Download Ch1 Algebra and functions Quadratic functions Equations and

Ch1
Algebra and
functions
Quadratic
functions
Integration
Differentiation
C1
Sequences
and Series
Equations and
Inequalities
Sketching
curves
Coordinate
geometry
Surds
Some numbers like√2, √3 cannot be written as a
decimal because they go on forever. Numbers like this
are called
surds.
For example √2 = 1.4142 . . .
2
You can manipulate surds by using the rule
√(a X b) = √a X √b
and vice versa
√a X √b = √(a X b)
√(4 X 5) = √4 X √5
√3 X √7 = √(3X7)
√12
=
√(4 X 3)
=
√4 X √3
=
2 X √3
=
2√3
3
5√6 – 2√24 + √294
=
4
5√6 – 2√(4X6) + √(49X6)
=
5√6 – 2 X √4 X √6 + √49 X √6
=
5√6 – 2 X 2 X √6 + 7 X √6
=
5√6 – 4√6 + 7√6
=
8√6
Another rule we can use to manipulate surds is
a
�b =
√a
√b
or vice versa
√a
√b
=
𝑎𝑎
�𝑏𝑏
√12
2
=
=
=
=
√(4X3)
2
√4 X √3
2
2 X √3
2
√3
5
√44
√11
=
6
44
�
11
=
√4
=
2
If we have a fraction with a surd on the bottom like
4
√11
we can rationalise the denominator. This means we
turn the bottom or denominator into a rational or non
surd. We might do this if we get a surd on the bottom
of our answer at the end of our question. This just
tidies the fraction up and it is better to give an answer
in this way.
1
√3
multiply top and bottom by √3
=
=
1 X √3
√3 X √3
√3
3
This is exactly the same number but now the surd is on
the top instead of the bottom.
7
1
2√5
multiply top and bottom by √5
=
=
=
1 X √5
2√5 X √5
√5
2 X 10
√5
20
This is exactly the same number but now the surd is on
the top instead of the bottom so it is tidier.
1
1 + √2
multiply top and bottom by (1 - √2)
=
=
=
1 X (1 - √2)
(1 + √2)(1 - √2)
1 - √2
1 + √2 - √2 + 2
1 - √2
3
Notice that this works because multiplying (1 +√2) by
(1 - √2) we end up with +√2 and -√2 which cancel
8
out.
1
1 + √2
multiply top and bottom by (1 - √2)
=
=
=
1 X (1 - √2)
(1 + √2)(1 - √2)
1 - √2
1 + √2 - √2 + 2
1 - √2
3
Notice that this works because multiplying (1 + √2) by
(1 - √2) we end up with +√2 and -√2 which cancel
out.
9
Chapter 2 – Quadratic equations
Quadratic equations are equations with ‘x2’s in.
An equation like 2x + 3 = 11 (a linear equation) has only
one answer that makes it true (x = 4).
Quadratic equations often (but not always) have two
solutions that make them true.
For example, take x2 – 5x + 6 = 0, there are two roots,
substituting x = 2 or x = 3 works.
x2
– 5x
+6
22
–5X2
+6
=0
√
32
–5X3
+6
=0
√
Notice that putting in any other number is not going to
get us zero and it is therefore not a root or solution.
There are three ways to solve quadratic equations
10
-
Factorisation
-
Completing the square
-
Using the formula.
Factorisation
Factorising means ‘solving by putting into brackets’.
If we multiply out two brackets
(x + 4)(x + 5)
=
x2 + 9x + 20
We notice that the number of ‘x’s is the same as the
two numbers added together and the number by itself
is the same as the two numbers added together.
4+5
=
9
4X5
=
20
This helps us when we go the other way and put a
quadratic equation into brackets.
11
The other thing we need to realise for solving
quadratic equations by factorisation is that if two
things multiply together and equal zero then either
the first thing or the second thing must equal zero. If
something X something else = 0
then either
something = 0
or
something else = 0
12
Solve the quadratic equation,
x2 – 5x + 6
=
0
-------------------------------------------------------------Factorise (put into brackets)
(x – 2)(x – 3) =
0
because (-2) + (-3) = -5 and (-2) X (-3) = +6
-------------------------------------------------------------Either
(x – 2)
=
0
or
(x – 3)
=
=
2
or
x
=
0
so
x
3
13
Completing the square
The second way of solving quadratic equations is by
completing the square.
