A rectangular plate has a length of (22.4 ± 0.2) cm and a width of (8.4 ± 0.1) cm. Calculate the area of the plate, including its uncertainty.

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A rectangular plate has a length of (22.4 ± 0.2) cm and a width of (8.4 ± 0.1) cm. Calculate the area of the plate, including its uncertainty.

Physics
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i understand how to get the area of the plate i just don't understand how to get the uncertainty.
probably has something to do with the tolerance in the dimensions the \[\pm.2cm \pm.1cm\]
What is maximum error in the area of the plate?

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The area could be 22.6 X 8,5=192.1 or it could be 22.2 X 8.3 = 184.26
looks like there could be a discrepancy of 7.84 sq cm
that's what i got and that's not the answer they want
Lets see what gogind has to offer.
For this kind of uncertainty (with 2 variables) you need to find the percentage uncertainties of each of the variable (sides of the rectangle). When you find them, just add them together to get the percentage uncertainty of the total area. From that you should be able to get the absolute uncertainty. I'll explain in latex if needed :D
oh ok thanks that helps a lot
Like one dimension if 22.4 with tolerance of .2 cm(+/2) To convert this to a percentage do we use 2 or 4 Would it be 2 or 4 divided by 22.4 to get the percentage?
* .2 or .4
If you have an uncertainty of +/- 2cm of a bar with the length lets say 200cm, the percentage uncertainty of this bar would be 200cm +/- (2/200)*100 %. Typo corrected :D
Thanks I am sure ncrosby can take it from there how about it ncrosby?
One more gogind, when you get it for both dimensions, how do you combine the +/- percentages, or will it be left as a +/- percentage?
You just sum them. In this example (question of this thread) you get 0.89% error for 1 side of the rectangle and 1.19% for the other. So the total percentage uncertainty for the area will be the sum of those two. Since it is a percentage, it already contains the information that the actual value of the area can be 2.08% smaller than 188.16 or 2.08% grater then 188.16(that is. a*b= 22.4*8.4= 188.16)
Thanks again gogind,I got around .9% a little less than you, but I understand the procedure. Thankfully I am not a student and this is not my problem. I am an ole timer who likes to play lol.
We could also use the Kline-McClintock uncertainty formula. \[A = bh\]\[U = \sqrt{ \left( {\partial A \over \partial b} w_b \right)^2 + \left( {\partial A \over \partial h} w_h \right)^2 }\]where \(w_b\) is the given uncertainty of the base dimension, and \(w_h\) is the given uncertainty of the height dimension.
thanks everyone I got it and it was right
Indeed we could. :D I would also like to note that what I said earlier : "For this kind of uncertainty (with 2 variables) you need to find the percentage uncertainties of each of the variable (sides of the rectangle)." I meant, two variables that are multiplied together, like eashmore wrote(A=b*h) .
Point taken. Good day to all.

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