Questions:

1

2

Error Propagation Solutions

1. Apply error propagation rules to the following:
1. Let A = 79.5 ± 0.6, B = 27.8 ± 0.4, and C = 54.6 ± 0.3. Evaluate F = A - B + C.


First we determine the principle value,

P(F) = P(A) - P(B) + P(C) = 79.5 - 27.8 + 54.6 = 106.3 .

Using the rule for addition and subtraction, the uncertainty is given by

δ(F)	= δ(A - B + C)
	= δA + δB + δC
	= 0.6 + 0.4 + 0.3
	= 1.3

Thus F = 106.3 ± 1.3. Note we keep two figures in the uncertainty since it begins with a 1.

2. Let A = -12.1 ± 0.2 and B = 3.45 ± 0.06. Evaluate F = A/B.

 

 
 
 

First we determine the principle value,

P(F) = P(A)/P(B) = -12.1/3.45 = -3.5072 .

Using the rule for multiplication and division, the uncertainty is given by

δ(F)	= δ(A/B)
	= P(A/B)[δ(A)/P(A) + δ(B)/P(B)]
	= P(F)[δ(A)/P(A) + δ(B)/P(B)]
	= |-3.5072|[0.02 / |-12.1| + 0.06 / 3.45]
	= 0.1190

Note that in our calculation, absolute and relative uncertainties must always be positive!

Thus F = -3.5072 ± 0.1190 = -3.51 ± 0.12. Note we keep two figures in the uncertainty since it begins with a 1.

3. Let A = 15.4 ± 0.2, B = 7.85 ± 0.03, and C = 6.24 ± 0.08. Evaluate F = A /(BC).

 

 
 
 

First we determine the principle value,

P(F) = P(A/BC) = 15.4/(7.85 × 6.24) = 0.3144 .

Using the rule for multiplication and division, the uncertainty is given by

δ(F)	= δ(A/BC)
	= P(F)[δ(A)/P(A) + δ(B)/P(B) + δ(B)/P(B)]
	= 0.3144(0.2 /15.4 + 0.03/7.85 + 0.08/6.24)
	= 0.0093

Thus F = 0.3144 ± 0.0093 = 0.314 ± 0.009 . Note we keep only one figure in the uncertainty.

4. Let R = 0.151 ± 0.005. Evaluate V = (4π/3)R³.

 

 
 
 

First we determine the principle value,

P(V) = (4π/3)P(R³) = (4/3)(0.151)³ = 0.01442 .

Using the rule for roots and powers, the uncertainty is given by

δ(V)	= (4π/3)δ(R3)
	= (4π/3)(3)[P(R)]2δ(R)
	= 4π(0.151)2(0.005)
	= 0.00143

Note that the exact factor 4π/3 just multiplied the uncertainty in R³. This makes sense. Suppose A = R³ and B = 2R³, surely δB = 2δA. Also see the answer to 2a below for further proof.

Thus V = 0.01442 ± 0.00143 = 0.0144 ± 0.0014. Note we keep two figures in the uncertainty since it begins with a 1.

5. Let X = 14.75 ± 0.09. Evaluate Y = 3X^½.

 

 
 
 

First we determine the principle value,

P(Y) = 3P(X^½) = 3(14.75)^½ = 11.522 .

Using the rule for roots and powers, the uncertainty is given by

δ(Y)	= 3δ(X½)
	= 3(½)[P(X)]-½δ(X)
	= (3/2)(0.09)/(14.75)½
	= 0.035

Again, note that the exact factor 3 just multiplied the uncertainty in X^½. This makes sense. Suppose A = X^½ and B = 3X^½, surely δB = 3δA. Also see the answer to 2a below for further proof.

Thus Y = 11.522 ± 0.035 = 11.52 ± 0.04 . Note we keep only one figure in the uncertainty.

6. Let θ = 27.5 ± 0.5°. Evaluate Y = sin(θ).

 

 
 
 

First we determine the principle value,

P(Y) = P{sin(θ)} = sin(27.5°) = 0.46175 .

