Pure FP3
Revision Notes
May 2013
2
FP3 09/02/2016 SDB
Contents
1
Hyperbolic functions ............................................................................................................ 3
Definitions and graphs.............................................................................................................................................. 3
Addition formulae, double angle formulae etc. ........................................................................................................ 3
Osborne’s rule ........................................................................................................................................................................3
Inverse hyperbolic functions .................................................................................................................................... 4
Graphs ....................................................................................................................................................................................4
Logarithmic form ....................................................................................................................................................................4
Equations involving hyperbolic functions ................................................................................................................ 5
2
Further coordinate systems .................................................................................................. 6
Ellipse ....................................................................................................................................................................... 6
Hyperbola ................................................................................................................................................................. 6
Parabola .................................................................................................................................................................... 7
Parametric differentiation ......................................................................................................................................... 7
Tangents and normals ............................................................................................................................................... 8
Finding a locus ......................................................................................................................................................... 9
3
Differentiation...................................................................................................................... 10
Derivatives of hyperbolic functions ....................................................................................................................... 10
Derivatives of inverse functions ............................................................................................................................. 10
4
Integration............................................................................................................................ 12
Standard techniques ................................................................................................................................................ 12
Recognise a standard function .............................................................................................................................................. 12
Using formulae to change the integrand ............................................................................................................................... 12
Reverse chain rule ................................................................................................................................................................ 12
Standard substitutions ........................................................................................................................................................... 13
Integration inverse functions and ln x ..................................................................................................................................14
Reduction formulae ................................................................................................................................................ 14
Arc length ............................................................................................................................................................... 18
Area of a surface of revolution ............................................................................................................................... 19
5
Vectors .................................................................................................................................. 22
Vector product ........................................................................................................................................................ 22
The vectors i, j and k ............................................................................................................................................. 22
Properties................................................................................................................................................................ 22
Component form .................................................................................................................................................... 23
Applications of the vector product ......................................................................................................................... 23
Volume of a parallelepiped .................................................................................................................................... 24
Triple scalar product ............................................................................................................................................... 24
Volume of a tetrahedron ......................................................................................................................................... 25
Equations of straight lines ...................................................................................................................................... 25
Vector equation of a line....................................................................................................................................................... 25
Cartesian equation of a line in 3-D ....................................................................................................................................... 26
Vector product equation of a line.......................................................................................................................................... 26
Equation of a plane ................................................................................................................................................. 27
Scalar product form .............................................................................................................................................................. 27
Cartesian form...................................................................................................................................................................... 27
Vector equation of a plane ................................................................................................................................................... 28
Distance from origin to plane............................................................................................................................................... 29
Distance between parallel planes ......................................................................................................................................... 30
Distance of a point from a plane ............................................................................................................................ 31
Line of intersection of two planes .......................................................................................................................... 32
Angle between line and plane ................................................................................................................................ 33
Angle between two planes ..................................................................................................................................... 34
Shortest distance between two skew lines .............................................................................................................. 34
6
Matrices................................................................................................................................ 35
Basic definitions..................................................................................................................................................... 35
Dimension of a matrix ......................................................................................................................................................... 35
Transpose and symmetric matrices ...................................................................................................................................... 35
Identity and zero matrices .................................................................................................................................................... 35
Determinant of a 3  3 matrix ................................................................................................................................ 35
Properties of the determinant ............................................................................................................................................... 36
Singular and non-singular matrices ........................................................................................................................ 36
Inverse of a 3  3 matrix ...................................................................................................................................... 36
Cofactors .............................................................................................................................................................................. 36
Finding the inverse............................................................................................................................................................... 37
Properties of the inverse....................................................................................................................................................... 37
Matrices and linear transformations ....................................................................................................................... 38
Linear transformations ......................................................................................................................................................... 38
Base vectors 𝒊, 𝒋, 𝒌 ............................................................................................................................................................ 38
Image of a line ....................................................................................................................................................... 40
Image of a plane 1 .................................................................................................................................................. 40
Image of a plane 2 .................................................................................................................................................. 40
7
Eigenvalues and eigenvectors ............................................................................................. 41
Definitions ............................................................................................................................................................. 41
2  2 matrices ......................................................................................................................................................... 41
Orthogonal matrices ............................................................................................................................................... 42
Normalised eigenvectors ...................................................................................................................................................... 42
Orthogonal vectors ............................................................................................................................................................... 42
Orthogonal matrices ............................................................................................................................................................. 42
Diagonalising a 2  2 matrix .................................................................................................................................. 43
Diagonalising 2  2 symmetric matrices ................................................................................................................ 44
Eigenvectors of symmetric matrices .................................................................................................................................... 44
Diagonalising a symmetric matrix ....................................................................................................................................... 44
3  3 matrices ......................................................................................................................................................... 46
Finding eigenvectors for 3  3 matrices. .............................................................................................................................. 46
Diagonalising 3  3 symmetric matrices .............................................................................................................................. 47
2
FP3 09/02/2016 SDB
1
Hyperbolic functions
y
Definitions and graphs
𝐬𝐢𝐧𝐡 𝒙 =
𝟏
𝟐
5
(𝒆𝒙 − 𝒆−𝒙 )
5
3
4
−4
−3
−2
(𝒆𝒙 + 𝒆−𝒙 )
𝐭𝐚𝐧𝐡 𝒙 =
y
−1
1
2
x
2
1
3
4
5
6
−5
−4
−3
−2
(𝒆𝒙 −𝒆−𝒙 )
4 𝒙
𝐬𝐢𝐧𝐡
=
𝐜𝐨𝐬𝐡 𝒙
(𝒆𝒙 +𝒆−𝒙 )
3
2
1
3
1
−5
𝟐
4 y
2
−6
𝟏
𝐜𝐨𝐬𝐡 𝒙 =
−6
−5
−4
−3
−2
−1
1
2
3
4
5
−1
−1
x
−6
−2
−1
1
2
3
4
−2
5
−1
−3
−2
−4
You should be able to draw the graphs of cosech x, sech x and coth x from the above:
cosech x
sech x
4 y
coth x
4 y
4 y
3
3
3
2
2
2
1
1
1
x
−6
−5
−4
−3
−2
−1
1
−1
2
3
−6
4
−5
5
−4
x
−6
−3
−2
−1
1
−5
2
−4
3
−3
4
−2
x
5
−1
1
2
3
4
−1
−1
−2
−2
−2
−3
−3
−3
−4
−4
−4
Addition formulae, double angle formulae etc.
The standard trigonometric formulae are very similar to the hyperbolic formulae.
Osborne’s rule
If a trigonometric identity involves the product of two sines, then we change the sign to write
down the corresponding hyperbolic identity.
Examples:
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
 sinh(𝐴 + 𝐵) = sinh 𝐴 cosh 𝐵 + cosh 𝐴 sinh 𝐵
no change
but cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
 cosh(𝐴 + 𝐵) = cosh 𝐴 cosh 𝐵 + sinh 𝐴 sinh 𝐵
and 1 + tan2 A

= sec2 A
1 – tanh2 A = sech2 A
FP3 09/02/2016 SDB
product of two sines, so change sign
tan2 𝐴 =
sin2 𝐴
cos2 𝐴
, product of two sines, so change sign
3
5
Inverse hyperbolic functions
Graphs
Remember that the graph of 𝑦 = 𝑓 −1 (𝑥) is the reflection of 𝑦 = 𝑓(𝑥) in 𝑦 = 𝑥.
y = arsinh x
4 y
y = arcosh x
4 y
sinh x
3
−4
−3
−2
−1
−1
−2
x
1
2
−6
3
−5
4
5
2
−4
−3
−2
−1
−3
x
1
−1
2
3
4
5
−6
−5
−4
−3
−2
−1
6
−1
x
1
2
3
4
5
6
−2
−3
−2
−4
tanh x
1
arcosh x
1
6
artanh x
3
2
arsinh x
1
−5
4 y
cosh x
3
2
−6
y = artanh x
−4
−3
Notice

−4
arcosh x is a function defined so that arcosh x  0.
there is only one value of arcosh x.
However, the equation cosh z = 2, has two solutions, +arcosh 2 and –arcosh 2.
Logarithmic form
1)
y = arsinh x
1

sinh 𝑦 =

𝑒 2𝑦 − 2𝑥𝑒 𝑦 − 1 = 0

𝑒𝑦 =
But
𝑒 𝑦 > 0 and 𝑥 − √𝑥 2 + 1 < 0

𝒚 = 𝐚𝐫𝐬𝐢𝐧𝐡 𝒙 = 𝐥𝐧(𝒙 + √𝒙𝟐 + 𝟏)
2)
2
(𝑒 𝑦 − 𝑒 −𝑦 ) = 𝑥
2𝑥±√4𝑥 2 +4
2
= 𝑥 + √𝑥 2 + 1 or 𝑥 − √𝑥 2 + 1
 𝑒 𝑦 = 𝑥 + √𝑥 2 + 1
only
y = arcosh x
1

cosh 𝑦 =

𝑒 2𝑦 − 2𝑥𝑒 𝑦 + 1 = 0

𝑒𝑦 =

𝑦 = arcosh 𝑥 = ln(𝑥 + √𝑥 2 − 1)
2
(𝑒 𝑦 + 𝑒 −𝑦 ) = 𝑥
2𝑥±√4𝑥 2 −4
2
= 𝑥 ± √𝑥 2 − 1
both roots are positive
or ln(𝑥 − √𝑥 2 − 1)
It can be shown that ln(𝑥 − √𝑥 2 − 1) = − ln(𝑥 + √𝑥 2 − 1)

𝑦 = arcosh 𝑥 = ± ln(𝑥 + √𝑥 2 − 1)
But arcosh x is a function and therefore has only one value (positive)

𝒚 = 𝐚𝐫𝐜𝐨𝐬𝐡 𝒙 = 𝐥𝐧(𝒙 + √𝒙𝟐 − 𝟏)
3) Similarly artan x =
4
1
2
1+𝑥
ln (
1−𝑥
)
( x  1)
(|𝑥| < 1)
FP3 09/02/2016 SDB
Equations involving hyperbolic functions
It would be possible to solve 6 sinh 𝑥 − 2 cosh 𝑥 = 7 using the 𝑅 sinh(𝑥 − 𝛼)
technique from trigonometry, but it is easier to use the exponential form.
Example: Solve 6 sinh 𝑥 − 2 cosh 𝑥 = 7
Solution:
6 sinh 𝑥 − 2 cosh 𝑥 = 7
1
1

