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 + δy y) / z .

(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) + Δθ Asin(θ).

(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 + dl t + δt λ }.

(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 + δB A/B² + δC/D + δD C/D² .

(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 + δa t + δt a.

(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 = t dl/λ + (δR₀/R₀ + δR/R)/λ .

(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 + ½δa t² + δt (v₀ + at) .

(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) + Δθ 2Rtan(θ) / cos²(θ) .

(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 + δg R]tan(θ) + Δθ Rg/cos²(θ)} / 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 + δv/v + δr/r + Δθ/tan(θ)} .

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Questions?mike.coombes@kwantlen.ca