Using the rule for functions, the uncertainty is given by

δ(Y)	= δ{sin(θ)}
	= Δθ cos(P(θ))
	= (0.5° × π/180°)cos(27.5°)
	= 0.00774

Note how the uncertainty in the angle had to be converted to radians.

Thus Y = 0.46175 ± 0.00774 = 0.462 ± 0.008 . Note we keep only one figure in the uncertainty.

7. Let λ = 3.51 ± 0.06. Evaluate N = e^-λ

First we determine the principle value,

P(N) = P(e^-λ) = e^-3.51 = 0.029897.

Using the rule for functions, the uncertainty is given by

δ(N)	= δ(e-λ)
	= dlP(e-λ)
	= (0.06)(0.029897)
	= 0.00179

Thus N = 0.029897 ± 0.00179 = 0.0299 ± 0.0018. Note we keep two figures in the uncertainty since it begins with a 1.

8. Let x = 0.75 ± 0.03. Evaluate φ = arctan(x).

First we determine the principle value,

P(φ) = arctan(0.75) = 36.870° or 0.64350 rad.

Since we are dealing with a function

δφ = δx / (1 + x²) = 0.03 / (1 + 0.75²) = 0.019 radians or δφ = 1.1°.

Thus, φ = 36.9 ± 1.1° or φ = 0.644 ± 0.019 rad where we keep two figures in the uncertainty since it begins with a 1.



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2. Find algebraic expressions for the absolute error in the following:

a.

First let A = R³. Using the power rule, δA/A = 3δR/R.

Let B = 4π/3. This is exact, so δB = 0.

Thus V = BA, where we now know how to get the uncertainties in A and B. To proceed further, we use the rule for multiplication.

δV/V	= δB/B + δA/A
	= 0 + 3δR/R

Cross-multiplying to get the equation for δV, we find

δV	= V × 3δR/R
	= 4πR3/3 ×3δR/R
	= 4πR2δR

b.

First let A = x². Using the power rule, δA/A = 2δx/x or δA = 2xδx if we cross-multiply to find the absolute uncertainty.

Similarly, let B = x². Hence, δB/B = 2δy/y or δB = 2yδy.

Next let C = A + B. Using the addition rule, this is

δC	= δA + δB
	= 2xδx + 2yδy

Now we have z = C^½. Using the rule for roots,

δz/z	= ½δC/C
	= ½[2xδx + 2yδy]
	= xδx + yδy

Cross-multiplying we have

δz = (x² + y²)^½ (xδx + yδy)

c.	R = Acos(θ)

First let X = cos(θ). Using the rule for functions δX = Δθsin(θ) where Δθ must be given in radians.

Now we have R = AX. Using the rule for multiplication.

δR/R	= δA/A + δX/X
	= δA/A + Δθsin(θ)/cos(θ)

Cross-multiplying to get the equation for δR, we find

δR	= Acos(θ)[δA/A + Δθ sin(θ) / cos(θ)]
	= δA cos(θ) + AΔθ sin(θ)

d.	N = N0e-λt

Let X = -λt. Using the rule for multiplication

δX/X = dl/l + δt/t.

We have ignored the minus sign because it is exact and the uncertainties must be positive.

Finding the absolute uncertainty next,

δX	= X(dl/l + δt/t)
	= λt(dl/l + δt/t)
	= t dl + l dt

Again we have ignored minus signs since uncertainties must be positive.

Now we have N = N₀e^X. We next let Y = e^X. Using the rule for powers,

δY = δXe^X.

So now N = N₀Y. We use the rule for multiplication to find

δN/N	= δN0/N0 + δY/Y
	= δN0/N0 + (δXeX/eX)
	= δN0/N0 + δX
	= δN0/N0 + t dl + l dt

We want an expression for the absolute error in N, so

δN	= N0e-λt [δN0/N0 + t dl + l dt]
	= δN0 e-λt + N0e-λt [t dl + l dt]

e.	F = A/B + C/D

Let X = A/B. Using the rule for division, δX/X = δA/A + δB/B. By cross-multiplying we can get the absolute uncertainty,

δX	=X [δA/A + δB/B]
	=δA/B + AδB/B2

Similary let Y = C/D. Hence, δY/Y = δC/C + δD/D and δY = δC/D + C δD/D².

