Differentials Solution 22

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SOLUTION 22: Consider a square of edge

$x$ . Then its perimeter is

$P=x+x+x+x = 4x \ \ \ \ \longrightarrow \ \ \ \ P'=4$ and its area is

$A= (length)(width) = x \cdot x = x^2 \ \ \ \ \longrightarrow \ \ \ \ A'=2x$

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GIVEN: The absolute percentage error of

$x$ is

$\displaystyle{ |\Delta x| \over x } \le \ 3\%$

$\ \ \ \$ a.) Use a Differential to estimate the absolute percentage error in perimter, i.e., estimate

$\ \displaystyle{ |\Delta P| \over P }.$ Then

$\displaystyle{ |\Delta P| \over P } \approx { | dP| \over P } = { | P' \cdot \Delta x| \over P } = { | 4 \cdot \Delta x| \over 4x } = { | \cdot \Delta x| \over x } \le \ 3\%$ i.e.,

$\displaystyle{ |\Delta P| \over P } \le \ 3\%$

$\ \ \ \$ b.) Use a Differential to estimate the absolute percentage error in area, i.e., estimate

$\ \displaystyle{ |\Delta A| \over A }.$ Then

$\displaystyle{ |\Delta A| \over A } \approx { | dA| \over A } = { | A' \cdot \Delta x| \over A } = { | 2x \cdot \Delta x| \over x^2 } = 2{ |\Delta x| \over x } \le \ 2 (3\%) = \ 6\%$ i.e.,

$\displaystyle{ |\Delta A| \over A } \le \ 6\%$

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