fp 3 revision notes

Pure FP3

Revision Notes

May 2013

Contents

1 Hyperbolic functions............................................................................................................3Definitions 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..................................................................................................6Ellipse.........................................................................................................................................................................6

Hyperbola...................................................................................................................................................................6

Parabola......................................................................................................................................................................7

Parametric differentiation..........................................................................................................................................7

Tangents and normals................................................................................................................................................8

Finding a locus...........................................................................................................................................................9

3 Differentiation......................................................................................................................10Derivatives of hyperbolic functions.........................................................................................................................10

Derivatives of inverse functions..............................................................................................................................10

4 Integration............................................................................................................................12Standard 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..................................................................................................................................22Vector 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

2 FP3 19/04/2023 SDB

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................................................................................................................................35Basic 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.............................................................................................41Definitions................................................................................................................................................................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

FP3 19/04/2023 SDB 3

1 Hyperbolic functions

Definitions and graphs

sinh x= 12

(e x−e−x ) cosh x=12

(ex+e− x) tanh x= sinh xcosh x

=( ex−e−x )( ex+e−x )

You should be able to draw the graphs of cosech x, sech x and coth x from the above:

cosech x sech x coth x

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 cosB+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 product of two sines, so change sign

and 1 + tan2 A = sec2 A

4 FP3 19/04/2023 SDB

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−4

−3

−2

−1

1

2

3

4

x

y

−6 −5 −4 −3 −2 −1 1 2 3 4 5

−2

−1

1

2

3

4

5

x

y

−6 −5 −4 −3 −2 −1 1 2 3 4 5

−2

−1

1

2

3

4

5

x

y

−6 −5 −4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

x

y

−6 −5 −4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

x

y

−6 −5 −4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

x

y

1 – tanh2 A = sech2 A tan2 A= sin2 Acos2 A

, product of two sines, so change

sign

Inverse hyperbolic functions

Graphs

Remember that the graph of y=f −1(x ) is the reflection of y=f ( x )∈ y=x.

y = arsinh x y = arcosh x y = artanh x

Notice 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

sinh y=12

( e y−e− y )=x

e2 y−2 xe y−1=0

e y=2 x ±√4 x2+42

=x+√x2+1 or x−√ x2+1

But e y>0 and x−√ x2+1<0 e y=x+√ x2+1 only

y=arsinh x=ln ( x+√x2+1 )

2) y = arcosh x

cosh y=12

(e y+e− y )=x

e2 y−2 xe y+1=0

e y=2 x ±√4 x2−42

=x±√ x2−1 both roots are positive

y=arcosh x=ln (x+√ x2−1 ) or ln (x−√x2−1 )

It can be shown that ln (x−√x2−1 )=−ln (x+√x2−1 ) y=arcosh x=± ln (x+√x2−1 )

FP3 19/04/2023 SDB 5

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−4

−3

−2

−1

1

2

3

4

x

ysinh x

arsinh x

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−4

−3

−2

−1

1

2

3

4

x

y

tanh x

artanh x

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−4

−3

−2

−1

1

2

3

4

x

ycosh x

arcosh x

But arcosh x is a function and therefore has only one value (positive)

y=arcosh x=ln (x+√ x2−1 ) ( x 1)

3) Similarly artan x = 12

ln( 1+x1−x ) (|x|<1 )

Equations involving hyperbolic functions

It would be possible to solve 6 sinh x−2 cosh x=7 using the R sinh ¿¿) 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

6 ×12

( ex−e− x)−2×12

(ex+e− x)=7

2 e2 x−7 ex−4=0

(2ex+1 ) (ex−4 )=0

ex=−12

(not possible) or 4

x = ln 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 note use of Osborn’s rule

2 sinh2 x + 5 sinh x – 3 = 0

(2sinh x – 1)(sinh x + 3) = 0

sinh x=12∨−3

x = arsinh 0.5 or arsinh (–3)

x=ln (0.5+√0.52+1 ) or ln ((−3)+√(−3)2+1) using log form of inverse

x= ln(1+√52 ) or ln ( √10−3 )

6 FP3 19/04/2023 SDB

x

y

xa−a

a–a

–b

b

x

y

ae–aex

y

x = x = –

Sx

y

x =

PN

y =

y =

2 Further coordinate systems

Ellipse

Cartesian equation

x2

a2 + y2

b2 =1

Parametric equations

x=a cosθ , y=b sin θ

Foci at S (ae, 0) and S (–ae, 0)

Directrices at x = ±ae

Eccentricity e < 1, b2 = a2(1 – e2)

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.

Hyperbola

Cartesian equation

x2

a2 −y2

b2 =1


x=a coshθ , y=b sinh θ(x=a sec θ , y=b tanθ also work)

Asymptotes xa=±

yb

Foci at S (ae, 0) and S (–ae, 0)

Directrices at x = ±ae

FP3 19/04/2023 SDB 7

P

Eccentricity e >1, b2 = a2(e2 – 1)

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.

y2

c2 −x2

d2 =1

is a hyperbola with foci on the y-axis,

Parabola

Cartesian equation

y2=4 ax


x=a t 2 , y=2 at

Focus at S (a, 0)

Directrix at 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 dydθ

=dydx

×dxdθ

dydx

=

dydθdxdθ

or dydx

=

dydtdxdt

using any parameter

8 FP3 19/04/2023 SDB

N

S

x=ae

.S

x = − a

PN

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

x = 5 cos , y = 8 sin at the point where = π3

Solution: When = π3

, cosθ=12

and sin θ=√32

x = 52

, y = 4 √3

anddydx

=

dydθdxdθ

=8cosθ

−5 sin θ=

−85√3

when θ= π3

gradient of normal is 5√3

8

equation of normal is y−4√3=5√38 ( x−5

2 )

5√3 x−8 y= 17√32

Sometimes normal, or implicit, differentiation is (slightly) easier.

