Calculus I:
functions of several variables
linear approximation
Error calculation for
several variable function

Error calculation for
a function of several variable

As usual we are going to generalize what we have just seen for functions of a single variable.

Concretely to pass from the differential df to the error Δf, we have:

• taken the absolute value of each term,
• replace the sign = by the sign ≈.

We will do the same for the functions of several variables.

• Proposition

Let x and y be two measurements, δx and δy the measurement errors and Δx, Δy the accuracy of the devices that measured x and y.

We can estimate the absolute error:

δf = |f(x + δx, y + δy) - f(x, y)|

of the following way:

|δf| ≈ |(∂f(x,y)/∂x)|Δx + |(∂f(x,y)/∂y)|Δy

|δf| = Δf

We have:

Δf ≈ |(∂f(x,y)/∂x)|Δx + |(∂f(x,y)/∂y)|Δy

Δf represents the order of magnitude of the absolute error.

We can also define the relative error:


• Definition

The relative error is the quotient: Δf/|f(x, y)|

This number is expressed in%.

We notice that here too we can get the relative error using the deriative of a logarithm. Indeed,

Δf / |f(x, y)| = |(∂f(x,y)/∂x)|Δx + |(∂f(x,y)/∂y)|Δy ]/ |f(x, y)|

(∂f(x,y)/∂x)/f(x, y) is the partial derivative of ln(f) with respect to x.

We have the same thing for the exit from y.

Thus, calculating the relative error amounts to calculating the differential of ln(f).


• Exercise

We measure a length l in meters and we obtain l = 50 ± 0.1 meters.

This means that the measured length is 50 meters and that the precision of the measure is 0.1 meter.

A runner travels this distance in t = 5.8 ± 0.01 seconds.

This means that the measured time is 5.8 seconds and the accuracy of this measurement is 0.01 seconds.

1. Calculate the average speed of the runner on this course.

2. Give an estimate of the absolute error made from these measurements.

3. Calculate the relative error made from these measurements.


Solution:

1. v = l/t = 50/5.8 = 8.6206

2. v (l,t) = l/t. So
dv = (1/t)dl - (l/t²)dt

Then
Δv = |(1/t)|Δl + |- (l/t²)|Δt

Δv = (1/5.8) x 0.1 + (50/5.8²) x 0.01 = 0.0321

The absolute error is therefore about 0.03 m.s-1.

Note : The thousandth figure is therefore meaningless in writing v = 8.6206 since the result is known at 0.03.

3. The relative error is:
Δv/|v| = 0.03/8.62 = 0.003

The relative error is therefore 0.3 %.