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MATH 302

iN α N∪T� SH∑LL

By

Matthew W. Haggard

- Math 302 In a Nutshell -

- 2 -

VECTORS DEFINITION

A vector is a quantity characterized by a magnitude and direction.

APPLICATION

Dot

Product

Work

W = F di

Reflection Reflected off of a plane with normal n , an

incident vector a results in the vector

= +b 2r a

ˆ ˆ( )= −r a n ni

Cross

Product

Torque

τ = ×r F�

Rotation of a rigid body ω= ×v r

NOTATION

Typed v (bold)

Written v�

Unit v

Cartesian

Unit Vectors x y z+ +i j k

Component , ,x y z

v v v

Graphic

OPERATIONS

Addition

, ,x x y y z z

a b a b a b + = + + + a b

Scalar Multiplication , ,x y z

k ka ka ka = a

Unit Vector ˆ =a

aa

Norm/Magnitude 2 2 2

x y za a a= = + + =a a a ai

Dot Product

(Inner Product)

x x y y z za b a b a b= + +a bi

cosθ=a b a bi

=a a ai

cosθ = =a b a b

a b a a b b

i i

i i

Cross Product

(Vector Product)

x y z

x y z

a a a

b b b

× =

i j k

a b (determinant)

sinθ× =a b a b

Right Hand Rule


- 3 -

Cross Product

continued…

Area of a parallelogram

A = ×a b

Volume of a parallelepiped

( )V = ×a b ci

CURVES, ETC.

Line 0( )t t= +r r d

y mx b= +

Ellipse

(circle for a=b)

[ ]( ) cos , sin ,0t a t b t=r

( ) cos sint a t b t= +r i j

Plane

( )0 0− =N r ri

ax by cz d+ + =

( ) ( ) ( )0 0 0A x x B y y C z z D− + − + − =

0Ax By Cz D+ + + =

0 0 0 ( )D Ax By Cz= − − − = −0

N ri

CURVE PROPERTIES

Tangent [ ]1 2 3( ) , ,t r r r′ ′ ′ ′=r

tan 0( )w w ′= +r r r

Arc-Length ( )t

as t dt′ ′= ∫ r ri

Motion and

Acceleration

( ) ( ) ( )t t t′ ′′= =a v r

tannorm= +a a a

tan =a v

a vv v

i

i

Angle of

Intersecting

Lines

cosθ =d D

d Di

Distance of

a line from a

point L

×=

0r P d

d

��


- 4 -

PLANE PROPERTIES

Defined by

Three

Noncollinear

Points

( )1 1 2 1 3 0PP PP PP× =�� i

Distance of a

plane from a

point.

( )1 0d

−−= =1 0

N r rN r N r

N N

ii i

1 1 1

2 2 2

Ax By Cz Dd

A B C

+ + +=

+ +

QUADRATIC SURFACES

Ellipsoid 2 2 2

2 2 21

x y z

a b c+ + =

Elliptic Cone 2 2

2

2 2

x yz

a b+ =

Hyperboloid of One Sheet 2 2 2

2 2 21

x y z

a b c+ − =

Elliptic Paraboloid 2 2

2 2

x yz

a b+ =

Hyperboloid of Two Sheets 2 2 2

2 2 21

x y z

a b c+ − = −

2 2 2

2 2 21

x y z

a b c− − + =

Hyperbolic

Paraboloid 2 2

2 2

x yz

a b− =

Parabolic Cylinder 2y ax=

Elliptic Cylinder 2 2

2 21

x y

a b+ =

Hyperbolic Cylinder 2 2

2 21

x y

a b− =


- 5 -

MULTIVARIABLE CALCULUS I – DIFFERENTIALS ETC…

LEVEL CURVES AND SURFACES Level curves and surfaces are “topographical” representations of higher dimension functions in a

lower, more comprehensible dimension. To think of higher dimensions, think not only of distances but

other “dimensions,” such as temperature, density or time. Level curves and surfaces are not only helpful

for visualizing higher dimension functions, but useful with gradients in determining other graph

properties.

Level

Curves A function of two variables ( , )f x y can be

graphed as a three-dimensional object (like

the sphere) or as level curves by setting

( , )f x y c= . For each c , there is a function

containing only two variables which, when

graphed, is a two-dimensional curve in the

plane. For each 1 1( , )x y on the level curve 1c

1 1 1( , )f x y c=

Level

Surfaces A function of three variables ( , , )f x y z can

only be graphed well as level surfaces. The

function depicted at right

2 2 2( , , )f x y z x y z= + +

is a fourth dimensional function that could

possibly represent the dimension of an

explosion over time. For each 1 1 1( , , )x y z of

the level surface 1c

( ) 2 2 2

1 1 1 11 1 1, ,f x y z cx y z = + + =

OPEN AND CLOSED SETS A set (whether two-dimensional or higher) may

have a boundary and an interior defined by boundary

points and interior points.

A point is a boundary point if all the

neighborhoods of that point include points both

inside and outside of the set.

A point is an interior point if some neighborhood

of that point is completely contained by the set.

A neighborhood is the set of points a certain

distance away (for two and three dimensions) or

closer from the reference point defined by

{ }: δ− <0

x x x where 0

x is the reference point

andδ is the distance.

