Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - Significant figures problem...

  1. #1
    Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    241
    Thanks
    21

    Significant figures problem...

    Hi, I put it in the physics forum, but it's applicable to this thread too, and I'm really confused

    A rectangular plate has a length of (17.4 ± 0.2) cm and a width of (9.4 ± 0.1) cm. Calculate the area of the plate, including its uncertainty.

    I proceed as follows:

    the % of error for Length = 0.2/17.4*100% = 1.15%
    the % of error for Width = 0.1/9.4*100% = 1.06%

    the area of the plate = W*L = 17.4*9.4 = 163.56cm^2

    the total uncertainty = [163.56cm^2 *(1.15%+1.06%)]/100% = 3.6 cm^2

    since the length has 2 significant figures, I have to write the result as
    1.6*10^2 ±4 cm^2 right,

    I should write the area in only 2 significant figures which is 1.6*10^2 , right, but still it gives me wrong...

    your input would be greatly appreciated,
    dokrbb
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,418
    Thanks
    1854

    Re: Significant figures problem...

    The most direct way to do this is to note that the largest the length and width could be is 17.6 cm and 9.5 cm, respectively. That means the largest possible area is (17.6)(9.5)= 167.2 sq. cm. The smallest possible length and width will be 17.2 cm and 9.3 cm. That means the smallest possible area is (17.2)(9.3)= 159.96 sq. cm. (17.4)(9.4)= 163.56 which is 167.2- 163.56= 3.65 smaller than the largest possible value and 163.56- 159.96= 3.6 sq. cm. larger than the smallest value. So we are certain the true area is 163.56\pm 3.6 sq. cm. Rounding that "error" to one decimal place would give your "4" but I don't see why you would do that.

    There is an engineer's "rule of thumb" that "when quantities are added, errors are add; when quantities are multiplied, relative errors are added". So you are correct that since the "relative errors" are \frac{0.2}{17.4}= 0.0115 (or 1.15%) and \frac{0.1}{9.4}= 0.0106 (or 1.06%), the relative error in there product is 1.15+ 1.06= 2.21% so the error would be (.0221)(163.56)= 3.62. I can see rounding that to 3.6 (the error I have, above) but not to "4".
    Thanks from dokrbb
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Significant Figures
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 11th 2011, 02:05 PM
  2. 900 to 1, 2, or 3 significant figures?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 7th 2011, 04:14 PM
  3. significant figures
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 16th 2009, 11:20 AM
  4. significant figures
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: September 6th 2008, 04:11 PM
  5. Significant Figures
    Posted in the Geometry Forum
    Replies: 1
    Last Post: September 29th 2007, 10:34 AM

Search Tags


/mathhelpforum @mathhelpforum