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Unit 3 - 1

Vectors

A vector may be considered as a set of
instructions for moving from one point to
another
...

The length of AB is proportional to the
magnitude of u and the arrow shows the
direction of u
...

The vector u can be represented in magnitude

C

AB and CD both represent the same vector u
A vector does not have a position – only magnitude and direction, so many
different directed line segments may represent this vector
...

thus

D

u

D

x

v

2

 2
 2
AB =   or u =  
 4
 4

Similarly:

 −3 
 −3 
CD =   or v =  
2
2

-3

C
x

Magnitude of a Vector in 2 dimensions:
We write the magnitude of u as | u |

The magnitude of vector u is |u| (the length of PQ)
y

 x
u =   then u = x 2 + y 2
 y

The length of PQ is written as

Q
u

2
8
PQ =   then PQ = 82 + 42
 4

4

The magnitude of a vector is the length of the
directed line segment which represents it
...

Examples:

Solutions:

 3
1
...


PQ = 82 + 42 = 80 = 8
...
PQ =   and P is (2, 1), find co-ordinates of Q
 3

2
...
P is (1, 3) and Q is (4, 1) find PQ

3
...


Examples:
Displacement, force, velocity, acceleration
...


Examples:
Temperature, work, width, height, length, time of day
...

Note that the z axis is the vertical axis
...
(z-component)

 xB − x A 


In general: AB = y B − y A ,


z −z 
A 
 B

 4
 
In component form: AB = 3
 
 2
 
Magnitude of a 3 dimensional vector

|u| =

( xB − x A )2 + ( y B − y A )2 + ( z B − z A ) 2

This is the length of the vector
...

AB2 = AR2 + BR2
= (AP2 + PR2) + BR2
= ( xB − x A ) + ( y B − y A ) + ( z B − z A )
2

2

2

and if u = AB
then the magnitude of u, | u | = length of AB
This is known as the
Distance formula for 3 dimensions

 xB − x A 


Recall that since: AB = y B − y A , then


z −z 
A 
 B
 x
 
2
2
2
if u =  y  then u = x + y + z
z
 

Since

x = xB – xA and y = yB – yA and z = zB – zA

Example:

1
...
If u = −2
  Find u
 2
 

| AB | =

u=

( 5 − 1)2 + ( 6 − 3)2 + ( 4 − 2 )2 = 42 + 32 + 22 = 29

( 3)2 + ( −2 )2 + ( 2 )2 = 9 + 4 + 4 = 17

-2-

Unit 3 - 1

Vectors

The components of a vector are unique
...
e
...


e
...


 2x   6 

  
if
 y + 3  =  8  then x = 3, y = 5 and z = 3
 z −1   2 

  

Addition and subtraction of vectors

b

a

Vectors are added ‘nose to tail’

a+b

b

This is known as the Triangle Rule
...


3a
a

e
...


multiply a by 3 - written 3 a
Vector 3a has three times the length
but is in the same direction as a

In component form, each component will be
multiplied by 3
...

Scalar Multiples

If a vector is a scalar multiple of another vector,
then the two vectors are parallel, and differ only
in magnitude
...


2
 
a= 1 
 −3
 

then

6
 
3a =  3 
 −9 
 

 12 
 
v =  16 
 −4 
 

⇒

3
 
v = 4 4 
 
 −1

2
 −6 
2
 
 
 
u =  −1 and v =  3  then v = −3  −1
3
 
 
 
 −9 
3
v is a scalar multiple of u and so v is parallel to u
...


y

B

e
...
PQ = q − p (note the order)
b

The component form of a position vector
corresponds to the co-ordinates of the point
...

For three points A, B, C - if the line AB is
parallel to BC, since B is common to both lines,
A, B and C are collinear
...


Show line segments are parallel
(ie
...


Example: A is (0, 1, 2), B is (1, 3, –1) and C is (3, 7, –7)
Show that A, B and C are collinear
...

Since B is a common point, then A, B and C are collinear

Ensure there is a COMMON point and
state it
...