A deleted neighborhood is a neighborhood that

does not contain the reference point.


- 6 -

A set is:

Closed iff it contains all of its boundary.

Open iff it contains a neighborhood of each

of its points.

PARTIAL DERIVATIVES Partial derivatives are calculated with respect to a certain

variable. All other variables are treated as constants during the

process of taking the derivative. A second-order partial is found

taking the derivative with respect to a variable again – holding

other variables as constants. A second-order partial taken of one

variable, then a different variable (xy

f oryx

f ) is called a mixed

partial.

Notation:

x

ff

x

∂=

∂

2 2

( )y x yx

f f zf f

yx x y x y

∂∂ ∂ ∂ = = = = ∂∂ ∂ ∂ ∂ ∂

Meaning:

The partial derivative gives the rate of change of f with respect

to the variable by which it was calculated. Partial differentials of

a multivariable function does not guarantee continuity, but if a

function is continuous, it will have equal mixed partials. Example:

3 2 2

2 2

3

( , ) 3 2 4

3 6 2

2 2

x

y

f x y x y x xy

f x y x y

f x y x

= + + +

= + +

= +

2

3

2

6 6

2

6 2

xx

yy

xy yx

f xy

f x

f x y f

= +

=

= + =

Application:

A tangent plane to a surface can be defined as

( )

( )

( )

1 0 0

2 0 0

0 0 0 0

,

,

,

x

y

f x y

f x y

x y f x y

= +

= +

= + +0

t i k

t j k

r i j k

( ) ( )1 2 0× − =0t t r ri

LIMITS A multivariable function can have different

apparent limits at one point depending upon the

angle of attack. This angle of attack can be any

parametric curve.

For 0y = , ( )0

lim 0,0x

f x→

=


- 7 -

To find a limit in a certain direction, make an

equation for a curve in that direction in terms of

only one variable. This curve could be a line or

parabola, or any other type of curves.

Here is an example with

( )3

2 2,

xy yf x y

x y

+=

+

For 0x = , ( )0

lim 00,y

f y→

=

For 2y x= , ( )2 3 2

2 20

2 8lim , 2

4x

x x xf x x

x x→

+= =

+

( )2

2 8

5

x

x

+ 2 8 2

5 5 5x= + =

∇ OPERATIONS x y z

∂ ∂ ∂∇ = + +

∂ ∂ ∂ i j k

Name Notation Input Output Calculation

gradient f∇ grad f Scalar → Vector f f f

x y z

∂ ∂ ∂+ +

∂ ∂ ∂i j k

divergence ∇ vi div v Vector → Scalar 31 2

vv v

x y z

∂∂ ∂+ +

∂ ∂ ∂

curl ∇× v curl v Vector → Vector

1 2 3

x y z

v v v

∂ ∂ ∂

∂ ∂ ∂

i j k

laplacian 2 f∇ ( )div grad f Scalar → Scalar

2 2 2

2 2 2

f f f

x y z

∂ ∂ ∂+ +

∂ ∂ ∂i j k

GRADIENT Meaning The gradient of a scalar function

( )f∇ returns a vector function

whose vectors point in the direction

of the greatest increase in f . The

resulting vector function resides

completely in the domain space of

f . If a function is differentiable at

a point, it is continuous at that point.

( ) 2 2,f x y x y= +

2 2f x y∇ = +i j

Calculation The Easy Way

Multiply each unit vector ( ), ,i j k by

the respective partial differential

( x∂ for i ,etc.)

Notice the a 2 variable function

gives a 2-dimensional vector

function and that a 3 variable gives a

3-dimensional.

( ) ( )

2 2

2

2

( , ) 2

2 2 4 1

2 2 4 1

f x y x xy xy y

f fx y y x xy

x y

f x y y x xy

= + + +

∂ ∂= + + = + +

∂ ∂

∇ = + + + + +i j

( ) ( ) ( )

2 2 2( , , )

2 2 2

g x y z x y z xyz

g x yz y xz z xy

= + + +

∇ = + + + + +i j k


- 8 -

The Mathematical Way

( ) ( ) ( ) ( )f f f o− = ∇ +x + h x x h hi

such that

( )

lim 0o

→=

h 0

h

h

( ) 2 2, 3f x y x y= +

( ) ( )

( )

( ) ( )

( )

1 2

2 2 2 2

1 2

2 2

1 2 1 2

f ?

1 2

( ) ( ) ( , ) ( , )

3 3

6 2 3

6 2 3

o

f f f x h y h f x y

x h y h x y

xh yh h h

x y h h

∇

+ − = + + −

= + + + − +

= + + +

= + + +

h

x h x

i j h i j h

�� i i

( ) 1 21 2

1 2

3 cos3

3

h hh h

h h

θ++=

+≤

i j hi j h

h h

i j h

i

h

1 23h h≤ +i j

1 2lim 3 0 6 2h h f x y→

+ = ∴∇ = +h 0

i j i j

Application Normal Vectors

f∇ gives the direction of greatest increase in f .

If a graph of f∇ is placed on top of a graph of the

level curves or surfaces of f , f∇ forms normal

vectors with the level curves or surfaces as depicted.

