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PHY2061 Enriched Physics 2 Lecture Notes
Refresher
Math and Physics Refresher
This course assumes that you have studied Newtonian mechanics in a previous
calculus-based physics course (i.e. PHY2060) and at least have co-registered in a vector
calculus course (Calc 3). Listed below are some of the concepts in basic math, calculus,
and physics that you are expected to know or to acquire during this course. This is not a
complete summary of introductory math and physics. It is only meant to be a refresher of
some of the concepts used in this course. Please report any inaccuracies to the professor.
Geometry
a
c
b
1. Pythagorean Theorem:
The square of the hypotenuse of a right triangle is the sum of the squares of the
two legs: c2  a2  b2
2. Circumference of a circle:
3. Volume of a sphere:
4. Surface area of sphere:
C  2R
4
V  R3
3
S  4R2
Calculus
1. Differentiation:
You are expected to be able to take simple derivatives:
d n
x  nx n 1
dx
d
sin x  cos x
dx
d
cos x   sin x
dx
d x
e  ex
dx
2. Product Rule
d
dg
df
f ( x )g( x )  f ( x )  g( x )
dx
dx
dx
d
Example:
x sin x  x cos x  sin x
dx
a f
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3. Chain Rule
bafg
d
df dg
f gx 
dx
dg dx
Examples:
d
sin 2 x  2 cos 2 x
dx
d
exp  x 2  2 x exp  x 2
dx
c h
c h
4. Integration
You are expected to be able to perform simple integrals:
1 n 1
x n dx 
x
n 1
z
z
z
z
dx cos x  sin x
dx sin x   cos x
dx e x  e x
A purist would note that a constant should be added to these indefinite integrals.
5. Change of Variables
To use integration tables correctly, you must be able to change variables. For
example:
L
nx
I  sin 2
0
L
nx
L
Let u 
 dx 
du
L
n
L n
I
du sin 2 u
0
n
x 1
Then use dx sin 2 x   sin 2 x
2 4
L n L
I

n 2
2
z
F I
HK
z
z
6. Integration By Parts
z
z
udv  uv  vdu
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Approximations
For small x, the following expansions are useful:
1.
1
 1  x  x 2 
1 x
a f  1  nx 
2. Binomial Expansion: 1  x
3. Taylor Expansion:
n
af af
f x  f 0 x
df
dx

x 0
x2 d2 f
2 ! dx 2

x 0
Differential Equations
We will study the solutions to several differential equations when we study circuits in this
class. Although you are not required to have taken a course in differential equations, we
will learn how to solve the simplest ones:
1. The exponential function is the only function whose derivative is the function
itself:
df
  f ( x)
dx
f ( x )  Ce x is the general solution, where C and  are constants
2. Two derivatives of the trigonometric functions give you back the same function
with a sign change:
d2 f
 k 2 f ( x)
dx 2
f ( x )  A sin kx  B cos kx is the general solution, where A, B, and k are constants
Using complex exponentials (see next section), you can also represent this
solution as:
f x  A  e i k x  B e  i k x
af
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Complex Numbers
The analysis of circuits in this class can be greatly simplified by the usage of complex
numbers.
Complex (or imaginary) numbers are based on i  1 . A complex number may be
represented by z  x  iy , where x and y are real numbers. It can be represented by a
point on a two-dimensional plane:
iy
r
z = x + iy

x
a
An alternative way to represent a complex number is z  rei   r cos  i sin 
f
To find the magnitude of a complex number (the length r), multiply the number by its
complex conjugate, then take the square root:
| z|  z * z  re  i re i  r
or,
| z|  z * z 
ax  iyfax  iyf 
x 2  y2
Note that to take the complex conjugate, replace i with –i
It is possible to represent the sine and cosine functions by complex exponentials:
e i  e  i
2
i
e  e  i
sin  
2i
cos  
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Vectors (see also Appendix H in HRK, vol.2)

Dot product: projection of one vector along another
a  b  b  a  axbx  a y by  az bz
a  b  ab cos 