6 × 2 (𝑒 𝑥 − 𝑒 −𝑥 ) − 2 × 2 (𝑒 𝑥 + 𝑒 −𝑥 ) = 7

2𝑒 2𝑥 − 7𝑒 𝑥 − 4 = 0

(2𝑒 𝑥 + 1)(𝑒 𝑥 − 4) = 0

𝑒𝑥 = − 2

x = ln 4
1
(not possible)
or 4
In other cases, the ‘trigonometric’ solution may be preferable
Example: Solve cosh 2x + 5 sinh x – 4 = 0
Solution:
cosh 2x + 5 sinh x – 4 = 0

1 + 2 sinh2 x + 5 sinh x – 4 = 0

2 sinh2 x + 5 sinh x – 3 = 0

(2sinh x – 1)(sinh x + 3) = 0

sinh 𝑥 =

x = arsinh 0.5 or arsinh (–3)

𝑥 = ln(0.5 + √0.52 + 1)
1
2
note use of Osborn’s rule
or − 3
or ln((−3) + √(−3)2 + 1)
using log form of
inverse

1+√5
𝑥 = ln (
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2
) or ln(√10 − 3)
5
2
Further coordinate systems
y
Ellipse
b
Cartesian equation
𝑥2
𝑎2
+
𝑦2
𝑏2
x
=1
–a
a
Parametric equations
–b
𝑥 = 𝑎 cos 𝜃 , 𝑦 = 𝑏 sin 𝜃
y
Foci at S (ae, 0) and S (–ae, 0)
Directrices at x = ±
x
𝑎
–ae
ae
𝑒
Eccentricity e < 1, b2 = a2(1 – e2)
x=–
𝑎
x=
𝑒
P
N
x
S
x=
Hyperbola
Cartesian equation
𝑎2
−
𝑦2
𝑏2
𝑒
y
An ellipse can be defined as the locus of a
point P which moves so that PS = e PN,
where S is the focus, e < 1
and N lies on the directrix.
𝑥2
𝑎
𝑎
𝑒
𝑏
y
y= 𝑥
𝑎
=1
Parametric equations
−a
x
a
𝑥 = 𝑎 cosh 𝜃 , 𝑦 = 𝑏 sinh 𝜃
(𝑥 = 𝑎 sec 𝜃 , 𝑦 = 𝑏 tan 𝜃 also work)
Asymptotes
𝑥
𝑎
= ±
𝑦
𝑏
y=− 𝑥
𝑎
𝑏
Foci at S (ae, 0) and S (–ae, 0)
Directrices at x = ±
𝑎
𝑒
P
Eccentricity e >1, b2 = a2(e2 – 1)
N
A hyperbola can be defined as the locus of a
point P which moves so that PS = e PN,
where S is the focus, e > 1 and N lies on the directrix.
6
S
FP3 09/02/2016 SDB
x
𝑦2
𝑥2
𝑐
𝑑2
−
2
=1
𝑥=
𝑎
𝑒
is a hyperbola with foci on the y-axis,
Parabola
Cartesian equation
𝑦 2 = 4𝑎𝑥
P
N
Parametric equations
𝑥 = 𝑎𝑡 2 , 𝑦 = 2𝑎𝑡
.
Focus at S (a, 0)
S
Directrix at x = – a
x=−a
A parabola can be defined as the locus of a
point P which moves so that PS = PN,
where S is the focus, N lies on the directrix
and eccentricity e = 1.
Parametric differentiation
From the chain rule

𝒅𝒚
𝒅𝒙
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=
𝑑𝑦
𝑑𝜃
𝒅𝒚
𝒅𝜽
𝒅𝒙
𝒅𝜽
=
𝑑𝑦
𝑑𝑥
or
×
𝑑𝑥
𝑑𝜃
𝒅𝒚
𝒅𝒙
=
𝒅𝒚
𝒅𝒕
𝒅𝒙
𝒅𝒕
using any parameter
7
Tangents and normals
It is now easy to find tangents and normals.
Example: Find the equation of the normal to the curve given by the parametric equations
𝜋
x = 5 cos  , y = 8 sin  at the point where  = 3
Solution:
𝜋
When  = 3 ,
5

x = 2,
𝑑𝑦
and
𝑑𝑥
cos 𝜃 =
1
and sin 𝜃 =
2
√3
2
y = 4√3
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
=
8 cos 𝜃
−5 sin 𝜃
=
−8
when 𝜃 =
5√3
𝜋
3
5√3

gradient of normal is

equation of normal is 𝑦 − 4√3 =

5√3 𝑥 − 8𝑦 =
8
5√3
8
5
(𝑥 − 2)
17√3
2
Sometimes normal, or implicit, differentiation is (slightly) easier.
6
Example: Find the equation of the tangent to xy = 36, or x = 6t, 𝑦 = , at the point
𝑡
where t = 3.
Solution:
𝑑𝑦
𝑑𝑥
When t = 3, x = 18 and y = 2.
can be found in two (or more!) ways:
𝑑𝑦
𝑑𝑥

𝑑𝑦
𝑑𝑥
=
=
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
−1
𝑡2
−6𝑡 −2
=
=
6
−1
9
,

xy = 36
when 𝑡 = 3
 equation of tangent is 𝑦 − 2 =
−1
9

𝑑𝑦

𝑑𝑦
𝑑𝑥
𝑑𝑥
=
=
𝑦=
36
𝑥
−36
𝑥2
−36
182
=
−1
9
,
when x = 18
(𝑥 − 18)
 x + 9y – 36 = 0
8
FP3 09/02/2016 SDB
Finding a locus
First find expressions for x and y coordinates in terms of a parameter, t or  , then
eliminate the parameter to give an expression involving only x and y, which will be the
equation of the locus.
𝑥2
Example: The tangent to the ellipse
9
+
𝑦2
16
= 1, at the point P, (3 cos  , 4 sin  ),
crosses the x-axis at A, and the y-axis at B.
Find an equation for the locus of the mid-point of AB as P moves round the
ellipse, or as  varies.
Solution:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
=
4 cos 𝜃
−3 sin 𝜃
 equation of tangent is 𝑦 − 4 sin 𝜃 =

4 cos 𝜃
−3 sin 𝜃
(𝑥 − 3 cos 𝜃)
3𝑦 sin 𝜃 + 4𝑥 cos 𝜃 = 12 cos2 + 12 sin2 = 12
Tangent crosses x-axis at A when y = 0,  x =
and crosses y-axis at B when x = 0,  𝑦 =
 mid-point of AB is (
Here x =
3
2 cos 𝜃
 cos 𝜃 =
3
2𝑥
2 cos 𝜃
and 𝑦 =
,
4
2sin 𝜃
,
4
sin 𝜃
)
2
sin 𝜃
and sin 𝜃 =
 equation of the locus is
FP3 09/02/2016 SDB
3
3
cos 𝜃
9
4𝑥 2
2
𝑦
+
4
𝑦2
= 1
since cos2 𝜃 + sin2 𝜃 = 1
9
3
Differentiation
Derivatives of hyperbolic functions
𝑦 = sinh 𝑥 =

𝑑𝑦
𝑑𝑥
=
1
2

𝑑𝑦
𝑑𝑥
2
(𝑒 𝑥 − 𝑒 −𝑥 )
(𝑒 𝑥 + 𝑒 −𝑥 ) = cosh x
𝑑(cosh 𝑥)
and, similarly,
Also,
1
𝑑𝑥
sinh 𝑥
𝑦 = tanh 𝑥 =
=
= sinh 𝑥
cosh 𝑥
cosh 𝑥 cosh 𝑥−sinh 𝑥 sinh 𝑥
cosh2 𝑥
=
1
cosh2 𝑥
= sech2 𝑥
In a similar way, all the derivatives of hyperbolic functions can be found.
𝒇(𝒙)
𝒇′ (𝒙)
sinh x
cosh x
tanh x
cosh x
sinh x
sech2x
all positive
coth x
cosech x
sech x
– cosech2x
– cosech x coth x
– sech x tanh x
all negative
Notice: these are similar to the results for sin x, cos x, tan x etc., but the minus signs do not
always agree.
The minus signs are ‘wrong’ only for cosh x and sech x
1
(= cosh 𝑥).
Derivatives of inverse functions
y = arsinh x
 sinh y = x

𝑑𝑦

𝑑(arsinh 𝑥)
10
𝑑𝑥
=
𝑑𝑥

1
cosh 𝑦
=
=
cosh 𝑦
𝑑𝑦
𝑑𝑥
= 1
1
√1+sinh2 𝑦
1
√1+𝑥 2
FP3 09/02/2016 SDB
The derivatives for other inverse hyperbolic functions can be found in a similar way.
You can also use integration by substitution to find the integrals of the 𝑓 ′ (𝑥) column
𝒇′ (𝒙)
𝒇(𝒙)
1
arcsin x
√1−𝑥 2
−1
arccos x
√1−𝑥 2
1
arctan x
1+𝑥 2
1
arsinh x
√1+𝑥 2
1
arcosh x
√𝑥 2 −1
1
artanh x
1
2
ln (
substitution needed for integration
1−𝑥
1
)
1
Note that ∫
𝑑𝑥 =
1−𝑥 2
2
1
∫ 1+𝑥 +
1 – cos2 u = sin2 u
 use x = cos u
1 + tan2 u = sec2 u
 use x = tan u
1 + sinh2 u = cosh2 u
 use x = sinh u
cosh2 u – 1 = sinh2 u
 use x = cosh u
partial fractions, see example below
1−𝑥 2
1
 use x = sin u
1 – tanh2 u = sech2 u  use x = tanh u
1−𝑥 2
1+𝑥
1 – sin2 u = cos2 u
1
1−𝑥
𝑑𝑥 =
1
2
ln (
1+𝑥
1−𝑥
) +c
With chain rule, product rule and quotient rule you should be able to handle a large variety of
combinations of functions.
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11
4
Integration
Standard techniques
Recognise a standard function
Examples: ∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + c
∫ sech 𝑥 tanh 𝑥 𝑑𝑥 = −sec 𝑥 + c
Using formulae to change the integrand
Examples: ∫ tan2 𝑥 𝑑𝑥 = ∫ 1 + sec 2 𝑥 𝑑𝑥 = 𝑥 + tan 𝑥 + c
∫ cos 2 𝑥 𝑑𝑥 =
1
2
∫ sinh2 𝑥 𝑑𝑥 =
1
2
∫ 1 + cos 2𝑥 𝑑𝑥 =
1
2
∫ cosh 2𝑥 − 1 𝑑𝑥 =
1
(𝑥 + 2 sin 2𝑥) + 𝑐
1
1
(2 sinh 2𝑥 − 𝑥) + 𝑐
2
Reverse chain rule
Notice the chain rule pattern, guess an answer and differentiate to find the constant.
Example: ∫ cos 2 𝑥 sin 𝑥 𝑑𝑥
𝑑(cos3 𝑥)
𝜕𝑥
so try u3  cos3 x
so divide by –3
1
∫ cos 2 𝑥 sin 𝑥 𝑑𝑥 = − 3 cos3 𝑥 + 𝑐
Example: ∫ 𝑥 2 (2𝑥 3 − 7)4 𝑑𝑥
𝑑(2𝑥 3 − 7)
‘looks like’ 𝑢4
𝜕𝑢
𝜕𝑥
so try u5  (2𝑥 3 − 7)5
5
𝑑𝑥

