Title: Default for 18.4 Errors and Percentage Change
Course: HELM Workbooks 2023-2024

Approximation and errors

In this section we consider functions of several variables for which takes some point as input. Often the point will be the result of some measurement and may be subject to measurement error, such as measuring distance to the nearest millimetre. Since the partial derivatives of the function tell us the instantaneous change at that point in each direction we can use these to approximate the function at nearby points, i.e. within the range of error for our point. As our input value is subject to some error we need to understand the resulting error in evaluating the function at that point: this is to understand the error, that is, the error in the value of the function given the error in the input values.

When working with Maclaurin and Taylor Series we saw in how to expand a function of a single variable \(f(x)\) as a Taylor series: \[f(x)=f(x_0)+(x-x_0)f'(x_0)+ \frac{(x-x_0)^2}{2!}f''(x_0)+ \dots \]

This can be written in the following alternative form (by replacing \(x − x_0\) by \(h\) so that \(x = x_0 + h\)): \[f(x_0+h)=f(x_0) + h f'(x_0)+\frac{h^2}{2!} f''(x_0)+\dots\]

This expansion can be generalised to functions of two or more variables, the following is taking the expansion with first derivatives only: \[f(x_0+h, y_0+k)\simeq f(x_0, y_0)+h f_x(x_0, y_0) + k f_y(x_0,y_0) \] where, assuming \(h\) and \(k\) to be small, we have ignored higher-order terms involving powers of \(h\) and \(k\) as these powers will typically be much smaller and so those terms do not contribute much. We define \(\delta f\) to be the change in \(f(x, y)\) resulting from small changes to \(x_0\) and \(y_0\), denoted by \(h\) and \(k\) respectively. Thus: \[\delta f = f(x_0 + h, y_0 + k) - f(x_0, y_0) \] and so \(\delta f \simeq h f_x(x_0, y_0)+k f_y(x_0,y_0)\). Using the notation \(\delta x\) and \(\delta y\) instead of \(h\) and \(k\) for small increments in \(x\) and \(y\) respectively we may write \[\delta f \simeq \delta x f_x(x_0,y_0) + \delta y f_y(x_0,y_0) \]

Finally, using the more common notation for partial derivatives, we write \[\delta f \simeq \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y}\ \delta y.\]

Informally, the term \(\delta f\) is referred to as the absolute error in \(f(x, y)\) resulting from errors \(\delta x, \delta y\) in the variables \(x\) and \(y\) respectively. Note that the right hand side is an approximation of the absolute error. Other measures of error are used. For example, the relative error in a variable \(f\) is defined as \(\frac{\delta f}{f}\) in order to account for the scale of the values, e.g. an absolute error of \(2\) might be considered small if the value was \(1,000\), but could be considered large if the value was \(10\). Finally the percentage relative error is when we convert the relative error to a percentage, i.e. \(\frac{\delta f}{f} \times 100\).

Key Point

Measures of Error

If \(\delta f\) is the change in \(f\) at \((x_0, y_0)\) resulting from small changes \(h, k\) to \(x_0\) and \(y_0\) respectively, then \(\delta f = f(x_0 + h, y_0 + k) − f(x_0, y_0)\), and

The absolute error in \(f\) is \(\delta f\).

The relative error in \(f\) is \(\dfrac{\delta f}{f}\).

The percentage relative error in \(f\) is \(\dfrac{\delta f}{f} \times 100\).

Note that to determine the error numerically we need to know not only the actual values of \(\delta x\) and \(\delta y\) but also the values of \(x\) and \(y\) at the point of interest.

Example

Estimate the absolute error for the function \(f(x, y) = x^2 + x^3 y\) for a general point \((x,y)\) and then for the point \((1,2)\) with \(\delta x = \delta y = 0.1\).

Solution

\(f_x = 2x+3x^2 y; \quad f_y = x^3\)

Then \(\delta f \simeq (2x+3x^2 y) \delta x + x^3 \delta y.\)

Now if \((x_0,y_0)=(1,2)\), \(\delta x = \delta y =0.1\) then we are using knowledge of the function and its derivates at \((1,2)\) to approximate the function at \((1.1,1.2)\). This is to serve as an estimate of the error in our function if our measured input is accurate to within \(0.1\).

