Chapter 3
Kinematics in Two Dimensions
3.1 Displacement, Velocity, and Acceleration
ro initial position
r final position
r r ro displacement
3.1 Displacement, Velocity, and Acceleration
Average velocity is the
displacement divided by
the elapsed time.
Chapter 2
Kinematics in One Dimension
Kinematics deals with the concepts that
are needed to describe motion.
Dynamics deals with the effect that forces
have on motion.
Together, kinematics and dynamics form
the branch of physics known as Mechanics.
2.1 Di
Chapter 1
Introduction and
Mathematical Concepts
1.1 The Nature of Physics
Physics has developed out of the efforts
of men and women to explain our physical
environment.
Physics encompasses a remarkable
variety of phenomena:
planetary orbits
radio and TV
Conceptual Physics
Lab #5: Hookes Law
Procedure:
For the first round of data collected, use the long spring.
First, preform a stretch test on the to determine how far the
springs full-elongation capacity, or its capacity to recover to
its original shape.
Conceptual Physics
Lab #7: Specific Heat
Data
Object Description
Mass of water in calorimeter,
25 mL = 25 g
Mass of metal object
Starting temperature of water
(room temperature)
Starting temperature of
object = Highest final
temperature of water & object
How do buffers work? Buffers work by reacting with any added acid or base to control the pH. a buffer
works by replacing a strong acid or base with a weak one.
(http:/www.sparknotes.com/chemistry/acidsbases/buffers/section1/page/2/ )
Buffers act as Shock
Mastering Physics
The creators describe it as:
"A groundbreaking, research-proven online tutorial and physics homework assignment system, provides a
variety of tutorial and problem types, with each problem type offering a different level of individualized
Calc III
Test 2
Name _
Ch. 12
SHOW YOUR WORK TO GET FULL CREDIT.
#1 Find the equation in parametric form of the line of intersection of two planes
x + 2= 3 and x 4 y + 3 z 5
y+z
=
#2. Sketch several level curves of the function f ( x, y=
)
#3.a) Compute t
Thomas Lands
Chapter 6.3 Problem #2
Now we need to simplify this problem so that we can integrate it. We would like to get it to the
following:
The best way to do this is to set the following:
cos x dx = dt
This will give us the result of:
Now that we kno
Question: Find the Maclaurin series and its integral of convergence.
()
()
First we have our basic equation for the taylor series
()
()
()
(
()
()
()
)
()
(
(
)
)
In a Maclaurin series the center is always at 0 so a=0
The series expansion will look like:
The downward velocity of a sky diver of mass m is ( )
(
). Show that
()
Answer:
We first find series expansion of
()
Let
()
()
()
()
()
()
()
()
()
()
So the expansion will be
()
()
()
()
()
(
()
)
()
So we have,
()
(
(
(
)
(
)
)
)
(
)
Mariah Einspahr
Eng. 090
Colleen weeks
02/21/12
The visit
I have never been so nervous in my life, I told my stepmom and daddy as we were
driving to the jail in Tallahassee Florida. Not only could my hands not stop sweating or
shaking, but my mind was lik
#4
This is a geometric series with first term 4 and common ratio so it is a convergent series.
So sum of the series
#16
So the series is not convergent.
#40
Write 0.18181818 as a geometric series and then write the sum of the geometric series as a
fractio
Research by Meadows, Meadows, Randers and Behrens
indicates that the earth has 3.2x10^9 acres of arable land
available. The world population of 1950 required 10^9 acres
to sustain it, and the population of 1980 required 2x10^9
acres. If the required acrea
Propositional Logic Symbols for Phi 113 (Logic)
Traditional
Operator
Our Symbol Name
~
*~
.
* .
v
v
horsehoe
Logical
Function
Used to Translate
tilde
>
* =
not, it is not the case that
dot
conjunction
and, also, moreover
wedge
or
negation
disjunction
or,
Section 8.8 Problem 29 Page 614.
Apply the Binomial Theorem to
binomial expansion for any positive integer n.
. Determine the number of nonzero terms in the
Which this comes out to:
This becomes:
Factor back in the
and we get:
Which Simplifies to:
This ch
section 8.1 problem #4
=
*
To find the first six terms, we simply needs to plug in n=1 6
=
*
=
=
*
=
*
=
=
*
=
=
*
=
=
*
=
=
Section 8.1 problem #18
Notice that
, so we can re-write the term as:
The sequence is unbounded, so it diverges.
Section 8.1 Probl
Section 8.9 Problem 12 Page 626
Find the Fourier series of the function on the given interval.
If
Since
, then
is odd we get:
This comes out to equal:
This comes out to:
Hence for -2<x<2 we get:
Thomas Lands
Problem 2 Section 8.5:
This comes out to:
By looking at this we can tell that:
Absolutely converges by the radio test.
CHECK
Problem 16 Section 8.5:
We know that
CHECK
Rackley, Adrienne
MAT202 C12 Calculus II, Louis Sass
April 8, 2013
Unit 5 Problem Solver, Section 8.6, Exercises 6 and 8, Pg. 593
Section 8.6, Exercises 6 and 8 ask you to determine the radius and interval of convergence.
6)
(
)
(
)
From the Ratio Test, w
T,[ ft
rck
eL = +z , ( )=?-
futE
d
LL&^
CLv-=
z
= t cfw_Xt
L^\Z
-f^.&) =
E
?
Y
<
4 xt n
, = U,-,
o w*d
4
= Ae
X
a:rh;Hs K = o
v-' +-
a
4*a = A e - z /.
cfw_=Ae'
tD cfw_= cfw_;-
6d",4
iT 4Fh>AF45/^s F*<=
fr:A^.rc.=E
|3aaq
,
>.b-z-Au- ealcr,r$- Y '= + L
4x
Exercise 8.9 #14
Find the Fourier Series of the function on the given interval:
()
cfw_
therefore
1. Find ao:
()
(
( ) (
( )
)
[
)
Find ao using Euler-Fourier formula. Separate integral for
piecewise function. Substitute
]
Evaluate integral to find ao =
2
Rackley, Adrienne
MAT202 C12 Calculus II, Louis Sass
February 24, 2013
Unit 2 Problem Solver, Section 6.1, Exercise 18, Pg. 451
Section 6.1, Exercise 18, Pg. 451 asks you to evaluate the integral
Rewrite the integrand to eliminate the fraction
Let
Substit
Rackley, Adrienne
MAT202 C12 Calculus II, Louis Sass
March 8, 2013
Unit 3 Problem Solver, Section 7.1, Exercise 30, Pg. 507
Section 7.1, Exercise 30, Pg. 507 asks you to find the value of a $1,000 investment at an annual interest
rate of 8% after 5 years
Problem 35, sec 8.7, pg. 605
2k
35. Use a Taylor series to verify the given formula = e 2
k = 0 k!
From the table on page 603 we find:
xk
x
k! = e
k =0
So, set x=2
xk
2k
k! = k! = e x = e 2
k