interphase edge calculus 3 lecture/recitation...
TRANSCRIPT
Interphase EDGE Calculus 3 Lecture/Recitation Notes
Jack-William Barotta
August 23, 2019
1
Contents
1 Introduction 12
2 Lecture I on July 1, 2019 17
2.1 How to Think About Math . . . . . . . . 17
2.2 n-dimensional space . . . . . . . . . . . . 18
2.3 Graph of an Equation in Rn . . . . . . . . 20
2.4 Functions . . . . . . . . . . . . . . . . . . 22
2.5 Level Curves of Functions . . . . . . . . . 26
3 Recitation I on July 2, 2019 27
4 Lecture II on July 3, 2019 30
4.1 Review . . . . . . . . . . . . . . . . . . . 30
4.2 Linear Transformation . . . . . . . . . . . 31
4.3 Determinants . . . . . . . . . . . . . . . . 36
2
4.4 2D Determinants . . . . . . . . . . . . . . 37
4.5 3D Determinants . . . . . . . . . . . . . . 42
4.6 Matrix Operations (ASE) . . . . . . . . . 44
4.6.1 Solving a Linear System . . . . . . 52
5 Lecture III on July 5, 2019 55
5.1 Introduction to Vectors . . . . . . . . . . 55
5.1.1 Addition . . . . . . . . . . . . . . 56
5.1.2 Scalar Multiplication . . . . . . . . 57
5.2 More Vector Properties . . . . . . . . . . 58
5.3 Applying these concepts . . . . . . . . . . 59
5.4 The Dot Product . . . . . . . . . . . . . . 62
5.4.1 Small Examples . . . . . . . . . . 64
5.5 The Cross Product . . . . . . . . . . . . . 67
5.5.1 Example Time . . . . . . . . . . . 70
5.6 Big Picture . . . . . . . . . . . . . . . . . 71
3
5.7 3D Geometry with Lines . . . . . . . . . . 72
5.8 3D Geometry and Planes . . . . . . . . . 77
5.8.1 TLDR Finding Equation of Plane . 82
6 Lecture IV on July 8, 2019 85
6.1 Review . . . . . . . . . . . . . . . . . . . 85
6.2 Planes in Space . . . . . . . . . . . . . . 87
6.3 Vector-Valued Functions . . . . . . . . . . 96
6.4 Quadric Surfaces . . . . . . . . . . . . . . 101
6.5 All other Path in Space Stuff (ASE) . . . 103
6.5.1 A Proof of Orthogonality . . . . . 104
7 Recitation II on July 9,2019 109
7.1 Point to Point . . . . . . . . . . . . . . . 110
7.2 Point to Line . . . . . . . . . . . . . . . . 110
7.3 Point to Plane . . . . . . . . . . . . . . . 113
4
7.4 Line to Line . . . . . . . . . . . . . . . . 116
7.5 Line to Plane and Plane to Plane . . . . . 120
8 Lecture V on July 10, 2019 123
8.1 Polar, Cylindrical, and Spherical Coordi-
nates . . . . . . . . . . . . . . . . . . . . 129
8.1.1 Polar Coordinates . . . . . . . . . 130
8.1.2 Cylindrical Coordinates . . . . . . 131
8.1.3 Spherical Coordinates . . . . . . . 135
9 Lecture VI on July 11, 2019 136
9.1 Limits . . . . . . . . . . . . . . . . . . . 136
9.2 Other tools for limits . . . . . . . . . . . . 146
9.2.1 Alternate Paths . . . . . . . . . . 146
9.2.2 Continuity . . . . . . . . . . . . . 147
5
9.2.3 Examples of Using the Further Tech-
niques . . . . . . . . . . . . . . . . 148
10 Recitation III on July 12, 2019 150
11 Lecture VII on July 15, 2019 155
11.1 Partial Derivatives . . . . . . . . . . . . . 155
11.2 A difficult Example . . . . . . . . . . . . 164
11.3 Linear Approximation . . . . . . . . . . . 168
12 Recitation IV on July 16, 2019 173
12.1 Partial Derivative Notation . . . . . . . . 173
12.2 Clarifying an Example in Class on Clairout’s
Theorem . . . . . . . . . . . . . . . . . . 176
12.3 Linear Approximation . . . . . . . . . . . 180
12.4 A Rigorous Proof of Clairout’s Theorem . 184
6
13 Lecture VIII on July 17, 2019 188
13.1 Review on Linear Approximations . . . . . 188
13.2 Multivariable Optimization . . . . . . . . 189
13.3 The Second Derivative Test (ASE) . . . . 198
13.3.1 An Example in Second Derivatives 201
13.4 Directional Derivative . . . . . . . . . . . 205
14 Recitation V on July 18, 2019 208
14.1 A Small Note on Multivariable Optimization209
14.2 Gradients and Directional Derivatives . . . 210
14.3 Following a Path of Max Increase . . . . . 212
15 Lecture IX on July 19, 2019 216
15.1 Review on Directional Derivatives . . . . . 216
15.2 Multivariable Chain Rule . . . . . . . . . 224
7
15.2.1 A Proof of the Multivariable Chain
Rule . . . . . . . . . . . . . . . . 228
16 Lecture X on July 23, 2019 230
16.1 Review on Partial Derivatives and Mixed
Partials . . . . . . . . . . . . . . . . . . . 230
16.2 Lagrange Multipliers . . . . . . . . . . . . 232
16.3 Integration . . . . . . . . . . . . . . . . . 239
17 Recitation VI on July 24, 2019 242
18 Lecture XI on July 25, 2019 255
18.1 Review on Ideas Behind Integration . . . . 255
18.2 Triple Integrals . . . . . . . . . . . . . . . 261
18.3 Integration in Other Coordinate Systems . 268
18.3.1 Polar Coordinates . . . . . . . . . 270
18.3.2 Cylindrical Coordinates . . . . . . 270
8
18.3.3 Spherical Coordinates . . . . . . . 271
19 Recitation VII on July 26, 2019 272
20 Lecture XII on July 29, 2019 281
20.1 Integration in Spherical and Cylindrical Co-
ordinates . . . . . . . . . . . . . . . . . . 281
20.2 Custom Coordinate Systems . . . . . . . . 289
20.3 Applications of Double and Triple Integrals
(ASE) . . . . . . . . . . . . . . . . . . . 295
20.3.1 Average Value of a Function . . . . 295
20.3.2 Center of Mass . . . . . . . . . . . 296
21 Recitation VIII on July 30, 2019 297
22 Lecture XIII on July 31, 2019 307
22.1 Vector Fields . . . . . . . . . . . . . . . . 307
9
22.2 Work in Vector Fields . . . . . . . . . . . 311
22.3 Fundamental Theorem of Vector Calculus 315
22.3.1 Conservative Vector Fields . . . . . 316
22.3.2 Checking Conservative Fields . . . 319
22.4 Green’s Theorem . . . . . . . . . . . . . . 320
23 Recitation IX on August 1, 2019 321
24 Lecture XIV on August 2, 2019 330
24.1 Green’s Theorem . . . . . . . . . . . . . . 330
24.2 Surface Integrals . . . . . . . . . . . . . . 331
24.3 Parametrizing Surfaces (ASE) . . . . . . . 333
24.3.1 A Better Treatment . . . . . . . . 338
24.4 Flux . . . . . . . . . . . . . . . . . . . . 344
24.5 Divergence . . . . . . . . . . . . . . . . . 346
25 Lecture XV on August 5, 2019 347
10
25.1 Divergence . . . . . . . . . . . . . . . . . 347
25.2 Curl . . . . . . . . . . . . . . . . . . . . 352
25.3 Divergence Theorem . . . . . . . . . . . . 358
25.4 Stokes’ Theorem . . . . . . . . . . . . . . 363
25.4.1 Same Border, Different Surface . . 366
26 Recitation X on August 6, 2019 370
27 Lecture XV on August 7, 2019 379
28 Thank You 380
These lecture notes are based off of the Interphase EDGE
2019 Iteration of Multivariable Calculus instructed by
Sam Watson. The introduction contains useful informa-
tion for those in my recitation section, and the rest of
the sections will be labeled in accordance with the lec-
ture/recitation that they are associated with. I would
11
greatly appreciate if you alerted me of any typos that
you may find. The more you help me, the more I can
help you. I hope this is of help to you!
1 Introduction
Hello and welcome to my recitation section of Calculus
3! I took this class two years ago with Professor Wat-
son, and I really thought it was a great help and aid in
18.02. In addition, I was a recitation instructor last year
for Professor Watson, which was even more fun! I am a
Mathematics Major (Course 18) and Economics (Course
14) here at MIT. I feel like in lecture sometimes it is hard
to write down all the little details, so I will be providing
these for you as an additional resource that is supposed
to be utilized to reinforce your very own notes you take
12
in lecture. Please get comfortable with my website for
resources because I will be updating it daily! All of the
items that are outlined in blue throughout this document
contain hyperlinks to my email in this case, but will also
have useful math resources, additional problems, photos,
or other things I decide to put in this. please bookmark
my website jack.mit.edu as I will be uploading all my re-
sources to that :). I, along with the rest of the Calc3 TAs
are currently writing solutions to the in-book exercises of
Professor Watson’s book. That being said, I would be
more than happy to go over those problems with you as
well for the additional practice as I am working through
them myself right now!
My ”official” Office Hours for the course will
be on:
13
• Thursdays 8-9
• Sundays 8-9
However, realistically, I will be having office
hours from:
•Wednesday 8-10
• Thursdays 8-10
• Sundays 8-101
I also would like to extend time to work individually
with students who may find the Office Hour setting a tad
too overwhelming, chaotic, and loud. (which it can defi-
nitely be at times). That being said, please email me, and
I would be happy to meet for an hour or so to go over
material related to the class. In my opinion, the most1Honestly we’ve been knowing that I have like office hours all the time
14
important thing you can learn this summer is how to use
your resources at MIT. They are just about everywhere
you look, and they are just waiting to be used by you.
This is your education, and you should be taking full ad-
vantage of the amazing opportunity you have in front of
you here. I hope that I can be one of those resources for
you over the summer and potentially forward into the fu-
ture through other organizations and things I am a part
of. I also love talking about things related or non re-
lated to the course, so always feel free to talk about MIT
life, math, or anything else you would like to know about.
I am very excited, and I hope that you can share my
excitement throughout our next seven weeks together. I
probably will make a lot of mistakes along the way, so
15
please yell at me and tell me to fix my mistakes!! As a
recitation section, I can promise that every single one of
us will make a handful of mistakes at the very least, so
lets us learn from our mistakes and try for better the next
time. Do not get bummed if things do not come super
quickly to you. MIT is a lot different than high school,
and it is always better to ask for help! I want to keep you
as engaged as possible, so we will probably do a lot of
activities such as board work, extensions to applications,
and maybe even some friendly competition and games.
I will try to incorporate all of your majors if I can into
problem solving, so that you can see how wonderful math
is and its ability to weave its way into almost everything.
My high school teacher had fun exam review games, and
I hope we can make some of our own. Well See. Seriously
16
though, overall I want you to love what youre learning
and have fun while doing it. Without further ado, let us
begin the actual material of the course!
2 Lecture I on July 1, 2019
Please use sswatson.com/interphase if you want to find
all of your Psets, Syllabus, and other material such as the
course textbook provided by Professor Watson. Home-
work is due Monday, quizzes are at the start of recitation
on Tuesdays, and Sam’s Office hours are 7-9 on Sunday.
2.1 How to Think About Math
There was a discussion on some meta math stuff. I will
say from experience that whether or not it is just for you
or the grader, writing down your thoughts in the solution
17
definitely helps. It allows you to organize your thoughts,
and it also gives insight to the grader on just how much
you do know about the material being tested/questioned
about.
2.2 n-dimensional space
We can express our understanding of Euclidean Space
(n-dimensional space) by a really fancy looking R, namely
R. So one-dimensional space is simply just R, the Real
number line, so we simply just have (x). Now, if we want
to go to R2 , we are now going to be representing the real
plane. This is how we start defining distance! For exam-
ple, in R, the signed distance from the origin to x is x.
We can apply this to higher dimensions! In, R2 , we now
have (x, y), and we see that the x-coordinate of a point is
18
the signed distance from the y-axis, and the y-coordinate
of a point is the signed distance to the x-axis. Finally, lets
move to R3 . This is where for me, stuff got visually and
geometrically difficult to follow. so now we have (x, y, z).
Now the x-coordinate is the signed distance to the yz
plane, the y-coordinate is the signed distance to the xz
plane, and the z-coordinate is the signed distance to the
xy plane. Fun fact for those who care: Distance does not
have to be Euclidean. Euclidean is just the most common
formulation of distance. However, there are three axioms
of distance and if your crazy, wack distance system abides
by the three axioms of distance, then it is considered a
distance! In Real Analysis, they like to demonstrate this
and give crazy problems that scared me and scarred my
view of distance. Lol rereading these notes I wrote from
19
last year I am so dramatic. Real Analysis is a great course
and many students in my section last year took and well...
tbh, had mixed reviews on Real Analysis, so take at your
own risk!
2.3 Graph of an Equation in Rn
The graph of an equation in Rn is the set of points that
satisfy the equation. A graph simply represents a visual
representation of the solution set. The biggest takeaway
from this section is your domain, range, and space you are
graphing in matter greatly. An equation by itself needs
the amount of dimensions specified in order to truly an-
swer the question correctly. For example asking to graph
x = 1 leads to 3 different representation depending on if
you are in R,R2, or R3. In R, we simply get that x = 1
20
is a point at x = 1. In R2, we get that x = 1 is a vertical
line to represent all (x, y) pairs that are (1, y) for all pos-
sible y. In R3, we get that x = 1 is a yz-plane at x = 1
Namely a plane to represent all (x, y, z) triples that are
of the form (1, y, z) for all possible y and z.
We can take this to more interesting cases. For exam-
ple, lets look at the equation, x2 + y2 = 1. Some of you
may recognize this as a unit circle. However, we need to
be careful. We need to be more precise with ourselves.
For example, in R2, we are correct. The set of points that
satisfy this equation are exactly the unit circle. Namely,
the squared distance from the origin is 1. However, lets
move to R3. In this case, we now have the unit cylin-
der. Since the equation puts no restriction on z, we do
21
not just have the unit circle. Instead, we have that the
squared distance from the z-axis is now 1. Thus, we get
a cylinder instead of a Circle.
2.4 Functions
A function is really cool, in my math nerd circle opin-
ion. A function is what I think of as a mutater, or a
changer of something. Basically, you put some stuff in
the function, the function then does what it needs to do,
and it will output the result! A function can be thought
of as a mapping. Youre mapping your input to an out-
put. So in the most common example, think of f (x) = x2
, every inputted x value is being squared and being out-
putted. Similar to graphing an equation, we can graph
22
also functions. How we can say this is:
y = f (x) ∀x ∈ R (1)
In words, this means y = f (x), which is our function,
for all x that are on the real line. The fancy symbols
are some math notation that is very important to know
in higher order math courses, so it is good to get a little
familiar :) However, as many of you won’t necessarily go
into mathematics, you may have functions that do more
than just what we generally think of as a function as. you
can have functions change colors, functions that count the
amount of times something occurs sometimes referred to
as indicator variables. Moving forward, we can graph
functions from Rn to Rm . That being said, the graph of
a function will exist in the sum of dimensions from the
23
input and the output! This is why we cant really graph
function that have a sum of input and output space that is
greater than or equal to four. So, to have an equation for
you to use that may be helpful, the number of dimensions
your graph should be is:
dim(graph) = dim(input) + dim(output) (2)
In words,the dimension of the graph you must use must
be equal to the sum of the dimension of the input space
plus the dimension of the output space. This is why we
can’t really graph when the sum of the input and output
is greater than three! We can write what are function
are doing in equations as well! For example, say your
function f is a function from R3 to R. We can write this
24
conveniently with an arrow as follows,
f : R3 −→ R (3)
What this means if the function, f takes in three inputs,
say for example (x, y, z) and outputs a single number,
persay (w).
Info unrelated to lecture, but for those interested: Func-
tion can also take in matrices and output matrices. For
example you might see something of the form, f : Rn,p −→
Rm,q That being said you don’t have to input n number
and output m numbers, you can instead output n by p
matrices and output m by q matrices. We can see this as
more motivation as to how extensive the field of functions
really is.
25
2.5 Level Curves of Functions
A Level set is the set of all points in the domain of
a function that map to the same output! When I first
learned about level sets, I did not really see the point,
but in higher dimensions it is definitely helpful. say you
have the function, f (x, y, z) = x2 + y2 + z2, using our
equation about dimensions necessary (Equation 2), we
see we cannot physically graph this function since four
dimensions are necessary (3 + 1 = 4). However, we def-
initely can graph the level sets. To do this, we set the
function equal to a number, lets set it equal to 1 just for
ease. Then we have 1 = x2 + y2 + z2, and this is the
unit sphere! We can graph that since its only in three
dimensions :)
26
3 Recitation I on July 2, 2019
Great work on the quiz! Also, I am really impressed by
all of the problem solving abilities and different techniques
I saw being deployed on the worksheet. There is not much
I have to other than a small discussion on level curves. As
many of you may have seen, the level curve of a function,
f , is the graph of an equation generated by picking a
specific value for f . In addition, it is important to make
sure that the level curves you are picking may sense and
are ”reasonable” for the function being analyzed. Let me
use an example for clarity. Suppose we have the function:
f (x, y) = e−(x2+y2) (4)
Lets first be very tedious with organizing our thoughts.
We note that f : R2 −→ R. Thus, lets figure out which
27
dimension the level curve is. For the case we have here,
dimensions necessary for the levels curves are exactly the
same as the dimensions of the input space. Thus, for
this case we need R2 in order to graph the level curves.
Now lets go on to what the graphs look like. In order to
construct a level curve, we are going to set our function
equal to some constant.
c = f (x, y) = e−(x2+y2) (5)
So, if we want to find the level curve, let us isolate the x
and y argument.
− ln c = x2 + y2 (6)
Okay great. But now we must ask, what values of c can
be utilize. notice that our function has a domain of all
28
of R2 and has a range of c ∈ (0, 1]. Lets make sure this
makes sense. If we plug in some specific values of c that
are in the co-domain, we get that, − ln c ≥ 0. Thus,
we get that the level curves of the function are simply
circles with radius equivalent to − ln c. This is called a
Gaussian Distribution curve for those who have seen
something similar before. It will be helpful later in the
course to realize that a level curve is perpendicular to a
function’s surface. Imagine the side of a mountain as the
function, and the level curve being a specific altitude of
the mountain.
29
4 Lecture II on July 3, 2019
4.1 Review
When we utilize the notation that:
f : R2 −→ R (7)
Means that the domain of f is R2 and the co-domain
of the function is R. Also for functions that have an
input plus output dimension greater than 3, we cannot
necessary work with graphing the function, but we in-
stead graph the level curves. In the case of, f (x, y, z) =
x2 +y2 +z2, we cannot visualize the function, but we can
visualize its level curves as spheres.
30
4.2 Linear Transformation
Definition: A function, f : R2 −→ R2 is linear if
and only if:
f (x, y) = (ax+by, cx+dy) where a, b, c, d ∈ R (8)
This is the definition of linear we will be utilizing through-
out this course, so please get it down! We can also conve-
niently express this in matrix notation that will be used
when discussing this concept both here and in linear al-
gebra courses you may take in the future:
f (x, y) =
a b
c d
xy
(9)
We can also introduce the 3-dimensional cousins that may
come up sometime in lecture and recitation below, al-
31
though it is not of the utmost importance:
f (x, y, z) = (ax+by+cz, dx+ey+fz, gx+hy+iz) where a, b, c, d, e, f, g, h, i ∈ R
(10)
With corresponding matrix notation:
f (x, y, z) =
a b c
d e f
g h i
x
y
z
(11)
Linear transformations are geometrically a collection of
scale, shear, rotate, and projection. One of the things
that people have trouble with getting their head around
is the fact that y = mx+ b is not linear. This is not liner
because the constant b is involved that will not map the
origin to the origin for all non-zero b. Some examples of
32
linear transformations are:
f (x, y) = (x + y, x + y) (12)
f (x, y) = (0, 0) (13)
f (x, y) = (x, y) (14)
Where we choose a, b, c, and d as some constants in order
to satisfy our expression. If we want to scale our input by
some factor c, we can represent this as the linear trans-
formation:
f (x, y) = (cx, cy) (15)
For some scalar c. We can also rotate a a linear transfor-
mation with the rotation found below:
f (x, y) = (cos θx− sin θy, sin θx + cos θy) (16)
33
So, for the case where we want to rotate the plane by π2 ,
then we can plug this in to the aforementioned equation
to obtain:
f (x, y) = (−y, x) (17)
We can project the plane, say onto the x − axis, with
the following transformation:
f (x, y) = (x, 0) (18)
Where we effectively compress all of the different y-values
for a given x. We also can have a shear. A classic example
of a shear that you may see in a complex variable and/or
engineering course is:
f (x, y) = (x + y, x) (19)
34
The interesting idea here is that the area is preserved with
this. Although the unit square is being transformed into
a parallelogram, we note that the base and the height
remain constant and as such, the area remains constant.
Also, if you think about the determinant as a means of
calculating the area, then we see that the determinant
does not change as ad − bc = 1 − 0 = 1 A convenient
way to present linear transformations is with the follow-
ing theorem.
Theorem A function from R2 −→ R2 is linear if and
only if it maps the origin to the origin and equally spaced
lines to equally spaced linear points. We can verify this by
looking through the different families of linear transfor-
mations (shear, project, scale, and rotate) that all satisfy
the theorem above.
35
4.3 Determinants
Determinants are really neat. I definitely did not know
there was a geometric meaning really to what a determi-
nant was. I just thought it was an annoying computa-
tion, so hopefully this section proves interesting! Deter-
minants are all about how linear transformations distort
areas. Lets consider a number line first. Consider the
linear transformation f (x) = 3x. We want to ask our-
selves, how does this distort areas/lengths/volumes. In
this case, we see that the length between two numbers
triples. The 3 in front of the x acts as a distorter of
the original lengths. Thus, we will call 3 the determi-
nant of the function since this is the distortion factor of
length. More generally, we have that given a function,
36
f : R −→ R,
f (x) = mx for m ∈ R (20)
we say that the determinant is m. m can be thought of as
the signed factor by which f transforms lengths. Again,
this would not work if we had f (x) = mx + b since the
origin would not be mapped to the origin!
4.4 2D Determinants
Now lets look at how area can be distorted. For the 2d
case, consider, f : R2 −→ R2,
f (x, y) = (ax + by, cx + dy) (21)
Lets try to figure out the area of the unit square under
this linear transformation. Lets start off by seeing where
37
each of the 4 vertices get mapped to with the linear trans-
formation at hand. We see that (0, 0), (1, 0), (0, 1), (1, 1)
gets mapped to (0, 0), (a, c), (b, d), (a + b, c + d) respec-
tively. We can calculate the area through some interesting
geometry. Lets start by calculating the massive rectangle
that has base, (a + c) and height (b + d). Then, we can
subtract off the excess that is not part of the parallelo-
gram! After some algebraic manipulation, we get that
the signed area is simply ad − bc. Thus, we can classify
the 2d determinant as being equivalent to:
area = det
a b
c d
= ad− bc (22)
Again, for ease, I will introduce the matrix notation of
the linear transformation that you will see all over the
38
place at classes at MIT.
f (x, y) =
a b
c d
xy
= (ax + by, cx + dy) (23)
Lets think about some interesting cases. In the case that
the determinant is −1, the the orientation of the area is
reversed! This is the 2D analog to reversing the length. In
addition, let us look at the case of when the linear trans-
formations turns area into a line. Well, a line has zero
area, and as such, the determinant of such a linear trans-
formation is exactly zero. You may notice while working
that this will occur when one row of the linear transfor-
mation matrix is a scalar multiple of the other row of the
linear transformation matrix.
39
Lets try some examples:
f (x, y) =
1 1
1 1
(24)
Then, the determinant of our linear transformation is
ad − bc = 1 − 1 = 0. What does this mean? This lin-
ear transformation smashes down everything into a line.
Thus, the area is zero. Lets look at the shear case:
f (x, y) =
1 1
0 1
(25)
This was the shear case. We see that the determinant is,
ad− bc = 1− 0 = 1. Thus, this confirms that the area of
the shear transformation is unchanged. Lets finally do an
arbitrary rotation matrix. We would not expect simply
40
rotating would change the area. Lets confirm this:
f (x, y) =
cos θ − sin θ
sin θ cos θ
xy
(26)
Lets calculate the determinant of this linear transforma-
tion. namely det = ad− bc = cos2 θ + sin2 θ = 1 Which
confirms our suspicion! Lets look at the linear trans-
formation that flips the unit square over the x − axis,
namely:
f (x, y) =
1 0
0 −1
xy
(27)
Then we can calculate the determinant as ad− bc = −1
which again checks out that the area is unchanged by
flipping over the axis, but the orientation flips leading to
the negative sign.
41
4.5 3D Determinants
In this course, the largest matrices we will do is 3D.
To be honest, I dont think any course makes you actually
compute determinants any higher than this. Anyways in
3D the determinant represents the signed factor by which
f transforms volumes. For a given matrix,
A =
a b c
d e f
g h i
(28)
The easiest way to compute the determinant is by de-
composing
the 3×3 matrix into smaller matrices. In order to this,
we pick a row. For convenience, I will choose the first row
of my matrix. Then the determinant can be expressed as
42
the following equation:
det(A) = det
a b c
d e f
g h i
= a det
e f
h i
−b det
d f
g i
+c
d e
g h
(29)
You may now just use the rule we know for a 2×2 matrix,
and then you can use scalar multiplication of the number
out front! This makes the three-dimensional case not as
daunting. Lets fully carry through the multiplication:
det(A) = a(ei− fh)− b(di− fg) + c(dh− eg) (30)
det(A) = aei− afh + bfg − bdi + cdh− ceg (31)
You may notice that for three dimensions there is a plus
minus pattern when I went across. This is because a
checkerboard pattern is in affect that is alternating be-
43
tween plus and minus. The checkerboard pattern for a 3
3 matrix looks like this:+ − +
− + −
+ − +
(32)
The general strategy should be that you assign a plus
to the first item in your matrix in the upper left hand
corner, and then you follow the checkerboard pattern!
The checkerboard pattern is very important so you dont
end up adding something you should subtract or vice-
versa.
4.6 Matrix Operations (ASE)
One of the most important things that you will proba-
bly be asked, with a high probability, is how to compute
44
the inverse of a matrix and utilize it to help solve a linear
system! Let me first walk you through a problem that
I put on the additional problem of Chapter 1 that will
help us compute the inverse of a matrix. The problem
statement is lengthy so try to stay awake reading it!
Problem:Solving for a matrices inverse is common
practice for an 18.02 ASE. I will now walk you through
solving such a problem given your current knowledge on
matrices as we have all the tools that are required. We
will just need to throw in some new jargon. First and
foremost, a matrix, A, has an inverse if det(A) 6= 0. This
must hold true in order for us to proceed. In linear algebra
speak, the columns of A must be linearly independent in
order for there to exist an inverse. Now suppose we have
45
the 3× 3 matrix provided below.
A =
1 1 0
1 0 2
0 0 1
(33)
You can quickly check that A has indeed det(A) 6= 0. We
will now compute the inverse. We will follow a recipe.
First and foremost, we will expand along cofactors. Do
not mind the word, but you may see it in other courses.
What this means is that say we are looking at the Aij
entry which denotes the ith row and jth column. We
now want to cross out this row and column, take the
determinant of what is left (which should be a 2×2 matrix
in our case), and put that in the ij spot of some newly
created 3 × 3 matrix. I will do the top one for you. In
A11 I will delete the first row and first column. I am now
46
left with a smaller matrix that has determinant equal to
zero. I will plug this in, lets call it B, B11 spot. You now
complete the rest. Okay, that’s the hard part. Now, we
follow a checkerboard pattern of changing the sign on our
entries. I will display the pattern below:+ − +
− + −
+ − +
(34)
Okay, so simply look at the matrix, B, that you cre-
ated and negate the entries that have negative signs in
the above checkerboard pattern. Alright! We are getting
closer. Lets call this new matrix that we switched the
sign of every other entry, C. Okay, we will finally now
take the transpose of C. All this means is that we will
swap the rows and the columns. Thus, column one is now
47
row one and so on. We commonly see this as CT . Boom!
