The circumference of a sphere was measured to be 72 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) 22.91 Correct: Your answer is correct. cm2 What is the relative error

dA(s) = 22,917 cm²     maximum error

dA(s) /A(s) = 0,01389   or 1,38 %    relative error

Step-by-step explanation:

The Volume of a sphere is:

V(s) = 4/3)*π*r³       where  r is the  radius of circumference

If the length of circumference is 72 cm then

L (c) = 72   = 2*π*r

r  = 72/2*π

r = 72/ 6,28      ⇒  r = 11,46 cm

And

L(c)  = 2*π*r

Differentiation on both sides of the equation give:

dL (c) = 2*π*dr    

dr  =  dL(c) /2*π

dr = 0,5 / 6,28     ⇒  dr = 0,07961

The surface area Is:

A (s) = 4*π*r²

And the maximum or absolute error is

dA(s)  = 8*π*r*dr

dA(s) = 22,917 cm²

The relative error is   dA(s) /A(s)  

dA(s) /A(s) = 8*π*r*dr/ 4*π*r²

dA(s) /A(s) = 2*(0,07961)/ (11,46)

dA(s) /A(s) = 0,01389   or 1,38 %

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