Posted by Steven on Friday, December 4, 2009 at 5:08am.
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, Friday, December 4, 2009 at 6:15am
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
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