Now \(f(1,2) = 3\), \(f_x(1,2) = 8\) and \(f_y(1,2) = 1\). So we approximate

\[f(1.1,2.1) \approx f(1,2)+f_x(1,2) \delta x + f_y(1,2)\delta y= 3+ 0.8+0.1=3.9.\]

Or equivalently the change or error in \(f\) is

\[\delta x = f(1.1,2.1) - f(1,2) \approx f_x(1,2) \delta x + f_y(1,2)\delta y= 0.8+0.1=0.9.\]

Of course we can calculate the actual change in \(f\) in this case by evaluating

\[ f(1.1,2.1) - f(1,2) = 4.0051 -3 = 1.0051 \]

so that our approximate change of \(0.9\) is close to the actual change between these two values. Smaller values of \(\delta x\) and \(\delta y\) would result in even better approximation as we look at points closer to \((x_0,y_0)\).

Note that what we are calculating is an approximate bound on the error. Better bounds can be calculated by careful consideration of the range of values for \(x\) and \(y\) but this tends to be time consuming for little benefit. We could also find better bounds by taking more terms in the Taylor expansion.

Exercise

Estimate the absolute error for \[f(x, y) = [value]\] at a general point \(([value], [value])\).

First, find \(f_x\) and \(f_y\), to obtain an expression for the absolute error.

\(\delta f \simeq\) \(\delta x\) + \(\delta y\)

Now obtain the estimated value of the absolute error with \((x_0,y_0)=([value],[value])\) and \(\delta x = [value]\) and \(\delta y = [value]\):

\(\delta f \simeq\) Keep 2 s.f. in your answer.

You should compare the estimate with the exact value of the error. I.e., how similar are \(f(x_0,y_0)-f(x_0+\delta x,y_0+\delta y)\) and the absolute error estimate found here.
Feedback: We are given the function \[f(x, y) = [value]\] and we are asked to estimate the absolute erroe at a general point \(([value], [value])\). First, find \(f_x\) and \(f_y\), to obtain an expression for the absolute error. \( f_x = [value] \\ f_y = [value] \) Hence, \(\delta f \simeq ([value]) \, \delta x +([value]) \, \delta y.\) Now obtain the estimated value of the absolute error with \((x_0,y_0)=([value],[value]),\) \(\delta x = [value]\) and \(\delta y = [value]\): \(\delta f \simeq ([value]) \, \delta x +([value]) \, \delta y = ([value])([value]) + ([value])([value]) = [value] .\) The actual error is calculated from: \[\begin{align} \delta f &= f( x_0 + \delta x, \, y_0+ \delta y) - f(x_0, y_0) = f ([value], \, [value]) - f ([value],[value]) \\&= [value] - ([value]) = [value] \end{align}\]

Approximations with more than two variables

If \(f\) is a function of several variables \(x,\, y,\, u,\, v,\, \dots\) the error induced in \(f\) as a result of making small errors \(\delta x\), \(\delta y,\, \delta u,\, \delta v,\, \dots\) in \(x,\, y,\, u,\, v,\, \dots\) is found by a simple generalisation of the expression for two variables given above: \[\delta f \simeq \dfrac{\partial f}{\partial x} \delta x + \dfrac{\partial f}{\partial y} \delta y + \dfrac{\partial f}{\partial u} \delta u + \dfrac{\partial f}{\partial v} \delta v + \cdots\]

Example

Suppose that the area of triangle \(ABC\) is to be calculated by measuring two sides and the included angle. Call the sides \(b\) and \(c\) and the angle \(A\).

Then the area \(S\) of the triangle is given by \(S= \dfrac{1}{2} b c \sin A\).

Now suppose that the side \(b\) is measured as \(4.00\text{m}\), \(c\) as \(3.00\text{m}\) and \(A\) as \(30 ^{\circ}\). Suppose also that the measurements of the sides could be in error by as much as \(\pm 0.005\text{m}\) and of the angle by \(\pm 0.01^{\circ}\). Calculate the likely maximum error induced in \(S\) as a result of the errors in the sides and angle.

Solution

Here \(S\) is a function of three variables \(b,c, A\). We calculate \(S= \dfrac{1}{2} \times 4 \times 3 \times \dfrac{1}{2} = 3\text{m}^2\).