And that is it! We will then just divide everything by
det(A) We will call our final product A−1
Solution: Lets first compute the matrix of A as we
will have to use it later. det(A) = −1. Okay now lets
hopp in. Lets do this expand by cofactor thing. I will do
this now:
B11 = 0 (35)
B12 = 1 (36)
B13 = 0 (37)
B21 = 1 (38)
B22 = 1 (39)
B23 = 0 (40)
48
B31 = 2 (41)
B32 = 2 (42)
B33 = −1 (43)
Okay great. Now we will implement the checkerboard
pattern displayed in the problem statement and as such
flip the signs of every other entry.
C11 = 0 (44)
C12 = −1 (45)
C13 = 0 (46)
C21 = −1 (47)
C22 = 1 (48)
C23 = 0 (49)
49
C31 = 2 (50)
C32 = −2 (51)
C33 = −1 (52)
Lets now put this together and make the matrix C:
C =
0 −1 0
−1 1 0
2 −2 −1
(53)
Lets now take the transpose of this matrix as said to by
swapping the rows and the columns
CT =
0 −1 2
−1 1 −2
0 0 −1
(54)
We will finally divide everything by the determinant, det =
50
−1 to finally get A−1 which is displayed below:
A−1 =
0 1 −2
1 −1 2
0 0 1
(55)
You can also check and confirm that AA−1 = A−1A = I
where I is the identity matrix denoted as:1 0 0
0 1 0
0 0 1
(56)
Well took some time, but this shows how to compute
a matrix inverse. I think that computing such a thing is
best shown just through an example. So hopefully that
51
was helpful. The good thing here is that you can always
check whether you made a mistake or not. How? Well
since AA−1 = A−1A = I , then we can always multiply
our two resulting matrices to obtain the identity.
4.6.1 Solving a Linear System
Consider you have the following equation:
Ax = b (57)
Where A is some 3 × 3 matrix, x is some 3 × 1 matrix
thought of as a vector, and b is some 3×1 matrix thought
of as a vector. Lets just show this in its full form so that
52
we know what we are referring to:a11 a12 a13
a21 a22 a23
a31 a32 a33
x
y
z
=
b1
b2
b3
(58)
In these types of problems, the question will generally
give you A and b and ask you to solve for x. They write
questions like this so that they first see if you can do some
matrix operations and then solve a system of equations.
Here is the important punchline of this section. If A is
invertible, then:
Ax = b (59)
A−1Ax = A−1b (60)
53
However, we already have discussed that for an invertible
matrix, AA−1 = I . As such,
Ix = A−1b (61)
x = A−1b (62)
Of course, this only works when the matrix is invertible.
However, now we have a quick way to solve for the x
vector that makes this true. This is equivalent to solving
a 3 equation system of equations. So, what I would expect
from this section is the ASE potentially asking you to first
solve for the inverse of some matrix and then use that to
solve for some x vector in part b that solves some system
of equations.
54
5 Lecture III on July 5, 2019
If your name wasn’t learned, then press F for you. Your
name is not learnt.
5.1 Introduction to Vectors
Vectors will be one of the main objects that we will con-
front in this course, whether we are calculating distances
in space or fluxes through surface. Lets get the basics
down today, so we can concern ourselves with all the ap-
plications later. A vector is an arrow from one point
to another in Rn. We wont need to concern ourselves
with all n-dimensions. We instead, should make sure we
are proficient in both R2 and R3 resulting in vectors like
(2, 1) and (1, 2, 1) respectively as some examples. A vec-
tor has a magnitude and direction. The length of the
55
vector is the distance from the head to tail. We can also
break down the vector into its x and y components. Two
vectors are equivalent if their components are equal. For
example (2, 1) and (4, 2) are not equivalent. While both
of these vectors are in the same direction, notice that the
first vector has a smaller magnitude in comparison to the
second vector.
5.1.1 Addition
Consider we have vectors, ~v = (v1, v2) and ~u = (u1, u2),
We can add the two components as:
~v + ~u = (v1, v2) + (u1, u2) = (v1 + u1, v2 + u2) (63)
If we want to represent this on the coordinate grid, we
would first draw ~v placing the tail of the vector on the
origin. Then, we will place the tail of ~u at the head of ~v.
56
We then draw a new vector, that represents the addition
of the two vectors from the tail of ~v to the head of ~u. We
sometimes denote this as the resultant vector. Do not get
lost in the jargon though. The conceptual understanding
is the most important aspect. The jargon can only add
once we are fluent in the concept.
5.1.2 Scalar Multiplication
Suppose we have ~v = (v1, v2), If we want to multiply
our vector by some scalar (constant), c, we get that:
c~v = c(v1, v2) = (cv1, cv2) (64)
What does this do? We see that this scales the original
vector whilst keeping the result parallel (or anti parallel
if the constant is negative) as the original vector, ~v.
57
5.2 More Vector Properties
We can only say that two vectors, ~v and ~u are parallel
if and only if,
~u = c~v for c ∈ R (65)
Again, given two vectors, ~v and ~u,
c(~u + ~v) = c~u + c~v (66)
This is sometimes considered the distributive property of
scalar multiplication. We can make a quick proof of this
in two dimensions,
c(~u + ~v) = c~u + c~v (67)
c(~u + ~v) = c(u1 + v1, v2 + u2) (68)
c(~u + ~v) = (cu1 + cv1, cu2 + cv2) (69)
58
c(~u + ~v) = (cu1, cu2) + (cv1, cv2) (70)
c(~u + ~v) = c(u1, u2) + c(v1, v2) (71)
c(~u + ~v) = c~u + c~v (72)
5.3 Applying these concepts
Problem Use vectors to show that the line segment
joining two midpoints of the sides of a triangle is parallel
to the third side and half its length.
Solution Lets start with some arbitrary triangle. Lets
first label the three vertices of the triangle as A. B, and
C labeling in a counterclockwise orientation. Lets now
construct a few vectors naming them with their respec-
tive two points involved. In addition to these points, lets
label the point, D, as the midpoint of ~AB and E as the
59
midpoint of ~AC. Immediately, we can say that:
1
2~BC = ~DE (73)
At this point, make sure your drawing is showing this so
that we are on the same page. In addition, by the way
we placed points D and E, we get that:
~AE = ~EC =1
2~AC (74)
as well as:
~AD = ~DB =1
2~AB (75)
Now lets combine some steps. Using vector addition we
can identity the smaller triangle, ADE , expressing in
vector notation as:
~AD + ~DE = ~AE (76)
60
Which, I can rearrange as a vector subtraction expres-
sion as:
~DE = ~AE − ~AD (77)
In addition, we can now look at the larger triangle, ABC,
and get a synonymous expression from vector subtrac-
tion
~BC = ~AC − ~AB (78)
Lets now plug in the previous expression we get relating
half length in equations 43− 45, to get that:
~DE = ~AE − ~AD =1
2~AC =
1
2~AB (79)
~DE = ~AE − ~AD =1
2( ~AC − ~AB) =
1
2~BC (80)
Therefore, we have shown that ~DE = 12~BC, thus by
equation 35, we see that since we can express ~DE as a
61
multiple of ~BC, then these two sides are parallel to one
another.
5.4 The Dot Product
Now we will move to the more interesting and useful
application of vectors that will be more extensively used
throughout the course. Consider we have two vectors, ~u
and ~v. Then we can express the dot product as:
~u · ~v = (u1, u2) · (v1, v2) = u1v1 + u2v2 (81)
Therefore, we essentially are multiplying together the re-
spective components, and then we are adding all of them
up together to give a scalar (number) value. We note
that the dot product is a measure of how parallel the two
vectors are to one another, and you can think of it as pro-
62
jecting one of the vectors along the direction of the other.
A fact that will help on your Homework (ooo hints in the
lecture notes, another bonus of reading) is that:
~u · ~u = |~u|2 (82)
The reason this is the case is because:
~u·~u = (u1, u2)·(u1, u2) = u21+u2
2 =√u2
1 + u22 = |~u|2(83)
Another really big key equation that we will have LARGE
amounts of time with is:
~u · ~v = |~u||~v| cos θ for θ ∈ [0, π] (84)
63
5.4.1 Small Examples
Lets now apply this to a small example. Consider you
see that:
~u · ~v = 0 (85)
What can we say about ~u and ~v, Well, since we can use
our alternative expression for the dot product involving
angle, we see that :
~u · ~v = |~u||~v| cos θ = 0 (86)
If this equal zero, and neither of the two vectors is just a
zero vector, then the cos θ = 0. Thus, the angle must be
θ = π2 . Therefore, the two vectors are perpendicular or
orthogonal.
Problem Find the angle between a face diagonal and
64
a space diagonal of a cube.
Solution Lets start off by making vector expressions
from the cube. The face diagonal, is basically moving
from the (0, 0) point of the (1, 1). Since we are going to
be using a cube and going into 3D, then lets expression
our first vectors as going from (0, 0, 0) to (1, 1, 0), Thus,
~v = (1, 1, 0) (87)
Now, lets go across the cube. first we need to cross to
the other side like the first vector we created, but then
we almost need to head up to the top corner. Namely,
we need to make it from the origin all the way up to the
point (1, 1, 1), Therefore, we get that:
~u = (1, 1, 1) (88)
65
Thus, let us now deploy our new equation for the dot
product to solve for the angle:
|~u||~v| cos θ = ~u · ~v (89)
|(1, 1, 0)||(1, 1, 1)| cos θ = (1, 1, 0) · (1, 1, 1) (90)
√2√
3 cos θ = 2 (91)
Therefore we can express cos θ as:
cos θ =2√6
(92)
which is enough to calculate what our angle is!
Why is ~i ·~j = 0 for the unit vectors i and j. Well, we
generally use i to represent the x-axis and j to represent
the y-axis. Therefore, the two objects are perpendicular
to one another that will ensure that the dot product is
66
equivalent to zero given our first example in this subsec-
tion.
5.5 The Cross Product
Cross Product is the more annoying brother of the dot
product, so brace yourselves. We will pretty much exclu-
sively calculate the cross product in R3. The formula for
the cross product is the following:
~u× ~v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1) (93)
67
Well2 this looks like a mess, A much more convenient way
to represent this is:
~u× ~v = det
~i ~j ~k
u1 u2 u3
v1 v2 v3
(94)
If you expand out this determinant you will see that equa-
tion 63 and 64 are equivalent, but you can trust me on it.
Now you may ask, what does the cross product do geo-
metrically? The cross product takes in two vectors, say ~u
and ~v and it produces a third vector, say ~w that is per-
pendicular to both ~u and ~v. This is extremely powerful,
and we will use it about the same as how much we use the
dot product throughout the course. In addition, we can2I had a typo before, I am so sorry for this. It has been cleared up now!, always confirm though using
the determinant formula
68
also represent the cross product in the following fashion
that is useful for angle calculation sometimes:
|~u× ~v| = |~u||~v| sin θ (95)
Therefore, if two vectors are parallel, then the cross prod-
uct is exactly zero. Why? Well, if two vectors are parallel,
or even anti parallel, then θ = 0 or θ = π. In either case,
sin θ will always be zero. Therefore, the cross product
must be zero. Since I see what I am about to say come
up a lot let me mention it briefly, in the above formula
(equation 65), note that the LHS is the magnitude of the
cross product! Therefore, look at the RHS. Maybe you
remember, that for a parallelogram with sides A and B,
the area of the parallelogram is AB sin θ. That being
said, this gives us insight that the RHS is representing
69
the area of a parallelogram. Thus, the magnitude of the
cross product, since the LHS = RHS, is also the area
of a parallelogram made by the two vectors. Keep this in
mind for sometime in the future :) The facts above are
extremely useful to know, and I recommend you do the
following problem without any computation below that I
have made for you:
5.5.1 Example Time
Problem Is the following statement True or False.
Please provide a sound argument as to why you think
that it is either true or false: (Take this as a good prac-
tice of your understanding before looking at the solution!)
~u · (~u× ~v) = |u|2 + ~u · ~v (96)
Solution FALSE! Note that the cross product in paren-
70
thesis creates a vector that is perpendicular to both ~v and
~u. Thus, taking the dot product between this vector and
~u must be equal to zero.
5.6 Big Picture
Lets compare our results for the dot product and cross
product. Note that the dot product has a result that
is a scalar quantity, namely just a number. However,
the cross product produces a vector, namely with our
case, a vector in R3, which just means a vectors with 3
components. As a check when you do answer a problem,
make sure that this always remains true. Of course, if it
is useful to work with a number once you calculate a cross
product, then take its magnitude like we see in equation
65. Also, Sam said the cross product is cooler, I disagree
71
#DotProductIsBetter.
5.7 3D Geometry with Lines
We are now edging closer to the calculus portion of
the course! This chapter is generally kind of difficult so
please feel free to always reach out to me. The distances
between object in space I think is very hard, and I will be
uploading a set of notes about them with the recitation
notes on Tuesday, July 9. Okay back to the course. Sup-
pose we want to represent a line in the coordinate grid,
R2, We can write this as:
y = mx + b (97)
We are going to stay in spirit of this by make is more
encompassing. Sorry that the notation is super wack, let
72
me break it down for you. Suppose with have a point on
a line (a, b), that points in direction ~u, lets say that the
slope of the line happens to be m, We can represent the
line as:
(a, b) + t~u t ∈ R (98)
What does this mean? it means that we can start at the
point (a, b), and we can move along the line by adding
~u. In addition, we can multiply by all multiples of ~u, and
we will still remain on the line. For example lets say the
direction of the line is ~u = (1,m). Then, we have the for
each unit of x, we move in the y direction by m, which
looks a lot like a slope right? Perfect! So we can say,
lets pick (0, 0) as a base point on our line, and lets even
73
choose m = 1 for simplicity, then:
(0, 0) + t(1,m) = (0, 0) + t(1, 1) for t ∈ R (99)
Which we can see as we plug in values of t, just gives us
the exact same thing as y = x. Hopefully this helped
clarify things. Lets bump it up to R3. Suppose we want
to write a line from (3,−4, 1) to (2,−1, 4). Lets pick the
first point as our base point, and lets say that at t = 1,
we make it to the second point. Then basically we need
to solve for ~u. This probably sounds a bit weird lets start
working it out,
(3,−4, 1) + t~u = (3,−4, 1) + 1~u = (2,−1, 4) (100)
Alright so we can rearrange this by basically, coming up
with the vector from the first point to the second point,
74
to get the direction. Just note that I picked t = 1 for
convenience, but we didn’t need to. There are an infinite
amount of ways to represent this. I just think always
picking t = 1 helps this out a lot. Okay so, we get that:
~u = (−1, 3, 3) (101)
Okay, so we can express this line as:
(3,−4, 1) + t(−1, 3, 3) = l = (x, y, z) (102)
for the line. Always check that the second point also lies
on the line afterwards. If I plug in t = 1, note that I
indeed get (2,−1, 4), which is exactly what we wanted.
In addition, I also get all the other points that are on this
line by plugging in different values of t. We can equally
75
represent this as:
x = 3− t, y = −4 + 3t, z = 1 + 3t for t ∈ R (103)
Hopefully this is starting to make sense. I struggled with
this section a lot as a student, so please always reach
out with questions. Lets move on to planes. All we are
saying here is that we want to represent all points on the
line. So what we do is we take two points, and we try
to write an equation from one point to the other. When
we include the t factor, we are essentially allowing for not
just a line from one point to the other, but for all points
in between and beyond the two points that run along the
lines existent between the two.
76
5.8 3D Geometry and Planes
Lets start off with a problem to figure this out. Lets
try to write an equation for a plane that passes through
(1, 0, 0), (0, 1, 1), (0, 0, 2). Lets really take our time with
this one. I will write up how to solve this, Saturday, and
add it. I am sorry that the first week is really overwhelm-
ing, I promise that I, and the rest of the TAs, will try our
best to demystify it for you. Lets proceed to do this now.
So we have three points, and you may think, intuitively
is this enough to clasify a plane? Do we need more? per-
haps 4. Lets test this with the real world case. Suppose
we have some plane, lets say the ground of your dorm
room. You are asked by the MIT facilities office whether
you want a desk with three legs or four legs. You say
you don’t really mind so long as the desk is not wobbily.
77
The facilities department immediately hands you a three-
legged desk. Lets explore why. Consider placing one leg
down at a time. The first one will make it crash. The sec-
ond one will make it somewhat more stable, maybe only
allowing it to crash in one or two directions. However,
placing the third leg causes it to be stable. In fact, plac-
ing the third leg is analogous to placing the third point
in R3 when defining a plane’s equation. Now lets add the
fourth leg. If the ground is purely even, then were set!
However, what if it is a little off? Well, I’m sure you have
experienced this in real life before, the table will wobble.
Why? The reason is because that three of the legs are
stablly making contact with the ground, the plane,and
the fourth leg does not necessarily lie on the plane any-
more. The reason is the three points, or legs, are defining
78
some flat space, the plane. Then, adding a fourth leg,
point, is in no guarantee going to lie on that plane! Lets
tackle the actual mathematics of this in order to get into
it.
When we are handed three points, and we want to find
the plane passes through all of them, we are going to
have to make two vectors. The reason is not direct, and
I will discuss it when it appears more apparent later in
the formulaic recipe. Okay, lets start at the point (1, 0, 0)
and write the vector to (0, 1, 1) and (0, 0, 2) as ~v and ~u
respectively.
~v = (−1, 1, 1) (104)
~u = (−1, 0, 2) (105)
Okay great. Now lets bring up why we did this. Well,
79
we want a nifty way to say, here is an equation that is
satisfied for all points that lie on the plane, these three
included, as well as is not satisfied for all points that do
not. So, we want to say that all vectors that lie in the
plane are perpendicular to some vector that is normal
(perpendicular) to the plane. Thus, we can deploy the
cross product now. We utilize the cross product because
the cross product takes two vectors, in this case two vec-
tors in the plane, and generate a vector that is perpen-
dicular to both vectors. Thus, lets take the cross product
of ~u and ~v
~n = ~v × ~u = (2, 1, 1) (106)
These will be called the coefficients of our plane. Thus,
given the previous discussion, we need all vectors that lie
in the plane to be perpendicular to this vector, ~n, namely
80
~n · ~w = 0, for some ~w in the plane. We can construct ~w
by using a similar strategy to getting ~v and ~u. Lets take
some arbitrary point (x, y, z) on the plane, and the point
(1, 0, 0), which we already know is on the plane. We then
can write ~w as:
~w = (x− 1, y − 0, z − 0) = (x− 1, y, z) (107)
Lets now take the dot product between this ’arbitrary’
vector, ~w and ~n to get an equation for the plane:
(2, 1, 1) · (x− 1, y, z) = 2x− 2 + y + z = 0 (108)
2x + y + z = 2 (109)
great! This is the equation for this plane. Lets just test
our three points very quickly to make sure that this does
81
work.
2(1) + 0 + 0 = 2 (110)
2(0) + 1 + 1 = 2 (111)
0 + 0 + 2 = 2 (112)
5.8.1 TLDR Finding Equation of Plane
Looks good. In the future, You can take an alternative
that I myself find more useful. I take the TLDR version
which is:
1. write two vectors, ~v and ~u from the three points in
the plane
2. take the cross product of the two vectors, making vec-
tor ~v × ~u = ~n = (a, b, c)
3. Write the equation ax+ by+ cz = d, where d is some
unknown value I will calculate in the next step of the
82
formula.
4. Plug in a point on the plane to the equation to solve
for the value of d
5. Write the equation ax+ by+ cz = d, where a, b, c are
found from step 2 and d is from step 4. Plug in an
extra point to make sure I didn’t mess up along the
way!
Let me do a quick example right now of the TLDR ver-
sion in action: Calculus the equation of the plane passing
through the points (1, 0, 0),(0, 1, 0), and (0, 0, 1).
1.
~v = (0, 1, 0)− (1, 0, 0) = (−1, 1, 0) (113)
~u = (0, 0, 1)− (1, 0, 0) = (−1, 0, 1) (114)
83
2.
~v × ~u = (1, 1, 1) (115)
3. The equation of this plane can be expressed, letting
a, b, c = 1, 1, 1 as:
x + y + z = d (116)
4. Plugging in the point, (1, 0, 0)
1 + 0 + 0 = d (117)
We get that d is just 1. Therefore we get that the
equation of the plane is:
x + y + z = 1 (118)
84
6 Lecture IV on July 8, 2019
6.1 Review
Press F for Sam’s Microphone. Lets have a quick re-
cap/review for things from last time. Reminder that the
dot product can be expressed as:
~v · ~u = |~v||~u| cos θ (119)
We also can use this to directly show that:
~u · ~u = |~u|2 (120)
In addition, we also mentioned the cross product. A cross
product takes in two vectors, and it creates a third vector
that is perpendicular to both of these vectors. In addi-
tion, the magnitude of the cross product is the area of
the parallelogram spanned by the two vectors. The cross
85
product equation is:
|~v × ~u| = |~v||~u| sin θ (121)
We Also mentioned lines in space. We can write an
equation for a line in space by taking in two points in
space. We need to define this line by some base point,
namely one of the base points, along with a vector that is
in the direction of the line. the magnitude of the vector
does not matter since we can scale it up and down to
reach all points on the line. If we have a base point P ,
and a direction of the line ~u, then we can express the line
as:
P + t~u for t ∈ R (122)
86
Which is just the mathematical representation of the idea
in the previous paragraph. Suppose we want to write a
line passing through the points (0, 0, 0) and (1, 1, 1). We
can then choose (0, 0, 0) as our base point, P , and we can
choose ~u to be the vector from the base point to the other
points, namely ~u = (1, 1, 1). Therefore, we can express
the equation of the line as:
l = (x, y, z) = (0, 0, 0) + t(1, 1, 1) for t ∈ R (123)
6.2 Planes in Space
We mentioned this briefly last time, and I wrote up
some notes over the weekend found in section 5 of my
lecture notes. They are detailed in solving the equation
of a plane. Lets carefully make our way through the prob-
lem mentioned at the end of class.
87
Problem: Find the equation of the plane passing
through A = (1, 0, 0),B = (0, 1, 1), and C = (0, 0, 2).
Solution: We want to start by first generating a vector
normal to the plane that has all three points contained
in it. We do this because all vectors that lie within the
plane will be perpendicular to the normal vector. Thus
lets first come up with a normal vector. We can do this
by making two vectors from the three points in the plane.
We can make a vector from ~AB and ~AC. Thus, we can
write the normal vector as:
~n = ~AB × ~AC (124)
Thus what we can do with this is say, lets pick an ar-
88
bitrary point D = (x, y, z). Then lets make a vector,
~AD that is a vector from the point A to the point D.
Notice that this vector must be be perpendicular to the
normal vector since the vector, AD, is in the plane, and
the normal vector3 is perpendicular to all vectors in the
plane. Therefore, it must be true that:
~AD · ~n = 0 (125)
(x− 1, y, z) · (2, 1, 1) = 0 (126)
2x + y + z = 2 (127)
This is our equation of the plane! Now, all points that
are on the plane will satisfy this equation, and all points
that are not on the plane will not satisfy this equation. A
question that came up in lecture was what happens if we3It was not explicitly calculate but the normal vector for this specific case happens to be ~n = (2, 1, 1)
89
have a parallel plane to the plane we just calculated. Well
lets think about it. The normal vector’s direction cannot
change since the planes are parallel. Therefore, only the
number on the right hand side, 2 is our specific case, will
change. Looking back, we can now read off the normal
vector to the plane looking at a final answer. Namely,
assume you have some plane with constants a, b, c, d ex-
pressed below:
ax + by + cz = d (128)
Then, the normal vector for this equation is:
~n = (a, b, c) (129)
Lets try out an example problem now, by the way, space
is big, lines are small - Sam :
90
Problem: Find the point where the line l = (x, y, z) =
(3 + t,−2t, 3) and x + y + z = 7
Solution Lets see what we can do here. Well lets check
if the line does intersect the plane since there is always
the chance that it doesn’t. Maybe the line passes by the
plane but doesn’t intersect it. In that case, there would
not be a point that is shared between the line and the
plane. We can try to plug in the parametric form of the
line, and we can plug it into the equation for the plane.
Lets try this because we can isolate the ”time” (the value
of t) that the line intersect the plane, and then we can
substitute back in the time to the line equation to find
the specific point that this occurs.
(3 + t) + (−2t) + 3 = 7 (130)
91
t = −1 (131)
We can now plug in this value into the parametric repre-
sentation of the line to find the point that they intersect.
Namely we get that:
(x, y, z) = (3− 1,−2(−1), 3) = (2, 2, 3) (132)
Lets continue on and try another example that will intro-
duce a massive portion of course content that is calculat-
ing distances between objects in space.
Problem: Find the distance from the point (9, 4, 1)
to the line l = (x, y, z) = (1− 2t, 3, t).
Solution: Lets start by extracting two pieces of in-
formation from our line; a base point and the direction
92
vector. We can read off the base point by setting t = 0.
Doing this, we get that P = (1, 3, 0) is a point on the
line. In addition, we can read off the direction vector by
looking at the coefficients in front of t, by equation 92.
So, we get that ~u = (−2, 0, 1). Great this is a lot of good
information that we will need. A naive solution would
be okay, I have a point on the line and another point in
space, I can just use the distance formula between them.
Well, this happens to not be the case. The reason being
is because we are interested in the shortest distance be-
tween the line and the point. While this idea represents
A distance, it is not the distance that we are looking for.
How about we try something else. Lets make a vector, ~v,
which will go from our base point on the line to the point
93
(9, 4, 1) If we do this we get that:
~v = (8, 1, 1) (133)
Great! Now look at what we have thus far. We have a
vector that goes from the line to the point in space. In
addition, we have a vector that points in the direction of
the line. From a purely trigonometric standpoint, we can
represent the distance between the line and the point as:
d = |~v| sin θ (134)
This is a good start. However, we don’t know all that
much about the angle sin θ since it is not very clear. In-
stead, allow me to introduce a small trick. That is, let
me multiply the top and bottom of the expression for
94
distance by |~u|. Doing so we get that:
d =|~v||~u| sin θ|~u|
(135)
Now take a moment and look at what we have here. The
top of this expression is something we are familiar with,
namely this is an expression for the magnitude of the
cross product. We can sub this in (This equation is Equa-
tion 91), to get:
d =|~v × ~u||~u|
(136)
Now we have an expression for distance in terms of the
two vectors that we begun the problem with! This is great
considering the fact that we can solve this just by getting
some magnitudes and solving for a cross product. For the
95
specific numbers utilized in this problem, we obtain that:
d =√
21 (137)
6.3 Vector-Valued Functions
I hope you had a good three-minute break. Lets now
concern ourselves with a class of functions that are con-
sidered vector-valued:
f : Rn −→ R (138)
~r : R −→ Rn (139)
We have previously been dealing with the first of the two
aforementioned functions. We will now deal with the sec-
ond type. Lets try to comprehend whats going on here
with a ”real-world” example.
96
Problem: Suppose you have a bug that is crawling
along the outside of a can. The bug is going to follow a
path that wraps around the can exactly once as it crawls
from the bottom to the top. Describe the path.
Solution: We may want to know about the velocity,
path, and even acceleration of the bug as it travels along
the surface of the can. Lets first try to describe the path.
We want to use a position vector for this that we will
denote as ~r(t). This is something we see all the time in
a physics classroom. We write the vector as a function
of time, namely the components of the vector change as
a function of time. Lets make some assumptions so that
we can come up with some path. Lets say that the height
of the cylindrical can is 1, the time it takes for the bug
97
to reach the top is 1, and finally the radius of the cylin-
drical can is also 1. lets also say that the bug starts at
the position, (1, 0, 0), and it makes it way up to the point
(1, 0, 1). Intuitively, the bug has one unit of time to travel
a distance 1 to the top of the cylinder. Therefore the z
component of the path should simply be t. This is a
steady rise up the can. We can get the x and y compo-
nents by going around a circle. The parametrization for
a circle will always be (~x(t), ~y(t)) = (cos 2πt, sin 2πt)4.
