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M O RE
E X A M PL E S
OF
E R R O R P R OP A GAT I ON W I T H P A RT I A L D E RI V AT I V E S
The following pages provide you with a summary of the concepts and expressions involved in
routine error analysis that you are expected to carry out in the lab. The tone here is more friendly
(hopefully) compared to the earlier sections. This brief document is supposed to serve as a quick
reference. For a more in-depth discussion of error analysis, please refer to sections 2.1-2.7.
Example 1.
w = (x+y)/2;
x = 16.7 ± 0.3 cm;
y = 3.4 ± 0.05 cm
w/x = 1/2; w/y = 1/2,
s2w = (sx w/x)2 + (sy w/y)2 = (0.3/2)2 + (0.05/2)2
w = (16.7 + 3.4)/2 ± sw = 9.55 ± 0.15 cm
-----------------------------------------------------------------------------------------------------------------Example 2.
w = x tan (y/1130 cm);
x = 10.4 ± .2 cm;
y = 87.3 ± .4 cm.
w/x = tan (y/1130 cm); w/y = (x/1130 cm)/cos2 (y/1130 cm)
s2w = = [(.2/1130)
tan (87.3/1130)]2 + [(.4) (10.4/1130)/cos2 (87.3/1130)]2
s2w = (.0155 cm)2 + (.0037 cm)2 ; sw = .016 cm
------------------------------------------------------------------w = 10.4 cm tan (87.3/1130) ± .016 cm = .081 ± .016 cm
-----------------------------------------------------------------------------------------------------------------Example 3.
w - 15.6 ln (x/y) - 15.6 ( ln x - ln y); x = 13.1± .3 sec;
w/x = 15.6/x;
y =5.4±.2 sec
w/y = -15.6/y
s2w = (.3/13.1)2 + (.2/5.4)2 ; sw = .68
--------------------------------------------------------------------w = 15.6 ln (13.1/5.4) ± .68 = 13.82 ± .68  13.8 ± 0.7
-----------------------------------------------------------------------------------------------------------------Example 4.
w = x sin ((y + z)/2); x= 19.3±2 cm; y= 57o15' ±3'; z =31o13'±2'
w/x = sin ((y+z)/2);
w/y = .5 cos ((y+z)/2) = w/z
Note that UNCERTAINTIES IN ANGLES MUST BE EXPRESSED IN RADIANS:
sy = 3' = (3/60)o [π radians/180o] = 8.7  10-4 radians
and sz = 5.8  10-4 radians
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s2w = [ .2 cm sin ((57o 13' +31o2')/2]2 + [ .00087(19.3 cm/2)cos((44o7.5')]2
+ [ .00058((19.3 cm/2)cos((44o7.5')]2
sw = [ (.1395 cm)2 + ( .0060 cm)2 + ( .0040 cm)2 ]
1/2
= .1397 cm
w = 13.46 ± .14 cm
E R R OR A NAL YS I S S U M M AR Y
-----------------------------------------------------------------------------------------------------------------General Rule for f(x, y, z).
[ = partial derivative]
-----------------------------------------------------------------------------------------------------------------We want to find the standard error in f(x, y, z) [sf], given standard error in z [sz].
GENERAL RULE: s2f = (sx f/x)2 + (sy f/y)2 + (sz f/z)2
-----------------------------------------------------------------------------------------------------------------From this general rule, we get the following useful special rules.
-----------------------------------------------------------------------------------------------------------------RULE 0:
When f = Kx, where K is a constant, sf = Ksx.
-------------------------------------------------------
Example 0.1: f = 3x,
sx = 0.2, then sf = 0.6 from rule 0.
(With x = 4.4 ± 0.2, f = 13.2 ± 0.6 .)
-----------------------------------------------------------------------------------------------------------------RULE 1:
Example 1.1:
When f = x ± y, then s2f = s2x + s2y
----------------------------------------------f = 3x-y, sx = 1.2, and sy = 0.4, then s2f = 3.6 2 + 0.42
So, if x = 8.1 ± 1.2, and y = 3.7 ± 0.4, then f = 20.6 ± 3.6. Note that for standard error in the 3x
term we used Rule 0 to get 3.6, and then applied rule 1 to get the standard error of f.
-----------------------------------------------------------------------------------------------------------------RULE 2: When f = Kxaybzc then (sf /f)2 = (as x/x)2 + (bsy/y)2 + (csz/z)2.
----------------------------------------------------------------------------------------------Example 2.1:
f = xy2, x = 11.4± 1.6, and y = 1.4±0.1, then f = 11.4(1.4)2
and (sf/f)2 = (1.6/11.4)2 + (0.2/1.4)2 + (0.2/1.4)2 , so f = 22.3±22.3(0.2) = 22.3±4.5
----------------------------------------------------------------------------------------------Example 2.2:
f = 3x + 6xy, x = 1.7± .1, y = 2.4± .3. f =5.1+24.5 = 29.6.
The squared FRACTIONAL error in the 6xy term is (.1/1.7)2 + (.3/2.4)2, of 3.4 for the 6xy term.
Since the standard error in the 3x term is 0.3 from rule 0, we use rule 1 to add the squares of the
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standard errors: (.3) (.3) + (3.4) (3.4) = 11.5, giving an overall standard error of 3.4, so f =
29.6±3.4.
Exercises: Calculate f ± sf in each case.
1)
f = 6x, x = 3.60± .15
2)
f = 3x + y , x = 2.1 ± .2, y = 4.3 ± .3
3)
f = x + y + 2z, x = 1.7 ± .1, y = 1.3 ± .2, z = 2.4 ± .2
4)
f = xy, x = 1.7 ± .4, y = 2.8 ± .2
5)
f = x1/3, x = 3.1 ± .3
6)
f = x3 + 2y, x = 2.3 ± .1, y = 3.4± .2
Ans:
1) 21.6 ± .9
2) 10.6 ± .7
3) 7.8 ± .5
4) 4.8 ± 1.2
5) 1.46 ± .05
6) 19.0 ± 1.6