If we multiply out a bracket
(x + 3)2
=
(x + 3)(x + 3)
x2 + 6x + 9
=
-------------------------------------------------------------(x + 3)2
x2 + 6x + 9
=
Now if we take away 9 from both sides
(x + 3)2 – 9
=
x2 + 6x
or
x2 + 6x
=
(x + 3)2 - 9
Notice that the number in the bracket is half of the
number in front of the x and the number by itself is
half that number squared.
14
We can complete the square on quadratic expressions
by looking at half the number in front of the x.
x2 + 8x
=
(x + 4)2 - 42
x2 + 8x
=
x2 - 10x
=
(x - 5)2 – (-5)2
x2 - 10x
=
(x - 5)2 - 25
(x + 4)2 - 16
Notice that it is – 25 not + 25!
x2 + 5x
=
x2 + 5x
=
5
X
2
5 2
5
(x + )2 – � �
2
2
5
25
2
4
(x + )2 -
5
2
=
25
4
15
Solve the quadratic equation x2 + 8x + 10 = 0 by
completing the square.
------------------------------------------------------------x2 + 8x + 10 = 0
Check the coefficient of x2 equals one
x2 + 8x + 10 = 0
Move the constant to the other side
x2 + 8x = - 10
Complete the square on the expression on the left hand
side.
(x + 4)2 – 42 = - 10
(x + 4)2 – 16 = - 10
Move the sixteen over to the other side
(x + 4)2 = - 10 + 16
(x + 4)2 = 6
Square root both sides and take away four
16
(x + 4) = +√6
or
(x + 4) = -√6
x = - 4 + √6
or
x = - 4 - √6
Solve the quadratic equation x2 - 12x + 7 = 0 by
completing the square.
-------------------------------------------------------------x2 – 12x + 7 = 0
Check the coefficient of x2 equals one
x2 – 12x + 7 = 0
Move the seven to the other side
x2 - 12x = - 7
Complete the square on the expression on the left hand
side
(x - 6)2 – 62 = - 7
(x - 6)2 – 36 = - 7
Move the 36 over to the other side and simplify
(x - 6)2 = - 7 + 36
(x - 6)2 = 29
Square root both sides and add six
(x - 6) = +√29
or
(x - 6) = - √29
x = + 6 + √29
or
x = + 6 - √29
17
Solve the quadratic equation 2x2 + 7x - 9 = 0 by completing the
square.
-----------------------------------------------------------------------2x2 + 7x - 9 = 0
Make the coefficient of x2 = 1 by dividing all the way through by two
7
9
2
2
x2 + x -
=0
Move the constant to the other side
7
9
2
2
x2 + x = +
Complete the square on the expression on the left hand side. Notice
7
that half of
2
7
is
.
4
7 2
7
(x - )2 – � � = +
4
4
7
49
4
16
(x - )2 –
Move the
49
16
=+
9
2
9
2
over to the other side and simplify
7
9
4
2
7
121
4
16
(x - )2 = +
(x - )2 =
+
49
16
Square root both sides and move the
7
121
4
16
(x - ) = + �
x=
18
7
4
121
+�
16
7
4
over
7
121
4
16
or
(x - ) = − �
or
x=
7
4
121
−�
16
To complete the square
1.
Make sure that the coefficient of x2 = 1, divide
through if necessary.
2.
Subtract the constant (number by itself) from both
sides.
3.
Complete the square for the quadratic expression.
4.
Get all the numbers on one side and simplify.
5.
Square root both sides.
19
Using the formula
The third way of solving quadratic equations is by using
the quadratic formula.
−b ± √ b 2 − 4ac
x=
2ac
Where a is the number in front of the x2, b is the
number in front of the x and c is the number by itself.
Notice that we will usually get two answers because of
the ± bit.
