About Calculus Linearization/ Maximum Error/ Relative Error?

Question

QUESTION: The circumference of a sphere was measured to be 81cm with a possible error of .5cm. Use differentials to estimate the maximum error in the calculated surface area.

Circumference=2pi*r Surface Area=4pi*r^2

if anyone knows the max error and relative error let me know thanks.

Answer 1

Error of the surface area is dA
Error of circumference is dC = .5 cm
C=2πr,
dC = 2π dr
dr = dC / 2π
dr= .5/2π
Since Area A = 4πr^2
dA = 8 πr dr
Since C = 81 cm. C=2πr
r=C/2π
r= 81/2π
With dA = 8 πr dr
dA = 8 π (81/2π) (.5/2π)= 25.78
Max error is 25.78 cm^2

Area = 4πr^2=4π(81/2π)^2
=2088 cm^2
Relative error= Max error/Area %
25.78./ 2088 %=1.23 %

Answer 2

Relative Error Calculus

Answer 3

C = 2πr, A = 4πr^2
dC = 2πdr, dA = 8πrdr
r = C/(2π), dA = 4Cdr
dr = dC/(2π), dA = 2CdC/π
dr = 0.5/(2π), dA = 2*81*0.5/π
dr ≈ 0.0796 cm, dA = 25.8 cm^2

dC/C = 2πdr/(2πr), dA/A = 8πrdr/4πr^2
(50/18)% = dr/r, dA/A = 2dr/r
dC/C = dr/r = 2.78%, dA/A = 5.56%

Answer 4

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