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Math Help - Percents?

  1. #1
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    Percents?

    Okay, this might be pretty basic, but I am in the process of approximating the percent error in computing the area of a square. One of the square's side is 19 cm and has a possible error of 0.05 cm.

    I got to where dA = +- 1.9cm2

    Now, I got dA/A = 1.9/19 which equals, about .0052631579

    With that being said, I have gone completely blank and have no idea on how to turn that decimal into a fraction of a percent. Thank you for your help!

    If my math is too sloppy, please explain how to make it look better(actually showing numerator on top of denominator, etc)
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    Re: Percents?

    Multiply the number by 100 and you have a percent. As a fraction, you already wrote it (although you missed the exponent in the denominator): \dfrac{1.9}{19^2} = \dfrac{19}{10\cdot 19\cdot 19} = \dfrac{1}{190}.
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  3. #3
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    Re: Percents?

    Quote Originally Posted by Devinhdl View Post
    Okay, this might be pretty basic, but I am in the process of approximating the percent error in computing the area of a square. One of the square's side is 19 cm and has a possible error of 0.05 cm.
    So each side could be as small as 19-.05= 18.95 cm or as large as 19+ .05= 19.05 cm. In the first case the area is (18.95)^2= 359.1025= 19^2- 1.8975 square centimeters. In the second case the area is (19.05)^2= 362.9025= 19^2+1.9025 square centimeters. We can write that approximately as 19^2\pm 1.9 square cm. We could also use a linear approximation from the start by taking dA= d(s^2)=  2s ds= 2(19)(.05)= 1.9 square centimeters. The "percent error" (also called "relative error" is \frac{dA}{A}= \frac{1.9}{361}.

    I got to where dA = +- 1.9cm2

    Now, I got dA/A = 1.9/19 which equals, about .0052631579

    With that being said, I have gone completely blank and have no idea on how to turn that decimal into a fraction of a percent. Thank you for your help!

    If my math is too sloppy, please explain how to make it look better(actually showing numerator on top of denominator, etc)
    Last edited by HallsofIvy; October 19th 2013 at 05:53 PM.
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    Re: Percents?

    Quote Originally Posted by Devinhdl View Post
    Okay, this might be pretty basic, but I am in the process of approximating the percent error in computing the area of a square. One of the square's side is 19 cm and has a possible error of 0.05 cm.

    I got to where dA = +- 1.9cm2

    Now, I got dA/A = 1.9/19 which equals, about .0052631579

    With that being said, I have gone completely blank and have no idea on how to turn that decimal into a fraction of a percent. Thank you for your help!

    If my math is too sloppy, please explain how to make it look better(actually showing numerator on top of denominator, etc)
    When you say the area of a square is x*y with the assumption that x=y, You will have two errors to consider for the area. Your question shows only one error dx. To find the error on the area, you have to find first the fractional error on x. this is dx/x. Since this error is repeated twice, then the error on the area is 2*(dx/x) =2*(0.05/19) = 0.1/19 = 0.053 ---> 5.3%
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    Re: Percents?

    Quote Originally Posted by votan View Post
    When you say the area of a square is x*y with the assumption that x=y, You will have two errors to consider for the area. Your question shows only one error dx. To find the error on the area, you have to find first the fractional error on x. this is dx/x. Since this error is repeated twice, then the error on the area is 2*(dx/x) =2*(0.05/19) = 0.1/19 = 0.053 ---> 5.3%
    That is wrong.
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    Re: Percents?

    Quote Originally Posted by SlipEternal View Post
    That is wrong.
    Sorry. All the books on error analysis say it is correct. Give me a moment to check my bookshelves.
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    Re: Percents?

    Quote Originally Posted by votan View Post
    Sorry. All the books on error analysis say it is correct. Give me a moment to check my bookshelves.
    The OP and HallsofIvy are both correct. Check their work.
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    Re: Percents?

    Yap. Just as I said. Take a look at "An introduction to Error Analysis", John Taylor. I taught that book in physics lab some years ago.

    If the errors dx and dy (delta x, delta y)are independent, then the error on the product xy would be sqrt((dx)^2 + (dy)^2). this one will gve about 0.7% error on the area. There are seral ways of calculating the errors. The error of the problem as asked is just what I said.
    Last edited by votan; October 19th 2013 at 08:35 PM.
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    Re: Percents?

    Quote Originally Posted by votan View Post
    Yap. Just as I said. Take a look at "An introduction to Error Analysis", John Taylor. I taught that book in physics lab some years ago.

    If the errors dx and dy (delta x, delta y)are independent, then the error on the product xy would be sqrt((dx)^2 + (dy)^2). this one will gve about 7% error on the area. There are seral ways of calculating the errors. The error of the problem as asked is just what I said.
    As I said, the OP and HallsofIvy both calculated the error correctly. The error you calculated is twice the relative error of a single side of the square which is not the relative error of the area. Check HallsofIvy's work, and you will see where you went wrong.
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    Re: Percents?

    Quote Originally Posted by SlipEternal View Post
    The OP and HallsofIvy are both correct. Check their work.
    Thanks for bringing this to my attention. HallsofIvy error is the same as mine 0.1/19. My number 0.053% should be 0.5%. My apology.
    Last edited by votan; October 19th 2013 at 08:55 PM.
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    Re: Percents?

    Quote Originally Posted by votan View Post
    Thanks for bringing this to my attention. HallsofIvy error is the same as mine 0.1/19. My number 0.053% should be 0.5%. My apology.
    Yes, that was the issue. I'm sorry, I am very tired. I just knew something looked off.
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