Sophhhowa
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I know how to do it in theory and I know the answer is 3.
Is it possible to do with rules
Z=x+y dz^2 = dx^2 + dy^2
Z=x^n y^m fz^2 = (n fx)^2 + (m fy)^2

Or do you *have to* go down the differentiation route. I know the method for differentiation but it gets a bit nasty.

How would you do it?


* fz is the fractional error in z
dz is the absolute error in z
Last edited by Sophhhowa; 1 month ago
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tej3141
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What subject is this and the year? I'm assuming its university
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Sophhhowa
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(Original post by tej3141)
What subject is this and the year? I'm assuming its university
First year physics
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tej3141
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(Original post by Sophhhowa)
First year physics
Thanks. I though so
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Stonebridge
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Unless you've been taught otherwise in your first year, the usual way to do this is similar to the A-Level method where you combine %age or fractional errors.
In this case, the % errors are very small, so it is a good method.
eg
R is 2%
w is 1% (10/1000)
C is 1% (0.05/5.00)

The absolute error in Z2 is equal to the absolute error in R2 plus the absolute error in 1/(wC)2
The % error in R2 is 2 x %error in R
Work out the absolute error in R2 from this % error
The % error in (1/wC)2 is 2 x (% error in w plus % error in C)
From that find the absolute error in (1/wC)2
Add those 2 absolute errors together as in step 1, to find the absolute error in Z2

Can you do the rest (absolute error in Z) yourself?
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