This can be useful to find the normal of any

function by simply turning it into a level

curve/surface.

Example:

2 2( , ) 1f x y z x y= = − −

Level-ize it!

2 2 2

2 2 2

0 1

( , , ) 1

x y z

g x y z x y z

= + + −

= + + −

f gives the level surfaces of g

2 2 2g x y z∇ = + +i j k

( , , )g x y z∇ gives the normal vector of f at ( , , )x y z

Tangent Lines and Planes Normal Vector to Curve

( )0 0,f x y= ∇N where

( ),f x y c=

Tangent Vector to Curve

If ( )0 0,f x y a b∇ = +i j

b a= −t i j or b a= − +t i j

Tangent Planet to Surface

( ) ( ) 0f∇ − =0 0r r ri

where ( ), ,f x y z gives the level

curves of the surface ( ),z x y


- 9 -

Directional Derivative:

One can find the rate of change of f with respect to any direction in the domain space

using directional derivatives. The directional derivative is found by taking the dot product

of a unit vector and the gradient of a function. In the direction b , the directional

derivative for f is

1

D f f f′= = ∇b b bbi

Proving if a function is a Gradient If a vector function is a gradient, then

the mixed partial derivatives of that

function will be equal.

( ) ?

,

yxff

x y

f x y y x

f y f x

∇ = −

= = −

i j

( )

1 1

for any

,

xy yxf f

f

f x y y x

= = −

∴

∇ ≠ −i j

Reconstructing from a Gradient If you are given a gradient and want to get the original function, use integration.

( ), , x y zf x y z f f f∇ = + +i j k

( ) ( )xf dx F y z= + φ + ϕ∫

(1) Integrate one of the partials with respect to

wherever the partial came from.

( ) ( ) ( )

d dFF y z y

dy dy′+ φ + ϕ = + φ

(2) Differentiate the result of the integration

with respect to the next variable.

( )

( )

y

y

dFy f

dy

dFy f

dy

′+ φ =

′φ = −

(3) Equate the result with the partial for that

variable ( )yf , solving for ′φ .

( ) ( )y dy y′φ = φ∫

(4) Integrate ′φ with respect to y to get φ to

substitute in (1).

(5) Repeat steps 2-4 using z

f and ( )zϕ

Substituting everything into (1) gives the

reconstructed gradient. Example

( ) ( ) ( ) ( )3 2

3 2

, , 4 3

, 4 , 3x y z

f x y z yz z xz y xy xz

f yz z f xz y f xy xz

∇ = + + + + +

= + = + = +

i j k

( ) ( )

( )

( )

3

3

4

x

y

f dx yz z dx

F xyz xz y z

dFxz y f xz y

dy

y xz

φ ϕ

φ

φ

= +

= + + +

′= + = = +

′ =

∫ ∫

4y xz+ −

( ) 24 2y dy y yφ= =∫

( )

( )

( )

2 3

2 2

2

3 3z

F xyz y xz z

dFxy xz z f xy xz

dz

z xy

ϕ

ϕ

ϕ

= + + +

′= + + = = +

′ = 23xz+ xy− 23xz−

( )2 3

0

0

2

z

F xyz y xz

ϕ

=

=

∴ = + +


- 10 -

DIVERGENCE Meaning The divergence of a vector function ( ∇ Fi ) yields a

scalar function which gives the flux at any point in the

vector function. The flux is the total amount of stuff

leaving a surface per unit area per unit time. To determine

the total flux of a body, integral calculus is usually used.

Calculation

31 2divFF F

x y z

∂∂ ∂= + +

∂ ∂ ∂F

( ) 2 3, , 2

div 4 0

5

x y z x y xz xyz

xy xy

xy

= + +

= + +

=

F i j k

F

Application If we have a fluid flow function

2z=F k and want to know the

total flux of that field through a

unit cube located in the first

octant with a vertex at the origin

(as depicted).

The only way for the fluid to

leave the object is through one of

the faces. With this particular

vector field, the fluid flow out of

the box is zero at every face

except the top face ( 1z = ). The

outflow for this surface is

2(1) 2= =F k k per unit time.

The surface area of the top of the

box is 1 so the flux through this

face is

amount leaving 2

2surface area 1

= =

As this is the only place where

fluid is leaving, the total flux is

2 as well.

Another way to find the total flux is through integral

calculus. This adjacent example would be:

( )

( )

( ) ( )( )

1 1 1

0 0 0

1 1 1

0 0 0

div 0 0 2

0 0 2

2

2

2 1 1 1

2

S S

S

dV z dV

z dVz

dx dy dz

dx dy dz

= ∇ + +

∂= + +

∂

=

=

=

=

∫∫∫ ∫∫∫

∫∫∫

∫ ∫ ∫

∫ ∫ ∫

F i j k

F

i

CURL Meaning The curl of a vector function ( ∇×F ) can be thought of as a rotational measurement of

some sort (much like the divergence is somewhat of a flow measurement).

