Estimating Tin Amount in Closed Tin Can with Differentials

[fk378]

Homework Statement

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 .04 cm thick.

Homework Equations

dz = (dz/dx) dx + (dz/dy) dy

The Attempt at a Solution

To find the area of the tin can we can see it as a rectangle. Since the diameter is given as 8cm, we can find the circumference 2(pi)r.
Surface Area (SA)=height(h) x circumference(C)
dSA=(dSA/dh) dh + (dSA/dC) dC
dh = (.04)(2) = dC
dSA=C(.08)+ h(.08)
With this I can find the error in finding the surface area, but I don't know how to figure out what the total surface area is.

 

[fk378]

Oh actually if i use the equation V=(pi)r^2(h) I get the right answer...but am I just finding the max error in the calculated volume here? I understand that the derivative of volume=area, but in this equation doesn't dV=total differential=error? How could the value of the error also be the value of the area?

 

[kkrizka]

but I don't know how to figure out what the total surface area is.

Try to layout the tin can as a map. That is, if you cut the edges and and layed everything flat. How would it look like?

Or lookup the equation for the surface area of a cylinder.

Hm, is the derivative of volume=area?

No, deriative of something is how much one thing changes in respect to another.

 

[Varnick]

It's just gone midnight, so I maybe misreading your post, but why is calculus necessary? Surely, with the correct formulae for the area of a cylinder, then you're sorted.

V