Now \(\dfrac{\partial S}{\partial b}= \dfrac{1}{2} c \sin A, \quad \dfrac{\partial S}{\partial c} = \dfrac{1}{2} b \sin A, \text{ and } \quad \dfrac{1}{2} b c \cos A,\) so \[\delta S \simeq \dfrac{\partial S}{\partial b} \delta b + \dfrac{\partial S}{\partial c} \delta c + \dfrac{\partial S}{\partial A} \delta A = \dfrac{1}{2} c \sin A \delta b + \dfrac{1}{2} b \sin A \delta c + \dfrac{1}{2} b c \cos A \delta A. \]

Here \(|\delta b|_{\text{max}} = |\delta c|_{\text{max}} = 0.005\) and \(|\delta A|_{\text{max}} = \dfrac{\pi}{180} \times 0.01\) (\(A\) must be measured in radians). Substituting these values we see that the maximum error in the calculated value of \(S\) is given by the approximation \[ \begin{aligned} |\delta S|_{\text{max}} & \simeq \left(\dfrac{1}{2} \times 3 \times \dfrac{1}{2}\right) \times 0.005 + \left(\dfrac{1}{2} \times 4 \times \dfrac{1}{2}\right) \times 0.005 + \left(\dfrac{1}{2} \times 4 \times 3 \times \dfrac{\sqrt{3}}{2}\right)\frac{\pi}{180} \times 0.01 \\ & \simeq 0.0097 \text{ m}^2 \end{aligned}\] Hence the estimated value of \(S\) is in error by up to about \(\pm 0.01\text{ m}^2\).

Engineering example

The height \(h\) of a building is estimated from

1. the known horizontal distance \(x\) between the point of observation \(M\) and the foot of the building and
2. the elevation angle \(\theta\) between the horizontal and the line joining the point of observation to the top of the building (see Figure below).

If the measured horizontal distance is \(x = 150 \text{m}\) and the elevation angle is \(\theta = 40^{\circ}\), estimate the error in measured building height due to an error of \(0.1^{\circ}\) degree in the measurement of the angle of elevation.

[Image]

The variables \(x, \theta\), and \(h\) are related by \[\tan \theta = \dfrac{h}{x}\] or \[x \tan \theta = h.\]

The error in \(h\) resulting from a measurement error in \(\theta\) can be deduced by differentiating \(x \tan \theta = h\) as follows. \[\dfrac{\mathrm{d}(x \tan \theta)}{\mathrm{d} \theta} = \dfrac{\mathrm{d} h}{\mathrm{d} \theta} \quad \implies \quad \tan \theta \dfrac{\mathrm{d} x}{\mathrm{d} \theta} + x \dfrac{\mathrm{d}(\tan \theta)}{\mathrm{d} \theta} = \dfrac{\mathrm{d} h}{\mathrm{d} \theta}.\] This can be written \[\tan \theta \dfrac{\mathrm{d} x}{\mathrm{d} \theta} + x \sec^2 \theta = \dfrac{\mathrm{d} h}{\mathrm{d} \theta}.\]

The equation above gives the relationship among the small variations in variables \(x, h\) and \(\theta\). Since \(x\) is assumed to be without error and independent of \(\theta\), \(\dfrac{\mathrm{d}x}{\mathrm{d} \theta} = 0\) and the equation becomes \[x \sec^2 \theta = \dfrac{\mathrm{d}h}{\mathrm{d} \theta}.\]

The above equation can be considered to relate the error in building height \(\delta h\) to the error in angle \(\delta \theta\): \[\dfrac{\delta h}{\delta \theta} \simeq x \sec^2 \theta.\]

It is given that \(x = 150 \text{m}\).

The incidence angle \(\theta = 40^{\circ}\) can be converted to radians i.e \(\theta = \dfrac{40 \pi}{180} \text{ rad} = \dfrac{2 \pi}{9} \text{ rad}\).

Then the error in angle \(\delta \theta = 0.1^{\circ}\) needs to be expressed in radians for consistency of the units.

So \(\delta \theta = \dfrac{0.1 \pi}{180} \text{ rad} = \dfrac{\pi}{1800} \text{ rad}\). Hence, \[\delta h = 150 \dfrac{\pi}{1800 \times \cos^2\left(\dfrac{2\pi}{9}\right)} \approx 0.45 \text{ m}. \]

So the error in building height resulting from an error in elevation angle of \(0.1^{\circ}\) is about \(0.45 \text{ m}\).