Calculation

1 2 3

curlx y z

F F F

∂ ∂ ∂=

∂ ∂ ∂

i j k

F

( ) 2, ,x y z yz x z= + +F i j k

( ) ( ) ( )

( )

2

curl 0 2

2

y x zx y z

yz x z

x z y

∂ ∂ ∂= = − + −

∂ ∂ ∂

= − −

i j k

F i j k

k j


- 11 -

Application The curl of a velocity field of a rigid, rotating

body gives a vector pointing along the axis of

rotation with a magnitude twice the value of the

angular velocity. Looking as if the curl vector is

pointing out of your eye, the body is rotating

clockwise.

= ×v w r by definition

ω=w k if rotating about the z-axis

curl =v 2w

CHAIN RULES To take the derivative of a function

with respect to a variable of which the

function is not directly dependent

as in f

t

∂

∂ where ( ),f f x y= and

( ) ( ), , ,x x t s y y t s= =

draw a diagram similar to the leftmost

one then trace every leg of the path from

the first point f to the variable by which

you are differentiating (in this case t ).

Tracing the paths, you end up with

, , ,f x f y

x t y t

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ Multiply the partials

found along the same path and add the

terms together.

f f x f y

t x t y t

∂ ∂ ∂ ∂ ∂= +

∂ ∂ ∂ ∂ ∂

One Ludicrous Example

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ), , , , , , , , , , , , , , , , ,r f g f x y z g m n x t u y t u v z t u m t u v n u

r r f x f y f z r g m g dn

u f x u y u z u g m u n du

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

EXTREME VALUES

At ( )0 0,f x y∇ = 0 , ( )0 0,x y is either a local extrema or a saddle point. You can use the second partial

test to usually tell what exactly the point is if 0f∇ = .

( ) ( ) ( )0 0 0 0 0 0, , , , ,xx xy yyA f x y B f x y C f x y= = =

2D B AC= −

If 0D > , ( )0 0,x y is a saddle point.

If 0D < , ( )0 0,x y is an extrema and…

if 0A > or 0C > , ( )0 0,x y is a local minimum.

if 0A < or 0C < , ( )0 0,x y is a local maximum.

Note: A local extrema can also exist if ( )0 0,f x y∇ does not exist.


- 12 -

Constrain

ed

Extremes

To maximize/minimize a function

( )f x whose domain is

constrained by a function

( ) 0g =x , try to get the function in

terms of at least one of the

variables.

Inside of Domain: Substitute the constraining

function into the function to be

maximized. If it is a function of a

single variable, take the derivative

and set it to zero – this will yield

where extrema are present.

If it is a function of several

variables, take the partial with

respect to each of the variables and

set them equal to zero. Solve in

any way you wish. Then substitute

backwards, finding one variable at

a time. Tada!

( ), , 2 2 2f x y z xy xz yz= + + constrained by

15x y z+ + = where f is the surface area of a

rectangular prism of dimensions x y z× × .

( ), , 15 0

15

g x y z x y z

z x y

= + + − =

= − −

( ) ( )2 2 15 2 15

2

f xy x x y y x y

xy

= + − − + − −

= 230 2 2x x xy+ − − 2

2 2

30 2 2

2 30 2 30 2

4 30 2 0, 4 30 2 0x y

y xy y

x x xy y y

f x y f y x

+ − −

= − + − + −

= − + − = = − + − =

15 2 2 30 4y x x y− = = = −

15 30 4

3 15

5

y y

y

y

− = −

=

=

2 15

2 10

5

x y

x

x

= −

=

=

15

15 5 5

5

z x y

z

z

= − −

= − −

=

( ) ( )( ) ( )( ) ( )( )5,5,5 2 5 5 2 5 5 2 5 5 150f = + + =

Boundary of Domain: To find the min/max on a

boundary, which boundary is

defined by the constraining

function(s), parameterize the

boundary and take the partials as

described for inside the domain.

The example at right is bound to

0, 0, 0x y z> > > because it is a

real object with positive

dimensions which yields the plane

in three-space depicted. It is

further confined to a plane in two-

space by putting the function to be

maximized in terms of only two

variables. The boundary points on

the x and y axes are discounted because that would

make one of the dimensions zero. The hypotenuse of

the triangular plane 15x y+ = also gives z a

dimension of zero.

Method of

Lagrange A function ( )f x constrained by ( ) 0g =x will have a

maximum where ( ) ( )λf gx x∇ = ∇ .

By changing the constraining function (which is a function of

one less dimension than f ) into ( ) 0g =x , the original

constraining function becomes a level curve of g . Because

the gradient of a function yields the normal vectors to the

level curves, the Method of Lagrange simply states that the

level curve of g will have a parallel normal and be tangent to

the maximum level curve of f . This is true because the two

level curves must be tangent at the maximum or minimum

value of f . If g is not tangent at a particular level curve of

f , it must be tangent at one “farther up the hill.”


- 13 -

Another way the Method of Lagrange can be used for functions of two variables (because

f g∇ ∇� ) is 0f g∇ ×∇ = .

This yields 0f g f g

x y y x

∂ ∂ ∂ ∂− =

∂ ∂ ∂ ∂

or for functions of several variables λyx z

x y z

ff f

g g g

∂∂ ∂= = = =

∂ ∂ ∂�

Example Maximize f with the given constraint.

