Posted by Steven on .
Suppose a cylindrical can is manufactured with radius r. The manufacturing process makes the length r imprecise; r can be off by ±0.03 cm. The manufacturing process makes the height of the can precisely equal to three times the radius r (including any error in r, but not introducing any new error).
a) Use ∆r= ±0.03 to approximate the corresponding error in the volume of the can by using differentials (or the linear approximation)
b) The fraction ∆r/r is called the relative error of percentage error in the radius. If the relative error in the radius is z, approximate the corresponding relative error in the volume.

Calculus 
drwls,
V = pi r^2 h = pi r^2 * 3r = 3 pi r^3
(a) If r becomes r + ∆r ,
V + ∆V = 3 pi (r + ∆r)^3
=3 pi [r^3+ 3r^2 ∆r + 3r(∆r)^2 + (∆r)^3]
V [1 +3(∆r)/r +3(∆r)^2/r^2 +(∆r)^3/r^3]
Since ∆r/r <<1, this linearizes to
V(1 + 3 ∆r/r)
So ∆V = 3V *(∆r/r)
(b) ∆V/V = 3z