Approximation of Volume using Differential

[Frank69]

Homework Statement

Use the differential to estimate the volume of tin a can of radius 4 cm., height 12 cm.
and thickness 0.04 cm is made of.

Homework Equations

(differential)

df = f'(x,y)dx + f'(x,y)dy

The approximated volume should be 14cm^3. However, I need the procedure to get to the result (since it was given).

The Attempt at a Solution

I know how to compute the differential. In order to do so I need the two derivatives in x and y. then the formula is straight forward: df = f'(x,y)dx + f'(x,y)dy
Unfortunately I have no clue how to apply this concept to solve the problem.
Any help would be appreciated.

 

[rock.freak667]

For a cylinder the volume is V=πr²h.

So in your formula it would be

dV= (∂V/∂r)dr + (∂V/∂h)dh

So when you put in ∂V/∂r and ∂V/∂h into the formula, 'dr' is the change in radius (your thickness) and 'dh' is your change in height (which is zero).

 

[Geezer]

Homework Statement

Use the differential to estimate the volume of tin a can of radius 4 cm., height 12 cm.
and thickness 0.04 cm is made of.

Homework Equations

(differential)

df = f'(x,y)dx + f'(x,y)dy

The approximated volume should be 14cm^3. However, I need the procedure to get to the result (since it was given).

The Attempt at a Solution

I know how to compute the differential. In order to do so I need the two derivatives in x and y. then the formula is straight forward: df = f'(x,y)dx + f'(x,y)dy
Unfortunately I have no clue how to apply this concept to solve the problem.
Any help would be appreciated.

You're given height (h), radius (r), and thickness (dr). You have all the info you need. Make the necessary substitutions and you'll have it.

If you get stuck, post what you have and we can provide further hints (if we've determined you're actually putting some effort into understanding the solution).

 

[Frank69]

Thanks to everyone. Geezer, you say that dh is zero. I instead set it to 0.04 (I considered also the bases).
And apparently I got the right result. My calculation was this:

V(r,h) = πr2h.

(∂V/∂r) = 2πrh

(∂V/∂h) = πr²

Then I plugged r = 4 and h = 12 in dV

dV = (∂V/∂r)(4,12)dr + (∂V/∂h)(4,12)dh

And I got: dV = 301.44*0.04 + 50.24 *0.04 = 12.0576 + 2.0096 = 14.0672

14 cm^3 was the result expected.

If instead I used dh = 0 as you suggest I would get:

dV = 301.44*0.04 + 50.24 *0 = 12.0576 + 0 = 12.0576

which is not the right answer.

Where's the catch?

 

[Geezer]

Thanks to everyone. Geezer, you say that dh is zero. I instead set it to 0.04 (I considered also the bases).

I didn't say dh=0. In fact, I never commented on dh (dh=dr since the thickness of the can is uniform).

 

[Frank69]

'dh' is your change in height (which is zero)

Sorry Gezeer, I meant what was written by rock.freak667 in the previous post.

Thanks to all of you anyway :)

 

[rock.freak667]

I don't think your answer should be 14 cm³ since if you calculate

dV as dV = πh(r₂²-r₁²) where r₂=4.04, r₁= 4 and h =12, you would get around 12.1 or so.

 

[Geezer]

Sorry Gezeer, I meant what was written by rock.freak667 in the previous post.

Thanks to all of you anyway :)

No biggie.