The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?

Solution:

Given, circumference of sphere = 84cm

Possible error, dC = ± 0.5cm or 1/2

We know, circumference, C = 2πr

84 = 2πr

r = 84/2π

r = 42/π

Now, dC = 2π dr

(1/2) = 2π dr

dr = ± 1/4π

Surface area = 4πr²

A = 4π(42/π)²

Use π = 22/7

A = 4(42)(42)(7)/22

A = 2245.09 cm²

Now, dA = 2[4πr](dr)

dA = 8πr dr

dA = 8π(42/π)(1/4π)

dA = 8(42)(1/4π)

Use π = 22/7

dA = 336(1/4(22/7))

dA = ± 26.72 cm

If dA = +26.72cm, A = 2245.09 + 26.72

A = 2271.81 cm²

If dA = -26.72cm,

A = 2245.09 - 26.72

A = 2218.37 cm²

Relative error = [(2245.09 - 2218.37)/2245.09](100)

= [26.72/2245.09](100)

= [0.0119](100)

= ± 1.19%

Therefore, the relative error is ±1.19%.

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The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. What is the relative error

Summary:

The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm. Using differentials to estimate the maximum error in the calculated surface area, the relative error is ±1.19%.