﻿ Unit 14 Section 4 : Error Propagation

# Unit 14 Section 4 : Error Propagation

If there is an error in a value that is used in a calculation, that error can become more significant when the calculation is made. For example, if the radius of a circle is rounded from 2.57 cm to 2.6 cm, an error of 0.49 cm² would be made when calculating the area of the circle.

In this section we consider how errors introduced by rounding can be increased (or propagated) in subsequent calculations.

## Example 1

The radius of a circle is given as 31 cm, correct to the nearest cm. What are the possible errors when calculating its area?

As the radius, r cm, is given as 31 cm to the nearest cm, we have

30.5 ≤ r < 31.5

 If r = 30.5, A = π × 30.5² = 2922.466566 cm² = 2920 cm² to 3 s.f.
 If r = 31, A = π × 31² = 3019.07054 cm² = 3020 cm² to 3 s.f.
 If r = 31.5, A = π × 31.5² = 3117.245311 cm² = 3120 cm² to 3 s.f.

If r = 30.5, then the error is

3019.07054 – 2922.466566 = 96.603974 cm²

whilst, if r = 31.5, the error is

3117.245311 – 3019.07054 = 98.17477043 cm²

Hence the maximum possible error occurs when r = 31.5, and is approximately 98.2 cm². In other words, there is a potential error here of almost 100 cm² in the area if we calculate it from the rounded radius.

## Example 2

A rectangular plot of land has sides with lengths of 38 m and 52 m correct to the nearest m.

Calculate the maximum and minimum possible values of:

(a)

the perimeter of the rectangle,

The sides have been given to the nearest metre, so

51.5 m ≤ length < 52.5 m
37.5 m ≤ width < 38.5 m

 Minimum perimeter = 2(37.5 + 51.5) = 178 m Maximum perimeter = 2(38.5 + 52.5) = 182 m
(b)

the area of the rectangle.

 Minimum area = 37.5 × 51.5 = 1931.25 m² Maximum area = 38.5 × 52.5 = 2021.25 m²

## Example 3

The values of x and y are given to 1 decimal place as x = 4.2 and y = 7.3

Determine the minimum and maximum values of:

(a)

x + y

First note that 4.15 ≤ x < 4.25 and 7.25 ≤ y < 7.35.

 Minimum value of x + y = 4.15 + 7.25 (minimum value of x + minimum value of y) = 11.4 Maximum value of x + y = 4.25 + 7.35 (maximum value of x + maximum value of y) = 11.6
(b)

yx

 Minimum value of y – x = 7.25 – 4.25 (minimum value of y – maximum value of x) = 3 Maximum value of y – x = 7.35 + 4.15 (maximum value of y – minimum value of x) = 3.2
(c)

 Minimum value of = = 0.56462585 (minimum value of x ÷ maximum value of y) = 0.565 to   3 s.f. Maximum value of = = 0.586206896 (maximum value of x ÷ minimum value of y) = 0.586 to   3 s.f.

Note that, for x and y both positive,

To find maximum value of x + y or xy,
use the largest value of x and largest value of y.

To find the minimum value of x + y or xy,
use the smallest value of x and smallest value of y.

To find maximum value of xy or ,
use the largest value of x and smallest value of y.

To find the minimum value of xy or ,
use the smallest value of x and largest value of y.

## Exercises

Question 1

A car park is a square with sides of length 48 m correct to the nearest metre.
Calculate the maximum and minimum possible values for:

(a)

the perimeter of the car park,

m ≤ perimeter < m
(b)

the area of the car park.

m² ≤ area <
Question 2

The radius of a circle is given as 9 cm correct to the nearest cm. Calculate the maximum and minimum possible values for the area of the circle. Round your upper and lower bounds to 2 decimal places.

cm² ≤ area < cm² (to 2 d.p.)
Question 3

A rectangle has sides of length 1.27 m and 2.43 m correct to 2 decimal places. Calculate the maximum and minimum possible values for:

(a)

the perimeter,

m ≤ perimeter < m

(b)

the area.

m² ≤ area <

Question 4

The formula V = is used to calculate the volume of a sphere of radius r. A sphere has radius 18 cm correct to the nearest cm.

Calculate the maximum and minimum possible values for the volume of the sphere. Round your upper and lower bounds to 2 decimal places.

cm³ ≤ volume < cm³ (to 2 d.p.)
Question 5

If x and y are given correct to 1 decimal place as x = 5.2 and y = 11.6, calculate the maximum and minimum possible values of:

(a)
≤  x²  <
(b)
≤  x + y  <
(c)
≤  xy  <
(d)
≤  yx  <
(e)
≤  x² + y²  <
(f)

≤   < (to 3 d.p.)

