# The circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm. (a) Use...

## Question:

The circumference of a sphere was measured to be {eq}\displaystyle 76\ cm {/eq} with a possible error of {eq}\displaystyle 0.5\ cm {/eq}.

(a) Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.). What is the relative error? (Round your answer to three decimal places.).

(b) Use differentials to estimate the maximum error in the calculates volume. (Round your answer to the nearest integer.). What is the relative error? (Round your answer to three decimal places.).

## Error Analysis:

This problem involves finding error in calculation of surface area and volume, given the error in measurement of the circumference. Note, the method of writing error in differential is taking logarithm both sides and then differentiating. For example:

{eq}\displaystyle C = \alpha \rho X^n \\ \ln C = \ln \alpha \rho X^n = \ln \alpha + \ln \rho + n\ln X {/eq}

Where, {eq}\displaystyle \alpha , \rho {/eq} are constants. Now we differentiate to find error -

{eq}\displaystyle \frac{dC}{C} = 0 + 0 + n \frac{dX}{X} {/eq}

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Given, the circumference of the sphere P was measured to be 76 cm. Thus, we can find the radius of the sphere -

{eq}\displaystyle P = 2 \pi R = 76...

Absolute & Relative Error: Definition & Formula

from

Chapter 6 / Lesson 7
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Did you know that measuring tools are not 100% accurate? Read this lesson to learn more about the accuracy of measuring tools and how this influences the absolute and relative error of measurements.