𝜕𝑢
= 3cos2 x (– sin x) = – 3cos2 x sin x
𝑑𝑥

‘looks like’ 𝑢2
= 5(2𝑥 3 − 7)4 × 6𝑥 2 = 30(2𝑥 3 − 7)4
∫ 𝑥 2 (2𝑥 3 − 7)4 𝑑𝑥 =
1
30
so divide by 30
(2𝑥 3 − 7)5 + 𝑐
Example: ∫ sech4 𝑥 tanh 𝑥 𝑑𝑥
=
∫ sech3 𝑥 (sech 𝑥 tanh 𝑥) 𝑑𝑥
𝑑(sech4 𝑥)
𝑑𝑥

12
= −4 sech3 𝑥 sech 𝑥 tanh 𝑥
‘looks like’
𝑢3
𝜕𝑢
𝜕𝑥
so try u4 = sech4 x
so divide by 4
1
∫ sech4 𝑥 tanh 𝑥 𝑑𝑥 = − 4 sech4 𝑥 + 𝑐
FP3 09/02/2016 SDB
Standard substitutions
1
∫ 𝑎2+𝑏2𝑥2 𝑑𝑥
1
∫ √𝑎2+𝑏2𝑥2 𝑑𝑥
1
∫ 𝑎2−𝑏2𝑥2 𝑑𝑥
1
∫ √𝑏2𝑥2 −𝑎2 𝑑𝑥
𝑏𝑥 = 𝑎 tan 𝑢
better than bx = a sinh u when there is no √
𝑏𝑥 = 𝑎 sinh 𝑢
better than bx = a tan u when there is √
𝑏𝑥 = 𝑎 tanh 𝑢
or use partial fractions
𝑏𝑥 = 𝑎 cosh 𝑢
better than bx = a sec u when there is √
For more complicated integrals like
1
∫ 𝑝𝑥2 +𝑞𝑥+𝑟 𝑑𝑥
or ∫
1
𝑑𝑥
√𝑝𝑥 2 +𝑞𝑥+𝑟
complete the square to give 𝑝(𝑥 + 𝑎)2 + 𝑏 and then use a substitution similar to one of the
four above.
1
Example: ∫
𝑑𝑥
√4𝑥 2 −8𝑥−5
=
4𝑥 2 − 8𝑥 − 5 = 4(𝑥 2 − 2𝑥 + 1) − 9
=
4(𝑥 − 1)2 − 9
1
∫ √4(𝑥−1)2 −9 𝑑𝑥
 2 dx = 3 sinh u du
Substitute 2(x – 1) = 3 cosh u
=
=
1
∫ √9(cosh2 𝑢−1)
1
∫ 𝑑𝑢
2
3 sinh 𝑢
2
= u+c =
𝑑𝑢
1
2𝑥−2
arcosh (
2
3
) + c
Important tip
∫√
𝑥𝑛
𝑎2 ±𝑥 2
𝑑𝑥
is best done with the substitution
u (or u2) = a2  x2, when n is odd,
or a trigonometric or hyperbolic function when n is even.
FP3 09/02/2016 SDB
13
Integration inverse functions and ln x
To integrate inverse trigonometric or hyperbolic functions and ln x we use integration by parts
with
𝑑𝑣
𝑑𝑥
=0
Example:
Find ∫ arctan 𝑥 𝑑𝑥
Solution:
I = ∫ arctan 𝑥 𝑑𝑥
take u = arctan x
and
𝑑𝑣
𝑑𝑥
=1

𝑑𝑢
𝑑𝑥
=
1
1+𝑥 2
 v = x
1

I = x arctan x  ∫ 𝑥 × 1+𝑥2 𝑑𝑥

I = ∫ arctan 𝑥 𝑑𝑥 = x arctan x 
Example:
Find ∫ arcosh 𝑥 𝑑𝑥
Solution:
I = ∫ arcosh 𝑥 𝑑𝑥
1
2
ln (1 + x 2) + c
take u = arcosh x
and
1

I = x arcosh x  ∫ 𝑥 ×
√

I = ∫ arcosh 𝑥 𝑑𝑥 = x arcosh x 
𝑥2 −1
𝑑𝑣
𝑑𝑥
=1

𝑑𝑢
𝑑𝑥
=
1
√𝑥 2 −1
 v = x
𝑑𝑥
√𝑥 2 − 1 + c
Reduction formulae
The first step in finding a reduction formula is usually often integration by parts (sometimes
twice). The following examples show a variety of techniques.
Example 1: 𝐼𝑛 = ∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥.
(a)
(b)
Find a reduction formula,
Find 𝐼0 , and (c) find 𝐼4
Solution:
(a)
Integrating by parts
𝑢 = 𝑥𝑛
and
14
𝑑𝑣
𝑑𝑥

𝑑𝑢
𝑑𝑥
= 𝑛𝑥 𝑛−1
= 𝑒𝑥  v = e x

𝐼𝑛 = 𝑥 𝑛 𝑒 𝑥 − ∫ 𝑛𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥

𝐼𝑛 = 𝑥 𝑛 𝑒 𝑥 − 𝑛𝐼𝑛−1
FP3 09/02/2016 SDB
(b)
𝐼0 = ∫ 𝑒 𝑥 𝑑𝑥
(c)
Using the reduction formula
= ex + c
𝐼4 = 𝑥 4 𝑒 𝑥 − 4𝐼3 = 𝑥 4 𝑒 𝑥 − 4(𝑥 3 𝑒 𝑥 − 3𝐼2 )
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12(𝑥 2 𝑒 𝑥 − 2𝐼1 )
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24(𝑥𝑒 𝑥 − 𝐼0 )
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24𝑥𝑒 𝑥 + 24𝑒 𝑥 + 𝑐
since 𝐼0 = 𝑒 𝑥 + 𝑐
π
Example 2: Find a reduction formula for 𝐼𝑛 = ∫0 ⁄2 sinn 𝑥 𝑑𝑥 .
π
Use the formula to find 𝐼6 = ∫0 ⁄2 sin6 𝑥 𝑑𝑥
Solution:
π⁄
2 sin𝑛 𝑥
𝐼𝑛 = ∫0
π
= ∫0 ⁄2 sin𝑛−1 𝑥 sin 𝑥 𝑑𝑥
𝑑𝑥
take u = sin n  1 x 
and

𝐼𝑛
π⁄
2
𝑑𝑣
𝑑𝑥
𝑑𝑢
𝑑𝑥
= (n  1) sin n  2 x cos x
= sin x  v =  cos x
π
 ∫0 ⁄2  cos 𝑥 (𝑛  1) sin𝑛  2 𝑥 cos 𝑥 𝑑𝑥
= [  cos 𝑥 sin𝑛  1 𝑥]0
π
= 0 + (𝑛  1) ∫0 ⁄2 cos2 𝑥 sin𝑛  2 𝑥 𝑑𝑥
π
= (𝑛  1) ∫0 ⁄2(1 − sin2 𝑥 ) sin𝑛  2 𝑥 𝑑𝑥
π
π
= (𝑛  1) ∫0 ⁄2 sin𝑛  2 𝑥 𝑑𝑥 − (𝑛  1) ∫0 ⁄2 sin𝑛 𝑥 𝑑𝑥

In
= (n  1) In2  (n  1) In

In
=
Now

𝑛−1
𝑛
I6
I6
FP3 09/02/2016 SDB
=
5
16
In2 .
=
5
6
π⁄
21
∫0
I4
=
𝑑𝑥
5
6
=
3
5
4
6
× I2 =
5𝜋
32
3
1
4
2
× × I0
.
15
Example 3: Find a reduction formula for
Solution:
𝐼𝑛 = ∫ sec 𝑛 𝑥 𝑑𝑥.
𝐼𝑛 = ∫ sec 𝑛 𝑥 𝑑𝑥 = ∫ sec 𝑛−2 𝑥 sec 2 𝑥 𝑑𝑥
𝑑𝑣
and

𝑑𝑢

take u = secn  2x
= sec2 x 
𝑑𝑥
= (n  2) secn  3x sec x tan x
𝑑𝑥
v = tan x
𝐼𝑛 = secn  2x tan x  ∫ tan 𝑥 (𝑛  2) sec 𝑛  3 𝑥 sec 𝑥 tan 𝑥 𝑑𝑥
= secn  2x tan x  (𝑛  2) ∫ tan2 𝑥 sec 𝑛  2 𝑥 𝑑𝑥
= secn  2x tan x  (𝑛  2) ∫(sec 2 𝑥 − 1) sec 𝑛  2 𝑥 𝑑𝑥
= secn  2x tan x  (𝑛  2)𝐼𝑛 + (𝑛  2)𝐼𝑛−2

(𝑛  1)𝐼𝑛 = secn  2x tan x + (𝑛  2)𝐼𝑛−2
Find a reduction formula for 𝐼𝑛 = ∫ tan𝑛 𝑥 𝑑𝑥.
Example 4:
Solution:

𝐼𝑛 = ∫ tan𝑛 𝑥 𝑑𝑥 = ∫ tan𝑛−2 𝑥 tan2 𝑥 𝑑𝑥
𝐼𝑛 = ∫ tan𝑛−2 𝑥 (sec 2 𝑥 − 1) 𝑑𝑥
= ∫ tan𝑛−2 𝑥 sec 2 𝑥 𝑑𝑥  ∫ tan𝑛−2 𝑥 𝑑𝑥

𝑛−1
tann  1x  In  2 .
0
Find a reduction formula for 𝐼𝑛 = ∫−1 𝑥 𝑛 (1 + 𝑥)2 𝑑𝑥.
Example 5:
Solution:
1
𝐼𝑛 =
0
take u = xn 
𝐼𝑛 = ∫−1 𝑥 𝑛 (1 + 𝑥)2 𝑑𝑥
and
16

