Using differentials to estimate the amount of tin

[tnutty]

Homework Statement

Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm, and height 12cm, if the tin is 0.04 cm thick.

The way I did this was to subtract the Max volume by the Enclosed volume.

Max Volume is pi(4)(12)
Enclosed Volume is pi ( 8 - 2(0.04) )/2 *12

I got about 12cm^3How do I do this with differentials ?V = pi*r^2*h

dv = 2pi*rh * dr + pi*r^2 dh

Then what?

I think dr = ( 8 - (8 - 2*0.4) ) / 2 = 0.4
Then what is dh ?

Do I have to compare it to the Total volume somehow?

 

[Mark44]

Homework Statement

Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm, and height 12cm, if the tin is 0.04 cm thick.

The way I did this was to subtract the Max volume by the Enclosed volume.

Max Volume is pi(4)(12)
Enclosed Volume is pi ( 8 - 2(0.04) )/2 *12

I got about 12cm^3


How do I do this with differentials ?


V = pi*r^2*h

dv = 2pi*r * dr + pi*r^2 dh

Then what?

I think dr = ( 8 - (8 - 2*0.4) ) / 2 = 0.4
Then what is dh ?

It seems to me that dh = 2* 0.4, since you have two lids (top and bottom) of thickness 0.4 cm each.

Do I have to compare it to the Total volume somehow?

No, your expression for dV should do the trick.

 

[tnutty]

No, the answer in the book says that its 16cm^3.

The first method I did, subtracting the max volume by the volume with radius 1/2( 8 - 2*0.04) was close because the radius will be 3.6, from the answer(16cm^3) the radius has to be close to 3.5.

I am not sure how to work with differentials with this problem though.

 

[Mark44]

Using dh = .08, I get
dV $\approx~4\pi(24(.04) + 4(.08))~\approx~16.08495$ cm^3
If I use dh = .04, I get an answer that is about 14.07 cm^3

My answer using dh = .08 is a lot closer to your book's answer.

 

[tnutty]

Using dh = .08, I get
dV $\approx~4\pi(24(.04) + 4(.08))~\approx~16.08495$ cm^3
If I use dh = .04, I get an answer that is about 14.07 cm^3

My answer using dh = .08 is a lot closer to your book's answer.

Sorry for my miscalculation, thanks a million, Mark.