Thus we can get the position of the bug as:
~r(t) = (cos 2πt, sin 2πt, t) (140)
If we want to then go on to calculate the velocity of
the bug at some time along its journey we can take the
derivative of the position vector component by compo-4Generally when you parametric a circle you will get just simply cos t, sin t where t ∈ [0, 2π]
98
nent. Namely,
~v(t) =d~r(t)
dt(141)
For our case, we can take the derivative of our bug’s po-
sition to obtain:
~v(t) = (−2π sin 2πt, 2π cos 2πt, 1) (142)
We can further differentiate velocity to obtain the accel-
eration by the equation:
~a(t) =d~v(t)
dt(143)
Lets do another example problem for a path to get more
familiar with the idea of time-varying vectors.
Problem: Find the path traced out by a point on a
rolling bike wheel with unit radius, and unit speed.
99
Solution: Lets have the point start at the bottom
of the bike wheel. In addition, lets start it at the origin.
Well first off, we don’t need all 3 dimensions. We can have
instead just x and y components. The first thing to notice
is that while the center of the wheel is constantly moving
down the block with a constant speed (unit speed of 1
in this case), the specific point on the wheel is oscillating
back and forth as it makes it way up and down. The
center of the wheel can be described for all times, t, as:
Center = (t, 1) (144)
Now, we have take out essentially the transitional motion.
Now the only motion that we have left is essentially the
rotational motion. We only have circular motion left.
100
Therefore, the point will adopt all of the properties of
the center’s translational motion whilst also including its
own circular motion as well. With the way we started the
picture, we need the point to start at (0, 0). Therefore,
we obtain that the path is:
~r(t) = (t, 1) + (− sin t,− cos t) = (t− sin t, 1− cos t)
(145)
6.4 Quadric Surfaces
Quadric surfaces are like quadratic surfaces, but in 3D.
They are the 3D analogs. You actually have come in
contact with some of them in our first recitation when we
were curves for things such as 1 = x2 + y2 + z2. We will
be graphing these in R3. What does this look like? It
looks like a sphere! However, this one comes off as simple
101
since we are probably somewhat familiar with this type.
Of course, there is a more formulaic way of dealing with
the general surface that will involve the level curves that
we covered. For example, what happens when we have
x2 + y2 − z2 = 1. This is something I sure do not know
what it looks like off the top of my head. However, lets
start taking slices of z for our equation. For example,
maybe we have that z = 0. Now we have a circle of
radius one centered at the origin. Now lets take a slice at
z = 1, Now, we have that x2 + y2 = 2. Now we have a
circle of radius,√
2 at Z=1. We also get the exact same
picture for z = −1. If we path these slices together, we
have what looks like an hourglass. See you tomorrow in
recitation!
102
6.5 All other Path in Space Stuff (ASE)
While I do not anticipate this section really coming up
on the exam, I will include it just so that we have some
record that it is taught at least the semester that I took
it! When we discuss paths in space, we sometimes refer
to concepts such as tangent vectors, normal vectors, bi-
normal vectors, and curvature. Each one is honestly just
a formula, and it doesn’t necessarily offer much other than
helping you solve problems that ask you to solve each of
these types. The tangent vector, is defined as:
~T (t) =~r′(t)
|~r′(t)|(146)
Which, upon first glance is just the velocity vector di-
vided through by its magnitude. The reason why this
is called the tangent vector is simply because it denotes
103
that direction of the velocity vector whilst omitting the
magnitude of the velocity vector. In addition to the tan-
gent vector, we can also discuss the normal vector that
is defined as:
~N(t) =~T ′(t)
|~T ′(t)|(147)
6.5.1 A Proof of Orthogonality
The normal vector as defined above is always perpen-
dicular to the tangent vector! We can quickly write up
a proof for this. Consider the tangent vector as defined
above. You can directly see that the tangent vector has a
constant magnitude for all time by definition. I will now
take use of this fact so that we obtain:
~T (t) · ~T (t) = |~T (t)|2 = 1 (148)
104
Since ~T (t) is a unit vector. As such, lets now take the
derivative of such a dot product:
d
dt
(~T (t) · ~T (t)
)= ~T ′(t)·~T (t)+~T (t)·~T ′(t) = 2~T (t)·~T ′(t)
(149)
However, remember that we have already shown that
~T (t)cdot~T (t) is a constant value. Therefore, this deriva-
tive must be equal to zero. As such we have that:
d
dt
(~T (t) · ~T (t)
)= 2~T (t) · ~T ′(t) = 0 (150)
2~T (t) · ~T ′(t) = 0→ ~T (t) · ~T ′(t) = 0 (151)
As such, I can simply divide this expression by |~T ′(t)|
without changing the fact that this will still be equivalent
to zero.
1
|~T ′(t)|~T (t) · ~T ′(t) = 0 (152)
105
~T (t) ·~T ′(t)
|~T ′(t)|= ~T (t) · ~N(t) = 0 (153)
Showing that given we way we have defined both the
normal and tangent vectors, they must be orthogonal for
all t. We can finally define our last vector that is rarely
asked about. But, in the case that it is, know that we
define the binormal vector as:
~B(t) = ~T (t)× ~N(t) (154)
If I were you, I would just be comfortable with tangent
and normal vectors. I think there is a pretty much zero
chance you are asked about a binormal vector. In the
case they do, I believe they would probably give you the
formula and make you compute it to see if you could com-
plete the cross product! The last application of the con-
106
cepts that we have just learned is curvature. Curvature
measures how smooth a curve it. It requires that ~r′(t) is
continuous and that the magnitude, |~r′(t)| 6= 0. The way
I think of curvature is that it is a measurement of how fast
we are changing direction like in circular motion. There
are two definitions that we will come in contact with. Use
whichever one is easier for the given problem. Here is the
formula, where we denote curvature by the greek letter,
κ:
κ =|~T ′(t)||~r′(t)|
=|~r′(t)× ~r′′(t)||~r′(t)|3
(155)
We utilize these two formulas when we are given some
~r(t). In the rare case that instead of providing use with
a ~r(t) expression, we are instead given y = f (x), so that
we could express ~r(t) = (x, f (x), we get the following
107
condensed form of the curvature expression:
κ =|f ′′(x)|
(1 + [f ′(x)]2)32
(156)
I wouldn’t think of this section as anymore than a col-
lection of new formulas that just utilize tools of paths
in space that we learned throughout the course. I do not
think that these formulas are commonplace even semester
by semester at MIT, so I wouldn’t bank on these on the
ASE. However, in the case that they are, these are the
formulas that represent the collection of.
Finally, I just wanted to add that if you are asked to
find the arc length of a curve, the following formula can
be utilized:
S =
ˆ b
a
||r′(t)||dt (157)
108
Where the path starts at time a and terminates at time
b
7 Recitation II on July 9,2019
First off, great job on the quiz! So, we covered a lot,
so I want to take some of this space to summarize how
to tackle a couple of the most common points, lines, and
planes in space questions that require utilizing a Copi-
ous amount of vector mathematics to solve. That being
said, lets starts going through them case by case in a
systematic approach. I will try to teach each with an
example. Here we go:
109
7.1 Point to Point
This is the most simple case and will seem like things
you are most likely familiar with given a point, (a, b, c),
and another point, (g, e, f ), Then the distance between
these two points is denoted as,
d =√
(a− g)2 + (b− e)2 + (c− f )2 (158)
This is just Pythagorean theorem in three dimensions!
7.2 Point to Line
Lets teach this one through a direct example. Suppose
that we have a point (1, 3, 4) out in space and the line
defined by the equation, l = (x, y, z) = (2− 2t, 3 + t, 4t).
We will then be asked if we can find the shortest distance
between this line and the point. This is referring to the
110
perpendicular distance that is the move for all of these
types of problem! Okay lets break this down into a recipe.
I will say, like with all of the distance formulas, you can
get creative and this is not the only way to do them.
1. Find a point on the line and the vector that denotes
the line’s direction. We can find a point on the line
by plugging in t = 0, with that, we get that the point
(2, 3, 0) is on the line. In addition, the direction of the
line is the coefficients of t. Doing so, we get that the
direction of the line is, ~u = (−2, 1, 4). We completed
step one!
2. Write a vector from a point on the line to the point
out in space. Okay so we already have both points
in question. The point on the line is (2, 3, 0) and the
point in space is (1, 3, 4). Therefore, the vector from
111
the first point to the second point is, ~v = (−1, 0, 4).
Perfect! Notice that the magnitude of this vector rep-
resents some distance from the line to the point, but
it does not represent the perpendicular distance be-
tween the line and the point.
3. Trigs and Tricks. Okay, so going off the rift at the
end of the second step, we can use some right triangle
trigonometry to calculate the perpendicular distance.
Namely,
d = |~v| sin θ (159)
So that was the trig. I diagram would be helpful to
draw out yourself, but I unfortunately don’t know how
to add that to the latex file. Here comes the trick. Be-
cause we don’t really know what the angle exactly is,
we want to get rid of it. We can do this by multiplying
112
and dividing our expression for distance by |~u|. Why,
because now the numerator of our function is the ex-
pression for the magnitude of the cross product. We
can represent this discussion in equation form as:
d =|~v||~u| sin θ|~u|
(160)
d =|~v × ~u||~u|
(161)
7.3 Point to Plane
Lets lead by example again. So planes are spoken about
in terms of their normal vector. Okay, so if we have a
plane denoted by the equation, 2x+ 3y− z = 6. We can
pluck off the coefficients of of the normal vector by looking
at the coefficients in front of x, y, and z. Therefore, for
this case we have that ~n = (2, 3,−1). Now lets say I want
113
to find the distance between this plane and the point,
(2, 2, 2). Lets do this systematically again.
1. Write a Unit normal vector from the normal vector
expression. So in order to change our normal vector, ~n
to the form of the unit normal vector by the following
equation:
~n =~n
|~n(162)
So, in our particular example we have that the magni-
tude of our normal vector is ]√
14. Thus, we get that
our unit normal vector is denoted as:
~n =1√14
(2, 3,−1) (163)
2. Write a vector from a point on the plane to the point
out in space. Okay so a point on the plane must
satisfy the plane’s equation. So there are a ton of
114
choices that are fine. I’ll just pick (1, 1,−1) because
that lies on the plane. So now we need to write a
vector from (1, 1,−1) to (2, 2, 2). The vector would
be:
~v = (1, 1, 3) (164)
3. Dot the vector, ~v with the unit normal vector to get
the distance. Why are we doing this? Well lets think
about it. We only want the component of ~v that lies
along the normal direction. Thus, if we take the dot
product of the ~v with the unit normal, we will simply
extract the components of the distance that lie along
the arbitrary vector ~v, and we only take the stuff in
the unit normal direction. This is, we get that:
d = ~v · ~n = (1, 1, 3) · 1√14
(2, 3,−1) =2√14
(165)
115
Of course, there are alternative ways to do this, let for
example stating that |~v| cos θ = d, then multiplying
the top and bottom by |~n|. Same exact results and
the same exact steps in all honesty. Just a different
approach.
7.4 Line to Line
Here we in my opinion one of the hardest to visual-
ize. Unlike in R2, we now have lines that can be skew.
Consider we have to lines, L1 = (2 − t, t, 4 + 3t) and
L2 = (1− 2t,−1 + t, 2t), and we want to know the per-
pendicular distance between the lines. Lets start off by
gaining some insight on the lines. Namely, lets define one
point on each line and also compute the lines direction.
L1 contains the point (2, 0, 4) and has a direction denoted
116
by the vector, ~v = (−1, 1, 3). Please see the section in
the notes if getting this part of the information is difficult
in section 6. In addition, L2 contains the point (1,−1, 0)
and has a direction denoted by the vector, ~u = (−2, 1, 2).
With all that information close enough to sniff, lets start
the process of getting the answer to this type of question.
1. Write a unit normal vector generated by the cross
product of the two line’s directions. So what does
this mean? It means that if we want to find the per-
pendicular direction that exists between the two lines
that are behaving as vectors, we can take the cross
product of line 1’s vector with line 2’s vector. For our
specific example we have that:
~n = (−1, 1, 3)× (−2, 1, 2) = (−1,−4, 1) (166)
117
In order to transform this into a unit vector, since it
will come to play later, lets compute the magnitude of
this cross product and divide through by it, namely:
~n =~n
|~n|(167)
With the magnitude of√
18, we can express the unit
normal vector as:
~n =~n
|~n|=
1√18
(−1,−4, 1) (168)
Great step one done.
2. Write a vector from a point on one line to a point on
the other line. If you remember back to the beginning
of this subsection, we found a point on each line. We
have that (2, 0, 4) is on line 1 5, and (−1, 1, 0) is on5Call center amiright
118
line 2. Therefore the vector that connects the two is:
~w = (−1,−1,−4) (169)
Dot the unit normal vector with the vector between
the two points. So this is really similar to the point
and the plane? Why might this be the case? Well lets
take a moment to try and internalize it. In step one
we took the cross product of the two vectors, which
essentially is creating a normal vector, a plane type
thing, from the two line vectors. We are then taking
this orthogonal vector to both of the line’s direction,
and we are dotting it with some vector from one line
to the other. What is this doing? The dot product is
essentially filtering out any of the distance that is not
strictly perpendicular between the two lines, and its
119
result is the distance between the two lines, namely:
d = ~hatn· ~w =1√18
(−1,−4, 1)·(−1,−1,−4) (170)
As always, if the distance turns out negative just take
the magnitude of this. This just means, since we are
working with vectors, that a direction we took hap-
pened to be the opposite, and there is not real mean-
ing besides this.
7.5 Line to Plane and Plane to Plane
For this one, we are at the are seemingly easier cases.
In both of these cases, we either are going to have the line
intersect the plane or it has to be parallel to the plane.
If it intersects the plane, then intuitively, the distance
between the two is zero. If the two are perpendicular. In
120
both cases, in order to not repeat myself, for both these
cases, simply pick a point on the line or plane and then
treat the problem like a point to plane problem!!!! The
exact same way and you should be perfect :))) Hope all
this helped! So yeah, that was a lot, but I hope it all
makes sense!! Also, sorry for the delay in the posting
of this! Let me do one with parallel planes just to have
one in the notes. Find the distance between the planes
x + y + z = 4 and x + y + z = 5. First off how do
we know that they are parallel. Well, since their normal
vectors are parallel, then it must be true that their plane
surfaces are also parallel.
1. Compute the unit normal vector of the plane. For this
plane we have that: ~n = (1, 1, 1). Thus if we would
like to calculate the unit normal of this, we would
121
obtain that:
~n =~n
|~n|=
1√3
(1, 1, 1) (171)
2. Calculate a vector from one point on the first plane
to a point on the second plane. For convieance, since
there are enourmous amount of options to choose from,
i’ll choose the point (4, 0, 0) from the first plane and
(5, 0, 0) from the second plane. Doing this, we obtain
that the vector from the first point to the second is:
~v = (1, 0, 0) (172)
3. Take the dot product between ~v and the unit normal.
We do this to compute the distance by essentially pro-
jecting ~v along the direction of the unit normal vec-
tors. Namely, we are extracting all of the perpendic-
122
ular distance from the ~v by dotting it with the unit
vector. For our case we have that:
d = ~v · ~n =1√3
(1, 1, 1) · (1, 0, 0) =1√3
(173)
Jeez that was a lot. Hopefully it is helpful throughout
the course :)
8 Lecture V on July 10, 2019
Lets kick off lecture 5 with a little bit of review from
last time. We ended lecture talking about a quadric sur-
face. A quadric surface is a 3D analog of parabolic type
curves, now we have quadratic surfaces. In order to best
draw quadric surfaces, we make it simpler for ourselves by
taking z = c for some constant c, and we look at how the
equation look in two variables, namely making some flat
123
shape. By us taking these slices, we get a conic section.
We place these flat shapes at the specific levels of z. We
then can construct the surface together by grouping all
of the levels curves together. Consider the example of:
x2 + y2 − z2 = 1 (174)
lets move over the z, and then set z = c and pick some
constants such as z = 0, 1, 4.
x2 + y2 = 1 + z2 (175)
x2 + y2 = 1 + c2 (176)
where we would then plug in our specific values of z, note
the resulting circle located at these specific values of c and
124
be able to graph these circles. Imagine now you have:
x2 + y2 = z2 (177)
Then notice, at the slice at z = 0, there is not necessar-
ily a circle, but instead, there is exactly a point. Lets
try to think about the exact shape of the aforementioned
equation. Note that if we take z slices, we actually re-
sult in what appears to be a cone. Moreso, we do not
necessarily have a cone just above the xy plane, but we
also have a cone below the xy plane. Indeed, the points
of each cones share the origin, and then expand either
above or below the plane into their conic shape. Lets
consider more interesting cases. Consider the case of:
x2 + y2 = −1 + z2 (178)
125
Now we need to do a bit more thinking. With this, imag-
ine we set z = 0, do we have a legitimate solution? no.
Why? because the smallest that x2 + y2 can be is zero.
As such, setting it equal to a negative value will not con-
struct any surface. Now, the first values where we start
to see a surface is at z = ±1. As such, instead of getting
a single surface across the space. We now have a surface
that has its lowest value at z = 1 and another surface
that has its largest value at z = −1. There exists a space
between them where there is no surface. Namely, there
are no surfaces for z ∈ (−1, 1). We’ll finally end with the
parabolic analog. Consider the function:
z = x2 + y2 (179)
126
Where you will get circles for each z = c slice for c ∈
[0,∞) that are increasing in radius, with radius equal to
√c. This looks just like a parabola but in 3-dimensional
space. In fact, for those that have seen polar coordinates
before, note that this is the function z = r2. So it is
a parabola in this type of coordinate system. But, if
you don’t know this, do not worry we’ll get a done of
practice with this very shortly. We can add all type of
transformations to the parabolic equation above, like
z = −x2 − y2 (180)
we just type the graph and flip it below the xy plane.
The final one we can look at is:
z = −x2 + y2 (181)
127
Lets try the technique we have been learning. Notice
that at the z = 0 slice, we obtain that y = ±x, we we get
a set of criss-crossing lines. And, as we start increasing
our z slices, we start to get hyperbolas. This is very hard
to see, but we are basically graphing a pringle! I attach
the following image for clarity.
Figure 1: Pringle 6
6Do not copy-strike me.
128
8.1 Polar, Cylindrical, and Spherical Coordinates
Now we are getting to a very important section. We
will be switching between all three coordinate systems,
Cartesian, Cylindrical, and Spherical all the time. Before
we introduce our new coordinate systems, lets take a step
back and take a bit more of a formal approach on our un-
derstanding of the Cartesian system. When we describe
x, we can define x as being the signed distance from the
y-axis. In 3D, we can represent x as the signed distance
from the yz- plane. A coordinate system in general is
an object that gives you enough specific location to find
the actual point that you are trying to describe. We can
make analogous arguments for both the y and z coordi-
nate by describing them as the signed distance from the
xz and xy plane respectively. Together, all three coor-
129
dinates together given enough information to specific a
point. More generally, a coordinate system on Rn is a
set of a function, f : Rn −→ R that can be used to
uniquely identify points in Rn. For example, in the case
of polar-coordinates, we note that (r, θ), where r(P ) is
the distance from the origin to the point, P . In addition,
θ(P ) is the signed angle between ~OP , the origin to the
point vector, and the positive x-axis. Lets start the table
of ”conversions” between Cartesian and other coordinate
systems.
8.1.1 Polar Coordinates
We can summarize the relationships between polar and
Cartesian coordinates as:
r2 = x2 + y2 (182)
130
θ = arctany
x7 (183)
We can also head in the opposite direction:
x = r cos θ (184)
y = r sin θ (185)
8.1.2 Cylindrical Coordinates
Cylindrical coordinates are the bigger brother to polar
coordinates. They adopt the same idea of polar coordi-
nates and add the z-direction. However, the z direction
is the same in both the Cartesian and Cylindrical coordi-
nate systems. In words, z is the distance to the xy-plane,
r is the distance to the z-axis, and θ(P ) is the signed
angle from the p-containing half plane whose boundary7This formula is pretty good. However, please make note of which quadrant the angle actually is in since
this function will not necessarily produce the correct one.
131
is along the z-axis.
r2 = x2 + y2 (186)
θ = arctany
x(187)
z = z (188)
We can also head in the opposite direction:
x = r cos θ (189)
y = r sin θ (190)
z = z (191)
Lets add an example here so that we can see how to graph
a system of inequalities, something that is a very powerful
tool that you will see come up all the time. Suppose we
have the following 3 inequalities that we are supposed to
132
graph in conjunction, a system.:
r ≤ 4 (192)
0 ≤ θ ≤ π
3(193)
0 ≤ z ≤ 2 (194)
Here is the resulting image. The strategy here is that we
want to say that any point, in three dimensional space
that satisfies all three of the above inequalities, then the
graph of all points that do this is the resulting graph found
below. I borrowed the illustration from Sam’s book, and
I do not own nor did I make this graph. We started this
by first graphing the first inequality which is a cylinder of
radius 4. However, now as we move to the second inequal-
ity, now we have to get rid of all the point in the cylinder
that do not have a theta coordinate that is, θ ∈ [0, π3 ].
133
Even here, we are not done! We now deploy the third
inequality that limits the values of z for, z ∈ [0, 2]. As
such, we cut off the points in our wedge that do not have
a z coordinate lying in the specified range for z.
Figure 2: Graph of Inequalities
A proper word explanation is that the points satisfying
r ≤ 4 are in a cylinder of radius 4 centered along the
z-axis. The points satisfying, 0 ≤ θ ≤ π3 are between the
two θ half planes, and the points satisfying, 0 ≤ z ≤ 2 are
between the z = 0 and z = 2 planes. Now its time for the
biggest and baddest of them all, spherical coordinates.
134
8.1.3 Spherical Coordinates
Lets first get the conversions up on the board so that
we have a good starting place:
ρ2 = x2 + y2 + z2 (195)
φ = arccosz√
x2 + y2 + z2(196)
θ = arctany
x(197)
And now, in the other direction:
x = ρ sinφ cos θ (198)
y = ρ sinφ sin θ (199)
z = ρ cosφ (200)
We can think, for some point P , that ρ(P ) is the distance
from P to the origin. The way we define θ is the same
135
as we do for cylindrical coordinates. Finally, φ(P ) is the
angle between ~OP , vector from the origin to P , and the
positive z-axis. An important note is that θ is bounded,
namely 0 ≤ θ ≤ 2π. φ is also bounded, 0 ≤ φ ≤ π.
9 Lecture VI on July 11, 2019
Today we are starting chapter 4, which is the first chap-
ter of multivariate calculus type stuff! Get hype! It does
start off with limits, which tend to be the most out there
of subjects. We are going to try today to get familiar
with the concept of multivariate limits.
9.1 Limits
Lets start off by talking about the single variable idea
of limits. We can think of it as what a function of doing
136
as a function approaches a specific value, arbitrarily close
to, but not at, the point. Lets get some vocabulary down
before we dive into the core idea. We state that a func-
tion is bounded below if its range is a subset of [a,∞) for
some a ∈ R What this is saying is that the function, f ,
does not have an output value that is smaller than a. The
greatest lower bound of a function, f , is the largest a such
that the range of f ⊂ [a,∞). This sideways U , is just a
sign for subset of. Imagine now, that you have a function
defined on the unit interval that is increasing. By strictly
increasing, I mean that f (a) ≤ f (b) for a < b. We can
write our function as: f : (0, 1) −→ R. Remember that
the function is always increasing. Therefore, if we want
to compute the limit at 0, even though the function is
not defined there, we can note that our function the way
137
it is drawn is bounded below by 2, then the limit as we
approach zero is 2.
Definition: If f is an increasing function on (0, 1),
then we say that limr→0 f (r) is the greatest lower bound
of f . We can have the same idea for decreasing func-
tion, namely if f is a decreasing function on (0, 1) then
we say that limr→0 f (r) is the smallest upper bound of
f . We aren’t encompassing everything though with this
idea. We are only looking at decreasing and increasing
functions, which is totally limiting a massive amount of
functions. We also are only dealing with single-variable
function that obviously may be a problem in a multivari-
ate class! Lets now drop these assumptions on f as we
trudge forth.
138
Definition: If ~a ∈ Rn and r > 0, the punctured ball,
B∗(~a, r) is the set of points, {~x ∈ Rn : 0 < |~x−~a| ≤ r}.
This is why we call this a punctured ball, let me break
down this notation. We take all the points that are con-
tained in a radius, r from some point a. We look at the
set of points in the ball, but we omit the point right at
the center, a. In the case of functions, f : R2 → R the
punctured ball is really just a punctured disk Lets con-
tinue with this:
Definition: Suppose D ∈ Rn and ~a ∈ Rn and that
f : D → R. We define, [m(r),M(r)], as the small-
est closed interval containing the image, range, of the
punctured ball, B∗(~a, r) ∩ D under f . The last thing,
139
B∗(~a, r) ∩ D means the points that are both in the im-
age of the punctured ball and the domain. Both of the
functions, M(r) and m(r) are f : R→ R, meaning that
they take in a radius value and they output another single
value. We say that the limit of f (~x) as ~x → ~a exists if
m(r) and M(r) converge to a common value L, we write:
lim~x→~a
f (~x) = L (201)
Wherever you see ~x note that this is a vector of values so
think (x, y) or (x, y, z) instead of the single variable case
of just x, that you came in contact with previously. Lets
consider the function
f (x, y) = x2 − y2 + 3 (202)
140
Well lets first convert this using polar coordinates:
x = t cos θ (203)
y = t sin θ (204)
f (x, y) = 3 + t2 cos2 θ − t2 sin2 θ (205)
f (r, θ) = 3 + t2(cos2 θ − sin2 θ) (206)
f (r, θ) = 3 + t2 cos(2θ) (207)
Therefore, since all cosine functions are bounded above
by 1 and below by -1, we can construct our m(r) and
M(r) by hitting the bounds for cosine since nothing else
is limiting it. I will say that this is the method we will
most likely doing throughout the rest of the limit section
because we cannot always ’guess’ what we think the two
m functions are going to be just by looking at it.Just a
141
reminder that t ≤ r. Therefore, as a last step, we es-
sentially sub out t with r since we are worried about the
biggest and smallest. Instead, we convert to polar coor-
dinates, and then we make sure that we pick a function
for m(r) and M(r) that are only function of r and not a
function of θ. It is important to know some of those trig
identities! Thus:
m(r) = 3− r2 (208)
M(r) = 3 + r2 (209)
Note that if we take the limit as r → 0, the two functions
do converge to the same value, namely 3. Here is a photo
of M(r) in purple ,m(r) pink in and then f (x, y) in blue.
142
.
Figure 3: Limit Functions
You can see that we approach the value of 3 at the same
point where they all meet! It does not need to be as
hand-wavy, you can see that I used the bounded nature
of trig functions to get the same values as Sam. Lets do
another example in order to try to get this down:
143
Problem: Determine whether :
lim(x,y)→(0,0)
(−xyx2 + y2
)(210)
Solution: We are going to be interested in whether as
we move towards the origin from multiple directions, if we
achieve the same limit. We cannot just simply complete
this problem right at the origin due to the fact that the
function is not even defined at the origin. Lets write that:
x = t cos θ (211)
y = t sin θ (212)
We can now substitute this into our function to achieve
144
that 8:
f (x, y) = f (t) =−t2 sin θ cos θ
t2= −1
2sin 2θ (213)
Now lets look at the biggest and smallest our function
f can be. Namely, we can achieve a largest value of 12
and a smallest value of −12 , As such we found our M(r)
and m(r) respectively. Therefore, as r → 0, we see that
M(r) and m(r) converge to different values, 12 and −1
2 .
Therefore, the limit does not exist. Lets have a couple of
other tools, in the back of our toolkit:8reminder that 2 sin θ cos θ = sin 2θ
145
9.2 Other tools for limits
9.2.1 Alternate Paths
Consider that you have two paths, ~r1 and ~r2 in Rn with
that property that:
limt→0
f (~r1(t)) 6= limt→0
f (~r2(t)) (214)
with ~r1(0) = ~r2(0) = ~a, Then we state that lim~x→~a does
not exist. This does not mean that if two paths do hap-
pen to have the same limit, that the limit does exist.
