Ice cube melting queation!!

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flaren5

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May 17, 2013
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Ice cube melting question!!

Good day all,

I'm working on a question that I seem to be getting stumped with. It's apparently a calculus question, but I'm not too sure on how to solve it as a calculus solution. I believe it's a rate of change question that I may need to find the derivative of, however that is not the way I solved it.

The question is....

"By about what percentage will the edge length of an ice cube decrease if the cube loses 7% of its volume by melting?"

The way that I attempted to solve it, is as follows...

100% - 7% = 93%
a3/b3 = 0.93
a/b = 0.931/3
= 0.9761
= 97.61%
(For some reason the answer doesn't make sense to me, which has lead me to believe that I'm not doing it correctly.)

Any insight to this question would be greatly appreciated. :)
 
Last edited:
flaren5 said:
Good day all,

I'm working on a question that I seem to be getting stumped with. It's apparently a calculus question, but I'm not too sure on how to solve it as a calculus solution. I believe it's a rate of change question that I may need to find the derivative of, however that is not the way I solved it.

The question is....

"By about what percentage will the edge length of an ice cube decrease if the cube loses 7% of its volume by melting?"

The way that I attempted to solve it, is as follows...

100% - 7% = 93%
a3/b3 = 0.93
a/b = 0.931/3
= 0.9761
= 97.61%
(For some reason the answer doesn't make sense to me, which has lead me to believe that I'm not doing it correctly.)

Any insight to this question would be greatly appreciated. :)
Click to expand...
The question has nothing to do with calculus. (Edit: I am wrong. It can be solved using calculus. Sort of like using a baseball bat to crush a gnat.)

It would help everyone (you included) if you adopted the habit of labeling your variables.

a = length of a side of the smaller cube.

b = length of a side of the bigger cube.

\(\displaystyle Given:\ a^3 = (1 - 0.07)b^3 = 0.93b^3 \implies a = b\sqrt[3]{93} \approx 0.9761 * b = (1 - 0.0239) * b.\)

The edge will shrink by approximately 2.39%. You just did not complete the thought.

Let's check. (Another good habit)

\(\displaystyle Edge\ of\ large\ cube = 10\ units \implies volume\ of\ large\ cube = 10^3 = 1000\ cubic\ units.\)

\(\displaystyle Edge\ of\ small\ cube = 10 - (10 * 0.0239) = 10 - 0.239 = 9.761\ units \implies volume\ of\ small\ cube = 9.761^3 \approx 930\ cubic\ units.\)

Looks good.
 
Last edited:
flaren5 said:
Good day all,

I'm working on a question that I seem to be getting stumped with. It's apparently a calculus question, but I'm not too sure on how to solve it as a calculus solution. I believe it's a rate of change question that I may need to find the derivative of, however that is not the way I solved it.

The question is....

"By about what percentage will the edge length of an ice cube decrease if the cube loses 7% of its volume by melting?"

The way that I attempted to solve it, is as follows...

100% - 7% = 93%
a3/b3 = 0.93
a/b = 0.931/3
= 0.9761
= 97.61%
(For some reason the answer doesn't make sense to me, which has lead me to believe that I'm not doing it correctly.)

Any insight to this question would be greatly appreciated. :)
Click to expand...

Calculus way:

V = a3

dV = 3 * a2 da

da/a = 1/3 * dV/V = 1/3 * 0.07 = 0.0233 = 2.33%

Now perform check....
 
mmm4444bot said:
But it could, if one chose calculus to solve it. :D
Click to expand...
Yes. You and Subhotosh are right. I have edited my post to admit that you can use the baseball bat of calculus to squash this gnat of a problem.
 
JeffM said:
Yes. You and Subhotosh are right. I have edited my post to admit that you can use the baseball bat of calculus to squash this gnat of a problem.
Click to expand...

No..no.. this is dainty badminton racquet - swatting at the bad old birdie...

Anyway, sometimes these problems are done to begin to inculcate the idea that the logarithmic changes lark around the percentage changes.

i.e

V = a3 → ln(V) = 3ln(a) → dV/V = 3 da/a ... and so on....
 
Subhotosh Khan said:
No..no.. this is dainty badminton racquet - swatting at the bad old birdie...

Anyway, sometimes these problems are done to begin to inculcate the idea that the logarithmic changes lark around the percentage changes.

i.e

V = a3 → ln(V) = 3ln(a) → dV/V = 3 da/a ... and so on....
Click to expand...
This proves the wisdom of my choosing a career other than teaching. My explications usually involved the simplest and most basic of mathematics that would do the trick: a horde of so-called educated people are suspicious of results justified even by elementary algebra and would look askance at anything involving calculus. When talking to my audience, I intended to persuade rather than to teach. I can see the pedagogical value in the calculus approach, particularly as a way of doing approximations.

\(\displaystyle \dfrac{dV}{V} = 3 * \dfrac{da}{a} \implies \dfrac{\Delta V}{V} \approx 3 * \dfrac{\Delta a}{a} \implies \dfrac{\Delta a}{a} = \dfrac{1}{3} * \dfrac{\Delta V}{V}.\)

Applied to this problem, it would give a quick approximation of \(\displaystyle \dfrac{7\%}{3} \approx 2.33\%\) rather than my slightly more exact algebraic

approximation of 2.39%. All this shows why students would benefit if they explained what they are currently studying.
 
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