# Calculating Uncertainty in multiplication/division (Volume&Ration)

by nut.the.dutch
Tags: uncertainty, volumeandration
HW Helper
P: 5,346
 Quote by nut.the.dutch 2. Find the uncertainty in the volume of a cookie. > Express your answer using one significant figure. 4. Find the uncertainty of the ratio. > Express your answer using one significant figure. uncertainity = sqrt [(uncertainty of A/A)^2 + (uncertainty of B/B)^2] I plugged in: sqrt [(.02/8.5)^2 + (.005/.07)^2] = .07146... = 7*10^-2 (one sig fig) ...and it was wrong.
Welcome to PF.

The formula from the Dartmouth site which uses the RSS of the relative errors should be OK. To apply it properly you need to account for the area uncertainty as the square of the dimension that you would have used. If you would have calculated the area as πd²/4, then you would want to add d's contribution to uncertainty twice.

Hence ((Δd/d)² + (Δd/d)² + (Δh/h)²)1/2
 HW Helper P: 5,346 On the other hand your course may also be treating uncertainty in a more conservative manner and the multiplication rule your teacher is using may only be based on the simple sum of the relative uncertainties. In which case you may be expected to use (Δd/d) + (Δd/d) + (Δh/h)
P: 7

## Calculating Uncertainty in multiplication/division (Volume&Ration)

Thank you! I just realized that to find the volume of the cylinder, I used the radius (1/2 diamter > 8.5*.5 = 4.25), not the diameter.

So sqrt[(.02/4.25)^2 + (.005/.07)^2] = .0715834209 - which would not make a difference in the final result since we need to round to one sig. figure, and it still works out to be 7*10^-2.

____

If I use ((Δd/d)² + (Δd/d)² + (Δh/h)²)1/2
= sqrt[(.02/4.25)² + (.02/4.25)² + (.005/.07)²]
= .0717379361 > which still gives the same answer with 1 sig. fig

If I use the diameter, and not the radius:
= sqrt[(.02/8.5)² + (.02/8.5)² + (.005/.07)²]
=.0715060381 > again, works out to the same number.

____

another thing, if I halve the diameter, do I also need to halve the uncertainty? (.02 > .01cm?) It seems logical in the sense that had you not halved the uncertainty, it would grow each time you reduce the actual measurment, to the point where the uncertainty will be greater than the measurement itself, and that's impossible, right?

And by the same logic, the uncertainty of the thickness would not change becuase it hasn't been manipulated:

If so, then I could use the formula again and:
=sqrt[(.01/4.25)² + (.01/4.25)² + (.005/.07)²]
=.0715060381 > Again. I guess this makes no difference.

= sqrt[(.01/8.5)² + (.01/8.5)² + (.005/.07)²]
=.071447946 > Yet again.

[EDITED TO ADD:] I tried adding the uncertainties (.005+.02) for both the ratio and volume, and it was wrong.
 HW Helper P: 5,346 This was why I suggested you use the πd²/4 for area, as it should clarify how you would take the relative uncertainty. (Δd/d)² ---> (.002/8.5)² Yes. Add it twice. Note your error ±.002 over 8.5 cm as per the original statement
 HW Helper P: 5,346 RSS = Root Sum of the Squares
P: 7
 Quote by LowlyPion This was why I suggested you use the πd²/4 for area, as it should clarify how you would take the relative uncertainty. (Δd/d)² ---> (.002/8.5)² Yes. Add it twice. Note your error ±.002 over 8.5 cm as per the original statement
Alright. I don't think I see what you're saying, to me it looks like you're just repeating to do what I've already done, which is:

 If I use the diameter, and not the radius: = sqrt[(.02/8.5)² + (.02/8.5)² + (.005/.07)²] =.0715060381 > again, works out to the same number
.

Are we speaking past each other? Becuase as I said, the .07....number I keep coming up with is incorrect. MasteringPhysics automatically checks answers, it's deemed that one wrong...
HW Helper
P: 5,346
 Quote by nut.the.dutch Alright. I don't think I see what you're saying, ...... If I use the diameter, and not the radius: = sqrt[(.02/8.5)² + (.02/8.5)² + (.005/.07)²] =.0715060381 > again, works out to the same number ...
Let me say it again:

Note your error ±.002 over 8.5 cm as per the original statement
 P: 7 Aagh - I'm so sorry, I copied the original problem wrong, it's not .002, but .02cm (and still .005cm for the other). Sorry for the confusion. :/
 HW Helper P: 5,346 Ok. That's fine. Then are you sure your instructor is expecting you to take the RSS to determine uncertainty propagation? Because regardless, the error of the diameter is so much smaller than the error of the thickness, that the RSS will be dominated by the thickness error. Might it simply be the sum of the relative errors in your course for taking multiplication/division uncertainties? In which case that sums to .076. But remember what you get from either method is a relative uncertainty result that would be expressed as a %. If they want the error expressed as an absolute volume then you still need to multiply that % by the nominal calculation.
 P: 7 No, I have no idea what he expects. Of the entire problem set for the section, this is the only problem with uncertainties. This is the first homework set though, but/and it's 'physics for engineering majors'. We haven't covered before, so maybe you're right and he's looking for a simpler solution. What you're saying, is that I take the 7*10^-2, and multiply it by the nominal calculation, so .076*4.0(volume) and .076*120(ratio). YEEEEEEEEESSSSSSS!!!!!! Thank you! That works, and I also get it.
 HW Helper P: 5,346 If the .076 is correct then the instructor is not using the RSS, he is using the simple sum of the relative errors as his multiplication/division rule, with the relative error of the diameter taken twice because of its effect on area. Good luck.
 P: 7 Yeah, .076 multiplied by the actual result (ratio, volume) was right. Thanks again!

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