class notes i

Calculus Class Notes for the Combined Calculus andPhysics Course

Semester I

Kelly Black

December 14, 2001

Support provided by the National Science Foundation - NSF-DUE-9752485

1

Section 0 2

Contents

1 Average Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 The Derivaitve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 Derivatives of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Anti-Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 Shifted Functions and Their Derivatives . . . . . . . . . . . . . . . . . . 277 Slope and Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . 379 Composition of Functions and the Chain Rule . . . . . . . . . . . . . . . 4210 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4611 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5212 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5813 Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6814 The Area Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7715 The Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8216 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . 8817 Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9518 Work Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10219 Unconstrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . 11020 Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . 11421 The Impulse-Momentum Theorem . . . . . . . . . . . . . . . . . . . . . . 12022 The Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12423 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . 13124 Working with Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13525 Inverse Trigonometric Functions and Their Derivatives . . . . . . . . . . 140

In Search of Newton Calculus and Physics

Section 1 3

1 Average Rate of Change

The definition of a function is given and explored on this day. The definition ofaverage velocity is also given, and the idea of the derivative is briefly explored. Almostall of the examples that are explored in the first third of this semester are functions givenin terms of time, t, not x. This is important since almost everything in the physics classis in terms of time.

In the first mini-lecture the definition of a function is given as well as the definitionfor the average velocity. In the second mini-lecture the idea of decreasing the change inthe domain is briefly explored leading up to the idea of the slope of the tangent line at apoint on a function. Finally, in the third mini-lecture an overview of the average velocityis given.

This is a relatively light day, and the activities are relatively straight-forward. As thecourse progresses the activities will get much more difficult, but in these first couple ofdays the activities are not too difficult so that students who are not familiar with thisapproach are not simply thrown to the wolves.

1.1 Mini-Lecture I The definition of function is given as well as the definition forthe average velocity. Note the the definition for the domain and the range is given in thethird mini-lecture.

Begin Class Notes

Definition: The word function is a generic term. A func-

tion is a method or a rule. A function defines the way to

provide one particular item given some initial item. Most of

the functions that we will deal with in this course will provide

a number given another number. For example, the following

method defines a function:

Given a number between 0 and 1, multiply the number

by 3 and add one to the result.

Mathematicians are big into compression and prefer simpler

representations. The definiton above could also be written as

f (t) = 3t + 1, for 0 < t < 1.

For example, given the number 12, the function returns52. Note


Section 1 4

that we gave this function a name, f (t). This is so that we can

refer to it later.

Definition: The average velocity is

change in distance

change in time.

Example: It takes 30 mintes to go 5 miles. What is the

average velocity?

5 miles12 hours

= 10miles

hour(Note the units.)

If x(t) is the position (in metres) at a given time, t (in hours),then we can also take a graphical view of the average rate ofchange.

The average velocity is the slope of the line through two points.End Class Notes


Section 1 5

1.2 Mini-Lecture II The average velocity is examined again, and the basic idea ofthe derivative is given.

Begin Class Notes

The average velocity is also called the average rate of change,

4x4t .

(The 4 is shorthand for change.)

We can also think of this as the slope of a secant line. (A secant

line is a line that goes through two specific points on a curve.)

The question that is being asked is what happens to 4x4t as 4tgets closer and closer to zero. Note that both4x and4t bothget closer and closer to zero, but their ratio does something

else.

Note: We have already seen two different ways to express a

function. The first is with a formula, and the second is using

a graph. There are many other ways. One way that we will


Section 1 6

make use of is a table.

t (sec.) x (m.)

0.0 0.0

0.5 2.0

1.0 3.0

1.5 2.0

Given a specific time, 0, 0.5, 1.0, or 1.5, the distance can be foundfrom the cooresponding value in the right column. Note that whenyou graph this function you do not connect the dots. The onlyinformation that we have is for specific points. If you connect thedots you are implying information that is not given.

End Class Notes

1.3 Mini-Lecture III The final mini-lecture provides an overview of function andaverage rate of change as well as some more formal definitions.

Begin Class Notes

Given the average velocity we can get the total change in dis-

ance.

Example: The average speed for a 3 hour trip was 45 miles

per hour.

Total distance = 45mph 3hr.= 135 miles.

We need some definitions and will refer to some things through-out the year. We assume that x(t) is the position given a time, t:


Section 2 7

Domain = Valid times that we can put into the function.

Range = Valid distances that can be returned by the function.

Total Change = Difference in a function over a given interval.

Average rate = Change in functions rangeChange in the domain .

Tangent Line = The line that just touches a curve at one point. (Locally)

End Class Notes


Section 2 8

2 Sequences

The definition of the derivative will be the limit definition. In order to motivatelimits we will first examine sequences. This day will begin by defining and working withsequences of numbers and will end with the formal definition of the limit.

The first mini-lecture will focus on sequences and will provide a small amount ofmotivation as to why we are looking at sequences. The second mini-lecture begins toexamine the limit, and the final mini-lecture relates this work back to the derivative.

2.1 Mini-Lecture I This is a relatively straight-forward introduction to sequences.Begin Class Notes

We will be looking at the average rate of change.

If we are finding the slopes of these lines we are really generating

a sequence of numbers:.

.333, .400, .444, .472, .485, . . .

The question that we will ask today is what does it mean to

examine a sequence of these slopes? How do we analyze the

sequence?

Definition: A sequence of numbes is an ordered list of num-

bers. A moreformaldefinitioncan be foundin the bookthatspecificallydefines thedomain asthe naturalnumbers, butthat isomitted herein order tokeep thingsmorestraight-forward forthe students.


Section 2 9

Example: nn2 1

n=2=

2

3,

3

8,

4

15,

5

24,

6

35, . . .

These numbers can be graphed:

Note that they can be easily graphed in matlab:

>> n = 2:1:10;

>> f = n./(n.^2-1);

>> plot(n,f,*)

End Class Notes

2.2 Mini-Lecture II Examples with the definition of the limit are given here.Begin Class Notes


Section 2 10

Given any value of > 0, if there is an N such that every

number in the sequence falls within the interval awe say that

the sequence converges.

Example:11

2

n

n=0= 0, ,

1

2,

3

4,

7

8,

15

16, . . .

This sequence converges to 1. Even though none of the num-

bers actually is 1, the numbers in the sequence get arbitrarily

close to 1.

Since none of the numbers in the sequence are 1 we have to

show that they get close. We ask if the numbers are close

to one 1

12

n 1 < .

We can simplify the left hand side(1 (12)n) 1 = 1 (12

)n 1 = (12)n = (12

)n.

Can I get this number to be smaller that ? If so, is it smaller

than for all numbers bigger than some value?

Given can I find a value of N such that this value is less than

for all values of n > N?12

n < ln

12

n < ln()In Search of Newton Calculus and Physics

Section 2 11

n ln

12

< ln()n >

ln()

ln(

12

).

So for any n > ln()ln(12)

we have that

1

12

n 1 < .

End Class Notes

2.3 Mini-Lecture III Limits are examined again, only this time it is in the contextof the derivative.

Begin Class Notes

Suppose that we have a function, f (t) = t2, and we want to

examine what happens to the average rate of change as the

change in time gets closer and closer to zero. The average rate

of change of f (t) is

f (t +4t) f (t)4t .

Does this look like 2t as4t gets close to zero? Let4t =(

12

)n.

From the activity we have that(t +

(12

)n) t2(12

)n 2t < ,

for every n > ln()ln(12)

. This means that as n gets large, the

average rate of change gets closer to 2t.

Note that the both the numerator and the denominator of the

average rate of change goes to zero, but the ratio does not.


Section 2 12

Given a parabola, what is the slope of the tangent line at any

time, t?

As 4t gets close to zero, the slope of the secant lines approach2t. For example at t = 5 the slope of the tangent line is 10.

End Class Notes


Section 3 13

3 The Derivaitve

The limit definition of the derivative is motivated but is not given on this day. Theaverage rate of change is examined at the beginning of the day using some numericalexamples. During the second mini-lecture the numerical values are related to the slopeof the tangent line. The principle aim of this approach is to demonstrate that the ideasbehind the limit are about the process of finding a limit. During the third mini-lecturea definition of the derivative is given and some examples are also given.

3.1 Mini-Lecture I The average rate of change is examined once again. The ideathat this is the slope of a secant line is emphasized.

Begin Class Notes

The average velocity is the change in the distance divided by

the change in the time. We have the following observations:

Tells us how distance changes as time changes. For us it has been discrete in nature. (This isnt really the

case in general.)

We would like to quantify what happens during one point intime.


Section 3 14

The average rate of change is the change in the range divided by

the change in the domain. This is the same as the slope of the

secant line.

To get at the rate of change at a point in time, we can look at

what happens as the change in the domain gets closer and closer

to zero.

