1. ## 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)

2. ## 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): $\displaystyle \dfrac{1.9}{19^2} = \dfrac{19}{10\cdot 19\cdot 19} = \dfrac{1}{190}$.

3. ## Re: Percents?

Originally Posted by Devinhdl
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 $\displaystyle (18.95)^2= 359.1025= 19^2- 1.8975$ square centimeters. In the second case the area is $\displaystyle (19.05)^2= 362.9025= 19^2+1.9025$ square centimeters. We can write that approximately as $\displaystyle 19^2\pm 1.9$ square cm. We could also use a linear approximation from the start by taking $\displaystyle dA= d(s^2)= 2s ds= 2(19)(.05)= 1.9$ square centimeters. The "percent error" (also called "relative error" is $\displaystyle \frac{dA}{A}= \frac{1.9}{361}$.

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)

4. ## Re: Percents?

Originally Posted by Devinhdl
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%

5. ## Re: Percents?

Originally Posted by votan
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.

6. ## Re: Percents?

Originally Posted by SlipEternal
That is wrong.
Sorry. All the books on error analysis say it is correct. Give me a moment to check my bookshelves.

7. ## Re: Percents?

Originally Posted by votan
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.

8. ## 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.

9. ## Re: Percents?

Originally Posted by votan
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.

10. ## Re: Percents?

Originally Posted by SlipEternal
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.

11. ## Re: Percents?

Originally Posted by votan
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.