𝐼𝑛 = [𝑥𝑛 
1

𝐼𝑛 = 0 
𝑛

𝐼𝑛 = 

𝐼𝑛 = 
𝑛

𝑛+3
= 

𝐼𝑛
3
𝐼𝑛
𝑛
3
(1 + 𝑥)3 ]
0
−1
= 
𝑑𝑥
= nxn 1
1
= (1 + x)2  v = 3 (1 + 𝑥)3
1
0
 ∫−1 𝑛𝑥 𝑛−1 × 3 (1 + 𝑥)3 𝑑𝑥.
0
∫ 𝑥𝑛−1 (1 + 𝑥)2 (1 + 𝑥)𝑑𝑥
3 −1
0
∫ 𝑥𝑛−1 (1 + 𝑥)2 𝑑𝑥 
3 −1
3
𝑑𝑣
𝑑𝑥
𝑑𝑢
𝐼𝑛−1
𝑛
𝑛+3
𝑛
3

𝑛
3
𝑛
0
∫ 𝑥𝑛 (1 + 𝑥)2 𝑑𝑥
3 −1
𝐼𝑛
𝐼𝑛−1
𝐼𝑛−1
FP3 09/02/2016 SDB
𝜋⁄
2
Find a reduction formula for 𝐼𝑛 = ∫0
Example 6:
Solution:
𝜋⁄
2
take u = xn 
𝑥 𝑛 cos 𝑥 𝑑𝑥
𝐼𝑛 = ∫0
𝑑𝑣
and

𝐼𝑛 =
𝜋⁄
2
[𝑥 𝑛 sin 𝑥]0
𝑥 𝑛 cos 𝑥 𝑑𝑥
𝜋⁄
2
 𝑛 ∫0
𝑑𝑥
𝜋 𝑛
(2 )

𝐼𝑛 =
(2 )

𝐼𝑛 =
(2 )
Example 7:
Solution:

𝜋 𝑛
𝜋 𝑛
𝜋⁄
2
{
 n [ 𝑥 𝑛1 (− cos 𝑥)]0
𝜋⁄
2
{
 n 0 + (𝑛 − 1) ∫0
𝑑𝑣
𝐼𝑛
FP3 09/02/2016 SDB
𝑑𝑥
𝜋⁄
2−
 ∫0
𝑑𝑢
𝑑𝑥
= (n1) xn 2
= sin x  v = cos x
cos 𝑥 × (𝑛 − 1)𝑥 𝑛−2 𝑑𝑥
𝑥 𝑛−2 cos 𝑥 𝑑𝑥
}
}
 n (𝑛 − 1) 𝐼𝑛−2
Find a reduction formula for
𝐼𝑛
= nxn 1
sin 𝑥 × 𝑥 𝑛−1 𝑑𝑥
and
𝐼𝑛 =
𝑑𝑥
= cos x  v = sin x
take u = xn1 

𝑑𝑢
sin[(𝑛−2)𝑥+2𝑥]
𝐼𝑛 = ∫
=
∫
=
∫
=
∫
=
∫
=
In  2 + 2∫ cos(𝑛 − 1)𝑥 dx
=
In  2 +
sin 𝑥
sin 𝑛𝑥
sin 𝑥
dx
dx
sin(𝑛−2)𝑥 cos 2𝑥 +cos(𝑛−2)𝑥 sin 2𝑥
sin 𝑥
dx
sin(𝑛−2)𝑥 (1−2sin2 𝑥) + cos(𝑛−2)𝑥 ×2 sin 𝑥 cos 𝑥
sin 𝑥
sin(𝑛−2)𝑥
sin 𝑥
2
𝑛−1
dx
dx + 2∫ cos(𝑛 − 2)𝑥 cos 𝑥 − sin(𝑛 − 2)𝑥 sin 𝑥 dx
using cos(A + B) = cos A cos B – sin A sin B
sin(n  1)x .
17
Arc length
All the formulae you need can be remembered from this diagram
Q
arc PQ  line segment PQ

( s)2  ( x)2 + ( y)2

(𝛿𝑥) ≈ 1 + (𝛿𝑥 )
𝛿𝑠 2
s
𝛿𝑦 2
y
P
x
and as  x  0
𝑑𝑠 2
𝑑𝑦 2
𝑑𝑥
𝑑𝑥

( ) = 1 + ( )

arc length = s =
𝑑𝑠

𝑑𝑥
2
𝑑𝑦
= √1 + ( )
𝑑𝑥
𝑑𝑦 2
∫ √1 + ( 𝑑𝑥 )
𝑑𝑥
Similarly
𝛿𝑠 2
𝛿𝑥 2
𝛿𝑠 2
𝛿𝑥 2
s = ∫ √(𝑑𝑦) + 1 𝑑𝑦
𝛿𝑦 2
and
(𝛿𝑡 ) ≈ ( 𝛿𝑡 ) + ( 𝛿𝑡 )

s = ∫ √( 𝑑𝑡 ) + ( 𝑑𝑡 )
𝑑𝑥 2
𝑑𝑥 2

(𝛿𝑦) ≈ (𝛿𝑦) + 1
𝑑𝑦 2
𝑑𝑡
or
s = ∫ √𝑥̇ 2 + 𝑦̇ 2 𝑑𝑡
for parametric equations.
2
Find the length of the curve 𝑦 = 3 𝑥
Example 1:
3⁄
2
, from
the point where x = 3 to the point where x = 8.
Solution:
15
The equation of the curve is in Cartesian form so we use
2
s
𝑑𝑦
= ∫ √1 + ( 𝑑𝑥 )
10
𝑑𝑥 .
5
𝑦 =
2
3
3
𝑥 ⁄2
𝑑𝑦

𝑑𝑥
= √𝑥
−5

s
=
8
∫3 √1
2

18
s
10
+ 𝑥 𝑑𝑥
= [ (1 + 𝑥 )
3
5
3⁄ 8
2]
3
=
2
3
× (9)
3⁄
2
−
2
3
× (4)
3⁄
2
2
= 123 .
FP3 09/02/2016 SDB
15
20
Example 2:
the cycloid
Solution:
Find the length of one arch of
x = a(t  sin t), y = a(1  cos t).
The curve is given in parametric form
so we use s = ∫ √𝑥̇ 2 + 𝑦̇ 2 𝑑𝑡 .
t=0
x = a(t  sin t), y = a(1  cos t)

𝑑𝑥

𝑥̇ 2 + 𝑦̇ 2 = a2(1  2 cos t + cos2t + sin2t) = 2a2(1  cos t)

√𝑥̇ 2 + 𝑦̇ 2 = 𝑎√2 (1 − [1 − 2 sin2 (2)]) = 2𝑎 sin ( )

s = ∫0 2𝑎 sin (2) 𝑑𝑡

s = [−4𝑎 cos (2)]
𝑑𝑡
𝑑𝑦
t=2π
= a(1  cos t), and
𝑑𝑡
= a sin t
𝑡
𝑡
2
𝑡
2𝜋
𝑡
2𝜋
0
= 4a   4a = 8a.
Area of a surface of revolution
A curve is rotated about the x-axis.
y
s
To find the area of the surface formed
between x = a and x = b, we consider a
small section of the curve, s, at a
distance of y from the x-axis.
y
x
a
b
When this small section is rotated about
the x-axis, the shape formed is
approximately a cylinder of radius y and
length s.
The surface area of this (cylindrical) shape  2 rl  2 ys
 The total surface area  ∑𝑏𝑎 2 𝑦𝑠
𝑏
and, as s  0, the area of the surface is A = ∫𝑎 2 𝑦 𝑑𝑠.
And so
We can use
𝑏
A = ∫𝑎 2 𝑦
𝑑𝑠
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑦
= √1 + ( )
remembering that
FP3 09/02/2016 SDB
𝑑𝑠
or
2
𝑑𝑥
or
𝑏
A = ∫𝑎 2 𝑦
𝑑𝑠
𝑑𝑡
𝑑𝑠
𝑑𝑡
𝑑𝑡
2
2
𝑑𝑥
𝑑𝑦
= √( ) + ( ) ,
𝑑𝑡
𝑑𝑡
as appropriate,
( s)2  ( x)2 + ( y)2
19
Example 1: Find the surface area of a sphere with radius r,
, between the planes x = a and x = b.
Solution:
The Cartesian form is most suitable here.
𝑏
A = ∫𝑎 2 𝑦
𝑑𝑠
𝑑𝑥
𝑑𝑥
x=a
x2 + y2 = r2

and
2𝑥 + 2𝑦
𝑑𝑠
𝑑𝑦
𝑑𝑥
𝑑𝑦
= √1 + ( )
𝑑𝑥

=0
𝑑𝑦
𝑑𝑥
=
−𝑥
𝑦
2
𝑑𝑥
 A =
𝑥2
𝑏
∫𝑎 2𝑦 √1 + 𝑦 2 𝑑𝑥
𝑏
= ∫𝑎 2 √𝑦 2 + 𝑥 2 𝑑𝑥
𝑏
= ∫𝑎 2𝑟 𝑑𝑥
 A
x=b
=
[2𝑟𝑥]𝑏𝑎 = 2𝑟(𝑏 − 𝑎)
since x2 + y2 = r2
since r is constant
Notice that the area of the whole sphere is from a = r to b = r giving
surface area of a sphere is 4 r2.
Historical note.
Archimedes showed that the area of a sphere is equal
to the area of the curved surface of the surrounding
cylinder.
h = 2r
Thus the area of the sphere is
A = 2 rh = 4 r2
since h = 2r.
r
20
FP3 09/02/2016 SDB
Example 2: The parabola, x = at2, y = 2at, between the
origin (t = 0) and P (t = 2) is rotated about the x-axis.
Find the surface area of the shape formed.
Solution:
The parametric form is suitable here.
𝑏
A = ∫𝑎 2 𝑦
and

𝑑𝑠

A
𝑑𝑡
𝑑𝑡
= 2𝑎𝑡 and
x
2
𝑑𝑡
𝑑𝑦
𝑑𝑡
= 2𝑎
= √(2𝑎𝑡)2 + (2𝑎)2 = 2𝑎√𝑡 2 + 1
=
2
∫0 2 2𝑎𝑡 × 2𝑎√𝑡 2 + 1 𝑑𝑡
1
= 8a2 × 3 [(𝑡 2 + 1)

