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C4
Ch 1Partial
fractions
Ch 2 Coordinate geometry
Ch 3 The binomial expansion
Ch 4Differentiation
Ch 5 Vectors
Ch 6 Integration
2
Chapter 1 – Partial fractions We can add (or take away) two fractions only if they have the same number on the bottom (denominator).
12 + 1
3
X 3 X 2
= 36 + 2
6
= 56
23 + 3
4
X 4 X 3
= 812
+ 912
= 1712
We can add (or take away) two fractions only of they have the same number on the bottom (denominator).
3
Similarly with algebraic fractions, we can only add or take them away if they have the same thing on the bottom.
We are going to make the bottoms the same on both fractions, multiply out the brackets, bring the fractions together and simplify the tops.
2(x + 3)
+ 1(x + 1)
X (x + 1) X (x + 3)
= 2 X (x + 1)(x + 1)(x + 3)
+ 1 X (x + 3)(x + 1)(x + 3)
= 2x + 2(x + 1)(x + 3)
+ 1x + 3(x + 1)(x + 3)
= 2x + 2 + 1x + 3(x + 1)(x + 3)
= 3x + 5(x + 1)(x + 3)
4
If we have two linear (not x2 terms) on the bottom then we can always split it up and write it like
6x - 2
(x – 3)(x + 1) ≡
A(x – 3)
+ B
(x + 1)
Notice that there is a ≡ in the middle, this means that both sides are always exactly the same no matter what value of x you put in, instead of sometimes the same when we have a normal = sign. Later we will see that both sides always have exactly the same number of ‘x’s, ‘x2’s and constants as well.
We can also go the other way and split a fraction up. This can be useful when we are integrating. There are two ways to solve these - Substitution. We put in a value of x that will make some of the brackets disappear. - Compare the coefficients. Both sides must have exactly the same number of everything, so look at how many ‘x2’s or ‘x’s or constants are on one side, the other side must have exactly the same.
5
We are going to multiply everything on both sides by the brackets on the bottom and then substitute ‘x’s that make one of the brackets disappear.
6x - 2(x – 3)(x + 1)
≡ A(x – 3)
+ B(x + 1)
X (x-3)(x+1) X (x-3)(x+1) X (x-3)(x+1) 6x – 2 ≡ A(x + 1) + B(x – 3) -------------------------------------------------------------------
When x = -1 6x – 2 ≡ A(x + 1) + B(x – 3) 6(-1) – 2 ≡ A(0) + B(-1 – 3)
-8 ≡ -4B B ≡ 2 ------------------------------------------------------------------- When x = +3 6x – 2 ≡ A(x + 1) + B(x – 3)
6(3) – 2 ≡ A(3 + 1) + B(0)
16 ≡ 4A A ≡ 4
6
We can do exactly the same thing if there are three linear (not x2 or squared) brackets on the bottom.
6x2 + 5x - 2
(x)(x - 1)(2x + 1) ≡
Ax +
B(x – 1)
+ C
(2x + 1)
X (x)(x-1)(2x+1) 6x2 + 5x – 2 ≡ A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) Notice that each letter now multiplies all the brackets it didn’t have on the bottom. ------------------------------------------------------------------------------------------------
When x = 0 6x2 + 5x – 2 ≡ A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) 6(0)2 – 5(0) - 2 ≡ A(0 - 1) + B(0)( ) + C(0)( ) – 2 ≡ -1A
A ≡ 2 ------------------------------------------------------------------------------------------------ When x = +1 6x2 + 5x – 2 ≡ A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1) 6(1)2 + 5(1) - 2 ≡ A(0)(-1) + B(1)(3) + C(-1)(0)
9 ≡ 3B
B ≡ 3 ------------------------------------------------------------------------------------------------
When x = − 12
6x2 + 5x – 2 ≡ A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1)
6(− 12)2 + 5(− 1
2) – 2 ≡ A(− 3
2)(0) + B(− 1
2)(0) + C(− 1
2)( (− 3
2)
-3 ≡ 34C
C ≡ -4
7
Chapter 2 – Coordinate geometry in the xy plane We can define a curve so that y and x are not defined in terms of each other anymore but a third parameter which we often call t.
8
t x = 2t y = t2
3 6 9
2 4 4
1 2 1
0 0 0
-1 -2 1
-2 -4 4
-3 -6 9
Draw the curve for the parametric equation x = 2t and y = t2 for -3 ≤ t ≤ 3
0
1
2
3
4
5
6
7
8
9
10
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
9
A Cartesian equation is the name for a curve in the form y = something to do with x. We can change a parametric equation back to the normal Cartesian equation. 1. Rearrange both equations until they both have the same ‘t thing’ 2. Set them equal to each other so that the ‘t’s disappear 3. Rearrange so that y = something
10
A curve has parametric equations, x = 2t, y = t2 Find the Cartesian equation of the curve. -------------------------------------------------------------- x = 2t y = t2
t = x2 y = (x
2)
The Cartesian equation is y = x2
4
11
Example The diagram shows a sketch of the curve with parametric equations x = t – 1 and y = 4 – t2. The curve meets the x axis at the points A and B. Find the coordinates of A and B.
