Transcript A3 Formulae - Haringeymath's Blog

KS4 Mathematics
A3 Formulae
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Contents
A3 Formulae
A A3.1 Substituting into formulae
A A3.2 Problems that lead to equations to solve
A A3.3 Changing the subject of a formula
A A3.4 Manipulating more difficult formulae
A A3.5 Generating formulae
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Formulae
A formula is a special type of equation that links two or more
physical variables.
For example in the formula,
P = 2(l + w)
P represents the perimeter of a rectangle and l and w
represent its length and width.
We can use this formula to work out the perimeter of any
rectangle given its length and width.
We do this by substituting the values we are given into the
formula.
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Formulae
Because formulae deal mainly with real-life quantities such as
length, mass, temperature or time, the given variables often
have units attached.
Units shouldn’t be included in the formula itself.
The units that have to be used are usually defined in the
formula. For example,
d
S=
t
This formula doesn’t mean much unless we say “S is the
average speed in m/s, d is the distance travelled in metres,
and t is the time taken in seconds”.
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Formulae
d
Use the formula S =
to find the speed of a car that
t
travels 2 km in 1 minute and 40 seconds.
Write the distance and the time using the correct units before
substituting them into the formula,
2 kilometres = 2000 metres
1 minute and 40 seconds = 100 seconds
Now substitute these numerical values into the formula,
2000
S=
100
We can write the
units at the end.
= 20 m/s
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Substituting into formulae
h
w
l
The surface area S of a cuboid is given by the formula
S = 2lw + 2lh + 2hw
where l is the length, w is the width and h is the height.
What is the surface area of a cuboid with a length of
1.5 m, a width of 32 cm and a height of 250 mm?
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Substituting into formulae
What is the surface area of a cuboid with a length of
1.5 m, a width of 32 cm and a height of 250 mm?
Before we can use the formula we must write all of the
amounts using the same units.
l = 150 cm, w = 32 cm and h = 25 cm
Next, substitute the values into the formula without the units.
S = 2lw + 2lh + 2hw
= (2 × 150 × 32) + (2 × 150 × 25) + (2 × 25 × 32)
= 9600 + 7500 + 1600
= 18,700 cm2
Don’t forget to write the units in at the end.
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Substituting into formulae
The distance d, in metres, that an object falls after being
dropped is given by the formula,
d = 4.9t2
where t is the time in seconds.
Suppose a boy drops a rock from a 100 metre high cliff.
How far will the rock have fallen after:
a) 2 seconds
b) 3 seconds
c) 5 seconds?
When t = 2,
When t = 3,
When t = 5,
d = 4.9 × 22
= 4.9 × 4
= 19.6 metres
d = 4.9 × 32
= 4.9 × 9
= 44.1 metres
d = 4.9 × 52
= 4.9 × 25
= 122.5 metres
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Substituting into formulae
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Contents
A3 Formulae
A A3.1 Substituting into formulae
A A3.2 Problems that lead to equations to solve
A A3.3 Changing the subject of a formula
A A3.4 Manipulating more difficult formulae
A A3.5 Generating formulae
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Problems that lead to equations to solve
Formulae are usually (but not always) arranged so that a
single variable is written on the left-hand side of the equals
sign. For example, in the formula
v = u + at
v is called the subject of the formula.
If we are given the values of u, a and t, we can find v by
substituting these values into the formula.
Suppose instead that we are given to values of v, u and a, and
asked to find t.
When these values are substituted we are left with an
equation to solve in t.
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Problems that lead to equations to solve
For example, suppose v = 20, u = 5 and a = 3. Find t.
Substituting these values into v = u + at gives us the equation,
20 = 5 + 3t
We can then solve this equation as usual.
swap both sides:
subtract 5 from both sides:
divide both sides by 3:
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5 + 3t = 20
3t = 15
t = 5 seconds
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Problems that lead to equations to solve
a
h
b
The formula used to find the area A of a trapezium with
parallel sides a and b and height h is:
A=
1
(a + b)h
2
What is the height of a trapezium with an area of 40 cm2
and parallel sides of length 7 cm and 9 cm?
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Problems that lead to equations to solve
Substituting A = 40, a = 7 and b = 9 into A =
Simplifying,
40 =
1
(7 + 9)h
2
40 =
1
×16h
2
1
(a + b)h gives
2
40 = 8h
swap both sides:
divide by 8:
8h = 40
h=5
So the height of the trapezium is 5 cm.
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Problems that lead to equations to solve
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Contents
A3 Formulae
A A3.1 Substituting into formulae
A A3.2 Problems that lead to equations to solve
A A3.3 Changing the subject of a formula
A A3.4 Manipulating more difficult formulae
A A3.5 Generating formulae
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The subject of a formula
Here is a formula you may know from physics:
V = IR
where V is voltage, I is current and R is resistance.
V is called the subject of the formula.
The subject of a formula always appears in front of the
equals sign without any other numbers or operations.
Sometimes it is useful to rearrange a formula so that one
of the other variables is the subject of the formula.
Suppose, for example, that we want to make I the subject
of the formula V = IR.
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Changing the subject of the formula
V is the subject
of this formula
The formula:
V = IR
can be written as a function diagram:
I
×R
V
÷R
V
The inverse of this is:
I
So:
V
I= R
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I is now the subject
of this formula
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Matchstick pattern
Look at this pattern made from matchsticks:
Pattern
Number, n
1
2
3
4
Number of
Matches, m
3
5
7
9
The formula for the number of matches, m, in pattern
number n is given by the formula:
m = 2n + 1
Which pattern number will contain 47 matches?