Solve the quadratic equation, 3x2 – 7x + 11 = 0
-------------------------------------------------------------a=+3
20
b = -7
c = +11
Solve the quadratic equation x2 - 12x + 7 = 0 by using the
quadratic formula.
x2 – 12x + 7 = 0
a=+1
b = - 12
c=+7
------------------------------------------------------------------
x=
−b ± √ b 2 −4ac
so
2ac
x=
x=
x=
−b + √ b 2 −4ac
2ac
x=
−b − √ b 2 −4ac
2ac
−(−12) + �(−12)2 − 4.1.7
2.1.7
−(−12) − �(−12)2 − 4.1.7
2.1.7
x=
(+12) + �(+144) − 28
x=
(+12) + �(+116)
x=
or
14
(+12) − �(+144) − 28
14
x = 1.63
x=
14
or
(+12) − �(+116)
14
x = − 0.09
21
A quadratic function will always have a distinctive
parabola shape. We can tell which way up it will be
when we draw it by looking at a, the number in front of
the x2.
f(x) = ax2 + bx + c
If
a
is
positive then
we will have
a
happy
graph.
Positive
people
SMILE!
If
a
is
negative then
we will have
a sad graph.
Negative
people
FROWN!
22
We can work out how many times the graph will cut
through the x axis and therefore how many roots we
have by looking at the discriminant or the b2 – 4ac bit
of the quadratic equation.
−b ± � (b 2 − 4ac)
x=
2a
If b2 – 4ac is
positive
then
we are adding
and
taking
away a positive
number so we
get
two
different
answers.
The
graph will cut
the x axis at
two
different
places
there
two
and
will
be
roots
or
solutions.
23
If
b2
–
equals
then
4ac
zero
we
are
adding
and
taking
away
nothing so we
get
the
same
answer
both
times.
The
graph will just
touch the x axis
once and there
will be only one
root
to
this
equation.
If b2 – 4ac is
less than zero
then we are
trying to
calculate the
square root of a
negative
number which
we can’t do, so
the graph won’t
go through the
x axis at all and
there will be no
solutions to this
equation.
24
Sketching quadratic functions
Sometimes it can be useful to complete the square on
quadratic functions especially if we are going to sketch
them.
Express the quadratic function f(x) = x2 + 4x + 9 in the
form
(x + A)2 + B.
-------------------------------------------------------------f(x) =
x2 + 4x + 9
f(x) =
{x2 + 4x} + 9
complete the square
f(x) =
{(x + 2)2 – 4} + 9
bring the numbers together
f(x) =
(x + 2)2 +5
25
Write the quadratic function y = 2x2 - 3x - 2 in the form
A(x + B)2 + C.
-------------------------------------------------------------f(x) =
2x2 – 3x - 2
take out a factor of 2
3
f(x) =
2[ x2 – x – 1 ]
f(x) =
2[ {x2 – x} - 1 ]
2
3
2
complete the square
f(x) =
3
9
4
16
2[ {(x - )2 –
bring the numbers together
3
9
4
16
3
25
4
16
f(x) =
2[ (x - )2 –
f(x) =
2[ (x - )2 –
get into the form asked for
f(x) =
26
3
50
4
16
2(x - )2 –
}–1]
–1]
]
This can help us when sketching quadratic graphs.
Take y = x2 – 3x - 10, rearranging by factorising
y = (x + 2)(x – 5)
tells us that the curve cuts through the x axis at the points
x = - 2 and x = + 5 (the roots). And rearranging by
completing the square
3
49
2
4
y = (x - )2 -
tells us that the lowest the graph goes is −
this happens when x = +
3
2
49
4
= - 12.25 and
. The smallest anything squared
can be is zero, so the smallest y can be happens when the
brackets equals zero. When the thing in the brackets is zero
y=−
49
4
and this happens when x = +
3
2
.
27
Chapter 3 – Equations and Inequalities
Simultaneous equations are equations with two letters
in like
2a + 3b
=
7
3a – 2b
=
4
There are an infinite number of solutions to just the
first equation by itself, (a = 5 and b = -1) works as
does (a = 0 and b = 7/3) but if we decide that our
solution needs to work for both equations at the same
time only one solution will work, (a = 2 and b = 1).