A
B
When it meets the x axis y = 0 y = 4 – t2 0 = 4 – t2 t2 = 4 t = ± 2 ------------------------------------------------------------------------ When t = +2 When t = -2 x = t – 1 x = t - 1 x = 2 -1 x = - 2 - 1 x = 1 x = - 3 The coordinates are (1, 0) and (-3, 0)
12
A curve has parametric equations x = at and y = a(t3 + 8) where a is a constant. The curve passes through the point (2, 0). Find the value of a.
y = a(t3 + 8) ÷ both sides by a 0 = a(t3 + 8) 0 = t3 + 8 t3 = -8 t = -2 -------------------------------------- x = at 2 = a X (-2) a = -1 -------------------------------------- The parametric equations are x = -1t and y = -1(t3 + 8)
We know that there
is a value of t for
which x = 2 when y
= 0 so we can find
this value of t.
Let y = 0 and then
solve for t.
Substitute this value
of t into the x
equation to find a
Substitute a into the
original equations
13
A curve is given parametrically by the equations x = t2 and y = 4t. The line x + y = - 4 meets the curve at A. Find the coordinates of A.
14
To find the coordinates where they meet we need to find t when they meet. Rearrange the equation of the line x + y + 4 = 0 Substitute x = t2 and y = 4t into this equation t2 + 4t + 4 = 0 (t + 2)2 = 0 t = - 2 -------------------------------------------------------------- Substitute back in to find the coordinates x = t2 = (-2)2 = 4 y = 4t = 4 X -2 = -8 The coordinate of A is (4, -8)
15
Converting trigonometric parametric equations into Cartesian equations.
Example - A curve has the parametric equations x = sin t + 2 y = cos t – 3 Find the Cartesian equation of the curve.
We need to get rid of any ‘t’s so rearrange the formulas. x = sin t + 2 y = cos t – 3 sin t = x – 2 cos t = y + 3 -------------------------------------------------------------- Now sin2 t + cos2 t = 1 so if we substitute in (x – 2)2 + (y + 3)2 = 1 This is the Cartesian equation of the curve. It is a circle with radius 1 and centre (+2, -3).
16
A circle with the equation
(x – a)2 + (y – b)2 = r2
has centre (a, b) and radius r. (a, b)
r
17
Example - A curve has the parametric equations x = sin t y = sin 2t Find the Cartesian equation of the curve.
We need to get rid of any ‘t’s so open up sin 2t and substitute in x = sin t y = sin 2t y = 2 sin t cos t y = 2x cos t ------------------------------------------------------------------------ Now sin2 t + cos2 t = 1 so rearrange and substitute in x = sin t so we get t in terms of x cos t2 t = 1 – sin2 t cos t = √(1 – sin2 t) cos t = √(1 – x2) ------------------------------------------------------------------------ Now substitute cos t = √(1 – x2) into the original equation and we have y in terms of x which is what we wanted. y = 2x cos t y = 2x √(1 – x2)
18
Finding the area under a parametric curve The area under a curve is given by ∫𝑦𝑦 𝑑𝑑𝑑𝑑 by using the
chain rule we can rewrite this as ∫𝑦𝑦 dxdt
dt (notice this
looks a bit like we could cancel the fractions and turn it back into the original). It is very important to realise that because we are integrating with respect to t the limits must be in terms of t as well.
19
If x = 2t + 4 and y = 3t – 1
find y dxdt
.
x = 2t + 4 y = 3t – 1 dxdt
= 2
y dxdt
= 2 (3t -1)
It is very important to realise that because we are integrating with respect to t the limits must be in terms of t as well.
20
The diagram shows a sketch of the curve with parametric equations x = t2, y = 2t(3 – t), t≥ 0. The curve meets the x axis at x = 0 and x = 9. The region R is bounded by the curve and the x axis. Find the area of R.
First find out what t is when x = 0 and x = 9 x = t2 x = t2 0 = t2 9 = t2 t = 0 t = 3 we ignore the negative square roots because t ≥ 0 --------------------------------------------------------------
∫𝑦𝑦 dxdt
dt = ∫ 2t(3 − t)2t dtt=3t=0 = ∫t=3
t=0 12t2 – 4t3 dt
= [4t3 – t4] = (108 – 81) – (0 – 0) = 27
0
1
2
3
4
5
0 1 2 3 4 5 6 7 8 9 10
21
Chapter 3 – The Binomial Expansion We can expand brackets a lot quicker using the binomial expansion.
(1 + x)n = 1 + nx + n(n – 1)2!
x2 + n(n – 1)(n – 2)3!
x3 + ...+ nCr xr
Remember that nCr is the number of of ways of picking a team of r from a total of n people
nCr = n!(n – r)! r!
5C2 = 5!3! 2!
= 5 X 4 X 3 X 2 X 1(3 x 2 X 1) (2 X 1)
= 5 X 42 X 1
22
Use the binominal expansion to find the first four terms of (1 + 2x)-1 ____________________________________________
(1 + x)n = 1 + nx + n(n – 1)2!
x2 + n(n – 1)(n – 2)3!
x3 + ...
(1 + 2x)-1 = 1 + (-1)(2x) + - 1 X - 22 X 1
(2x)2 + -1 X - 2 X - 33 X 2 X 1
(2x)3 + ... (1 + 2x)-1 = 1 - 2x + 4x2 – 8x3 + ...
Notice that if the power is negative then the expansion will go on forever so we normally stop after the first few terms. This is because it converges very quickly as if x is a small number then x4 is a very, very small number and can be ignored.
The infinite expansion is only valid if l2xl < 1 i.e. if ½<x<½ .
23
Use the binominal expansion to find the first four terms of (1 - 3x)½ . ____________________________________________
(1 + x)n = 1 + nx + n(n – 1)2!
x2 + n(n – 1)(n – 2)3!
x3 + ...