( ) ( )22, 1f x y x y= + − constrained by

22 1

4

xy+ = . ( )

22, 1 0

4

xg x y y= + − =

( ) 12 2 1 , 2

2f x y g x y∇ = + − ∇ = +i j i j

(1a)

( )

λ

λ2 2 1 λ2

2

f g

x y x yi j i j

∇ = ∇

+ − = +

(1b)

0f g f g

x y y x

∂ ∂ ∂ ∂− =

∂ ∂ ∂ ∂

( ) ( ) 12 2 2 1 0

2x y y x

− − =

(2b)

4 0

3

xy yx x

xy x

− + =

− =

(2a)

λ2

2

4 λ

4 λ

x x

x x

xy xy

=

=

=

( )

( )

2 1 λ2

1 λ

1 λ

y y

y y

x y xy

− =

− =

− =

( )4 1

4

3

xy x y

xy xy x

xy x

= −

= −− =

(3)

3 x− y x=

1

3y=−

(4) 221 1

, 0 13 4 3

14 1

9

32

9

xg x

x

x

− = = + − −

=± −

=±

(5) 2 2

32 1 32 1, 1

9 3 9 3

32 16 485.33

9 9 9

f ± − = ± + − −

= + = ≈

( ),f x y has maximum value 5.33 at

32 1 32 1, , ,

9 3 9 3

− − −


- 14 -

MULTIVARIABLE CALCULUS II – INTEGRALS ETC…

INTEGRALS BY ∑∑∑ ∑ PROPERTIES

( )

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1 1 1

m n m n

ij ij

i j i j

m n m n

ij ij

i j i j

m n m n

i j i j

i j i j

m n m n m n

ij ij ij ij

i j i j i j

a a

ka k a

a b a b

a b a b

= = = =

= = = =

= = = =

= = = = = =

=

=

=

+ = +

∑∑ ∑ ∑

∑∑ ∑∑

∑∑ ∑ ∑

∑∑ ∑∑ ∑∑

Numerically, the integral of a function ( )f x over a

certain region is the number I that satisfies the

inequality ( ) ( )f fL P I U P≤ ≤ where ( )fL P is

the lower sum and ( )fU P is the upper sum.

By dividing a region P into arbitrary rectangular

pieces (we’ll call partitions denoted ij

R ), we can

define:

( )1 1

m n

f ij i j

i j

L P m x y= =

= ∆ ∆∑∑

( )1 1

m n

f ij i j

i j

U P M x y= =

= ∆ ∆∑∑

ijm is the minimum value of f on a partition

ijR

ijM is the maximum value of f on a partition

ijR

1i i ix x x −∆ = − 1j j j

y y y −∆ = −

i jx y∆ ∆ is the area of a partition

ijR

Algorithm:

Example:

Approximate ( )3 4

0 0,f x y dxdy∫ ∫ where ( )

2 2

, 14 9

x yf x y = + +

using { } { }0,1,3,4 , 0,1,2,3x y∈ ∈

(Note that the function is always increasing in x and y for this region)

(1) Find ij

m andij

M 2211 1

4 9

jiij

yxm

−−= + + ,

22

14 9

jiij

yxM = + +

(2) Using inequalities, try to

get ij ij

m M≤∗≤ .

Averaging the separate x and

y values usually helps. You

can see in the middle terms

that the average of i

x and 1ix −

were substituted for x

( )

( )

2

1

2 2

1

2

1

2 2

1

2

4 4 4

2

9 9 9

i i

i i

j j

j j

x x

x x

y y

y y

−

−

−

−

+ ≤ ≤

+ ≤ ≤

( ) ( )22

2 22 21111 1 1 1

4 9 4 9

11

22

4 9

j ji ij ji i

ij ijm M

y yx xy yx x −−

−− + + ≤ + ≤ + +

++ +

��


- 15 -

(3) Multiply by i j

x y∆ ∆ throughout, then take the double sum of everything. This should give

( ) ( )f fL P I U P≤ ≤ , so the integral is the middle portion of the inequality.

( ) ( )

( )( ) ( )

22

11

1 1 1 1 1 1

22

11

1 1

11

22 14 9

11

221

4 9

j jm n m n m ni i

ij i j i j ij i j

i j i j i j

j jm n i i

f i j

i j

y yx x

m x y x y M x y

y yx x

L x yP

−−

= = = = = =

−−

= =

++∆ ∆ + + ∆ ∆ ∆ ∆

≤ ≤

++ ≤ + + ∆ ∆

∑∑ ∑∑ ∑∑

∑∑ ( )f

U P≤

(4) Simplify the middle inequality (separating into different sums and looking for telescoping sums)

( ) ( ) ( )22

11

1 1 1 1 1 1

1 1 111

4 9 22

m n m n m n

i j j j i j i ji i

i j i j i j

x y y y x y x yx x −−= = = = = =

∆ ∆ + + ∆ ∆ + ∆ ∆+ ∑∑ ∑∑ ∑∑

( )( )( )

( )

1 1 1

1 1

2

1

1 1 1 1

1 1

4 4

1 1

9 4

m n

i i i i i i j

i j

m n m n

j j i ji j

i j i j

x x x x x x y

y y x yx y

− − −= =

−= = = =

= ++ + − ∆

+ + + ∆ ∆∆ ∆

∑∑

∑∑ ∑ ∑

�

Here, I will just show how to solve the first term; the second term is similar; the third term is easy:

( )( )( )( )

( )( )

( ) ( )( ) ( )( ) ( )( )

( ) ( )( ) ( )( ) ( )( )

2 21

3 3

1 1 1

1 1

3 32 2

1 1

1 1

2 2 2 2 2 2

3 0 1 0 1 0 2 1 2 1 3 2 3 2

2 2 2

1144

1

16

1

16

13 0 1 0 1 0 3 1 3 1 4 3 4 3

16

i i

i i i i i i j

i j x x

j i i i i

j i

x x x x x x y

y x x x x

y y x x x x x x x x x x x x

−

− − −

= = −

− −= =

+ + − ∆ =

= ∆ + −

= − + − + + − + + −

= − + − + + − + + −

∑∑

∑ ∑

��

41

8 =

Solving all the two other terms, you end up with

41 35 1513

12 218 9 72+ + = ≈

The real integral yields 28. This is because the approximation leaves out a lot of volume.


- 16 -

MATRICES AND LINEAR ALGEBRA DEFINITION OF A MATRIX

A matrix is a rectangular set of elements used to compact

calculation involving systems of linear equations, transformations,

etc. It’s size is denoted m n× where m is the number of rows and

n is the number of columns.

Matrix

Multiplication

( ) ( ) ( )( ) ( ) ( )

2 2

v yv u xu

u xx c d c d c dc d v

a b a b a ba b t wt

t yww y

× × ×

+ + + = + + +

2 3 2 3

��

jk j kc = a bi for j row= , k column=

Properties

≠AB BA in general

=AC AD

does not imply =C D

( ) ( ) ( )k k k= =A B AB A B

( ) ( )=A BC AB C

( )+ = +A B C AC BC

( )+ = +C A B CA CB

( )Τ Τ Τ=AB B A

det∗ ∗ ∗ ∗

= ∗ ∗ ∗ ∗ for 2 2× ,

a bad bc

c d= −

(1) Pick any one row or column

Determinant

(2) For each element (reference element) of the

chosen row/column, cross out the row and column of

which it is a member. This will yield a matrix of

( 1) ( 1)n n− × − size for each reference element.

b c

h i

a c

g i

a b

g h

NOTATION

Typed A (bold)

Written A

Identity I

Augmented

� ∗ ∗ ∗ = ∗ ∗ ∗

A

[ ] [ ]= A BA B

Transpose TA

Inverse −1A

Double

Subscript

11 12 1

21 22 2

1 2

n

n

m m mn

a a a

a a a

a a a

�

�

� � � �

�

jka for

j row=

k column=

OPERATIONS, ETC…

Addition a b t u a t b u

c d v w c v d w

+ + + = + +

Scalar

Multiplication

a b ka kbk

c d kc kd

=

Transposition

g

h

g

b c

b

fh

d

c

d

f

Τ =

a

i i

e e

a

Identity

Matrix

1 0 0

0 1 0

0 0 1

�

�

� � � �

�

Rank

is the number of linearly

independent rows or

columns.

rank rankΤ=A A


- 17 -

(3) Multiply the determinant of these matrices (called

minor matrices) with the reference element used to

get them (the element at the intersection of the lines).

b cd

h i

a ce

g i

a bf

g h

(Determinant

…) (4) Multiply each of the resulting constant-

determinant couples by a 1± depending on the

position of the reference element given by the formula

( 1) j k+− . The minor multiplied by the sign give what

is called the cofactor. Add all the cofactors together

to get the determinant.

Note: You can also find the sign of a cofactor

visually by choosing the corresponding element of a

checkerboard, sign matrix.

b c a c a bd e f

h i g i g h− + −

Sign Matrix:

+ − + − + − + − +

�

�

�

� � � �

1 1 2 2det

j j j j jn jna C a C a C= + + +A �

1 1 2 2detk k k k nk nk

a C a C a C= + + +A �

for ( 1) j k

jk jkC M+= − ,

jkM is the

minor

Properties

( ) ( )det det det det= =AB BA A B

det( ) detnk k=A A

1 1

detdet

− =AA

( )rank n n n× = iff

det 0≠A

Changes from Manipulation det

det

D

d

=

=

A

B change= +B A

Row Interchange d D= − Transposition d D=

Addition of Rows d D= Zero Row/Column 0d =

Scalar Multiplication d kD= Proportional Row/Column 0d =

Inverse

Matrix

− −= =1 1AA A A I

1

detjk

AΤ− =

1A

Awhere

jkA is the cofactor (see determinant above) of

jka .