Why? There are an infinite amount of paths, so just hav-
ing two approach the same value does not actually allow
us to say it exists. We would need to turn to our M(r)
and m(r) notation used before that. This is useful trick
in the case that the limit does not exist. For example,
sometimes you may try plugging in paths like y = x or
146
y = 0 to show that the limit does not exist at the origin
perhaps if they lead to different coordinates.
9.2.2 Continuity
Definition: f is continuous if its values equal its limit.
Theorem
1. x, y, z are continuous
2. sums and products of continuous functions are con-
tinuous like (x + y + z, xyz, x2 + y2, etc)
3. Compositions of continuous functions are continuous
(exy+z) for example. Most functions that we are deal-
ing with should be continuous.
lets now turn to an example for the alternative paths
example:
147
9.2.3 Examples of Using the Further Techniques
Suppose we want to consider all directions at the exact
same time. We can let the angle of approach be θ. We
can make a substitution that:
x = t cos θ (215)
y = t sin θ (216)
When we do this we have to be very careful. Why? Be-
cause by making the substitution only takes into account
straight paths, along a specific value of θ, but we are not
taking into account any curvy path. Like maybe a pos-
sible candidate could be y = x2 or even y = x3. These
are curvy paths that were not tested by just making a
polar coordinate conversion. We will cover this problem
in lecture tomorrow so be on the lookout for that! Also
148
just a reminder, Sam is still teaching at the moment, and
his plane is leaving in 56 minutes lol. I have attached an
image for clarity on the subject. We see that the limit
appears to exist along all straight lines, but if you take
the y = x2 curvy path the origin, you will reach a differ-
ent limit value. As Sam said, you have to surf your way
to the origin along the curvy paths.
Figure 4: Different Curves to the Origin
149
10 Recitation III on July 12, 2019
Great work today in recitation, and thank you to Kla-
jdi’s half section for joining us. I just thought it would be
useful to put some examples I made from the worksheet
straight into the lecture notes. So there is not necesasily
anything new in this section that isn’t from the work-
sheets, but it hopefully helps in organizing your studying.
That being said lets just go through a few of the problems
for us to get more comfortable with limits.
Problem: Show that lim(x,y)→(0,0)
(x2 + y2)32(1− sin2(x2 + y3)) = 0.
By showing that M(r) and m(r) converge to 0.
Solution: Lets start as we have in the past by plugging
150
in polar coordinates:
f (x, y) = f (t, θ) = (t2)32(1−sin2(t2 cos2 θ+t3 sin3 θ)) = t3 cos2(t2 cos2 θ+t3 sin3 θ))
(217)
Now we are close. Notice now that the cos2 curve is
bounded above by 1 and below by 0, irrespective of the
argument. therefore, we can bound f (x, y).
0 ≤ f (r, θ) ≤ r3 (218)
As such we arrive on the fact that M(r) = r3 and m(r) =
0. Therefore, as we take the limit as r −→ 0, we get that
the limit exists and is equal to zero.
Problem: Show that lim(x,y)→(0,0)
(x2 + y2) sin1
x2 + y2= 0.
By showing that M(r) and m(r) converge to 0.
151
Solution: Lets start as we have in the past by plugging
in polar coordinates:
f (x, y) = f (t, θ) = t2(sin2 θ + cos2 θ) sin1
t2= t2 sin
1
t2
(219)
Now we are close. Notice now that the sine curve is
bounded above by 1 and below by -1. therefore, we can
bound f (x, y).
− 1r2 ≤ f (r, θ) ≤ 1r2 (220)
And as such, we get that M(r) = r2 and m(r) = −r2.
Therefore, as we take the limit as r −→ 0, we see that
the limit on both sides approaches zero, and as such the
limit exists and is zero.
Problem: Show that lim(x,y)→(0,0)−x2yx4+y2 does not exist
152
even though the limits along every line through the origin
exist and are equal.
Solution: First, let us prove that the limit converges
to a value if we approach it through any line y = mx by
doing a substitution:
lim(x,y)→(0,0)
−x2y
x4 + y2= lim
x→0
−x2mx
x4 + (mx)2(221)
= limx→0
−mx3
x2(x2 + m2)(222)
= limx→0
−mx(x2 + m2)
(223)
=0
0 + m2= 0 (224)
We’ve proved that the limit converges for any linear
approach to the origin, however, that doesn’t guarantee
that the limit will converge to the same value for any type
153
of approach. For instance, we could approach the origin
through a parabolic track of the form y = ax2 in which
case the limit becomes:
lim(x,y)→(0,0)
−x2y
x4 + y2= lim
x→0
−x2ax2
x4 + (ax2)2(225)
= limx→0
−ax4
x4(1 + a2)(226)
= limx→0
−a(1 + a2)
=−a
(1 + a2)(227)
which, similar to that previous problem, depends on the
specific parabola we use to approach the origin (in this
case determined by the value of a). We therefore con-
clude that the limit does not exist.
Hopefully the limit stuff is all down. The most impor-
tant thing to get out of it, in my opinion is M(r) and
154
m(r) which really comes down to picking a floor and ceil-
ing for your function. Basically, we are saying that our
function is never larger thanM(r) and never smaller than
m(r). Then, if the floor and the ceiling are converging
towards the same values, namely closing in on the center
of the room, we get the limit exists at that point, and it is
equal to that said value. See you on Monday! One week
until the exam :)
11 Lecture VII on July 15, 2019
11.1 Partial Derivatives
Today we are getting to derivatives finally! Lets take
a step back and generalize the derivative from single-
variable calculus. Derivatives are really just seeing how
much the function output changes as we change the input
155
slightly. We see that:
f (a + h)− f (a) ≈ 0 (228)
for some small h. This captures the idea mentioned above
that the function can increase or decrease as you move
a small amount away from a but for really small h the
change is not very large. If we want to gain more infor-
mation, we can instead look at:
f (a + h)− f (a)
h(229)
Which, as we take the limit as h goes to zero, becomes
the formula that is used for a derivatives of the function
f, namely:
f ′(a) = limh→0
f (a + h)− f (a)
h(230)
156
If we zoom in at this at this point, we will see a straight
line. While the function itself might be curvy all over the
place, if we zoom in so much, we see that the function
appears to be linear, and as such, we can think of the
derivative of f at a is just the slope at that point. It
tells us how sensitive f is to small changes in the input.
Lets take a step up into 2 variables so that we can handle
multivariable differentiation.
If f : R2 → R, then we define the partial derivative of
f with respect to x and y respectively as:
∂xf (a, b) =∂f
∂x(a, b) = lim
h→0
f (a + h, b)− f (a, b)
h(231)
∂yf (a, b) =∂f
∂y(a, b) = lim
h→0
f (a, b + y)− f (a, b)
h(232)
Effectively what we are doing is saying let me hold one of
157
my variables constant and only look at changes in the
other. We see that for ∂xf (a, b), we are just holding the
y-variable constant and looking at a small change in x to
remark on how sensitive f is with respect to changes in
x. Lets try an example of taking partial derivatives:
Problem: Differentiate ex sinxy with respect to (w.r.t)
x and y.
Solution:
∂x(f (x, y) = ∂x(ex sinxy) = ∂x(e
x) sinxy+ex∂x(sinxy)9
(233)
∂x(f (x, y) = ex sinxy + yex cosxy (234)
∂y(f (x, y) = ∂y(ex sinxy) (235)
9I am using product rule here since x comes up in both of the terms!
158
∂y(f (x, y) = ex∂y(sinxy) = xex cosxy (236)
Theres not much new here. We are just holding one vari-
able constant and taking the derivatives with it one of the
variables. The actual application of the partial derivatives
is what is going to be fun. So we can put our tools of par-
tial derivatives to work looking at graphs. Suppose we
want to calculate the sign of the partial derivative at a
specific point. Suppose we look at the point (1, 1) on the
graph below.
Figure 5: f (x, y)
159
At this point, we want to determine if the partial deriva-
tive with respect to x and y is positive or negative. Lets
first look at the partial derivative with respect to x. Graph-
ically, what this means is that if we scoot a little bit away
from (1, 1) in the positive x direction, what direction are
we heading? We can see that we would be heading down-
wards, looking like rolling down the hill, therefore we ex-
pect the partial derivative at this point to be negative.
Now lets look at the y direction. If we scoot out just a
little bit forwards in the positive y-axis. We see that if
we were to take a step forward in the positive y direc-
tion, we would have to walk a little bit uphill since the
function is increasing. Therefore, since we would walking
upwards, the partial derivative with respect to y at this
point is positive. Lets continue with even more examples:
160
Problem: Given these 3 graphs, decide which of these
graphs is f , ∂xf , and ∂yf . Here is the picture of the three
graphs:
Figure 6: Graphs of f , ∂xf , and ∂yf in no particular
order.
Solution: Maybe we want to start by guessing that
the first graph is f . This is nothing more than a guess.
Suppose we choose to look at the x axis. Well if the first
graph is indeed f , then we would expect that since f re-
mains flat along the x-axis, we would expect the deriva-
tive with respect to f along this to be zero. This happens
161
to not be the case for either of the two other graphs. As
such, there is no way that this can be the function itself.
Maybe now lets choose the second graph to be the func-
tion f . Lets look at the rightmost edge towards us. Note
that as we move along the edge from back to front, along
the +x-direction, we see that the shape initially increases,
and then it decreases. As such, we would expect a graph
of the derivative with respect to x to first start off as
positive, to match the initial increase, and turn negative,
about halfway through to match the decrease. Therefore,
the third graph captures this, so we say that the third
graph is the graph of ∂xf . Finally, we can label the first
graph as ∂yf . We can look at the y-axis edge to match
the behavior of the two.
162
Here is a neat little theorem that will come in handy
throughout the course, it is not all too powerful in the
grander scheme of things but something to mention nonethe-
less.
Theorem Clairout’s Theorem states that if fxy and fyx
exist and are continuous, then:
fxy = fyx (237)
I have a cool proof for this that I will include in the recita-
tion notes tomorrow for those that are interested in com-
pleting a mathematics major, so be on the lookout for
that :). Anyways, back to the course. Lets do the follow-
ing example:
163
11.2 A difficult Example
Problem: Given the values of f shown, approximate
fxy(P ). Let be be the bottom left corner that has value 2.
0.1 to the right of this point, is another point with value
3. 0.1 above the point P , lets have a point Q that has
value 4. In the top right corner which is 0.1 away from
the point Q and 0.1 above the bottom corner (the four
points form a square) has value 6. Sorry I didn’t get a
picture of the drawing. If any of you have it email it to me.
Solution: Lets thinking about what we have to do
here. A reminder that gy measures the change in the y-
direction of g. Here, we are doing this for g = fx. For
those that haven’t seen it, I know I haven’t, fxy means
to first take the derivative with respect to x and then
164
take the derivative with respect to y. Therefore, lets first
concern ourselves with the inside derivative, the partial
derivative with respect to x of f . If we have just the pic-
ture available to us, then if we scoot over 0.1, our function
changes value by 1 on the bottom left (lets call this point
P ) points Therefore, fx can be viewed as the change in
the function value over the change in the movement over.
Therefore, we would get that:
∂xf (P ) ≈ 1
0.1= 10 (238)
Now lets do the same thing for the upper left point. We
note that the function, when scooted over from the up-
per left point to the right has a function value change
of 2 in the space of 0.1. Therefore, we can get a similar
165
expression, calling this point Q.
∂xf (Q) ≈ 2
0.1= 20 (239)
Okay so that takes care of the first derivative. Now lets
take ∂y of ∂xf . at each of the points. Now we can look
at the points P and Q that we have been looking at
throughout the problem. We see that value of ∂xf goes
from 10 to 20 as we move up from point P to Q changes
our y value by 0.1. Therefore, We have our function, ∂xf
changing value by 10 in the space of 0.1 scooting up in
the y-direction. Therefore, we can calculate ∂y(∂xf ), fxy
as:
∂y(∂xf ) ≈ 20− 10
0.1= 100 (240)
Resulting in the answer of 100. Lets just take a recap as
to what we did since I probably made some spelling errors
166
and weird sentences trying to catch up. I first started by
look at the point P in the bottom left and the point Q
in the top left. I then said, lemme scoot over from each
point a little bit to the left, seeing how much the function
changed each time over the amount of space I scooted
over. This represented my ∂xf at each of the points. Now
I want to calculate ∂yg of my function g which is g = ∂xf .
Therefore, I start at the point, P , scoot up along the y-
direction to the point Q. I see that my function changes
by 10 whilst making a scoot of only 0.1. Thus, I get that
the ∂yg = ∂y(∂xf ) = fxy ≈ 100. Please email me with
any questions you may have in this section because I know
that this problem got some confusion as an exercise, let
me add, if we instead did fyx which would be ∂x(∂yf ).
Similar to single-variable, lets see how we can linearly
167
approximate function at a specific point. This was added
under the recitation notes.
11.3 Linear Approximation
If we have a well-behaved, i.e, a function that doesn’t
blow up, have asymptotes, or slope of 4000000000, we
can make a linear approximation to the function a spe-
cific point. This is similar to solving for the tangent line
at a point in single-variable calculus. However, now since
we are over in higher dimensions, we will approximate our
surface, functions, with a tangent plane. Lets start with
a definition involving differentiability:
Definition: f is differentiable at a point a if there
168
exists a linear function L such that:
limx→a
f (x)− L(x)
|x− a|= 0 (241)
The differentiable clause in this is much necessary. If
the function is not differentiable at the point in question,
then we cannot say that there exists some linear func-
tion, L(x). Look at the x = 0 point of the absolute value
function f (x) = |x|. The function is not differentiable
at this point, and as such, we do not have the ability to
come up with a linear function that can approximate the
function at this point. The slope of this linear approxima-
tion is going to equal the derivative value at that point.
Now for two variables, we can generalize the above defi-
169
nition so that we can make linear functions for function,
f : R2 → R. Lets restart the definition for multivariable
case:
Definition: A function, f : R2 → R is differentiable
at a point (a, b) if there exists a linear function L such
that:
lim(x,y)→(a,b)
f (x, y)− L(x, y)
|(x, y)− (a, b)|= 0 (242)
lim(x,y)→(a,b)
f (x, y)− L(x, y)√(x− a)2 + (y − b)2
= 0 (243)
In words, this is saying that if we zoom really far in around
that point (a, b), the function f (x, y) strongly resembles
the linear approximation of f, L(x, y). Within the func-
tion L(x, y), we make an analog to the single-variable
case by making the coefficients of x and y10 in L(x, y)10In the single variable case, we had the coefficient of x being the derivative with respect to x
170
are simply the partial derivatives with respect to x and y
respectively. If we want a closed form expression (all this
means in having an equation to represent this idea), we
can express L(x, y) around the point (a, b) as:
L(x, y) = f (a, b) +∂f
∂x(x− a) +
∂f
∂y(y − b) (244)
In order to make sure our functions in question are differ-
entiable, lets throw a theorem into the notes that we can
cite to ensure that our function is differentiable at a point.
Theorem If both ∂xf and ∂yf exist and are continu-
ous throughout a disk around the point in question (think
a small neighborhood around the point), then we say that
f is differentiable at each point in the disk. To put this
theorem into action lets illuminate it with an example:
171
Problem: Show exy sin(x2 + y2) is differentiable ev-
erywhere.
Solution: It is clear that the partial derivatives with
respect to x and y are just combinations of continuous
function like exponential, trigonometric, and polynomial
functions, so the theorem above says that f is differen-
tiable. In order to really show this, we would have to take
the partial derivatives. I will say that since the original
function is made up of trigonometric, polynomial, and
exponential functions, the partial derivatives will also be
made of this. As such, since we just have a composition
(multiplication, addition, etc.) of continuous functions,
then the overall function is continuous. Lets now move
172
to writing the equation of the tangent plane:
Definition: The linear approximation of f : R2 → R
at (a, b) is the function:
L(x, y) = f (a, b) +∂f
∂x(x− a) +
∂f
∂y(y − b) (245)
12 Recitation IV on July 16, 2019
12.1 Partial Derivative Notation
So, there was so mystery about what went on in lecture
today. I want to clarify a few things ahead of time so that
we are familiar with what is going on in the course. I
heard from a few that some notation is quite funky, so let
me show all of the partial derivative stuff briefly through
173
an example. Let,
f (x, y) = x2y + x3 (246)
Suppose we first want to take the partial derivatives with
respect to x and y. Let me now do this below:
∂f
∂x= ∂xf = fx = 2xy + 3x2 (247)
∂f
∂y= ∂yf = fy = x2 (248)
Okay great. This is just the first partial derivatives. Now
we can introduce the second partial derivative. So, we
have a few more options here, 4. We can, for example,
compute all of the following combinations:
fxy = ∂y(∂xf ) =∂
∂y(∂f
∂x) =
∂2f
∂y∂x(249)
fxx = ∂x(∂xf ) =∂
∂x(∂f
∂x) =
∂2f
∂x2(250)
174
fyx = ∂x(∂yf ) =∂
∂x(∂f
∂y) =
∂2f
∂x∂y(251)
fyy = ∂y(∂yf ) =∂
∂y(∂f
∂y) =
∂2f
∂y2(252)
Unfortunately, there are just so many ways to write these
things, so we are forced to move around with all of these
notations. I like the last one the best in each row, but
that is just me. Lets move on to compute each of these
for our example problem above.
fxy = ∂y(∂xf ) =∂
∂y(∂f
∂x) =
∂
∂y(2xy+ 3x2) = 2x (253)
fxx = ∂x(∂xf ) =∂
∂x(∂f
∂x) =
∂
∂x(2xy + 3x2) = 2y + 6x
(254)
fyx = ∂x(∂yf ) =∂
∂x(∂f
∂y) =
∂
∂x(x2) = 2x (255)
175
fyy = ∂y(∂yf ) =∂
∂y(∂f
∂y) =
∂
∂y(x2) = 0 (256)
It11 appeared that the order of the differentiation wasn’t
all that clear today during lecture so I wanted to clear
that up. In addition, I want to show the exercise that
was left for at home from today in class.
12.2 Clarifying an Example in Class on Clairout’s Theorem
Problem: Given the values of f shown, approximate
fxy(P ). Let be be the bottom left corner that has value
2. 0.1 to the right of this point, is another point with
value 3. 0.1 above the point P , lets have a point Q that
has value 4. In the top right corner which is 0.1 away
from the point Q and 0.1 above the bottom corner (the
four points form a square) has value 6. Sorry I didn’t get11Notice that equation (213) gives the same result as equation (211). This shows Clairout’s Theorem!
176
a picture of the drawing. If any of you have it email it to
me. In class, we did fxy. Now lets do fyx, and show that
it is actually equal to fxy
Solution: First off lets clear up the notation. If we
are trying to fine fyx, we are first going to compute fy,
and then we are going to compute the partial derivative
of fy with respect to x. Lets get on with this now. Okay,
so first we want to compute fy. Lets start at the point P
at the bottom left and scoot up to the point in the top
left. If we do this, note that we are going to scoot up 0.1
units while having a function value change from 2 to 4
for a net change of 2. Therefore, we can approximate the
partial derivative here as a change in the functions value
177
over the change in y. Namely,
fy(P ) ≈ 4− 2
0.1= 20 (257)
We can also calculate this idea on the right hand side of
our little box. Lets perhaps compute the partial deriva-
tive of f with respect to y on the right side of the box.
We see that the bottom right corner has a function value
of 3 and the top right corner has a function value of 6.
Therefore if we scoot up by 0.1 units, we bring about a
change of 3 in the function value. Therefore, we can again
compute the partial derivative at the bottom right corner
point, lets call G as:
fy(G) ≈ 6− 3
0.1= 30 (258)
Okay great. So now we need to apply the next partial
178
derivative. Remember, we are trying to compute fyx
namely we are first scooting up from point P and then
we are scooting to the right of this function essentially.
Thus, note that our new function is not just f, but it is
instead fy. Therefore, we are scooting to the right of the
fy function. Thus, lets look at our function values of fy
Well in the bottom left corner at point P we have that the
function value is 20. In addition, we have in the bottom
right corner at point G, we have the function value is 30.
Therefore, if we scoot over 0.1 to the right we bring about
a net change of 10 on the function, fy value therefore, we
can approximate:
(fy)x ≈30− 20
0.1= 100 (259)
As such, we have show that whether we take fyx or fxy
179
we end up both having a value of 100 verifying Clairout’s
theorem that states that the two quantities are equal for
continuous functions. Hopefully this clarifies things.
12.3 Linear Approximation
Linear approximations are just really the multivariable
analog to tangent lines in single-variable calculus. What
is going on here is that we are saying, okay, I have a
function that is defined and has derivatives at some point.
The function might be a bit peculiar and difficult, so let
me approximate the function with a tangent plane. Okay
so the formula for the tangent plane is as follows at the
point (a, b):
z = L(x, y) = f (a, b)+∂f (a, b)
∂x(x−a)+
∂f (a, b)
∂y(y−b)
(260)
180
So lets explain this. What are we saying? We’re saying is,
let me pick a point that is quite close to the point (a, b).
Then the first term tells me, well, the value at the point
is probably pretty close the value of the function at the
point (a, b). However, maybe it is not quite that. Thus,
we add in the partial derivative terms. What these are
saying is that the value may vary at these points close
to the base point, (a, b) by a bit. Namely, we say that
the slopes in both directions multiplied by how much you
move in each direction will tell you how much to add and
subtract from the base value found at the point (a, b).
Think of the terms involving partial derivatives at a di-
mensional analysis standpoing. We are effectively taking
∂f(a,b)∂x ∆x which has units, air quotes, of ∆f . This is not
rigorous but it captures the essence of what is going on.
181
So perhaps, you function is increasing in both the x and
y variables around the point (a, b). Then, this is saying
that if the function is increasing, we would expect that
both ∂f∂x and ∂f
∂y would be positive. Thus, we start at the
value f (a, b) and we add in the small amount of changes,
(x − a) and (y − b) multiplied by the slopes of each of
the variables at that point. Lets illustrate this with an
example since I am probably rambling.
Problem: Consider the function,
f (x, y) = x ln y (261)
Compute the linear approximation of the function f (x, y)
around the point (1, e).
182
Solution: We can construct a linear approximation
with the following equation:
f (x, y) ≈ f (a, b) +∂f (a, b)
∂x(x− a) +
∂f (a, b)
∂y(y − b)
(262)
Therefore, we can directly compute this as:
f (x, y) ≈ f (1, e)+∂f (1, e)
∂x(x−1)+
∂f (1, e
∂y(y−e) (263)
f (x, y) ≈ 1 + ln(e)(x− 1) +1
e(y − e) (264)
f (x, y) ≈ 1 + (x− 1) +1
e(y − e) (265)
Please email me with any questions and if there are any
errors. I typed this up very quickly so that you could all
look over it if necessary, so please alert me ASAP. You
will be rewarded with candy!
183
12.4 A Rigorous Proof of Clairout’s Theorem
Totally unnecessary for the course, but cool nonethe-
less. So, I know there are a few people that are actually
quite interested in getting a degree in mathematics. In do-
ing so, many of you will take analysis courses that seek to
prove many of the things we use everyday in calculus. In
class this week, we have learned about Clairout’s theorem
that states that, for a continuous function, f : R2 → R,
∂
∂x
∂f
∂y=
∂
∂y
∂f
∂x(266)
Lets now go on to prove this with rigor. Lets start off by
stating the theorem we seek to prove:
Theorem Theorem Given f : [a, b] × [c, d] → R
has continuous second-order partial derivatives. Then,
184
fxy = fyx on (a, b)× (c, d).
In order to prove the theorem, I want to cite a Theorem
in Arthur Mattuck’s, Real Analysis textbook. I will now
state it here:
Theorem 12.6: Let g ∈ C([a, b] × [c, d]). Then
there exists a sequence pn(x, y) of two-variable polyno-
mials such that pn → g uniformly. We will now utilize
this Theorem, 12.6, for our continuous function fxy gen-
erating a sequence of polynomials such that pn,
|pn(x, y)− fxy(x, y)| < ε(n) ∀(x, y) ∈ [a, b]× [c, d]
(267)
185
under the condition that:
limn→∞
= 0 (268)
Then, for any rectangle, D = [x1, x2]× [y1, y2] ⊂ [a, b]×
[c, d],
|¨
D
pndxdy −¨
D
fxydxdy| < ε(n)A(D) (269)
Where A(D) = (x2 − x1)(y2 − y1) is the area of our
predescribed rectangle, D. Note:
¨D
fxydxdy =
¨D
fyxdydx (270)
Since these double integrals are equivalent to,
f (x2, y2)− f (x2, y1)− f (x1, y2) + f (x1, y1) (271)
Consequently, since pn is a polynomial, then we can also
186
the fact that:
¨D
pndxdy =
¨D
pndydx (272)
which stands true for each n ∈ N. Thus, we generate the
equation:
|¨
D
pndydx−¨
D
fyxdydx| < ε(n)A(D) (273)
Finally, we take the limit as n → ∞ to achieve the fol-
lowing equation:
¨D
fxy − fyxdydx = 0 (274)
Which implies that for a function with continuous partial
derivatives that:
fxy = fyx (275)
187
13 Lecture VIII on July 17, 2019
13.1 Review on Linear Approximations
Lets start off with a bit of review. We covered linear
approximations in a jiffy, so maybe lets go back and clar-
ify. If f is differentiable, then f be linearly approximate
as:
L(x, y) = f (a, b) + ∂xf (a, b)(x− a) + ∂yf (a, b)(y − b)
(276)
So now, close to the point (a, b), the linear approxima-
tion is having values that are similar to the function’s
values. So, sometimes we will use the linear approxima-
tion instead of the actual function to approximate the
function’s value around (a, b). There are not approxima-
tions beyond second order on the ASE, so please do not
188
spend time on this if you are planning to do this.
13.2 Multivariable Optimization
Now we are going to move forward to optimization. The
only difference we have as we move forward in dimensions
is that, we used to set our first derivative equal to zero
back in single-variable calculus. The only difference here
is that we set our partial derivatives, namely with respect
to x and y both equivalent to zero. Lets consider an ex-
ercise back from single-variable calculus:
Exercise: Find the maximum and minimum value of
f (x) = |(1− x)(x− 3)| over the interval [0, 3].
Solution: Here we go. So, we will want to find the
189
critical points of the function along the interval. In addi-
tion, we are going to want to check the ends of the inter-
val! Since we are not just looking at the entire space, and
we are instead only looking at a smaller interval, we are
going to not just check the derivatives equal to zero, but
we are also going to check the edges of the interval, where
x = 0 and where x = 3. The Extreme Value Theorem
tells us that f has a maximum and a minimum (since
f is continuous and defined on a closed interval. Also
such a maximum and minimum must occur at a critical
point or an endpoint. So, we final all critical points and
endpoints and check 12. Since we have to deal with the
absolute value bars, we actually get a graph that is a tad
more funkier than it would have been without. Here is a12This is a very-well worded answer. However, on exams you would not have to state all of these statements
unless otherwise asked.
190
graph of the function in question:
Figure 7: Graph of |(1− x)(x− 3)|
It appears right from the graph that the functions seems
to have its largest value 3 at x = 0. In addition, we see
that at x = 1 and x = 3 both achieve the function’s
smallest value on the interval of 0. Does this make sense
though. Well, lets think about it. Since we are using some
form of an absolute value function, then we should never
get a function value that is less than zero. Therefore,
we would expect both the maximum and minimum to be
191
greater than or equal to zero. Also, you can see why it
is important to check the endpoints when we are working
on some closed interval since our maximum was at one of
the endpoints. Lets now move forward into two dimen-
sions to see if we can take the ideas of single-variable and
move it into multivariable.
Exercise: Let f (x, y) = −x2 − y2 + x + 23y + 23
36 on
[0, 1]2. Just a note. Seeing [0, 1]2 is just the unit square
and it means we are letting both x and y be in [0, 1].
Solution: Again, we can start with the Extreme Value
Theorem. WE can state that if f : D → R is continuous
and D is closed (includes all boundary points). So, in
our case, we have a closed square since we are including
192
the boundary in our domain and bounded (contained in
some large box). Bounded simply means that our func-
tion doesn’t run off to infinity somewhere in the domain.