There are two ways to solve simultaneous equations
Elimination
and
Substitution
28
Elimination
To solve a simultaneous equation by elimination we
need to multiply and get rid of one of the letters by
adding or taking away the two equations.
Solve the simultaneous equation
2a + 3b
=
8
=
23
2a + 3b
=
9a – 3b
=
11a
=
1a
=
8
3a – 1b
X3
69
77
7
---------------------------------------------------------Find the other letter by substituting back in to one of
the original equations, it doesn’t matter which one.
2a + 3b
=
8
2 X 7 + 3b
=
8
14 + 3b
=
8
3b
=
-6
b
=
-2
29
To solve a simultaneous equation by substitution we
need to rearrange one of the equations till we get one
letter by itself then substitute this into the other
equation.
Solve the simultaneous equation
2a + 3b
=
8
3a – 1b
=
23
-----------------------------------------------------------------------Rearrange the 2nd equation to get b by itself
1b
=
3a – 23
-----------------------------------------------------------------------Substitute 1b = 3a – 23 into the 1st equation
2a + 3b
=
8
2a + 3(3a – 23) =
8
2a + 9a – 69
=
8
11a
=
77
a
=
11
-----------------------------------------------------------------------Substitute (a = 11) back into one of the original equations to
find the other letter.
b
30
=
-2
We can also use substitution to solve simultaneous
equations where one is linear and the other a quadratic.
Solve the simultaneous equation
c + 2d =
3
c2 + 3cd
=
10
--------------------------------------------------------------------------------------Rearrange the 1st equation to get c by itself
c
=
3 – 2d
--------------------------------------------------------------------------------------Substitute c = 3 – 2d into the 2nd equation
c2 + 3cd
=
10
(3 – 2d)2 + 3(3 – 2d)d
=
10
--------------------------------------------------------------------------------------Multiply out the brackets, add common things and bring together on one side
so the d2 are positive.
2d2 + 3d + 1 =
0
This is a quadratic equation which can be factorised (put into brackets) and
solved.
(2d + 1)(d + 1)
d=
d=
=0
1
2
1
2
or
d=-1
31
Inequalities
Find the set of values for which x2 – 4x + 5 < 0.
Factorise it (put into two brackets) as normal.
(x - 5)(x + 1) = 0
x = 5 and x = -1 are the critical values. Draw a sketch.
it will be a happy graph because it is positive x2.
y = x2 - 4x - 5
The values for which x2 – 4x – 5 is less than zero is
when it is below the x axis, i.e. when -1 < x < 4. Notice
that this is just < not≤ because the original question
was only <.
32
Find the set of values for which x2 – 4x + 5 > 0.
Find the critical values exactly the same as before.
X2 – 4x + 5 = 0
(x + 1)(x – 4) = 0
The critical values are x = - 1 and x = +4.
y = x2 - 4x - 5
The graph is positive i.e. above the x axis when x < -1
or when x > 5.
33
Find the set of values for which 3 – 5x – 2x2 > 0.
Find the critical values exactly the same as before.
3 – 5x – 2x2 = 0
(2x - 1)(x + 3) = 0
The critical values are x =
1
2
and x = -3.
If we sketch the graph out it will be a sad face because
the x2 is negative.
y = 3 - 5x - 2x2
The graph is positive i.e. above the x axis when -3 < x
1
< .
2
34
Find the set of values for which 12 + 4x > x2
Rearrange and continue exactly as before.
Find the critical values.
3 – 5x – 2x2 = 0
(2x - 1)(x + 3) = 0
The critical values are x =
1
2
and x = -3.
If we sketch the graph out it will be a sad face because
the x2 is negative.
35
Chapter 4 – Sketching curves
When we sketch a curve we do not need to draw it
perfectly but just give an idea of the main shape and
points of interest.
Some things we might be interested in are
- What is the general shape?
- Where does the graph cross the x axis?
- Where does it cross the y axis?
- Where are the maximum and minimum points?
- What happens as x gets very big?
- Are there any asymptotes?
36
Remember the general shape of a positive x3 +
something curve.
And a negative x3 + something curve.
Notice they often cut through the x axis in three places
and y gets very big or very small when x gets very big
or very small or vice versa.