1a b c

d e f

g h i

− =

1 1

det det

e f d f d e e f b c b c

h i g h g h h i h i e f

b c a c a b d f a c a c

h i g i g h g h g i d f

b c a c a b d e a b a b

e f d f d e g h g h d e

Τ

− − = − − = − − − −

A A

Gauss-Jordan Method (defined below)

[ ] − → 1

A I I A

Example of Gauss-Jordan:

( )2 1 2 1 2 1

2 2

1 16

2 22

2 3 1 0 2 3 1 0 1 0 2 3

1 2 0 1 0 .5 .5 1 0 1 1 2

R R R R R R

R R

− → − →

→

− → → − −


- 18 -

Inverse of a 2 2×

1

1

det

a b d b

c d c a

−−

= − A

Properties

1 1 1( )− − −=AC C A

2 1 1 2( ) ( )− −=A A

1 1( ) ( )− Τ Τ −=A A

LINEAR SYSTEMS MANIPULATION

When representing systems of linear equations with augmented

matrices, several row operations are available which will maintain the

integrity of the system. Any combination of the three methods of

manipulation can be used together…

REPRESENTATION Augmented matrices

provide a compact way to

represent a system of linear

equations and is based on

matrix multiplication.

Row

Interchange 1 2

1 0 0 1

0 1 1 0

R Ra b t u

t u a b

↔ →

Addition of

Rows 1 2 2

1 0 1 0

0 1 1 1

R R Ra b a b

t u a t b u

+ → → + +

Multiplication

of Constants 1 1

2 2

5

1

2

1 0 5 5 5 0

0 1 2 2 0 1 2

R R

R R

a b a b

t u t u

→

→

→

ax by t+ =

cx dy u+ =

a b x t

c d y u

⇒ =

a b t

c d u

⇒

Gauss Elimination Gauss-Jordan Elimination

1

0 1

0 0 1

∗ ∗ ∗ ∗ ∗ ⋅ ⋅ ⋅ ⋅

∗

�

�

�

�

Reduce the augmented matrix

to 1’s along the main diagonal

(these positions are called pivot

positions) with 0’s below the

main diagonal and any numbers

above the diagonal.

1 0 0

0 1 0

0 0 1

∗ ∗ ⋅ ⋅ ⋅ ⋅

∗

�

�

�

�

The same as a

Gauss except that

above the main

diagonal, all the

numbers are 0.

SOLUTION VECTORS When solving a linear system of

equations, if you end up with

zero rows, the variables for

which there are not pivot

positions are considered free

variables and can range

anywhere.

Simply put the variables with

pivot positions in terms of the

free variables for a solution

vector.

If the system is homogeneous,

the solution vector will be a

vector space.

12 13 1 112 13 1

23 2 223 2

3

1 1

0 1 0 1

0 0 0 0 0 0 0 0

a a b xa a b

a b xa b

x

⇒ =

3x is a free variable (it lacks a pivot position)

1 1 13 12 2 23 13 12 23 1 12 2

2 2 23 23 2

3

( )

1 0

x b a t a b a t a a a b a b

x tb a t a b

x t

− − − − + − = = +− −

for =b 0

1 13 12 23

2 23

3 1

x a a a

x t a

x

− + = −

CRAMER’S RULE

2 2× 1 2

1 2

a

c

x

x

tb

x u

x

d

+ =

+ =

c d

a b =

A

1det

t

u dx

b

=A

, 2det

a

c ux

t

=A

This pattern of solution

replacement works for

n n× matrices as well.


- 19 -

Cramer’s Theorem:

If, for a linear system of n equations in the same

number of unknowns, det 0D = ≠A , the system

has precisely one solution given by the formulas:

1 21 2, , , n

n

DD Dx x x

D D D= = =�

where k

D is the determinant obtained from D by

replacing in D the kth column by the column with

the entries 1, ,n

b b� (the solution matrix)

Hence if the system is homogeneous and 0D ≠ ,

it has only trivial solution [ ]0 0Τ

� . If 0D = ,

the homogeneous system also has nontrivial

solutions.

It is by this theorem that eigenvalues can be

found.

LINEAR TRANSFORMATIONS =y Ax

1 1 1 2

2 2 1 2

y x ax bxa b

y x cx dxc d

+ = = +

A transformation from x space to y space by the transformation

matrix A .

cosθ sinθ

sinθ cosθ

−

Rotation ofθ in the plane 1 0

0 1

− −

Reflection in the origin

0 1

1 0

Reflection over line 1 2x x= 0

0 1

a

Stretch along 1x of a

1 0

0 1

−

Reflection over 1x axis

FUNDAMENTAL THEOREM FOR LINEAR SYSTEMS

Existence For A is m n× , solutions exist iff �( ) ( )rank rank=A A

Uniqueness A has precisely one solution iff �( ) ( )rank rank n= =A A

Infinitely many solutions �( ) ( )rank rank r= =A A If r n< , A has infinitely many solutions

Gauss elimination If solutions exist, they can all be found by Gauss elimination.

EIGENVALUES AND EIGENVECTORS Definition λ=Ax x , λ is an eigenvalue of A if ≠x 0

The solutions of x for λn

=Ax x or ( λ )n

− =A I x 0 are called the eigenvectors of A

for λn

.

Solution ( λ )− =A I x 0 This is a homogeneous system, so in order for it to have a non-trivial

solution det( λ )− =A I 0 by Cramer’s theorem.

11 12 1

21 22 2

1 2

λ

λ0

λ

n

n

n n nn

a a a

a a a

a a a

−

−=

⋅ ⋅ ⋅

−

�

�

�

�

Taking this characteristic determinant yields a characteristic polynomial from which

multiple λmay be found.

