Relative error mathematics question:? - tabletop computing
The rectangle represents a table. They measure the length and width specified (48.5 x 36.5). relative error of this size is 0.04 phases. What is the relative error in the calculation of the surface of the table?
Thursday, February 18, 2010
Tabletop Computing Relative Error Mathematics Question:?
3 comments:
A = xy
=> A = n + ln ln
=> DA / A = dx / dy + x / y
=> Error in the region
Relative error = relative error of length + width
= 0.04 + 0.04
= 0.08.
Default is Absoulte location "a" given.
relative error = A / DA
Area = L * B
= (L + dL) (b + PP)
= Lb + b * L * DL * DL + PP + PP
Well, we neglect PP * dl, because it is very small
becomes a zone lb + b * L * DL + PP
Mode errors in pounds = + b * L * DL + PP - lbs
Dl = b * L + * PP
in the areas now relative error = (b * L * DL + PP) / lb
Db = DL + L / B
= Relative error on the relative error of length + width
= 0.04 0.04 = 0.08.
ErrorInArea = CalculatedArea * SqrRoot [(.04/48.5) ^ 2 + (.04/36.5) ^ 2)]
or at least, as we in the physics program is not at my university.
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