37
Sketch the curve with equation y = (x – 2)(x – 1)(x + 1)
General shape looking at how just the ‘x’s multiply each other, +x X +x X +x
= positive x3.
--------------------------------------------------------------------------------------Cuts through x axis when y = 0
0 = (x – 2)(x – 1)(x + 1)
x = +2
x = +1
x = -1
--------------------------------------------------------------------------------------Cuts through y axis when x = 0
y = (0 – 2)(0 - 1)(0 + 1) = -2 X -1 X 1 = +2
--------------------------------------------------------------------------------------As x gets very big and positive, y gets very big and positive
As x » ∞+ , y » ∞+
As x gets very big and negative, y gets very big and negative
As x » ∞- , y » ∞-
38
Sketch the curve with equation y = (x – 3)(2 - x)(x + 4)
General shape looking at how just the ‘x’s multiply each other, +x X -x X +x
= negative x3.
--------------------------------------------------------------------------------------Cuts through x axis when y = 0
0 = (x – 3)(2 - x)(x + 4)
x = +3
x = +2
x = -4
--------------------------------------------------------------------------------------Cuts through y axis when x = 0
y = (0 – 3)(2 - 0)(0 + 4) = -3 X +2 X +4 = -24
--------------------------------------------------------------------------------------As x gets very big and positive, y gets very big and negative
As x » ∞+ , y » ∞As x gets very big and negative, y gets very big and positive
As x » ∞- , y » ∞+
39
Sketch the curve with equation y = (x +2)(x – 1)2
General shape looking at how just the ‘x’s multiply each other, +x X +x X +x =
positive x3.
------------------------------------------------------------------------------------------------Cuts through x axis when y = 0
0 = (x + 2)(x – 1)(x – 1)
x = -2
x = +1
x = +1
Notice we get the same answer twice.
------------------------------------------------------------------------------------------------Cuts through y axis when x = 0
y = (0 + 2)(0 - 1)(0 - 1) = +2 X -1 X -1 = +2
-----------------------------------------------------------------------------------------------As x gets very big and positive, y gets very big and positive
As x » ∞+ , y » ∞+
As x gets very big and negative, y gets very big and negative
As x » ∞- , y » ∞-
40
Sketch the curve with equation y = x3 + x2 – 2x
Start off by factorising
y
=
x(x2 + x – 2)
y
=
x(x + 2)(x – 1)
Once we have got it into brackets we can continue as normal...
Notice that any curve of this type will always go through the (0,0) point
because of the x at the front.
41
We can draw transformations of straightforward y = x3
graphs.
42
And y = - x3
43
We can sketch reciprocal (y = number over x) graphs.
The number on top can be positive like y =
or negative like y =
44
- 12
x
12
x
We can solve quadratic equations by seeing where the
graph cuts through the x axis. In the same way we can
solve other equations by plotting two curves and seeing
where they cross over.
The places the curves intersect is the same as the
solutions.
We can tell how many solutions an equation has by
looking at how many times the curves cross.
45
a.
On the same diagram sketch the curves y = x(x –
3) and y = x2 (1 – x).
b.
Find
the
coordinates
of
the
points
of
intersection.
The points of intersection are the same as the
solutions of the equation
x(x - 3)
=
x2(1 – x)
If we multiply out and rearrange on one side
x3 – 3x
=
x(x2 – 3)
=0
0
x(x + √3)(x - √3) = 0
so x = 0, x = -√3 or x = +√3 are the three
46
solutions.
We can transform a curve f(x) by a translation.
Notice that which way the curve moves depends on if
the number is inside or outside of the bracket.
f(x) + 7 = (x)2 + 7
f(x+4) = (x+4)2
f(x) = (x)2
f(x-3) = (x-3)2
f(x) - 4 = (x)2 - 4
47
f(x) + 2 = 1/x + 2
f(x+5) = 1/(x+5)
f(x) = 1/x
f(x-4) = 1/(x-4)
f(x) - 3 = 1/x - 3
The lines that the curves tend towards (but never
actually reach) are called asymptotes.