(1 - 3x)½ = 1 + (½)(-3x) + 12 X -122 X 1
(-3x)2 + 12 X -12 X -323 X 2 X 1
(-3x)3
+ ...
(1 - 3x)½ = 1 - 32 x - 9
8 x2 – 27
16 x3 + ...
Again this infinite expansion would go on forever and is
only valid if l-3xl < 1 i.e. if − 13 < x < 1
3 .
24
Use the binominal expansion to find the first four terms of (4 + x)½. ____________________________________________
(4 + x)½ = [4(1 + x)]½ = 4½(1 + x4)½ = 2(1 + x
4)½
expand (1 + x4)½ as normal
2(1 + x4)½ = 2 [1 + 1
8 x - 1
128 x2 + 1
1024 x3 + ...]
= 2 + 14 x - 1
64 x2 + 1
512 x3
--------------------------------------------------------------
valid for l x4 l < 1 i.e l x l < 4
If the number at the front is bigger than a one then take it out down the front.
25
Use the binominal expansion to find 1 + x1 + 3x
as far as
the term in x3. ____________________________________________
1 + x1 + 3x
= (1 + x) X (1 + 3x)-1
--------------------------------------------------------------
Expand (1 + 3x)-1 as normal (1 + 3x)-1 = 1 – 3x + 9x2 – 27x3 -------------------------------------------------------------- Multiply out the brackets (1 + x) X ( 1 – 3x + 9x2 – 27x3) Notice that you only actually need to multiply the terms that are going to result in terms less than x4.
26
Use partial fractions to find the binominal expansion of (4 – 5x)(1 + x)(2 – x)
as far as the term in x3. ____________________________________________________
(4 – 5x)(1 + x)(2 – x)
= 3(1 + x)
+ -2(2 – x)
by partial fractions
------------------------------------------------------------------------
3(1 + x)
= 3 x (1 + x)-1
expand (1 + x)-1 as normal = 1 – x + x2 – x3 + ...
3(1 + x)
= 3 x (1 + x)-1 = 3[1 – x + x2 – x3] = 3 – 3x + 3x2 – 3x3
------------------------------------------------------------------------
-2(2 - x)
= -2 x (2 - x)-1
expand (2 - x)-1 as normal = 12 + 1
4 x + 1
8 x2 + 1
16 x3 + ...
-2
(2 - x) = -2 x (2 - x)-1 = -2[1
2 + 1
4 x + 1
8 x2 + 1
16 x3] = -1 - 1
2 x - 1
4 x2
- 18 x3
------------------------------------------------------------------------
3(1 + x)
+ -2(2 – x)
= [3 – 3x + 3x2 – 3x3] + [-1 - 12 x - 1
4 x2 - 1
8 x3]
27
Chapter 4 – Differentiation
28
We can differentiate parametric curves (ones given in terms of t) by using the fact that dydtdxdt
= dydt
÷ dxdt
=dydt
X dtdx
= dydx
so to find dydx
we need to differentiate y with respect to t,
differentiate x with respect to t and divide one by the other.
29
Differentiate x and y with respect to t x = t3 + 1 y = t2 + 1 dxdt
= 3t2 dydt
= 2t
--------------------------------------------------------------
dydx
= dydtdxdt
= 2t3t2 = 2
3t
-------------------------------------------------------------- when t = 2 dydx
= 23 X 2
= 13
Find the gradient at the point P where t = 2, on the curve given parametrically by x = t3 + t, y = t2 + 1
30
Sometimes we need to find dydx
for curves that can’t be
rearranged into y = something like x2 + y2 = 8x. These are called implicit relationships. The differential with respect to x of anything to do with y is what it would be if you differentiated with respect
to y times by dydx
.
The differential of y2 = 2y dydx
d(y2)dx
= 2y dydx
Sometimes we may need to use the product rule.
The differential of x2y3 = x23y2 dydx
+ 2xy3
d(x2y3)
dx = x23y2 dy
dx + 2xy3
31
To find a gradient of a curve, differentiate everything
with respect to x and then rearrange for dydx
.
Find the gradient of the curve x2 + y3 = 4x at the point (1, 2). ____________________________________________ Differentiate everything with respect to x and rearrange
to get dydx
= something
2x + 3y2 dydx
= 4
3y2 dydx
= 4 – 2x
dydx
= 4 – 2x3y2
--------------------------------------------------------------
Substitute x = 1, y = 2 into dydx
. Notice that we need to
substitute both x and y now. dydx
= 4 – 2x3y2 = 4 – 2(1)
3(2)2 = 16
32
We can differentiate something like y = 2x where x is a power rather than a multiple.
In general if y = ax then dydx
= ax Ln a.
If y = 2x find dydx
.
____________________________________________ y = 2x Ln both sides Ln y = Ln 2x Bring x down the front by log laws Ln y = x Ln 2 Differentiate both sides 1y dydx
= Ln 2
Multiply y up dydx
= y Ln 2
Substitute y = 2x in from the beginning dydx
= 2x Ln 2
33
The radius of a circle is expanding at the rate of 5 cm per sec. Find the rate at which its Area is expanding when the radius is 3. ______________________________________________________________ Rate means with respect to time, we want the rate of change of its Area (with
respect to time) which is dAdt
.
We know the rate of change of the radius with respect to time which is drdt
= 5.
and by the chain rule dAdt
= dAdr
X drdt
so we need to work out dAdr
.