Example: Find the equation of the tangent to xy = 36, or x = 6t, y=6t

, at the point where

t = 3.

Solution: When t = 3, x = 18 and y = 2.

dydx

can be found in two (or more!) ways:

FP3 19/04/2023 SDB 9

dydx

=

dydtdxdt

=−6 t−2

6

dydx

=−1

t 2=−1

9 , when t=3

xy = 36 y=36x

dydx

=−36

x2

dydx

=−36

182 = −19

, when x = 18

equation of tangent is y−2=−19

( x−18 )

x + 9y – 36 = 0

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.

Example: The tangent to the ellipse x2

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:dydx

=

dydθdxdθ

=4cosθ

−3 sin θ

equation of tangent is y−4 sin θ= 4 cosθ−3 sin θ

( x−3 cosθ )

3 y sinθ+4 x cosθ=12 cos2 + 12 sin2 = 12

Tangent crosses x-axis at A when y = 0, x = 3

cosθ,

and crosses y-axis at B when x = 0, y= 4sin θ

mid-point of AB is ( 32cosθ

,4

2sinθ )

10 FP3 19/04/2023 SDB

Here x = 3

2cos θ and y= 2

sin θ

cosθ= 32 x

∧sin θ= 2y

equation of the locus is 9

4 x2+ 4

y2=1 since cos2θ+sin2θ=1

3 Differentiation

Derivatives of hyperbolic functions

y=sinh x=12

( ex−e− x )

dydx

=12

(ex+e− x) = cosh x

and, similarly, d (cosh x )

dx=sinh x

Also, y=tanh x= sinh xcosh x

dydx

= 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 xcosh xtanh x

cosh xsinh xsech2x

all positive

FP3 19/04/2023 SDB 11

coth xcosech xsech 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 (¿ 1cosh x ).

Derivatives of inverse functions

y = arsinh x

sinh y = x cosh ydydx

=1

dydx

= 1cosh y

= 1

√1+s inh2 y

d (arsinh x )

dx= 1

√1+x2

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) substitution needed for integration

arcsin x1

√1−x2 1 – sin2 u = cos2 u use x = sin u

arccos x−1

√1−x2 1 – cos2 u = sin2 u use x = cos u

arctan x1

1+ x2 1 + tan2 u = sec2 u use x = tan u

arsinh x1

√1+x2 1 + sinh2 u = cosh2 u use x = sinh u

arcosh x1

√x2−1 cosh2 u – 1 = sinh2 u use x = cosh u

artanh x1

1−x2 1 – tanh2 u = sech2 u use x = tanh u

12

ln( 1+x1−x ) 1

1−x2 partial fractions, see example below

12 FP3 19/04/2023 SDB

Note that ∫ 1

1−x2dx=1

2∫ 1

1+x+ 1

1−xdx=1

2ln( 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.

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 xdx=∫1+sec2 x dx=x+ tan x + c

∫cos2 x dx=12∫1+cos 2 x dx=1

2 ( x+ 12

sin 2 x)+c

∫sinh2 x dx=12∫cosh 2x−1 dx=1

2 ( 12

sinh 2x−x)+c

FP3 19/04/2023 SDB 13

Reverse chain rule

Notice the chain rule pattern, guess an answer and differentiate to find the constant.

Example: ∫cos2 x sin x dx ‘looks like’ u2 ∂u

∂ x so try u3 cos3 x

d ( cos3 x )dx

= 3cos2 x (– sin x) = – 3cos2 x sin x so divide by –3

∫cos2 x sin x dx = −13

cos3 x+c

Example: ∫ x2 (2 x3−7 )4 dx ‘looks like’ u4 ∂ u

∂ x so try u5 (2 x3−7 )5

d ( 2x3−7 )5

dx ¿5 (2 x3−7 )4 × 6 x2 = 30(2 x3−7 )4

so divide by 30

∫ x2 (2 x3−7 )4 dx = 1

30(2 x3−7 )5+c

Example: ∫ sech4 x tanh x dx

= ∫ sech3 x (sech x tanh x ) dx ‘looks like’ u3 ∂ u

∂ x so try u4 = sech4 x

d ( sech4 x )dx

¿−4 sech3 x sech x tanh x so divide by 4

∫ sech4 x tanh x dx=−14

sech4 x+c

Standard substitutions

∫ 1

a2+b2 x2dx bx=a tan u better than bx = a sinh u when there is no √❑

∫ 1

√a2+b2 x2dx bx=a sinh u better than bx = a tan u when there is √❑

∫ 1

a2−b2 x2dx bx=a tanh u or use partial fractions

∫ 1

√b2 x2−a2dx bx=acoshu better than bx = a sec u when there is √❑

14 FP3 19/04/2023 SDB

For more complicated integrals like

∫ 1

p x2+qx+rdx or ∫ 1

√ p x2+qx+rdx

complete the square to give p ( x+a )2+b and then use a substitution similar to one of the four

above.

Example: ∫ 1

√4 x2−8 x−5dx 4 x2−8 x−5 =

4 ( x2−2 x+1 )−9=4 ( x−1 )2−9 = ∫ 1

√4 ( x−1 )2−9dx

Substitute 2(x – 1) = 3 cosh u 2 dx = 3 sinh u du

= ∫ 1

√9 (cosh2u−1 )3 sinh u

2du

=12∫ du = u + c =

12

arcosh (2 x−23 ) + c

Important tip

∫ xn

√a2 ± x2dx 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.