C (12, 7, -1)

Example:

Find the ratio in which B divides AC
...


2
2
AB
3
 
 
AB = 3  2  and BC = 2  2  So,
=
or AB : BC = 3 : 2
2
BC
 −1
 −1
 
 

A (2, -3, 4)

Solution: B divides AC in ratio of 3 : 2

Points dividing lines in given ratios
...
If A is (2, 1, –3) and B
is (16, 15, 11), find the co-ordinates of P
...

Note that QN is shown as –2 because the two line segments are MQ and QN, and
QN is in the opposite direction to MQ
...

M is (–3, –2, –1) and N is (0, –5, 2)
...


Q
3

-2
N (0, -5, 2)

M (-3, -2, -1)
Solution:

MQ
3
=
−2
QN

so − 2 MQ = 3QN

∴ –2(q – m) = 3(n – q)
–2q + 2m = 3n – 3q
q = 3n – 2m

 0 
 −3 
 0 
 −6 
 6 
 
 


 


q = 3  −5  − 2  −2  =  −15  −  −4  =  −11 
 2
 −1 


 


 
 
 6 
 −2 
 8 

Q is Q(6, –11, 8)

Example:
If P divides AB in the ratio m : n, show that p, the
position vector of P is given by:

p=

n

mb + na
m+ n

P

m

B

AP m
=
n
PB

so n AP = mPB

∴ n (p – a) = m (b – p)

np–na=mb–mp

A

np+mp =mb+na
(n + m) p = m b + n a

p=

m b+n a
m+ n
q
...
d

-5-

Unit 3 - 1

Vectors

Unit Vectors
Definition:

a
 
If AB =  b  then a2 + b2 + c2 = 1
c
 

A unit vector has a magnitude of 1

Unit Vectors i, j, k

z

The unit vectors in the directions of the axes,
y

OX, OY and OZ are denoted by:
1
 
i =  0 ,
 0
 

0
 
j =  1 ,
 
0

(0, 0, 1)

0
 
k = 0
 
1

k

j

(0, 1, 0)

i

(1, 0, 0)

O

x

Every vector can be expressed in terms of the
unit vectors i, j, k
...


Calculate a + b

Add the components:

2
...


Calculate | a |

| a | = √(3 + 2 + (–1) ) = √( 9 + 4 + 1 ) = √14

4
...


Express 2 a + 3 b in component form

2a+3b=

6
...

Find the relation between a and b

a2 + b2 + ( ½ ) 2 = 1

7
...


A practical explanation of this comes from physics
...


The scalar product is written as a
...


and defined as:

a
...

where θ is the angle between the vectors
...
b is a real number, the sign of which is

b

b

b

determined by the size of angle θ
...
b
An alternative form for the scalar product can
be derived using components
...
b = x1 x2 + y1 y2 + z1 z2

Where a = x1i + y1 j + z1k

b = x2i + y2 j + z2k

and

 x1 
 
a =  y1 
z 
 1
 x2 
 
b =  y2 
z 
 2

Perpendicular Vectors a
...
b = 0
then if neither a nor b are zero,
cos θ must be zero, so θ = 90°

a
θ = 90°
b

The vectors a and b are perpendicular

Examples:
1
...
b for |a| = 2, |b| = 5, θ = π/6
 2

 1

 −3 
 

a
...
Calculate a
...
Calculate p
...
b = |a| |b| cos θ

 4
 −1 
 
 
p =  −3  and q =  4 
 2
8
 
 

What can you deduce about p and q ?

a
...
q = x1x2 + y1y2 + z1z2 = 4 x (–1) + (-3) x 4 + 2 x 8 = 0
Since neither p nor q are zero, then p and q are perpendicular
...
b = a b cos θ

a
...
b = x1 x2 + y1 y2 + z1 z2

Assuming that neither a nor b are zero
...
b = 0 ⇔ θ = 90° or π/2
i
...
a is perpendicular to b
assuming a ≠ 0, b ≠ 0

hence

cos θ =

a
...
Choose your
vectors carefully
...


