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Question: If \[f = {x^2}\], then the relative error in \[f\]is A) \[\dfrac{{2\Delta x}}{x}\] B) \[\dfrac{...

If f=x2f = {x^2}, then the relative error in ffis
A) 2Δxx\dfrac{{2\Delta x}}{x}
B) (Δx)2x\dfrac{{{{\left( {\Delta x} \right)}^2}}}{x}
C) Δxx\dfrac{{\Delta x}}{x}
D) (Δx)2{\left( {\Delta x} \right)^2}

Explanation

Solution

There are two types of experimental error : Absolute error and relative error.
Absolute error is defined as a measure of how far 'off' a measurement is from a real value or an indication of the uncertainty in a measurement. Here absolute error is expressed in terms of the difference between the expected value and actual values.
Absolute Error = Actual Value  Measured ValueAbsolute{\text{ }}Error{\text{ }} = {\text{ }}Actual{\text{ }}Value{\text{ }} - {\text{ }}Measured{\text{ }}Value
To calculate relative error, you need to determine an absolute error first. Relative error expresses how large the value of absolute error as compared to the total size of the object you’re measuring.
Relative Error = Absolute Error Known ValueRelative{\text{ }}Error{\text{ }} = {\text{ }}\dfrac{{Absolute{\text{ }}Error{\text{ }}}}{{Known{\text{ }}Value}}

Complete step-by-step answer:
We have been given within the problem f=x2f = {x^2} and we need to find the relative error for ff.
If   x\;x is the actual value or real value of a quantity, x0{x_0} is the measured value of the quantity and Δx\Delta x is the absolute error, then the relative error is expressed using the below formula.
Relative error = (x0x)x = (Δx)xRelative{\text{ }}error{\text{ }} = {\text{ }}\dfrac{{\left( {{x_0} - x} \right)}}{x}{\text{ }} = {\text{ }}\dfrac{{\left( {\Delta x} \right)}}{x}
So, here we will first differentiate the equationf=x2f = {x^2} with respect to  x\;x. We get:
dfdx=2x\Rightarrow \dfrac{{df}}{{dx}} = 2x
And we can also write the equation as ΔfΔx=dfdx\dfrac{{\Delta f}}{{\Delta x}} = \dfrac{{df}}{{dx}}
Δf=dfdxΔx\Rightarrow \Delta f = \dfrac{{df}}{{dx}}\Delta x
Now by substituting the value of dfdx\dfrac{{df}}{{dx}} from the equation dfdx=2x\dfrac{{df}}{{dx}} = 2x. We get:
Δf=2xΔx\Rightarrow \Delta f = 2x\Delta x
The relative error in f is:
Δff=2xΔxx2=2Δxx\dfrac{{\Delta f}}{f} = \dfrac{{2x\Delta x}}{{{x^2}}} = \dfrac{{2\Delta x}}{x}
So, if f=x2f = {x^2}, then the relative error in ffis 2Δxx\dfrac{{2\Delta x}}{x}.

Therefore, the option (A) is the correct answer.

Note: In this given problem, you are asked to seek out the relative error in f=x2f = {x^2}. For this, you would be required to differentiate the ff with respect to x to find the relative error.
An important note that relative errors are dimensionless. When we are expressing relative errors, it is usual to multiply the fractional error by hundred and express it as a percentage.