P, (t = 2)
𝑑𝑡
𝑑𝑥
𝑑𝑦
= √( ) + ( )
𝑑𝑡
𝑑𝑡
𝑑𝑡
2
𝑑𝑠
𝑑𝑥
𝑑𝑠
y
A
FP3 09/02/2016 SDB
=
8𝜋𝑎2
3
(5
3⁄
2
3⁄ 2
2]
0
− 1)
21
5
Vectors
Vector product
The vector, or cross, product of a and b is
̂
𝒏
b
̂
a  b = ab sin 𝒏
̂ is a unit (length 1) vector which is
where 𝒏
perpendicular to both a and b, and  is the angle
between a and b.

a
̂ is that in which a right hand corkscrew would move when turned
The direction of 𝒏
through the angle  from a to b.
b
̂ , where −𝒏
̂ is in the
Notice that b  a = ab sin −𝒏
̂ , since the corkscrew would move in
opposite direction to 𝒏
the opposite direction when moving from b to a.

a
̂
−𝒏
Thus b  a =  a  b.
The vectors i, j and k
For unit vectors, i, j and k, in the directions of the axes
i  j = k,
j  k = i,
i  k = j,
̂
𝒌
k  i = j,
j  i = k,
j
k  j = i.
i
Properties
aa = 0
ab = 0
since  = 0

a is parallel to b
since sin  = 0   = 0 or 
or a or b = 0
a  (b + c) = a  b + a  c
a  b is perpendicular to both a and b
22
remember the brilliant demo with the straws!
from the definition
FP3 09/02/2016 SDB
Component form
Using the above we can show that
𝑎1
𝒊
𝑎2 𝑏3 − 𝑎3 𝑏2
𝑏1
a  b = (𝑎2 ) × (𝑏2 ) = (−𝑎1 𝑏3 + 𝑎3 𝑏1 ) = |𝑎1
𝑎3
𝑏3
𝑎1 𝑏2 − 𝑎2 𝑏1
𝑏1
𝒋
𝑎2
𝑏2
𝒌
𝑎3 |
𝑏3
Applications of the vector product
Area of triangle OAB =

1
2
𝑎𝑏 sin 𝜃
B
b
area of triangle OAB =
1
2
a  b

O
a
Area of parallelogram OADB is twice the area
of the triangle OAB
 area of parallelogram OADB = a  b
B
b
O
A

D
a
A
Example: A is (1, 2, 1), B is (2, 3, 0) and C is (3, 4, 2).
Find the area of the triangle ABC.
Solution:
C
1
The area of the triangle ABC = |2 ⃗⃗⃗⃗⃗
𝐴𝐵 × ⃗⃗⃗⃗⃗
𝐴𝐶 |
⃗⃗⃗⃗⃗ = b  a =
𝐴𝐵
3
(1)
−1
⃗⃗⃗⃗⃗ = c  a =
and 𝐴𝐶
𝒊
𝒋
𝒌
4
2 −3
4
(2)
−3
A
B
−1
 ⃗⃗⃗⃗⃗
𝐴𝐵 × ⃗⃗⃗⃗⃗
𝐴𝐶 = |3 1 −1| = ( 5 )
 area ABC =
FP3 09/02/2016 SDB
1
2
2
√12 + 52 + 32 =
1
2
√35
23
Volume of a parallelepiped
𝒏
̂
In the parallelepiped
h = a cos 
h
and area of base = bc sin 
 volume
a
c

= h  bc sin 

b
= a  bc sin   cos 
̂ and a
 is the angle between 𝒏
and (i)
bc sin  is the magnitude of b  c
(ii)
 a . (b  c) = a  bc sin   cos 
 volume of parallelepiped = a . (b  c)
Triple scalar product
a . (b  c)
=
𝑎1
𝑏2 𝑐3 − 𝑏3 𝑐2
𝑎
( 2 ) . (−𝑏1 𝑐3 + 𝑏3 𝑐1 )
𝑎3
𝑏1 𝑐2 − 𝑏2 𝑐1
=
𝑎1 (𝑏2 𝑐3 − 𝑏3 𝑐2 ) + 𝑎2 (−𝑏1 𝑐3 + 𝑏3 𝑐1 ) + 𝑎3 (𝑏1 𝑐2 − 𝑏2 𝑐1 )
=
𝑎1
| 𝑏1
𝑐1
𝑎2
𝑏2
𝑐2
𝑎3
𝑏3 |
𝑐3
By expanding the determinants we can show that
a . (b  c) = (a  b) . c
keep the order of a, b, c but change the order of the  and .
For this reason the triple scalar product is written as {a, b, c}
{a, b, c} = a . (b  c) = (a  b) . c
It can also be shown that a cyclic change of the order of a, b, c does not change the
value, but interchanging two of the vectors multiplies the value by 1.

24
{a, b, c} = {c, a, b} = {b, c, a } = {a, c, b} = {c, b, a} = {b, a, c}
FP3 09/02/2016 SDB
Volume of a tetrahedron
The volume of a tetrahedron is
1
3
Area of base × ℎ
The height of the tetrahedron is the same as the
height of the parallelepiped, but its base has half
the area
1

volume of tetrahedron =

volume of tetrahedron = |
6
h
a
c
b
volume of parallelepiped
1
6
{𝒂, 𝒃, 𝒄}|
Example: Find the volume of the tetrahedron ABCD,
given that A is (1, 0, 2), B is (1, 2, 2), C is (1, 1, 3) and D is (4, 0, 3).
Solution:
1
⃗⃗⃗⃗⃗ , 𝐴𝐶
⃗⃗⃗⃗⃗ , 𝐴𝐵
⃗⃗⃗⃗⃗ }|
Volume = |6 {𝐴𝐷
3
⃗⃗⃗⃗⃗
𝐴𝐷 = 𝒅 − 𝒂 = (0) ,
1
3
0
1
−2 2
⃗⃗⃗⃗⃗ , 𝐴𝐶
⃗⃗⃗⃗⃗ , ⃗⃗⃗⃗⃗
 {𝐴𝐷
𝐴𝐵 } = | 0
 volume of tetrahedron is
1
6
0
⃗⃗⃗⃗⃗
𝐴𝐶 = ( 1 ) ,
−5
−2
⃗⃗⃗⃗⃗
𝐴𝐵 = ( 2 )
0
1
−5| = 3 × 10 + 2 = 32
0
1
 32 = 5 3
Equations of straight lines
Vector equation of a line
P
r = a +  b is the equation of a line through
the point A and parallel to the vector b,
𝛼
𝑥
𝑙
𝛽
or (𝑦) = (𝑚) + 𝜆 ( ) .
𝛾
𝑧
𝑛
l
A
a
r
b
FP3 09/02/2016 SDB
25
Cartesian equation of a line in 3-D
Eliminating  from the above equation we obtain
𝑥−𝑙
𝛼
𝑦−𝑚
=
𝛽
𝑧−𝑛
=
𝛾
(= )
𝛼
𝛽
is the equation of a line through the point (l, m, n) and parallel to the vector ( ) .
𝛾
This strange form of equation is really the intersection of the planes
𝑥−𝑙
𝛼
=
𝑦−𝑚
𝑦−𝑚
and
𝛽
=
𝛽
𝑧−𝑛
𝛾
(and
𝑥−𝑙
𝛼
=
𝑧−𝑛
𝛾
).
Vector product equation of a line
P
⃗⃗⃗⃗⃗
𝐴𝑃 = r  a and is parallel to the vector b

l
A
⃗⃗⃗⃗⃗  b = 0
𝐴𝑃

(r  a)  b = 0 is the equation of a line
through A and parallel to b.
or
r  b = a  b = c is the equation of a
line parallel to b.
a
r
b
Notice that all three forms of equation refer to a line through the point A and parallel to the
vector b.
Example: A straight line has Cartesian equation
𝑥 =
2𝑦+4
5
=
3−𝑧
2
.
Find its equation (i) in the form r = a +  b, (ii) in the form r  b = c .
Solution:
First re-write the equation in the standard manner
26

𝑥−0

the line passes through A, (0, 2, 3), and is parallel to b, (2.5) or ( 5 )
1
=
𝑦−−2
2.5
=
𝑧−3
−2
1
2
−2
−4
FP3 09/02/2016 SDB
0
(i)
2
r = (−2) + 𝜆 ( 5 )
3
−4
0
2
(𝒓 − (−2)) × ( 5 ) = 0
(ii)


𝒓
3
−4
1
× (2.5)
−2
= (−2) × ( 5 ) = |0 −2
2
0
3
2
−4
𝒊
𝒋
2
5
𝒌
−7
3|
−4
= (6)
4
−7
𝒓 ×( 5 )= ( 6 ) .
−4
4
𝒏
Equation of a plane
Let 𝒏 be a vector perpendicular to the plane .
a

r
Let A be a fixed point in the plane, and P be a
general point, (x, y, z), in the plane.
Then is parallel to the plane, and therefore
perpendicular to n
⃗⃗⃗⃗⃗ . n = 0
 𝐴𝑃

P
A
Scalar product form
O
(r  a) . n = 0
 r. n = a . n = a constant, d
 r. n = d is the equation of a plane perpendicular to the vector n .
Cartesian form
𝛼
If n = (𝛽) then r . n =
𝛾
𝛼
. ( 𝛽) =  x +  y +  z
𝛾
𝑧
𝑥
(𝑦)
𝜶
  x +  y +  z = d is the Cartesian equation of a plane perpendicular to (𝜷) .
𝜸
FP3 09/02/2016 SDB
27
Example: Find the scalar product form and the Cartesian equation of the plane through the
points A, (1, 2, 5), B, (1, 0, 3) and C, (2, 1, 2).
Solution:
We first need a vector perpendicular to the plane.
A, (3, 2, 5), B, (1, 0, 3) and C, (2, 1, 2) lie in the plane
−4
−1
−2
−7
⃗⃗⃗⃗⃗ = (−2) and 𝐴𝐶
⃗⃗⃗⃗⃗ = (−1) are parallel to the plane
 𝐴𝐵
 ⃗⃗⃗⃗⃗
𝐴𝐵  ⃗⃗⃗⃗⃗
𝐴𝐶 is perpendicular to the plane
𝒊
𝒋
𝒌
12
6
−7
2
1
⃗⃗⃗⃗⃗  𝐴𝐶
⃗⃗⃗⃗⃗ = |−4 −2 −2| = (−26) = 2  (−13)
𝐴𝐵
−1
−1
using smaller numbers
 6x  13y + z = d
but A, (3, 2, 5) lies in the plane
 d = 6  3  13  2 + 5 = 3
 Cartesian equation is 6x  13y + z = 3
6
and scalar product equation is r . (−13) = 3.
1
Vector equation of a plane
P
A
r = a +  b +  c is the equation of a plane,  ,
through A and parallel to the vectors b and c.

