Questions: 12

Error Propagation Using Calculus Solutions

1. Apply error propagation rules to the following single variable functions:

 

 

(a) Let x = 14.75 ± 0.09. Evaluate F = 3x^½.

First the principle value of F is

P(F) = 3(14.75)^½ = 11.5217 .

Next, we take the derivative of F with respect to x

dF/dx = 3 / 2x^½ .

The uncertainty in F is thus

δF = δx dF/dx = 0.09 × 3 / 2(14.75)^½ = 0.0352 .

Keeping one figure in the uncertainty, the result is F = 11.52 ± 0.04.

(b) Let θ = 27.5 ± 0.5°. Evaluate F = sin(θ) - cos(θ).

First the principle value of F is

P(F) = sin(27.5°) - cos(27.5°) = -0.42526 .

Next, we take the derivative of F with respect to θ

dF/dθ = cos(θ) + sin(θ) .

The uncertainty in F is thus

δF = Δθ dF/dθ = (0.5° × π/180°) {cos(27.5)° + sin(27.5°)} = 0.0118 .

Keeping one figure in the uncertainty, the result is F = -0.43 ± 0.01.

(c) Let θ = 27.5 ± 0.5°. Evaluate F = sin(θ) + cos(θ).

First the principle value of F is

P(F) = sin(27.5°) + cos(27.5°) = 1.34876.

Next, we take the derivative of F with respect to

dF/dθ = cos(θ) - sin(θ) .

The uncertainty in F is thus

δF = Δθ dF/dθ = (0.5° × π/180°) |cos(27.5°) - sin(27.5°)| = 0.00371 .

Keeping one figure in the uncertainty, the result is F = 1.349 ± 0.004.

(d) Let t = 2.35 ± 0.06 s. Evaluate F = 5t² - 3t + 2.

First the principle value of F is

P(F) = 5(2.35)² -3(2.35) + 2 = 22.5625 .

Next, we take the derivative of F with respect to t

dF/dt = 10t - 3 .

The uncertainty in F is thus

δF = δt dF/dt = 0.06 {10(2.35) - 3} = 1.23 .

Keeping one figure in the uncertainty, the result is F = 23 ± 1.

(e) Let = 0.754 ± 0.004 rad. Evaluate F = [tan(θ)]^½.

First the principle value of F is

P(F) = [tan(0.754)]^½ = 0.96907 .

Next, we take the derivative of F with respect to t

dF/dθ = ½[tan(θ)]^-½ / cos²(θ) .

The uncertainty in F is thus

δF = δt dF/dt = 0.004 × ½ /{ [tan(.754)]^½ cos²(.754)} = 0.00388 .

Keeping one figure in the uncertainty, the result is F = 0.969 ± 0.004.



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

 

 

(a) 

We need to do two partial derivatives

∂ z/∂ x = ½ (1/z) (2x) = x / z , and

∂ z/∂ y = ½ (1/z) (2y) = y / z .

The uncertainty in z is thus

δz = { [δx x/z]² + [δy y/z]² }^½ .

(b) R = Acos(θ)

We need to do two partial derivatives

∂ R/∂ A = cos(θ) , and

∂ R/∂q = -Asin(θ) .

The uncertainty in R is thus

δR ={ [δA cos(θ)]]² + [δθ Asin(θ)]² }^½.

(c) N = N₀e^-λt

We need to do three partial derivatives

∂ N/∂ N₀ = e^-λt = N/N₀ ,

∂ N/∂l = -tN₀ e^-λt = -tN , and

∂ N/∂ t = -λN₀ e^-λt = -λN .

The uncertainty in N is thus

δN = N{ [δN₀/N₀]² + [dl t]² + [δt λ]² }^½.

(d) F = A/B + C/D

We need to do four partial derivatives

∂ F/∂ A = 1/B ,

∂ F/∂ B = -A/B² ,

∂ F/∂ C = 1/D , and

∂ F/∂ D = -C/D² .

The uncertainty in F is thus

δF = { [δA/B]² + [δB A/B²]² + [δC/D]² + [δD C/D²]² }^½.

(e) v = v₀ + at

We need to do three partial derivatives

∂ v/∂ v₀ = 1 ,

∂ v/∂ a = t , and

∂ v/∂ t = a .

The uncertainty in v is thus

δv = { [δv₀]² + [δa t]² + [δt a]² }^½.

(f) 

We need to do three partial derivatives

∂ t/∂l = -(1/λ²)ln(R₀/R) = -t/λ ,

∂ t/∂ R₀ = (1 / λR₀) , and

∂ t/∂ R = - (1 / λR).

The uncertainty in t is thus

δt = (1/λ){ [t dl]² + [δR₀/R₀]² + [δR/R]² }^½.

(g) d = v₀t + ½at²

We need to do three partial derivatives

∂ d/∂ v₀ = t ,

∂ d/∂ a = ½t² , and

∂ d/∂ t = v₀ + at .

The uncertainty in d is thus

δd = { [δv₀ t]² + [½δa t²]]² + [δt (v₀ + at)]² }^½ .

(h) X = Rtan²(θ)

We need to do two partial derivatives

∂ X/∂ R = tan²(θ) , and

∂ X/∂q = 2R tan(θ) / cos²(θ) .

The uncertainty in X is thus

δX = { [δR tan²(θ)]² + [Δθ 2Rtan(θ) / cos²(θ)]² }^½ .

(i) 

We need to do three partial derivatives

∂ v/∂ R = ½[Rgtan(θ)]^-½ gtan(θ) = gtan(θ) / 2v ,

∂ v/∂ g = ½[Rgtan(θ)]^-½ Rtan(θ) = Rtan(θ) / 2v, and

∂ v/∂q = ½[Rgtan(θ)]^-½ Rg/cos²(θ) = Rg / 2vcos²(θ)

The uncertainty in v is thus

δv = { [δR g]² + [δg Rtan(θ)]² + [δθ Rg/cos²(θ)]² }^½ / 2v .

(j) L = mvrsin(φ)

We need to do four partial derivatives

∂ L/∂ m = vrsin(φ) = L/m ,

∂ L/∂ v = mrsin(φ) = L/v ,

∂ L/∂ r = mvsin(φ) = L/r , and

∂ L/∂f = mvrcos(φ) = L/tan(φ) .

The uncertainty in L is thus

δv = L{ [δm/m]² + [δv/v]² + [δr/r]² + [Δθ/tan(θ)]² }^½ .

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Questions?mike.coombes@kpu.ca