Lesson 5: Application of formulae

Introduction

You'll be pleased to hear that your new substitution skills will prove extremely useful within maths. Solving equations, plotting graphs, calculating areas, volumes and surface areas of shapes, and working out compound interest all involve algebraic substitution.

As far as formulae are concerned, they're everywhere! We use them to calculate loads of important and useful stuff in areas such as science and engineering, computing, manufacturing and finance. You'll find some examples below, though we barely scratch the surface here. Enjoy!

SECTION A

Using formulae within maths

You're expected to remember common formulae off by heart, though some may be given to you in exams. If the formula is not given, always write it out before you do your substitution.

Example 1

The formula for the area of a triangle is \(~A=\Large\frac{bh}{2}~\),
where \(~b~\) is the length of the base and \(~h~\) is the perpendicular height.
Work out the area of a triangle that has a base of 3 metres and a perpendicular height of 2.8 metres.

Assign values to variables: \(~b=3~\) and \(~h=2.8~\)

Substitute: \(~A=\Large\frac{(3)(2.8)}{2}~\)

Evaluate: \(~A=4.2~\) square metres

When you study geometry, you'll understand how some of these formulae come about and how to choose units of measurement. Values to be assigned to the variables are usually shown on a diagram in geometry questions.

Example 2

The compound interest formula is \(~P\times\Big(1+\Large\frac{i}{100}\normalsize~\Big)^n\)

\(P~\) is the initial (principle) amount, \(~i~\) is the rate of interest and \(~n~\) is the number of years.

The sum of £2000 in a bank account earns interest at 3.5% per annum for 4 years. No withdrawals are made from the account during this period. What is the balance on the account after 4 years?

Assign values to variables: \(~P=2000~\), \(~i=3.5~\) and \(~n=4~\)

Substitute: \(~2000\times\Big(1+\Large\frac{3.5}{100}\normalsize~\Big)^4\)

Division inside brackets: \(~2000\times(1+0.035)^4\)

Addition inside brackets: \(~2000\times(1.035)^4\)

Indices then multiplication: \(~£2295.05\)

This can be worked out on a calculator in one step as all but the most basic calculators follow the rules of BIDMAS. You still have to be a bit careful, though. If in doubt, break it down into steps using your own knowledge of the order of operations.

Again, you need to study repeated percentage change in order to properly understand it. But at least you know you can do it and you're more aware of how areas of maths are connected, which is really important.

Practise to master

SECTION A

Solve these problems using your knowledge of substitution and formulae.

Question 1
The formula for the circumference of a circle is \(~C=\pi d~\), where \(~\pi~\) is roughly equal to 3.14 and \(~d~\) is the diameter of the circle. Calculate the circumference of a circular flower bed which has a diameter of 4 metres.

\(C=3.14\times 4=12.56\) metres
A more accurate value for \(~\pi~\) can be found on your calculator.

Question 2
The formula for calculating compound interest is \(~P\times\Big(1+\Large\frac{i}{100}\normalsize~\Big)^n\)
\(P~\) is the initial (principle) amount, \(~i~\) is the rate of interest and \(~n~\) is the number of years.
An investor buys stocks and shares for £6000. If they increase in value at a rate of 7% per year, how much will they be worth after 5 years?

\(~6000\times\Big(1+\Large\frac{7}{100}\normalsize~\Big)^5\)
\(~6000\times(1+0.07)^5\)
\(~6000\times(1.07)^5\)
\(£8415.31\)

Question 3
The formula for the area of a circle is \(~A=\pi r^2~\), where \(~\pi~\) is roughly equal to 3.14 and \(~r~\) is the radius of the circle. Calculate the area of a flat circular roof which has a radius of 5 metres.

\(A=3.14\times 5^2=78.5\) square metres
Again, you'll learn about units of measurement in geometry.

Question 4
The formula for calculating percentage increase is
\(\large\frac{New~amount~-~Original~amount}{Original~amount}~\normalsize \times 100\)
The price of a telescope increases from £34.99 to £39.99
What is this as a percentage increase (to the nearest 1%)?