At this point I like to do the pre-class work on an overheademphasizing the difference between total change and averagerate of change.

End Class Notes

3.2 Mini-Lecture II The average rate of change for f(t) = t2 is derived, and theidea of the tangent line is introduced.

Begin Class Notes

The average rate of change for f (t) = t2 is

(t + h)2 t2h

=t2 + 2th + h2 t2

h

=2th + h2

h= 2t + h.

What happens as h gets closer and closer to zero?


Section 3 15

Graphically, we are finding secant lines.

As h gets closer and closer to zero the secant lines get closer

and closer to the tangent line. The tangent line at a point,

(f (t0), t0), is the straight line that just touches the curve at

the point but does not cross through the curve. Somestudents getconfusedbetween thedifferencebetween thetangent lineand thederivative.This is thereason that Itry toemphasizethedifferencebetween thetangent lineand theslopes of thesecant lines.

The secant lines above get closer and closer to the tangent line,and the average rates of changes get closer and closer to the slopeof the tangent line.

End Class Notes

3.3 Mini-Lecture III A definition of the derivative is given and several examplesare given.

Begin Class Notes

If f (t) = t2 then

f (t + h) f (t)h

= 2t + h.

The slope of the tangent line at any point is 2t.


Section 4 16

If g(t) = t3 then

g(t + h) g(t)h

= 3t2 + 3th + h2,

The slope of the tangent line at any point is 3t2.

In general, the slope of the tangent line to the function tn is

ntn1.

Definition: The derivative of a function at t is the slope of

the tangent line at t.

Notation: The derivative of a function, x(t) is denoted

x(t),

x(t),dx

dt(t).

it is important to note that this is just notation. For example,

the notation dxdt is not dx divided by dt. This is just a way to

indicate that we are finding the slope of the derivative. Also,

the notation ddt (g(t)) indicates that you are being asked to

find the derivative of the function g(t).

Examples:What is ddt

(t4)? What is ddt

(t7)?

What is ddt(t10

)? What is ddt

(t15

)?

End Class Notes


Section 4 17

4 Derivatives of Polynomials

The derivative of a polynomial is found on this day. During the first mini-lecture thegeneral notion of the derivative is reinforced. During the second mini-lecture the ideathat the derivative of the sum of two functions is the sum of the derivatives of the twofunctions is examined. Finally, in the final mini-lecture the general form of the derivativeof a polynomial is given.

4.1 Mini-Lecture I The graphical nature of the derivative is re-examined. Anexample of a function that is not differentiable is also given.

Begin Class Notes

The average rate of change of f (t) from t to t + h is

f (t + h) f (t)h

.

The average rate of change is the change in the range divided

by the change in the domain. This is the same as the slope of

the secant line. Moreover, as h gets close to zero, the secant

line approaches the tangent line.


Section 4 18

Does a tangent line always exist?

Does a unique tangent line exist at t = 0?

In the next activity we will look at what happens when you try

to find the derivative of the sum of two functions, f (t) + g(t).

We will start out by looking at a specific example. Recall that

the averarge rate of change for a function is

f (t + h) f (t)h

.

Quick Note: The factorial notation is defined to be

n! = n(n 1)(n 2)(n 3) 3 2 1.For example 5! = 5 4 3 2 1 = 120.

End Class Notes

4.2 Mini-Lecture II The idea that the derivative of the sum of two functions isthe sum of the derivative of both functions.

Begin Class Notes

What is the derivative of t2 + t?


Section 4 19

If I have f (t) + g(t) what is ddt (f (t) + g(t))?

d

dt(f (t) + g(t)) = f (t) + g(t).

Note that this implies that

d

dt(f1(t) + f2(t) + f3(t) + fn(t)) = f 1(t) + f 2(t) + f 3(t) + f n(t).

Also note that when looking at the average rate of change for a

function multiplied by a constant, cf (t), we get that

cf (t + h) cf (t)h

= cf (t + h) f (t)

h.

This implies that the derivative of cf (t) is cf (t).

Example: ddt(4t2

)= 8t.

Example: ddt(t + 4t5

)= 1 + 20t4.

Example: ddt(1 + 8t9

)= 72t8.End Class Notes

4.3 Mini-Lecture III Derivatives of non-trivial polynomials are examined.Begin Class Notes

In the activity we looked at the polynomial

Q(t) = a0 + a1t + a2t2 + a3t

3 + + antn.Note that Q(0) = a0.


Section 4 20

What happens to the derivatives?

Q(t) = a1 + 2a2t + 3a3t2 + 4a4t3 + 5a5t4 + + nantn1.So Q(0) = a1.

Q(t) = 2a2 + 3 2a3t + 4 3a4t2 + 5 4a5t3 + + n (n 1)antn2Q(0) = 2a2.

Q(t) = 3 2a3 + 4 3 2a4t + 5 4 3a5t2 + +n (n 1) (n 2) antn3,

Q(0) = 3 2a2.

QIV (t) = 4 3 2a4 + 5 4 3 2a5t + +n (n 1) (n 2) (n 3) antn4,

QIV (0) = 4 3 2a4.If I know that

Q(0) = 1,

Q(0) = 4,Q(0) = 3,Q(0) = 12,QIV (0) = 6,

can I find a formula for Q(t)?

If there is time I like to ask the students to graph out and

play with the function

1 + t +t2

2+t3

3!+t4

4!+ + t

9

9!.


Section 4 21

It is interesting to note that for t close to zero the derivativeof this function looks a lot like the original function.

End Class Notes


Section 5 22

5 Anti-Derivatives

The anti-derivative is introduced on this day. It is important to introduce the anti-derivative as soon as possible because the physics class is making use of the kinematicequations.

The first mini-lecture introduces the idea of going backwards given that you know thederivative of a function but not the function itself. This idea is addressed again in thesecond mini-lecture and the definition of the anti-derivative is given. An example of thederivation of the kinematics equations is given during the second mini-lecture. Finally,an introduction to differential equations is given in the third mini-lecture.

it is unusual to discuss differential equations so soon in a calculus class, but this is oneof the continuing themes that brings the physics and the calculus class together. We willemphasize the role that Newtons second law has on developing mathematical models forphysical phenomenon. Newtons second law is used to develop the differential equationsthat describe how things move.

5.1 Mini-Lecture I A brief introduction to the anti-derivative is given. The basicrule is not given, rather this is just a brief way to introduce the idea that it is possible togo backwards. The idea is re-examined in the second mini-lecture so this is just a briefoverview of the basic idea.

Begin Class Notes

What are the following derivatives?

d

dt

(t6 + t3 t

)= ?

d

dt

(t

12 t2 + 1

)= ?

d

dt

(t

23 4t + 7

)= ?

d

dt

(t

43 t + 323

)= ?

Note that the addition of a constant does not change the slope

of the graph, it only shifts the graph up or down.

Suppose that I know that

d

dt(f (t)) = 1.


Section 5 23

What is f (t)? Notice that

d

dt(t + c) = 1,

where c is any constant.

Suppose that I know that

d

dt(f (t)) = t + 1,

what is f (t)? Notice that

d

dt

12t2 + t + c

= t + 1,where c is any constant.

This idea is explored in the activity, and we will discuss how tofind f (t) given its derivative in the second mini-lecture.

End Class Notes

5.2 Mini-Lecture II The anti-derivative is defined and further explored at thistime. An example is given in which the kinematic equations for constant acceleration arederived.

Begin Class Notes

If you know that ddt (f (t)) is a given function the process of

funding f (t) itself is called finding the anti-derivative.

For example, if

d

dt(f (t)) = t3,

then

f (t) =4

t

4

+ c


Section 5 24

where c is a constant.

Ifd

dt(f (t)) = t6,

then

f (t) =7

t7+ c,


In general if

d

dt(f (t)) = tn,

then

f (t) =1

n + 1tn+1 + c,


Example: Assume that an object has a constant downward

acceleration. Can I find its velocity and position if its initial

velocity is v0 and its initial position is y0?

From Netwons second law we get that

ma = mg,a = g.


Section 5 25

Finding the anti-derivative we have that

v(t) = gt + c1, butv(0) = v0 = g(0) + c1 c1 = v0,v(t) = gt + v0.

Finding the anti-derivative again we have that

y(t) = 12gt2 + v0t + c2 buty(0) = y0 = 12g(0)2 + v0(0) + c2 c2 = y0,y(t) = 12gt2 + v0t + y0.

These formulas are the kinematic equations for constant accel-eration.

End Class Notes

5.3 Mini-Lecture III An overview of Newtons second law is given. The connectionbetween the second law and differential equations is discussed as well.

Begin Class Notes

Newtons Second Law: The mass times the acceleration of

an object is equal to the sum of all of the forces (Note that the

acceleration and the forces are all vectors.) This is expressed

symbolically as

m~a =i

~Fi.