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Changing the subject of the formula
m is the subject
of this formula
The formula:
m = 2n + 1
can be written as a function diagram:
n
×2
+1
m
The inverse of this is:
n
÷2
–1
or
m–1
n=
2
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m
n is the subject of
this formula
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Changing the subject of the formula
To find out which pattern will contain 47 matches,
substitute 47 into the rearranged formula.
m–1
n=
2
47 – 1
n=
2
n=
46
2
n = 23
So, the 23rd pattern will contain 47 matches.
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Changing the subject of the formula
We can also change the subject by performing the same
operations on both sides of the equals sign. For example, to
make C the subject of
9C
F=
+ 32
5
subtract 32:
multiply by 5:
divide by 9:
9C
F – 32 =
5
5(F – 32) = 9C
5(F – 32)
=C
9
5(F – 32)
C=
9
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Change the subject of the formula 1
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Find the equivalent formulae
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Contents
A3 Formulae
A A3.1 Substituting into formulae
A A3.2 Problems that lead to equations to solve
A A3.3 Changing the subject of a formula
A A3.4 Manipulating more difficult formulae
A A3.5 Generating formulae
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Formulae where the subject appears twice
Sometimes the variable that we are making the subject of a
formula appears twice. For example,
S = 2lw + 2lh + 2hw
where S is the surface area of a cuboid, l is its length, w is its
width and h is its height.
Make w the subject of the formula.
To do this we must collect all terms containing w on the same
side of the equals sign.
We can then isolate w by factorizing.
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Formulae where the subject appears twice
S = 2lw + 2lh + 2hw
Let’s start by swapping the left-hand side and the right-hand
side so that the terms with w’s are on the left.
2lw + 2lh + 2hw = S
subtract 2lh from both sides:
2lw + 2hw = S – 2lh
factorize:
w(2l + 2h) = S – 2lh
divide by 2l + 2h:
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S – 2lh
w =
2l + 2h
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Formulae involving fractions
When a formula involves fractions we usually remove these
by multiplying before changing the subject.
For example, if two resistors with a resistance a and b ohms
respectively, are arranged in parallel their total resistance R
ohms can be found using the formula,
1
1
1
=
+
a
b
R
aΩ
bΩ
Make R the subject of the formula
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Formulae involving fractions
1
1
1
=
+
a
b
R
multiply through by Rab:
Rab Rab Rab
=
+
a
b
R
simplify:
ab = Rb + Ra
factorize:
ab = R(b + a)
divide both sides by a + b:
ab
=R
a+b
ab
R=
a+b
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Formulae involving powers and roots
The length c of the hypotenuse of a right-angled triangle is
given by
c = √a2 + b2
where a and b are the lengths of the shorter sides.
Make a the subject of the formula
square both sides:
subtract b2 from both sides:
square root both sides:
c2 = a2 + b2
c2 – b2 = a2
√c2 – b2 = a
a = √c2 – b2
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Formulae involving powers and roots
The time T needed for a pendulum to make a complete swing
is
T = 2π gl
where l is the length of the pendulum and g is acceleration
due to gravity.
Make l the subject of the formula
When the variable that we wish to make the subject appears
under a square root sign, we should isolate it on one side of
the equation and then square both sides.
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Formulae involving powers and roots
l
g
T = 2π
divide both sides by 2π:
T
=
2π
square both sides:
T2
l
=
g
4π2
multiply both sides by g:
T2g
= l
2
4π
l
g
T2g
l=
4π2
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Change the subject of the formula 2
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Contents
A3 Formulae
A A3.1 Substituting into formulae
A A3.2 Problems that lead to equations to solve
A A3.3 Changing the subject of a formula
A A3.4 Manipulating more difficult formulae
A A3.5 Generating formulae
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Writing formulae
Write a formula to work out,
1) the cost, c, of b boxes of crisps at £3 each
c = 3b
2) the distance left, d, of a 500 km journey after travelling k km
d = 500 – k
3) the cost per person, c, if a meal costing m pounds is shared
between p people
m
c=
p
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Writing formulae
4) the number of seats in a theatre, n, with 25 seats in each
row, r
n = 25r
5) the age of a boy Andy, a, if he is 5 years older than his sister
Betty, b
a=b+5
6) the average weight, w, of Alex who weighs a kg, Bob who
weighs b kg and Claire who weighs c kg.
a+b+c
w=
3
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Writing formulae
A window cleaner charges a £10 call-out fee plus £7
for every window that he cleans. Write a formula to
find the total cost C when n windows are cleaned.
C = 7n + 10
Using this formula, how much would it cost to
clean all 105 windows of Formula Mansion?
We substitute the value into the formula,
C = 7 × 105 + 10
= 735 + 10
= 745
It will cost £745.
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Writing formulae
C = 7n + 10
At another house the window cleaner made £94.
How many windows did he have to clean?
We substitute this value into the formula to give an equation,
94 = 7n + 10
swap both sides:
7n + 10 = 94
subtract 10 from both sides:
7n = 84
n = 12
He cleaned 12 windows.
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Using formulae to write mathematical rules
When conducting a mathematical investigation, it is usually
necessary to use formulae to write rules and generalizations.
For example, Sophie is investigating patterns of shaded
squares on a numbered grid.
She starts by looking at arrangements of numbers in two by
two squares on a 100 square grid. For example,
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34
35
44
45
She notices that the sum of the numbers
in a two by two square is always equal
to four times the number in the top lefthand square plus 22. Sophie writes this
as a formula.
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Number grid patterns
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