48
Multiplying outside the brackets will stretch or squash
the curve towards the x axis.
f(x) = sin x
½ f(x) = ½ sin x
2
1
0
-1
-2
0
90
180
270
360
2 f(x) = 2 sin x
2
2
1
1
0
0
-1
-2
0
90
180
270
360
-1
0
90
180
270
-2
49
360
Multiplying inside the brackets will stretch or squash
the curve towards or away from the y axis.
f(2x) = sin 2x
1
0
-360
-270
-180
-90
0
90
180
270
360
90
180
270
360
90
180
270
360
-1
f(x) = sin x
1
0
-360
-270
-180
-90
0
-1
f(½x) = sin ½x
1
0
-360
-270
-180
-90
0
-1
50
- f(x) is a reflection in the x axis.
f(-x) is a reflection in the y axis.
f(x)
Notice that – f(x) and f(-x) are not the same.
51
Chapter 5 – Coordinate geometry
in the xy plane
ibvwhciwc
52
53
Chapter 6 – Sequences and Series
iopwdpo
54
55
Chapter 7 - Differentiation
The gradient of a line is how steep it is. We pick any
two points on the line and divide the change in the y
coordinates by the change in the x coordinates.
gradient =
difference in y coordinates
difference in x coordinates
y = 2x - 1
9
A = (5, 9)
8
7
6
B = (2, 3)
5
4
3
difference in y
2
difference in x
1
1
2
3
4
5
-4
5-2
=
6
-2
-3
9–3
2
0
-3 -2 -1-1 0
=
The gradient is
2.
-5
The gradient of a straight line is the same at any point.
56
The gradient of a curve constantly changes as we
change the point we are looking at.
f(x) = x2
25
20
15
10
5
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5
We can find out the gradient at any point on a curve by
differentiating and substituting in the x value.
We normally write
dy
dx
or f’(x) for the gradient.
57
y
dy
dx
y
dy
dx
3x
=
x7
f(x)
f’(x)
6
=
7x
=
3x8
=
8
f(x)
X
7
dy
=
dx
y
x11
=
11x10
=
5x4
f’(x) =
x9
+
4
X
5x3
f’(x) =
24x7
=
=
f(x)
20x3
= 2x5 - 3x2
x5
f’(x) = 10x4 - 6x1
dy
dx
= 9x8 + 5x4
The power drops down the front and the power
decreases by one.
58
y
=
7x
f(x)
y
=
7x1
f’(x) =
dy
= 1 X 7x0
dx
dy
=
dx
=
9x
9
7
‘x’s turn into just their number.
y
6
=
y
6x0
=
dy
dx
f(x)
=
3x5
-7
=
f’(x) =
15x4
0X
6x-1
Constants disappear!
59
y
dy
dx
=
x-3
f(x)
=
=
-3x-4
f(x)
=
1
x2
x-2
-2x-3
f’(x) =
y
dy
dx
=
=
x1/3
f(x)
=
1 -2/3
x
3
f(x)
=
1
√x
x1/2
1 -1/2
x
2
f’(x) =
Multiply
brackets
out
before
you differentiate.
y =
y=
dy
dx
x2(x – 4)
Divide
out
fractions
before
you differentiate.
x6 – 5x
f(x)=
x2
x3 – 4x
x6
-
5x
f(x)
=
f(x)
= x4 – 5x-1
2
= 3x - 4
x2
x2
f’(x) =4x3 + 5x-2
60
We can find the gradient at any point on the curve
by substituting the x value at that point.
Find the gradient of the curve y = x3 when x = 4.
-------------------------------------------------------------y
dy
dx
dy
dx
dy
dx
=
x3
=
3x2
=
3 X (4)2
=
48
Find the gradient of the curve y = 3x2 – 4x + 5
when x = 2.
-------------------------------------------------------------y
dy
dx
dy
dx
dy
dx
=
=
6x - 4
=
6X2-4
=
8
3x2 – 4x + 5
61
Find the gradient of the curve y =
4
x2
- √x when x =
1.