The thing to notice is that the derivative of the velocity is the

acceleration.

Example: A car of mass m kg has a force of t2 t N actingon it. Find the position of the car if its initial velocity is 2

metres per second, and its initial position is 0 m.

max = t2 t,

d

dtv(t) =

1

m

(t2 t

)


Section 6 26

v(t) =1

m

13t3 1

2t2 + c1

v(0) =1

m(0 0) + c1

2 = c1

v(t) =1

m

13t3 1

2t2 + 2.

The position is found in the same manner

d

dtx(t) =

1

m

13t3 1

2t2 + 2

x(t) =1

m

112t4 1

6t3 + 2t + c2

x(0) =1

m(0 0) + 0 + c2

0 = c2

x(t) =1

m

112t4 1

6t3 + 2t

When you have a relation that includes the derivative of a func-

tion the relationship is called a differential equation. The

differential equation that we started out with above was

max = t2 t,

d

dtv(t) =

1

m

(t2 t

).

Newtons second law related the derivative of the velocity with theforces acting on the system. Newtons second law is a differentialequation.

End Class Notes


Section 6 27

6 Shifted Functions and Their Derivatives

The derivative of functions that are shifted to the left or right in their domain isexplored. The main goal for this day is to find the derivatives and anti-derivatives ofP (t a) where P (t) is a polynomial.

The first mini-lecture is a very simply overview of what happens to a function whenthere is a shift in the domain. This is mainly a graphical interpretation. The secondmini-lecture covers the derivative of a polynomial that has been shifted while the thirdmini-lecture focuses on the anti-derivative of a shifted polynomial.

6.1 Mini-Lecture I A graphical overview of what it means to shift a functionis given. This is a very short and very straightforward discussion. The students areexpected to graph the tangent lines to the shifted graphs in the activity and deduce thatddtf(t a) = f (t a).

Begin Class Notes

Given a function, f (t), how do you find f (t a)?

The graph of f (t a) is just like the graph of f (t) only it isshifted to the right a units.

End Class Notes

6.2 Mini-Lecture II This is another unusually short mini-lecture. The basic ideais given and several examples are stated.

Begin Class Notes

Given f (t) the graph of f (t a) is just shifted a units to theright.


Section 6 28

Given f (t) how can we find ddt (f (t))? It is just f evaluated at

t a.

Examples: Find the following derivatives:

1. ddt

(t 3

).

2. ddt

((t + 4)75

).

3. ddt

((t 14)2

).

End Class Notes

6.3 Mini-Lecture III This is another straight-forward mini-lecture. The anti-derivative for general polynomials is given.

Begin Class Notes

Note thatd

dt

(3 + (t 1)2 + (t 1)15 + 6t4

)= 2(t 1) + 5(t 1)4 + 24t3.

Example from the activity: v(t) = (t 1)2 + t, x(0) = 3x(t) =

1

3(t 1)3 + 1

2t2 + c.

At t = 0, x(0) = 3.

x(0) =1

3(1)3 + 0 + c,

= 13

+ c.


Section 6 29

To satisfy the boundary conditions

3 = 13

+ c,

c = 3 +1

3,

=10

3.

x(t) =1

3(t 1)3 + 1

2t2 +

10

3.

Example from the activity: v(t) =t + 4, x(0) = 1.

x(t) =(t + 4)3/2

3/2+ c.

At t = 0, x(0) = 1.

x(0) =43/2

3/2+ c,

=8

3/2+ c,

=16

3+ c.

To satisfy the boundary conditions

1 =16

3+ c,

c = 1 163,

= 133.


Section 6 30

x(t) =(t + 4)3/2

3/2 13

3.

In General: If

d

dt(f (t)) = c0 + c1(t a) + c2(t a)2 + c3(t a)3 + . . . + cn(t a)n,

where cn and a are all constants, then

f (t) = k + c0(t a) + c1(t a)2

2+ c2

(t a)33

+ c3(t a)3

4+ . . . +

cn(t a)n+1n + 1

,

where k is a constant.

Example: What is the antiderivative of

(t 1)4 + (t 7)18 6t 3

2

43?

End Class Notes


Section 7 31

7 Slope and Concavity

The derivative is examined on this day as well as concavity. The discussions in classfocus on the meaning that can be attributed to the values of the derivatives. In particularthe shape of a curve given knowledge about the derivatives of the function is one of theimportant aspect of this days activities.

During the first mini-lecture the idea that a positive derivative means that the functionincreases is discussed as well as how the concavity affects the shape of the graph of afunction. The second mini-lecture focuses on how derivative information can be used todetermine the shape of the original function. The third mini-lecture a broad overview ofthe days activities is given.

7.1 Mini-Lecture I The relationship between the derivative and whether or not afunction is increasing or decreasing is given. These ramificiations are examined in termsof the concavity.

Begin Class Notes

What is happening to a function if the derivative is positive?

What is happening to a function if the derivative is negative?

Note that if f (t) goes from positive to negative the functionreaches a local max.

If f (t) goes from negative to positive the function reaches alocal min.


Section 7 32

If f (t) is positive then the function is increasing. If f (t) isdecreasing then the function is decreasing. If f (t) is positivethen f (t) is increasing so...

if the derivative is positive then the function is increasing ata faster rate,

if the derivative is negative then the function is decreasing ata slower rate.

These things imply that the graph of the function is concave up.

If f (t) is negative then the function is concave down.


Section 7 33

End Class Notes

7.2 Mini-Lecture II The idea that the derivaitve gives you a great deal about theoriginal function is explored during this mini-lecture. This is mainly a graphical view ofwhat information the derivaitve gives you.

Begin Class Notes

Given the derivative you know a lot about the original function.

If the derivative is postive the function is increasing. If the

derivative is negative the function is decreasing. If the deriva-

tive is increasing the function is concave up. If the derivative

is decreasing the function is concave down. If the derivative

changes sign you have a local max or local min.


Section 7 34

Note that if there is a cusp in the original function the deriva-tive is discontinuous.

End Class Notes

7.3 Mini-Lecture III An overview of the days activities is given. The main goal is

to simply reinforce some of the ideas of what the derivative is and the difference between

the derivative and concavity.

Begin Class Notes

There is a difference between concavity and the derivative.


Section 7 35

If there is time do the following graphical problem. Plot thevelocity and then discuss how to get the acceleration and theposition. Provide hints such as do the acceleration first sinceit can help with the shape of the position.


Section 7 36

End Class Notes


Section 8 37

8 Product and Quotient Rules

The product and quotient rules are introduced and examined on this day. By the endof the day the students should have some experience using both rules and should haveworked out several examples. The examples include graphical and analytic views of thetwo methods of differentiation.

During the first mini-lecture the product rule is given. The introduction relies on thestudents to finish the pre-class activity. The second mini-lecture provides an example ofthe product rule, and the quotient rule is derived from the product rule. The focus ofthe final mini-lecture is the quotient rule as well as what happens when the denominatorapproaches zero.

8.1 Mini-Lecture I The product rule is introduced. This mini-lecture relies heavilyon the pre-class activity.

Begin Class Notes

Suppose we multiply two functions

h(t) = f (t)g(t).

From the pre-class work we have that

h(t2) h(t1)t2 t1 = g(t2)

f (t2) f (t1)t2 t1 + f (t1)

g(t2) g(t1)t2 t1 .

As t2 gets closer and closer to t1 what happens?

If f (t) and g(t) are continuous and differentiable then

d

dt(f (t)g(t)) = f (t)g(t) + f (t)g(t).

This is called the Product Rule.

Examples:

1. ddt ((t + 37)(t + 47))


Section 8 38

2. ddt((4t + 7t2)(t + 300)17

)3. ddt

(t + 3

(t2 + 1

))

4. ddt

((18t)t + 5

)

End Class Notes

8.2 Mini-Lecture II Examples from the product rule are given, and the quotientrule is derived. Note that the case when the denominator goes to zero is not examinedhere. That case will be examined in the third mini-lecture.

Begin Class Notes

Product Rule: The hard part is recognizing when to use it.

For example:

1. ddt((t2 3t + 1)(t2 + 5t + 1))

2. ddt((t + 4)(t2 + 3t + 2)(t + 1)500

)

Note that the product rule implies that we can do the last ex-

ample out in one step, ddt (f (t)g(t)h(t)) = f(t)g(t)h(t) +

f (t)g(t)h(t) + f (t)g(t)h(t).

