the melting time) in terms of the quantity s1/s0.

6-taking into account that 1/4 of the cube melted in 1 hour, find an approximation to s1/s0 using the function you wrote in #1 for V in terms of s.

7-find tmelt. how much longer will you have to wait for the ice cube to melt completely .
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Ugh! Too many words.  Let's try to extract the good stuff.

Yes,  dV/dt = kA, or  6ks^2.  [You do NOT really need the 6; it can be subsumed in the k, but...]

and, since you will be concluding  k < 0, you don't have to write - kA.

So: Your question amounts to:

Find t such that V(t) = 0, given that  V(1) = 3/4 V(0)
....................
Your variables are:

Side length = s,  and  ds/dt is to be found.

Volume = V,  and  dV/dt = 6k s^2

Area = A,  and dV/dt is to be found.

.............
Relations:

V = s^3
A = 6s^2

Implicit diff:

dA/dt = 12 s ds/dt

dV/dt = 3s^2 ds/dt
********* BUT *********
dV/dt (also) = 6k s^2

Set  6k s^2 = 3s^2 ds/dt

6k = 3 ds/dt

2k = ds/dt

ds/dt IS A CONSTANT!

(it's equal to 2k)

So we can solve the equation:

ds/dt = 2k

s = 2kt + s0


To find s0, use the facts:

V = s^3 = (2kt + s0)^3

V(1) = 3/4 V(0)



Now
V(1) = (2k + s0)^3

V(1) = (s1)^3

V(0) = (s0)^3

(s1)^3 = 3/4(s0)^3

(s1)^3/(s0)^3 = 3/4

s1/s0 = (3/4)^(1/3)    <<<< cube root, that is.

= 0.91 (about)

Now  s1 = 0.91 s0, or:

s0 + 2k = 0.91 s0

2k = - 0.09 s0

k = - 0.045 s0

Ok, that does it, more or less.

The length of time needed to melt the whole thing, depends on s0 [a bigger cube takes longer, as your grandmother could have guessed]  and on k, which, I suppose, depends on the state of global warming or something.

s = 2kt + s0

s = 2(- 0.045 s0)t + s0

Now t[all ice is melted] is obtained by putting  s = 0 and solving for t.

............................................
Suggestion for related rates:  I have done (more than) a few of them here:

http://en.allexperts.com/q/Calculus-2063/2009/11/Related-Rates-87.htm