It means that, like in the limits, we can put a roof and
a ceiling around the function boxing it in, or if you will,
bounding the function. Now, we can say that our critical
points in R2 as:
∂f
∂x=∂f
∂y= 0 (277)
And, like the other case, we will have to check the end-
points, but in this case, the borders to see if the maximum
or minimum lies along this. What are our edges in the
case of the unit square? Well, it seems like we have 4
edges with the following equations describing them:
y = 0 (278)
193
y = 1 (279)
x = 0 (280)
x = 1 (281)
We call these are boundary critical points checkers. And,
we call the places where both ∂xf and ∂yf equal zero
or the function is not differentiable, the interior critical
points. The thing about boundary critical points is that
they suck? Why, well lets see. Suppose I plug in the
border on the bottom of the unit square where y = 0.
Well now, what happens to our function? We now just
have a function of one variable between [0, 1]. Thus, we
basically have a smaller single-variable sub-problem that
we find the absolute maximum and minimum along each
of the four boundaries. As such, since we have four of
194
the boundaries, we would expect to have 4 smaller single
variable absolute maximum and minimum problems as
we check the boundary conditions. Lets actually go on to
try one of the borders out. When f (t, 0):
f (t) = −t2 + t +23
36(282)
Now we have a function of 1 variable. Note that I used t
to parametric x along this edge since x can vary between
0 and 1. Done forget to check the corners! However,
note that you only need to check each corner once since
it will pop up as a edge of the interval for two of your
parametrizations. I will write a detailed solution to this
problem in the recitation notes for tomorrow. So, if you
are reading along, go on to that to see all the work. For
195
now, lets move onto the interior critical points.
∂f
∂x= 0 = −2x + 1 (283)
∂f
∂y= 0 = −2y +
2
3(284)
Which results in the point of (x, y) = (12,
13) being the crit-
ical point yielding the function value of 1. At this point,
we would check all of boundary points, that I will add in
later. Assuming we do that, we get that the maximum
of the function is 1 and the minimum of the function is
1136. So, knowing how to solve the equations is extremely
important, and I have to help people with this through-
out the year as a tutor. So, it was a great question and
deserves a full answer. Check the recitation notes for to-
morrow for a follow up.
196
Problem: Find the critical points of
f (x, y) = (2x2 + 3y2)e−x2−y2
(285)
Solution. Taking the partial derivatives of x and y
respectively and setting each equal to zero:
∂xf = 2x(−2x2 − 3y2 + 2)e−x2−y2
(286)
∂yf = 2y(−2x2 − 3y2 + 3)e−x2−y2
(287)
Thus, if we look at the part in the front of the partial
derivative with respect to x, we achieve that x = 0. Then,
we can plug this into our second equation and see that
y can be either 0 or 1 or −1, and we set fx = fy = 0.
If y = 0 utilizing the same process, then x = 0, 1,−1.
197
Finally if x 6= 0 and y 6= 0, then we must have the
following two equations being true:
− 2x2 + 3y3 = 0 (288)
− 2x2 + 3y2 + 2 = 0 (289)
Which yields no solutions actually. Therefore, we only
get the points that were discussed prior yielding (0, 0),
(0, 1), (0,−1), (1, 0), and finally (−1, 0).
13.3 The Second Derivative Test (ASE)
The second derivative test is a way that we can clas-
sify the critical points of a function similar to that in
single-variable calculus. Since, we have 4 different partial
derivatives, the general formula and conditions are a lit-
tle bit more extensive than previously. Let me call the
198
quantity we are going to use to organize all the second
order partials D and define D as:
D = fxxfyy − fxyfyx = fxxfyy − f 2xy (290)
We will always skip the second equality. since we know
that by Clairout’s Theorem, fxy = fyx for at least twice-
differentiable continuous functions. Now, in multivari-
able, we have three potential classifications. We have a
max, min, and a saddle. The maximum and minimum
are similar to those in single-variable, but the saddle is
the new type of classification that we have here. For a
saddle, the function f has fxx and fyy being of oppo-
site sign, namely moving around the point in questions
doesn’t exhibit uniform behavior of moving either up or
down as you would get at an mix or max. Lets look at
199
the conditions for each point. Consider D(a, b), namely
for the critical point (a, b):
D(a, b) = fxx(a, b)fyy(a, b)− fxy(a, b)2 (291)
We obtain a:
1. Relative Minimum: If D > 0 and fxx(a, b) > 0
2. Relative Maximum: If D > 0 and fxx(a, b)
3. Saddle: If D < 0
4. Unknown if D(a, b) = 0 We basically do not have
enough information to determine the nature of this
critical
Just a sidenote because I am asked this question a lot
throughout the year. By symmetry, if D > 0, then it
is always the case that both fxx and fyy must be of the
200
same sign. As such, everywhere you see a fxx condition
for the relative min and max conditions above, you can
replace that with a fyy condition is that is what suites
your fancy. If D > 0 and the −fxy will always contribute
something non-positive, then it must be the case that fxx
and fyy be of the same sign! Lets just do a quick example
to reinforce all that was covered.
13.3.1 An Example in Second Derivatives
Problem: Find and classify the critical points of the
function:
f (x, y) = 3x2y + y3 − 3x2 − 3y2 + 7 (292)
Solution: Lets start off by taking all of the partial
201
derivatives and second-order partial derivatives as they
will all come in play throughout the problem:
fx = 6xy − 6x (293)
fy = 3x2 + 3y2 − 6y (294)
fxx = 6y − 6 (295)
fyy = 6y − 6 (296)
fxy = 6x (297)
Okay, now lets find the critical points of this function
so that we can classify each of them. We can find the
critical points of the function by setting both the partial
derivative with respect and the partial derivative with
respect to y equal to zero. Therefore:
fx = 6xy − 6x = 0 = 6x(y − 1) = 0 (298)
202
Therefore, we obtain that either x = 0 or y = 1 from the
partial derivative with respect to x. Lets now plug these
in, one at a time into our partial derivative with respect
to y set equal to zero. For the case of x = 0,
fy = 3x2 + 3y2 − 6y = 0 = 3y2 − 6y = 3y(y − 2) = 0
(299)
yielding the result that when x = 0, y = 0 or y = 2. Now
lets plug in the y = 1 case into our partial derivative with
respect to y.
fy = 3x2 + 3y2 − 6y = 0 = 3x2 − 3 = 0 (300)
Yielding that when y = 1, x = 1 or x = −1. As such, we
have a total of four critical points located at (0, 0), (0, 2),
(1, 1) and finally (−1, 1). Lets now plug each of these
points into our second derivative test. By definition, the
203
second derivative test is:
D(x, y) = fxxfyy−f 2xy = (6y−6)(6y−6)−(6x)2 = (6y−6)2−(6x)2
(301)
Lets now evaluate, and classify each point.
D(0, 0) = (−6)2 = 36 > 0 (302)
So, immediately we know that (0, 0) is either a relative
minimum of maximum. Since fxx = −6 < 0, (0, 0) must
be a relative max.
D(0, 2) = (6)2 − 0 = 36 > 0 (303)
So, immediately we know that (0, 2) is either a relative
minimum of maximum. Since fxx = 6 > 0, (0, 0) must
204
be a relative min.
D(1, 1) = 02 − 36 = −36 < 0 (304)
Therefore, (1, 1) must be a saddle point.
D(−1, 1) = 02 − 36 = −36 < 0 (305)
Therefore, (−1, 1) must be a saddle point. Hopefully this
all makes sense because this will most definitely be on the
ASE!
13.4 Directional Derivative
This is a pretty neat section. Suppose we want to cal-
culate a derivative off at some direction that is not either
strictly in the x or the y direction. If that were the case,
then we could just use partial derivatives with respect to
x and y. Now, lets suppose we want to calculate in some
205
arbitrary direction ~u, where ~u is a Unit Vector by def-
inition. Then we can express the directional derivative in
the direction of ~u at the point (a, b) as:
(D~uf )(a, b) = limh→0
(f ((a, b) + h~u)− f (a, b)
h
)(306)
If we are at the point (a, b) maybe we want to deploy what
we learned last time with regard to linear approximations:
(D~uf )(a, b) = limh→0
(L(a + hu1, b + hu2)− L(a + hu1, b + hu2) + f (a + hu1, b + hu2)− f (a, b)
h
)(307)
(D~uf )(a, b) = limh→0
(f ((a, b) + h~u)− L((a, b) + h~u)
h
)+limh→0
(fx(a, b)hu1 + fy(a, b)hu2
h
)(308)
Now look at the first limit. This goes to zero by our defini-
tion of the linear approximation from last class. In addi-
tion, the second limit has the h get divided out, therefore
just leaving the expression without any of the h’s being
206
present. A much more convenient form that will be uti-
lized when we are actually calculating such a thing is:
(D~uf )(a, b) =
(∂f (a, b)
∂x,∂f (a, b)
∂y
)·~u =
(∂f (a, b)
∂x,∂f (a, b)
∂y
)·(u1, u2)
(309)
Where |~u| = 1 Again, I repeat, ~u is a Unit Vector.
This is one of the most common mistakes I see as a TA
when people are working through such problems. This
is a great formula that we will be in contact with. The
vector of the partial derivatives has a name. It is called
the gradient, and it is defined as below:
~∇f (a, b) =
(∂f (a, b)
∂x,∂f (a, b)
∂y
)(310)
Which allows us to write our direction derivative as:
(D~uf )(a, b) = ~∇f (a, b) · ~u = |~∇f (a, b)| cos θ (311)
207
Therefore, our directional derivative is the largest when
the unit vector, ~u, points in the same direction as the
gradient. This is the case of walking in the direction of
max increase, i.e. walking the steepest path up a hill.
In the case that the gradient and the unit vector are or-
thogonal, the directional derivative is zero. This is the
equivalent of walking along a certain level. The smallest
the directional derivative can be is when the unit vector
is anti-parallel to the gradient. This is the equivalent of
taking the steepest path down the hill.
14 Recitation V on July 18, 2019
Great work today in recitation. The problems were
quite difficult, and we seemed to have a pretty good un-
derstanding of what was going on. Let me give a quick
208
recap of some of the highlights from both me talking,
questions, and things I think would be relevant.
14.1 A Small Note on Multivariable Optimization
One thing that I noticed while working through the
annoying problems on the worksheet is that it is both
beneficial and important to check that the critical points,
points in question, are within the boundary. Like, for
example, if you are working in the unit square, and you
calculate that there exists a critical point at (2, 3), then
we must immediately omit this. Even if this is a critical
point on the function, it is not within our region that we
are optimizing by. As such, we will not include it in trying
to find our absolute maximum and minimum.
209
14.2 Gradients and Directional Derivatives
The gradient at a point (a, b) points in the direction of
maximum increase. So, picture yourself on a mountain.
~∇f (a, b) =
(∂f (a, b)
∂x,∂f (a, b)
∂y
)(312)
Here is the equation for reference. You calculate that
at where you are standing, the direction denoted by the
vector, 15(3, 4) is the gradient of the function. As such,
if you wanted to get to the top of the mountain as fast
as possible, you would take a step forward in this said
direction. It is not necessarily true that once you reach
the new point, that the direction of maximal increase is
the same as the previous point. This was a great question
in class! The gradient that is evaluated at each point. It
tells you, given you are at this point, this is the direction
210
you should head in order to ascend in the quickest way
possible. The directional derivative comes in place in the
following equation:
(D~uf )(a, b) = ~∇f (a, b) · ~u = |~∇f (a, b)| cos θ (313)
Where ~u is a unit vector in some arbitary direction. We
can see that the direction of maximal increase should be
in the same direction as the gradient. As a matter of fact,
we can define it as:
~uGreatestInc =~∇f| ~∇f |
(314)
As such if we want to head in the direction of maximal
decrease, we can express this as:
~uGreatestDec =− ~∇f| ~∇f |
(315)
211
14.3 Following a Path of Max Increase
So now, thanks to a combination of questions and con-
versation with Jordan, Hector, Ivan, Grace, and Raima,
I wanted to include this section. Suppose we want to fol-
low a path along the gradient. How can we compose this
path? We seen before that calculating the gradient at a
specific point tells you what direction to head given you’re
at that point. However, suppose now we want to calcu-
late the whole path of travel. How could we do this. Well,
we could calculate the gradient for an arbitrary (x, y)13.
Then we could think, well if I am following the path of
greatest increase then I better have it that my velocity
always points in the same direction of the gradient. As
such, the velocity of the particle should be a scalar mul-13could simply be (x, y)
212
tiple of the gradient of the function. Therefore we get
that:
~∇f (x, y) = c~v(t) (316)
Where c is some constant. Therefore, we can just take
c = 1 for convenience:
~∇f (x, y) = ~v(t) = (~x′(t), ~y′(t)) (317)
As such, we can match component by component in order
to try and craft back some ~r(t) function. Let me illumi-
nate this with an example so that we have something to
follow along with :)
Problem: Suppose that the temperature in a room
[0, 5]3 is given as a function of position by T (x, y, z) =
50 + x2 + (y − 3)2 + 2z. You are a bug starting at po-
213
sition (3, 2, 2), and you are cold. You decide to move in
the direction of greatest temperature increase at all times.
First find the direction that the bug initially wants to fly
in. Then calculate the path of the bug, ~r(t).
Solution: We have that
(∇f )(x, y, z) = 〈2x, 2y − 6, 2〉 (318)
and thus that we will initially move in the direction
(∇f )(3, 2, 2) = 〈6,−2, 2〉. (319)
We wish to find our position in space as a function of time
at an arbitrary speed if we follow the direction of greatest
increasing temperature, which we will call −→r (t). Because
we will always point in the direction of the gradient, we
214
know that
−→r ′(t) = 〈x′(t), y′(t), z′(t)〉 = λ〈2x(t), 2y(t)− 6, 2〉.
(320)
If we let λ = 1 (its exact value does not matter) we can
then solve for −→r (t). We have that
−→r (t) = 〈c1e2t, c2e
2t + 3, 2t + c3〉. (321)
If we let −→r (0) = (3, 2, 2), then we have that c1 = 3,
c2 = −1, and c3 = 2, giving us that
−→r (t) = 〈3e2t,−e2t + 3, 2t + 2〉. (322)
I thought that this was a really cool example problem
that is a great application of the gradient and ideas from
the paths in space chapter.
215
15 Lecture IX on July 19, 2019
We are officially halfway through the summer! I hope
that you have had a great experience thus far! Let me
know how my notes are please so that I can make them
better for those that use them. Lets kick off lecture with
a review:
15.1 Review on Directional Derivatives
We ended class with:
(D~uf )(a, b) = ~∇f (a, b) · ~u = |~∇f (a, b)| cos θ (323)
Where ~u is a unit vector, |~u|. This represents the sensi-
tivity of f to small changes in the ~u direction from (a, b).
~∇f (a, b) represents the gradient of f at the point (a, b).
Remember, that the gradient points in the direction of
216
maximal increase. Thus, if we would want to find the
direction of maximal increase, then the unit vector:
~uGreatestInc =~∇f| ~∇f |
(324)
represents this direction. In addition, the direction of
maximal decrease would be:
~uGreatestDec =− ~∇f| ~∇f |
(325)
The directions orthogonal to the gradient would have a di-
rectional derivative equivalent to zero. This is the equiv-
alent of walking around a level curve instead of walking
up or down a function. Given that ~∇f (a, b) = (α, β),
two vectors that are orthogonal to the gradient, and as
217
such have a directional derivative equivalent to zero are:
~uorthog1 =1√
α2 + β2(−β, α) (326)
~uorthog2 =1√
α2 + β2(β,−α) (327)
Lets now consider a level set of a function f , that we will
assume is a differentiable function. For example, maybe
we have f : R2 → R. Then, the gradient, by defini-
tion will always be Perpendicular to the level curves of
f . Since, moving in a direction along the level curve will
produce a directional derivative equivalent to zero, then
moving perpendicular to this will either point in the gra-
dient’s direction, of max increase, or in the direction op-
posite the gradient’s direction, of max decrease. Cheers
to Victor for answering this question in class :0.
It is important to remember that the gradient
218
is orthogonal to the level curves of a function,
f . Lets do a quick example to reinforce our learning:
Problem: Find an equation of a plane tangent to
x2 + y2 + 2z2 = 4 at (1, 1, 1).
Solution: We have solved problems similar to this uti-
lizing a linear approximation method that ends up creat-
ing a tangent plane. This problem is a bit different. Note,
this is an equation. We were using tangent planes to ap-
proximate functions. We used to be looking at f (x, y),
looking at the graph of f . Now we have an equation
though. There is no function clearly seen here. If we
wanted, we could solve this equation for z, but this is
problematic? Why, well, if we solve this equation for z,
219
we get two different surfaces for the square root, and we
have to choose which surface to use. So, lets try a dif-
ferent way to solve this problem. Instead, we can think
of the following. Maybe, our equation is a level set of a
function. Namely, Consider the function, f : R3 → R.
In fact consider the following function:
f (x, y, z) = x2 + y2 + 2z2 (328)
Now, we can see that if we look at the level set of f (x, y, z) =
c where c = 4, we can envision our equation in the
problem statement as simply a level set of the function
mentioned above. As a quick reminder to the conversa-
tions in chapter one, we cannot graph the actual function
f (x, y, z) since it would require four dimensions. How-
ever, we definitely can graph its level sets which happen
220
to be ellipsoids, like the equation given. We can now take
the gradient of this function at the point (1, 1, 1) because
we know that the gradient of this function will be per-
pendicular to the level set of the function, our original
equation we were given. Namely,
~∇(x2 + y2 + 2z2) = (2x, 2y, 4z) (329)
Which, at the point (1, 1, 1) results in (2, 2, 4) being the
gradient. As such, since this vector is the gradient, and
the gradient is by nature perpendicular to the level sur-
face, then the vector ~n = (2, 2, 4) is indeed perpendicular
to our surface. However, where before have we seen nor-
mal vectors coming into play? Planes! We note that
a tangent plane at the point (1, 1, 1) will be defined by
its normal vector in the form, ax + by + cz = d where
221
~n = (a, b, c). Thus,
2x + 2y + 4z = d (330)
plugging in the information available to us. We can solve
for d by plugging a point in on our plane, (1, 1, 1). As
such, we obtain that d = 8, and the equation for the
plane tangent to the surface at (1, 1, 1) is equivalent to:
2x + 2y + 4z = 8 (331)
It is quite confusing to get all of this. The biggest point
of confusion for me was this. Planes are defined by their
normal vector. Therefore, although this vector itself is
normal to the level curve that we solved for, a normal
vector defines a tangent plane! To recap, what we did
was say, okay, I am given an equation. I am going to
222
think of my equation as a level set of a function f . I am
then going to take the gradient of f at the point noting
that the gradient is tangent to the function f and it is
orthogonal to the level sets of the function f . As such,
the gradient points orthogonal to the level set. So, for
a tangent plane, this would be the normal vector that
defines the plane. We then wanted a full equation for our
tangent plane, so we plugged in the point to solve for d,
resulting in the equation for the tangent plane. I know
there is a lot of flipping between tangent and normal, so
read this over a few times to make sure you have it all
down. Lets have a rapid flip over to the chain rule.
223
15.2 Multivariable Chain Rule
The chain rule was first taught to us in single-variable
calculus. We used it when we had compositions of func-
tions such as f (g(t)). Lets start off with an example.
Lets compute the derivative of f (g(t)):
limh→0
f (g(h + t))− f (g(t))
h= f ′(g(t))
(g(t + h)− g(t)
h
)= f ′(g(t))g′(t)
(332)
Now lets see how we can take the idea of the chain rule and
apply it over in the multivariable setting. I will introduce
the subject by stating the formulas just for reference. I
will then go on to explain it afterwards. The multivariable
chain rule is expressed succinctly as:
df
dt=∂f
∂x
dx
dt+∂f
∂y
dy
dt(333)
224
However, lets look a little bit more at this expression.
It seems like, upon first glance we are matching a par-
tial derivative with respect to a variable and multiplying
it with the derivative of this said variable with respect
to time. We are then summing over all of the variables,
essentially getting the contribution from each of the vari-
ables to the overall change in f with respect to time. In
fact, this component by component sum actually is a hid-
den dot product. We can write the multivariable chain
rule alternatively as:
df
dt= ~∇f ·~r′(t) =
(∂f
∂x,∂f
∂y
)·(~x′(t), ~y′(t)) =
∂f
∂x
dx
dt+∂f
∂y
dy
dt
(334)
At a very hand-wavy level, we are having each of the
terms having a cancellation’ of the dx terms to be left
with each term of f over t. In addition, we want that
225
our final answer is in terms of t since we are taking the
derivative with respect to time. Let me show you a quick
example of this.
Problem: Compute dfdt for f (x, y) = x2 +y for ~r(t) =
(t, t2)
Solution: We can solve this in two ways. I will high-
light both of them now. We can solve it
1. we can plug in our expressions for x and y into the
function so that we have f only in terms of t. Lets
do that now. Note that this is not really utilizing
anything new here in terms of taking the mix of partial
derivatives and full derivatives.
f (t) = (t)2 + t2 = 2t2 (335)
226
df
dt= 4t (336)
Now lets try it our new way and see if we can get the
exact same expression for dfdt
2. We will now take use of the multivariable chain rule.
df
dt=∂f
∂x
dx
dt+∂f
∂y
dy
dt(337)
df
dt= (2x, 1) · (1, 2t) = 2x + 2t (338)
However, now we need to plug in our expression for
x since our final answer should only be in terms of t.
As such, we obtain that:
df
dt= (2x, 1) · (1, 2t) = 2t + 2t = 4t (339)
Leading to the same exact answer validating our claim.
Lets nove to the proof
227
15.2.1 A Proof of the Multivariable Chain Rule
Lets try to come up with a proof for this:
d
dt(f (~r(t))) = lim
h→0
(f (~r(t + h))− f (~r(t))
h
)(340)
Lets now introduce a linear approximation of f at the
point ~r(t):
d
dt(f (~r(t))) = lim
h→0
(L(~r(t + h))− L(~r(t + h)) + f (~r(t + h))− f (~r(t))
h
)(341)
Due to L being a linear approximation of f (~r(t). As such,
we can group the two middle terms in our numerator,
which goe to zero as h→ 0. Therefore, we are left with:
d
dt(f (~r(t))) = lim
h→0
(L(~r(t + h))− f (~r(t))
h
)(342)
228
Now, since we at the point, both L and f are the exact
same value, we can substitute our second term in the
numerator with L(~r(t)).
d
dt(f (~r(t))) = lim
h→0
(L(~r(t + h))− L(~r(t))
h
)(343)
d
dt(f (~r(t))) = lim
h→0
(fx(r1(t + h)− r1(t)) + fy(r2(t + h)− r2(t))
h
)(344)
d
dt(f (~r(t))) = fx(r
′1) + fy(r
′2) = ~∇f (~r(t)) · ~r′(t) (345)
Where ~r = (r1(t), r2(t)) = (x(t), y(t)). This proof is
not necessary to ever use, but it is the proof behind the
pudding of the formula that we are going to be working
with. It is much more important to know how to compute
the multivariable chain rule like the example that I did
prior to the proof. Unlike directional derivatives, our dot
product does not need to include a unit vector. The ~r′(t)
229
has a magnitude that can be anything. As such, it would
make sense to incorporate its magnitude since objects
moving faster, namely having a higher magnitude should
have a larger change in the overall derivative.
16 Lecture X on July 23, 2019
Great job on the midterm yesterday. You all did really
well. Lets start off today with a review of a few topics.
16.1 Review on Partial Derivatives and Mixed Partials
What is the interpretation of each of our partial deriva-
tives. Consider fx and fy. fx and fy represent scooting
a bit to the right and up respectively and noting how the
function value changes. In addition, we can discuss fxx.
This represents how much the function fx changes if we
230
scoot a bit to the right. Namely, we are looking at the
rate of change in the x direction. We can discuss fyy.
This represents how much the function fy changes if we
scoot a bit to the right. Namely, we are looking at the
rate of change in the y direction. Finally, there are also
mixed partial derivatives. Consider fxy. This is finding
the function fx. Then, we can compare how fx changes
with respect to y. Namely, we look at the rate of change
of fx in the y direction. The third problem on the home-
work best illustrates this. If you draw functions in Rn,
are you an R-tist? In addition, lets just again, state the
multivariable chain rule. We can express this as:
df
dt= ~∇f ·~r′(t) =
(∂f
∂x,∂f
∂y
)·(~x′(t), ~y′(t)) =
∂f
∂x
dx
dt+∂f
∂y
dy
dt
(346)
231
16.2 Lagrange Multipliers
Lagrange multipliers are a new way to solve optimiza-
tion problems along the boundary. Consider the following
function:
f (x, y) = −x2 − y2 + x +2
3y +
23
36(347)
That we wish to optimize on the following region:
D = (x− 1
2)2 + (y − 1
2)2 ≤ 1
4(348)
Last week, we learned how to solve this type of function
with some really lengthy process of first checking the inte-
rior critical points, then points that aren’t differentiable,
then writing equations for the borders, completing a slew
of single variable problems, and finally evaluating all of
these points to see which was the absolute smallest and
232
largest. We are now going to take a quicker more, ”so-
phisticated” approach. We note that along the border of
the region, the gradient of f , ~∇f , must be perpendicular
to the boundary of the region D, ∂D at a point if f is
to have a local optimum there. As such, for an optimum
along the boundary of the region D, it must be true that:
~∇f = λ~∇g (349)
Thinking of the boundary of the region D, as a level set
of g, then we recall that the gradient of g, ~∇g will point
orthogonal to the boundary. As such, we would expect
~∇g to point in the same direction of f . Namely, we would
expect that ~∇f and ~∇g to point in the same direction as
one another. Since both ~∇f and ~∇g are vectors, we can
capture this idea by using the vector identity that ~∇f is
233
a scalar multiple of ~∇g. So, for some constant λ ∈ R, it
must be true that:
~∇f = λ~∇g (350)
This is the punchline. Seeing why this equation is true is
what we completed with the discussion of the level sets of
g, but the equation that we will be utilizing to find local
optimum around the boundary of our region that we are
optimizing over, we will just be using this equation to
help us solve for what we do not know. Lets now try to
solve this. We were given an inequality expressing D. We
now want to have that g would be that:
g(x, y) = (x− 1
2)2 + (y − 1
2)2 (351)
234
Where we just took the border function that was equal
to 14, and noted that this was a level set, and we called
this the function g that this was a level set of. Now we
have some partial derivatives to take:
~∇f = λ~∇g (352)
(−2x + 1,−2y +2
3) = λ(2(x− 1
2), 2(y − 1
2)) (353)
Nice! So, if we equate each of the components, we obtain
that:
− 2x + 1 = λ2(x− 1
2) (354)
− 2y +2
3= λ2(y − 1
2) (355)
This is looking good. However, we have one problem. We
have two equations with three unknowns. How can we
introduce a third equation that must be satisfied? Well,
235
we already noted that all of this mess is only true on
the boundary. Thus, we can obtain a third equation be
writing the equation for the boundary. We have that:
(x− 1
2)2 + (y − 1
2)2 =
1
4(356)
Solving the first equation we obtain that, λ = −1 and
x = 12. However, when we look at the case of λ = −1,
the second equation in our set cannot hold. As such, we
only obtain one critical point along the boundary where
x = 12, yielding the y-coordinate of 1 or 0. Inside, we also
obtain a critical point of (12,
13). We can now evaluate our
function at all three of the critical points:(1
2, 0
)=
8
9(357)
(1
2, 1
)=
5
9(358)
236
(1
2,1
3
)= 1 (359)
Lets move on to a more practical example.
Problem: Find the maximum volume of a lidless box
with a surface area of 72.
Solution: So, lets turn this sentence into a type of La-
grange multiplier idea. We want to maximize our function
f = V , the volume, subject to the constraint of the sur-
face area of the lidless box g = S. Since the box is going
to be rectangular, we can express the volume function
f (x, y, z) = xyz. In addition, we can express the surface
area function S = g(x, y, z) = xy + 2xz + 2yz = 72. It
actually doesn’t matter which side the lid is taken off of,
so it does not necessarily matter that we took a xy side
237
off, we could have equivalently taken off a yz perhaps.