Example:

d

dth(t) =

d

dt

f (t)

g(t)

Yikes! We cant do this. We can rearrange things so that we can

use the product rule, though,

h(t)g(t) = f (t),


Section 8 39

d

dt(h(t)g(t)) =

d

dtf (t),

h(t)g(t) + h(t)g(t) = f (t),h(t)g(t) = f (t) h(t)g(t),h(t)g(t) = f (t) f (t)

g(t)g(t),

h(t) =f (t) f(t)g(t)g(t)

g(t),

h(t) =f (t)g(t) f (t)g(t)

g2(t).


d

dt

f (t)g(t)

= f (t)g(t) f (t)g(t)g2(t)

.

This is called the Quotient Rule.

Examples:

1. ddt

(3t2+5t

4+t

).

2. ddt

(t+1t1

).

End Class Notes


Section 8 40

8.3 Mini-Lecture III Several examples are given that require the quotient rule.Begin Class Notes

The hard part about the quotient rule is to know how to recog-

nize it:

1. ddt

(t23t+2t3+6

).

2. ddt

(34tt

).

3. ddt

(1 1t

)= ddt

t1t =

ddt

(t1t

).

Note: What happens when the denominator gets close to zero?

ddt

(tt1

)= (t1)t(1)

(t1)2 =1

(t1)2 .

As t gets close to 1, what happens?

Moral of the story: be careful! Things break down when the

denominator gets close to zero.


Section 9 41

Note: The anti-derivative of (t 1)2 is 1t1 + c. How canthis be consistant with the example above?

1

t 1 + c =1

t 1 +ct ct 1

=ct c + 1t 1 .

If you let c = 1 you get the result from the example above. Moralof the story: (1) be careful and (2) the constant matters!

End Class Notes


Section 9 42

9 Composition of Functions and the Chain Rule

The main focus of this day is the composition of functions. The first activity focuseson compositions, and the second activity provides an example that demonstrates howcomposition of functions can be used to simplify a difficult problem.

The first mini-lecture provides a definition of composition and is relatively straight-forward. The second mini-lecture focuses on how to recognize which functions are beingcomposed given a composition, and in the final mini-lecture the chain rule is stated.

9.1 Mini-Lecture I The definition for composition of functions is given. Severalexamples are given.

Begin Class Notes

The composition of two functions, f (t) and g(t), is defined to

be

g f (t) = g(f (t)).

Example:

g(t) =t + 3,

f (t) = t3 + 4,

g f (t) = g(f (t)) = t3 + 4 + 3 = t3 + 7.

Example:

g(t) = t2 + t + 1,

f (t) =t,

g f (t) = g(f (t)) =(t)2

+t + 1 = t +

t + 1.

Note that

f g(t) = t2 + t2 + 1 = t + t2 + 1.A couple of things to note about this. First, f (g(t)) is not equal

to g(f (t)). Also, sometimes their can be a bit of ambiguity about


Section 9 43

what the result is. In the example above, I can put in a + or a- in front of the t and the answer is correct.

Given that t2 = 4 is t = 2 or is t = 2?End Class Notes

9.2 Mini-Lecture II Compositions are examined once again. At this time the ideaof identifying the functions that are being composed is examined.

Begin Class Notes

Given that

h(t) =t2 + 3t + 5,

can we find an f (t) and a g(t) so that h(t) = f (g(t))?

It is important to know which operation is done last. This

will identify the f (t) and everything else is g(t).

You can think about this in terms of your calculator. If you

were to program a calculator what would you have to at the

very end?

In this example the last thing that is done is to take the square

root of some number. This indicates that

f (t) =t and

g(t) = t2 + 3t + 5.


Section 9 44

Example:

h(t) =(t + 4t + t2

)3,

Outside Function = t3,Inside Function =

t + 4t + t2.

Example:

h(t) =(15t2 4t +t 1

)4,

Outside Function = t4,Inside Function = 15t2 4t +t 1.

Again, you have to be very careful here because the order mat-ters.

End Class Notes

9.3 Mini-Lecture III The chain rule is introduced and several examples are given.Begin Class Notes

Be careful about units! If f (t) has a domain that is mea-

sured in seconds and a range that is measured in metres, and

g(t) has a domain that is measured in meteres and a range

that is measured in metres per second, then the composition

g(f (t)) makes sense, but it cannot be done the other way.

Note: Given that h(t) = f (g(t)) how can we find the deriva-

tive?

h(t+h)h(t)h =

f(g(t+h))f(g(t))h =

f(g(t+h))f(g(t))g(t+h)g(t) g(t+h)g(t)h .


Section 9 45


d

dt(f (g(t))) = f (g(t))g(t).

This is called the Chain Rule.

Examples:

1. ddt

(t2 + 1

)

2. ddt

((t2 + 4t + 1

)6)

3. ddt

((4t + 1)18

)

4. ddt

(tt2 1

)

5. ddt

((1 t)30pi

)

6. ddt

(2t (1 4t)10)13

End Class Notes


Section 10 46

10 Exponential Functions

On this day exponential functions are introduced and examined. The basic propertiesare not discussed in great detail. It is assumed that some students know these and othershave forgotten them. Students should be able to work on these on their own.

The first mini-lecture is very brief. It is a simple introduction to exponentials. Thesecond mini-lecture examines differential equations and slope fields. The graphs fromthis exercise should look a lot like the plots done in the first activity.

The second activity and the second mini-lecture can take more time than usual. Forthis reason the first mini-lecture is kept very short, and it is a good thing to cut off all ofthe students on this first activity. The last activity is not critical. If most of the studentscan do most of the work on the third problem the activity can be stopped.

The final mini-lecture just focuses on the solutions approximated from the slope fields.The similarity between the resulting approximations and exponential functions is givenat the very end.

10.1 Mini-Lecture I This is an extremely brief mini-lecture because the secondactivity and the second mini-lecture are longer than usual. This mini-lecture is literally2 minutes long. This is a very brief introduction to the exponential function. The mainthing is that most of the students at least start on the third problem in the first activity.

Begin Class Notes

In polynomial function we take a number and raise that number

to a fixed degree.

In an exponential function we take a fixed number and raise it

to a variable power. For example

f (t) = 2t.

In this function f (1) = 2, f (2) = 4, f (3) = 9. Note thatf (3.5) = 23.5 = 2321/2. Since we can find integer powers andfind roots, it is no problem to just find the exponential for anynumber.

End Class Notes

10.2 Mini-Lecture II Differential equations are introduced and the general ideaabout slope fields is introduced.

Begin Class Notes

Newtons Law is really a differential equation

m~a =i

~F .


Section 10 47

What makes this a differential equation is that the acceleration

is really the derivative of the velocity. If you have an equation

that has the derivative of a function in it, the equation is called

a differential equation. The basic idea is to find an unkown

function that satisfies the equation.

There are times when the acceleration depends on the velocity.

For example, if there is air resistance, the force due to the air

resistance is proportional to the velocity squared. So if the

velocity is doubled the force due to air friction goes up by a

factor of four.

The question is how do we find the function to the differential

equation if we do not know the original equation? There are

two different things that we do:

1. Qualitative Aspect. We can try to get a feel for the be-

havior of the solution based on the equation.

2. Quantitative Aspect. Use the insight gained in the first

step and our knowledge of calculus to find the analytic formula

for the function satisfying the differential equation.

Today we will concentrate on the first aspect.

Example: Suppose that we are given that

a(t) = v2(t).In Search of Newton Calculus and Physics

Section 10 48

Given the velocity at any time we can find the acceleration. So

if v(t0) is equal to 2, then the acceleration is -4. Note that this

means that the slope of the velocity is -4 if v(t0) is 2. A tangent

line to the curve at this time is

y 2 = 4(t t0).

Locally, we know the shape of the function, v(t), near this point.

Why not draw a potential tangent line at every point in the

plane? If you can do that, you have the tangent line at any point

which will tell you the shape of the curve at any point. The way


Section 10 49

that you do this is to start at some initial point and then move to

the right in the direction indicated by the slope field.

I like to go through and do a couple of examples on an over-head with the next graph.

End Class Notes

10.3 Mini-Lecture III Some of the acitivities are examined.Begin Class Notes

The slope field gives you the idea of what shape the solutions

might have. If you have an idea of what the functions look

like, then you can get an idea of what kind of function to look

for in the solution of the original differential equation.

Example: A ball is dropped, and the force of friction is pro-

portional to one-third of its velocity.


Section 10 50

From Newtons Second Law we get

ma = mg Ff ,a = g 1

3mv.

If the acceleration is zero what is the velocity?

0 = g 13m

v,

v = 3mg.

If the velocity approachs 3mg then the acceleration gets close

to zero. This is the steady state for the ball.

Sketch different solutions on the following graph on an over-

head:


Section 11 51

Note that the solutions increase or decrease to the steady state.

Is this consistant with the equation?

a = g 13m

v.

If v > 3mg then the acceleration is negative so the velocity de-creases. Likewise, if v < 3mg then the acceleration is positive sothe velocity increases. All solutions get closer to the steady state.

