# Homework Help: Using differentials for stuff?

1. Feb 3, 2012

### ElijahRockers

EDIT:

I just found this thread, which handily has the same exact problem. The OP says that dh is = 0 though, and I don't quite understand why he does that.

1. The problem statement, all variables and given/known data

Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. (Round the answer to two decimal places.)

3. The attempt at a solution

I wasn't really sure how to start this one, but I took a stab at it.

$V=\pi*r^{2}*h$ where r = D/2. I then found the differential of this.

$dV=\frac{\pi}{4}D^{2}*dh + \frac{\pi}{2}hD*dD$

I put in 8 for D, 12 for h, and .04 for dH and dD. I got 8.04cm3, but that is apparently wrong, according to the homework.

So I tried using the area instead.

$A=hD\pi + \frac{\pi}{2}D^{2}*dD$

Differentiating, I get

$dA=\pi*Ddh + \pi(h+D)*dD$

I put in 8 for D, 12 for h, and .04 for dH and dD. I got 3.52, but this answer is also incorrect.

Under normal circumstances I would calculate the area of the can and multiply by .04, but my gut feeling tells me this is wrong as well. Am I supposed to multiply dA by dV?

I only have one chance left to get it right, so I was hoping to figure it out here first. Thanks.

Last edited: Feb 3, 2012
2. Feb 3, 2012

### Dick

Think about it. If the thickness of the tin is 0.04cm is dH really 0.04cm? There are two thicknesses of tin adding to dH. Same thinking about it advice for dD.

3. Feb 3, 2012

### ElijahRockers

If the tin is .04 thicker, then why wouldn't dh be .04? I mean, thicker tin = taller can, right?

I'm not really sure what you mean by this.

4. Feb 3, 2012

### Dick

The can is thicker on the top and thicker on the bottom, right? The total change in h is the sum of the two isn't it? I'm not sure how to be clearer than that. This isn't really a deep concept.

Last edited: Feb 3, 2012
5. Feb 4, 2012

### ElijahRockers

Ok so yes, the can is thicker on top and on bottom. .04 thicker on top, and .04 thicker on bottom. Right?

So dh is .08 then? I'm sorry I'm having a hard time visualizing this problem it seems poorly written to me, or maybe I just don't understand the concept of a differential...

Well anyway I got the question wrong again and that was my last try so blah. :(

6. Feb 4, 2012

### Dick

The problem was probably that dD is also 0.08 for the same reason.

7. Feb 4, 2012

### HallsofIvy

$V= \pi r^2h$ so $dV= \pi(2rh dr+ r^2dh)$
dh is .08 while dr= 0.04 (if D is the diameter, dD= 2dr= 0.08 as Dick says).