Exercise

Estimate the maximum error in \[f(x, y) = [value]\] at the point \(x=[value], \, y=[value]\) if maximum errors \(\pm [value]\) and \(\pm [value]\) are made in \(x\) and \(y\) respectively.

First, find \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial f}{\partial y}\):

\(\dfrac{\partial f}{\partial x} = \)

\(\dfrac{\partial f}{\partial y} = \)

Now for \(x = [value]\) and \(y=[value]\), calculate the value of \(f(x,y)\):

\(f([value],[value]) = \)

Now since the error in the measured value of \(x\) is \(\pm [value]\) and in \(y\) is \(\pm [value]\) we have \( |\delta x|_{\text{max}} = [value], |\delta y|_{\text{max}}= [value] \).

Write down an expression to approximate to \(|\delta f|_{\text{max}}\):

\(|\delta f|_{\text{max}} \simeq \big|\) \(\big| |\delta x|_{\text{max}} + \big|\) \(\big| |\delta y|_{\text{max}}\)

Calculate \(|\delta f|_{\text{max}}\) at the point \(x = [value], y = [value]\) and give bounds for \(f([value], [value])\)

\(|\delta f|_{\text{max}} \simeq\) Keep 2 s.f.

Hence \(f([value], [value])\) can be expressed as:

\(\leq f([value], [value]) \leq\)
Feedback: Differentiating \(f(x,y)\) with respect to \(x\) and \(y\) we obtain \[\frac{\partial f}{\partial x}= [value] \quad \frac{\partial f}{\partial y} = [value].\] Input the coordinates \(x=[value]\) and \(y=[value]\) into the formula for \(f(x,y)\) to obtain \(f([value],[value])=[value]\). Now, since the error in the measured value of \(x\) is \(\pm [value]\) and in \(y\) is \(\pm [value]\) we have \( |\delta x|_{\text{max}} = [value], |\delta y|_{\text{max}}= [value] \). Therefore, the expression to approximate \(|\delta f|_{\text{max}}\) is \[ |\delta f|_{\text{max}} \simeq |\frac{\partial f}{\partial x}||\delta x|_{\text{max}}+|\frac{\partial f}{\partial y}||\delta y|_{\text{max}}=|[value]||\delta x|_{\text{max}}+|[value]||\delta y|_{\text{max}} \] We can then calculate \(|\delta f|_{\text{max}}\) at the point \(x=[value], y=[value]\) and give bounds for \(f([value],[value])\):\[\begin{aligned} |\delta f|_{\text{max}} & \simeq [value]\times [value] + [value] \times [value] \\ & = [value] + [value] \\ & = [value] \end{aligned} \] Hence we quote \(f([value], [value]) = [value] \pm [value]\) which can be expressed as \([value] \leq f([value], [value]) \geq [value]\).

Relative error and percentage relative error

Two other measures of error can be obtained from a knowledge of the expression for the absolute error. As mentioned earlier, the relative error in \(f\) is \(\dfrac{\delta f}{f}\) and the percentage relative error is \(\left(\dfrac{\delta f}{f} \times 100 \right)%\). Suppose that \(f(x,y)=x^2+y^2+xy\) then \[\begin{aligned} \delta f & \simeq \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y} \delta y \\ & = (2x+y) \delta x + (2y+x) \delta y. \end{aligned} \]

The relative error is \[\begin{aligned} \frac{\delta f}{f} & \simeq \frac{1}{f} \frac{\partial f}{\partial x} \delta x + \frac{1}{f} \frac{\partial f}{\partial y} \delta y \\ & = \frac{(2x+y)}{x^2+y^2+xy} \delta x + \frac{(2y+x)}{x^2+y^2+xy} \delta y \end{aligned}\]

