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Uncertainty in area of a circle 
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#1
Oct110, 07:53 PM

P: 7

1. The problem statement, all variables and given/known data
The radius of a circle is measured to be 14.3+0.3cm. Find the circle's area and the uncertainty in the area. I don't understand how to correctly apply uncertainty equations with sigma and partial derivatives to these types of problems. 2. Relevant equations A=(pi)(r^2) (pi)(r^2)=642.4cm 


#2
Oct110, 08:34 PM

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Well then, we have A=πr^{2}. If we take ln of both sides we will get
lnA=ln(πr^{2})=lnπ+2lnr Now just take differentials dA/A = 2*dr/r dA is nothing but the error in A. Same with dr. Just substitute the numbers. I really could not explain it properly without showing you the differentials. 


#3
Oct110, 08:47 PM

P: 7

You said to take the ln of both sides. As in the natural log? I didn't know these had anything to logs or am I reading something wrong.



#4
Oct110, 10:20 PM

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P: 6,207

Uncertainty in area of a circle
So if you had A=r^{3} then dA/A = 3*dr/r It comes out the same if you just differentiate it normally. 


#5
Oct210, 10:47 AM

P: 7

I know at the beginning I asked how to use sigma and partial derivatives to solve this type of problem but I don't really know much about them yet. We haven't gotten to them in my math class. This problem is coming from an intro to physics lab course that focuses on propagation of error and uncertainty in measurements made and then using Excel functions like STDEV and (chi^2) to figure out stuff related to uncertainties.
Is there a standard formula to use if given a measurement or multiple measurements and the uncertainity in them? "dA/A", is that supposed to be a partial derivative? 


#6
Oct210, 02:34 PM

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dA would be the error in A. A would be the actual measurement. The relative error in A would then be dA/A 


#7
Oct210, 03:08 PM

P: 7

Actually I think I got it worked out. Let me know if this looks right.
A=(∏)(r)^2 ∂(A)/∂(r) = 2(∏)(r) sigma_A=√(((∂A/∂r)^2)(sigma_r)^2)) sigma_A=√(((2∏(14.3))^2)(0.3)^2))= 26.9cm Area = 642.4cm Uncertainty = 26.9cm 


#8
Oct210, 03:57 PM

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∂A/∂r= 2πr or ∂A=2πr ∂r Now if we divide both sides by A (which is πr^{2} as well) ∂A/A = 2πr/πr^{2} ∂r or ∂A/A = 2∂r/r 


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