Substituting the resulting eigenvalues back into ( λ )− =A I x 0 and solving the system

yields corresponding eigenvectors.


- 20 -

Example 5 3

3 5

=

A

For what λ λ=Ax x ?

( λ )− =A I x 0

2

0 det( λ )

5 λ 3

3 5 λ

(5 λ)(5 λ) 3 3

λ 10λ 16

(λ 8)(λ 2)

= −

−=

−

= − − − ⋅

= − +

= − −

A I

2λ 8, λ 21 = =

(Eigenvalues of A )

For λ λ 81= =

( 8 )− =A I x 0

5 8 3 0

3 5 8 0

1 1 0

0 0 0

− −

−

1

2

1

1

x tt

x t

= =

(Eigenvector for λ 8= )

For 2λ λ 2= =

( 2 )− =A I x 0

5 2 3 0

3 5 2 0

1 1 0

0 0 0

− −

1

2

1

1

x tt

x t

− − = =

(Eigenvector for λ = 2 )

EIGENVALUE APPLICATIONS Stretching /

Transforming =y Ax where A is the transformation matrix

mapping x onto y .

The eigenvalues of A tell the amount of

distortion in the direction of the

corresponding eigenvectors. These directions

are called the principle directions.

This picture is similar to the transformation

caused by the matrix

5 3

3 5

=

A

(the one used in the example above)

Check the unit vectors to see what

transformation has occurred.

Markov

Processes

In a region redistribution system (things

flowing from one area to another) of the form

=Ax y , the limit state (all changes in region

are in equilibrium) is found by =Ax x which

is simply an eigenvalue problem with λ 1= .

In order to multiply as =Ax y where A is

the redistribution matrix, x is the initial state

as a column vector and y is the resulting state

as a column vector, the columns of A need to

add up to 1. Transpose as necessary.

From From FromI II III

1 1

2 2

3 3

To I

To II

To III i

x yb c

x yd f

x yg h

e

a =

From the solution vector [ ]1 2 3t k k kΤ

, a

percentage solution vector is easily calculated

by solving for t in 1 2 3 100tk tk tk+ + =

These letters correspond to the matrix

at left.


- 21 -

Leslie Model A Leslie model is similar to a Markov

process, but is specific to age-changing

population growth. The life span of a species

is divided into groups and the predicted

transition between groups is projected in a

matrix (as shown at right).

The limit state (if there is one) is found in the

same manner as with other Markov processes.

0

0

0

0

0 0

0 0

f

f

f

IIa b

IIIIc

IIIIIId

=

number born from II per woman

number born from III per woman

fraction of I going to II

fraction of II going to III

a

b

c

d

=

=

=

=

MATRIX TERMS basis: A set of linearly independent vectors in a

vector space equal in number to the dimension of

the space.(360)

characterisitc polynomial: A polynomial that

results from a characteristic determinant.(372)

characteristic determinant: det( λ )− =A I 0

(372)

characteristic value: Another name for an

eigenvalue.(371)

characteristic vector: Another name for an

eigenvector.(371)

consistent: A system that has at least one

solution.(326)

determined: If there are an equal number of

equations and unknowns in a system.(326)

dimension: The maximum number of linearly

independent vectors.(360)

eigenspace: The vector space defined by

eigenvectors corresponding to an eigenvalue plus

the zero vector. (371)

homogeneous: If all of the equations in the

system equal zero (all of the numbers on the

solution side of an augmented matrix are

zero).(340)

image: From a mapping/transformation, it is the

result (in the new space) of the original.(362)

inconsistent: A system with not solutions.(326)

linear combination: A combination of vectors in

a vector space multiplied by any scalar, added

together.(360)

linear dependence: When one or more of the

equations is/are a linear combination of other

equations in the system.(332)

linear independence: When none of the

equations is/are a linear combination of other

equations in the system.(332)

nonhomogeneous: If at least one of the equations

in the system is not equal to zero (at least one of

the numbers on the solution side of an augmented

matrix not being zero).(341)

nonsingular: A matrix with an inverse.(350)

orthogonal: Τ =A A A transformation that is

only a rotation (with a possible reflection), but no

size distortion.. Orthogonal matrices have

determinants of -1 or 1 and eigenvalues with

absolute value of 1, though they could be

complex.(381)

overdetermined: If there are more equations than

unknowns in a system.(326)

singular: A matrix with no inverse.(350)

skew-symmetric: Τ = −A A The main diagonal

is made up entirely of zeros. Eigenvalues are

purely imaginary or zero.(381)

span: The set of all linear combinations of given

vectors (or vector space) with the same number of

components.(335)

spectral radius: The largest absolute value of all

the eigenvalues.(371)

spectrum: The set of eigenvalues for a given

matrix.(371)

stochastic matrix: A square matrix with

nonnegative values where all rows add up to 1.

Used for region transition probabilities.(377)

symmetric: 1Τ −=A A Eigenvalues are real.(381)

underdetermined: If there are more unknowns

than equations in a system.(326)

vector space: A nonempty set of vectors formed

by a basis such that any linear combination of the

basis vectors lies within the set.(334)

(Page numbers are for Advanced Engineering Mathematics; Kreyszig, Erwin: 8th ed.)

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