Cross product: product of vectors, with direction given by right-hand rule
i
j k
a  b  b  a  det ax a y az   a y bz  by az  i   axbz  bx az  j   axby  bx a y  k
bx
by
bz
a  b  ab sin 
Vector Calculus
The gradient is the rate of change of a function along each direction:



  xˆ  yˆ  zˆ
x
y
z
F  grad  f   f
It yields a vector function when applied to a scalar function.
The Laplacian operator is:
2
2
2
2
     2  2  2
x y z
When applied to a scalar function, it yields a scalar result.
The line integral is the integral of a vector function projected
along a one-dimensional path:
b
dr
F

d
r

C
a F r  t   dt dt
The curve C is the path of function r(t), where t is a variable
that parameterizes the path.
P2
The line integral is the inverse of the gradient:
P2
f ( P2 )  f ( P1 )   f  dr
P1
This means that for conservative forces F  f , it doesn’t matter what path one takes to
go from P1 to P2, the line integral is just the function evaluated at the endpoints. Also, if
the path is a closed loop, the line integral equals zero.
P1
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The surface integral is the integral of a vector function projected onto a two-dimensional
surface:
 dr dr 
S F  ds  S F r  u, v    du  dv  du dv
The surface S is parameterized by two variables, u and v.
The divergence of a vector function:
F F F
  F  div  F   x  y  z
x
y
z
The divergence is the flux out of a volume V, per unit volume, as V  0 :
1
  F  lim  F  dA
S
V 0 V
Divergence Theorem:
  F dV   F  dΑ
V
S
The curl of a vector function:
 F F   F F   F F 
 F  curl  F    z  y  i   z  x  j   y  x  k
z   x z   x y 
 y
The curl (when projected along the normal to a loop) is the circulation of a vector
function around a closed loop of area A, per unit area, as A  0 :
1
(  F)  nˆ  lim  F  ds
A 0 A C
Stokes’ Theorem:
  F   dA   F  ds
S
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Physics – Mechanics
Newton’s Laws:
1. An object maintains constant velocity unless acted upon by an external force
2. The acceleration of an object is proportional to the applied external force divided
by the mass of the object (the inertia)
F  ma
This is a vector equation. It can also be written as:
dp
F
where p  mv is the momentum
dt
dv d 2 x
This is your first differential equation! Note that a 
also.

dt dt 2
3. Force exerted by body 1 on body 2 is equal and opposite to the force that body 2
exerts on body 1. In other words, the force of gravity acting on you is balanced
by an opposite force applied by the floor so that you do not fall to the center of the
Earth!
If the force exerted on a body is constant, then so is the acceleration. In that case
vat
x  vt 
1 2
at
2
Some examples of forces:
Gravitational force near the surface of the Earth: F  mg where g  9.8 m / s 2
mm
where G  6.6726  10 11 N - m 2 / kg 2
Newton’s law of Gravity: F  G 1 2 2
r
Hooke’s Law: F  kx where k  spring constant, x  displacement
M2
r
Conservation of Momentum:
Momentum is defined by p  mv . It is a vector quantity. The vector sum of the
momentum of all particles before an interaction is the same as it is afterwards. In
other words, it is conserved. For two masses which collide (but do not stick):
final
m1v1  m2 v2  m1v1  m2 v2 
initial
This is an elastic collision. If they do stick, it is inelastic. In that case:
m1v1  m2 v2  m1  m2 v
b
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PHY2061 Enriched Physics 2 Lecture Notes
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Work and Energy:
Applying a force over a distance requires work:
W  F d
W12  
2
1
F  ds
The work done is opposite to the change in the potential energy:
W  U
The force is derived from the rate of change of the potential energy:
F  U
Conservation of Energy:
In Newtonian physics, we learn that the kinetic energy is conserved for elastic
collisions (but not inelastic collisions, where some of the energy goes into mass!) The
kinetic energy is given by
1
K  mv 2
2
For the elastic collision of two objects, conservation of energy implies:
2
2
1
1
1
1
m1v12  m2 v22  m1v1  m2 v2 
2
2
2
2
Special Relativity
Separate lecture notes will accompany the class for this material. With regards to the
above, in Special Relativity, energy of any form is always conserved.
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Circular motion:
In uniform circular motion, the magnitude of the velocity of an object is constant,
though its components are not. The magnitude of the centripetal acceleration to
achieve uniform circular motion is
ac 
v2
r
The centripetal force responsible for this acceleration is just the mass times this
quantity.
V
d
ac
r
dv dt
v dt
v dt dv dt dv
d 


r
v dt
v

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dv
v2
a
dt
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