-------------------------------------------------------------y
=
y
=
dy
dx
dy
dx
dy
dx
dy
dx
62
=
=
4
x2
4x-2 – x1/2
1
-8x-3 - x-1/2
2
-8 X (1)-3 -
1
X (1)-1/2
2
1
=
=
- √x
-8
2
−
17
2
We can use the gradient to find the equation of the
tangent line at a point.
f(x) = x2
25
20
15
10
5
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5
63
Find the equation of the tangent to the curve
y = x3 – 3x2 + 2x - 1 at the point (3, 5).
First we are going to differentiate and substitute x =
3 to find the gradient of the curve at this point which
is the same as the gradient of the tangent to the
curve at this point. Then we are going to put this
gradient and x1 = 3, y1 = 5 into the equation we
learned in Chapter 5 and rearrange to find the
equation of the line.
-------------------------------------------------------------y
dy
=
dx
dy
dx
x3 – 3x2 + 2x - 1
=
=
dy
3x2 – 6x + 2
3 X (3)2 – 6 X (3) + 2
=
dx
11
-------------------------------------------------------------y – y1
=
y–5
=
y
64
m(x – x1)
11(x – 3)
=
11x - 28
We can also use the gradient to find the equation of
the normal line at a point.
Remember the gradient of the normal is always
minus one over the gradient of the tangent. So if the
gradient of the tangent is 3 then the gradient of the
normal is −
1
3
.
(gradient of tangent) X (gradient of normal) = 1
gradient of normal =
-1
gradient of tangent
f(x) = x2
25
20
15
10
5
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5
65
Find the equation of the normal to the curve
y = 8 - 3√x at the point x = 4.
First we are going to differentiate and substitute x = 4 to find the gradient
of the curve at this point.
Then we are going to calculate the gradient of the normal by working out
minus one over this.
Next we need to work out the value of y when x = 4 which we will find out
by substituting into the original equation of the curve.
And finally we are going to put the gradient, x1 and y1 into the equation we
learned in Chapter 5 and rearrange to find the equation of the line.
--------------------------------------------------------------------------------------y
=
y
dy
dx
dy
dx
dy
dx
8 - 3√x
=
8 – 3x½
=
− x-½
=
− (4)-½
=
−
3
2
3
2
3
4
The gradient of the tangent is −
the normal
is
4
3
3
4
so this means that the gradient of
.
--------------------------------------------------------------------------------------When x = 4
y
=
8 - 3√4
y
=
2
--------------------------------------------------------------------------------------y – y1
y–2
y
66
=
m (x – x1)
4
=
3
=
(x – 4)
4
10
3
3
x-
Up till now we have been looking at
dy
dx
as the
gradient, which it is, but another way to look at it is
‘the rate of change of y with respect to x’.
Or if we increase x by ‘a bit’, how much does y
increase by?
We don’t just have to look at ‘y’s and ‘x’s we can use
any variables.
If we draw a graph of a train setting off on a journey
we plot velocity against time. We can find
dv
dt
, ‘the
rate of change of velocity with respect to time’, or the
acceleration.
We could look at
dp
dt
, ‘the rate of change of prices
with respect to time’ or inflation.
We can observe a balloon that is being blown up and
calculate
dV
dr
, ‘the rate of change of the Volume with
respect to the radius’.
67
If the velocity of a car at any time is
v
3t2 + 7t
=
find
dV
dt
.
v
=
dV
dt
=
3t2 + 7t
6t + 7
If the Volume of a balloon at any time can be found
using the formula
V
find
V
dV
dr
68
=
dV
dr
.
=
=
4
πr3
3
4πr2
4
πr3
3
We can differentiate twice to find the second
derivative written as
d2y
or
dx2
y
dy
dx
d2y
dx2
f’’(x).
=
5x3 – 4x2 + 7x - 9
=
15x2 – 8x + 7
=
30x - 8
69
Now let’s see numerically why differentiation works. We
are going to find the gradient at x = 1 for the curve y =
x2.
4
f(x) = x2
10
3
5
2
0
1
0
1
2
3
1
We are going to calculate the gradient between our start
point (1, 1) and our end point. Our first end point is going
to be (2, 4), our next end point will be (1.5, 2.25) and
the one after that (1.25, 1.5625). We will keep halving
the x distance between the start point and the end point
and see what happens to the gradient.
