Small Change Differentiation Question

[Mertance]

Hi everyone,
I've got a question from the end of a chapter on small change that I can't quite work out and would appreciate if someone could give me some working out for it.

A hollow metal cube has an exterior edge length of 10cm. Use differentiation to find the approximate volume of metal required to make the cube if the walls are 2mm thick.

The answer given is 120cm³.

Thanks in advance.

 

[Mertance]

Hi everyone,
I've got a question from the end of a chapter on small change that I can't quite work out and would appreciate if someone could give me some working out for it.

A hollow metal cube has an exterior edge length of 10cm. Use differentiation to find the approximate volume of metal required to make the cube if the walls are 2mm thick.

The answer given is 120cm³.

Thanks in advance.

I forgot to add that I could get the equation for Volume of metal required without any differentiation which I got as [MATH]8t^3+6l^2t-12lt^2[/MATH] with t(thickness), l(edge length). That's not using differentiation or small change so I'm a little stuck. The only other way I can really think of to solve it is by taking an original cube with thickness 1 and subtracting the new thickness (2mm) cube which would provide a small change which would make sense with the chapter but I'm not sure if that would work.

 

[Deleted member 4993]

I forgot to add that I could get the equation for Volume of metal required without any differentiation which I got as [MATH]8t^3+6l^2t-12lt^2[/MATH] with t(thickness), l(edge length). That's not using differentiation or small change so I'm a little stuck. The only other way I can really think of to solve it is by taking an original cube with thickness 1 and subtracting the new thickness (2mm) cube which would provide a small change which would make sense with the chapter but I'm not sure if that would work.

In these types of problems - you need to assume that (t/l) is very-very small (like h in limit problems).

So:

V = l^3 * [6 * (t/l) - 12 * (t/l)^2 + 8 * (t/l)^3] ...............~ l^3 * 6 * (t/l) ..................... [assuming (t/l)^2 and (t/l)^3 ~ 0]

So:

V ~ 6 * l^2 * t

Continue....

 

[JeffM]

I forgot to add that I could get the equation for Volume of metal required without any differentiation which I got as [MATH]8t^3+6l^2t-12lt^2[/MATH] with t(thickness), l(edge length). That's not using differentiation or small change so I'm a little stuck. The only other way I can really think of to solve it is by taking an original cube with thickness 1 and subtracting the new thickness (2mm) cube which would provide a small change which would make sense with the chapter but I'm not sure if that would work.

You did not say explicitly how you calculated your formula, but it is correct and exact.

[MATH]x = s^3 - (s - 2t)^3 = 8t^3 + 6s^2t - 12st^2.[/MATH]
So first you had to think about using the difference of two volumes, then figure out the dimensions of the interior cube correctly (2t rather than t), and then do the algebra. Now you can do the arithmetic in mm^3 to avoid fractions.

[MATH]8(2)^3 + 6(100^2)(2) - 12(100)(2^2) = 64 + 120000 - 2400 = 117,664.[/MATH]
So the exact answer is 117.664 cm^3.

Or you could go

[MATH]x = s^3 - (s - 2t)^2 \approx s^3 - \left \{ s^3 - 2t * \left ( \dfrac{d}{ds} \ s^3 \right ) \right \} = s^3 - s^3 + 2t(3s^2) = 6s^2 * t[/MATH]......... Corrected

[MATH]\therefore x = 6(100^2)(2) = 120000.[/MATH]
So 120 cm^3.

The relative error is approximately 2%.

 

[Steven G]

If you don't use differentials here is how I would do it.
There are six sides. Let's concentrate on one side for now. The volume of a side is 10cm*10cm*2mm=20cm^3. So the volume of all sides is 120cm^3.

I thought that I could also compute v(10cm+2mm=10.2cm)-v(10cm)= [1061.208-1000]cm^3=61.208cm^3. I guess I am wrong. What exactly have I calculated and can I modify it to get the correct answer?

 

[JeffM]

If you don't use differentials here is how I would do it.
There are six sides. Let's concentrate on one side for now. The volume of a side is 10cm*10cm*2mm=20cm^3. So the volume of all sides is 120cm^3.

I thought that I could also compute v(10cm+2mm=10.2cm)-v(10cm)= [1061.208-1000]cm^3=61.208cm^3. I guess I am wrong. What exactly have I calculated and can I modify it to get the correct answer?

Jomo

If you calculate the volume of the sides, you can treat each side as being of the same dimensions, but then have to add for the corners. It may be easier to treat the sides as having different dimensions to account for the corners. I do not have time right now to work it all out.

 

[Deleted member 4993]

If you don't use differentials here is how I would do it.
There are six sides. Let's concentrate on one side for now. The volume of a side is 10cm*10cm*2mm=20cm^3. So the volume of all sides is 120cm^3.

I thought that I could also compute v(10cm+2mm=10.2cm)-v(10cm)= [1061.208-1000]cm^3=61.208cm^3. I guess I am wrong. What exactly have I calculated and can I modify it to get the correct answer?

For first order of calculation, that should be 4 mm.(top and bottom - etc.). Then the outer volume becomes ~1125 cm³

 

[MarkFL]

I would begin with:

[MATH]\frac{\Delta V}{\Delta s}\approx\frac{dV}{ds}[/MATH]
Or:

[MATH]\Delta V\approx\frac{dV}{ds}\Delta s[/MATH]
We have:

[MATH]V=s^3\implies \frac{dV}{ds}=3s^2[/MATH]
[MATH]s=10\text{ cm}[/MATH]
[MATH]\Delta s=-\frac{2}{5}\text{ cm}[/MATH]
Hence:

[MATH]\Delta V\approx3(10\text{ cm})^2\left(-\frac{2}{5}\text{ cm}\right)=-120\text{ cm}^3[/MATH]
This approximate decrease in volume represents the approximate volume of material needed to make the box.