--------------------------------------------------------------------------------------- For a circle A = πr2 so dAdr
= 2πr
--------------------------------------------------------------------------------------- dAdt
= dAdr
X drdt
dAdt
= 2πr X 5
when r = 3 dAdt
= 30π cm2/sec
When the radius is 3cm the area of the circle is increasing by 30π cm2 per second.
34
We can set up Differential equations for many real life situations but they especially occur in the cases of radioactive decay, population growth and cooling objects.
In the decay of radioactive particles the rate at which particles decay is proportional to how many particles are left. dNdt
α N
which can be rewritten as dNdt
= -kN
the minus sign arises because the number of particles is going down.
35
The population of a town is growing at a rate proportional to the size of the population. dPdt
α P
which can be rewritten as dPdt
= kP
Newton’s Law of Cooling says that an object loses temperature at a rate that is proportional to the difference between the object and the surrounding air temperature. dΘdt
α (Θ – Θ0)
where Θ0 is the surrounding air temperature dΘdt
= (Θ – Θ0)
36
37
Chapter 5 – Vectors
38
If we want to describe where an ant walking on a classroom floor is we need to use 2 coordinates, (x, y), that tell us that how far along and how far across the ant is
This ant is at the point (4, 3).
x
y
39
If we want to describe where a fly is in a classroom we need to use three numbers, (x, y, z) to tell us how far along, how far across and how far up it is.
This fly is at the point (2, 3, 4).
x
y
z
40
To find out how far a point (x, y) is from the origin in two dimensions we use Pythagoras in two dimensions.
Distance = √ (x2 + y2)
Find the distance of the point (4, 3) from the origin. Distance = √ (x2 + y2) Distance = √ (42 + 32 Distance = √ (16 + 9 Distance = √ (25 Distance = 5
41
To find out how far a point (x, y, z) is from the origin in three dimensions we use Pythagoras in three dimensions.
Distance = √ (x2 + y2 + z2)
Find the distance of the point (6, 7, 8) from the origin.
Distance = √ (x2 + y2 + z2)
= √ (62 + 72 + 82)
= √ (36 + 49 + 64)
= √ (149)
= 12.2 (3sf)
42
To find out the distance between two points (x1, y1) (x2, y2) we use Pythagoras.
Distance = √ { (x2 – x1)2 + (y2 – y1)2 }
Find the distance between the points (9, 3) and (6, -1).
Distance = √ { (x2 – x1)2 + (y2 – y1)2 }
= √ { (9 – 6)2 + (3 – -1)2 }
= √ { (3)2 + (4)2 }
= √ { 9 + 16 }
= √ { 25 }
Distance = 5
43
To find out the distance between two points (x1, y1, z1) (x2 , y2 , z2) we use Pythagoras in three dimensions.
Distance = √ { (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
Find the distance between the points (2, 7, 4) and (9, -3, 8). ____________________________________________
Distance = √ { (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2 }
= √ { (9 - 2)2 + (-3 - 7)2 + (8 - 4)2 }
= √ { (7)2 + (-10)2 + (4)2 }
= √ { 49 + 100 + 16}
= √ { 165 }
= 12.8 (3sf)
44
We can also use a position vector to say where a fly is. We use i, j and k where i, j and k are unit vectors (i.e. of length one) pointing in the x, y and z directions. Vectors are normally written in bold. This is nearly the same idea as coordinates but a position vector has a direction as well as a length now. We can write a position vector in terms of i, j and k or as a column vector. Instead of saying the fly is at coordinates (7, 3, 8) we
would say it has position vector 7i + 3j + 8k or �738�.
45
We can find the modulus (or magnitude) of a vector v = xi + yj + zk. lvl = √(x2 + y2 + z2)
Find the modulus of the vector v = 3i + 8j - 5k. lvl = √(x2 + y2 + z2)
lvl = √(32 + 82 + (-5)2)
lvl = √(9 + 64 + 25)
lvl = √98
lvl = 9.89
46
The points A and B have position vectors 4i + 2j + 7k and 3i + 4j – 1k. Find the vector AB. ____________________________________________ AB = b – a = (4i - 2j + 7k) – (3i + 4j -1k) = 1i – 6j + 8k
Exactly the same as in two dimensions if we know the position vectors a and b for points A and B we can find the vector AB by calculating b – a.
a
b
b - a
A
B
O
OA
OB
AB
47
The points A and B have position vectors a = �t5
t − 1� and b = �
2tt3�.
a. Find AB. b. By differentiating lABl2, find the value of t for which lABl is at a minimum. c. Find the minimum value of lABl. _____________________________________________________________________ AB = b – a
= �2tt3� – �
t5
t − 1�.
= �t
t − 54 − t
�
------------------------------------------------------------------------------------------------
If lABl = √(x2 + y2 + z2) then lABl2 = (x2 + y2 + z2) so
lABl2 = t2 + (t – 5)2 + (4 – t)2
multiplying out brackets and simplifying
lABl2 = 3t2 – 18t + 41
If we call lABl2 y then
y = 3t2 – 18t + 41
𝑑𝑑𝑦𝑦𝑑𝑑𝑑𝑑
= 6t – 18
The minimum of lABl2 occurs when dydt
= 0 and this is exactly the same as the minimum
of lABl so lABl is at a minimum at the point t = 3.
