Transcript Math Tools / Math Review

Math Tools / Math Review
1
Resultant Vector
Problem: We wish to find the vector sum of vectors A
and B
Pictorially, this is shown in the figure on the right.
A

A+B
B
Mathematically, we want to break the vector into x, y, and
maybe z components and find the resultant vector

Let A  6 xˆ  7 yˆ

And B  12 xˆ  6 yˆ
 
A  B  (6  12) 2  (7  6) 2
y (7  6)
tan   
x (6  12)
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System of equations

Let 6x+7y=15
 And 4x-3y=9
 Now find x and y which satisfy these
equations
 I use “method of minors”
6 7
Determinat e 
 (6)( 3)  (7)( 4)  18  28  46
4 3
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System of Equations cont’d
The solution for x is found by creating a “minor” wherein the constants in
the equation are substituted in place of the “x” value and the value of the
minor is found. It is then divided by the value of the determinant.
15
x
9
7
 3 (15)( 3)  (7)(9)

 2.347
det
 46
I then “back-substitute” the value of x into my initial equation and solve for y.
6(2.347)+7y=15 and solve for y (y=.1304)
4
Or I could use the minors again
6 15
9 (6)(9)  (4)(15)
y

 0.1304
det
 46
4
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Larger Systems
Larger systems are broken down into their resultant minors.
For example:
6x + 7y +10z =12
-9x+15y+2z=60
5x +12y-10z=15
6
7
  
  
  
10
15 2
9 2
 9 15
det   9 15 2  6
7
 10
 3434
12  10
5  10
5 12
5 12  10
12 7
60 15
x
10
2
15 12  10
det
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Spherical Coordinates
Cartesian coordinates: x, y, z
Spherical coordinates: r, , f
Math Majors NOTE Theta!
x  r cos f sin 
y  r sin f sin 
z  r cos 
r  x  y  z
2
2

1
2 2
 x2  y2  z 2
 y
x
z
  cos 1  
r
  tan 1  
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Cylindrical Coordinates
Cartesian coordinates: x, y, z
Spherical coordinates: r, f, z
Math Majors NOTE Phi!
x  r cos f
y  r sin f
zz
r  x  y
2

1
2 2
 x2  y2
 y
x
  tan 1  
zz
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Showing my age
In the old days, I would tell you to use
your integral tables
 Now, I say use your calculators to
integrate
IF YOU DARE!*

*
I only say this since I have seen some integrals which are easily found in
the tables being integrated incorrectly by the calculator.
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Partial Derivatives



So what is the difference
between “d” and  ?
“d” like d/dx means the
function only contains
the variable x.
When the function
contains not just x but
may be y and z, we use
the partial differential, 
For example :
f ( x, y, z )  xyz
f

 ( xyz)  yz
x x
Note that the variables y and z
are held constant when the
differential operator acts on the
function
What is the solution to?

( xyz)  ?
y

( xyz)  ?
z
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Introduction to “Del”

We can now make a special differential
operator called “del”. Del is defined as



  xˆ 
yˆ  zˆ
x
y
z
We treat “del” as a vector and thus, we can
apply the “dot” and “cross” products to them.
 But first, let’s recall the “dot” and “cross”
product

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The “dot” or scalar product

The scalar product is
defined as the
multiplication of two
vectors in such a
way that result is a
vector
 is the angle between
A and B

A  a x xˆ  a y yˆ  a z zˆ
and

B  bx xˆ  by yˆ  bz zˆ
Then
 
A  B  a x bx  a y by  a z bz
or
 
 
A  B  A B cos 
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“Cross” or vector product


The vector product is the multiplication of two vectors such
that the result is a vector and furthermore, the resulting vector
is perpendicular to the either of the two original vectors
The best way to find a vector product is to set it up as a
determinant as shown on the right
xˆ
 
A  B  ax
yˆ
zˆ
ay
az
bx
by
bz
 
A  B  (a y bz  a z by ) xˆ  (a x bz  a z bx ) yˆ  (a x by  a y bx ) zˆ
 
 
A  B  A B sin 
 is the angle between A and B
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First application of “del”: gradient


The gradient is defined as the shortest or steepest path up a
mountain or down into a valley.
Let’s go back to f=xyz then
f  ( xyz)  yzxˆ  xzyˆ  xyzˆ



You see that “grad(f)” makes a vector which points in a
particular direction.
Also, note that grad(f) takes a scalar function and makes a
vector of it
A particle which travels through a region of space wherein the
potential energy, U(x,y,z), varies as a function of space has a
force exerted on it equivalent to

 U   U   U 
 yˆ   
F  U   
 xˆ   
 zˆ
 x   y   z 
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The scalar product and 

We can apply  to the scalar product i.e.

·A where A is some vector
·A is called the “divergence” of A or
“div(A)”.
 Geometrically, we are discussing if A is
diverging from some central point.A is not

A is diverging
from a central
point so
Div(A) is equal to
some value
diverging from
a central point
so
Div(A) is equal
to zero
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The vector product and 

We can apply  to the vector product i.e.

xA where A is some vector
xA is called the “curl” of A or “curl(A)”.
 Geometrically, we are discussing if A is
curling around some central point.

A is not curling
around a central
point so
curl(A) is equal to
zero.
A is curling
around a
central point
so curl(A) is
equal to some
value
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What about A ·  and A x ?
These two products do not describe the
geometrical properties
 A ·  is not equal to  · A due to the
nature of the differential operator
 -(A · )U would be equivalent to A ·F,
where F is a force described by -U
 Likewise for -(A x )U

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Two Special Integrals


Integrating over a closed loop:
 
 B  ds
Integrating over a closed surface:
 
B

d
a

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Integrating over a closed loop

The loop can be circular or
rectangular.

ds  rdˆ
 
 B  ds  B(2r )
From 0 to 2
A
D
B
C
Looping from Point A to Point D
using straight line segments

ds  ds A xˆ  dsB yˆ  dsC xˆ  dsD yˆ
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Closed Surface Integral
 
 E  da
The vector n-hat is
normal to the surface.
First

da  da nˆ
This means that “da”
must consist of the
differential distance in
the phi direction
multiplied by the
differential distance in
the theta direction so
Theta is integrated from 0 to
 and phi is integrated from
0 to 2
Therefore if E depends only
on R, then

da  ( R sin df )( Rd f )
 R 2 sin  d df


2
E

d
a

E
4

R


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