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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
- Math 302 In a Nutshell -
- 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
����
- Math 302 In a Nutshell -
- 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− =
- Math 302 In a Nutshell -
- 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.
- Math 302 In a Nutshell -
- 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→
=
- Math 302 In a Nutshell -
- 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
- Math 302 In a Nutshell -
- 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
- Math 302 In a Nutshell -
- 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
=
=
∴ = + +
- Math 302 In a Nutshell -
- 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
- Math 302 In a Nutshell -
- 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.
- Math 302 In a Nutshell -
- 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.”
- Math 302 In a Nutshell -
- 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
− − −
- Math 302 In a Nutshell -
- 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 −−
−− + + ≤ + ≤ + +
++ +
������������� �����������
- Math 302 In a Nutshell -
- 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.
- Math 302 In a Nutshell -
- 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
- Math 302 In a Nutshell -
- 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
− → − →
→
− → → − −
- Math 302 In a Nutshell -
- 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.
- Math 302 In a Nutshell -
- 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.
- Math 302 In a Nutshell -
- 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.
- Math 302 In a Nutshell -
- 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.)