The actual value of the relative error can be obtained if the actual errors of the independent variables are known and the values of \(x\) and \(y\) at the point of interest. In the special case where the function is a combination of powers of the input variables then there is a short cut to finding the relative error in the value of the function. For example, if \(f(x,y,u)=\dfrac{x^2 y^4}{u^3}\) then \[\frac{\partial f}{\partial x} = \frac{2xy^4}{u^3}, \quad \frac{\partial f}{\partial y} = \frac{4x^2y^3}{u^3}, \quad \frac{\partial f}{\partial u} = -\frac{3x^2y^4}{u^4}.\] Hence \[\delta f \simeq \frac{2xy^4}{u^3} \delta x + \frac{4x^2 y^3}{u^3} \delta y- \frac{3x^2y^4}{u^4} \delta u\] Finally, \[\frac{\delta f}{f} \simeq \frac{2xy^4}{u^3} \times \frac{u^3}{x^2y^4} \delta x + \frac{4x^2 y^3}{u^3} \times \frac{u^3}{x^2 y^4} \delta y - \frac{3x^2 y^4}{u^4} \times \frac{u^3}{x^2 y^4} \delta u\] Cancelling down the fractions, \[\frac{\delta f}{f} \simeq 2 \frac{\delta x}{x} + 4 \frac{\delta y}{y} - 3 \frac{\delta u}{u} \qquad \qquad (1)\] so that \[\text{rel. error in } f \simeq 2 \times (\text{rel. error in } x) + 4 \times (\text{rel. error in } y) - 3 \times (\text{rel. error in } u).\] Note that if we write \[f(x,y,u)=x^2 y^4 u^{-3}\] we see that the coefficients of the relative errors on the right-hand sie of \((1)\) are the powers of the appropriate variable.

To find the percentage relative error we simply multiply the relative error by \(100\).

Exercise

If \(\displaystyle f=[value]\) and \(x, y, u\) are subject to percentage relative errors of \([value]\)%, \([value]\)% and \([value]\)% respectively, find the approximate percentage relative error in \(f\).

First, find \(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\) and \(\dfrac{\partial f}{\partial u}:\)

\(\dfrac{\partial f}{\partial x} = \)

\(\dfrac{\partial f}{\partial y} = \)

\(\dfrac{\partial f}{\partial u} = \)

Now write down an expression for the percentage relative error in \(f\):

\(\dfrac{\delta f}{f} \times 100 \simeq \) \(\dfrac{\delta x}{x} \times 100 +\)  \(\dfrac{\delta y}{y} \times 100 - \) \(\dfrac{\delta u}{u} \times 100\)

Inputting the the percentage relative errors of \([value]\)%, \([value]\)%, \([value]\)% in \(x,y,u\), the value of the percentage relative error is %.
Feedback: We are given \[ \displaystyle f=[value]\] and \(x, y, u\) are subject to percentage relative errors of \([value]\)%, \([value]\)% and \([value]\)% respectively. We are asked to find the approximate percentage relative error in \(f\). First, find \(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\) and \(\dfrac{\partial f}{\partial u}:\) \(\dfrac{\partial f}{\partial x} = \displaystyle [value]\) \(\dfrac{\partial f}{\partial y} = \displaystyle [value]\) \(\dfrac{\partial f}{\partial u} = \displaystyle [value]\) Now we write down the expression of \( \delta f .\) \[ \delta f \simeq [value] \delta x +[value] \delta y [value] \delta u \] Hence we write down an expression for the percentage relative error in \(f\): \[ \dfrac{\delta f}{f} \times 100 \simeq [value] \times [value] \delta x \times 100 +[value] \times [value] \delta y \times 100 [value] \times [value] \delta u \times 100 \] Then we calculate the values of the percentage relative error: \[\dfrac{\delta f}{f} \times 100 \simeq [value] \dfrac{\delta x}{x} \times 100 + [value] \dfrac{\delta y}{y} \times 100 [value] \dfrac{\delta u}{u} \times 100\] Inputting the the percentage relative errors of \(\displaystyle [value]\)%, \([value]\)%, \([value]\)% in \(x,y,u\), the value of the percentage relative error is \([value] \%.\)

Engineering example

Error in power to a load resistance

Introduction

The power required by an electrical circuit depends upon its components. However, the specification of the rating of the individual components is subject to some uncertainity. This Example concerns the calculation of the error in the power required by a circuit shown in the Figure below given a formula for the power, the values of the individual components and the percentage errors in them.

