See Error, Uncertainty and Graphs: Propagation of Errors .
Perimeter: Is your propagation for an addition/subtraction or multiplication/didvision ? Are the errors given for length and width absolute or relative ?
Area: Is your propagation for an addition/subtraction or multiplication/didvision ? Can you use the errors given in your propagation formula as they are or do you have to convert them ?
22 comments:
I figured out the perimeter and even the area and its error, but can't seem to get the perimeter error. If I take the error in the Area problem, and divide it by the Area calculated, then I should have the error used in the perimeter, correct?
Re 9/3 10:58 am:
No, you get the relative error of the area.
Use equation (1.6) for the error of the perimeter, which is an addition of the length and width.
for the area part i keep getting it wrong. the length on mine is 7.3 m and the width is 5.5 m. A=lw=7.3*5.5=40.15
but when i put in 40.15 or any rounding of it (40, 40.1, 40.2, or 41) it says its wrong and i need the area to find the error of the area
did you check the units
I am looking at part II of this question and I am confused by delta A and A and delta B and B. Do you put the "supposed error" for each dimension respectively in delta A and delta B or do you calculate the area first and then put in the error?
Re 9/6 4:30 pm:
The area is a product of l and w. Go into the error writeup to "multiplication and division".
Your given delta l = delta A and delta w = delta B. Use the equation given for the error of the product. Is the result an absolute error or a relative error.
In the quiz you are asked to give a unit. Does it ask for an absolute or relative error. Do you have to convert the result ?
If, yes with which equation in the error writeup ?
I can not get the area right. I am doing A=l*w but it is comming out wrong and I checked the units.
never mind i found the problem
Area = LxW I have 8.8m and 7m so I get 61.6 m. It keeps coming up as incorrect. Am i missing something?
What was the problem?
Area is in m^2 not m.
thanks. I was going crazy
I'm confused as to finding the error of the perimeter using equation 1.6. do you take the length and width and input that into 1.6?
To calculate the error for the area is it not the sq rt of the deltas divided by l and w.?
To calculate the error in the perimeter with an accuracy of +/-0.05m that means that for a length of 10.9m, 10.9-.05=10.85 to 10.9+.05=10.95 are possible. The difference in those two number is 0.1 so 0.1 would be the deltaA and deltaB, Not .05. Plug .1 in to equation 1.6 and you get:
Delta S= Sq.rt (0.1^2)+(0.1^2) = .14m
Can someone please explain/give an example of how to get the error in the area?!?!
I feel like to get the error of the area, you do the formula √(.1/A)^2 + (.1/B)^2) which gives you ∆S/S which is the relative error. According to 1.4 you have change relative error to absolute error. (∆S/S) * S gives you that. S is supposed to be the central value. What is this central value? The area which you calculated?
how do you get the error for the area???
I came to believe that the reason we use .1 in the equation for the error of the perimeter instead of .05 is not because of Erica's reasoning of subtracting the two values you get when you add and subtract .5 (unless I misunderstood you), but rather the hint that stated to use 2L and 2W, and to look at equation 1.1. That's where you will end up doing sqrt((.1)^2+(.1)^2) as the error for the perimeter.
i cant get the area error!! plz help
To get the absolute error of the area you first have to find the relative error. Your Delta A and Delta B will both be .05 (or whatever was given in the problem as your accuracy). The length can be your A and the width can be your B. Plug those numbers into equation 1.7. That will give you the relative error. Then times that number by your area to get the absolute error. Remember to put your answer in 2 decimals since that is what you have for area. Also remember that your units will be the same as for your area.
Re 9/10 11:48 am and 9/10 12:31 pm:
Is the perimeter (l*w) or 2(l*w).
(1.6) gives the error of the product(l*w). What do you have to to to get the error of the perimeter from it (see (1.1)). The absolute errors of l and w are 0.05and NOT 0.1 (9/10 6:29 pm is
correct)
Re others see Sarah above.
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