Integration inverse functions and ln x

To integrate inverse trigonometric or hyperbolic functions and ln x we use integration by parts

with dvdx

= 0

Example: Find ∫ arctan xdx

FP3 19/04/2023 SDB 15

Solution: I = ∫ arctan xdx take u = arctan x dudx

= 1

1+ x2

and dvdx

= 1 v = x

I = x arctan x ∫ x ×1

1+ x2dx

I = ∫ arctan xdx = x arctan x 12

ln (1 + x 2) + c

Example: Find ∫ arcosh x dx

Solution: I = ∫ arcosh x dx take u = arcosh x dudx

= 1

√x2−1

and dvdx

= 1 v = x

I = x arcosh x ∫ x ×1

√x2−1dx

I = ∫ arcosh x dx = x arcosh x √ x2−1 + 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: I n=∫ xn ex dx.

(a) Find a reduction formula, (b) Find I 0, and (c) find I 4

Solution:

(a) Integrating by parts

u=xn dudx

¿n xn−1

and dvdx

=ex v = e x

I n=xn ex−∫ n xn−1ex dx

I n=xn ex−n I n−1

16 FP3 19/04/2023 SDB

(b) I 0=∫ex dx = e x + c

(c) Using the reduction formula

I 4=x4 ex−4 I 3=x4 ex−4 ( x3 ex−3 I 2 )

¿ x4 ex−4 x3e x+12 (x2 ex−2 I 1 )

¿ x4 ex−4 x3e x+12 x2ex−24 (x ex−I 0 )

¿ x4 ex−4 x3e x+12 x2ex−24 xe x+24 ex+c since I 0=ex+c

Example 2: Find a reduction formula for I n=∫0

π2

sinn x dx.

Use the formula to find I 6=∫0

π2

sin6 x dx

Solution: I n=∫0

π2

sinn x dx = ∫0

π2

sinn−1 x sin x dx

take u = sin n 1 x dudx

= (n 1) sin n 2 x cos x

and dvdx

= sin x v = cos x

I n=[cos x sinn 1 x ]0π2 ∫

0

π2

cos x (n 1)sinn 2 x cos x dx

= 0 + (n1)∫0

π2

cos2 x sinn 2 x dx

= (n1)∫0

π2

(1−sin2 x ) sinn2 x dx

= (n1 )∫0

π2

sinn 2 x dx−(n1)∫0

π2

sinn xdx

In = (n 1) In2 (n 1) In

FP3 19/04/2023 SDB 17

In = n−1

nIn2 .

Now I6 = 56

I4 = 56

×34

I2 = 56

×34

×12

I0

I6 = 516

∫0

π2

1 dx = 5 π32

.

Example 3: Find a reduction formula for I n=∫ secn x dx.

Solution: I n=∫ secn x dx=∫ secn−2 x sec2 xdx

take u = secn 2x dudx

= (n 2) secn 3x sec x tan x

and dvdx

= sec2 x v = tan x

I n = secn 2x tan x ∫ tan x (n2 ) sec n3 x sec x tan x dx

= secn 2x tan x (n 2 )∫ tan2 x secn 2 x dx

= secn 2x tan x (n2 )∫ (sec 2 x−1 ) sec n2 x dx

= secn 2x tan x (n2 ) In +(n2 ) In−2

(n1 ) I n = secn 2x tan x + (n2 ) In−2

Example 4: Find a reduction formula for I n=∫ tann x dx.

Solution: I n=∫ tann x dx=∫ tann−2 x tan2 x dx

I n=∫ tann−2 x (sec2 x−1 ) d x

= ∫ tann−2 x sec2 x d x ∫ tann−2 xd x

I n = 1

n−1 tann 1x In 2 .

18 FP3 19/04/2023 SDB

Example 5: Find a reduction formula for I n = ∫−1

0

xn (1+x )2 dx.

Solution: I n = ∫−1

0

xn (1+x )2 dx take u = xn dudx

= nxn 1

and dvdx

= (1 + x)2 v = 13

(1+x )3

I n = [ xn 13(1+ x)3]

−1

0

∫−1

0

nxn−1×13

(1+x )3

dx .

I n = 0 n3∫−1

0

xn−1(1+x )2 (1+x ) dx

I n = n3∫−1

0

xn−1(1+x )2dx n3∫−1

0

xn(1+x )2dx

I n = n3

I n−1 n3

I n

n+3

3 I n =

n3

I n−1

I n = n

n+3 I n−1

Example 6: Find a reduction formula for I n = ∫0

π2

xn cos x dx

Solution: I n = ∫0

π2

xn cos x dx take u = xn dudx

= nxn 1

and dvdx

= cos x v = sin x

I n = [ xn sin x ]0π2 n∫

0

π2

sin x× xn−1 dx

take u = xn1 dudx

= (n1) xn 2

and dvdx

= sin x v = cos x

FP3 19/04/2023 SDB 19

P

Q

x

y s

I n = ( π2 )

n

n{¿¿ ∫0

π2

−cos x×(n−1) xn−2dx}

I n = ( π2 )

n

n{0 + (n−1)∫

0

π2

xn−2cos x dx}

I n = ( π2 )

n

n (n−1) I n−2

Example 7: Find a reduction formula for I n = ∫ sin nxsin x

dx

Solution: I n = ∫ sin [ (n−2 ) x+2 x ]sin x

dx

= ∫ sin (n−2 ) x cos2 x+cos (n−2 ) x sin 2xsin x

dx

= ∫ sin (n−2 ) x (1−2sin2 x )+cos (n−2 ) x× 2sin x cos xsin x

dx

= ∫ sin (n−2 ) xsin x

dx + 2∫cos (n−2 ) x cos x−sin (n−2 ) x sin x dx

= In 2 + 2∫cos (n−1 ) x dx using cos(A + B) = cos A cos B – sin A sin B

I n = In 2 + 2

n−1 sin(n 1)x .

