Approximation Principle: For a differentiable function f(x) with x a number in the domain of f, and with Δx a small change in the value of x, set Δy equal to f(x + Δx) − f(x) as the corresponding change in the value of the function. Then Δy ≈ f'(x) • Δx. That is to say, Δy is very close to f'(x) • Δx for small values of x.
Take the Volume of the tank without Interior Copper Cladding as l(l − 2)(l − 3) or V = l3 − 5l2 + 6l.
Then Dl = 3l2 − 10l + 6. Conditions of the problem given above have l as 6 feet and Δl as 2(1/4 inch) or
1/24 of a foot.
Then obtain ΔV as (3(6)2 − 10(6) + 6) times 1/24 or 2.25 cubic feet of Copper used.
If this tank is considered to have a top side or "lid", then the actual value is given by
[(6)3 − 5(6)2 + 6(6)] cubic feet minus [(6 − 1/24)3 − 5(6 − 1/24)2 + 6(6 − 1/24)]
cubic feet which gives 2.227502893 cubic feet of Copper used, which is very close to
2.25 cubic feet of Copper found above.
Treating the tank as open to the sky or without a 6th "top side" would give the amount of
Copper used as [(6)3 − 5(6)2 + 6(6)] minus [(6 − 1/24)(4 − 1/24)(3 − 1/24 ÷ 2)] or
1.73614728 cubic feet.
Adding the product of the dimensions of the eliminated top [(6 − 1/24)(4 − 1/24)(1/24 ÷ 2)]
back to 1.73614728 cubic feet will again give 2.227502893 cubic feet of Copper used.