Calculate the size of the angle between the

vectors:

3
 2
 
 
p =  −1  and q =  3 
5
 
 
 2

p
...
q = 6 – 3 + 10 = 13

|p| = √(32 + (–1) 2 + 52) = √35
So

cos θ =

|q| = √(22 + 3 2 + 22) = √17

13
= 0
...
5329…)
35 17

Hence θ = 57
...
p
...


Calculate the size of the angle between vectors:

u = i + 3j – k

u
...
v = 2 – 9 + 5 = –2

and v = 2i – 3j - 5k
|u| = √(12 + 3 2 + (–1)2) = √11
So

cos θ =

|v| = √(22 + (–3) 2 + (–5)2) = √38

−2
= -0
...
0978…)
11 38

Hence θacute = 84
...
p
...
4° = 95
...
b < 0 ⇒ θ is obtuse (2nd quadrant) – because cos θ < 0
Example:
3
...


Calculate the size of angle ABC:

BA and BC

We need the scalar product of
z

 −2 −1   −3 

  
BA= a -b =  0−6  =  −6 
 5−( −8)   13 

  

y

Α (−2, 0, 5)

 7 −1   6 

  
BC = c - b =  9−6  =  3 
 4−( −8)   12 

  

B (1, 6, -8)

O
C (7, 9, 4)

 −3   6 
  
BA
...
 3  = −18−18+156=120
 13   12 
  

x

cos θ =

BA
...
5967…

So θ = 53
...
4°

Unit 3 - 1

Vectors

Some Results of the Scalar Product

Using: |a| |b| cos θ

a
...
a = a 2
i
...
j = k
...
i = | i | | i | cos 0° = | i| | i | x 1 = 1 x 1 x 1 = 1 where | i | = 1

or

Obtain equivalent result for j
...
k

i2 = j2 = k2 = 1

i
...
j = i
...
k = 0

Obtain equivalent result for j
...
k

Distributive Law
a
...
b + a
...


b

|a| = 3, |b| = 2, |c| = 4
...
(a + b + c )

2
60° 3
a

c
60°

a
...
a + a
...
c
= 32 + 3 x 2 x cos 60° + 3 x 4 x cos 60° (since |a||a|= a2 = 3 x 3 )
= 9 + 6 x ½ + 12 x ½
= 18
Example:
The vectors

a, b and c are defined as:
a = 3i + j + 4k
b = –2i + j – k
c = –i + 4j + 2k

a)

Solution:

a
...
b + a
...
c = 3 x (–1) + 1 x 4 + 4 x 2 = – 3 + 4 + 8 = 9
a
...
c = –9 + 9 = 0 But a
...
c = a
...
(b + c) = 0

hence b + c is perpendicular to a

Solutions:

Evaluate:
1
...
(i + j)

1
...
(i + j) = i
...
j = 1 + 0 = 1

2
...
(i + k)

2
...
(i + k) = j
...
k = 0 + 0 = 0

3
...


i2 + j2 + k2 = 1 + 1 + 1 = 3

4
...
(i + j + k)

4
...
(i + j + k) = i
...
j + i
...

Similarly we can deduce that the graph of the derived function from
y = cos x is y = - sin x

d
(cos x) = − sin x
dx

y

y

y= sin x

y= cos x

x

x

y= cos x

x

x

y= - sin x

y = sin x ⇒ d/dx (sin x) = cos x

y = cos x ⇒ d/dx (cos x) = –sin x

We can of course prove this using the limit formula
...


y = 2sin x

1
...


y = 1 – sin x

2
...


y = 1 + cos x

3
...


y = ½ cos x

4
...


y = sin x – cos x

5
...


y = 3sin x + 2cos x

6
...


y = x + cos x

7
...


y = √x – cos x

8
...


y = x2 + 2x – 3sin x

9
...


y =

1− x cos x
x

10
...
)

y = u3

d
n
n −1 d
(
...
)
(
...