------------------------------------------------------------------------------------------------
Putting t = 3 back in to find out the actual magnitude of lABl
lABl = √(3t2 – 18t + 41)
= √(3X32 – 18 X 3 + 41)
= √41
48
If we have two vectors a = ui + vj + wk b = xi + yj + zk then the scalar product = a.b = ux + vy + wz
Find the scalar product of the vectors a = 8i - 5j - 4k b = 5i + 4j - k ___________________________________________ a.b = (8 X 5) + (-5 X 4) + (-4 Xa -1) = 40 - 20 + 4 = 24
49
50
We can use the scalar product to find the angle between two vectors.
cos (angle) = a.blal lbl
where a.b is the scalar product and lal is the magnitude of the vector a.
O
a
b
51
Find the angle between the vectors a = - i + j + 3k b = 7i - 2j + 2k _______________________________________ a.b = (-1 X 7) + (1 X -2) + (3 X 2) = -7 - 2 + 6 = - 3 --------------------------------------------------------------
lal = √[(-1)2 + (1)2 + (3)2 ] = √11 -------------------------------------------------------------- lbl = √57 --------------------------------------------------------------
cos (angle) = a.blal lbl
= -3√11 X √57
angle = 96.9°
52
We can use just the scalar product to prove that two vectors are perpendicular.
Because cos (angle) = a.blal lbl
and cos 90° = 0 We only need to prove that the top is = 0 because 0 divided by anything is sill zero. So if the scalar product, a.b = 0 then the two vectors cross at right angles.
Given that the vectors a = 2i - 6j + 1k b = 5i + 2j + λk are perpendicular find the value of λ. -------------------------------------------------------------- a.b = (2 X 5) + (-6 X 2) + (1 X λ) = -2 + λ to be perpendicular a.b = 0 so λ = 2.
53
Given the vectors a = - 2i + 5j - 4k b = 4i - 8j + 5k Find a vector that is perpendicular to both a and b. _____________________________________________________________________ Call the vector perpendicular to a and b v = xi + yj + zk then a.v = 0 and b.v = 0 because they are perpendicular ------------------------------------------------------------------------------------------------- If a.v = 0 then -2x + 5y – 4z = 0
------------------------------------------------------------------------------------------------- If b.v = 0 then 4x – 8y + 5z = 0 ------------------------------------------------------------------------------------------------- Now we can pick any one dimension and choose it to be a constant. So if we pick the z dimension and let z = 1 so if if we substitute z = 1 we get the simultaneous equation -2x + 5y = 4 4x - 8y = -5 solving simultaneous equations x = 7
4 and y = 3
2 and we have already chosen z to be = 1
------------------------------------------------------------------------------------------------- One possible vector that is perpendicular to a and b is v = 7
4 i + 7
4 j + 1k.
we can find another one by just multiplying by 4 to get v = 7i + 7j + 4k
54
We can write the equation of a straight line in which passes through a point a and is parallel to a vector b. r = a + tb t is the parameter. You can think about this as saying go to point a and continue in the direction b. This is similar to the equation of a line y = mx + c.
a
O
b
r = a + tb
55
Find a vector equation of the straight line which passes through the point A, with position vector 3i – 5j + 4k, and is parallel to the vector 7i – 3k. ____________________________________________
r = a + tb r = 3i – 5j + 4k +t(7i – 3j + 0k)
or r = (3 +7t)i + (-5 -3t)j + (4 +0t)k
or
r = �3 + 7t−5 − 3t
4�
56
We can also write the equation of a straight line which passes through two points c and d. r = c + t(d – c) Where t is the parameter. You can think about this as saying start at point c and continue in the direction (d – c) which is the same as vector CD. Again the form of this is similar to the equation of a line y = m(x2 – x1)+ y1 in two dimensions.
O
d
r = c + t(d – c)
c
(d – c) = CD
We do not have to use c as our start point we could use d or indeed any point on the line so the line
r = d + t(d – c)
57
Find a vector equation of the straight line which passes through the points c = 4i + 5j - 1k and d = 6i + 3j + 2k
r = c + t(d – c) --------------------------------------------------------------
(d – c) = 2i - 2j + 3k
-------------------------------------------------------------- r = 4i + 5j -1k + t(2i - 2j +3k)
or r = (4 -2t)i + (5 -2t)j + (-1 +3t)k or
r = �4 − 2t5 − 2t−1 + 3t
�
or
r = �45−1
� + t �2−23�
58
In two dimensions two lines will definitely cross over one another unless they are parallel. In three dimensions two lines will generally not intersect because there are so many more ways that they can miss each other rather than intersect. If they do intersect then each of the three coordinates must be equal in both lines at exactly the same time.
59
Show that the lines with vector equations r = (3i + 8j – 2k) + μ(2i - 1j + 3k) r = (7i + 4j +3k) + λ(2i +1j + 4k) intersect and find the position vector of the point of intersection. ______________________________________________________________ If the lines intersect then each of the three coordinates must be the same at exactly the same time so taking simultaneous equations we get 3 + 2μ = 7 + 2λ 8 - μ = 4 + λ -2 + 3μ = 3 + 4λ We can choose any two of these but choosing the top two and solving simultaneously we get μ = 3 λ = 1 --------------------------------------------------------------------------------------- Check that this also works for the z coordinate -2 + 3X3 = 3 + 4X1 It does work!
---------------------------------------------------------------------------------------
So subbing back into all three coordinates we get the point of intersection
p = 9i + 5j +7k or
p = �957�
60
We can find the angle between two straight lines by using
cos (angle) = a.blal lbl
where a and b are only the direction bit of the lines.