Arc length

All the formulae you need can be remembered from this diagram

arc PQ line segment PQ

(d s)2 (d x)2 + (d y)2

20 FP3 19/04/2023 SDB

( δsδx )

2

≈ 1+( δyδx )

2

and as d x 0

( dsdx )

2

=1+( dydx )

2

dsdx

=√1+( dydx )

2

arc length = s = ∫√1+( dydx )

2

dx

Similarly

( δsδy )

2

≈ ( δxδy )

2

+1 s = ∫√( dxdy )

2

+1dy

and ( δsδt )

2

≈( δxδt )

2

+( δyδt )

2

s = ∫√( dxdt )

2

+( dydt )

2

dt or s = ∫√ x2+ y2dt

for parametric equations.

Example 1: Find the length of the curve y=23

x32 , from the

point where x = 3 to the point where x = 8.

Solution: The equation of the curve is in Cartesian form so we use

s = ∫√1+( dydx )

2

dx .

y=23

x32

dydx

= √ x

s = ∫3

8

√1+x dx

= [ 23

(1+ x )32 ]

3

8

= 23

× (9 )32−

23

× ( 4 )32

s = 1223

.

FP3 19/04/2023 SDB 21

−5 5 10 15 20 25 30 35

5

10

15

x

x

ba

y

y

s

Example 2: Find the length of one arch of the cycloid x = a(t sin t), y = a(1 cos t).

Solution: The curve is given in parametric form

so we use s = ∫√ x2+ y2dt .

x = a(t sin t), y = a(1 cos t)

dxdt

= a(1 cos t), and dydt

= a sin t

x2+ y2 = a2(1 2 cos t + cos2t + sin2t) = 2a2(1 cos t)

√ x2+ y2 = a√2(1−[1−2 sin2( t2 )]) = 2 a sin( t

2 )

s = ∫0

2 π

2 a sin( t2 )dt

s = [−4 a cos( t2 )]0

2 π

= 4a 4a = 8a.

Area of a surface of revolution

A curve is rotated about the x-axis.

To find the area of the surface formed between x = a and x = b, we consider a small section of the curve, ds, at a distance of y from the x-axis.

When this small section is rotated about the x-axis, the shape formed is approximately a cylinder of radius y and length ds.

The surface area of this (cylindrical) shape 2 rl 2 yds

The total surface area ∑a

b

2 y s

and, as ds 0, the area of the surface is A = ∫a

b

2 y ds.

And so A = ∫a

b

2 ydsdx

dx or A = ∫a

b

2 ydsdt

dt

22 FP3 19/04/2023 SDB

t=0 t=2π

h = 2r

r

We can use dsdx

=√1+( dydx )

2

or dsdt

=√( dxdt )

2

+( dydt )

2

, as appropriate,

remembering that (d s)2 (d x)2 + (d y)2

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.

A = ∫a

b

2 ydsdx

dx

x2 + y2 = r2

2 x+2 ydydx

=0 dydx

=−xy

and dsdx

=√1+( dydx )

2

A = ∫a

b

2 y √1+ x2

y2dx = ∫

a

b

2√ y2+ x2 dx

= ∫a

b

2 r dx since x2 + y2 = r2

A = [ 2 rx ]ab=2 r (b−a) 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.

Thus the area of the sphere is

A = 2 rh = 4 r2 since h = 2r.

FP3 19/04/2023 SDB 23

x=a x=b

P, (t = 2)

x

y

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.

A = ∫a

b

2 ydsdt

dt

and dsdt

=√( dxdt )

2

+( dydt )

2

dxdt

=2 at and dydt

=2 a

dsdt

=√(2at )2+ (2a )2 = 2 a√ t2+1

A = ∫0

2

2 2 at × 2 a√t 2+1 dt

= 8a2 ×13

[( t2+1 )32 ]0

2

A = 8 π a2

3 (5

32−1)

24 FP3 19/04/2023 SDB

a

b

a

b

i

j

5 Vectors

Vector product

The vector, or cross, product of a and b is

a ´ b = ab sin n

where n is a unit (length 1) vector which is perpendicular to both a and b, and is the angle between a and b.

The direction of n is that in which a right hand corkscrew would move when turned through the angle from a to b.

Notice that b ´ a = ab sin −n, where −n is in the opposite direction to n, since the corkscrew would move in the opposite direction when moving from b to a.

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, k ´ i = j,

i ´ k = j, j ´ i = k, k ´ j = i.

Properties

a ´ a = 0 since = 0

a ´ b = 0 a is parallel to b since sin = 0 = 0 or

or a or b = 0

a ´ (b + c) = a ´ b + a ´ c remember the brilliant demo with the straws!

a ´ b is perpendicular to both a and b from the definition

FP3 19/04/2023 SDB 25

a

b

A

B

O

a

b

A

B

O D

AB

C

Component form

Using the above we can show that

a ´ b = (a1

a2

a3)×(b1

b2

b3)=(−a2b3−a3 b2

a1b3+a3 b1

a1b2−a2b1) = | i j k

a1 a2 a3

b1 b2 b3|

Applications of the vector product

Area of triangle OAB = 12

ab sin θ

area of triangle OAB = 12a ´ b

Area of parallelogram OADB is twice the area of the triangle OAB

area of parallelogram OADB = a ´ b

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: The area of the triangle ABC = |12

AB× AC|

AB = b a = ( 31

−1) and AC = c a = ( 42

−3)