a
r
b
c
O
Example: Find the vector equation of the plane through the points A, (1, 4, 2), B, (1, 5, 3)
and C, (4, 7, 2).
Solution:
0
3
5
4
We want the plane through A, (1, 4, 2), parallel to ⃗⃗⃗⃗⃗
𝐴𝐵 = (1) and ⃗⃗⃗⃗⃗
𝐴𝐶 = (3)
1
0
3
−2
5
4
 vector equation is r = ( 4 ) + 𝜆 (1) + 𝜇 (3) .
28
FP3 09/02/2016 SDB
Distance from origin to plane
Example: Find the distance from the origin to the plane
4x  2y + 4z = 7.
Solution: Let M be the foot of the perpendicular from the
origin to the plane. The distance of the origin from the
plane is OM.
M
We must first find the intersection of the line OM with
the plane.
𝒏
O
OM is perpendicular to the plane
4
and so is parallel to n = (−2) .
4
0
4
4
0
4
4
 the line OM is r = (0) +  (−2) =  (−2) ,
since it passes through (0, 0, 0)
and the point of intersection of OM with the plane is given by
4(4)  2( 2) + 4 (4 ) = 7
 16 + 4 + 16 = 7
  =
7
36
 ⃗⃗⃗⃗⃗⃗
𝑂𝑀 =
4
7
(−2)
36
4
⃗⃗⃗⃗⃗⃗ | = 7 √42 + 22 + 42 =
 distance = |𝑂𝑀
36
7
6
1
The distance of the origin from the plane is 1 6.
FP3 09/02/2016 SDB
29
Distance between parallel planes
1
Example: Find the distance between the parallel planes
L
 1: 2x – 6y + 3z = 9 and  2: 2x – 6y + 3z = 5
Solution: Let l be the line which passes through the
origin and which is perpendicular to both planes.
2
M
The distance between the planes will be the length of
LM, where L and M are the points of intersection of the
line l and the planes  1 and  2.
l
𝒏
2
𝒏 = (−6)
3
O
2
 the equation of l is 𝒓 = 𝜆 (−6)
3
L is given by 2  2 – 6  (6) + 3  3 = 9

⃗⃗⃗⃗⃗ =
𝑂𝐿
2
(
)
−6
49
3
9
Similarly ⃗⃗⃗⃗⃗⃗
𝑂𝑀 =
2
(−6)
49
3
5
2
(
) −
−6
49
3
9

⃗⃗⃗⃗⃗⃗ =
𝑀𝐿

⃗⃗⃗⃗⃗⃗ | =
ML = |𝑀𝐿
4
49
2
(
) =
−6
49
3
5
2
(
)
−6
49
3
4
√22 + 62 + 32 =
The distance between the planes is
30
9
 = 49

4
7
4
49
√49 =
4
7
.
FP3 09/02/2016 SDB
Distance of a point from a plane
Example: Find the distance from the point K, (5, 4, 7), to
the plane 3x  2y + z = 2.
Solution: Let M be the foot of the perpendicular from K
to the plane. The distance from the point, K, to the plane
is KM.
M
We must first find the intersection of the line KM with
the plane.
𝒏
K
KM is perpendicular to the plane
3
and so is parallel to n = (−2) .
1
It also passes through K, (5, 4, 7),
5
3
7
1
 the line KM is r = (−4) +  (−2), which meets the plane when
3(5 + 3)  2(4  2) + (7 + ) = 2
 15 + 9 + 8 +4 + 7 +  = 2
  = –2
5
3
7
1
 m = = (−4) – 2 (−2)
5
3
5
7
1
7
⃗⃗⃗⃗⃗⃗⃗ = m  k = (−4) – 2 (−2) − (−4)
 𝐾𝑀
3
= –2 (−2)
1
⃗⃗⃗⃗⃗⃗⃗ | = 2 × √32 + 22 + 12 = 2 × √14
 distance = |𝐾𝑀
For the general case, the above method gives –
The distance from the point (, ,  ) to the plane n1x + n2y + n3z + d = 0 is
|𝑛1 𝛼+𝑛2 𝛽+𝑛3 𝛾+𝑑|
√𝑛1 2 +𝑛2 2 +𝑛3 2
This formula is in the formula booklet, but is not mentioned in the text book‼
In the above example the formula gives the distance as
|3×5−2×(−4)+1×7−2|
√32 +22 +12
FP3 09/02/2016 SDB
=
28
√14
= 2√14
notice that the d is on the L.H.S. of the equation
31
Line of intersection of two planes
Example: Find an equation for the line of intersection of the planes
and
Solution:
x + y + 2z = 4
I
2x  y + 3z = 4
II
Eliminate one variable –
I + II 
3x + 5z = 8
We are not expecting a unique solution, so put one variable, z say, equal to  and find
the other variables in terms of .
z=  x=
8−5𝜆
3
8−5𝜆
I 
y = 4  x – 2z = 4 

8⁄
−5⁄
𝑥
3
3
(𝑦) = (4⁄ ) + 𝜆 (−1⁄ )
3
3
𝑧
0
1
or
8⁄
𝑥
−5
3
(𝑦) = (4⁄ ) + 𝜆 (−1)
3
3
𝑧
0
3
 2 =
4−𝜆
3
making the numbers nicer in the direction vector only
8 4
−5
which is the equation of a line through ( , , 0) and parallel to (−1) .
3 3
32
3
FP3 09/02/2016 SDB
Angle between line and plane
Let the acute angle between the line and the plane be .
First find the angle between the line and the normal vector,  .
There are two possibilities  as shown below:
l
l
n




n
(i)
n and the angle  are on the same

side of the plane
 = 90  
Example: Find the angle between the line
(ii)
n and the angle  are on opposite

sides of the plane
 =   90
𝑥+1
2
=
𝑦−2
1
=
𝑧−3
−2
and the plane 2x + 3y  7z = 5.
2
Solution:
2
The line is parallel to ( 1 ), and the normal vector to the plane is ( 3 ).
−2
a . b = ab cos 
 cos  =
7
√62

−7
21 = √22 + 12 + 22 √22 + 32 + 72 cos 
  = 27.3o
 the angle between the line and the plane,  = 90  27.3 = 62.7o
FP3 09/02/2016 SDB
33
Angle between two planes
𝒏𝟐
𝒏𝟏
If we look ‘end-on’ at the two planes, we can see
that the angle between the planes, , equals the angle
between the normal vectors.


plane 1
Example: Find the angle between the planes
2x + y + 3z = 5
and
2x + 3y + z = 7
2
Solution:
The normal vectors are (1)
2
and (3)
3
a . b = ab cos 
 cos  =
10
14

plane 2
1
10 = √22 + 12 + 32 × √22 + 12 + 32 cos 
  = 44.4o
Shortest distance between two skew lines
l1
A
It can be shown that there must be a line
joining two skew lines which is
perpendicular to both lines.
Y
This is XY and is the shortest distance
between the lines.
X
C
l2
r=a+b
r=c+d
Suppose that A and C are points on l1 and l2.
The projection of AC onto XY is of length AC cos  , where  is the angle between AC
and XY
̂
 XY = AC cos  = ⃗⃗⃗⃗⃗
𝐴𝐶 . 𝒏
perpendicular to l1 and l2.
̂ is a unit vector (length 1) parallel to XY and so
where 𝒏
But b  d is perpendicular to l1 and l2
̂ =
 𝒏
𝒃𝒅
|𝒃  𝒅|
̂
 shortest distance between l1 and l2 is d = XY = ⃗⃗⃗⃗⃗
𝐴𝐶 . 𝒏
 d = |(𝒄  𝒂)
.
𝒃𝒅
|
|𝒃  𝒅|
This result is not in your formula booklet, SO LEARN IT
34

please
FP3 09/02/2016 SDB
6
Matrices
Basic definitions
Dimension of a matrix
A matrix with r rows and c columns has dimension r  c.
Transpose and symmetric matrices
The transpose, AT, of a matrix, A, is found by interchanging rows and columns
𝑎 𝑏 𝑐
A = ( 𝑑 𝑒 𝑓)
𝑔 ℎ 𝑖

𝑎 𝑑 𝑔
AT = (𝑏 𝑒 ℎ )
𝑐 𝑓 𝑖
(AB)T = BTAT
- note the change of order of A and B.
A matrix, S, is symmetric if the elements are symmetrically placed about the leading
diagonal,
or if S = ST.
𝑎 𝑏 𝑐
Thus, S = (𝑏 𝑑 𝑒) is a symmetric matrix.
𝑐 𝑒 𝑓
Identity and zero matrices
The identity matrix
1 0 0
I = (0 1 0)
0 0 1
and the zero matrix is
0 = (0 0 0)
0 0 0
0 0 0
Determinant of a 3  3 matrix
The determinant of a 3  3 matrix, A, is
det (A) = 

=
=
𝑎 𝑏 𝑐
𝑒 𝑓| = 𝑎 |𝑒
ℎ
𝑔 ℎ 𝑖
|𝑑
𝑑
𝑓
| − 𝑏|
𝑔
𝑖
𝑓
𝑑
| + 𝑐|
𝑔
𝑖
𝑒
|
ℎ
aei  afh  bdi + bfg + cdh  ceg
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35
Properties of the determinant
1) A determinant can be expanded by any row or column using
+ − +
|− + −|
+ − +
𝑎 𝑏 𝑐
𝑎
𝑏 𝑐
e.g  = |𝑑 𝑒 𝑓| = −𝑑 |
| + 𝑒 |𝑔
ℎ 𝑖
𝑔 ℎ 𝑖
2)
𝑐
𝑎
|
−
𝑓
|
𝑖
𝑔
𝑏
|
ℎ
using the middle row and
. leaving the value unchanged
Interchanging two rows changes the sign of the determinant
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
|𝑑 𝑒 𝑓| =  |𝑎 𝑏 𝑐 |
𝑔 ℎ 𝑖
𝑔 ℎ 𝑖
which can be shown by evaluating both determinants
3) A determinant with two identical rows (or columns) has value 0.
𝑎 𝑏 𝑐
𝑏 𝑐|
𝑔 ℎ 𝑖
 = |𝑎
interchanging the two identical rows gives  =    = 0
4) det(AB) = det(A)  det(B)
this can be shown by multiplying out
Singular and non-singular matrices
A matrix, A, is singular if its determinant is zero, det(A) = 0
A matrix, A, is non-singular if its determinant is not zero, det(A) ≠ 0
Inverse of a 3  3 matrix
This is tedious, but no reason to make a mistake if you are careful.
Cofactors
𝑎 𝑏 𝑐
𝑑
𝑒 𝑓) the cofactors of a, b, c, etc. are A, B, C etc., where
In (
𝑔 ℎ 𝑖
𝑒 𝑓
A =+|
|,
ℎ 𝑖
D = |𝑏
𝑑 𝑓
|,
𝑔 𝑖
C = +|
𝑎 𝑐
𝑖|
F = |
𝑎 𝑐
𝑓| ,
I = + |𝑎
B = |
𝑐| ,
ℎ 𝑖
E = + |𝑔
𝑏 𝑐
|,
𝑒 𝑓
H =  |𝑑
G = +|
𝑑 𝑒
|,
𝑔 ℎ
𝑎 𝑏
|,
𝑔 ℎ
𝑏|
𝑑 𝑒
These are the 2  2 matrices used in finding the determinant, together with the correct
sign from
+ − +
|− + −|
+ − +
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FP3 09/02/2016 SDB
Finding the inverse
1) Find the determinant, det(A).
If det(A) = 0, then A is singular and has no inverse.
𝐴 𝐵 𝐶
2) Find the matrix of cofactors C = (𝐷 𝐸 𝐹)
𝐺 𝐻 𝐼
𝐴 𝐷 𝐺
3) Find the transpose of C, CT = (𝐵 𝐸 𝐻)
𝐶 𝐹 𝐼
4) Divide CT by det(A) to give A–1 =
𝐴 𝐷 𝐺
𝐸 𝐻)
det(𝑨)
𝐶 𝐹 𝐼
1
(𝐵
See example 10 on page 148.
Properties of the inverse
1) A1A = AA1 = I
2) (AB)1 = B1A1
Proof (AB)1 AB = I
- note the change of order of A and B.
from definition of inverse