61
Find the acute angle between the two straight lines with vector equations a = (2i + 1j + 1k) + μ(3i - 8j - 1k) b = (7i + 4j + k) + λ(2i + 2j + 3k) ____________________________________________ We only need the direction bits of the lines! lal = √[(3)2 + (-8)2 + (-1)2] = √74 lbl = √[(2)2 + (2)2 + (3)2 ] = √17 -------------------------------------------------------------- a.b = 3X2 + -8X2 + -1X3 = -13
--------------------------------------------------------------
cos (angle) = a.blal lbl
cos (angle) = -13√74 X √17
angle = 68.5°
62
63
Chapter 6 – Integration
∫ xn = xn+1
n + 1 + C
∫ ex = ex + C
∫ 1x = Ln lxl + C
∫ cos x = sin x + C
∫ sin x = - cos x + C
∫ sec2 x = tan x + C
∫ cosec x cot x = - cosec x + C
∫ cosec2 x = - cot x + C
∫ sec x tan x = sec x + C
64
Integrate the following with respect to x
2 cos x + 3x - √x
_______________________________________ Separately ∫ 2 cos x dx= 2 sin x
∫ 3x dx = 3 Ln lxl
∫ √x dx = ∫ x½ = 23x3/2
-------------------------------------------------------------- so
∫ (2 cos x + 3x - √x )dx = 2 sin x + 3 Ln lxl - 2
3x3/2 + C
65
Find ∫ (cos xsin2 x
- 2ex) dx
____________________________________________ cos xsin2 x
= cos xsin x
X 1sin x
= cot x cosec x
∫ cot x cosec x dx = - cosec x ∫ 2 ex dx = 2ex -------------------------------------------------------------- so
∫ (cos xsin2 x
- 2ex) dx = - cosec x – 2ex + C
66
∫ f’(ax +b) dx = 1a f(ax + b) + C
∫ (ax +b)n dx = 1a (ax +b)n+1
n + 1 + C
∫ eax +b dx = 1a eax +b + C
∫ 1ax +b
dx = 1a Ln lax +bl + C
∫ cos (ax +b) dx = 1a sin (ax + b) + C
∫ sin (ax +b) dx = − 1a cos (ax + b) + C
∫ sec2 (ax +b) dx = 1a tan (ax + b) + C
∫ cosec (ax +b) cot (ax + b) dx = − 1a cosec (ax + b) + C
∫ cosec2 (ax +b) dx = − 1a cot (ax + b) + C
∫ sec (ax +b) tan (ax +b) dx= 1a sec (ax + b) + C
67
∫ (3x +4)5 dx = 13 . 1
6 . (3x + 4)6 + C = 1
18 (3x + 4)6 + C
∫ sin (2x +1) dx = − 12 cos (2x +1) + C
∫ cosec (3x +2) cot (3x +2) dx = − 13 cosec (3x +2) + C
∫ 15x + 4
dx = 15 . Ln l 5x +4 l + C
68
We may not be able to integrate it in the form that we are given. We may need to change it using trigonometric identities before it can be integrated.
Find ∫ (tan2 x) dx ____________________________________________ ∫ (tan2 x) dx tan2 x + 1 = sec2 x = ∫ (sec2 x – 1) dx = tan x – x + C
69
Find ∫ sin2 x dx ____________________________________________ ∫ (sin2 x) dx cos 2x = 1 – 2sin2 x
sin2 x = 12 - 1
2cos 2x
= ∫(12 − 1
2cos 2x) dx
= 12 x - 1
4 sin 2x + C
70
Sometimes it can be useful to use partial fractions .
Find ∫ x - 5(x +1)(x – 2)
dx
x - 5(x +1)(x – 2)
= A(x +1)
+ B(x – 2)
solving partial fractions as normal we find that A = 2 and B = - 1
x - 5(x +1)(x – 2)
= 2(x +1)
- 1(x – 2)
∫ 2(x +1)
- 1(x – 2)
= 2 Ln lx + 1l – Ln lx – 2l
71
If the power of the top is bigger than the bottom then we may need to do long division on the improper fraction.
Find ∫ 9x2 – 3x + 29x2 - 4
dx
____________________________________________ (9x2 – 4) (9x2 – 3x + 4) = 1 rem (– 3x + 6) this means that
9x2 – 3x + 29x2 - 4
= 1 + 6 – 3x(9x2 – 4)
next we split 6 – 3x(9x2 – 4)
using partial fractions
6 – 3x(9x2 – 4)
= 1(3x + 2)
- 2(3x – 2)
so
9x2 – 3x + 29x2 - 4
= 1 + 1(3x + 2)
- 2(3x – 2)
∫ 9x2 – 3x + 29x2 - 4
= ∫ 1 + 1(3x + 2)
- 2(3x – 2)
= x + 13 Ln l3x – 2l - 2
3 Ln l3x + 2l + C
72
Sometimes we can simplify an integral by changing the variable by choosing a good substitution.
Use the substitution u = 2x + 5 to find ∫ x√(2x +5) dx ____________________________________________________
If u = 2x + 5 then x = u - 52
-------------------------------------------------------------- ---------- If u = 2x +5 then √(2x +5) = √u = u½ ------------------------------------------------------------------------
If u = 2x + 5 then dudx
= 2 so rearranging dx = 12 du
------------------------------------------------------------------------
∫ x√(2x +5) dx = ∫ (u – 5)2
X u½ X 12 du
this multiplies out and simplifies to
∫ 12 u3/2 – 5 u1/2 du
= 110
u5/2 - 56 u3/2 + C
Don’t forget to substitute back in to give the answer in ‘x’s.
