Try our Free Online Math Solver!

Calculus Notes
I. Questions
The most commonly asked question in a mathematics course is:
Will this be on the exam?
Well not everything will be on the exam, and so this is certainly a
legitimate question.
On the other hand, there are lots of things that will not be on the
exam that work in your favor if you understand them and can work
with them. The calculus course is based on a few very simple and
powerful ideas.
In order to make use of ideas and make them work for you it is
necessary to think and recall and comprehend.
You need to use curiosity and imagination. You need to be willing to
look for meaning and logic . You need to appreciate structure.
You need to be willing to calculate and work and ask questions.
The question with which we began this section is not sufficient!
In a very real sense , you need to question everything.
Why does mathematics work the way it does?
What is the nature of number and geometry ? How does mathematics
manage to model complex phenomena in the world? What is the
internal structure of mathematics? How do questions in
mathematics become interesting? What sort of mathematical
problems do you find easy? Which ones are hard? What is
mathematics? Is it calculation, reasoning, problem solving, a
language, an art form, a mental discipline, a form of philosophical
investigation?
What are the largest mathematical ideas that you know?
What is a number?
What is the meaning of continuity?
What is the nature of the discrete and the continuous?
What beliefs are in back of mathematics as we do it?
What beliefs are in back of the physical modeling that is so
tightly related to mathematics?
What is the nature of time and space?
How does mathematics have anything to do with time and space?
We want you to ask questions.
We want you to ask questions whose answers really mean a lot to
you. We want you to ask questions that have nothing to do with the
exams and everything to do with understanding.
What should I do in order to ace this course?
Ask questions.
Exercise curiosity, attention, intelligence and imagination.
Work hard.
Use logic.
Think about the material in the course.
Teach the material in the course!
Teach it to another student.
Teach it to yourself.
Teach it to the professor by asking him really tough questions about
how calculus works and how it is applied!
II. Integration and Differentiation
Calculus is based on a small number of key ideas. The purpose of
this section is to give you a quick introduction to these ideas that
will be useful as a start and as a reference as we progress in the
course.
Three Big Ideas
Calculus is all about three basic mathematical notions.
1. Deriviatives = DIFFERENCES
and
2. Integrals = SUMMATIONS
and
3. LIMITS.
Please come back to this remark after you have read the rest of the
section!
Certainly differences and sums are related to one another.
For example,
(21)+(32) +(43) + (54) + (65) + ... +(99  98) + (10099)
= 100 1 = 99.
A sum of differences can be a bigger difference because all the little
differences cancel each other out.
Taking differences can reveal a rule.
What is the rule for the following sequence?
1,6,11,16,21,26,31,36,41,...
Limits are interesting to think about. For example, you may be able
to see at once that
1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ...
has limiting value equal to 2 as the number of terms in this sum
becomes arbitrarily large. What is the geometric interpretation of
this claim?
Another example:
π^{2}/6 = 1/1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/49 + ...
π^{2}/6 is equal to the sum of the reciprocals of all the squares.
This last example is a deeper result than the first series. Try
checking it on your calculator.
Differentiation
Consider the following problems:
1. You drive a car for one hour and cover a distance of 60 miles.
What is your average velocity?
Answer: 60 miles/hour.
2. In driving a car for one hour you cover 21 miles in the first 1 5
minutes. How fast should you drive in the remaining 45 minutes in
order to cover a total of 60 miles in the hour? What is your average
velocity for the whole hour?
Answer: You need to cover 39 miles in 45 minutes. That is a
velocity of 39 miles/(3/4 hour) = (4/3)39miles/hour
= 5 2
miles/hour.
Your velocity in the first 15 minutes was
21miles/(1/4 hour) = 84 miles/hour. (You were speeding!)
Your average velocity for the hour was
(1 / 4 ) (8 4 mph) + (3 / 4 ) (5 2 mph) = (84 + 156) /4 m p
h
= 240/4 mph = 60 mph.
Notice that this average velocity is the same as the total number of
miles covered divided by the total time. Can you figure out why it
worked out this way? I will leave the question of why it works in
general to class discussion, but the arithmetic of the situation
reveals a pattern:
AvgVelocity = (1/4)(84) + (3/4)(52 )
= ( 1 / 4 ) (2 1 / (1 / 4 )) + (3 / 4 ) (60 21) / (3 / 4 ) )
=21 + (6021)
=6 0 .
Discuss what happened here. Will it work in other problems about
average velocity?
3. In driving a car you start off by accelerating your car.
The distance you travel while you are accelerating is given by
the formula X(t) = t^{2} where t is in minutes and X(t)
is in miles.
You only accelerate in this way for two minutes. How far do you
travel in one minute? If you kept on accelerating, how far would you
go in 10 minutes?
What is your velocity at time t = 1 minute?
Answer: X(1) = 1, so you travel one mile in one minute.
X(10) = 100 miles, so you would go 100 miles in ten minutes.
Now to see approximately how fast your are going at one minute,
lets first see how far you go in a time Δt after 1 minute. We have
X(1) = 1 and
X(1 + Δ t) = (1 + Δ t)^{2} = 1 + 2Δ t + (Δ t)^{2} .
This means that the distance you travel in time Δ t, after already
traveling one mile is
Δ X = X(1 + Δ t)  X(1) = 1 + 2Δ t + (Δ t)^{2} 1 = 2Δ t + (Δ t)^{2}.
So your average velocity in the time interval Δ t is
Δ X/Δt = (2Δ t + (Δ t)^{2})/Δ t = 2 + Δ t miles per minute.
This is 120 + 60Δ t miles per hour.
Your velocity is varying as a function of time, but if we make the
time interval Δt very short, we can estimate your "instantaneous
velocity" at time t = 1 as the limiting value of 120 + 60Δ t as
Δt goes to zero. This limiting value is clearly 120. So we can say
that
you are going at 120 miles per hour after one minute.
4. You would be crazy to keep on accelerating at the rate given in
the previous problem . But lets calculate some other limiting
velocities. Given that your distance function is X(t) = t^{2} ,
what is your instantaneous velocity at an arbitrary positive time t?
Answer:
Δ X(t) = X(t + Δ t)  X(t) = (t + Δ t)^{2}  t^{2
}= (t^{2} + 2tΔ t + (Δ t )^{2} )  t^{2}
= 2tΔ t + (Δ t)^{2} .
Thus
Δ X(t)/Δ t = 2t + Δ t .
We conclude from this that the instantaneous velocity of
our car at time t is 2t miles per minute, which is equal
to 120t miles per hour where t is measured in minutes.
At the 30 second point you are going 60 miles per hour.
If you keep accelerating at this rate, you will be going 240 miles per
hour after two minutes and 1200 miles per hour after 10 minutes!
The method by which we have solved these problems is called
differentiation. The general pattern is just the same as in our
example for velocity. We have a function F(t) (it was X(t) = t^{2}
in
our example) and we define the difference quotient by the formula
Δ F(t)/Δ t = (F(t + Δ t)  F(t))/Δ t
The derivative of F(t) at t is defined by taking the limit of
Δ F/Δt
as Δ t goes to zero. We denote this limit by dF(t)/dt
and sometimes by F'(t). Thus we write
d ( t ^{2} )/dt = 2t
via the calculations that we did in the problem solving above .
Prev  Next 