We now can deploy our Lagrange Multiplier problem:
~∇f = λ~∇g (360)
(yz, xz, xy) = λ(y + 2z, x + 2z, 2y + 2x) (361)
Now we have three equations above with 4 unknowns. So
we must introduce the constraint as well that:
xy + 2xz + 2yz = 72 (362)
So that we have the set of equations that:
yz = λ(y + 2z) (363)
xz = λ(x + 2z) (364)
xy = λ(2x + 2y) (365)
238
xy + 2xz + 2yz = 72 (366)
Which I will solve later and put into the notes right here.
We can now finally move on to integration
16.3 Integration
First lets think of a single-variable case to realign our-
selves with integration after so much differentiation. We
can think of integration back in single-variable as the
signed area that lies under the graph of f . We can think
of an integral as splitting the interval of which we are tak-
ing the integral over into a bunch of tiny pieces. We then
see how much each piece contributes its volume times the
value of the function on that tiny piece. We are effec-
tively, in the single-variable case , taking the signed area
of each one of these really skinny rectangles. We then
239
add each one of these contributions. Then, we take the
number of pieces to infinity, making them infinitely thin!
Lets hop right into an example of an integral.
Example: Integrate,
f (x, y) = y sin(πyx) (367)
over the unit square, [0, 1]2.
Solution: Instead now of some interval, note that we
are now integrating f over a 2D region, namely a square.
Now we want to divide our little region into a bunch of
little squares! Then, for each of these little squares that
we cut the unit square into, we are going to take the
function’s value at that square. Then, we can express the
240
volume as the function value at the really tiny square
multiplier by the area of the super tiny square. We are
then going to sum up each of these contributions from
each tiny square. The formal expression for this is lets
first sum up each row of squares and then each column
of these contributions that we previously hypothetically
computed. The actual integral expression for this idea is:
ˆ 1
0
ˆ 1
0
f (x, y)dxdy =
ˆ 1
0
ˆ 1
0
y sin(πyx)dxdy (368)
Where, ˆ 1
0
f (x, y)dx (369)
represents summing up along each row, and then:
ˆ 1
0
(ˆ 1
0
f (x, y)dx
)dy (370)
represents taking each of these total row contributions
241
and summing those up. When first taking the derivative
with respect to x, similar to partial differentiation, we
hold y constant and treat it as such.
17 Recitation VI on July 24, 2019
I will take this space in order to just reach out with
some further discussion points with Lagrange Multipliers.
Firstly, Lagrange multiplier only work along the bound-
ary of your region. For example, suppose you have some
region D, that is the unit disk. That utilizing the La-
grange multiplier method will only work along x2+y2 = 1,
otherwise known as the unit circle, which happens to
the border of the disk. That being said, the Lagrange
multiplier relies heavily on the direction of the gradient.
Namely, we should think of a Lagrange multiplier prob-
242
lem as a process. We first decide, I want to maximize
and/or minimize some function f . In addition, I want
to compute this Optimization of f over some boundary
which we can refer to as g. Then, what the Lagrange
multiplier formula tells us is that:
~∇f = λ~∇g (371)
Lets walk through some problems in order to really get
down the overarching idea. And, we can even some some
really neat problems along the way!
Problem: Find the maximum and minimum of the
function f (x, y) = 2x − 3y. subject the constraint,
x2 + y2 = 64
243
Solution:We can utilize the method of Lagrange mul-
tipliers in order to solve this problem where we classify
g(x, y) = x2 + y2
~∇f = λ~∇g (372)
We have that:
(2,−3) = λ(2x, 2y) (373)
With the third equation of:
x2 + y2 = 64 (374)
We can write x and y in terms of λ by solving the first
two equations, namely:
x =1
λ(375)
y =−3
2λ(376)
244
Plugging into our third equation, we obtain that:
1
λ2+
9
4λ2= 64 (377)
Leading to:
λ = ±√
13
16(378)
Which, if we plug back into our equations, obtain two
points:
(16√13,−24√
13) & (
−16√13,
24√13
) (379)
Which, we can evaluate our function at these two points
leading to:
f (16√13,−24√
13) =
32√13
+72√13
=104√
13(380)
f (−16√
13,
24√13
) =−32√
13+−72√
13=−104√
13(381)
resulting in the max and min respectively along the con-
245
straint.
Problem: A right cylindrical can is to have a volume
of 0.25 cubic feet (approximately 2 gallons): Find the
height h and radius r that will minimize surface area of
the can. What is the relationship between the resulting r
and h?
Solution: Lets first get down equations for both the
surface area and the volume.
f (r, h) = S = 2πr2 + 2πrh (382)
g(r, h) = V = πr2h = 0.25ft3 (383)
246
Which can deploy the Lagrange multiplier equation:
∇f = λ∇g (384)
(4πr + 2πh, 2πr) = λ(2πrh, πr2) (385)
Lets now divide through by component:
2r + h
r=
2h
r(386)
rh = 2r2 (387)
Since r cannot be not be negative, we can express h in
terms of r.
h = 2r (388)
We can then plug this in to our constraint equation:
2πr3 = 0.25 (389)
247
r =
(0.25
2π
)13
(390)
h =
(1
π
)13
(391)
Problem: Assume there are two commodities with
amounts x and y with respective prices of px and py. In
addition, suppose that you have a utility function that
you wish to maximize of the form: U(x, y) = xαy1−α
for some constant α ∈ (0, 1). You maximize your utility
function14 subject to some budget constraint. Namely,
suppose that in total you have m dollars to spend on both
your products. Maximize the utility function to boom up
the ’merican economy by:14This is called a Cobb-Douglass Utility Function
248
1. Writing a constraint equation relating the amount and
prices of the two goods along with your total amount
of money, m.
2. Setting up a function f and g and utilizing the La-
grange multiplier equation to find the amounts of x
and y you should purchase.
Solution:
1. Lets start off by writing a budget constraint. We note
that we can buy both x and y at prices px and py that
has to be less than or equal to the amount of money
we have m
pxx + pyy ≤ m (392)
2. We can now construct our function that we wish to
maximize u(x, y) subject to our constraint pxx+pyy ≤
249
m. Therefore, we can make the function g(x, y) as:
f (x, y) = xαy1−α (393)
g(x, y) = pxx + pyy (394)
We can now utilize methods of Lagrange multipliers
to solve this problem:
~∇f = λ~∇g (395)
(αxα−1y1−α, (1− α)xαy−α) = λ(px, py) (396)
We can add in the constraint as well:
pxx + pyy = m (397)
If we divide the first equation by the second equation
250
we obtain that:
αxα−1y1−α
(1− α)xαy−α=pxpy
(398)
αy
(1− α)x=pxpy
(399)
Lets now solve this equation for y
y =1− αα
pxx
py(400)
We can now plug this into our constraint equation:
pxx+pyy = pxx+py1− αα
pxx
py= x(px+
1− αα
px) = m
(401)
As such, we obtain that:
x =m
(px + 1−αα px)
=mα
px(402)
Which we can now plug into our expression for y to
251
obtain:
y =1− αα
pxpy
m
(px + 1−αα px)
=(1− α)m
αpy(1 + 1−αα )
=m(1− α)
py(403)
Problem: Deriving Snell’s Law. Snell’s Law is known
to man as n1 sin(α) = n2 sin(β). Please reference the
picture on the next piece of paper for the drawing setup
of the problem. Now, given a beam of light starting at
the point A, a distance A above the horizontal, passing
through the interface at the middle, and reaching point
B, a distance B below the horizontal, moving from space
with a refractive index of n1 to a region of space with re-
fractive index n2 respectively. Now, light normally moves
at the speed of c. However, in a refractive index n, light
takes on the velocity of v = cn. With this knowledge, de-
rive Snell’s law.
252
Solution: So, light always follows the path that takes
the shortest time. Therefore we can make our f function
be time over the journey since we are seeking to mini-
mize this. In addition, We can make the length in the x
direction our constraint, since we cannot change the ac-
tual position of the two points in question. Therefore, we
can simply take that distance over time is velocity, rear-
range and get a formula for time over the entire journey,
namely:
f (α, β) = t =
∑d∑v
=a
v1 cosα+
b
v2 cos β(404)
And our accompanying constraint:
g(α, β) = L = a tanα + b tan β (405)
253
We now have an f and g with two unknown variables that
happen to be our two angles so it is time to have some
fun!
∇f = λ∇g (406)
(a
v1secα tan,
b
v2sec β tan β) = λ(a sec2 α, b sec2 β)
(407)
Lets now solve each of our component equations for λ or
otherwise we just divide through:
λ =tanα
v1 secα=
tan β
v2secβ(408)
Lets now plug in our expressions for the velocity that were
provided in the image:
n1
csinα =
n2
csin β (409)
n1 sinα = n2 sin β (410)
254
Which is the equation that is known as snells law!
18 Lecture XI on July 25, 2019
18.1 Review on Ideas Behind Integration
Last week, we ended with an example of an integral
over a unit square. We can try to generalize our idea of
the integral by thinking, at the core, what an integral is.
If f : D → R where D is some shape, then:
ˆD
f (411)
is the result of a process that is:
1. Split D into many small pieces
2. Total the products, ”piece volume” × ”value of func-
tion at piece
3. take numbers of pieces to ∞255
4. Add up all of the contributions
Last time, we ended by considering:
¨D
f (x, y)dA (412)
Where D is the unit square. The idea here was that we
would look at the contributions along a row at height y,
totaling up to, as we first sum along each row and then
sum all rows together, we obtain:
∑y
(∑x
f (x, y)∆x∆y
)(413)
Then, If we take the limit as ∆x and ∆y → 0, we can
back out the double integral, namely:
lim∆y→0
lim∆x→0
∑y
∑x
f (x, y)∆x
∆y =
¨f (x, y)dxdy
(414)
256
Great now we can utilize the idea of single-variable cal-
culus to actually compute the double integral here. Lets
actually compute the integral in our case here:
ˆ 1
0
ˆ 1
0
f (x, y)dxdy =
ˆ 1
0
ˆ 1
0
y sin(πyx)dxdy (415)
ˆ 1
0
[−cos(πyx)
π
]1
0
dy =
ˆ 1
0
1− cos(πy)
πdy (416)
ˆ 1
0
1− cos(πy)
πdy =
[y − 1
π sin(πy)
π
]1
0
=1
π(417)
Great! However, lets think about more complicated re-
gions. The main difference in my opinion between single
variable and multivariable integration is the fact that your
regions can be a whole slew of funky shapes. As we move
away from squares, we can now try to integrate function
over other shapes, like perhaps a triangle in the next ex-
ample.
257
Problem: Suppose f : D → R is defined by:
f (x, y) = x2y (418)
Where D is the triangle with vertices at (0, 0), (2, 0) and
(0, 3). Calculate˜D fdA
Solution: We have here another integration problem.
Now however, we are going to integrate over a triangle
instead of a square. The main difference here is that now,
our region is a little more convoluted since our variables
are not necessarily bounded by just constants. We can
express this first as sums, by first getting the contribution
from one row for some fixed y and then summing over all
258
rows. Namely, we have that:
∑y
(ˆx2ydx
)∆y (419)
From here, we can note that x is bounded between x = 0
and the side of the triangle. This is the main difference
here. Why? Because we can write an x = equation that
describes the side of the triangle. As such, we can write
the equation for that part of the triangle as y = 3− 32x.
Solving for x, we obtain that:
x = 2− 2
3y (420)
We can now put our inner bounds for x into our double
integral. ∑y
(ˆ 2−23y
0
x2ydx
)∆y (421)
We can now take the limit as ∆y → 0. Which leads to
259
our second double integral. Given our region, we note
that y can vary between y = 0 as well as y = 3. As such,
our bounds for y are just two numbers, and there are no
functions of x involved. Lets put that into the problem:
ˆ 3
0
(ˆ 2−23y
0
x2ydx
)dy (422)
You could also do the integrals in the opposite order if
you would like, and you would obtain:
ˆ 2
0
(ˆ 3−32y
0
x2ydy
)dx (423)
This is honestly the hardest part of double and even
triple integrals... the Bounds! I highly encourage you to
take the time and draw the region. Then, you would try
to come up with your bounds by utilizing the picture. One
thing to note as you are making you’re making your way
260
through double and triple integrals is that the outermost
integral must only have bounds that are numbers. In
addition, each inner integral can only be a function of the
outer integrand variables. For example, in our triangle
problems x was the variable of the inner integral and
y is the variables of the outer integral. As such, it was
totally valid that our inner variable with respect to x was
a function of our outer integral y. In addition, we are in
the clear with our outer integral since our outer integral
was only a function of numbers, and not y nor x.
18.2 Triple Integrals
We are stepping it up a notch now! Suppose that now,
f : D → R, where D ⊂ R3. This notation means
that D is a Subset of R3. Mathematically, we are now
261
calculating a 4D volume. We are getting the by letting
the region represent the three first dimensions and then f
representing the fourth dimension. However, this doesn’t
need to be the case. Instead of trying to wrap our head
around 4D volumes in the 4D chess version of life, we can
instead interpret f as a density value, then:
ˆD
f = M (424)
where M , can be thought of as a Mass since f is the den-
sity andD would be the volume, so we obtain mass! How-
ever, you can also think of this with respect to any densi-
ties that you may have come in contact with throughout
your physics class. Like, for example, you could interpret
f as a volumetric charge density, and as such, you would
be calculating the total charge by taking the integral of
262
f . Since f can be a function, this can represent all types
of densities! Particularly, you don’t need to have uniform
densities as many of you are probably used to seeing. Lets
illuminate this topic with an example.
Problem: Integrate f (x, y, z) = x2 + y2 + z2 over
[1, 2]3.
Solution: Lets think of the function, f (x, y, z) as the
density. Therefore, we would expect the furthest point
of the cube to be the densest given our function of f and
its interpretation. As such, if we wanted to compute the
total mass of the cube that we have defined, we can do
263
so as:
M =
˚f (x, y, z)dV =
ˆ 2
1
ˆ 2
1
ˆ 2
1
(x2 + y2 + z2
)dxdydz
(425)
Before we hop into this integral, lets do a sanity check as
to a number for the mass that is definitely larger than the
actual mass. Well, the density at the top back corner is
12. This is the largest the density can ever be. Therefore,
since we are working with a unit cube, and we have a max
density of 12, we could guess that the mass of the cube
is no more than 12. This is a mental upper bound we
can put on this. In 3D, we can carry the same ideas that
guided us in double integrals and apply it here. We can
take a tiny slice of our cube. Given a particular z constant
slice, we now have a double integral. We can now use our
methodology that we utilized in our double integrals, by
264
first looking at a particular y constant slice compute the
contribution along this row, then sum over all rows, and
now sum over all z slices to get the final result. As such,
it is like we are just adding a dimension and calculating
another subproblem! As such, we obtain:
M =
˚f (x, y, z)dV =
ˆ 2
1
ˆ 2
1
ˆ 2
1
(x2 + y2 + z2
)dxdydz
(426)
M =
ˆ 2
1
ˆ 2
1
[x2z + y2z +
z3
3
]2
1
dydz (427)
M =
ˆ 2
1
ˆ 2
1
(x2 + y2 +
7
3
)dydz (428)
M =
ˆ 2
1
[x2y +
y3
3+
7
3y
]2
1
dz (429)
M =
ˆ 2
1
(14
3+ x2
)dz (430)
M =
[14
3x +
x3
3
]2
1
=14
3+
7
3= 7 (431)
265
Since the actual integral was not evaluated in class, I have
provided it here in case you wanted to see it. Lets con-
tinue with a last example related to finding the volume
of a tetrahedron.
Problem: Find the volume utilizing a triple integral.
Here is an image of the shape provided.
Figure 8: Tetrahedron Volume Problem
Solution: Lets define a function f (x, y, z) = 1. Then,
the volume equals the mass. Lets start off by looking at
slices. We can first note that 0 ≤ z ≤ 4. However, now,
266
as we make some y constant slices. Now our bounds on
the integral get a bit more challenging since these slices
will be a function of z. As such we can say that 0 ≤
y ≤ 3 − 34z. Finally, we can have the x slices. For the
x bounds, we note that x is a function of both y and z
since we already have them fixed in space. We can derive
the bounds for x as 0 ≤ x ≤ 2 − 23y −
12z. As such, we
can set up our triple integral as:
M =
ˆ 4
0
ˆ 3−34z
0
ˆ 2−23y−
12z
0
f (x, y, z)dxdydz (432)
M = V =
ˆ 4
0
ˆ 3−34z
0
ˆ 2−23y−
12z
0
1dxdydz = 4 (433)
Where we took advantage of that fact that when f (x, y, z) =
1 then the mass and the volume are equivalent.
267
18.3 Integration in Other Coordinate Systems
Earlier in the course, we did quite briefly go over alter-
native coordinate systems like cylindrical and spherical
coordinates. Luckily, we introduced them because they
will come in handy greatly when discussing integrating
regions that are circular or even spherical in nature. We
can rethink back to polar coordinates at the back end of
a single-variable calculus course. For example image in-
tegrating f (x, y) = x + y over the region of a cartoid.
15. To quickly put them in my notes, in case you want
to know we can express the following double and triple
integrals respectively in the different coordinate systems
as, over some region D, where D ⊂ R2 for polar and15This is the heart-shaped graph’s official name that we see in polar graphs
268
D ⊂ R3 in cylindrical and spherical coordinates.
¨D
f (x, y)dA =
¨D
f (x, y)dxdy =
¨D
f (r, θ)rdrdθ
(434)
which represent the polar coordinate double integral where
we pick up this r factor in addition to our infinitesimal
pieces. In addition, in cylindrical coordinates, we have
that:
˚D
f (x, y, z)dV =
˚D
f (x, y, z)dxdydz =
˚D
f (r, θ, z)rdrdθdz
(435)
And finally in spherical coordinates we obtain that:
˚D
f (x, y, z)dV =
˚D
f (x, y, z)dxdydz =
˚D
f (ρ, θ, φ)ρ2 sinφdρdθdφ
(436)
Where just as a reminder, the conversions between carte-
sian and polar, cylindrical, and spherical respectively are:
269
18.3.1 Polar Coordinates
r2 = x2 + y2 (437)
θ = arctany
x(438)
We can also head in the opposite direction:
x = r cos θ (439)
y = r sin θ (440)
18.3.2 Cylindrical Coordinates
r2 = x2 + y2 (441)
θ = arctany
x(442)
z = z (443)
270
We can also head in the opposite direction:
x = r cos θ (444)
y = r sin θ (445)
z = z (446)
18.3.3 Spherical Coordinates
ρ2 = x2 + y2 + z2 (447)
φ = arccosz√
x2 + y2 + z2(448)
θ = arctany
x(449)
And now, in the other direction:
x = ρ sinφ cos θ (450)
271
y = ρ sinφ sin θ (451)
z = ρ cosφ (452)
19 Recitation VII on July 26, 2019
I think that for integration, the best thing that you can
do is more and more practice. That being said, let me
add to the this as well as the next set of recitation notes
set examples that have to do with integration. Note that
these will be familiar from the worksheets.
Problem: Given the density of some unit cube of mass
in the first octant is given by:
ρ(x, y, z) = xyz (453)
Find the total mass of the cube.
272
Solution: The mass of the cube is given by:
M =
ˆ 1
0
ˆ 1
0
ˆ 1
0
xyzdxdydzM =1
8(454)
What we have here is a triple integral with all numeri-
cal bounds. However, we may not be as lucky sometimes
and have to compute integrals that have more complex.
Namely, we have to come in contact with integrals that
contains bounds that are functions of other variables.
Lets explore an example that illuminates this.
Problem: An application of the average value of a
function is center of mass. We can define the center of
273
mass of an object by the following equation:
(xCOM , yCOM) =
(˜R xdA˜R 1dA
,
˜R ydA˜R 1dA
)(455)
Where we are essentially calculating the average value of
the x and y components, applying the function from the
previous problem. As such, calculate the center of mass of
a right isosceles triangle with the vertices at (0, 0), (1, 0),
and (1, 1).
Solution: We can start by calculating just the area,
the denominator of both terms, of the triangle that we
are working with. We could actually compute the area
utilizing a double integral, which I will show, but we could
also just get the area by drawing the region and using that
the area of a triangle is A = 12Bh = 1
2 in our case. Lets
274
show this using the double integral method.
¨R
dA =
ˆ 1
0
ˆ x
0
dydx =
ˆ 1
0
[y]x0 dx =
ˆ 1
0
xdx =
[x2
2
]1
0
=1
2
(456)
Now we can compute, utilizing the same bounds for the
integrals both,˜R xdA and
˜R ydA
¨R
ydA =
ˆ 1
0
ˆ x
0
xdydx =
ˆ 1
0
[xy]x0 dx =
ˆ 1
0
x2dx =1
3
(457)¨R
ydA =
ˆ 1
0
ˆ x
0
ydydx =
ˆ 1
0
[y2
2
]x0
dx =
ˆ 1
0
x2
2dx =
1
6
(458)
Then, we can use our center of mass formula to obtain
that:
(xCOM , yCOM) =
(˜R xdA˜R 1dA
,
˜R ydA˜R 1dA
)=
(1312
,1612
)=
(2
3,1
3
)(459)
This is a great application of utilizing double integrals to
275
gain insight on other meaningful quantities found through-
out the course. In this case, we have bounds that are ac-
tually functions of x. Namely, given our triangular region,
we note that if we take an x = c for some constant c slice,
the height that we move up from y = 0 depends on x. As
such, our upper bound on the y-variable also depends on
x! Lets trudge forth with some more examples. We do
not need to limit ourselves to just working with cartesian
coordinates, but we can take our first steps into polar co-
ordinates through the following two examples that I will
first work out and then explain the methodology behind
it.
276
Problem: Compute:
ˆ ∞−∞
ˆ ∞−∞
(1
πe−(x2+y2)
)dxdy (460)
By converting to polar coordinates, and integrating with
respect to r and θ
Solution: Since we’re integrating over the entire xy
plane, we would integrate over every possible r and θ
which means that our integral becomes:
ˆ ∞−∞
ˆ ∞−∞
(1
πe−(x2+y2)
)dxdy =
ˆ 2π
0
ˆ ∞0
e−r2
πrdrdθ
(461)
= 2π
ˆ ∞0
e−u1
2πdu =
[−e−u
]∞0
= 1 (462)
277
Problem: The surface area of a function, f (x, y), over
some region R, is defined as:
S =
¨R
√f 2x + f 2
y + 1dA (463)
Compute the surface area of the function f (x, y) = 12x
2 +
12y
2 over the unit disk.
Solution: Lets first take our partial derivatives that
we will have to utilize in our surface area formula:
fx = x (464)
fy = y (465)
Plugging this into our formula, we obtain that:
S =
¨R
√x2 + y2 + 1dA (466)
278
Where R is the unit disk. At this point, the integral
smells a lot like a conversion to polar coordinates. As
such, lets us make the conversion now, noting that the
unit disk has bounds of 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. We
obtain that:
S =
ˆ 2π
0
ˆ 1
0
√r2 + 1rdrdθ (467)
We can now solve this integral using u-substitution. Let-
ting u = r2 + 1, we obtain that:
du
2= rdr (468)
Plugging this into our expression, noting that the bounds
we have on u are 1 ≤ u ≤ 2 we have that:
S =1
2
ˆ 2π
0
ˆ 2
1
√ududθ =
1
2
ˆ 2π
0
[2
3u
32
]2
1
dθ =1
3
ˆ 2π
0
232−1dθ =
2π
3
(2
32 − 1
)(469)
279
The major piece of advice to get out of this small collec-
tion of exercises is that it can be very beneficial to spend
a decent amount of time thinking about what coordinate
thinks best for solving the problem and then what exactly
the bounds are. The tricky part with integrals is knowing
what to put on your bounds. If we can master knowing
what, in each coordinate systems, bounding θ for example
looks like graphically, we will be able to cruise through
this section. The next lecture will focus on moving into
even more complex, and even custom!, coordinate sys-
tems that take the same ideas and apply them forward.
The motivation behind this is that in many cases, simply
sticking to cartesian and polar won’t work either because
we need more dimension in the case of polar, or we have
280
spherical like objects that are quite difficult to work with
under cartesian like systems.
20 Lecture XII on July 29, 2019
20.1 Integration in Spherical and Cylindrical Coordinates
Last time, we ended off at Polar coordinates! Today
we are going to be going to the 3D spaces in order to
compute functions over regions that are volumes. This is
really exciting, and you will be need to master this. It
will show up countless times throughout the rest of the
course as well as the ASE if you plan on taking that. As
such, I would encourage you to complete the homework
problems that were unassigned as they are great practice
for both. Suppose we have the following region:
281
Figure 9: Spherical Region
And, we want to integrate some function f (x, y, z) over
this region. That being said, we are going to need to
use integration techniques to compute this. This entails
that we are going to need to figure out our bounds for
this spherical like region. The keyword here is spherical.
Regions that appear spherical are going to be easier to in-
tegrate over spherical coordinates! Same goes with cylin-
drical type shapes using cylindrical coordinates. That
being said, let us introduce a general form of the integral
that we want to look at, calling this spherical-like region
282
R: ˚R
f (x, y, z)dV (470)
Where, given our conversions found in section 18.3, we
can write as:
˚R
f (ρ, θ, φ)ρ2 sinφdρdφdθ (471)
Where for our specific shape in figure 9, it appears that
the following bounds successfully describe the picture:
ˆ π2
π3
ˆ π2
0
ˆ 1
0
f (ρ, θ, φ)ρ2 sinφdρdφdθ (472)
Notice that we get a ρ2 sinφ when we use spherical co-
ordinates. We can justify this in multiple ways. What
we should think is that this factor is a correction fac-
tor since we are now using a funkier coordinate system!
A quick justification is the following. Consider just the
283
dρdφdθ. If this were to represent dV we would have a
problem! Why? Well, volume should have some sort of
length cubed from a purely dimensional analysis point of
view. However, looking at this term, we only have length
instead of length cubed. As such, we need to incorporate
a length squared somewhere, which is part of the reason
as to why we include the ρ2 sinφ into the expression. The
exact form of this comes from a Jacobean that we will
discuss later. This comes from the distortions you can
think of of the unit square as we move towards a curvy
coordinate system such as spherical coordinates. I think
the Jacobean greatly shows this which I promise we will
do during recitation once we cover this particular topic,
which I will then add back into the notes right here! Re-
mind me if I forget please. Lets move onto an example of
284
this now:
Problem: Consider a solid with density δ(x, y, z) =
(x2 + y2 + z2)−1 and which occupies the cone:
Figure 10: Cone
Find its mass.
Solution: We have some flexibility for solving this.
For example, in cylindrical coordinates, we see that we
have z = r to describe the equation of the cone. The
caviat here however is not necessarily the region, but it
is the density. If we look at our density function, we
285
do not have the nicest looking thing! In this case the
density is barking spherical coordinates and the shape
is barking cylindrical coordinates. Lets trudge forth on
this one with spherical coordinates. We are going to use
spherical coordinates due to δ. As such, we can represent
our mass, M , as:
M =
˚δdV (473)
M =
ˆ 2π
0
ˆ π4
0
ˆ secφ
0
δ(ρ, θ, φ)ρ2 sinφdρdφdθ (474)
To put in words, we sweep out all of the possible theta
values since our shape is circular in nature, some people
maybe even call it isometric (not important but you may
see the word flashed around). Next, we have that φ can
at most be π4 . We have this because remember that the
lateral sides of the cone are where z = r, and as such,
286
since the two are exactly equal to each other, we get that
π4 is the largest this can be. In addition, we sweep out
all the other φ values that are less than this since the
shape is filled. We finally get the ρ slices, We can start
off saying that ρ must be greater than or equal to zero to
get the lower bound. Then, we can use the definition of
phi to obtain that:
cosφ =z
ρ=
1
ρ(475)
Leading to the fact that:
ρ =1
cosφ= secφ (476)
As such, we can represent our mass, as, converting our
287
density to spherical coordinates and obtaining that δ = 1ρ2
M =
ˆ 2π
0
ˆ π4
0
ˆ secφ
0
1
ρ2ρ2 sinφdρdφdθ =
ˆ 2π
0
ˆ π4
0
ˆ secφ
0
sinφdρdφdθ = π ln 2
(477)
We just focused on spherical coordinates! Now, lets just
have a quick discussion on cylindrical coordinates. Since
cylindrical coordinates are really just polar coordinates
with the addition of the z-coordinate from the cartesian
coordinates, cylindrical coordinates are not a large step
away from polar coordinates. In fact, we can express the
conversion of to cylindrical coordinates as:
˚D
f (r, θ, z)rdrdθdz (478)
Over some region D. In discussion of the dimensional
analysis, lets do a quick check that the units of our vol-
ume element are in length cubed. We pick up a length
288
squared from drdz noting that dθ does not have any
length elements. Then, the addition r sitting out front of
the infinitesimals, allows us to have the volume element
represented by length cubed.
