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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 -

    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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