AB × AC = |i j k3 1 −14 2 −3| = (−1

52 )

area ABC = 12

√12+52+32 =

12

√35

26 FP3 19/04/2023 SDB

a c

b

h

Volume of a parallelepiped

In the parallelepiped

h = a cos

and area of base = bc sin

volume = h ´ bc sin

= a ´ bc sin ´ cos

and (i) is the angle between n and a

(ii) bc sin is the magnitude of b ´ c

a . (b ´ c) = a ´ bc sin ´ cos

volume of parallelepiped = a . (b ´ c)

Triple scalar product

a . (b ´ c) = (a1

a2

a3) . (−b2 c3−b3 c2

b1 c3+b3 c1

b1 c2−b2 c1)

= a1 (b2c3−b3 c2)+a2 (−b1 c3+b3 c1 )+a3(b¿¿1 c2−b2 c1)¿

= |a1 a2 a3

b1 b2 b3

c1 c2 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.

FP3 19/04/2023 SDB 27

h a

b

c

l

A

a

b

P

r

{a, b, c} = {c, a, b} = {b, c, a } = {a, c, b} = {c, b, a} = {b, a, c}

Volume of a tetrahedron

The volume of a tetrahedron is

13

Areaof base ×h

The height of the tetrahedron is the same as the height of the parallelepiped, but its base has half the area

volume of tetrahedron = 16

volume of parallelepiped

volume of tetrahedron ¿

|LINK Word . Document .12 C:\\Users\\simon\\Documents\\Maths\\Revision Notes\\FP 3 Revision Notes 2012.docx OLE LINK 1¿¿ 16

{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: Volume = |16

{ AD , AC , AB }| AD=d−a=(3

01) , AC=( 0

1−5) , AB=(−2

20 )

{ AD , AC , AB }=| 3 0 10 1 −5

−2 2 0 |=3× 10+2=32

volume of tetrahedron is 16

´ 32 = 513

Equations of straight lines

Vector equation of a line

28 FP3 19/04/2023 SDB

l

A

a

b

P

r

r = a + b is the equation of a line through the point A and parallel to the vector b,

or ( xyz )=( l

mn )+ λ(α

βγ ) .

Cartesian equation of a line in 3-D

Eliminating from the above equation we obtain

x−lα

= y−mβ

= z−nγ

(= )

is the equation of a line through the point (l, m, n) and parallel to the vector (αβγ ) .

This strange form of equation is really the intersection of the planes

x−lα

= y−mβ and

y−mβ

= z−nγ (¿ x−l

α= z−n

γ ) .

Vector product equation of a line

AP = r a and is parallel to the vector b

AP ´ b = 0

(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.

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= 2 y+45

= 3−z2

.

Find its equation (i) in the form r = a + b, (ii) in the form r ´ b = c .

FP3 19/04/2023 SDB 29

AP

O

a

r

Solution:

First re-write the equation in the standard manner

x−0

1= y−−2

2.5= z−3

−2

the line passes through A, (0, 2, 3), and is parallel to b, ( 12.5−2) or ( 2

5−4)

(i) r ¿( 0−23 )+ λ( 2

5−4)

(ii) (r−( 0−23 ))×( 2

5−4) = 0

r ×( 12.5−2) = ( 0

−23 )×( 2

5−4) = |i j k

0 −2 32 5 −4| = (−7

64 )

r ×( 25

−4 ) = (−764 ) .

Equation of a plane

Scalar product form

Let n be a vector perpendicular to the plane .

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

AP . n = 0 (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 .

30 FP3 19/04/2023 SDB

AP

O

a

r

b

c

Cartesian form

If n = (αβγ ) then r . n = ( x

yz ) .(αβγ ) = a x + y + z

a x + y + z = d is the Cartesian equation of a plane perpendicular to (αβγ ) .

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

AB = (−4−2−2) and AC = (−1

−1−7) are parallel to the plane

AB ´ AC is perpendicular to the plane

AB ´ AC = | i j k−4 −2 −2−1 −1 −7| = ( 12

−262 ) = 2 ´ ( 6

−131 ) 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

and scalar product equation is r . ( 6−13

1 ) = 3.

Vector equation of a plane

r = a + b + c is the equation of a plane, , through A and parallel to the vectors b and c.

FP3 19/04/2023 SDB 31

O

M

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: We want the plane through A, (1, 4, 2), parallel to AB = (015) and AC = (334)

vector equation is r = ( 14

−2)+λ (015)+μ(334).

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.

We must first find the intersection of the line OM with the plane.

OM is perpendicular to the plane

and so is parallel to n = ( 4−24 ) .

the line OM is r = (000) + ( 4−24 )=( 4

−24 ), 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 = 7

36 ( 4−24 )

32 FP3 19/04/2023 SDB

O

M

L

1

2

l

distance = |OM| = 7

36√42+22+42= 7

6

The distance of the origin from the plane is 1 16

.

Distance between parallel planes

Example: Find the distance between the parallel planes

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.

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.

n=( 2−6

3 ) the equation of l is r=λ( 2

−63 )

L is given by 2 ´ 2 – 6 ´ (6) + 3 ´ 3 = 9 = 9

49

OL = 9

49 ( 2−63 )

FP3 19/04/2023 SDB 33

K

M

Similarly OM = 5

49 ( 2−63 )

ML = 9

49 ( 2−63 )− 5

49 ( 2−63 )= 4

49 ( 2−6

3 ) ML = |ML| =

449

√22+62+32 =

449

√49 = 47

The distance between the planes is 47

.

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.

We must first find the intersection of the line KM with the plane.

KM is perpendicular to the plane

and so is parallel to n = ( 3−21 ) .