Problem in words

The power delivered to the load resistance \(R_L\) for the circuit shown in the Figure below is given by \[P = \dfrac{25 R_L}{(R+R_L)^2}\]

[Image]
Circuit with a load resistance

If \(R=2000 \Omega\) and \(R_L = 1000 \Omega\) with a maximum possible error of \(5\)% in either, find \(P\) and estimate the maximum error in \(P\).

Mathematical statement of the problem

We can calculate \(P\) by substituting \(R=2000\) and \(R_L=1000\) into \(P=\dfrac{25 R_L}{(R+R_L)^2}\).

We need to calculate the absolute errors in \(R\) and \(R_L\) and use these in the approximation \(\delta P \approx \dfrac{P}{R} \delta R + \dfrac{P}{R_L} \delta R_L\) to calculate the error in \(P\).

Mathematical analysis

At \(R=2000\) and \(R_L=1000\)

\(P=\dfrac{25 \times 1000}{(1000+2000)^2} = \dfrac{25}{9000} = \dfrac{25}{9} \times 10^{-3} \approx 2.77 \times 10^{-3}\) watts.

A \(5\)% error in \(R\) gives \(|\delta R|_{\text{max}}= \dfrac{5}{100} \times 2000 = 100\) and \(|\delta R_{L}|_{\text{max}} = \dfrac{5}{100} \times 1000 = 50\)

\(|\delta P|_{\text{max}} \approx \dfrac{P}{R} |\delta R|_{\text{max}} + \dfrac{P}{R_L}|\delta R_L|_{\text{max}}\)

We need to calculate the values of the partial derivatives at \(R = 2000\) and \(R_L = 1000\).

\[P = \dfrac{25R_L}{(R+R_L)^2} = 25 R_L (R+R_L)^{-2}\]

\[\dfrac{P}{R} = -50 R_L (R+R_L)^{-3}\]

\[\dfrac{P}{R_L}= 25(R+R_L)^{-2} - 50 R_L(R+R_L)^{-3}\]

So \[\dfrac{P}{R}(2000,1000)=-50(1000)(3000)^{-3}=\dfrac{-50}{1000^2\times 27}=-\dfrac{50}{27} \times 10^{-6}\]

\[\begin{aligned} \frac{P}{R_L}(2000,1000) & =25(3000)^{-2}-50(1000)(3000)^{-3} \\ & =\left(\frac{25}{9} - \frac{50}{27}\right) \times 10^{-6} \\ & = \left(\frac{75-50}{27}\right) \times 10^{-6} \\ & = \frac{25}{27} \times 10^{-6} \end{aligned}\]

Substituting these values into \(|\delta P|_{\text{max}} \approx \dfrac{P}{R} |\delta R|_{\text{max}} + \dfrac{P}{R_L}|\delta R_L|_{\text{max}}\) we get: \[|\delta P|_{\text{max}} = \dfrac{50}{27} \times 10^{-6} \times 100 + \dfrac{25}{27} \times 10^{-6} \times 50 = \left(\dfrac{5000}{27} + \dfrac{25 \times 50}{27}\right) \times 10^{-6} \approx 2.315 \times 10^{-4}\]

Interpretation

At \(R=2000\) and \(R_L=1000\), \(P\) will be \(2.77 \times 10^{-3}\)W and, assuming \(5\)% errors in the values of the resistors, then the error in \(P \approx 2.315 \times 10^{-4}\)W. This represents about \(8.4\)% error. So the error in the power is greater than that in the individual components.

Exercise

The sides of a right-angled triangle enclosing the right-angle are measured as \([value] \, \textrm{m}\) and \([value] \, \textrm{m}.\) The maximum errors in each measurement are \(\pm [value] \, \textrm{m}.\) Find the maximum error in the calculated area.

The maximum error is \(\text{m}^2\).