End Class Notes


Section 11 52

11 Exponential Functions

This day represents a continuation of the previous days activities. The focus on thisday is on exponential functions. Special attention is paid to the range and domain ofthese functions as well as their inverses.

The first mini-lecture focuses on exponential functions and provides several examples.The second mini-lecture goes back to the differential equation that is motivating the useof exponential functions, and the last mini-lecture focuses on derivatives of exponentialfunctions.

11.1 Mini-Lecture I Exponential functions are given and explored by looking atseveral examples. The main focus is on the difference between exponential decay andexponential growth. In terms of decay, we look at functions that include an exponentialdecay term and ask about the long-term behavior of the function. This is done becauseof students confusion when faced with functions that have only one decay term in thelater activities.

Begin Class Notes

Exponential Functions:

f (t) = 2t

Notice that f (1) = 2, f (2) = 4, f (3) = 8, f (4) = 16. The

function grows really fast in time. We call this exponential

growth.

f (t) = 3t


Section 11 53

Notice that f (1) = 13, f (2) =19, f (3) =

127, f (4) =

181. The

function decays really fast in time. We call this exponential decay.

Properties of exponential functions:

24 23 = 23+4 = 27.2t 3t = (2 3)t = 6t.

5t+7 = 5t 57.7t+c = 7t 7c.

6t

2t =(

62

)t= 3t.

Note that with exponential decay, anything that looks like ctgets closer to zero as t gets large. For example, what happens tothe function f (t) = 5 7t in time? The function gets closer andcloser to 5.


Section 11 54

End Class Notes

11.2 Mini-Lecture II The differential equation that is used to motivate exponentialfunctions is re-examined. This is a very brief overview that includes a reminder abouthow derivatives are found. The actual derivatives of exponential functions are given inthe last mini-lecture. This is just an overview to motivate how we find the derivatives.

Begin Class Notes

Newtons Second Law:

m~a =i

~F , or

md

dtv(t) =

iFi.

Newtons Second Law gives us a differential equation that pro-

vides a mathematical model of some physical phenomena. The

goal is to find a function, v(t), that satisfies the equation.

Example: The force due to friction on a car rolling across a

flat surface is one half its velocity, and this is the only force

acting on the object,

ma = 12v,

a =12m

v,


Section 11 55

d

dtv(t) =

12m

v(t).

The goal here is to find a function, v(t), whose derivative is

directly proportional to the function itself! Can we find such

a function? It turns out that exponential functions hold the

key to this question.

Before starting the activity we need a couple of notes for the

next activity. The average rate of change for a function from

a to b is defined to be

f (b) f (a)b a .

To get the derivative of the function we look at

f (t + h) f (t)h

and let h get closer and closer to zero.End Class Notes

11.3 Mini-Lecture III The derivatives of exponential functions are examined.Begin Class Notes

Let f (t) = 2t. Then

f(t+h)f(t)h =

2t+h2th =

2t2h2th = 2

t[

2h1h

].

Note that the fraction on the right does not depend on t

in any way! The derivative looks like the original function

multiplied by some constant.

In general,

bt+hbth =

btbhbth = b

t[bh1h

].


Section 11 56

The derivative of an exponential is the original exponential

multiplied by some constant.

Lets see what bh1h is for different values of b:

h 2h1h

1 1

1/2 .83

1/4 .76

1/8 .724

1/16 .708

1/32 .701

1/64 .700

h 3h1h

1 2

1/2 1.46

1/4 1.26

1/8 1.18

1/16 1.14

1/32 1.12

1/64 1.10

h 2.7h1h

1 1.7

1/2 1.29

1/4 1.13

1/8 1.06

1/16 1.03

1/32 1.01

1/64 1.00

There is a number, approximately 2.718 - oh heck lets just call

it e, such that

d

dtet = et.

Examples:

d

dte2t

d

dt

(et

)d

dt

(tet

)d

dt

(t2 + 7epit

)d

dt

(et

2)


Section 11 57

Note to the lecturer: we do not give the general formulafor ddtb

t just yet. It will be examined once the natural log isfound.

End Class Notes


Section 12 58

12 Inverse Functions

The definition of a function is revisited as well as the range and domain of a function.The main thrust of the days activities is to relate these ideas to the inverse of a function.The examples focus on the inverses of exponential functions. Note that the activities onthis day are shorter than usual and there is more lecture time than normal.

The first mini-lecture focuses on the definition of a function and its inverse. Thesecond mini-lecture focuses on inverse functions with examples of exponentials, and thelast mini-lecture focuses on derivatives of inverse functions.

12.1 Mini-Lecture I The definition of a function and its inverse are given. Therange and domain of a function are formally defined.

Begin Class Notes

Suppose that f (t) is a function. Then for every number, t, that

is acceptable to the function there is another number, y, for

which y = f (t).

Example: f (t) is the square of t. (Given a number, t, the

function returns t2).

Example: f (t) is the number, b, that satisfies b2 = t.


Section 12 59

For any given positive number, t, there are two numbers that

satisfy b2 = t. This is not a function.

We will have to be careful. So far we have played things fast

and loose and should have been more precise. We will have to

clean up the mess now or else we will be in big trouble.

Definition: The set of all possible values of t that can be used

in f (t) is called the domain of f (t).

Definition: The set of all possible values that can be returned

by f (t) is called the range of f (t).

Example: f (t) = t2 + 1


Section 12 60

The domain is the set of all real numbers. The range is all

numbers equal to or bigger than one.

Example: f (t) = 3t


Section 12 61

The domain is the set of all real numbers. The range is all

numbers strictly bigger than zero.

Example: f (t) =t


Section 12 62

The domain is the set of all non-negative numbers. The range

is all numbers equal to or bigger than zero.

Definition: A function, f1(t), is the inverse of f (t) if thefollowing are true:

f (f1(t)) = t,f1(f (t)) = t.

Example: The inverse oft is t2. (We will see later that we

have to be more careful in terms of the range and domain of thesetwo functions, though.)

End Class Notes

12.2 Mini-Lecture II The inverse is explored further. The logarithm is definedduring this mini-lecture.

Begin Class Notes

Is the function f (t) = t2 + 5t + 2 invertible?


Section 12 63

For almost any value of f , there are two different values of

t that return the same number. The inverse is not a function!

Note: At the bottom of the parabola the derivative of the

function is zero:

f (t) = 2t + 5 t = 52.

The domain of the function is all of the real numbers. The

range of the function is all numbers greater than or equal to

f(52

).

If I limit the domain to all t 52 then on this new domain Ican find an inverse since I am only looking at the right half of

the parabola:

y = t2 + 5t + 2

t =5+

254(2y)2 = f

1(y).

The last equation is the inverse of f when the domain is

restricted.


Section 12 64

Example: The exponential function y = et.

It is invertible, but how do we find the inverse of this monster?

Definition: Given that y = et the function to find t given a

value for y is called the natural logarithm. We will denote

this function as ln(y). From this definition we have that

ln(et)

= t,

eln(t) = t.

Note thatf (t) = et Domain is all real numbers.

Range is all positive numbers.

f1(t) = ln(t) Domain is all positive numbers.Range is all real numbers.

Properities of the natural log.

ln(a b) = ? = y, theney = eln(ab) = a b = eln(a)eln(b) = eln(a)+ln(b),

ln(a b) = ln(a) + ln(b).In Search of Newton Calculus and Physics

Section 12 65

ln(ab)

= ? = y, then

ey = eln(ab) = ab =

(eln(a)

)b= eb ln(a),

ln(ab)

= b ln(a).

These two results imply that

ln(ab

)= ln(a) ln(b).

Note:

ddte

f(t) =?

Do not forget the chain rule!

Finally, in the activity it is important to ask if

ef(t) = t,

then what is f (t)?End Class Notes

12.3 Mini-Lecture III The derivative of the natural log is explored and severalexamples are given.

Begin Class Notes

We have that if

eln(t) = t,


Section 12 66

then

d

dteln(t) = 1, or

eln(t)d

dtln(t) = 1,

td

dtln(t) = 1,

d

dtln(t) =

1

t.

In general we have that

at =(eln(a)

)t= et ln(a)

ddtat = et ln(a) ln(a)ddta

t = at ln(a).

Examples:

d

dt4t = ?

d

dt(18.7)t = ?

d

dt(.25)t = ?

Why should the derivative be negative for that last one?

The general formula for logs can also be found:

y(t) = loga t

ay(t) = t,In Search of Newton Calculus and Physics

Section 13 67

(ay ln(a))d

dty(t) = 1,

d

dty(t) =

1

ay(t) ln(a),

=1

t ln(a).