It also passes through K, (5, 4, 7),

the line KM is r = ( 5−4

7 ) + ( 3−21 ), which meets the plane when

3(5 + 3) 2(4 2) + (7 + ) = 2

15 + 9 + 8 +4 + 7 + = 2

= –2

34 FP3 19/04/2023 SDB

m = = ( 5−4

7 ) – 2( 3−21 )

KM = m k = ( 5−4

7 ) – 2( 3−21 )−( 5

−47 ) = –2 ( 3

−21 )

distance = |KM| = 2 ×√32+22+12=2×√14

For the general case, the above method gives –

The distance from the point (a, , ) to the plane n1x + n2y + n3z + d = 0 is

|n1α +n2 β+n3 γ+d|√n1

2+n22+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

= 28

√14=2√14 notice that the d is on the L.H.S. of the

equation

Line of intersection of two planes

Example: Find an equation for the line of intersection of the planes

x + y + 2z = 4 I

and 2x y + 3z = 4 II

Solution: 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 = 8−5 λ

3

FP3 19/04/2023 SDB 35

n

l

n

l

I y = 4 x – 2z = 4 8−5 λ

3 2 =

4−λ3

( xyz )=¿ (

83430

)+λ (−53

−131

) or ( x

yz )=¿ (

83430

)+λ (−5−13 ) making the numbers nicer in the direction vector only

which is the equation of a line through ( 83

,43

, 0) and parallel to (−5−13 ) .

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:

36 FP3 19/04/2023 SDB

plane 1 plane 2

(i) n and the angle are on the same side of the plane

= 90

(ii) n and the angle are on opposite sides of the plane

= 90

Example: Find the angle between the line x+1

2= y−2

1= z−3

−2

and the plane 2x + 3y 7z = 5.

Solution: The line is parallel to ( 21

−2), and the normal vector to the plane is ( 23

−7).

a . b = ab cos 21 = √22+12+22√22+32+72 cos

cos = 7

√62 = 27.3o

the angle between the line and the plane, = 90 27.3 = 62.7o

Angle between two planes

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.

Example: Find the angle between the planes

2x + y + 3z = 5 and 2x + 3y + z = 7

FP3 19/04/2023 SDB 37

l1 r = a + b

l2 r = c + d X

Y

C

A

Solution: The normal vectors are (213) and (231)a . b = ab cos 10 = √22+12+32×√22+12+32 cos

cos = 1014

= 44.4o

Shortest distance between two skew lines

It can be shown that there must be a line joining two skew lines which is perpendicular to both lines.

This is XY and is the shortest distance between the lines.

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 where n is a unit vector (length 1) parallel to XY and so perpendicular to l1 and l2.

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 please

6 Matrices

Basic definitions

Dimension of a matrix

A matrix with r rows and c columns has dimension r ´ c.

38 FP3 19/04/2023 SDB

Transpose and symmetric matrices

The transpose, AT, of a matrix, A, is found by interchanging rows and columns

A = (a b cd e fg h i ) AT = (a d g

b e hc 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.

Thus, S = (a b cb d ec e f ) is a symmetric matrix.

Identity and zero matrices

The identity matrix I = (1 0 00 1 00 0 1)

and the zero matrix is 0 = (0 0 00 0 00 0 0)

Determinant of a 3 ´ 3 matrix

The determinant of a 3 ´ 3 matrix, A, is

det (A) = = |a b cd e fg h i| = a|e f

h i|−b|d fg i|+c|d e

g h| = aei afh bdi + bfg + cdh ceg

Properties of the determinant

1) A determinant can be expanded by any row or column using ¿

FP3 19/04/2023 SDB 39

1) Find the determinant, det(A).If det(A) = 0, then A is singular and has no inverse.

2) Find the matrix of cofactors C = ( A B CD E FG H I )

3) Find the transpose of C, CT = ( A D GB E HC F I )

4) Divide CT by det(A) to give A–1 = 1

det ( A) ( A D GB E HC F I )

See example 10 on page 148.

Properties of the inverse

1) A1A = AA1 = I

2) (AB)1 = B1A1 - note the change of order of A and B.

Proof (AB)1 AB = I from definition of inverse

(AB)1 AB (B1A1) = I (B1A1)

(AB)1 A (BB1)A1 = B1A1 (AB)1 A I A1 = B1A1

(AB)1 A A1 = B1A1 (AB)1 = B1A1

3) det(A1) = 1

det ( A)

Matrices and linear transformations

Linear transformations

T is a linear transformation on a set of vectors if

FP3 19/04/2023 SDB 41

(i) T (x1 + x2) = T (x1) + T (x2) for all vectors x and y

(ii) T (kx) = kT (x) for all vectors x

Example: Show that T ( xyz ) = ( 2 x

x+ y−z ) is a linear transformation.

Solution: (i) T (x1 + x2) = T ((x1

y1

z1)+( x2

y2

z2)) = T (( x1+x2

y1+ y2

z1+z2))

= ( 2 ( x1+x2 )x1+x2+ y1+ y2

−z1−z2) = ( 2x1

x1+ y1

−z1) + ( 2x2

x2+ y2

−z2) = T (x1) + T (x2)

T (x1 + x2) = T (x1) + T (x2)

(ii) T (kx) = T (k ( xyz )) = T (kx

kykz ) = ( 2kx

kx+ky−kz ) = k ( 2x1

x1+ y1

−z1) = kT (x)

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

i=(100) , j=(010) ,k=(0

01)

Under the transformation with matrix (a b cd e fg h i )

(100)→(adg) the first column of the matrix

(010)→(beh) the second column of the matrix

42 FP3 19/04/2023 SDB

(001)→(cfi ) the third column of the matrix

This is an important result, as it allows us to find the matrix for given transformations.