Note: Keep 1 s.f. in your answer.
Feedback: The area of a right-angled triangle is \[A=\frac{1}{2}xy\] where \(x, \, y\) are the sides enclosing the right-angle. First we calculate \(\dfrac{\partial A}{\partial x}\) and \(\dfrac{\partial A}{\partial y}.\) \[ \displaystyle\frac{\partial A}{\partial x} = [value] \quad \text{and} \quad \displaystyle \frac{\partial A}{\partial y} = [value] \] Therefore, \[\delta A \approx \frac{\partial A}{\partial x} \delta x + \frac{\partial A}{\partial y} \delta y = [value] \delta x + [value] \delta y\] Letting \(x=[value], \, y=[value]\) and \( \delta x = \delta y = [value]\), we obtain \[\delta A \approx \frac{[value]}{2} \delta x + \frac{[value]}{2} \delta y = [value]+[value] = [value]\text{m}^2.\]

Exercise

The sides of a right-angled triangle enclosing the right-angle are measured as \([value] \, \textrm{m} \) and \([value] \, \textrm{m}.\) The maximum errors in the measurement of each side are \(\pm [value] \, \textrm{m}.\)

The angle opposite the \([value] \, \textrm{m} \) side is calculated from \(\tan\theta = \frac{[value]}{[value]}\) as \(\theta = [value]\).

Calculate the approximate maximum error in that angle, giving your answer in radians.

The maximum error in \(\theta\) is .

Note: Keep 1 s.f. in your answer.
Feedback: We know that \( \tan \theta = \dfrac{y}{x}\) so \(\theta = \tan^{-1}\dfrac{y}{x}.\) Therefore: \[\delta \theta = \frac{\partial \theta}{\partial x} \delta x + \frac{\partial \theta}{\partial y} \delta y = - \frac{y}{x^2+y^2} \delta x + \frac{x}{x^2+y^2} \delta y\] We have \( y= [value], \, x = [value] \) and \( \delta y = \delta x= [value]. \) So the maximum error in \(\theta\) is: \[\left| \frac{-[value]}{[value]^2+[value]^2}([value]) \right| + \left| \frac{[value]}{[value]^2+[value]^2} ([value]) \right| = [value]\]

Exercise

If \(\displaystyle{v=\sqrt{\frac[value][value]}}\) find the maximum percentage error in \(v\) due to errors of \([value] \%\) in \(x\) and \([value] \% \) in \(y\).

The maximum percentage error in \(v\) is %.
Feedback: We are given the funtcion \(\displaystyle{v=\sqrt{\frac[value][value]}}\) and \(x, y\) are subject to percentage relative errors of \([value] \%\) and \([value] \%\) respectively. First, find \(\dfrac{\partial v}{\partial x}\) and \(\dfrac{\partial x}{\partial y}.\) \(\dfrac{\partial v}{\partial x} = \displaystyle{ [value]}\) \(\dfrac{\partial v}{\partial y} = \displaystyle{ [value]}\) Now we write down the expression of \( \delta v .\) \[ \left|\delta v \right|_{\text{max}} \simeq \left| \frac{\partial v}{\partial x} \right| \left| \delta x \right|+ \left| \frac{\partial v}{\partial y} \right| \left| \delta y \right| \simeq \left| [value] \right| \left| \delta x \right| + \left| [value] \right| \left| \delta y \right| \] Hence we write down an expression for the percentage relative error in \(f\): \[ \begin{align} \dfrac{\delta v}{v} \times 100 & \simeq \left| [value] \times \sqrt{\frac[value][value]} \, \delta x \times 100 \right| + \left| [value] \times \sqrt{\frac[value][value]} \, \delta y \times 100 \right| \\ & \simeq \left| [value] \delta x \times 100 \right| + \left| [value] \delta y \times 100 \right| \end{align}\] The percentage relative errors of \(x,y \) are \([value] \%\) and \([value] \%\) respectivelly so: \( \displaystyle \frac{\delta x}{x} = [value] \) and \( \displaystyle \frac{\delta y}{y} = [value] \) Substituting them into the previous relation we find: \[\dfrac{\delta v}{v} \times 100 \simeq \left| [value] \delta x \times 100 \right| + \left| [value] \delta y \times 100 \right| = \frac{[value]}{2} +\frac{[value]}{2} =[value] \%\] Alterantive method: If we first take logarithms of both sides \[\ln v = \frac{1}{2}\ln [value] + \frac{1}{2} \ln x - \frac{1}{2} \ln [value] - \frac{1}{2} \ln y\] We can then apply the total derivative (implicitly) to obtain \[\frac1v \mathrm{d}v= \frac{1}{2x}\mathrm{d}x-\frac{1}{2y}\mathrm{d}y\] So \(\dfrac{\delta v}{v} \approx \dfrac{\delta x}{2x}- \dfrac{\delta y}{2y}\) The maximum percentage error in \(v\) is therefore \[v=\left|\frac{\delta x}{2x}\right|+\left|-\frac{\delta y}{2y}\right|= \frac{[value]}{2} +\frac{[value]}{2} = [value] \% \]

Exercise

If \(\displaystyle{n=\frac{1}{2L}\sqrt{\frac{E}{d}}}\) and \(L, E\) and \(d\) can be measured correct to within \(1\)%, how accurate is the calculated value of \(n\)?