In function notation:

If y = f(g(x)), a composite function,
then y = f(u) and u = g(x)

dy dy du
and
=
×
dx du dx
d
d
d
( f ( g ( x) ) = f ′(u )× ( g ( x) ) = f ′( g ( x)) ( g ( x))
dx
dx
dx
d
d
f (
...
)
dx
dx

that is

u = 3x + 1

n

dy dy du
=
×
⇒
dx du dx

so
so:

This is just another way of stating the rule above
...

Practical Application

Differentiate the bracket – with respect to the bracket
then multiply by the derivative of the bracket with respect to x
...
)
×
d (
...
) is the same function in each case
– the contents of the bracket
...


Solutions:

1
...


dy/dx = 4(x – 1)3 × 1 = 4(x – 1)3

2
...


dy/dx = 2(5x + 1)1 × 5 = 10(5x + 1)

3
...


dy/du = 3(4 – u2)2 × (–2u) = -6(4 – u2)2

4
...


dy/dt = –3(t3 – 5)–4 × 3t2 = –9t2(t3– 5)-4

5
...


y = (2x + 3)–1

6
...


dy/dx = –1(x2 + 2x)–2 × (2x + 2) = –(2x + 2)(x2 + 2x)–2

7
...


y = (t2 – t – 2)½

dy/dx = –1(2x + 3)–2 = –(2x + 3)–2

dy/dt = ½ (t2 – t – 2)–½ × (2t – 1)
dy/dt = ½ (2t – 1) (t2 – t – 2)–½

- 11 -

Unit 3 - 2

Further Differentiation and Integration

Chain Rule – Trigonometric functions
The Chain Rule also applies to trigonometric functions
...


1
...


2
...
)

or

y = ( …
...
)
y = ( …
...


Example:

1
...
) where (…) = 2x
So, dy/dx = cos 2x × 2

dy
d
(
...
)
dx
dx

1
...
)
= − sin(
...


y = (1 + cos x)3

This is ( … cos x)3
where (…) = 1 + cos x
So, dy/dx = 3(…)2 × (–sin x)

2
...
sin x )

dy/dx = –3 sin x (1 + cos x)2

dy
n −1 d
(
...
sin x )
dx
dx

n

3
...
cos x )
(
...
cos x )n

dy/dx = 3(sin x)2 × cos x
dy/dx = 3 cos x sin2 x

There will only be two functions at most, all you have to
do is identify them, and use the above rules
...


y = cos 5x

1
...


y = sin (2x – 3)

2
...


y = cos (x2 – 1)

3
...


y = √(sin x)

Hint: write as y = (sin x)

4
...


y = cos2 x

Hint: write as y = (cos x)2

5
...


y =

1
sin t

Hint: write as y = (sin t)–1

6
...


y =

3
4cost

Hint: write as ¾ (cos t)–1

7
...


y = sin 2x + cos 3x

8
...


y = √(1 + cos x)

9

dy/dx = ½(1 + cos x)–½ × (–sin x) = –½ sin x (1 + cos x)– ½

10
...


y = 2 sin x cos x Hint: write as y = sin 2x

1
x

½

Hint: write as y = (1 + cos x)½

1
Hint: write as y = x–1 – (sin x)–½
sinx

–½

× cos x = ½ cos × (sin x)

–½

10
...
dy/dx = 2 cos 2x

- 12 -

Unit 3 - 2

Further Differentiation and Integration

Integration – Standard Integrals - 1
We will be able to integrate functions that we recognise as
the result of a Chain Rule differentiation
...


∫ ( ax + b )

In principle –
1
...

3
...


Recognise the function as a Chain Rule derivative
...

Put in any necessary multipliers/divisors
Check the result by differentiation
...

So we need to have these two multipliers
in the denominator of the integrated function,
in order to cancel out upon differentiation
...

Examples:

Solutions: (Check by differentiation)

Integrate these functions: (Don’t forget the constant)

In each case consider what function it came from:

1
...


(x + 1)5 × (1/5) = 1/5 (x + 1)5 + c

2
...


(3x – 2)3 × (1/3) × (1/3) =

3
...


(x – 5)–1 × (-1)

4
...