20.2 Custom Coordinate Systems
This section is really cool! Consider we have a region
bounded by xy = 1, xy = 3, y = 12x and y = 2x. This
region is really funky! There is not really a nice way for
us to use either cylindrical, spherical, or even cartesian
coordinates. That being said, we can introduce a custom
coordinate system. If you notice it appears that we are
taking y = cx cuts for c ∈ [1, 2]. In addition, we are also
taking slices along yx = d for d ∈ [12, 3]. This is looking
and smelling quite familiar at this point! It seems that
289
we are taking some new variable, say u = yx and v = xy
and giving them bounds! Namely, we note that, given the
way I have defined these new variables, u and v, we note
that u is bounded between 12 and 2, while v is bounded
between 1 and 3. Therefore, we can integrate a function
y2 say over this region bordered by these functions as:
¨D
y2dA =
ˆ 3
1
ˆ 2
12
y
xyxJdudv =
ˆ 3
1
ˆ 2
12
uvJdudv
(479)
Great this looks really nice! All we have here is some
function bounded by numbers, not even functions! The
tricky part is that J sitting right out front of dudv. That
J is called the Jacobean. For polar and cylindrical coor-
dinates, we say that J = r and for spherical coordinates
we say that J = ρ2 sinφ. Now, lets get a formula for a
general Jacobean that we will receive by transforming to
290
some custom coordinates. The Theorem getting all the
formality of this section is:
Theorem: Suppose that T : R2 → R2 is a differen-
tiable transformation that maps a region R one-to-one
onto a region D. Then, for any continuous function f ,
we have that:
¨D
f (x, y)dxdy =
¨R
f (T−1(x, y))
∣∣∣∣∂(x, y)
∂(u, v)
∣∣∣∣ dudv(480)
Where,
J =
∣∣∣∣∂(x, y)
∂(u, v)
∣∣∣∣ (481)
It should be noted that it is more-so useful to work with
J−1 a lot since we generally write our function u and v
291
as functions of x and y. As such, we can define J−1 as:
J−1 =1
J=
∣∣∣∣∂(u, v)
∂(x, y)
∣∣∣∣ (482)
For all of these, note that we are taking the absolute
value of the Jacobean. We do this so that changing the
orientation of our area doesn’t have any effect since we
only care about the area distortion not the orientation
distortion. We define:
J =
∣∣∣∣∂(x, y)
∂(u, v)
∣∣∣∣ = | det
∂x∂u
∂x∂v
∂y∂u
∂y∂v
| (483)
Similarly, we can define J−1 as:
J−1 =1
J=
∣∣∣∣∂(u, v)
∂(x, y)
∣∣∣∣ = | det
∂u∂x
∂u∂y
∂v∂x
∂v∂y
| (484)
As such, for our case, we see that since we have written
292
u and v as function of x and y, maybe we should solve
for J−1 and then take the reciprocal of this in order to
calculate J from this in our expression. Lets get on to do
this now for our example where u = yx and v = xy:
J−1 =1
J=
∣∣∣∣∂(u, v)
∂(x, y)
∣∣∣∣ = | det
∂u∂x
∂u∂y
∂v∂x
∂v∂y
| = | det
− yx2
1x
y x
| = |−2y
x| = 2u
(485)
Therefore, we take say that:
J−1 =1
J= 2u (486)
J =1
2u(487)
Lets now actually plug this into our example problem that
293
we encountered what seems like a page and a half ago!:
¨D
y2dA =
ˆ 3
1
ˆ 2
12
y
xyxJdudv =
ˆ 3
1
ˆ 2
12
uvJdudv
(488)¨D
y2dA =
ˆ 3
1
ˆ 2
12
y
xyxJdudv =
ˆ 3
1
ˆ 2
12
uv1
2ududv =
ˆ 3
1
ˆ 2
12
v
2dudv = 3
(489)
The point here is that the trouble, as is the moral with
this entire chapter, is not in the actual integral. The
problem is in setting the bounds for double and triple
integrals. The problem here is trying to find what u and
v that is a nice custom coordinate system to transfer over
to in order to best integrate over some region! Once we
choose these u and v variables that will be variables of
both x and y, we then can compute the Jacobean that
will allow to account for the area distortion.
294
20.3 Applications of Double and Triple Integrals (ASE)
There are two quantities that will most likely be refer-
enced by name on the ASE relating to double and triple
integrals. I will put them here for reference, and we have
done problems in workshop that relate to them so be sure
to check them out for additional practice.
20.3.1 Average Value of a Function
The average value of a function over some region D is:
Avg(f ) =
˜D f (x, y)dA˜
D 1dA(490)
In addition, we can also compute the average value over
some 3D region to get the analog that:
Avg(f ) =
˝D f (x, y, z)dV˜
D 1dV(491)
295
Some common things that pop up throughout the 18.02
course is things such as the average value of x or y over
the unit disk. By symmetry, both of these are exactly
zero. However, if it helps, I would definitely go on to
compute both of these utilizing the formula given above.
Very closely tied to this, we can compute the center of
mass of an object.
20.3.2 Center of Mass
Given a density function δ(x, y), the center of mass of
an object that occupies some region D is:
(xCOM , yCOM) =
(˜D xδ(x, y)dA˜D δ(x, y)dA
,
˜D yδ(x, y)dA˜D δ(x, y)dA
)(492)
Interpret this as finding the average value of the x and y
respectively over the mass of the object. In many cases,
296
the δ(x, y) = 1, and we do not even need to worry about
this. In this case, we are literally just taking the average
value of x and y over the region D. We have the 3D
analog of this as well:
(xCOM , yCOM , zCOM) =
(˝D xδ(x, y, z)dV˝D δ(x, y, z)dV
,
˝D yδ(x, y, z)dV˝D δ(x, y, z)dV
,
˝D zδ(x, y, z)dV˝D δ(x, y, z)dV
)(493)
21 Recitation VIII on July 30, 2019
In spirit of the last two recitations that I typed up
notes for, let me add some problems that I think are use-
ful to see as part of the lecture notes that come from the
worksheets! I will say that the overarching idea of what
we went over in class is that we can use more creative
coordinate systems in order to compute double or triple
integrals. Fortunately, we are not just limited to the co-
297
ordinate grid, but we have the freedom to branch out and
use spherical and cylindrical coordinates, and even cus-
tom coordinates with a Jacobian, in order to compute. I
will put some examples here that I liked from the work-
sheet :)
Problem: Back in the day, Archimedes (without any
knowledge of calculus) calculated both the surface area
and volume of two intersecting cylinders on their axis.
This is known as a ”Groin Vault”. Given the infinite
cylinder x2 +y2 = 1 and the infinite cylinder x2 +z2 = 1,
calculate the volume that is encompassed in the intersec-
tion of the two cylinders.
Solution: We can try and find bounds for all three of
298
our variables. We can first note that −1 ≤ x ≤ 1. Since
this is the case, we have that −√
1− x2 ≤ y ≤√
1− x2
as well as −√
1− x2 ≤ z ≤√
1− x2. As such, we can
set up our triple integral in order to try and compute the
volume of the groin vault.
˚dV =
ˆ 1
−1
ˆ √1−x2
−√
1−x2
ˆ √1−x2
−√
1−x21dzdydx (494)
=
ˆ 1
−1
ˆ √1−x2
−√
1−x2[z]√
1−x2
−√
1−x2 dydx =
ˆ 1
−1
ˆ √1−x2
−√
1−x22√
1− x2dydx
(495)
=
ˆ 1
−1
2√
1− x2 [y]√
1−x2
−√
1−x2 dx =
ˆ 1
−1
(2√
1− x2)2
dx
(496)˚dV =
ˆ 1
−1
4(1− x2)dx =
[4x− 4x3
3
]1
−1
= 2
(4− 4
3
)=
16
3
(497)
Surprisingly there is no π in the answer which comes as
299
a shock to many since the integral is essentially barking
to utilize some form of cylindrical coordinates perhaps.
Problem: Evaluate˝
16zdV over the region E, where
E is the upper half of the sphere x2 + y2 + z2 = 1.
Solution: Converting everything to spherical coordi-
nates our integral becomes:
˚16zdV =
ˆ 2π
0
ˆ π2
0
ˆ 1
0
16ρ cosφ(ρ2 sinφ)dρdφdθ
= 2π
ˆ π2
0
2 sin 2φdφ
= 2π [− cos 2φ]π20 = 4π
Problem: Evaluate˝ √
3x2 + 3y2 for the Solid bounded
300
by z = 2x2 + 2y2 and the plane z = 8
Solution: We note that the intersection of z = 2x2 +
2y2 and z = 8 is a circle satisfying the equation
4 = x2 + y2. (498)
This means that we will be integrating over a circular re-
gion. For that reason, switching over to polar cylindrical
coordinates is a good idea. In polar cylindrical coordi-
nates, our two functions bounding z above and below
become
z = 2r2 and z = 8. (499)
We are interested in a circular intersection of radius 2, so
θ ranges from 0 to 2π and r ranges from 0 to 2. Thus,
301
the triple integral over this 3d domain D is given by
˚D
√3x2 + 3y2dV =
˚D
√3r2rdzdrdθ (500)
=
ˆ 2π
0
ˆ 2
0
ˆ 8
2r2r2√
3dzdrdθ
(501)
At this point, we evaluate
=√
3
ˆ 2π
0
ˆ 2
0
r2(8− 2r2)drdθ (502)
=√
3
ˆ 2π
0
8(2)3
3− 2(2)5
5dθ (503)
= 2π√
3
(64
3− 64
5
)(504)
= 128π√
3
(1
3− 1
5
). (505)
Problem: Evaluate˜x2 + 2xy+y2dA over R where
R is the region bounded by the curves x+y = 2, x+y = 4,
y − x = 1 and y − x = −1.
302
Solution: We can start by defining custom coordi-
nates, u and v as follows:
u = x + y (506)
v = y − x (507)
Where u ∈ [2, 4] and v ∈ [−1, 1]. We can compute the
inverse Jacobian matrix:
J−1 =1
J= det
∣∣∣∣∂(u, v)
∂(x, y)
∣∣∣∣ =
1 1
−1 1
= 2 (508)
Therefore, we have that the Jacobian of this transforma-
tion is J = 12. We can implement the transformation and
303
compute the integral over u and v:
ˆ 4
2
ˆ 1
−1
(x2+2xy+y2)1
2dvdu =
ˆ 4
2
ˆ 1
−1
(x+y)21
2dvdu =
1
2
ˆ 4
2
ˆ 1
−1
u2dvdu =
ˆ 4
2
u2du =56
3
(509)
Problem: Verify that dV = ρ2 sin(φ)dρdφdθ
Solution: We can compute a 3-dimensional Jacobian
as follows:
J =
∣∣∣∣∂(x, y, z)
∂(ρ, φ, θ)
∣∣∣∣ (510)
Lets now write our expressions that represent conversions
between Cartesian and spherical coordinates:
x = ρ sinφ cos θ (511)
304
y = ρ sinφ sin θ (512)
z = ρ cosφ (513)
The Jacobian matrix can be expressed as:
J =
∣∣∣∣∣∣∣∣∣∣det
∂x∂ρ
∂x∂φ
∂x∂θ
∂y∂ρ
∂y∂φ
∂y∂θ
∂z∂ρ
∂z∂φ
∂z∂θ
∣∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣det
sinφ cos θ ρ cosφ cos θ −ρ sinφ sin θ
sinφ sin θ ρ cosφ sin θ ρ sinφ cos θ
cosφ −ρ sinφ 0
∣∣∣∣∣∣∣∣∣∣
(514)
Lets now compute the determinant of this matrix:
J = sinφ cos θ det
ρ cosφ sin θ ρ sinφ cos θ
−ρ sinφ 0
(515)
− ρ cosφ cos θ det
sinφ sin θ ρ sinφ cos θ
cosφ 0
(516)
305
− ρ sinφ sin θ det
sinφ sin θ ρ cosφ sin θ
cosφ −ρ sinφ
(517)
J =∣∣+ρ2 sin3 φ cos2 θ + ρ2 cos2 φ sinφ cos2 θ − ρ sinφ sin θ
(−ρ sin2 φ sin θ − ρ cos2 φ sin θ
)∣∣(518)
J =∣∣+ρ2 sin3 φ cos2 θ + ρ2 cos2 φ sinφ cos2 θ + ρ sinφ sin θ
(ρ sin2 φ sin θ + ρ cos2 φ sin θ
)∣∣(519)
J =∣∣+ρ2 sin3 φ cos2 θ + ρ2 cos2 φ sinφ cos2 θ + ρ sinφ sin θ (ρ sin θ)
∣∣(520)
J = ρ2 sinφ∣∣sin2 φ cos2 θ + cos2 φ cos2 θ + sin2 θ
∣∣ (521)
J = ρ2 sinφ∣∣(sin2 φ + cos2 φ) cos2 θ + sin2 θ
∣∣ (522)
J = ρ2 sinφ∣∣cos2 θ + sin2 θ
∣∣ (523)
J = ρ2 sinφ (524)
Therefore, since our Jacobian is equal to: J = ρ2 sinφ,
We can express the volume element in spherical coordi-
306
nates as:
dV = ρ2 sinφdρdφdθ (525)
22 Lecture XIII on July 31, 2019
22.1 Vector Fields
Welcome back! Today we are going to be diving into
the vector calculus portion of the course. This will prob-
ably seem a bit new to all of you, so I recommend reading
through these portions a bit more slowly. This is what
we have been building up for, as well as all applications
to look at. We start by defining the vector field.
Definition: A vector field in R3 is a function, f :
R3 → R3. A vector field in R2 is a function, f : R2 → R2.
307
We probably have all come in contact with a few vec-
tor fields before. One can think of a gravitational field,
electric field, and even flows fields of fluids all as great ex-
amples of vector fields. We draw a vector field by drawing
a vector representing the output of the function, and we
place this at the point that is the input to the function.
As such, the location of each arrow can be thought of
as the domain, and the arrow itself can be thought of as
the co domain as a way of visually representing it. Given
that we can write our vector field, ~F (x, y, z), we can try
to construct a way to represent the vector field. Lets see,
suppose we have the following image,
308
Figure 11: Vector Field 1, Gravitational Attraction.
and we are trying to write a vector field to this point. All
the arrows are pointing towards the source point. We
want to write a vector field that has arrows of increasing
size as we approach (1, 1, 1) pointing towards (1, 1, 1). As
such, we can obtain that the vector field can be expressed
as:
~F (x, y, z) = − GMm
((x− 1)2 + (y − 1)2 + (z − 1)2)32
(x−1, y−1, z−1)
(526)
The constants GMm are just for the physical interpreta-
309
tion. However, we could omit those and still capture the
vector field. Lets move on to an easier example.
Problem: Find a vector field, ~F : [0, 2]× [0, 1]→ R2
whose plot looks like this:
Figure 12: Vector Field 2, Shear Flow.
Solution: Lets see the patterns we have here. As y
increases, the vector that appears to point totally in the
x direction decreases in magnitude. Notice all the vectors
in our little square all are pointing in the positive x di-
rection, and the magnitude does not change as we move
along any horizontal line. Since there is no y component,
310
and the magnitude of the arrow pointing in the x direc-
tion shrinks as we move up, leading to the vector field
being expressed as:
~F (x, y) = (1− y, 0) (527)
Sometimes in classes that you may take in the future,
you will see this type of vector field referred to as a shear
flow! For those interested, this is Not curl-free, which
generally catches students by surprise. Lets move on to
start to talk about work.
22.2 Work in Vector Fields
We recall from physics that:
W = ~F · ~d (528)
311
This tells us that work can be expressed as the dot prod-
uct between force and the displacement of the object.
Now we want to generalize for paths that are not neces-
sarily straight. Specifically, lets discuss the work it takes
to move a particle along a path C, in a vector field ~F is:
n∑k=1
(~F (~r(tu))
)· (~r(tu)− ~r(tu−1) (529)
As we take the limit of k →∞, we can replace the sum,
of over-dramatic notation and proof stuff that distracts
from the point, and we simply boil this expression down
to the real expression for work that is:
W =
ˆ r2
r1
~F (~r(t)) · dr (530)
A much more workable way to utilize this expression is
through a parametrization. Suppose that you parametrize
312
~r(t). Then, you can express the above expression for work
as:
W =
ˆ t2
t1
(~F (~r(t)) · d~r
dt
)dt (531)
Where we just get a function of t that is just a single vari-
able integral that we can handle with our knowledge of
single-variable calculus. The above expressions is what we
will be extensively working with throughout the course, so
I encourage you to get this down pat. Lets get a theorem
to justify why it does not matter what parametrizations
we take of the same path C.
Theorem: If ~r1 and ~r2 are parametrizations of the
313
path C, and ~F is a vector field,
W =
ˆ t2
t1
(~F (~r1(t)) · d~r1
dt
)dt =
ˆ t2
t1
(~F (~r2(t)) · d~r2
dt
)dt
(532)
All this is saying in English is that if I can write a ~r(t)
that describes the curve C, then this is a totally valid
parametrization. There really is not any special parametriza-
tion that you must utilize. Lets do an example. Suppose
we have the curve y2 = x between (0, 0) and (1, 1). Then
a totally valid paramatrization as:
~r(t) = (t2, t) for t ∈ [0, 1] (533)
Another totally valid parametrization is:
~r(t) = (2t2, 2t) for t ∈ [0,1
2] (534)
You can see that basically sticking a constant out front
314
does not matter since we just adjust the time in the in-
terval. Another parametrization is that:
~r(t) = (t,√t) for t ∈ [0, 1] (535)
Also valid! The difference between this and the first
parametrization is that the one will start off much faster
and finish slower than the first.
22.3 Fundamental Theorem of Vector Calculus
In general it is not true that given we have two paths,
C1 and C2 with the same start and end points do not
have the same work being done. Namely:
ˆC1
~F (~r(t)) · d~r 6=ˆC2
~F (~r(t)) · d~r (536)
315
22.3.1 Conservative Vector Fields
However, this is true for conservative vector fields.
Conservative vector fields are defined as, given ~F is a
conservative vector field,
~F = ~∇f (537)
where f is just some function, f : Rn → R, that we
will refer to as a potential function. For a conservative
vector field, it is true that given we have two paths, C1
and C2 with the same start and end points do have the
same work being done. Namely:
ˆC1
~∇f (~r(t))·d~r =
ˆC1
~F (~r(t))·d~r =
ˆC2
~F (~r(t))·d~r =
ˆC2
~∇f (~r(t))·d~r
(538)
316
In general, conservative vector do not depend on the path
taken between points ~r(a), the start point, and point ~r(b),
the end point. As such, the climactic piece of information
is that for a conservative vector field, ~F = ~∇f , the fol-
lowing equation holds, that is denoted as the fundamental
theorem of vector calculus. Namely,
W =
ˆC
~F (~r(t))·d~r =
ˆC
~∇f (~r(t))·d~r = f (~r(b))−f (~r(a))
(539)
Why is this true you may ask? Well, let me provide you
with a better back of the envelope proof. Given that
~F = ~∇f , we can express a work integral as:
W =
ˆC1
~F (~r(t))·d~r =
ˆC1
~∇f (~r(t))·d~r =
ˆC1
(~∇f (~r(t)) · d~r
dt
)dt
(540)
317
Notice that in the last equality, the portion in parentheses
is just the multivariable chain rule that represents dfdt ! As
such, we obtain that:
W =
ˆC1
(~∇f (~r(t)) · d~r
dt
)dt =
ˆC1
df
dtdt =
ˆC1
df = f (~r(b))−f (~r(a))
(541)
Where ~r(a) is the starting point and ~r(b) is the ending
point.
Just to have it stated in all its glory, The Fundamental
Theorem of Vector Calculus states that if C is a path
from ~a to ~b and f is a differentiable function, then:
ˆC
~∇f · dr = f (~b)− f (~a) (542)
318
22.3.2 Checking Conservative Fields
Note that if ~F = (M,N) = ~∇f , where M is the first
component of the vector field and N is the second com-
ponent of the vector field, then the following relations
between M,N , and ~∇f is that:
M =∂f
∂x(543)
N =∂f
∂y(544)
Then, we say that a field is conservative if:
∂M
∂y=∂N
∂x(545)
This is another neat application of Clairout’s Theorem!
This is totally sufficient to see if ~F actually is conserva-
tive or not conservative. The Theorem simply states that:
319
Theorem: If My = Nx for a vector field ~F = (M,N),
then ~F is a conservative vector field, under the assump-
tion that ~F is differentiable everywhere on R2.
22.4 Green’s Theorem
This section is super neat and is actually just a spe-
cific case of the general Stoke’s Theorem that we will
encounter later.
˛~F · d~r =
¨D
(Nx −My) dA (546)
Where¸
simply means a closed integral. You may see this
a ton for the rest of the class. Consider first a conservative
vector field. If a conservative vector field has that ~F =
~∇f , then the work of this vector field is simply equal to
f (b)− f (a) where b is the endpoint and a is the starting
320
point. However, for a closed loop, the starting and end
points are exactly the same! As such, the work around
any closed loop for a conservative vector field is exactly
zero. We can also see this by looking at the right hand
side of the above equation. Since for a conservative vector
field, Nx = My, then the RHS will always evaluate to
zero for a conservative vector field over a closed loop!
We covered a lot today, so hopefully it was not all too
overwhelming.
23 Recitation IX on August 1, 2019
Let me add in a few examples from workshop that will
prove helpful on Exam’s and the ASE.
Problem: Evaluate the work being done by the vector
321
field, F = 〈2x, 3y, 4z〉 along a helical path that starts at
(0, 0, 0) and stops at (0, 0, 1).
Solution: We can begin by finding a potential func-
tion f for F, that is, a scalar function f whose gradient,
−→∇f = F. By inspection (taking into account that each
component of F should be a corresponding partial deriva-
tive of f, we can come to the conclusion that a possible f
is:
f (x, y, z) = x2 +3
2y2 + 2z2
Once we’ve accomplished this, we can employ the FTOVC
322
which states that:
ˆ (0,0,1)
(0,0,0)
F · dr =
ˆ (0,0,1)
(0,0,0)
−→∇f · dr (547)
= f (0, 0, 1)− f (0, 0, 0) (548)
= 2− 0 (549)
= 2 (550)
Problem: Suppose an object is moving in a vector
field, F, such that:
F =
⟨−x
(x2 + y2 + z2)32
,−y
(x2 + y2 + z2)32
,−z
(x2 + y2 + z2)32
⟩(551)
along the path r(t) = 〈1 + t, t3, t cos(πt)〉 from t = 0 to
t = 1. Find the work done by this vector field on the
323
object.
Solution: We begin in the same manner as our previ-
ous problem where we can come to the conclusion that a
possible f is:
f (x, y, z) =1
(x2 + y2 + z2)12
Which means that the question now is what is the differ-
ence in the value of f at t = 1 vs at t = 0, that is from
(1,0,0) to (2,1,-1). So we have
W = f (2, 1,−1)− f (1, 0, 0) (552)
=1√6− 1 (553)
Problem: Find the Work done on a particle that goes
324
through a Force field F = 〈y,−x〉 through a triangular
path starting and ending at the origin and going through
the points (1,0) and (1,1) with a) a line integral and b)
Green’s Theorem
Solution: We begin by noting that the vector field
that we’re given is not a conservative vector field and
thus, we cannot use the FTOVC. However, we can think
of our path as a superposition of 3 line segments and
evaluate the work done on our particle through each of
those line segments and add the results up to get the total
work done. The paths can be parameterized as follows:
1. r1(t) = 〈t, 0〉 for t ε [0,1].
2. r2(t) = 〈1, t〉 for t ε [0,1].
3. r3(t) = 〈t, t〉 for t ε [0,1] (but from 1 to 0.
325
So then our work calculation reduces to:
W = W1 + W2 + W3 (554)
=
ˆF · dr1 +
ˆF · dr2 +
ˆF · dr3 (555)
=
ˆF(r1(t)) · r′1dt +
ˆF(r2(t)) · r′2dt +
ˆF(r3(t)) · r′3dt
(556)
=
ˆ 1
0
〈0,−t〉 · 〈1, 0〉dt +
ˆ 1
0
〈t,−1〉 · 〈0, 1〉dt +
ˆ 0
1
〈t,−t〉 · 〈1, 1〉dt
(557)
= 0− 1 + 0 (558)
= −1 (559)
Now, using green’s theorem, the first thing to do is take
the corresponding partial derivatives and take their dif-
ference to find the integrand:
Nx −My = −1− 1 = −2
326
Now that we have our integrand we just need to integrate
that function over the triangle to get our work which turns
into -2 times the area of our triangle (which is 12) and thus,
our work is −1 .
Problem: Suppose you have a force field, F = 〈x3,−y4〉.
In addition, assume an object is moving through the force
field along a circular path where the path is described by
r(t) = 〈cos(2πt), sin(2πt)〉 from t = 0 to t = 1. Show
that the work done is zero through means of a line inte-
gral (math), the fundamental theorem of calculus (math),
and through a direct statement (all words).
Solution: We first show this through a direct compu-
327
tation:
˛C
F · dr =
ˆ 1
0
〈x3,−y4〉 · 〈−2π sin(2πt), 2π cos(2πt)〉dt
(560)
=
ˆ 1
0
〈cos3(2πt),− sin4(2πt)〉 · 〈−2π sin(2πt), 2π cos(2πt)〉dt
(561)
= −2π
ˆ 1
0
cos3(2πt) sin(2πt) + sin4(2πt) cos(2πt)dt
(562)
= −2π
ˆ 1
0
cos3(2πt) sin(2πt)dt− 2π
ˆ 1
0
sin4(2πt) cos(2πt)dt
(563)
1 (564)
Each of these integrals can be evaluated through a u-
328
substitution. Let
u1 = cos(2πt) and u2 = sin(2πt). (565)
Then, we have
du1 = −2π sin(2πt)dt and du2 = 2π cos(2πt)dt.
(566)
Making this change of variables we realize that we would
be integrating from 1 to 1 and from 0 to 0 respectively so
the value of our integral is 0 .
Using the FTOVC we quickly release that the function
is conservative and that the starting and ending points
are the same and thus we obtain 0 as a result.
329
24 Lecture XIV on August 2, 2019
Last class, we ended with Green’s Theorem. Lets pick
up right where we left of:
24.1 Green’s Theorem
˛C
~F · d~r =
¨D
(Nx −My) dA (567)
Where ~F = (M,N) Where C is a closed loop, and D
is the region enclosed by the curve C. We, on the left
hand side, have an expression that tells us to calculate
the work as I go around some path that starts and ends
at the same point. The right-hand side can be thought
of as the curl of the vector field integrated over the area.
The reason why we must have a closed loop, is because
we must trap some area. Well, apparently an intuition
330
for Green’s Theorem is doing an approximation of a work
integral over an infinitesimal square’s border... who knew!
Anyways, we are finally done with that proof, lets move
on to Surface Integrals.
24.2 Surface Integrals
If S is the surface and f : S → mathbbR is a function,
then we can make the equation that:
ˆS
fdA = limnumber of patches→∞
∑patch
f (patch)area(patch)
(568)
A way we can try to conceptualize what is going on is
thinking about trying to find the average global temper-
ature. Namely, we are setting lets have some function,
temperature, that we integrate over the surface of the
earth. Then, we can divide through by the total surface
331
area to extract the average temperature. This is an ap-
plication of average value that we went over in workshop
with some example problems!
Avg(f ) =
´S fdA´S dA
(569)
where S is the surface, and f : S → R. Lets try to polish
off this idea with an example:
Problem: Find´S fdA, where f (x, y, z) = 2x2 +
2y2 + 2z2 over the unit sphere.
