Question

The circumference of a sphere was measured to be 76 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.) cm2 What is the relative error

Answer 1

The maximum error in the calculated surface area is 24.19cm² and the relative error is 0.0132.

Given that the circumference of a sphere is 76cm and error is 0.5cm.

The formula of the surface area of a sphere is A=4πr².

Differentiate both sides with respect to r and get

dA÷dr=2×4πr

dA÷dr=8πr

dA=8πr×dr

The circumference of a sphere is C=2πr.

From above the find the value of r is

r=C÷(2π)

By using the error in circumference relation to error in radius by:

Differentiate both sides with respect to r as

dr÷dr=dC÷(2πdr)

1=dC÷(2πdr)

dr=dC÷(2π)

The maximum error in surface area is simplified as:

Substitute the value of dr in dA as

dA=8πr×(dC÷(2π))

Cancel π from both numerator and denominator and simplify it

dA=4rdC

Substitute the value of r=C÷(2π) in above and get

dA=4dC×(C÷2π)

dA=(2CdC)÷π

Here, C=76cm and dC=0.5cm.

Substitute this in above as

dA=(2×76×0.5)÷π

dA=76÷π

dA=24.19cm².

Find relative error as the relative error is between the value of the Area and the maximum error, therefore:

$\begin{aligned}\frac{dA}{A}&=\frac{8\pi rdr}{4\pi r^2}\\ \frac{dA}{A}&=\frac{2dr}{r}\end$

As above its found that r=C÷(2π) and r=dC÷(2π).

Substitute this in the above

$\begin{aligned}\frac{dA}{A}&=\frac{\frac{2dC}{2\pi}}{\frac{C}{2\pi}}\\ &=\frac{2dC}{C}\\ &=\frac{2\times 0.5}{76}\\ &=0.0132\end$

Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.

Learn about relative error from here brainly.com/question/13106593

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