(5 – 2x)–2 × (–1/2) × (–1/2) = 1/4 (5 – 2x)–2 + c

5
...


(2x + 1)

6
...


(1 – 4x)–1/2 × (–2) × (– ¼ ) = ½ (1 – 4x)–1/2 + c

7
...


(v + 4)3/2 × 2/3 = 2/3 (v + 4)3/2 + c

Straight line form is: 3(2t + 3)-4

8
...


(2x + 3) × 2 × ½ = (2x + 3)

Straight line form is 2(1 – t)–1/3

10
...


½

Straight line form is: (v + 4)½

3

( 2t + 3)

4

1

9
...

3

(1 − t )

- 13 -

3/2

½

1

/9 (3x – 2)3 + c

= – (x – 5)–1 + c

3/2

× 2/3 × ½ = 2/6 (2x + 1)

½

+c

= 1/3 (2x + 1)3/2 + c

Unit 3 - 2

Further Differentiation and Integration

Integration – Standard Integrals - 2
Integration of trigonometric functions, is just the reverse
of differentiation:

Since:

d
( sin x ) = cos x
dx

∫ cos x dx = sin x + c

and

and

d
( cos x ) = − sin x
dx

∫ sin x dx = − cos x + c
We can also integrate trigonometric functions that we
recognise as the result of a Chain Rule differentiation
...


So we need to have the multiplier
in the denominator of the integrated function,
in order to cancel out upon differentiation
...

2
...

4
...

Work out what it must have come from
...


Sounds complicated, but again, with a little practice, it
becomes second nature
...


3cos x

1
...


5sin x

2
...


sin 4x

3
...


5 cos 2x

4
...


3 sin ½ x

5
...


cos(x + 2)

6
...


sin(3x + 4)

7
...


sin 2x + cos 3x

8
...


t2 + 2cos 2t

9
...


y
π /2

∫0

sin x dx

π

The integral represents the area between the
curve and the x-axis
...


π /2

∫0

π

/2

The above integral
represents the shaded
area on the graph
...


y = sin x

1

2π

x

−1

π /2

sin x dx = [ − cos x ]0

π

=  − cos  − ( − cos 0 )
2


= – 0 – (–1) = 1

Examples:
π /4

1
...


∫

π

π

= [ − cos t + sin t ]0 = ( − cos π + sin π ) − ( − cos 0 + sin 0 )

(sin t + cos t ) dt

0

=

3
...

y

π

/2

−1

= 1+1= 2

We cannot integrate between 0 and π because the areas above and below
the x-axis will cancel out to zero
...


y = sin 2x

1

( −(−1) + 0 ) − ( −1 + 0 )

The second integral will be negative (below the x-axis) so we ignore the
negative sign (since an area is always positive)
...
However, by symmetry, the area below
the x-axis is the same as that above the x-axis, apart from the sign
...
Hence total shaded area = 2

- 15 -

Unit 3 - 3

The Exponential and Logarithmic Functions

Growth Function

Examples of growth functions:

This is of the form:

Bank Account – compound interest

A(n) = ka

n

£200 at 7% for 6 years
...
07

with a > 1

6

Population growth
9

A

Now 47,000 growth 3% per year
...
03

A=ka n

Appreciation

k

House cost £55 000 when purchased
...

0

Value after 25 years A = 55 000 × 1
...
95

Population decline
A
k

Was 20,000 declines 3% per year
...
97

A=kan

20

Depreciation
0

n

Car cost £23 000 depreciates 20% each year
...
8

Examples:

Solutions:

1
...


An open can is filled with 2 litres of cleaning
fluid, which evaporates at the rate of 30% per
week
...


After 1 week
After t weeks

Calculate how much fluid remains after 6
weeks
...


A population of 100 cells increases by 60%
per hour
...


After 6 weeks
2
...
This
means that a given mass of radium will decay
steadily and be halved in 1600 years
...
7 mls remain
...
7 mls

3
...
5)t / 1600 then put t = 1600 (half life)

which gives R(t) = 5 × 0
...
5 g which is correct
...
Using C0
for the initial amount of carbon-14 present
...
6

R(400) = 5(0
...
5)0
...
2 g

Calculate the mass remaining after 400 years
...