(AB)1 AB (B1A1) = I (B1A1)

(AB)1 A (BB1)A1 = B1A1

(AB)1 A I A1 = B1A1

(AB)1 A A1 = B1A1

(AB)1 = B1A1
3) det(A1) =
FP3 09/02/2016 SDB
1
det(𝑨)
37
Matrices and linear transformations
Linear transformations
T is a linear transformation on a set of vectors if
(i)
T (x1 + x2) = T (x1) + T (x2)
(ii)
T (kx) = kT (x)
𝑥
Example:
for all vectors x
2𝑥
Show that T (𝑦) = (𝑥 + 𝑦) is a linear transformation.
𝑧
Solution:
for all vectors x and y
−𝑧
𝑥1
𝑥2
𝑥1 + 𝑥2
𝑧1
𝑧2
𝑧1 + 𝑧2
T (x1 + x2) = T ((𝑦1 ) + (𝑦2 )) = T ((𝑦1 + 𝑦2 ) )
(i)
=
2(𝑥1 + 𝑥2 )
(𝑥1 + 𝑥2 + 𝑦1 + 𝑦2 )
−𝑧1 − 𝑧2
2𝑥1
2𝑥2
−𝑧1
−𝑧2
(𝑥1 +𝑦1 ) + (𝑥2 +𝑦2 )
=

T (x1 + x2) = T (x1) + T (x2)
(ii)
T (kx) = T (𝑘 (𝑦)) = T (𝑘𝑦) =
𝑥
𝑘𝑥
𝑧

𝑘𝑧
= T (x1) + T (x2)
2𝑘𝑥
(𝑘𝑥 + 𝑘𝑦)
−𝑘𝑧
2𝑥1
= k (𝑥1 +𝑦1 ) = kT (x)
−𝑧1
T (kx) = kT (x)
Both (i) and (ii) are satisfied, and so T is a linear transformation.
All matrices can represent linear transformations.
Base vectors 𝒊, 𝒋, 𝒌
1
𝑖 = (0),
0
0
𝒋 = (1),
0
0
𝒌 = (0)
1
Under the transformation with matrix
𝑎 𝑏 𝑐
𝑒 𝑓)
𝑔 ℎ 𝑖
(𝑑
𝑎
1
(0) → (𝑑 )
𝑔
0
the first column of the matrix
0
𝑏
(1) → (𝑒 )
0
ℎ
the second column of the matrix
𝑐
0
(0) → (𝑓)
𝑖
1
the third column of the matrix
This is an important result, as it allows us to find the matrix for given transformations.
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FP3 09/02/2016 SDB
Example: Find the matrix for a reflection in the plane y = x
0
Solution:
0
The z-axis lies in the plane y = x so (0) → (0)
1
1
0

the third column of the matrix is (0)
1
Also
1
0
( 0) → ( 1 )
0
0

0
1
( 1) → ( 0)
0
0


0
the first column of the matrix is (1)
0
1
the second column of the matrix is (0)
0
0 1 0
the matrix for a reflection in y = x is (1 0 0) .
0 0 1
Example: Find the matrix of the linear transformation, T, which maps (1, 0, 0) → (3, 4, 2),
(1, 1, 0) → (6, 1, 5) and (2, 1, 4) → (1, 1, 1).
Solution:
Firstly
Secondly



→ ( 4)
1
( 1)
0
→ (1)
3

first column is (4)
2
2
6
1
1
0
3
0
0
0
0
2
0
but (1) = (0) + (1) → (4) + T (1)
5
0
6
3
3
0
5
2
3
2
(1)
−4
2
(1)
−4
3
 second column is (−3)
3
1
→ (1)
−1
1
0
0
0
0
1
= 2 (0) + (1)  4(0) →
3
3
0
1
2
3
1
−1
3
3
0
2
3
1
2(4) + (−3)  4T (0)
2(4) + (−3)  4T (0) = ( 1 )
0
2
2
 third column is (1)
T ( 0) = ( 1)
1

3
T (1) = (1)  (4) = (−3)
Thirdly
but
1
( 0)
0
2
3
3
2
2
T = (4 −3 1) .
2
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3
2
39
Image of a line
2
3
−3
1
Example: Find the image of the line r = ( 0 ) +  (−2) under T,
where T =
Solution:
3
(1
2
−2
3
−1
2
(( 0 )
−3
3
−2
3
−1
T (r) = (1
2

.
As T is a linear transformation, we can find
T (r) = T

1
4)
1
3
(−2))
1
+ 
2
1
4) ( 0 )
1
−3
3
14
1
9
2
3
−3
1
= T ( 0 ) + T(−2)
3
+ ( 1
2
−2
3
−1
1
3
4) (−2)
1
1
T (r) = (−10) +  ( 1 ) and so the vector equation of the new line is
3
14
1
9
r = (−10) +  ( 1 ) .
Image of a plane 1
Similarly the image of a plane r = a +  b +  c , under a linear transformation, T, is
T (r) = T (a +  b +  c) = T (a) +  T(b) +  T (c).
Image of a plane 2
To find the image of a plane with equation of the form ax + by + cz = d, first construct a
vector equation.
Example: Find the image of the plane 3x  2y + 4z = 7 under a linear transformation, T.
Solution:
To construct a vector equation, put x = , y =  and find z in terms of  and .
7−3𝜆+2𝜇

3  2 + 4z = 7

𝜆
1
0
0
𝑥
𝜇
(𝑦) = (
) = ( 0 ) + 𝜆( 0 ) + 𝜇( 1 )
7−3𝜆+2𝜇
−3⁄
1⁄
7⁄
𝑧
2
4
4
4

0
𝑥
4
0
(𝑦) = ( 0 ) + 𝜆 ( 0 ) + 𝜇 (2)
7⁄
𝑧
−3
1
4

z =
4
making the numbers nicer in the ‘parallel’ vectors
and now continue as for in image for a plane 1.
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FP3 09/02/2016 SDB
7
Eigenvalues and eigenvectors
Definitions
1) An eigenvector of a linear transformation, T, is a non-zero vector whose direction is
unchanged by T.
So, if e is an eigenvector of T then its image e is parallel to e, or e =  e

e = T (e) =  e.
e defines a line which maps onto itself and so is invariant as a whole line.
If  = 1 each point on the line remains in the same place, and we have a line of invariant
points.
2) The characteristic equation of a matrix A is det(A   I) = 0
Ae =  e

(A   I) e = 0

A   I is a singular matrix

det(A   I) = 0

the solutions of the characteristic equation are the eigenvalues.
has non-zero solutions
eigenvectors are non-zero
2  2 matrices
Example: Find the eigenvalues and eigenvectors for the transformation with matrix
A = (1
−2
Solution:
1
)
4
.
The characteristic equation is det(A   I) = 0
1−𝜆
−2
1
| = 0
4−𝜆

|

(1  )(4  ) + 2 = 0

 2  5 + 6 = 0

1 = 2 and 2 = 3
For 1 = 2
(
1
−2
1 𝑥
)( )
4 𝑦
𝑥
= 2(𝑦)

x + y = 2x

x=y
and
2x + 4y = 2y

x=y

eigenvector e1 = (1)
FP3 09/02/2016 SDB
1
we could use (
3.7
), but why make things nasty
3.7
41
For 2 = 3
(
1 𝑥
)( )
4 𝑦
1
−2
𝑥
= 3(𝑦)

x + y = 3x

2x = y
and
2x + 4y = 3y

2x = y

eigenvector e2 = (1)
choosing easy numbers.
2
Orthogonal matrices
Normalised eigenvectors
A normalised eigenvector is an eigenvector of length 1.
In the above example, the normalized eigenvectors are e1 =
1⁄
( √2),
1⁄
√2
and e2 =
1⁄
( √5)
2⁄
√5
.
Orthogonal vectors
A posh way of saying perpendicular, scalar product will be zero.
Orthogonal matrices
If the columns of a matrix form vectors which are
(i)
(ii)
mutually orthogonal (or perpendicular)
each of length 1
then the matrix is an orthogonal matrix.
Example:
42
1⁄
( √5)
2⁄
√5
and
1⁄
( √5)
2⁄
√5
−2⁄
√5)
(
1⁄
√5
.
−2⁄
√5)
(
1⁄
√5
=
−2
5
are both unit vectors, and
+
2
5
= 0,  the vectors are orthogonal
FP3 09/02/2016 SDB
1
⁄ 5
 M = (2 √
⁄ 5
√
−2
⁄ 5
√)
1
⁄ 5
√
is an orthogonal matrix
Notice that
1
M TM
and so
⁄ 5
= (−2 √
⁄ 5
√
2
⁄ 5
√)
1
⁄ 5
√
1
⁄ 5
(2 √
⁄ 5
√
−2
⁄ 5
√)
1
⁄ 5
√
1 0
)
0 1
= (
the transpose of an orthogonal matrix is also its inverse.
This is true for all orthogonal matrices
think of any set of perpendicular unit vectors
Another definition of an orthogonal matrix is
 M TM = I
M is orthogonal
 M 1 = M T
Diagonalising a 2  2 matrix
Let A be a 2  2 matrix with eigenvalues 1 and 2 ,
𝑢1
𝑢2
and eigenvectors
e1 = (𝑣 ) and e2 = (𝑣 )
1
2
𝑢1
𝜆 𝑢
then A e1 = (𝑣 ) = ( 1 1 )
𝜆1 𝑣1
1

