Questions: 1 2

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.
   

   

   ---

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/∂θ = -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/∂ λ = -tN₀ e^-λt = -tN , and
   ∂ N/∂ t = -λN₀ e^-λt = -λN .

   The uncertainty in N is thus∂ λ

   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/∂ λ = -(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/∂θ = 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/∂θ = 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/∂ φ = 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 .

   

Questions? mike.coombes@kpu.ca