= 235
...
6

Check that, starting with 5g of radium, the
decay function for the mass after t years is

R(t ) = 5(0
...
7 mls remain

After h hours number of cells N(h) = 100 × 1
...


30% evaporation, means that 70% remains

4
...
5)t / 5720

- 16 -

Unit 3 - 3

The Exponential and Logarithmic Functions

The exponential function
An exponential function is of the form

a
where
If
If

Note: In general an exponential function will take the form:

x

A( x) = ab x

a is a constant
...


a > 0, the function is increasing (growth)
a < 0, the function is decreasing (decay)

a will represent an initial value
b will represent the multiplier
x will represent the variable

a may take any positive value
depends on situation function is modelling
...


ex

The number e crops up on many occasions in the natural world
...
718 282 828 …
...


Linking the exponential and
logarithmic functions
...


1 = a0

⇔

loga 1 = 0

Use these two relationships to simplify and evaluate logarithmic and

a = a1

⇔

loga a = 1

Examples:

exponential functions and expressions
...


log3 81 = 4

2
...


logy 20 = 4

3
...


logz 10 = ½

5
...
form: log2 4 = 2

5
...
Write in exp
...


102 = 100

7
...
form: log9 3 = ½

7
...
Write in exp
...


82/3 = 4

9
...
form: loga c = b

9
...
Solve: logx 9 = 2

10
...
Solve: log4 x = 0
...


40
...
Solve: log3 81 = x

12
...
Write in log form: 81 = 3
1

3
...
Write in log form: z

13
...
Solve: log10 x = 0
...

14
...
5

(3√8 )2 = 4

i
...


so x = 3
½

4 =x

√4 = x

x=2

x=7

=x

- 17 -

Use calculator

10 yx 0
...
162…

x = 3
...
p)

Unit 3 - 3

The Exponential and Logarithmic Functions

Rules of Logarithms

These are derived from the corresponding Rules of Indices

log a xy = log a x + log a y

a m × a n = a m+ n

x
= log a x − log a y
y

a m ÷ a n = a m−n

log a

(a )

log a x p = p log a x

m

p

= a mp

Proofs:

Let

loga x = m

loga y = n

1
...


this will enable you to simplify expressions

x/y = am ÷ an = am–n

so xy = am–n

⇒

loga x/y = m – n

loga x/y = m – n ⇒ loga x/y = loga x – loga y
3
...
log 7 + log 2

1
...
log 12 – log 2

2
...
log 6 + log 2 – log 3

3
...
log 2 + 2 log 3

4
...
2 log 3 + 3 log 2

5
...
log8 2 + log8 4

6
...
log5 100 – log5 4

7
...
log4 18 – log4 9

8
...
2 log10 5 + 2 log10 2

9
...
3 log3 3 + ½ log3 9

10
...
5 log8 2 + log8 4 – log8 16

11
...
log2 ( ½ ) – log2 ( ¼ )

12
...
loga x + loga 2 = loga 10

13
...
loga x – loga 5 = loga 20

14
...


loga x + 3loga 3 = loga 9

2

2

= log10 ( 25 × 4) = log10 100 = log10 102 = 2

= –1 – (–2) = 1

Solve for x:

15
...
3010…
...
6094 …
...
2

=

1
...


2nd Fn

ln/ex

1
...
4739…

50
...
7

e
You will need to use the above keys, when
solving exponential or logarithmic equations
...
3010
or

⇒

2nd Fn

log/10x

0
...
99986…

10

yx

0
...
99986…

Solving exponential equations
...


Take log10 of both sides
...
8613 …

20 = et

2
...


Changes to log form:

loge 20 = loge et
loge 20 = t loge e

(Always choose loge when dealing with
growth or decay functions with e as the base
because loge e, makes calculation simpler)

( but loge e = 1)

t = loge 20 = 2
...

