What is the approximate uncertainty in r for the given formula V=\pi r^{2} h?

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The discussion centers on calculating the approximate uncertainty in the radius (r) from the volume (V) formula V=πr²h, given uncertainties in V (7%) and h (3%). The initial calculation suggests a 5% uncertainty in r, derived from combining the uncertainties, but this leads to inconsistencies when checking against the original equation. Participants clarify that the correct interpretation depends on whether r is measured or calculated, with a 2% uncertainty in r being valid if considering how it affects the calculated volume. Ultimately, the confusion arises from differing assumptions about which variables are measured versus calculated. The conversation highlights the importance of clearly defining the context of measurements and calculations in uncertainty analysis.
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Homework Statement


V=\pi r^{2} h

The uncertainties in V and h are shown below
V 7%
h 3%

The approximate uncertainty in r is:
A. 10% B. 5% C. 4% D. 2%

2. The attempt at a solution
According to the key the correct answer is B. This answer can be calculated by:
r^{2}=\frac{V}{\pi h}
which gives 3%+7% = 10% uncertainty
and subsequently
r=\sqrt{\frac{V}{\pi h}}
giving 10% * 1/2 = 5%

However:
if the value of 5% uncertainty for r is inserted in the original equation it all doesn't make sense:
V=\pi r^{2} h should then give an uncertainty for V of 3%+5%+5%=13%

Can anyone please explain this?
 
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I'd say the answer was 2%, but I can't really explain how your book got to that answer.

\frac{dV}{V}=2 \frac{dr}{r}+\frac{dh}{h}
 
That was my first thought as well. But the key says no.

And as proved in my previous post, the 5% uncertainty does make sense, however if used in the original equation (or by simply reversing the second) it doesn't sum up.
 
Bump. Any ideas?
 
you must be clear about what is measured and what is calculated.

if you measure V and h and are wanting to calculate r, then 5% is the correct error. but then your check does not make sense because there you are assuming r and h are measured and V is calculated.

If the question asked what error in r (which is measured) would give rise to a 7% error in V (which is calculated) given a 3% error in h (which is measured), then the answer would be 2%.

hope this helps
 
Yes, that makes sense. Thanks!
 
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