the circumference of a sphere was measured to be 84 cm with possible error of 0.5 cm

Question

the circumference of a sphere was measured to be 84 cm with possible error of 0.5 cm

 

(a) use differentials to estimate the max error in the calculated surface area. what is relative error?

 

(b) use differentials to estimate the max error calculated in volume. what is relative error? 

Answer 1

Circumference (C) = 2*pi*r, C(measured) = 84 +- .5 cm Inferring that r = (84+-.5)/(2*pi) = 13.45 or 13.29 r = 13.369 if C = 84

Surface Area (S) = 4*pi*r^2 S = 4*pi*(13.369)^2 = 2,246 cm^2

Volume (V) = (4/3)*pi*r^3 V = (4/3)*pi*(13.369)^3 = 10,008.88 cm^3

Since C = 2*pi*r dC = 2*pi dr or dr = dC/(2*pi) = .5/(2*pi) = +- .0796

a) S = 4*pi*r^2 dS = 8*pi*r dr = +- 26.75 so S could range from 2272.75 to 2219.25

The relative error could be {(2246-2272.75)/2246}(100) or {(2246-2219.25)/2246}(100)

giving -1.19% or 1.19%

b) dV = 4*pi*r^2 dr dV = 4*pi*(13.369)^2 (+- .0796) = +- 178.78

So V could range from 10187.66 to 9830.1

The relative error could be {(10008.88-10187.66)/10008.88}(100) or {(10008.88-9830.1)/10008.88)(100)

giving -1.79% or 1.79%