𝑢1
A (𝑣
1
and
𝑢2
𝜆1 𝑢1
𝑣2 ) = (𝜆1 𝑣1
𝑢2
𝜆 𝑢
A e2 = (𝑣 ) = ( 2 2 )
𝜆2 𝑣2
2
𝜆2 𝑢2
) =
𝜆2 𝑣2
𝑢1
(𝑣
1
𝑢2
𝜆1
𝑣2 ) ( 0
0
)
𝜆2
………. I
Define P as the matrix whose columns are eigenvectors of A, and D as the diagonal
matrix, whose entries are the eigenvalues of A
I


𝑢1
P = (𝑣
1
AP = PD

𝑢2
𝑣2 )
and D = (
𝜆1
0
0
)
𝜆2
P 1AP = D
The above is the general case for diagonalising any matrix.
In this course we consider only diagonalising symmetric matrices.
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43
Diagonalising 2  2 symmetric matrices
Eigenvectors of symmetric matrices
Preliminary result:
𝑦
𝑥
𝒙 = (𝑥1 ) and 𝒚 = (𝑦1 )
2
2
𝑥
𝑦
The scalar product 𝒙 . 𝒚 = (𝑥1 ) . (𝑦1 ) = x1y1 + x2y2
2
𝑦
but (x1 x2) (𝑦1 ) = x1y1 + x2y2
2

2
𝒙𝑇 𝒚 = 𝒙 . 𝒚
This result allows us to use matrix multiplication for the scalar product.
Theorem: Eigenvectors, for different eigenvalues, of a symmetric matrix are orthogonal.
Proof:
Let A be a symmetric matrix, then A T = A
Let A e1 = 1 e1 , and A e2 = 2 e2 ,
1 ≠ 2.
1 e1T = (1 e1) T = (A e1)T = e1T A T = e1T A
since AT = A

1 e1T = e1T A

1 e1T e2 = e1T A e2 = e1T 2 e2 = 2 e1T e2

1 e1T e2 = 2 e1T e2

(1  2) e1T e2 = 0
But
1  2 ≠ 0 

the eigenvectors are orthogonal
e1T e2 = 0
 e1
. e2
= 0
or perpendicular
Diagonalising a symmetric matrix
The above theorem makes diagonalising a symmetric matrix, A, easy.
44
1)
Find eigenvalues, 1 and 2, and eigenvectors, e1 and e2
2)
Normalise the eigenvectors, to give 𝒆̂𝟏 and 𝒆̂𝟐 .
3)
Write down the matrix P with 𝒆̂𝟏 and 𝒆̂𝟐 as columns.
P will now be an orthogonal matrix since 𝒆̂𝟏 and 𝒆̂𝟐 are orthogonal

P 1 = P T
4)
P TA P will be the diagonal matrix D = (
𝜆1
0
0
).
𝜆2
FP3 09/02/2016 SDB
Example: Diagonalise the symmetric matrix A = ( 6
−2
Solution:
The characteristic equation is

(6  )(9  )  4 = 0

2  15 + 50 = 0

 = 5 or 10
|
6−𝜆
−2
−2
).
9
−2
| = 0
9−𝜆

(  5)(  10) = 0
For 1 = 5
(
𝑥
−2 𝑥
) (𝑦 ) = 5 (𝑦 )
9
6
−2

6x  2y = 5x

x = 2y
and
2x + 9y = 5y

x = 2y
 e1 = (2)
1
and normalising  𝒆̂𝟏 =
2⁄
( √5)
1⁄
√5
For 2 = 10
(
6
−2
𝑥
−2 𝑥
) (𝑦) = 10 (𝑦 )
9

6x  2y = 10x

and
2x + 9y = 10y

2x = y
2x = y
 e2 = ( 1 )
−2
and normalising  𝒆̂𝟐 =
1⁄
( √5 )
−2⁄
√5
Notice that the eigenvectors are orthogonal
2
1

⁄ 5
P = (1 √
⁄ 5
√

D = P TA P = ( 01 𝜆 ) = (5 0 ).
0 10
2
FP3 09/02/2016 SDB
⁄ 5
√ )
−2
⁄ 5
√
𝜆
0
45
3  3 matrices
All the results for 2  2 matrices are also true for 3  3 matrices (or n  n matrices). The
proofs are either the same, or similar in a higher number of dimensions.
Finding eigenvectors for 3  3 matrices.
Example: Given that  = 5 is an eigenvector of the matrix
3 −1 2
𝑀 = (−2 1 −1), find the corresponding eigenvector.
4 −1 −2
Solution: Consider Me = 5e

𝑥
𝑥
3 −1 2
(−2 1 −1) (𝑦) = 5 (𝑦)
𝑧
𝑧
4 −1 −2

3x – y + 2z = 5x

–2x – y + 2z = 0
I
–2x + y – z = 5y

–2x – 4y – z = 0
II
4x – y – 2z = 5z

4x – y – 7z = 0
III
Now eliminate one variable, say x:
I – II  3y + 3z = 0

y = –z
We are not expecting to find unique solutions, so put z = t, and then find x and y in terms
of t.

from I,


46
y = –t, and,
2x = 2z – y = 2t + t = 3t
x = 15t
1 ∙ 5𝑡
e = ( −𝑡 )
𝑡
3
or (−2) choosing t = 2
2
check in II and III O.K.
FP3 09/02/2016 SDB
Diagonalising 3  3 symmetric matrices
2
Example: A = (−2
0
−2
1
2
0
2)
0
.
Find an orthogonal matrix P such that P TAP is a diagonal matrix.
Solution:
1)
Find eigenvalues
The characteristic equation is det(A  I) = 0

2−𝜆
| −2
0
−2
1−𝜆
2
0
2|
−𝜆

(2  )[ (1  )  4] + 2  [2  0] + 0 = 0

3  32  6 + 8 = 0
= 0
By inspection  = 2 is a root  ( + 2) is a factor
2)

( + 2)(2  5 + 4) = 0

( + 2) (  1) (  4) = 0

 = 2, 1 or 4.
Find normalized eigenvectors
1 = 2 

2
(−2
0
−2
1
2
2x  2y = –2x
2x + y + 2z = –2y
2y = –2z
I  y = 2x,

2 = 1 

FP3 09/02/2016 SDB
𝑥
0 𝑥
2) (𝑦) = −2 (𝑦)
𝑧
0 𝑧
and III
1
e1 = ( 2 )
−2
2
(−2
0
−2
1
2
 y = z
I
II
III
choose x = 1 and find y and z
and e1 = e1 = √9 = 3  𝒆̂𝟏 =
1⁄
3
2⁄
3
−2
( ⁄3)
𝑥
0 𝑥
2) ( 𝑦 ) = 1 ( 𝑦 )
𝑧
0 𝑧
2x  2y = x
2x + y + 2z = y
2y = z
I
II
III
47
I  x = 2y, and II  z = 2y

2
choose y = 1 and find x and z
and e2 = e2 = √9 = 3
e2 = (1)
2
 𝒆̂𝟐 =
3 = 4 

2
(−2
0
2⁄
3
1⁄
3
2
( ⁄3)
𝑥
0 𝑥
2) ( 𝑦 ) = 4 ( 𝑦 )
𝑧
0 𝑧
−2
1
2
2x  2y = 4x
2x + y + 2z = 4y
2y = 4z
I
II
III
I  x = y, and III  y = 2z

−2
choose z = 1 and find x and y
and e3 = e3 = √9 = 3
e3 = ( 2 )
1
 𝒆̂𝟑 =
3)
Find orthogonal matrix, P


4)
−2⁄
3
2⁄
3
1
( ⁄3 )
P = (𝒆̂𝟏
P =
𝒆̂𝟐
1⁄
3
2
( ⁄3
−2⁄
3
𝒆̂𝟑 )
2⁄
3
1⁄
3
2⁄
3
−2⁄
3
2⁄ )
3
1⁄
3
is required orthogonal matrix
Find diagonal matrix, D

𝜆1
0
0
0
P TAP = D = ( 0 𝜆2
0
0) =
𝜆3
−2
(0
0
0
1
0
0
0)
4
A nice long question! But, although you will not be asked to do a complete
problem, the examiners can test every step above!
48
FP3 09/02/2016 SDB
Index
Differentiation
hyperbolic functions, 10
inverse hyperbolic functions, 10
Further coordinate systems
hyperbola, 6
locus problems, 9
parabola, 7
parametric differentiation, 7
tangents and normals, 8
Hyperbolic functions
addition formulae, 3
definitions and graphs, 3
double angle formulae, 3
inverse functions, 4
inverse functions, logarithmic form, 4
Osborne’s rule, 3
solving equations, 5
Integration
area of a surface, 19
inverse hyperbolic functions, 14
inverse trig functions, 14
ln x, 14
reduction formulae, 14
standard techniques, 12
Matrices
cofactors matrix, 36
determinant of 33 matrix, 35
diagonalising 22 matrices, 43
diagonalising symmetric 22 matrices, 44
diagonalising symmetric 33 matrices, 47
dimension, 35
identity matrix, 35
inverse of 33 matrix, 36
non-singular matrix, 36
singular matrix, 36
symmetric matrix, 35
transpose matrix, 35
zero matrix, 35
FP3 09/02/2016 SDB
Matrices - eigenvalues and eigenvectors
22 matrices, 41
characteristic equation, 41
for 3 x 3 matrix, 46
normalised eigenvectors, 42
orthogonal matrices, 42
orthogonal vectors, 42
Matrices - linear transformations
base vectors, 38
image of a line, 40
image of a plane, 40
Vectors
triple scalar product, 24
vector product, 22
volume of parallelepiped, 24
volume of tetrahedron, 25
Vectors - lines
angle between line and plane, 33
cartesian equation of line in 3-D, 26
distance between skew lines, 34
line of intersection of two planes, 32
vector equation of a line, 25
vector product equation of a line, 26
Vectors - planes
angle between line and plane, 33
angle between two planes, 34
Cartesian equation, 27
distance between parallel planes, 30
distance of origin from a plane, 29
distance of point from a plane, 31
line of intersection of two planes, 32
vector equation, 28
49