Question

The side of a square is measured to be 12 ft with a possible error of ±0.1 ft. Use differentials to estimate the error in the calculated area. Include units in your answer

Answer 1

Write the Area function: A(x)=x^2.

Differentiate this with respect to x.

Multiply both sides of the resulting equation by dx.

Your result?

Now replace x in your result with 12 ft, and replace dx with (plus or minus) 0.1 ft.

Answer 2

The resulting equation, for dA, represents the approx error in the area.

Topic: "Differentials."

Answer 3

so its A(x)=2x?

Answer 4

then d(x)A(x)=2x(d(x))

Answer 5

Actually, you want A '(x) = 2x dx, or\[\frac{ dA }{ dx }=2x\] Multiply both sides of this equation by dx and simplify the result.

Answer 6

so dxDa/dx=2xdx
dA=2xdx
dA=\[24\pm.1\]

Answer 7

is that correct?

Answer 8

@TheSmartOne

Answer 9

@3mar

Answer 10

this, in differential language, is:

\(A = x^2\)

\(dA = 2x ~ dx\)

A very cheap but ubiquitous trick is this - take logs of \(A = x^2\):

\(\implies \ln A = \ln x^2 = 2 \ln x\)

then \(d ( \ln A = 2 \ln x) \implies \dfrac{dA}{A} = 2 \dfrac{dx}{x} \implies dA =2A \dfrac{dx}{x}\)

if \(A = x^2\) then you get the same result.


but really it's all about Taylor Expansions



\(A(x + \delta x) = A(x) +\delta x A'(x) + \mathcal O (\delta x^2)\)

\(\delta A = A(x + \delta x) - A(x) = \delta x f'(x) + \mathcal O (\delta x)\)

\(\implies \delta A = \delta x (2x) + \mathcal O (\delta x)\)

\(\implies \delta A \approx (\pm 1) . (24) \)

Answer 11

it seems really unlikely the error would be 14.4 isn't that a really high number

Answer 12

typo

\(\delta A \approx (\pm \color{red}{0.}1) . (24) = \pm 2.4\)

Answer 13

so would the final answer be plus or minus 2.4 feet squared @IrishBoy123

Answer 14

144 square feet plus or minus 2.4 square feet.