Examples:

d

dt

(t3t

)=?

d

dt

(t + 1 log10

(t2))

= ?

d

dt

(14t log37.1(t)

)= ?

One last note:

The anti-derivative of et is et + c.

The anti-derivative of 1t is ln(t) + c.

End Class Notes


Section 13 68

13 Taylor Polynomials

The focus for this day is on the derivation of Taylor polynomials. The day begins bygraphing polynomials whose derivatives match the logarithm at t = 1, and by the end ofthe day the general form for the Taylor series is derived.

The first mini-lecture only covers the derivatives of polynomials. This is done in sucha way as to motivate the use of polynomials in a form that is useful in the later parts ofthis days class. The second mini-lecture is a review of the first activity, and the generalform for the Taylor polynomial is given. In the final mini-lecture the idea of linearizationsare examined with an eye towards the use of Newtons method.

13.1 Mini-Lecture I The derivatives of a particular polynomial are examined. Thisis used to motivate the general form for the Taylor Series.

Begin Class Notes

Define a polynomial to be

P (t) = 5 + 6(t 1) + 712

(t 1)2 + 8 13!

(t 1)3

+91

4!(t 1)4 + 10 1

5!(t 1)5.

We will make note of some of the properties of this polynomials.

First what is the value of the polynomial at t = 1?

P (1) = 5.

What about the first derivative?

P (t) = 6 + 7(t 1) + 8 12!

(t 1)2 + 9 13!

(t 1)3 + 10 14!

(t 1)4,P (1) = 6.

Okay, what about the second derivative?

P (t) = 7 + 81

1!(t 1) + 9 1

2!(t 1)2 + 10 1

3!(t 1)3,

P (1) = 7.


Section 13 69

The third derivative?

P (t) = +8 + 91

1!(t 1)1 + 10 1

2!(t 1)2,

P (1) = 8.

The fourth derivative?

P (t) = 9 + 101

1!(t 1)1,

P (1) = 8.

Finally, the fifth derivative?

P V (t) = 10,

P V (1) = 10.

Wow! Notice how those derivatives match up with the coefi-cients above?

End Class Notes

13.2 Mini-Lecture II An overview of the first activity is given, and the generalform for Taylor polynomials is given.

Begin Class Notes

I like to put up the following plots:


Section 13 70


Section 13 71


Section 13 72


Section 13 73

Definition:

The Taylor polynomial of degree N for a function f (t)

around the fixed point t0 is

PN(t) = f (t0) + f(t0)(t t0) + f (t0)1

2(t t0)2

+f (t0)1

3!(t t0)3 + f IV (t0) 1

4!(t t0)4

+fV (t0)1

5!(t t0)5 + + f (N)(t0) 1

N !(t t0)N

This is a polynomial. The function and its derivatives are

evaluated at some constant, t0, so the coefficients are all con-

stants. If t0 = 0 then the polynomial is also called the

Maclaren polynomial.


Section 13 74

Example: Find the Taylor polynomial for the differential

equation

d

dty = 2y(t),

y(1) = 1.

y(t) = 2y(t) y(0) = 2,y(t) = 2y(t) = 4y(t) y(0) = 4,y(t) = 4y(t) = 8y(t) y(0) = 8,y(4)(t) = 8y(t) = 16y(t) y(4)(0) = 16,

...

y(n)(t) = 2N1y(t) = 2Ny(t) y(n)(0) = 1,

P6(t) = 1 + t + 12t2 +

1

3!t3 +

1

4!t4 +

1

5!t5 +

1

6!t6

Finally, note that if N = 1 you get a line. This line is calledthe linearization of the function at a point.

End Class Notes

13.3 Mini-Lecture III The second activity is discussed with an emphasis on lin-earizations and an introduction to Newtons method. I like to have everybody runningMatlab or other software package at this time and step through the calculations as I dothem on the board.

Begin Class Notes

The linearization of of f (t) at a point t0 is the first Taylor

polynomial at the given point,

P1(t) = f (t0) + f(t0)(t t0).

This is the point-slope form for a line!


Section 13 75

We can use this to find the points at which a function crosses

the t-axis. For example, suppose I want to find a value of t

that satisfies t3/27t2 = 0? I cannot solve this analytically,so I have to play a game:

t3/2 = 7t + 2,

t3 = (7t + 2)2, let

g(t) = t3 (7t + 2)2.I can find the linearization by finding the derivative of g

g(t) = 3t 14(7t + 2)3.The linearization about a point can now be found.

We will try to find the point, t, which satisfies the original

equation. First graph g(t) and make an intial estimate. Here

we will let t = 48 and proceed:

>> t = 48;

>> g = t.^3 - (7*t+2).^2

g =

-3652

>> gp = 3*t.^2-14*(7*t+2)

gp =


Section 13 76

2180

>> t = t - g/gp

t =

49.6752

Cycle through this until t gets close to the root which should

only take two more iterations.

This process is called the Newton-Raphson Method.End Class Notes


Section 14 77

14 The Area Function

The main idea for this idea is the area function. The integral is motivated in termsof calculating change in distance given the velocity. There is not an explicit connectionmade to the anti-derivative, though. That will happen in the following class.

The first mini-lecture is very short and introduces the way to find the change indistance given the average velocity. This idea is expanded upon in the second mini-lecture in which the Riemann sum is introduced. Finally, the integral is defined in thelast mini-lecture.

14.1 Mini-Lecture I The total change in distance is found by using the averagevelocity. The basic question that is examined is how do we find the change in distancegiven the velocity?

Begin Class Notes

The average velocity is defined to be

Average Velocity =Change in distance

Change in time.

Given the average velocity we can find the change in distance

by multiplying by the change in time.

Example: The times and velocities for a trip are given below:

t (sec.) v (m/sec.)

0 1

1 3

2 4

3 5

We do not have the average velocities here, but we can ap-

proximate them by pretending that the velocities are constant

over each time interval.


Section 14 78

Estimate of the change in distance:

Low Estimate = 1 1 + 3 1 + 4 1 = 8m.High Estimate = 3 1 + 4 1 + 5 1 = 12m.

End Class Notes

14.2 Mini-Lecture II The connection between the change in distance and the areaunder the velocity graph is made explicit. The basic idea behind Riemann sums is alsodiscussed.

Begin Class Notes

Notice! If the velocity is a piecewise constant function, then

the change in distance is equal to the area under the graph.

Is this true in general? We will ask a different question. Given a

velocity plot, how can we approximate the change in distance?

We will break up the domain of the function, the time, into small

pieces and try to make an approximation over each small piece.

To do this we will assume that the velocity is about constant

over each little interval.


Section 14 79

Example: If the velocity of a car is v(t) = tet from t = 1 to

t = 3, approximate the distance traveled using 15 intervals.

The width of each time interval is 3115 =215:

t0 = 1

t1 = 1 +215

t2 = 1 +415

t3 = 1 +615

tn = 1 + n

2

15.

The change in distance can be found using two different ap-

proximations. The first assumes that the velocity over each

interval is the same as the left side of each panel,

4x 14n=0 (tnetn) 215 36.4m.(This is called the left hand Riemann sum.)

The other approximation assumes that the velocity over each

interval is the same as the right side of each panel,

4x 15n=1 (tnetn) 215 44.1m.In Search of Newton Calculus and Physics

Section 14 80

(This is called the right hand Riemann sum.)

If we want to get a better approximation we can make the size of

the interval smaller (increase the number of intervals). We can

keep doing this and making the size of each interval extremely

small.

The change in position is the area under a curve!End Class Notes

14.3 Mini-Lecture III An example from the activity is examined in order todemonstrate how the area can be used to simplify a given problem, and the definition ofthe area function is given.

Begin Class Notes

Given the velocity one way to find the total change in po-

sition is to find the area under the velocity curve.

Example:

Which of the two cars is ahead after 2.5 seconds. The area in

Area 1 represents the distance that car 2 is ahead of car 1

after the first 1.1 seconds. The area in Area 2 represent the


Section 14 81

distance that car 2 is ahead of car 1 from t = 1.1 seconds to

2.5 seconds. Since Area 2 is bigger than Area 1 car 1 is

ahead after 2.5 seconds.

Definition: The area under a function, f (t), from t1 to a time

t is called the integral and is denoted

A(t) = tt1f (s) ds.

The s is called a dummy variable.End Class Notes


Section 15 82

15 The Integral

The area function is revisited, and the areas under more complicated functions areexamined. The first mini-lecture focuses on how to approximate the area function usingRiemann sums. In the second mini-lecture we examine the difference between left andright Riemann sums with a couple of notes about what happens when the function isnegative. In the third mini-lecture the relationship between a discontinuous accelerationand the resulting velocity is examined.

15.1 Mini-Lecture I The area function is revisited with an emphasis on calculatingthe Riemann sums.

Begin Class Notes

Given the velocity, the change in position is found by finding

the area under the graph.