Percentage increase \(=~\large\frac{39.99-34.99}{34.99}\normalsize \times 100~=~14\)%
Mathematicians are not crazy about wordy formulae but they do crop up.

Question 5
The formula for the volume of a cylinder is \(~V=\pi r^2h\)
\(\pi~\) is roughly equal to 3.14, \(~r~\) is the radius of the base and \(~h~\) is the height.
Calculate (to 1 decimal place) the volume of a cylindrical barrel with a base radius of 0.8 metres and a height of 1.2 metres.

\(V=3.14\times 0.8^2\times 1.2~=~2.4\) cubic metres

Question 6
The formula for calculating depreciation is \(~P\times\Big(1-\Large\frac{d}{100}\normalsize~\Big)^n\)
\(P~\) is the initial (principle) amount, \(~d~\) is the rate of depreciation and \(~n~\) is the number of years. A car is purchased for £2500 at the end of 2011. It depreciates in value at a rate of 15% per annum. What is its value at the end of 2015 (to the nearest pound)?

\(~2500\times\Big(1-\Large\frac{15}{100}\normalsize~\Big)^4\)
\(~2500\times(1-0.15)^4\)
\(~2500\times(0.85)^4\)
\(£1305\)

Question 7
The formula for the volume of a sphere is \(~V=\frac{4}{3}~\pi r^3\)
\(\pi~\) is roughly equal to 3.14, \(~r~\) is the radius of the sphere.
Calculate (to 1 decimal place) the volume of a giant beach ball with a radius of 2 metres.

\(~V=\frac{4}{3}~\pi (2)^3~=~33.5\) cubic metres

Question 8
The formula for calculating repeated percentage increase is
\(P\times\Big(1+\Large\frac{i}{100}\normalsize~\Big)^n\)
\(P~\) is the initial amount, \(~i~\) is the rate of increase and \(~n~\) is the number of years.
The population of a remote island grows by 2% per year. If the population is 25,000 now, estimate what it will be in 10 years time.

\(~25000\times\Big(1+\Large\frac{2}{100}\normalsize~\Big)^{10}\)
\(~25000\times(1+0.02)^{10}\)
\(~25000\times(1.02)^{10}\)
\(30,475\)

Question 9
The formula for the surface area of a sphere is \(~A=4\pi r^2\)
\(\pi~\) is roughly equal to 3.14, \(~r~\) is the radius of the sphere.
Assuming the earth is a sphere with a radius of about 6400 kilometres, calculate its surface area in square kilometres.

\(~A=4(3.14)(6400)^2~=~514,457,600\) square kilometres
That's over half a billion square kilometres!

Question 10
The formula for calculating percentage decrease is
\(\large\frac{Actual~decrease}{Original~amount}~\normalsize \times 100\)
The price of a dress is reduced by £5. It was originally £29.50.
What percentage discount has been applied (to the nearest 1%)?

Percentage discount \(=~\large\frac{5}{29.5}\normalsize \times 100~=~17\)%

SECTION B

Using formulae within science

As you progress in science, you will come across all sorts of useful formulae. With each formula you use, you will need to know what each variable represents and what its value is. From there, it's basically just a case of applying your substitution skills. I'll give you one example but what an example it is! It's the world's most famous formula.

Example

Albert Einstein's \(~E=mc^2\),
 \(~E~\) is the amount of energy contained within stuff (in Joules)
 \(~m~\) is the mass of that stuff (in kg)
 \(~c~\) is the speed of light (300,000,000 metres per second!)
Now I weigh 85 kg. I'm quite tall! So let's work out the amount of energy contained within me.

Assign values to variables: \(~m=85~\) and \(~c=300,000,000~\)

Substitute: \(~E=(85)(300,000,000)^2~=~7.65\times 10^{18}\) Joules

That's 765 with 16 zeros after it!
If only I could harness that energy, this website would be finished in no time!