Example: Find the matrix for a reflection in the plane y = x

Solution: The z-axis lies in the plane y = x so (001)→(001) the third column of the matrix is (001)Also (100)→(010) the first column of the matrix is (010)

(010)→(100) the second column of the matrix is (100) the matrix for a reflection in y = x is (0 1 0

1 0 00 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 (100) → (342) first column is (3

42)

Secondly (110) → (615) but (110) = (100) + (010) → (342) + T (010)

T (010) = (615) (342) = ( 3

−33 ) second column is ( 3

−33 )

Thirdly ( 21

−4) → ( 11

−1)

FP3 19/04/2023 SDB 43

but ( 21

−4) = 2 (100) + (010) 4(001) → 2(342) + ( 3

−33 ) 4T (001)

2(342) + ( 3

−33 ) 4T (001) = ( 1

1−1)

T (001) = (212) third column is (212) T = (3 3 2

4 −3 12 3 2) .

Image of a line

Example: Find the image of the line r = ( 20

−3) + ( 3−21 ) under T,

where T = (3 −2 11 3 42 −1 1) .

Solution: As T is a linear transformation, we can find

T (r) = T (( 20

−3)+( 3−21 )) = T ( 2

0−3) + T( 3

−21 )

T (r) = (3 −2 11 3 42 −1 1) ( 2

0−3) + (3 −2 1

1 3 42 −1 1)( 3

−21 )

T (r) = ( 3−10

1 ) + (1419 ) and so the vector equation of the new line is

r = ( 3−10

1 ) + (1419 ) .

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).

44 FP3 19/04/2023 SDB

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 .

3 2 + 4z = 7 z = 7−3 λ+2 μ

4

( xyz )=(

λμ

7−3 λ+2 μ4

) = (0074

)+λ (10

−34

)+μ(0112)

( xyz )= (

0074

)+λ ( 40

−3)+μ(021) making the numbers nicer in the ‘parallel’ vectors

and now continue as for in image for a plane 1.

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 has non-zero solutions eigenvectors are non-zero

A I is a singular matrix

det(A I) = 0

the solutions of the characteristic equation are the eigenvalues.

FP3 19/04/2023 SDB 45

2 ´ 2 matrices

Example: Find the eigenvalues and eigenvectors for the transformation with matrix

A = ( 1 1−2 4) .

Solution: The characteristic equation is det(A I) = 0

|1−λ 1−2 4−λ|=0

(1 )(4 ) + 2 = 0

2 5 + 6 = 0 1 = 2 and 2 = 3

For 1 = 2

( 1 1−2 4)(x

y) = 2( xy )

x + y = 2x x = y

and 2x + 4y = 2y x = y

eigenvector e1 = (11) we could use (3.73.7), but why make things nasty

For 2 = 3

( 1 1−2 4)(x

y) = 3( xy )

x + y = 3x 2x = y

and 2x + 4y = 3y 2x = y

eigenvector e2 = (12) choosing easy numbers.

Orthogonal matrices

Normalised eigenvectors

A normalised eigenvector is an eigenvector of length 1.

46 FP3 19/04/2023 SDB

In the above example, the normalized eigenvectors are e1 = (1

√21√2

), and e2 = (1

√52√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) mutually orthogonal (or perpendicular)(ii) each of length 1

then the matrix is an orthogonal matrix.

Example:

(1

√52√5

) and (−2

√51√5

) are both unit vectors, and

(1

√52√5

) . (−2

√51√5

) = −25

+ 25 = 0, the vectors are orthogonal

M = (1

√52√5

−2

√51√5

) is an orthogonal matrix

Notice that

M TM = (1

√5−2√5

2

√51√5

) (1

√52√5

−2

√51√5

) = (1 00 1)

and so the transpose of an orthogonal matrix is also its inverse.

FP3 19/04/2023 SDB 47

This is true for all orthogonal matrices think of any set of perpendicular unit vectors

Another definition of an orthogonal matrix is

M is orthogonal M TM = I M 1 = M T

Diagonalising a 2 ´ 2 matrix

Let A be a 2 ´ 2 matrix with eigenvalues 1 and 2 ,

and eigenvectors e1 = (u1

v1) and e2 = (u2

v2)

then A e1 = (u1

v1) = ( λ1u1

λ1 v1) and A e2 = (u2

v2) = ( λ2u2

λ2 v2)

A (u1

v1

u2

v2) = ( λ1u1

λ1 v1

λ2u2

λ2v2) = (u1

v1

u2

v2) (λ1 0

0 λ2) ……….

Define P as the matrix whose columns are eigenvectors of A, and D as the diagonal matrix, whose entries are the eigenvalues of A

P = (u1

v1

u2

v2) and D = (λ1 0

0 λ2)

AP = PD P 1AP = D

The above is the general case for diagonalising any matrix.

In this course we consider only diagonalising symmetric matrices.

Diagonalising 2 ´ 2 symmetric matrices

Eigenvectors of symmetric matrices

Preliminary result:

x = (x1

x2) and y = ( y1

y2)

48 FP3 19/04/2023 SDB

I

I

The scalar product x . y = (x1

x2) . ( y1

y2) = x1y1 + x2y2

but (x1 x2) ( y1

y2) = x1y1 + x2y2

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 = e1

T A since AT = A

1 e1T = e1

T A

1 e1T e2 = e1

T A e2 = e1T 2 e2 = 2 e1

T e2

1 e1T e2 = 2 e1

T e2

(1 2) e1T e2 = 0

But 1 2 ≠ 0 e1T e2 = 0 e1 . e2 = 0

the eigenvectors are orthogonal or perpendicular

Diagonalising a symmetric matrix

The above theorem makes diagonalising a symmetric matrix, A, easy.