= 110
(2x + 5)5/2 - 56 (2x + 5)3/2 + C
73
Use the substitution u = 1 + sin x to find ∫ cos x sin x (1 + sin x)3 dx ____________________________________________
If u = 1 + sin x then dudx
= cos x i.e. cos x dx = du
-------------------------------------------------------------- If u = 1 + sin x then sin x = u - 1 -------------------------------------------------------------- ∫ cos x sin x (1 + sin x)3 dx = ∫ (u – 1)u3 du this multiplies out and simplifies to ∫ u4 – u3 du
= 15 u5 - 1
4 u4 + C
Substitute back in to give the answer in terms of the original ‘x’s.
= 15(1 + sin x)5 - 1
4(1 + sin x)4 + C
74
Use the substitution u2 = 2x + 5 to find ∫ x√(2x + 5) dx
If u2 = 2x + 5 then x = u2 - 52
------------------------------------------------------------------- If u2 = 2x + 5 then √(2x + 5) = u -------------------------------------------------------------------
If u2 = 2x + 5 then 2u dudx
= 2 by implicit differentiation
i.e. dx = u du -------------------------------------------------------------------
∫ x√(2x + 5) dx = ∫ (u2 - 52
) X u X u du
this multiplies out and simplifies to
∫ 12 u4 – 5
2 u2 du
= 110
u5 - 56 u3 + C
Substitute back in to give the answer in terms of the original ‘x’s.
= 110
(2x + 5)5/2 - 56 (2x + 5)3/2 + C
75
Use the substitution u = x + 1 to find ∫ 20 x(x +1)3 dx
______________________________________________________________
If u = x + 1 then when x = 0 u = 1 x = 2 u = 3
--------------------------------------------------------------------------------------- If u = x + 1 then x = u - 1 --------------------------------------------------------------------------------------- If u = x + 1 then (x + 1)3 = u3 ---------------------------------------------------------------------------------------
If u = x + 1 then dudx
= 1 i.e. dx = du
---------------------------------------------------------------------------------------
∫ 𝑑𝑑= 2𝑑𝑑 = 0 x(x + 1)3 dx = ∫ 𝑢𝑢 = 3
𝑢𝑢 =1 (u – 1)u3 du
this multiplies out and simplifies to
∫ 𝑢𝑢=3𝑢𝑢= 1 u4 – u3 du
= [ 15 u5 - 1
4 u4 ]u = 1
= [ 2435
- 814 ] - [ 1
5 - 1
4 ]
= 48.4 – 20 = 28.4
If we have limits then we need to change these as well.
76
We can use integration by parts to integrate some expressions.
∫ u dvdx
dx = uv - ∫ v dudx
dx
We use integration by parts to change an integral that is hard to do into one that is easier. We would look to use this for an expression that has one bit that would be easier to integrate if it was differentiated times by another bit that would not be any harder to integrate if it is integrated. These will often be of the form ∫ x cos x dx because the x differentiates to 1 which is easier and cos integrates to sin which is no harder. Another example is ∫ x ex dx
x differentiates to 1 and ex integrated is just ex again which is no harder.
77
Use integration by parts to find ∫ x cos x dx ____________________________________________ Choose
u = x dvdx
= cos x then
dudx
= 1 v = sin x
--------------------------------------------------------------
∫ u dvdx
dx = uv - ∫ v dudx
dx
∫ x cos x dx = x sin x - ∫ sin x . 1 dx = x sin x - (- cos x) = x sin x + cos x + C
78
Use integration by parts to find ∫ x Ln x d x ____________________________________________ Choose
u = Ln x dvdx
= x then
dudx
= 1x v = 1
2 x2
--------------------------------------------------------------
∫ u dvdx
dx = uv - ∫ v dudx
dx
∫ x Ln x dx = 12 x2 Ln x - ∫ 1
2 x2 . 1
x dx
= 12 x2 Ln x - ∫ 1
2 x dx
= 12 x2 Ln x - 1
4 x2 + C
Usually if we have an x or a power of x at the front (x2,x 3 etc) we will choose that to be the u as this will (possibly eventually) turn into 1 which is easier to integrate.
The only time we don’t choose x to be the u is when we have an Ln term in which case we let u = the Ln term.
79
If we have a power of x bigger than one, i.e. x 2 , x 3 etc, we may need to do integration by parts twice in a
row before we get rid of the x term.
Use integration by parts to find ∫ x2 ex dx _____________________________________________________________________ Choose
u = x2 dvdx
= ex then
dudx
= 2x v = ex
---------------------------------------------------------------------------------------
∫ u dvdx
dx = uv - ∫ v dudx
dx
∫ x2 ex dx = x2 e x - ∫ 2x ex dx
---------------------------------------------------------------------------------------
Now ∫ 2x ex dx still can’t be integrated so we need to do integration by parts again on just this bit --------------------------------------------------------------------------------------- Choose
u = 2x dvdx
= ex then
dudx
= 2 v = ex
---------------------------------------------------------------------------------------
∫ u dvdx
dx = uv - ∫ v dudx
dx
= 2x ex - ∫ 2 ex dx = 2x ex - 2 ex
--------------------------------------------------------------------------------------- Substituting this back into our original integral we end up with
∫ x2 ex dx = x2 ex - 2x ex + ex
80
∫ tan x dx = Ln lsec xl +C
∫ sec x dx = Ln lsec x + tan xl +C
∫ cot x dx = Ln lsin xl +C
∫ tan x dx = - Ln lcosec x + cot xl +C
81
We can use standard patterns to integrate some expressions.