Example:

The change in distance is

x(t) x(0) = Area under graph = t + 12(

12t)t = t + 14t

2.

For more complicated problems we have to approximate the

area using Riemann sums. This process proceeds in the fol-

lowing order:


Section 15 83

1. Identify the domain. (So far we have used the time for specific

values.)

2. Determine the width of each little interval.

3. Approximate the area under each small interval by assuming

that the velocity is constant over each interval.

4. Add up all of the areas.

Example: Approximate the following area using 20 intervals: 61t2 dt.

The domain is the time from 1 to 6. The width of each interval

is the length of the domain divided by the number of intervals,6120 . A Matlab script to find the approximation is the following:

>> h = (6-1)/20;

>> t = 1:h:6;

>> f = t.^2;


Section 15 84

>> size(f)

ans =

1 21

>> lhs = sum(f(1:20))*h

lhs =

67.3438

>> rhs = sum(f(2:21))*h

rhs =

76.0938

End Class Notes

15.2 Mini-Lecture II The Riemann sum is examined once again.Begin Class Notes

When we use the left hand Riemann sum we assume that the

velocity is constant over each interval and that the velocity is

equal to the value on the left side of each interval. When we

use the right hand Riemann sum we assume that the velocity

is equal tothe value on the right side of each interval. It is

important to remember that these are only approximations

and not the actual area.


Section 15 85

If a function is increasing then how does the left hand Riemann

sum compare to the actual area? Lets look at our example

from the first mini-lecture:

The true area is bigger than our approximation. Likewise what

if we take the right hand Riemann sum? The true area is less

than our approximation. In this situation we cannot get the

true area, yet, but we can get a bound on it.

What happens when we have a decreasing function?

Finally, what happens when the function is negative? Does the

integral still do the job for us? Yes! If the velocity is negative

then the change in position is negative.


Section 15 86

If the velocity is negative over an interval then the integral givesthe change in position which is negative. (We can have negativeintegrals.) In the graph above, the object passes through its start-ing position at 6 seconds.

End Class Notes

15.3 Mini-Lecture III More examples are given, but this time the graphs arediscontinuous. The graphs use discontinuous accelerations because some students willnot accept a discontinuous velocity but will accept a discontinuous acceleration. It iseasier to motivate discontinuous accelerations because it is easier to imagine forces thatare suddenly applied to an object.

Begin Class Notes

Example: Suppose that the acceleration is given by

a(t) =

1 0 < t < 1,1 1 t < 2.


Section 16 87

The change in velocity is given by the area under the accelera-

tion curve. The change in velocity is

v(t) v(0) = 1 t 0 < t < 1,1 (t 1) 1 t < 2.

If v(0) = 0 then the function looks like the following graph.

The velocity is a continuous function! This is an important note.Because we are adding up areas and not looking at slopes, the pro-cess of finding integrals returns a function that is smoother thanthe original function.

End Class Notes


Section 16 88

16 The Fundamental Theorem of Calculus

The fundamental Theorem of Calculus is given. This should not be much of a shockto the students. This is simply a matter of reminding the students that another way toget the position given the velocity is to find the anti-derivative. Since we have been usingintegrals to do this, it should not be much of a surprise.

The fundamential theorem is derived in the first mini-lecture. The discussion n thenext two mini-lectures focus on how to use the fundamental theorem to help us solvedifferential equations. The first mini-lecture is longer than usual, and the other twomini-lectures are very short.

16.1 Mini-Lecture I The Fundamental Theorem of Calculus is derived. Thisderivation makes use of the mean value theorem for integrals which is derived.

Begin Class Notes

So far we have used two different ways to fnd the position given

the velocity:

1. Given v(t) we can find the position by finding the anti-derivative.

2. Given a graph of v(t) we can find the change in position by

finding the area under the curve.

Are these two the same thing or are they different? There must

be some kind of connection. It turns out that the connection

is through something called the Fundamental Theorem of

Calculus. Homework assignment: look up the word

fundamental.

Before we can show the connection we will need an intermediate

result: (This is mostly graphical and not as rigorous as some would

like.)

Mean Value Theorem for Integrals

Suppose that f (t) is a continuous function.


Section 16 89

(min f ) (t2 t1) t2t1 f (t) dt (max f ) (t2 t1), (min f ) 1t2t1

t2t1 f (t) dt (max f ) .

If the function f (t) is continuous then there must be a point

t1 c t2 wheref (c) =

1

t2 t1 t2t1f (t) dt.

Fundamental Theorem of Calculus: Define the area

function to be

Area(T ) = T0v(t) dt.

We will assume that v(t) is continuous and the Mean Value The-

orem for Integrals holds. Then we can find the average rate of

change of the area function

Area(T + h) Area(T )h

=T+hT v(t) dt

h

=1

h

T+hT

v(t) dt

= v(c) for some T + h c T .In Search of Newton Calculus and Physics

Section 16 90

As h gets closer and closer to zero, c must get close to T . This

implies that

d

dTArea(T ) =

d

dT

T0v(t) dt

= v(T ).

The anti-derivative of the integral is the integrand.

So what? Suppose that v(t) = t2. What is the change in

distance between t = 0 and t = 3?

Method 1

x(t) =1

3t3 + c.

This implies that

x(3) x(0) = 13

33 + c (0 + c)= 9.

Method 2

30 t

2 dt = 13t330

= 1333 0 = 9.

End Class Notes

16.2 Mini-Lecture II Riemann sums are revisited. The fundamental theoremis motivated in the context of solving differential equations. This is a simple look atRiemann sums, and a more comprehensive view is presented in the final mini-lecture.

Begin Class Notes

The idea behind Riemann sums is that you assume that the

velocity is relatively constant over a pre-defined time interval.


Section 16 91

Over a given interval from tn to tn+1 you assume that the ve-

locity is relatively constant. For the left hand Riemann sum

we have

4x v(tn) (tn+1 tn) .For the left hand Riemann sum we have

4x v(tn1) (tn+1 tn)

If x(t) = f (x, t) and x(0) = 1 then we can use the same idea.

The function f (x, t) is treated in the same way as we treat

the velocity above. If we know the velocity at a single point

in time, we can approximate the change in distance using the

Riemann Sum.


Section 16 92

Since4x v(tn)4t we can approximate the value of the posi-tion at the next time step. Once we have the position at this newtime level we can repeat the process to approximate the positionat the following time level.

End Class Notes

16.3 Mini-Lecture III The role that the fundamental theorem plays in finding thesolution to differential equations is explored in this last mini-lecture.

Begin Class Notes

Integrals are used to undo derivatives because you have to

find the anti-derivative to find an integral.


Section 16 93

Example:

d

dtx(t) = f (x, t).

We can take the integral of both sides to get t1t0

d

dtx(t) dt =

t1t0f (x, t) dt,

x(t1) x(t0) = t1t0f (x, t) dt.

The integral on the right can be approximated using Riemann

sums.

Integrals give you total change. Derivatives give

you the rate of change.

If there is time I like to go over the following example.

Suppose that you know that a system follows the differential

equation

d

dtx(t) = 2x(t),

x(0) = 5.

What is the function x(t)?

I can divide both sides by x(t) to getddtx(t)

x(t)= 2.

I can integrate both sides with respect to t to get

T0

ddtx(t)

x(t)dt =

T0

2 dt.


Section 16 94

This equation can be solved

ln |x(t)|T0

= 2tT0,

ln |x(T )| ln |x(0)| = 2T,ln |x(T )| = ln |x(0)| + 2T,

x(T ) = eln |x(0)|+2T ,x(T ) = x(0)e2T .

End Class Notes


Section 17 95

17 Center of Mass

The main focus of this days activities is on how Riemann sums are used to convertcontinuous problems into a discrete problem. The process is examined in terms of howthe fundamental theorem is used to convert the problem into one that we can solveanalytically.

The first mini-lecture focuses on linear density and how we can find the mass of a rodwhose density is not a constant. The second activity extends this idea and examines howto find the center of mass of a rod. Finally, in the third mini-lecture the center of massfor a lamina is found by looking at one quick example. The students will be expected tolook up this method in their books and do some of this work on their own.

17.1 Mini-Lecture I Riemann sums and their role in approximating the mass of arod with non-constant density is examined. The fundamental theorem is not mentionedhere but will play a part in the first activity.

Begin Class Notes

Imagine a thin rod of length L.

I can cut out a very thin piece, and the piece must have some

mass. The mass per unit length is defined to be

(x) =mass

length

at a point x.

To approximate the total mass of the original rod I can break

the mass into a whole bunch of very thin slices. The mass of

each little slice can be approximated, and the total mass can

be found by adding up the results for each little piece.