Practise to master

SECTION B

Solve the following science problems.

Question 1
Consider an object travelling at a constant speed, \(~S=\large\frac{D}{T}~\)
\(S~=~\)speed, \(~D~=~\)distance travelled, \(~T~=~\)time taken
Calculate the speed of a car which travels 81 miles in one and a half hours.

\(D=81~\) and \(~T=1.5~\)
\(S=\large\frac{81}{1.5}\normalsize~=~54\) miles per hour

Question 2
Consider an object moving with constant acceleration.
\(v=u+at~\)
\(u~=~\)initial speed, \(~v~=~\)final speed, \(~a~=~\)rate of acceleration, \(~t~=~\)time taken
A motorbike passes a sign at a speed of 35 metres per second then accelerates at a rate of 2 metres per second per second. What is its speed 12 seconds after passing the sign?

\(u=35~\), \(~a=2~\) and \(~t=12~\)
\(v=35+2(12)=35+24=59~\) metres per second

Question 3
The density of a material is given by the formula \(~D=\large\frac{M}{V}~\)
\(D~=~\)density, \(~M~=~\)mass, \(~V~=~\)volume
Calculate the density of an ice cube which has a mass of 2.79 grams and a volume of 3 cubic centimetres.

\(M=2.79~\) and \(~V=3~\)
\(D=\large\frac{2.79}{3}\normalsize~=~0.93\) grams per cubic centimetre
Since water has a higher density of 1 gram per cubic centimetre, ice floats in it!

Question 4
Consider an object moving with constant acceleration.
\(s=ut+\frac{1}{2}~at^2~\)
\(s~=~\)displacement (distance travelled), the others you know.
A man jumps out of a plane, accelerating at 10 metres per second per second.
How far has he fallen after 30 seconds?

\(u=0~\), \(~a=10~\) and \(~t=30~\)
\(s=(0)(30)+\frac{1}{2}~(10)(30)^2~= 4500\) metres
Oooh, but why would you do that?!

Question 5
\(W=M\times G~\)
\(W~=~\)weight, \(~M~=~\)mass, \(~G~=~\)gravitational field strength
Gravitational field strength on the moon is 1.6 N/Kg
What is an astonaut's weight (in Newtons) on the moon if his mass is 85 kg?

\(M=85~\) and \(~G=1.6~\)
\(W=85\times 1.6~=~136~\) N
You've no idea what this means but it feels good to be able to work it out, right?!

Question 6
Consider an object moving with constant acceleration.
\(s=\frac{1}{2}(u+v)t~\)
A cyclist travelling at a speed of 3 metres per second realises he's late for work and starts to accelerate. After 15 seconds, he reaches a speed of 5 metres per second. How far did he travel in these 15 seconds?

\(u=3~\), \(~v=5~\) and \(~t=15~\)
\(s=\frac{1}{2}(3+5)\times 15~=~60~\) metres
The formulae in questions 2, 4 and 6 are called the suvat equations (you can probably guess why). They're supposed to be quite advanced ... really?!

Question 7
Kinetic energy\(~=\frac{1}{2}~mv^2\)
\(m~=~\)mass, \(~v~=~\)velocity
Work out the kinetic energy (the units will be J for Joules) of a bowling ball which has a mass of 5 kg and is travelling at a velocity of 6 metres per second.

\(m=5~\) and \(~v=6~\)
Kinetic energy\(~=\frac{1}{2}~(5)(6)^2~=90\) J
Kinetic energy is basically the energy of a moving object.

Question 8
Consider an object moving with constant acceleration.
\(v=\sqrt{u^2+2as}\)
A sports car is travelling on a German motorway at 45 metres per second when it accelerates at 7 metres per second per second to overtake a boy racer. What speed does the sports car reach after accelerating over a distance of 85 metres?

\(u=45~\), \(~a=7~\) and \(~s=85~\)
\(v=\sqrt{45^2+2(7)(85)}~=~56.7\) metres per second
Science tends to use metres and seconds rather than miles and hours.

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