Differential Question

Question

A metal fabricating company produces a cubical box with 5-inch sides, made from sheet aluminum. The standard tolerance for work of this kind is +/- (1/35) inch. What is the resulting possible (approximate) variation in the volume of the box?

Answer 1

For a cube specifically, V = s³ so dV/ds = 3s²

 

∴ dV = 3s² ds

 

ds = (1/35)"

s = 5 "

therefore dV = 3(5")²((1/35)")

 = 75/35 cubic inches or simplifying

 

= 2 & (1/7) cubic inches or as a decimal: 2.142 cubic inches.

 

It's a bit of an approximation because it assume that the variation in volume changes at the same constant rate with s as the current rate of change when s is exactly 5" and ignores the cross-terms that result from taking small (1/35") parts to the second and third powers.

 

However V = lwh, and l = lmeasured+ Δl, and w = wmeasured + Δw and h = hmeasured+Δh,
(so just for your reference) another way to determine the fractional variation in V is to compute the product of the fractional variations of each of the dimensions.

ΔV = 3 V ( (Δs)/s) = 3 (125 cubic inches) ((1/35 ")/(5") = 375/(175) cubic inches = 75/35 cubic inches! How about that? The same answer, without using calculus.