Example: Approximate the density of a rod that is of length

1m using six panels.


Section 17 96

The length of each piece is 16, and the mass of each piece is

approximated by assuming that the density does not change

too much over each interval. Here we will assume that the

density is approximately equal to the value of the density at

the left end of each piece,

mass (0)16

+

16

16

+

26

16

+

36

16

+

46

16

+

56

16.

Wait a minute... this is a left hand Riemann sum!

We can look at this another way.

We are finding the Riemann sum for the density. It is importantto realize that we started out with a physical system and convertedit into a mathematical model. In terms of the mathematics we arenow concentrating on this function, called the density, and are do-ing the same old math operations that we were doing with velocityand position.

End Class Notes

17.2 Mini-Lecture II The center of mass for a rod of non-uniform density is found.Begin Class Notes

The center of mass of an object is the point at which I can apply

a force equal to the weight, and the object will balance.


Section 17 97

How do we find this point? Lets play a game. Suppose that I

have a wrench on a bolt at the origin, and the length of the

wrench is one meter. Now suppose that I have a wrench on the

same bolt, but it is two meters long. If I apply the same force

on each of the wrenchs which one will twist harder around

the origin? The longer wrench will have twice the torque on

the bolt.

The idea behind finding the center of mass depends on this idea.

We want to find the place where the torques balance.

Definition: The torque due to a force, F , that is perpindicular

to the x-axis and a distance x from the origin is xF .

If a rod is balanced then all of the torques must equal zero.

Suppose that I push up on the rod at x with a force equal to

the weight of the rod. Then the torque due to this is xMg

where M is the total mass of the rod.

How do we find the torque about the origin for a rod? Again,

we break it up into pieces.


Section 17 98

For example if the rod is broken up into six pieces, the torque

due to each piece is approximated and added up:

Torque g0 (0)16

+ g1

6

16

16

+ g2

6

26

16

+

g3

6

36

16

+ g4

6

46

16

+ g5

6

56

16.

In general the torque can be approximated using a Riemann

sum and set equal to the torque above

xMg N1i=0

xi(xi)g4x.The center of mass can be isolated

x N1i=0 xi(xi)4x

M.

End Class Notes

17.3 Mini-Lecture III The center of mass for a lamina is found, and the generalformulae are given. One quick example for the center of mass for a lamina is given. Thestudents are expected to read their calculus book to fill in the blanks.

Begin Class Notes

The center of mass for a rod of density (x) from x1 to x2 is

x =

x2x1x(x) dxx2

x1(x) dx

.

What happens if the mass is not a rod but a flat object whose

density is uniform but has a non-trivial shape? We play the

same game. We break the piece into little strips, find the mass,

and find the center of mass.

Example: Find the center of mass for a flat piece of metal

whose density is uniform. The top of the piece follows the for-


Section 17 99

mula y =x and the bottom of the strip follows the formual

y = x2. The values of x go from 0 to 1.

To find the total mass the piece is broken up into thin strips.

The domain, 0 x 1, is divided into small pieces

M = 10

(x x2) dx.

The first moment from the y-axis can then be found by mul-

tiplying the mass of each strip by its distance from the y-axis,

which is xn,

My = 10x(x x2) dx.

The moment for the y-direction is found by dividing up the

piece into horizontal strips.


Section 17 100

The first moment from the x-axis can then be found by mul-

tiplying the mass of each horizontal strip by its distance from

the x-axis,

Mx = 10y(y y2) dy.

The center of mass can now be calculated

x =MyM,

y =MxM

.

This is a very quick and dirty introduction into finding the

center of mass for a lamina. You (the students) should read

the book and figure this one out for yourselves.

Note that if you have a big complicated thing, you can break it

up into separate pieces.


Section 18 101

Given the triangle above, one way to tackle the problem is todivide it up into two pieces, 1 and 2. You can find the formula foreach of the lines and can find the center of mass for piece 1 andpiece 2. Once you have the two masses and their center of mass,you can use the standard formula that you use in your physicsclass to get the total center of mass.

End Class Notes


Section 18 102

18 Work Integrals

The work done by moving an object along a curve is discussed. The first mini-lecturefocuses on the general idea for an object being moved in one dimension. The secondmini-lecture focuses on paths in two dimensions. Finally, the third mini-lecture is anoverview of the days activities.

18.1 Mini-Lecture I The role of Riemann sums and work integrals is examined.The main focus is on how to find the work for an object moving in one dimension.

Begin Class Notes

Riemann Sums:

1. Take the domain and break it into small pieces.

2. Find the desired quantitity for each small piece.

3. Add up all of the quantities for every small piece. We need to

find the following:

Nn=0

( )4x

where x comes from the domain of the function.

4. Convert the sum to an integral so that the Fundamental Tho-

erem of Calculus can be used to form an analytic result.

So what? We have a simple definition from physics,

Work = Force Distance.

Example: An object is moved from the origin to the north-

east at a speed of 1 metre per second. After a time, t, the distance

traveled is 1 t metres.


Section 18 103

x(t) = t cos(pi4

)t =

2

2 t

y(t) = t sin(pi4

)t =

2

2 t

The work done is the force times the distance. If the mass is

80 kg then the normal force is 80g and the force due to friction is

80g. After 4 seconds the object is moved 4 metres, and the work

done is 320g.

What happens if the force changes in time? We will first ask

what happens when the force is known as a function of the position.

We will worry about the time varying case later.

Suppose I pull on object from x = 0 to x = 4 along the direction

of travel with a force ofxN. What is the work done?


Section 18 104

The work from xn to xn+1 is approximately equal toxn (xn xn+1).

The total work can be approximated by adding up the work over

each interval

Total work N1n=0

xn4x.

The total work is 40

x dx =

2

3x

3240.

End Class Notes

18.2 Mini-Lecture II The work for a force applied along a path in two dimensionsis examined. The work is for a force that is changing in time.

Begin Class Notes

Suppose that an object is moved along some path. At any par-

ticular time, t, the coordinates of the object are x(t) t andy(t) = t(t 1). The path over the times t = 0 through t = 1looks like a parabola.

Suppose that the force acting on the object is 2N, and the di-

rection of the force is always along the direction of the path.

Find the work from t = 0 to t = 1.


Section 18 105

The domain is the time from t = 0 to t = 1. If we want to use

the fundamental theorem we have to divide up the time into

small intervals, find the work over each interval, and add up

the work for each piece. Keep in mind that we have to have a

sum that looks like

Nn=0

( )4t.

We need the change in the domain, 4t, in order to use thefundamental theorem.

Once the time is broken up into small pieces, we find the posi-

tion at the endpoints of each time interval. The first question

that we ask is what distance have we traveled over one small

time interval?

The distance traveled is approximately4x2 +4y2,


Section 18 106

which menas that the work done is approximately

24x2 +4y2.

The problem is that I need something times 4t. To get thisI will multiply and divide by 4t

24x2 +4y2 = 2

4x2 +4y24t4t

= 2

(4x2 +4y2) 14t24t,

= 2

4x4t

2 +4y4t

24t.The total work can now be approximated as

N1n=0

2

4x4t

2 +4y4t

24t.As the time interval approaches zero we get 1

02x2(t) + y2(t) dt =

10

2

12 + (2t 1)2 dt.

End Class Notes

18.3 Mini-Lecture III An overview of the work along a path in two dimensions isprovided.

Begin Class Notes

A mass m is falling along a curve.


Section 18 107

We can break the y interval into small pieces. Find the work

done over each piece.

The key things are

distance 4x2 +4y2

cos() 4y4x2 +4y2work mg 4y4x2 +4y2

4x2 +4y2

= mg4y.

The total work is found by adding the work up over each seg-

ment,

N1n=0

mg4y.

As 4y gets close to zero we have y2y1mg dy = mg(y2 y1).


Section 18 108

Note: For the 2D problem we have the force

~F = ~ + ~,

which is 2

2N at pi4 radians.

The distance traveled is approximately4x2 +4y2.

The cosine of the angle between the force and the direction of

movement is approximately

cos() = cos( )= cos() cos() + sin() sin()

= cos(pi

4

) 4x4x2 +4y2 + sin(pi

4

) 4y4x2 +4y2 .The work is approximately equal to

2

2

cos(pi

4

) 4x4x2 +4y2 + sin(pi

4

) 4y4x2 +4y2 4x2 +4y2,

= 2

2

2

24x +

2

24y

,In Search of Newton Calculus and Physics

Section 18 109

= 2

2

2

2+

2

2

4y4x

4x,=

2 + 24x4y4x.

The total work is approximately

N1n=0

2 + 24x4y4x,

which approaches an integral as 4x gets close to zero x2x1

2 + 2y dx = 2x + 2y(x)x2x+1

.

End Class Notes


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