Examples:

Solutions:

1
...
6 = 16
x

1
...
6x = 2

log10 0
...
6
2
...
65)t

2
...
65)t

x log10 0
...
36

0
...
65t

log10 0
...
65t

log10 0
...
65 t = log10 0
...
65

For what value of t does D(t) = 2

t = 12
...
Solve: e = 120

3
...
S(t) = 225 e–0
...


70 = 225 e–0
...
3111 = –0
...
596

70 ÷225 = e–0
...
3111 = loge e–0
...
3111 = e–0
...
3111 = –0
...
3111 ÷ (–0
...
243

Unit 3 - 3

The Exponential and Logarithmic Functions

Experiment and Theory

y

y = ax n

(polynomial)
or

y=2e x

2
0

y = ab x (exponential)

0

x

x

polynomial graph

both are similar
...


y

y=3x2

In experimental work, data can often be modelled
by equations of the form:

exponential graph

Proof:

y = axn
log y = log axn

log y = log a + x log b

This looks like:

This looks like:

Y = log a + n X

Y = log a + X log b

where n is the gradient
and log a is the y-intercept
...


log y = log a + log bx

log y = log a + n log x

An exponential graph is a straight line when x is
plotted against log y

log y = log abx

log y = log a + log xn

A polynomial graph is a straight line when log x
is plotted against log y

y = abx

where log b is the gradient
and log a is the y-intercept
...

Take logs of both sides of the given relationship
(base 10 or base e according to the question)
Equate log a to the y-intercept
...


Example:
The following data was obtained from an experiment
Logs were taken of
data as shown below

the

n

Suggested relation is y = ax

Take log10 of both sides log10 y = log10 ax

n

⇒
⇒

a graph was plotted – the line of best fit showing a
n
straight line
...

Find the values of

log10 y = log10 a + log10 xn
log10 y = log10 a + n log10 x ……
...
e
...
31

log10 a = 0
...
29 = 0
...
p
...
31

and gradient = 0
...
0 (1 d
...
)

So relationship is:

y = 2x0
...
e
...
04, 0
...
18, 0
...
31 = 0
...
35 = 0
...
29, log10 a = 0
...
3 = 2 (1 d
...
)
Again this gives the relationship of:

- 20 -

y = 2x0
...

The diameter (x millimetres) and gain in weight (y grams) were measured and recorded for each sponge
...


X (= log e x)

2
...
31

2
...
65

2
...
10

Y (= log e y )

7
...
60

7
...
70

9
...
00

A graph was drawn and is shown here
...
(2
...
0) and (3
...
0)
Substitute into log e y = log e a + b log e x
giving:

7
...
1 b …
...
0 = loge a + 3
...
0 = b

Hence relationship is:

Note:

…
...
e
...
7 so a = e0
...
0

and

a = 2
...
0 ( 1 d
...
)

You should be confident in applying the method in part (b) rather than relying on the
gradient and y-intercept, as in this case, you cannot determine the y-intercept
...
15

2
...
00

1
...
95

1
...
33

79
...
89

62
...
70

57
...
94
1
...
90

log10 y = log10 ab

1
...
86

log10 y = log10 a + log10 b x

1
...
82
1
...
78

Add a row to the table showing log10 y

1
...
74
1
...
95
2
...


2
...
10

2
...
15

2
...
00

1
...
95

1
...
33

79
...
89

62
...
70

57
...
92

1
...
81

1
...
78

1
...
93, 1
...
15, 1
...
92 = log10 a + 2
...
( 1 )

and:

1
...
93 log10 b …
...
16 = 2
...
93 log10 b
0
...
22 log10 b
log10 b = 0
...
727 = 5
...
p
...
92 – 2
...
3
log10a = 1
...
56
log10a = 0
...
36 = 2
...
3 (1 d
...
)

Hence relationship is:

y = 2
...
3) x

- 22 -

2
...


The wave function

π/
2

We can express a cos x + b sin x in the form of a
single wave
...


y = sin x

This single wave is called the wave function
...
You will always be given the
appropriate form in the question
...

Expand R cos ( x - α )

R cos ( x - α ) = R cos x cos α + R sin x sin α

Step 2
...