The gradient of the first line is 3. The gradient of the
second line is 2.5, the gradient of the third line is 2.25. If
we keep halving the distance we get 2.125, 2.0625,
2.03125 as the gradient for the next three lines. Every
time we halve the x distance between the start point and
the end point the gradient gets closer and closer to 2. If
we keep going, many. many times the x distance will
eventually be practically nothing and the gradient will be
practically 2.
70
2
Now we are going to differentiate using algebra.
We are going to look at two points on the graph y = x2. The start point is
at (x, x2) and the end point is a tiny bit further in the x direction, we call
this tiny bit extra δx (delta x). This is not a multiple, δ X x, it is all one
thing, δx, the increase in x. This means the end point is at (x+δx, (x+δx)2
).
Now remember that the gradient between two lines
difference in y coordinates
=
difference in x coordinates
end y – start y
=
end x – start x
=
(x+δx)2 – x2
x – (x + δx)
=
x2 + 2xδx + δx2– x2
=
2xδx + δx2
=
2x + δx
x – (x + δx)
δx
As δx, the tiny bit extra in the x direction, tends to zero, the gradient tends
to 2x.
End point
(x+δx, (x+δx)2)
Start point
(x, x2)
δx
71
Chapter 8 - Integration
Integration is the opposite of differentiating.
We say if
dy
dx
= something then what was y equal to
before we differentiated?
We increase the power by one and divide by the new
power.
dy
If
dx
= xn then
y=
If we are just given
dy
dx
xn+1
n+1
then we never know if there
was a constant (number by itself) there before we
differentiated. For example
y
=
x2 + 7
x2 + 3
y
y
=
=
x2 - 4
will all differentiate to
dy
dx
=
2x
so to show we know there could have been a number
there before we differentiated we add a C for
72
constant at the end.
dy
dx
dy
dx
dy
dx
y
=
f’(x) =
x7
8
=
x
=
1 8
x
8
+
C
=
x6
+
x3
x4
=
=
C
8
x7
7
1 7
x
7
+
f(x)
C
+
=
+
x5
5
+
C
=
1 5
x
5
+
C
f’(x)
=
x2
-
x5
f(x)
=
4
1
+ x4
4
x4
=
C
x3
x6
-
3
1 3
x
3
6
1
- x6 +
6
Remember that when we divide by a fraction we can
flip it and multiply.
dy
dx
y
=
=
=
f’(x) =
x-3
x
-2
-2
+
1
− x-2+
2
C
C
f(x)
=
=
x1/2
x3/2
3
2
+C
2 3/2
x
3
+
C
73
When we differentiate, ‘x’s turn into numbers.
y
=
dy
3x
=
dx
3
So when we integrate, numbers turn into ‘x’s.
dy
dy
dx
y
=
7
=
3x2
+
=
x3
+
y
9
9x
+
C
A curve has the gradient at any point of
dy
3x2
=
dx
+
2x
and passes through the point (2, 10). Find the equation
of the curve.
-----------------------------------------------------dy
3x2
=
dx
y
=
x3
+
2x
+
x2 + C
When x = 2, y = 10 so substitute to find C
10
C
74
3(2)3
=
=
+
- 18
The curve is y = x3 + x2 - 18
2(2) + C
The proper way to write integration is with the
integral sign (called the summa).
3x2
∫
+
dx
x3
=
C
The integral sign says integrate this thing.
dx does not mean d X x, it means ‘do with respect
to x’. We will normally be integrating with respect
to x but we will occasionally integrate with respect
to other things as well.
∫
4x3
dx
5x2 dx
∫
∫
7x5
∫
5t4
5x3
=
+ 2x4
dt
x4
=
+
3
dx
=
=
+C
7x6
6
t5
+
C
2x5
5
+
+C
C
75
Always get rid of brackets and fractions before you
integrate.
∫
=
(x + 1)(x + 2)
x2 + 3x + 2
∫
x3
=
3
dx
3x2
+
+2
2
x6 – 5x
∫
dx
x2
=
∫
=
∫
=
76
dx
x6
x2
-
5x
dx
x3
x4 - 5x-3
x4
4
dx
-
5x-2
-2
77
78
79