Now F = X + Y. Using the rule for addition

δF	= δX + δY
	= δA/B + B δC/C2 + δC/D + C δD/D2

f.	v = v0 + at

Let X = at. Using the rule for multiplication, δX/X = δa/a + δt/t. By cross-multiplying we can get the absolute uncertainty,

δX	= X [δA/A + δB/B]
	= at [δa/a + δt/t]
	= t δa + a δt

Now v = v₀ + X. Using the rule for addition,

δv	= δv0 + δX
	= δv0 + t δa + a δt

g.

Let X = R₀/R. Using the rule for multiplication, δX/X = δR₀/R₀ + δR/R.

Now we have t = (1/λ)ln(X). Next let Y = ln(X). Using the rule for functions, δY = δX/X.

Now we have t = Y/λ. We use the rule for division,

δt/t	= δY/Y+ dl/l
	= (δX/X) / ln(X)+ dl/l
	= (δR0/R0 + δR/R) / ln(R0/R)+ dl/l

We want an expression for the absolute error in t, so

δt	= t[(δR0/R0 + δR/R) / ln(R0/R)+ dl/l]
	= (1/λ)ln(R0/R) [(δR0/R0 + δR/R) / ln(R0/R)+ dl/l]
	= (1/λ)(δR0/R0 + δR/R)+ (dl/l2)ln(R0/R)

h.	d = v0t + ½at2

Let X = ½at². Treating this as X = ½att, we can use the rule for multiplication and find δX/X = δa/a + 2δt/t. Note that factor ½ is exact and thus has no relative uncertainty.

Converting to an absolute uncertainty

δX	= ½at2(δa/a + 2δt/t)
	= ½t2δa + atδt

Next we let Y = v₀t. Again using the rule for multiplication, δY/Y = δv₀/v₀ + δt/t.

Converting to an absolute uncertainty

δY	= v0t(δv0/v0 + δt/t)
	= t δv0 + v0δt

As a result v = X + Y. Using the rule for addition,

δv	= δX + δY
	= ½t2δa + atδt + t δv0 + v0δt
	= ½t2δa + tδv0 + (v0 + at) δt

i.	X = R tan2(θ)

Let A = tan(θ). Using the rule for functions, δA = Δθ/cos²(θ).

We now have X = R A². Thinking of this as X = RAA, we can use the rule for multiplication to find

δX/X	= δR/R+ 2δA/A
	= δR/R + [2Δθ/cos2(θ)]/tan(θ)
	= δR/R + 2Δθ/[cos2(θ) tan(θ)]

Converting to an absolute uncertainty

δX	= X {δR/R + 2Δθ/[cos2(θ) tan(θ)]}
	= R tan2(θ) {δR/R + 2Δθ/[cos2(θ) tan(θ)]}
	= δR tan2(θ) + 2Δθ R tan(θ)/cos2(θ)
	= δR tan2(θ) + 2Δθ R sin(θ)/cos3(θ)

j.

Let A = tan(θ). Using the rule for functions, δA = Δθ/cos²(θ).

We now have v = (RgA)^½.

Let B = RgA. Using the rule for multiplication, δB/B = δR/R + δg/g + δA/A.

So now, v = B^½. Following the rule for roots,

δv/v	= ½δB/B
	= ½(δR/R + δg/g + δA/A)

Converting to absolute uncertainty,

δv	= ½v (δR/R + δg/g + δA/A)
	= ½(RgA)½ (δR/R + δg/g + δA/A)
	= ½(Rgtan(θ))½ (δR/R + δg/g + [Δθ/cos2(θ)]/tan(θ))
	= ½(Rgtan(θ))½ (δR/R + δg/g + Δθ/[sin(θ)cos(θ)])

Questions?mike.coombes@kwantlen.ca