Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Approximating uncertainty

  1. Sep 1, 2004 #1
    Can someone get me started with finding the approximate uncertainty of the area of a circle with a radius of 2.0 cm?


    Thanks,
    Yours
     
  2. jcsd
  3. Sep 1, 2004 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Ask yourself by what amount will the area of the circle change if you introduced a small error in the radius of the circle?
     
  4. Sep 2, 2004 #3
    um...

    Can you give me the answer so I know if I'm doing this right? (I need to show work for full credit anyway)
     
  5. Sep 2, 2004 #4

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Tell us what you did so far and I'm sure someone will provide further guidance! :-)
     
  6. Sep 2, 2004 #5
    I found the area of both pi*(2.0+0.1)^2 and pi*(2.0-0.1)^2, which is 13.9 and 11.3, respectively. I also found the area of pi*(2.0)^2, which is 12.6. Then I got stuck.
     
  7. Sep 2, 2004 #6

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    That's a good start! Generally, the error in area will be related to the error in measuring the radius by [itex]\delta A = 2\pi r \delta r[/tex] which is essentially what you have for the specific case that [itex]\delta r = 0.1[/itex] and [itex]r = 2[/itex]. ([itex]\delta[/itex] means "error" or "deviation.")
     
  8. Sep 2, 2004 #7
    So the final answer is 2.8 X 10^9 +/- 1.3?
     
    Last edited: Sep 2, 2004
  9. Sep 2, 2004 #8

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    I would refine it just a little. The radius was given to two significant digits meaning that it is given to the nearest 0.1 centimeters from which I would use [itex]\pm 0.5[/itex] for the error in r.
     
  10. Sep 2, 2004 #9
    Thank you for your help, Tide!
     
  11. Sep 2, 2004 #10
    I'm being asked to compute the uncertantity of air density that i calculated using pv=nrt. I have a uncertanty of .1 on the barometer, and .5 degrees for temperature, but the equation in my fluid mechanics book is too complicated. I can't draw what they're trying to say. The formula has to do with like 3 terms squared to the ^1/2 power. I just can't get over all the complicated lingo, I know its really easy. My recitation teacher explained it, but I'm having a brain freeze, anybody?
     
  12. Sep 2, 2004 #11

    Chronos

    User Avatar
    Science Advisor
    Gold Member

    It appears you should be using the formula
    [tex]D =\frac{P}{RT}[/tex]
    where:
    D = density, kg/m3
    P = pressure, Pascals ( multiply mb by 100 to get Pascals)
    R = gas constant , J/(kg*degK) = 287.05 for dry air
    T = temperature, degK = deg C + 273.15
     
  13. Sep 2, 2004 #12

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    I don't know whether I can make it any easier but essentially you just need to substitute numbers into the equation

    [tex]\frac {\delta n}{n} = \sqrt{ \left( \frac{\delta v}{v} \right)^2 + \left( \frac{\delta p}{p} \right)^2 + \left( \frac{\delta T}{T} \right)^2}[/tex]

    The symbol [itex]\delta[/itex] means "deviation" or "error" of a given quantity. Note that generally a quantity such as [itex]\frac {\delta n}{n}[/itex] is called the "relative error."
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook