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Homework Help: Maximum possible percentage error

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    The maximum error in the measurement of the mass and length of the side of a cube are 3% and 2% respectively. What is the maximum error in the measurement of density? (The questions has options as:- (a)3% (b)5% (c)6% (d)9% (e)12%

    2. Relevant equations
    Density(ρ)= Mass/Volume, Volume(cube)= (Length of cube)^3

    3. The attempt at a solution
    I have thought of two ways to do this:-
    1) Just by adding the percentage errors, I get the answer as 5%.

    2)Or by adding them like this:- 3% + 3*2%=9%

    Please tell me which one of them is the correct one. If both are wrong, please tell me how to solve this problem.
  2. jcsd
  3. Jul 9, 2014 #2
    I dont think you have the correct approach here.
    Think about working with actual numbers.
    Say you have a 1M^3 block with a mass of 1000kg
    What is it's desnity?
    Now what is the worst cast scenario on the errors and figure out what that density would be
  4. Jul 9, 2014 #3


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    You can approach it with either relative or absolute errors, the final result will not matter. However, just adding them is not correct unless the measurements are correlated, which is probably a bad assumption in this case.

    Independent errors should be added in quadrature. Note that when you compute the volume of the cube, the error on the side length enters 3 times and these are correlated. This gives an error on the volume of the cube, which a priori is uncorrelated to the measurement and of its mass.
  5. Jul 9, 2014 #4


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    Correlated errors is a required assumption in this case. The problem statement asks for the "maximum error". The maximum error will occur when the errors are correlated in just the right way so that error is maximized.

    Given this fact, the applicable rule of thumb is that small percentage errors add when you multiply or divide.

    edit: [And multiply when you exponentiate]
    Last edited: Jul 9, 2014
  6. Jul 9, 2014 #5


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    Well I'm not sure if what I did is legit or not, I never spent a whole lot of time with statistal things.

    I basically brute forced it.

    The exact density is ##\frac{M}{V} = \frac{M}{L^3}##
    if the margin of error on mass is 3% then the maximum measured value for M is 1.03M, minimum being .97M.
    For L it would be 1.02L and .98L

    I don't want to give too much away, see if that helps you at all.
  7. Jul 9, 2014 #6


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    While this is true, I do not see how the measurement of weight would correlate with the measurement of a length (unless you already know the density). Of course, this is more related to the formulation of the problem not coinciding with what you would expect in a real situation. In the end, I find that such things lead to misconception and confusion.
  8. Jul 11, 2014 #7
    Got the answer!!!

    The answer was 9%. When I asked my teacher about this, he did it using the second method that I mentioned as my approach to the solution. Anyways, thanks for the help guys, really like to discuss questions again!!!!
  9. Jul 11, 2014 #8


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    There is an 'engineer's rule of thumb' that when measurements are added (or subtracted), their errors add and when measurements are multiplied (or divided), their percentage errors add so, yes, since density = mass/length^3, the error in calculating the density is (approximately) 3%+ 2%+ 2%+ 2%= 9%.

    That is approximate because the "engineer's rule of thumb" is just that- an approximation. To get a more accurate value you could instead do it this way: a fraction is largest when the numerator is largest and the denominator is smallest. Here, the numerator is the mass. Calling the measured mass "M", with error 3%, the largest it could be is 1.03M. Calling the length "L", with error 2%, the smallest it could be is .98L so that the volume could be a little as [itex](.98L)^3= 0.8337L^3[/itex]. The largest the density could be is [itex]\frac{1.03M}{0.8337L^3}= 1.0943(M/L^3)[/itex] which is approximately 9% larger than the calculated density.

    (That "engineer's rule of thumb" can be derived using Calculus: if F= x+ y, then dF= dx+ dy. If F= xy then dF= xdy+ ydx. Dividing both sides by F= xy,
    [tex]\frac{dF}{F}= \frac{xdy}{xy}+ \frac{ydx}{xy}= \frac{dy}{y}+ \frac{dx}{x}[/tex])
    Last edited by a moderator: Jul 11, 2014
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