Square and add to obtain R

…
...
(2)

R2 sin2 α + R2 cos2 α = 52 + 32
R2(sin2 α + cos2 α ) = 52 + 32
Note: sin2 α + cos2 α = 1 so, R2 = 52 + 32

Step 4
...

This gives you tan α
...

Identify the quadrant for α by looking at the two
equations obtained in step 2
...


Step 6
...

Put it all together

R = √34

so tan α =

5
3

S
+

A
++

T

+
C

∴ 3 cos x + 5 sin x = √34 cos (x – 59)°

5
⇒ α = 59
...

3

α = 59°

Always use this method of setting out your working
...
Work it out !

- 23 -

Unit 3 - 4

The Wave Function a cos x + b sin x

We have shown that:
3 cos x + 5 sin x = √34 cos (x – 59)°

√34

5
y = √34 cos (x - 59)°

The combined waveform is a cosine wave, of

3

y = 3cos x

amplitude √34
periodicity – same as original waves (2π )
phase shift is 59° to the right
...


ii)

y = 5sin x

solve the equation
3 cos x + 5 sin x = constant

-5

Maximum and minimum values
Example:

Solution:

Find the maximum and minimum values of:

Express the two functions as a single function
– in the form of R cos (x ± α ) or R sin (x ± α )

3 cos x + 5 sin x
for 0 ≤ x ≤ 360°
and state the values of x at which they occur
...
e
...
e
...


Hence maximum value is √34 when x = 59°
and
minimum value is –√34 when x = 239°

Solving Equations
Solution:

Example:

Solve the equation: 3 cos x + 5 sin x – 2 = 0
for 0 ≤ x ≤ 360°

Express 3 cos x + 5 sin x in the form of R cos (x ± α ) or R sin (x ± α )
Since we have already done this above, we shall use the above result:
and express 3 cos x + 5 sin x as √34 cos (x – 59)°
The equation we have to solve becomes:
√34 cos (x – 59) = 2
∴ cos (x – 59) = 2/√34
∴ cos (x – 59) = 0
...
9°
cosine is positive, so angle lies in 1st or 4th quadrants
...
9° or 349
...
9 or x – 59 = 360 – 69
...


Solve for 0 ≤ x ≤ 180

6 cos (3x + 60) – 3 = 0

6 cos (3x + 60) = 3
cos (3x + 60) = 0
...


i)
ii)

i)

Express √3 cos x – sin x in the form k sin(x – α)
and hence solve the equation √3 cos x – sin x = 0 for 0 ≤ x ≤ 360
k sin (x – α) = k sin x cos α – k cos x sin α
comparing coefficients:

–k sin α = √3
k cos α = –1

k sin α = –√3
k cos α = –1

… (1)
… (2)

squaring and adding:

k2 = (√3) 2 + 12

k2 = 3 + 1 = 4

k=2

dividing:

tan α = √3

acute α = 60°

from (1) and (2)

sin α and cos α both negative, so α lies in 3rd quadrant
∴ α = 180 + 60° = 240°

√3 cos x – sin x = 2 sin (x – 240)

Hence:

ii)

Using

2 sin (x – 240) = 0

sin (x – 240) = 0

(x – 240) = – 180°, 0° , 180°, or 360°

∴ x = 60° or x = 240°

(because we are adding 240°, we need to make sure we cover all the range, so we need to consider the
solution –180° as well, we do not need to go any further back, since we would be then out of the range)
3
...


4 cos 2x + 3 sin 2x + 5

R cos(2x - α) = R cos 2x cos α + R sin 2x sin α
compare coefficients:

R sin α = 3
R cos α = 4

squaring and adding:

R2 = 32 + 42

dividing:

tan α = ¾

R2 = 25

R=5

acute α = 0
...
643) + 5
Maximum value is:

10 when (2x – 0
...
32 rad, 3
...
60 rad – discard – out of range)

Minimum value is:

0 when (2x – 0
...
89 rad or 5
...


- 25 -

- 26 -



---