The maximum percentage error in \(n\) is %.
Feedback: We are given the funtcion \(\displaystyle{n=\frac{1}{2L}\sqrt{\frac{E}{d}}}\) and \(L, E, d\) are subject to percentage relative errors of \([value] \%\). First, find \(\dfrac{\partial n}{\partial L},\) \(\dfrac{\partial n}{\partial E}\) and \(\dfrac{\partial n}{\partial d}\) \(\dfrac{\partial n}{\partial L} = \displaystyle [value]\) \(\dfrac{\partial n}{\partial E} = \displaystyle [value]\) \(\dfrac{\partial n}{\partial d} = \displaystyle [value]\) Now we write down the expression of \( \delta n .\) \[ \begin{align} \left|\delta n \right|_{\text{max}} & \simeq \left| \frac{\partial n}{\partial L} \right| \left| \delta L \right|+ \left| \frac{\partial n}{\partial E} \right| \left| \delta E \right| + \left| \frac{\partial n}{\partial d} \right| \left| \delta d \right| \\ & \simeq \left| [value] \right| \left| \delta L \right| + \left| [value] \right| \left| \delta E \right| + \left| [value] \right| \left| \delta d \right| \end{align}\] Hence we write down an expression for the percentage relative error in \(f\): \[ \begin{align} \dfrac{\delta n}{n} \times 100 & \simeq \left| [value] \times 2L \sqrt{\frac{d}{E}} \, \delta L \times 100 \right| + \left| [value] \times 2L \sqrt{\frac{d}{E}} \, \delta E \times 100 \right| + \left| [value] \times 2L \sqrt{\frac{d}{E}} \, \delta d \times 100 \right| \\ & \simeq \left| [value] \delta L \times 100 \right| + \left| [value] \delta E \times 100 \right| + \left| [value] \delta d \times 100 \right| \end{align}\] The percentage relative errors of \(L,E, d\) are \([value] \%\) so: \[ \frac{\delta L}{L} = [value], \quad \frac{\delta E}{E} = [value], \quad \frac{\delta d}{d} = [value] \] Substituting them into the previous relation we find: \[\dfrac{\delta v}{v} \times 100 \simeq [value] +\frac{[value]}{2} + \frac{[value]}{2} =[value] \%\] Alterantive method: Take logarithms of both sides \[\ln n = -\ln 2 - \ln L + \frac{1}{2} \ln E - \frac{1}{2} \ln d\] So \(\displaystyle \frac{\delta n}{n} = -\frac{\delta L}{L}+ \frac{\delta E}{2E} - \frac{\delta d}{2d}\) The maximum percentage error in \(n\) is therefore \[v=\left|\frac{\delta L}{L}\right|+\left|-\frac{\delta E}{2E}\right| + \left|-\frac{\delta d}{2d}\right|= [value] +\frac{[value]}{2} + \frac{[value]}{2} = [value] \% \]

Exercise

The area of a segment of a circle which subtends an angle \(\theta\) is given by \(A= \dfrac{1}{2}r^2(\theta-\sin \theta)\). The radius \(r\) is measured with a percentage error of \(+0.2\)% and \(\theta\) is measured as \(45^\circ\) with an error of \(+0.1^\circ\). Find the percentage error in the calculated area.

The percentage error in the calculated area is
Feedback: \(A=\dfrac{1}{2}r^2(\theta-\sin \theta\) so \(\dfrac{\delta A}{A} = \dfrac{2 \delta r}{r}+ \dfrac{1-\cos \theta}{\theta - \sin \theta} \delta \theta\) \(\dfrac{\delta A}{A} = 2(0.2)\)% + \(\left(\dfrac{1-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}-\dfrac{1}{\sqrt{2}}}\right)\dfrac{\pi}{1800} \times 100\)%\(=(0.4+0.65)\)%=\(1.05\)%