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A question in Giancoli's Physics for Scientists and Engineers (2. ed) has me confused. Here it is (Ch 1, Problem 3):

What is the area, and its approximate uncertainty, of a circle of radius $2.7 \times 10^4$ cm?

I got the correct answer of $2.3 \times 10^9 \text{ cm}^2$, but the uncertainty provided in the answer was $0.2 \times 10^9 \text{ cm}^2$. How was this uncertainty calculated? As far as I can tell, it is not possible to determine the uncertainty of a measure just from its value, because I have no idea with what device/technology the radius was measured. All we can know is that the doubtful figure is in the $10^3$ position, and therefore the uncertainty in the area will be $x \times 10^9$, but $x$ could be anything? Why is $x = 0.2$?

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Most likely, the authors assume Gauss error propagation in which the error on a function $f(x)$ of a variable $x$ is calculated as $$ \Delta f = \frac{\partial f}{\partial x} \Delta x~.$$ In your case, $f(x) = \pi x^2$ and $\frac{\partial f}{\partial x} = 2\pi x$, and $\Delta x = 0.1\times 10^4$cm (this is the worst case - realistically, we could also assume $\Delta x = 0.05\times 10^4$cm). Taking $\Delta x = 0.1\times 10^4$cm leads to $$\Delta f = 2\pi x \Delta x = 2\pi \times 2.7\times 10^4 \text{cm} \times 0.1\times 10^4\text{cm} = 0.1696 \times 10^9\text{cm}\simeq 0.2 \times 10^9\text{cm}~.$$

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  • $\begingroup$ I appreciate your answer. However, I don't think this would be prior knowledge for someone starting an undergraduate physics course. The only mathematics of uncertainty covered in this book is significance arithmetic. $\endgroup$ Sep 8, 2021 at 23:47
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A rough and ready way is to say that the radius is $2.7\pm 0.1 \times 10^4\,\rm cm$ which is approximately a $3.7\%$ error.

The area is the radius squared so the error in the area is twice the error in the radius, $2\times 3.7 = 6.4\%$ which then translates into an error of $2.3 \times 10^9 \times 0.064 \approx 0.2 \times 10^9\,\rm cm^2$ in the area.

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  • $\begingroup$ Thanks, I found a paragraph in the book that said to assume the error is '1 or 2' in the least significant figure, which I must have missed before. I guess I shouldn't get hung up too much on the details, it seems sig fig arithmetic is vague and inconsistent amongst texts, and is only 'rough and ready' anyway :) $\endgroup$ Sep 9, 2021 at 0:21

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