Find ∫ 2xx2 + 1
dx
____________________________________________ We notice that the top is the differential of the bottom
∫ 2xx2 + 1
dx = Ln lx2 + 1l + C
82
Find ∫ cos x sin2x dx ____________________________________________ Try
y = sin3 x = (sin x)3 then dydx
= 3 cos x sin2 x
it is 3 times too big
so if y = 13 sin3 x then dy
dx = cos x sin2 x
so ∫ 13 sin3 x dx = cos x sin2 x
83
Find ∫ x(x2 + 5)3 dx
Try
y = (x2 + 5)4 then
dy
dx = 4 X 2x X (x2 + 5)3 = 8x(x2 + 5)3
which is 8 times too big
so if y = 1
8 (x2 + 5)4 then
dy
dx = x(x2 + 5)3
so ∫ x(x2 + 5)3 x dx = 1
8 (x2 + 5)4
84
Remember the Trapezium rule
∫ y dx ≈ 12 h X [y0 + yn + 2(y1 + y2 + ... + yn-1 ]
where h = b - an
85
Find ∫ 𝜋𝜋/30 sec x dx using the chain rules with four strips.
h = π/3 - 04
= π12
X
0
π12
π6
π4
π3
Y
1
1.035
1.155
1.414
2
I ≈ 12 h X [y0 + yn + 2(y1 + y2 + ... + yn-1 ]
≈ 12 X π
12 X [1 + 2 + 2(1.035 + 1.115 + 1.414)]
≈ 1.34 (3sf)
86
We can create a solid shape by revolving a curve, between the limits a and b, 360 degrees (2π radians) around the x axis. This is called a Volume of revolution. We can find the volume of this curve by integrating y squared and multiplying by pi (notice this is very similar to πr2 the area of a circle).
a
b
Volume = π ∫ 𝑏𝑏𝑎𝑎 y2 dx
87
The region R is bounded by the curve with equation y =
sin 2x, the x axis and the lines x = 0 and x = π2 .
Find the volume of the solid formed when the region is rotated 2π radians about the x axis. ____________________________________________________
Volume of revolution = π ∫ ba y2 dx
---------------------------------------------------------------------
Remember that
sin2 x = 12 (1 – cos 2x)
so sin2 2x = 12 (1 – cos 4x)
---------------------------------------------------------------------
π ∫ ba y2 dx
= π ∫ π/20 sin2 2x dx
= π ∫ π/20
12 (1 – cos 4x) dx
= π [ 12x - 1
2 sin 4x ]π/2
= π [ (π4 – 0) – (0 - 0) ]
88
The curve C has parametric equations x = t(1 + t)
y = 11 + t
where t is the parameter and t ≥ 0. The region R is bounded by the curve C, the x axis and the lines x = 0 and x = 2. Find the volume of the solid formed when the region R is rotated 2π radians about the x axis. ________________________________________________________
Volume of revolution = π ∫ ba y2 dx
------------------------------------------------------------------------------- If x = t + t2 then when x = 2 t = 1 dxdt
= 1 + 2t When x = 0 t = 0
dx = (1 + 2t) dt
-------------------------------------------------------------------------------
π ∫ ba y2 dx
= π ∫ 10
1(1 + t)2 (1 + 2t) dt
= π ∫ 10
2(1 + t)
- 1(1 + t)2 dt by partial fractions
= π [ 2 Ln l 1 + t l + 1(1 + t)
]1
= π [ ( 2 Ln 2 + 12 ) – (0 + 1) ]
= π ( 2 Ln 2 - 12 )
89
We can use integration to solve differential equations. We always get all the y things on one side, the x things on the other side and then integrate.
Find a general equation to the differential equation
(1 + x2) dydx
= x tan y
________________________________________________
Get all the x things on one side and the y things on the other and then integrate.
∫ 1tan y
dy = ∫ x1 + x2 dx
∫ cot y dy = 12 ∫ 2x
1 + x2 dx
Ln lsin yl = 12 Ln l1 +x2l + C
-------------------------------------------------------------------
Another way we could have finished off is by choosing our constant to be = Ln k instead (it doesn’t change anything it’s still a constant!)
Ln lsin yl = 12 Ln l1 +x2l + Ln k
Ln lsin yl = Ln l k√(1 + x2 ) l so sin y = k√(1 + x2)
this is just an alternative way which makes the final answer neater.
90
Sometimes we are given boundary conditions which allow us to calculate exactly what the equation is without any need for C.
Find a particular solution to the differential equation dydx
= -3(y – 2)(2x + 1)(x + 2)
if x = 1 when y = 4.
__________________________________________________________________
Get all the x things on one side and the y things on the other and then integrate.
∫ 1(y – 2)
dy = ∫ -3(2x + 1)(x + 2)
dx
taking partial fractions
∫ 1(y – 2)
dy = ∫ 1(x + 2)
- 2(2x + 1)
dx
Ln ly – 2l = Ln lx + 2l - Ln l2x + 1l + Ln k
Ln ly -2l = Ln l k� 𝑑𝑑+22𝑑𝑑 + 1
� l
remove ‘Ln’s
y – 2 = k� 𝑑𝑑+22𝑑𝑑 + 1
�
Substitute x = 1 when y = 4
4 – 2 = k� 1 + 22.1 + 1
�
k = 2
y = 2 + 2� 𝑑𝑑+22𝑑𝑑 + 1
�
which can be simplified to
y = 3 + � 32𝑑𝑑 + 1
�
91
92