Solution: Well lets think about this a bit before we
dive in. Since we are trying to integrate over the unit
sphere, by definition, x2 + y2 + z2 = 1 always! This is
the surface of the unit sphere. Therefore, we simply have
332
that:
2
ˆS
x2 + y2 + z2dA = 2
ˆS
1dA = 2(4πR2) = 8π (570)
since R = 1 for the unit sphere. If we persay wanted to
calculate the average value, we could simply divide our
answer by the total surface area, 4π, leaving us with just
an average value of 2 We kind of expected this to happen
considering that the function f is always equal to 2 over
the surface of the unit sphere.
24.3 Parametrizing Surfaces (ASE)
This section is useful for the ASE, but it is not neces-
sarily tested in this course. Consider the points (x, y, z)
with y = sin 7z for x ∈ [0, 1] and z ∈ [0, π2 ]. Although
there does exist a formulaic way to parametrize a surface,
333
in this case, we can parametrize this surface by inspection
as:
~r(u, v) = (v, sin 7u, u) (571)
The idea here is we are trying to express a surface of 3
variables but only use two variables, u and v. Lets move
forward with another example:
Example: Parametrize the part of the unit sphere in
the first octant
Solution: We can bring in spherical coordinates if we
are talking about spherical-like shapes! we can think of
our parametrization of the unit sphere in terms of θ and
φ for our parametrization, where for the unit sphere in
the first octant, we obtain that: θ ∈ [0, π2 ] and φ ∈ [0, π2 .
334
As such, we can express our x, y, and z coordinates as:
~r(θ, φ) = (sinφ cos θ, sinφ sin θ, cosφ) (572)
where we basically have expressed spherical coordinates
by letting ρ = 1 in the case of the unit sphere. We can
express the formula for surface area as:
¨S
fdA =
¨D
f (~r(u, v))|~ru × ~rv|dudv (573)
Where S is the surface that we are integrating over, and
D is the region, below the surface. The thing that we are
taking the magnitude of can be thought of as a jacobian
that is transforming our funky surface region into a nice
flat region D that is just a rectangle with the bounds on
θ and φ. Lets break this down piece by piece. WE are
first saying, okay f was given to be in terms of x, y, and
335
z. I am going to hell with these coordinates and plug
in my parametrization of ~r into f . In the case that we
are simply trying to find the surface area itself, we set f
equal to one. However, in the case it was equal to a func-
tion, we would plug in our parametrization in terms of θ
and φ for our x, y, and z. Then we are going to calculate
this Jacobian looking object sitting our front of our dudv.
What this notion means is we are going to take the partial
derivatives of our parametrization with respect to u and
v, cross those, and then take the magnitude of this cross
product to represent the area distortion. Then, our dudv
is going to be the two variables that we are parametrizing
with respect to. So, for the case of this problem specifi-
cally, we have that we are parametrizing with respect to
θ and φ where each of these two variables are bounded
336
between [0, π2 ] giving us our bounds for the integral. Let
me compute this problem thoroughly so that you can see
this after class. Lets move on to another example in the
meantime.
Problem: Find the average value of z over the upper
half of the unit sphere.
Solution: we have that, we can take use of the parametriza-
tion of the unit sphere, and simply limit φ[0, π2 and have
θ ∈ [0, 2π], with the parametrization that:
~r(θ, φ) = (sinφ cos θ, sinφ sin θ, cosφ) (574)
337
yielding the integral that:
ˆf
dA =
ˆ 2π
0
ˆ π2
0
z|~rθ × ~rφ|dφdθ = π (575)
Then, from here, we can divide through by the surface
area, 2π in order to obtain the average value of 12.
24.3.1 A Better Treatment
So, we touched on surface integrals in class but lets get
down a more formulaic way of attack. Consider that I
have a surface that exists in 3D space, but I want to write
my 3-dimensional object just with only two variables, this
is how we will be approaching surfaces. Lets start off with
how to parametrize a surface and then move forward to
how to set up a surface integral. So perhaps you want to
parametrize, the part of the surface, z = 1−x2−y2 that
lies above the xy plane. Well our task is two be able to ex-
338
press this surface in terms of only two variables, which we
maybe shall call u and v to stay in line with the notation
you will see. well, similar to how we parametrize a line
whilst doing work integrals, lets now try to parametrize
this surface. Luckily, all three of our variables are written
in terms of only x and y! Therefore, we can express the
surface in the following parametrization.
~r(u, v) = (u, v, 1− u2 − v2) (576)
where the bounds of our u and v will trace out the unit
disk given that this is the ’shadow’ if you will of our sur-
face on the xy plane. Great! Okay, now that we have
parametrized this surface, lets now compute the surface
area!
339
The formula for surface area is:
S =
¨A
|~ru × ~rv|dudv (577)
The thing that we need to first compute is |~ru× ~rv|. We
can do so with the following:
~ru = (1, 0,−2u) (578)
~rv = (0, 1,−2v) (579)
Which when we take the cross product and the take the
magnitude of this cross product obtain that:
|~ru × ~rv| =√
4(u2 + v2) + 1 (580)
which I can then plug into my formula for the surface
340
area to obtain that:
S =
¨A
√4(u2 + v2) + 1dudv (581)
Where now it will be advantageous to switch over to cyl-
nidrical coordinates given our ’shadow’ region of the unit
disk and integrand. Namely, we obtain that:
S =
ˆ 2π
0
ˆ 1
0
√4r2 + 1rdrdθ (582)
Which we can solve to compute the surface area. I will
say that we the worksheets throughout the course alluded
to this formula in a way. That is that we can alternately,
given we have z = f (x, y), express surface area of a sur-
face over a region as:
S =
¨A
√f 2x + f 2
y + 1dA (583)
341
Which you can see exactly fits the bill as what we ended
up obtaining doing the more formal parametrization! Both
work, so just choose which one that you like! The most
common thing you will probably be asked it to parametrize
some surface of a sphere of radius a. Well, you might
think, okay, I have spherical coordinates to hopefully parametrize
this, but my spherical coordinates are written in terms
of 3 variables, and I can only write my parametrization
in terms of 2 coordinates! However, since the radius, ρ
is fixed, since we are concerned with the surface of the
sphere, we actually only have 2 variables, θ and φ and as
such, we can express the parametrization of a sphere of
radius a as:
~r(u, v) = (a sinu cos v, a sinu sin v, a cosu) (584)
342
Nice. Now we can do this exact same thing as previously
done, that is, compute ~ru and ~rv to then take the cross
product and magnitude of. The last shape that I see
parametrized a lot is a cone! A cone has the surface,
z =√x2 + y2. Which we can conveniently parametrize
as:
~r(u, v) = (u, v,√u2 + v2) (585)
In all honesty, there is a bit of ugly and tedious labor that
comes with the cross product being involved, but the over-
arching idea is not all too bad. Namely, we are integrating
over the region that lies underneath, some shadow region,
of our surface that exists in 3-dimensional space. We are
then parametrizing or surface to find some correctional
factor, |~ru× ~rv to then integrate over the shadow region!
With that I think that is all that is necessary for ASE
343
prep.
24.4 Flux
A natural application of surface area is flux. We can
think, suppose we have some vector field that is flowing
our in space, perhaps think of water flowing through a
net. That being said, suppose we want to see how much
water actually does flow through that net. Well, we can
think lets see how much stuff goes through our surface S,
namely we can say that:
ˆS
~F · d ~A =
¨~f (~r(u, v)) · (~ru × ~rv)dudv (586)
Where:
d ~A = (~ru × ~rv)dudv (587)
344
We utilize the dot product here because we want to get
only the stuff that penetrates the surface S. Think of
that as motivation for taking the dot product since we
are essentially filtering out all the stuff, vector field, that
is not going through the surface. I find it helps if you
think of a wall with holes punctured out. Now, consider
that you have a hose. First, you decide let me point the
hose right at some of these small holes. Well then, we
would expect a ton of water to make it through the holes
and to the other side. Then suppose I start to veer off
to an angle from the wall, and aim the hose. Well now,
water is still going through the holes, but not as much
since since we are at an angle, some is not making it
through like it normally would. Namely, the components
of the water from my hose not in the direction of the hole
345
are making it through. Now finally, consider the last case
where I am shooting the water in line with the face of the
wall, namely perpendicular to the normal vector of the
wall. Well now, none of the water is going through! All
my water is running along the sides of the wall and as
such, none can really come through! Hopefully this helps
internalize this a bit!
24.5 Divergence
Divergence is a measure of the flow density! We com-
pute the divergence of a vector field, ~F as:
~∇ · ~F =
(∂
∂x,∂
∂y,∂
∂z
)· (M,N,P ) (588)
Where F = (M,N,P )
346
25 Lecture XV on August 5, 2019
Hey all! Sorry I couldn’t be in attendance today, I had
to do something for the Office of Minority Education.
Anyways, I still want to ensure that you have all the
tools that you need in order to do great on the Final
exam this Thursday. I want to start off with reviewing
Divergence, go over curl, and then introduce the final
two big equations of the course; Divergence and Stokes’
Theorem.
25.1 Divergence
We define the divergence of a vector field, ~F = (M,N,P )
as:
div(~F ) = ~∇ · ~F =
(∂
∂x,∂
∂y,∂
∂z
)· (M,N,P ) (589)
347
The equation by itself is not all too bad. Lets do an exam-
ple and then complete a discussion on what this quantity
represents.
Problem: Compute the divergence of the vector field
~F = (x2y, yx, z) at the point (1, 1, 0).
Solution: We can utilize the aforementioned equation
to express the divergence of the vector field ~F as:
div(~F ) = ~∇·~F =
(∂
∂x,∂
∂y,∂
∂z
)·(x2y, yx, z) =
∂
∂x(x2y)+
∂
∂y(yx)+
∂
∂z(z)
(590)
div(~F = 2xy + x + 1 (591)
Where, we can calculate the divergence at the point (1, 1, 0)
348
as:
div(~F (1, 1, 0)) = 2 + 1 + 1 = 4 (592)
Great! So that is the computation necessary to compute
the divergence of a vector field, ~F , at a point. Now, lets
see what it represents. Conceptually, the divergence rep-
resents for us how much ”stuff” (vector field) is flowing
in and flowing out a specific point. Namely, if we look
at an infinitesimal volume surrounding a point, so in our
previous case, (1, 1, 0), we would see, given the answer
is positive, that there is more stuff (vector field) flowing
out from the point than there is flowing in! Lets look
at another pictorial example now. Consider the following
two vector fields that we have seen before, that I will now
display below:
349
Figure 13: Divergence of Vector Field 1
In this case suppose I pick a point, that is not (1, 1, 1).
Then, I can conclude that in this case the divergence at
this some random point should be negative! Why? Well,
lets see. If we look at a point, lets say (1, 12,
12), Then we
notice that the size of the arrows representing the vector
field flowing in are larger than the size of the vectors flow-
ing out, we get a net negative vector field. Lets turn to
another example now.
350
Figure 14: Divergence of Vector Field 1
Here we have an interesting case. Lets look at perhaps the
point (12,
12). At this point, we see that the flow coming
in from left to right is exactly the same size as the flow
coming out from left to right. As such, this is what we
refer to as zero divergence! The overarching point here is
that we want to isolate one point in our mind, and then
pictorially ask ourselves whether the flow in is greater
than, less than, or equal to the flow out by looking at the
vector field that the point is exposed to. If the flow in is
greater than the flow out, we get a negative divergence.
If the flow out is greater than the flow in, we achieve a
351
positive divergence. Finally if the flow equal the flow out,
we achieve zero divergence. Of course, it is much easier to
simply compute the divergence by utilizing the equation,
but it also important that we conceptually master how to
look at a graph of flow and be able to say what sign the
divergence will have. Lets move on to curl now. Finally,
note that the divergence is a scalar. It is simply some
function that is not a vector! We can think of divergence
as a function, f : Rn → R.
25.2 Curl
Next we will discuss the Curl of a vector field. Lets
first present the formula, similar to divergence and then
follow through with the conceptual understanding. The
formula for curl, given that we have some vector field,
352
~F = (M,N,P ) is computed as :
Curl(~F ) = ~∇× ~F = det
i j k
∂∂x
∂∂y
∂∂z
M N P
(593)
Curl(~F ) = (Py −Nz,Mz − Px, Nx −My) (594)
Where Mx, for example, represents the partial derivative
of M with respect to x. Lets first note that taking the
curl of a vector field produces another vector. Namely,
the curl of a vector field can be thought of as a function,
f : Rn → Rn, which is this case, f : R3 → R3. Lets just
go through an example real quickly of an actual compu-
tation of the Curl of a vector field.
Problem: Compute the Curl of the vector field, ~F =
353
((x2y, yx, xz)
Solution: We can turn to our formula of the curl that
states that:
Curl(~F ) = ~∇× ~F = det
i j k
∂∂x
∂∂y
∂∂z
M N P
(595)
Curl(~F ) = ~∇× ~F = det
i j k
∂∂x
∂∂y
∂∂z
x2y yx xz
(596)
Curl(~F ) = (Py −Nz,Mz − Px, Nx −My) (597)
Curl(~F ) =(0− 0, 0− z, y − x2
)(598)
So perhaps, if we wanted to compute the curl at the point
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(2, 1, 0), we would obtain that:
Curl(~F (2, 1, 0)) = (0, 0,−3) (599)
Okay great! Computing it is quite annoying due to the
cross product, but we can trudge on through that with-
out too much worry. Lets try to get down the conceptual
understanding now. Curl can be described as a circu-
lation density. Namely, we want to think, If I place a
little paddle-wheel, something that will rotate in my vec-
tor field, at a specific point in my vector field, how much
and it what direction will it rotate. We define that if the
little paddle wheel we place at a point rotates counter-
clockwise due to the vector field, we have a positive curl.
In addition, if the little paddle wheel would rotate clock-
wise at the point, we would state that we have a negative
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curl. Lets try this out with a picture we have seen a few
times by now. Consider the vector field below:
Figure 15: Curl of Shear Vector Field
Imagine we want to compute the curl at the point (12,
12).
Well, lets try this conceptual method that we laid out.
Consider we place a little paddle wheel, baby box even if
you prefer, at the point in question. Then consider, how
the vector field at this point will make the box move, given
it is locked at the point. We see that the vector on the
lower end of the box will be larger than the vectors at the
upper end of the box. As such, we would expect that the
arrows would cause the box to rotate counterclockwise.
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As such, we would expect the curl at this point to be
positive. Lets test this out given the exact form of the
vector field. It turns out that we can express the vector
field of the aforementioned picture as:
~F = (1− y, 0, 0) (600)
Therefore, if we compute the curl of ~F , we obtain that:
curl(~F ) = (0, 0, 1) (601)
Namely, point your right thumb in the direction of the
curl vector, straight up, and the wrap your fingers around,
this will indicate the curl. Since your fingers will wrap
around counterclockwise, we obtain that the curl is posi-
tive, and even more so, the curl is always positive. So, for
any point in this little rectangle, we get a positive circula-
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tion density that will result in a positive curl. So, we can
use a formulaic way of deciding whether the curl is posi-
tive, leading to a counterclockwise rotation, or negative,
leading to a clockwise rotation by imagining placing a lit-
tle paddle-wheel at a specific point, and noting how the
vector field would spin around that little paddle wheel.
Lets move on to the applications of both of these ideas.
25.3 Divergence Theorem
The Divergence Theorem is extremely powerful and
comes up all over the place in physics and applied maths.
The Divergence Theorem states that Let E be a sim-
ple solid region and S is the boundary surface of E with
positive orientation. Let F be a vector field whose com-
ponents have continuous first order partial derivatives.
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Then, ‹S
F · ndA =
˚E
Div(F )dV (602)
Lets break this down because it looks quite hefty. On
the left hand side, we are saying that given we have some
closed surface, namely our surface represents a shell per-
haps that is trapping some 3 dimensional volume, then I
can compute the flux of my vector field ~F through this
shell. Basically, we are just saying compute the flux of ~F
through some closed surface. The loop in the double inte-
gral indicates that the surface is closed similar to the loop
in the single integral means that the loop is closed. Now,
the right hand side says that this flux is actually going to
be equal to the divergence of ~F through the volume that
is trapped by the surface. Lets do a quick example to see
what I am saying. In general, we are going to be asked
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to compute the flux through say unit cube from inside to
outside. And, instead of computing some nasty surface
integrals like the left-hand side would lead us to, we are
going to use the trick on the RHS to make out lives easier.
Lets do the example now to see what I mean.
Problem: Compute the flux through the unit cube of
the vector field, ~F = (x, x, x)
Solution: Well, if we saw this problem last week, per-
haps we would try to calculate the flux, flow, whichever
word you like, through each of the six faces. Doing this
would give us the LHS of divergence theorem. However,
since the unit cube is a closed surface, enclosing some
volume, we can instead compute the divergence of F and
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then integrate this over the volume of the unit cube. Lets
do that now:
‹S
F ·ndA =
˚E
Div(F )dV =
ˆ 1
0
ˆ 1
0
ˆ 1
0
(1+0+0)dxdydz = 1
(603)
Not too bad right? Lets do a bit more of a difficult ex-
ample now.
Problem: Compute the flux of the vector field ~F =
(x2, 2, 3) through the surface bounded by z = 0 and
z = 4− x2 − y2.
Solution: Now we can really see the power of Diver-
gence Theorem. Namely, it would really suck to compute
the surface integral over this shape. Instead we can turn
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to utilize the RHS of Divergence Theorem again:
‹S
F · ndA =
˚E
Div(F )dV =
˚(2x + 0 + 0)dV
(604)
Now, we can take use of our knowledge of triple integrals,
and convert this to cylindrical coordinates to solve that:
‹S
F · ndA = 2
ˆ 2π
0
ˆ 2
0
ˆ 4−r2
0
r cos θrdzdrdθ = 0
(605)
The result of zero means that the net flux through the
surface is exactly zero. We interpret that as the amount
of crap that flows into our cereal bowl looking thing is the
exact same amount of crap 16 that flows out of the cereal
bowl looking shape. I will add more and more examples
underneath the recitation section form tomorrow. For
now, lets move on to our last theorem of the class, Stokes’16Crap is just the Vector Field
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Theorem.
25.4 Stokes’ Theorem
Lets just start off with the definition and then unpack
all that is going on within the daunting definition, Let
S be an oriented smooth surface that is bounded by a
simple, closed, smooth boundary curve C with positive
orientation, namely we move around it counterclockwise.
Also let F be a vector field then,
˛~F · d~r =
¨S
(~∇× ~F
)· d ~A (606)
Okay so what does all of this mean. Lets give this a
similar analysis to divergence theorem. We first look at
the left-hand side. The left-hand side is a line integral.
More specifically, it is a closed line integral meaning that
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we perhaps have the boundary that could be a circle, a
square, some loop that is closed. We are then saying that
the work done by the vector field, ~F , as I walk around my
closed loop is going to be equal to the curl of ~F along the
surface who has a boundary that I initially walked upon.
Lets think about a trash bag for visualization. Consider
a trash bag. We have an opening, where we throw in
the trash, we have a border that surrounds then opening
and then we have the bag where all the trash goes in.
We are saying that the work it takes to move around the
hole of the trashbag the border, is equal to the amount
of ~∇× ~F that goes through the surface of the trashbag.
As a matter of fact, we have already seen an example of
Stokes’ Theorem whilst doing Green’s Theorem. If you
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recall, Green’s Theorem tells us that:
˛~F · d~r =
¨Nx −MydA (607)
This looks pretty similar to Stokes’ Theorem. In Fact it
is just a special case of Stokes Theorem. Consider the
following. Consider that you have a vector field, ~F =
(M,N,P ). In addition, you have that your surface is
just some area on the xy plane that would be definition
have a normal vector that is equal to ~n = (0, 0, 1). Think
of the area vector as just being similar to a plane vector.
Namely, the area vector is the vector that points normal
to the actual area surface. Therefore, if we compute ~∇×
~F , and then dot this with our area vector, we obtain that:
˛~F ·d~r =
¨S
(~∇× ~F
)·(0, 0, 1)dA =
¨Nx−MydA
(608)
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Thus, we basically use Stokes’ Theorem so that we never
have to compute an actual surface integral. Namely, the
RHS can be pretty rough if we have some surface that has
a really hard area vector, and as such, we can instead just
compute the work done around the border of the surface
in order to compute, the RHS of the equation.
25.4.1 Same Border, Different Surface
Another interesting trick that I find extremely helpful
is the following. Suppose you have two surface where the
unit circle is the border of the surface. The first surface,
S1 happens to be z = 1 − x2 − y2, that lies above the
xy plane and the second surface is just the unit disk, S2.
Now suppose I want to calculate
¨S1
~∇× ~F · d ~A (609)
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as well as: ¨S2
~∇× ~F · d ~A (610)
Each of these alone is a surface integral over the respective
surfaces, S1 and S2. However, we just learned that by
Stokes Theorem, each of these is equivalent to the work
is takes to move around the unit circle as the unit circle
is acting as the border of each of these shapes. Therefore,
we can conclude by transitivity that:
˛C
~F · d~r =
¨S1
~∇× ~F · d ~A =
¨S2
~∇× ~F · d ~A (611)
Which more importantly means that:
¨S1
~∇× ~F · d ~A =
¨S2
~∇× ~F · d ~A (612)
meaning that if we have two surfaces that share the same
border, then it must be true that˜S~∇× ~F ·d ~A are equal.
367
Lets see how we can use this to our advantage with the
example mentioned.
Problem: Compute˜S~∇ × ~F · d ~A for the surface
z = 1 − x2 − y2 that lies above the xy-plane, where
~F = (3, x, 4)
Solution: Well, parametrizing this surface would kind
of stink. So instead, we can use this idea of surface In-
dependence to compute this integral. We can utilize the
fact that we can replace our current surface with the unit
disk that has a much nicer normal vector of ~n = (0, 0, 1),
and compute the integral. We first compute the curl of
~F which happens to be:
~∇× ~F = (0, 0, 1) (613)
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I can then compute the integral by dotting this with my
normal vector ~n and integrating the area of the unit disk.
Doing so, we obtain that:
¨S
~∇× ~F · d ~A =
¨(0, 0, 1) · (0, 0, 1)dA =
¨dA = π
(614)
In addition, we can confirm this by taking the line integral
as well for good luck. We can start by parametrizing our
path,
~r(t) = (cos t, sin t, 0)) for t ∈ [0, 2π] (615)
Then we can set up the work integral as:
˛~F ·d~r =
ˆ 2π
0
(3, cos t, 4)·(− sin t, cos t, 0)dt =
ˆ 2π
0
cos2 t = π
(616)
Yay! We get the same answer! Hopefully this clears up
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Stokes’ Theorem a bit. I know this one tends to be tricky
this is why I recommend honestly never calculating some
nasty surface integral. Instead, either compute a line in-
tegral over a closed path, or use the ideas of surface In-
dependence in order to find a flat surface that shares the
border with the crazy surface that can easily be utilized
to calculate the work done. Last lecture of material we
made it!
26 Recitation X on August 6, 2019
In recitation, we computed a few great problems on
Stokes and Divergence Theorem. I’ll attach a few here
that we computed that can be used later for reference.
Problem: Divergence Theorem: Use the divergence the-
orem to evaluate˜S F ·dS where F = 〈xy,−1
2y2, z〉 and
370
the surface consists the paraboloid z = 4− (x2 + y2) and
the circle in the xy plane it encloses.
Solution: Note that this setup is symmetrical in cylin-
drical coordinates so we will use them for this problem.
Therefore, the bounds for the region are reduced to:
0 ≤ z ≤ 4− r2
0 ≤ r ≤ 2
0 ≤ θ ≤ 2π
Next, we calculate the divergence of the vector field which
is given by: (∇ · F) = ∂M∂x + ∂N
∂y + ∂P∂z = y − y + 1 = 1
The integral is then,
371
¨S
F · dS =
˚R
(∇ · F)dV (617)
=
ˆ 2π
0
ˆ 2
0
ˆ 4−r2
0
rdzdrdθ (618)
= 2π
ˆ 2
0
4r − r3dr (619)
= (2π)(8− 4) (620)
= 8π (621)
(622)
Problem: Given that for the groin vault, encountered
last week, we computed the volume to be,˝
dV = 163 ,
compute the flux through the groin vault given that ~F =
〈x, y, z〉.
372
Solution: We want to compute the flux through a
closed surface. Therefore, we can use divergence theorem,
which states
‹∂D
F · dA =
˚D
∇ · FdV, (623)
where ∂D is the surface that bounds a 3 dimensional
domain D. But note that ∇ · F = 3, so
˚∇ · FdV =
˚3dV (624)
= 3 · 16
3(625)
= 16. (626)
Problem: Stoke’s Theorem: Use Stokes’ theorem to
373
evaluate´C F · dr, where C is the triangle with vertices
(1, 0, 1), (0, 1, 1), and (0, 0, 1), oriented counterclockwise
when viewed from above, and F = (x+y2, y+z2, z+x2).
Solution: With our surface being this triangle lying
on the z = 1 plane, we simply get a normal vector that
is pointing straight upwards. Therefore, we can utilize
Stokes’ Theorem over this region to obtain that:
¨S
∇× F · dA =
˛C
F · dr.
Thus, we need only find˜S∇× F · dA.
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¨S
∇× F · dA =
ˆ 1
0
ˆ 1−x
0
〈−2z,−2x,−2y〉 · 〈0, 0, 1〉, dydx
=
ˆ 1
0
ˆ 1−x
0
−2ydydx
= −1
ˆ 1
0
(1− x)2dx
=(1− x)3
3|10
= −1
3
Problem: Use Stoke’s Theorem to compute
¨ (∇× ~F
)·
d ~A where ~F = (3, x2, 4) through the surface z = 4−x2−
y2 that sits above the xy-plane.
Solution: We can do this by computing both the left
375
and the right hand sides of Stokes’ Theorem. We first
should note that we are not limited to the surface z =
4 − x2 − y2. Specifically, we could choose any surface
that shares the border on the xy-plane that the surface
z = 4 − x2 − y2 does. To make matters simple. I will
choose the flat surface, with area vector in the k direction.
and have my surface be the disk of radius 2 since this
shares the border of x2 + y2 = 4 on the xy plane. Lets
go ahead now and compute the surface integral.
¨ (∇× ~F
)=
¨((0, 0, 2x) · (0, 0, 1)) dA =
ˆ 2π
0
ˆ 2
0
2r cos θrdrdθ = 0
(627)
Okay, so as of now, we expect that if we take the closed
loop integral around the boundary of our surface, the LHS
of Stokes’ Theorem, we also obtain that zero work has
been done. Lets try that now utilizing the parametriza-
376
tion that:
~r(t) = (2 cos t, 2 sin t, 0) for t ∈ [0, 2π] (628)
Doing so, we can compute the line integral as:
˛~F ·d~r =
ˆ 2π
0
(3, 4 cos2 t, 4)·(−2, 2 cos t, 0) dt =
ˆ 2π
0
8 cos3 tdt = 0
(629)
Which shows that we got the same value for each side of
stokes’ theorem. Each by itself would have been valid to
fully answer the question. I showed both just so that one
could see that they are equivalent.
Problem: Gauss’ Law in Electricity and Magnetism
is known as:
‹S
~E · ~ndA =1
ε0
˚ρdV (630)
377
Where ~E is the electric field, ε0 is the permitivity of free
space, and ρ is the charge density of your object. Using
you knowledge of Divergence Theorem, show that the
above expression is equivalent to:
∇ · ~E =ρ
ε0(631)
Solution: We can start by using the Divergence The-
orem on the LHS of Gauss’ Law. Doing so, we have the
equation:
‹S
~E · ~ndA =
˚ (∇ · ~E
)dV (632)
378
Which we can now plug into our original equation:
‹S
~E · ~ndA =
˚ (∇ · ~E
)dV =
1
ε0
˚ρdV (633)
Therefore, since we are integrating over the same exact
volume region, we can set the two arguments equal to one
another to obtain that:
∇ · ~E =ρ
ε0(634)
Which is referred to as the differential form of Gauss’
Law.
27 Lecture XV on August 7, 2019
Today we are going to be having a review day! Lets try
to get through as many problems as possible.
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28 Thank You
Thank you all for such a great summer! I hope you all
learned a lot from the course, and you feel prepared to
apply the concepts and principles learnt in the last few
weeks in all of your majors :)
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