Question 6

Hannah measures with a ruler that has marks every 0.5 cm. She says that the lengths of the side of a rectangle are 8.5 cm and 12 cm correct to the nearest 0.5 cm.
Calculate the minimum possible values of:

(a)

the perimeter of the rectangle,

cm
(b)

the area of the rectangle.

cm²
Question 7

The radius of a circle is said to be 18 m ± 0.5 m.

(a)

What is the maximum possible radius of the circle?

m
(b)

Write the circumference of the circle in the form a m ± b m, giving the values of a and b correct to 2 decimal places.

m ± m (to 2 d.p.)

109.96 m ≤ circumference < 116.24 m (to 2 d.p.)

a = the average of 109.96 and 116.24 = 113.10

b = 113.10 – 109.96 = 3.14

so circumference = 113.10 m ± 3.14 m (to 2 d.p.)

Question 8

A square has sides with lengths that are given as 1.8 cm ± 0.05 cm.

(a)

Write the perimeter of the square in the form a cm ± b cm.

cm ± cm
(b)

Write the area of the square in the form a cm² ± b cm².

cm² ± cm² (to 2 d.p.)
Question 9

The area of a square is given as 14 m² correct to the nearest m².
Calculate the maximum and minimum possible values for the length of the side of the square, rounding your answers to 3 significant figures.

m ≤  side of square  < m (to 3 s.f.)
Question 10

A circle has area 50 cm² ± 2 cm². What are the maximum and minimum possible values of the radius of the circle?
Round your upper and lower bounds to 2 decimal places.

cm ≤  radius of the circle  < cm (to 2 d.p.)
Question 11

Mary, Arvind and Nesta wrote a mass to different numbers of decimal places.

Mary wrote 1.7 to 1 decimal place.
Arvind wrote 1.748 to 3 decimal places.

(a)

Can both of them be correct?

 Mary's figure means 1.65 ≤ mass < 1.75 Arvind's figure means 1.7475 ≤ mass < 1.7485
Since these intervals overlap, Mary and Arvind can both be correct (for example, if the mass was 1.7478).
(b)

Nesta wrote 1.7474 to 4 decimal places.

Can both Nesta and Arvind be correct?

 Nesta's figure means 1.74735 ≤ mass < 1.74745 Arvind's figure means 1.7475 ≤ mass < 1.7485
Since these intervals do not overlap, Nesta and Arvind cannot both be correct.
(c)

They wrote a volume to different numbers of decimal places.

Mary wrote 2.6 to 1 decimal place.
Arvind wrote 2.65 to 2 decimal places.

They are both correct.

What are the lower and upper bounds of the volume?

≤ volume <
(d)

The mass of a block is 0.10 kilograms to 2 decimal places.

Its volume is 0.02 litres to 2 decimal places.

density =

Work out the minimum and maximum possible density of the block.

kg/litre ≤ density < kg/litre
Question 12

The exact formula to convert a temperature measured in degrees Celsius (C) to degrees Fahrenheit (F) is

F = + 32

An approximate formula often used for this conversion is

F = 2C + 30

(a)

The temperature is 20.0 °C, measured to the nearest tenth of a degree.
What would you obtain as the upper and lower bounds of the temperature in degrees Fahrenheit if you used the approximate formula for the conversion?

°F ≤ temperature < °F
(b)

What are the upper and lower bounds of the temperature in degrees Fahrenheit obtained by using the exact formula to convert 20.0 °C to Fahrenheit?

°F ≤ temperature < °F
(c)

Calculate the percentage error when the approximate formula, rather than the exact formula, is used to convert 30.0 °C to Fahrenheit.

(to 2 d.p.)

Using the exact formula, 30.0 °C converts to 86 °F.
Using the approximate formula, 30.0 °C converts to 90 °F.

The percentage error = × 100 = 4.65% (to 2 d.p.)

(d)

For what range of values of C does the approximate formula give an answer in degrees Fahrenheit which differs by a percentage error of no more than 5% from the exact formula?
Give your answers to one decimal place.

≤ C < (to 1 d.p.)
 –5% ≤ × 100 < 5%

–0.05(1.8C + 32) ≤ 0.2C – 2 < 0.05(1.8C + 32)

–0.09C – 1.6 ≤ 0.2C – 2 < 0.09C + 1.6

0.4 < 0.29C and 0.11C < 3.6

1.379310345 ≤ C < 32.7272727272

1.4 ≤ C < 32.7 (to 1 d.p.)