1) Find eigenvalues, 1 and 2, and eigenvectors, e1 and e2

2) Normalise the eigenvectors, to give e1 and e2 .

3) Write down the matrix P with e1 and e2 as columns.P will now be an orthogonal matrix since e1 and e2 are orthogonal P 1 = P T

4) P TA P will be the diagonal matrix D = (λ1 00 λ2

).Example: Diagonalise the symmetric matrix A = ( 6 −2

−2 9 ).

Solution: The characteristic equation is |6−λ −2−2 9−λ|=0

FP3 19/04/2023 SDB 49

(6 )(9 ) 4 = 0

2 15 + 50 = 0 ( 5)( 10) = 0

= 5 or 10

For 1 = 5

( 6 −2−2 9 )(x

y )=5( xy )

6x 2y = 5x x = 2y

and 2x + 9y = 5y x = 2y

e1 = (21)

and normalising e1 = (2

√51√5

)For 2 = 10

( 6 −2−2 9 )(x

y )=10( xy )

6x 2y = 10x 2x = y

and 2x + 9y = 10y 2x = y

e2 = ( 1−2)

and normalising e2 = (1

√5−2√5

)Notice that the eigenvectors are orthogonal

P = (2

√51√5

1

√5−2√5

) 50 FP3 19/04/2023 SDB

D = P TA P = (λ1 00 λ2

) = (5 00 10).

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

M=( 3 −1 2−2 1 −14 −1 −2), find the corresponding eigenvector.

Solution: Consider Me = 5e

( 3 −1 2−2 1 −14 −1 −2)( x

yz )=5 (x

yz )

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.

y = –t, and,

from I, 2x = 2z – y = 2t + t = 3t

x = 15t

e = (1 ∙5 t−tt ) or ( 3

−22 ) choosing t = 2 check in II and III O.K.

FP3 19/04/2023 SDB 51

Diagonalising 3 ´ 3 symmetric matrices

Example: A = ( 2 −2 0−2 1 20 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 20 2 −λ| = 0

(2 )[ (1 ) 4] + 2 ´ [2 0] + 0 = 0

3 32 6 + 8 = 0

By inspection = 2 is a root ( + 2) is a factor

( + 2)(2 5 + 4) = 0

( + 2) ( 1) ( 4) = 0

= 2, 1 or 4.

2) Find normalized eigenvectors

1 = 2 ( 2 −2 0−2 1 20 2 0)( x

yz )=−2( x

yz )

2x 2y = –2x I2x + y + 2z = –2y II2y = –2z III

I y = 2x, and III y = z choose x = 1 and find y and z

52 FP3 19/04/2023 SDB

e1 = ( 12

−2) and e1 = e1 = √9 = 3 e1 = (1323

−23

)2 = 1 ( 2 −2 0

−2 1 20 2 0)( x

yz )=1(x

yz )

2x 2y = x I2x + y + 2z = y II2y = z III

I x = 2y, and II z = 2y choose y = 1 and find x and z

e2 = (212) and e2 = e2 = √9 = 3

e2 = (231323)

3 = 4 ( 2 −2 0−2 1 20 2 0)( x

yz )=4 ( x

yz )

2x 2y = 4x I2x + y + 2z = 4y II 2y = 4z III

I x = y, and III y = 2z choose z = 1 and find x and y

e3 = (−221 ) and e3 = e3 = √9 = 3

e3 = (−232313

) FP3 19/04/2023 SDB 53

3) Find orthogonal matrix, P

P = ( e1 e2 e3 )

P = (1323

−23

231323

−232313

) is required orthogonal matrix

4) Find diagonal matrix, D

P TAP = D = (λ1 0 00 λ2 00 0 λ3

)=(−2 0 00 1 00 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!

Index

Differentiationhyperbolic functions, 10inverse hyperbolic functions, 10

Further coordinate systemshyperbola, 6locus problems, 9parabola, 7parametric differentiation, 7tangents and normals, 8

Hyperbolic functionsaddition formulae, 3definitions and graphs, 3double angle formulae, 3inverse functions, 4inverse functions, logarithmic form, 4Osborne’s rule, 3solving equations, 5

Integrationarea of a surface, 19inverse hyperbolic functions, 14

inverse trig functions, 14ln x, 14reduction formulae, 14standard techniques, 12

Matricescofactors matrix, 36determinant of 3´3 matrix, 35diagonalising 2´2 matrices, 43diagonalising symmetric 2´2 matrices, 44diagonalising symmetric 3´3 matrices, 47dimension, 35identity matrix, 35inverse of 3´3 matrix, 36non-singular matrix, 36singular matrix, 36symmetric matrix, 35transpose matrix, 35zero matrix, 35

Matrices - eigenvalues and eigenvectors2´2 matrices, 41

54 FP3 19/04/2023 SDB

characteristic equation, 41for 3 x 3 matrix, 46normalised eigenvectors, 42orthogonal matrices, 42orthogonal vectors, 42

Matrices - linear transformationsbase vectors, 38image of a line, 40image of a plane, 40

Vectorstriple scalar product, 24vector product, 22volume of parallelepiped, 24volume of tetrahedron, 25

Vectors - linesangle between line and plane, 33cartesian equation of line in 3-D, 26distance between skew lines, 34

line of intersection of two planes, 32vector equation of a line, 25vector product equation of a line, 26

Vectors - planesangle between line and plane, 33angle between two planes, 34Cartesian equation, 27distance between parallel planes, 30distance of origin from a plane, 29distance of point from a plane, 31line of intersection of two planes, 32vector equation, 28

FP3 19/04/2023 SDB 55