The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm² What is the relative error?

Solution:

It is given that

Circumference of a sphere = 74 cm

Error = 0.5 cm

We know that

Circumference of a sphere C = 2πr

It can be written as

74 = 2πr

r = 74/2π = 37/π

C = 2πr

By differentiating with respect to r

dC / dr = 2π

dC = 2π dr

dr = dC / 2π = 0.5 / 2π

The formula to find the surface area of a sphere is 

S = 4πr² 

The maximum error in the surface area is

dS = 8πrdr

Here 

On substituting the value of 

dS = 8π × 37 / π × 0.5 / 2π = 23.5 = 24 cm² 

The maximum error in the surface area is 24 cm²

We know that

Relative error = ΔS/S

ΔS/S = 8πrdr / 4πr² 

ΔS/S = 2 dr/r

We can write it as

ΔS / S = 2 × [0.5 / 2π] / [37 / π] 

ΔS / S = 2 × 0.5 / 2π × π / 37

ΔS/S = 0.014

Therefore, the maximum error in the calculated surface area is 24 cm² and the relative error is 0.014.

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The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm² What is the relative error?

Summary:
The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. Using differentials the maximum error in the calculated surface area is 24 cm² and the relative error is 0.014.