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Interphase EDGE Calculus 3 Lecture/Recitation Notes
Jack-William Barotta
August 23, 2019
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Contents
1 Introduction 7
2 Lecture I on July 1, 2019 9
2.1 How to Think About Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 n-dimensional space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Graph of an Equation in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.5 Level Curves of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Recitation I on July 2, 2019 13
4 Lecture II on July 3, 2019 15
4.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2 Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.3 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4.4 2D Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4.5 3D Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.6 Matrix Operations (ASE) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.6.1 Solving a Linear System . . . . . . . . . . . . . . . . . . . . . . . . . 26
5 Lecture III on July 5, 2019 27
5.1 Introduction to Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5.1.1 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.1.2 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.2 More Vector Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
5.3 Applying these concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2
5.4 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5.4.1 Small Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
5.5 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
5.5.1 Example Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5.6 Big Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5.7 3D Geometry with Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5.8 3D Geometry and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
5.8.1 TLDR Finding Equation of Plane . . . . . . . . . . . . . . . . . . . . 40
6 Lecture IV on July 8, 2019 41
6.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6.2 Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.3 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.4 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
6.5 All other Path in Space Stuff (ASE) . . . . . . . . . . . . . . . . . . . . . . . 49
6.5.1 A Proof of Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . 50
7 Recitation II on July 9,2019 52
7.1 Point to Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
7.2 Point to Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
7.3 Point to Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
7.4 Line to Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
7.5 Line to Plane and Plane to Plane . . . . . . . . . . . . . . . . . . . . . . . . 58
8 Lecture V on July 10, 2019 59
8.1 Polar, Cylindrical, and Spherical Coordinates . . . . . . . . . . . . . . . . . 62
8.1.1 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
3
8.1.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 63
8.1.3 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 65
9 Lecture VI on July 11, 2019 65
9.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
9.2 Other tools for limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
9.2.1 Alternate Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
9.2.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
9.2.3 Examples of Using the Further Techniques . . . . . . . . . . . . . . . 71
10 Recitation III on July 12, 2019 72
11 Lecture VII on July 15, 2019 75
11.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
11.2 A difficult Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
11.3 Linear Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
12 Recitation IV on July 16, 2019 83
12.1 Partial Derivative Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
12.2 Clarifying an Example in Class on Clairout’s Theorem . . . . . . . . . . . . 85
12.3 Linear Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
12.4 A Rigorous Proof of Clairout’s Theorem . . . . . . . . . . . . . . . . . . . . 88
13 Lecture VIII on July 17, 2019 90
13.1 Review on Linear Approximations . . . . . . . . . . . . . . . . . . . . . . . . 90
13.2 Multivariable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
13.3 The Second Derivative Test (ASE) . . . . . . . . . . . . . . . . . . . . . . . 95
13.3.1 An Example in Second Derivatives . . . . . . . . . . . . . . . . . . . 96
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13.4 Directional Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
14 Recitation V on July 18, 2019 100
14.1 A Small Note on Multivariable Optimization . . . . . . . . . . . . . . . . . . 100
14.2 Gradients and Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . 101
14.3 Following a Path of Max Increase . . . . . . . . . . . . . . . . . . . . . . . . 102
15 Lecture IX on July 19, 2019 104
15.1 Review on Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 104
15.2 Multivariable Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
15.2.1 A Proof of the Multivariable Chain Rule . . . . . . . . . . . . . . . . 109
16 Lecture X on July 23, 2019 110
16.1 Review on Partial Derivatives and Mixed Partials . . . . . . . . . . . . . . . 110
16.2 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
16.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
17 Recitation VI on July 24, 2019 116
18 Lecture XI on July 25, 2019 123
18.1 Review on Ideas Behind Integration . . . . . . . . . . . . . . . . . . . . . . . 123
18.2 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
18.3 Integration in Other Coordinate Systems . . . . . . . . . . . . . . . . . . . . 129
18.3.1 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
18.3.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 130
18.3.3 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 131
19 Recitation VII on July 26, 2019 131
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20 Lecture XII on July 29, 2019 136
20.1 Integration in Spherical and Cylindrical Coordinates . . . . . . . . . . . . . . 136
20.2 Custom Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
20.3 Applications of Double and Triple Integrals (ASE) . . . . . . . . . . . . . . . 142
20.3.1 Average Value of a Function . . . . . . . . . . . . . . . . . . . . . . . 142
20.3.2 Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
21 Recitation VIII on July 30, 2019 143
22 Lecture XIII on July 31, 2019 148
22.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
22.2 Work in Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
22.3 Fundamental Theorem of Vector Calculus . . . . . . . . . . . . . . . . . . . 152
22.3.1 Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . 153
22.3.2 Checking Conservative Fields . . . . . . . . . . . . . . . . . . . . . . 154
22.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
23 Recitation IX on August 1, 2019 155
24 Lecture XIV on August 2, 2019 159
24.1 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
24.2 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
24.3 Parametrizing Surfaces (ASE) . . . . . . . . . . . . . . . . . . . . . . . . . . 161
24.3.1 A Better Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
24.4 Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
24.5 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
25 Lecture XV on August 5, 2019 167
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25.1 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
25.2 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
25.3 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
25.4 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
25.4.1 Same Border, Different Surface . . . . . . . . . . . . . . . . . . . . . 176
26 Recitation X on August 6, 2019 178
27 Lecture XV on August 7, 2019 183
28 Thank You 183
These lecture notes are based off of the Interphase EDGE 2019 Iteration of Multivariable
Calculus instructed by Sam Watson. The introduction contains useful information for those
in my recitation section, and the rest of the sections will be labeled in accordance with the
lecture/recitation that they are associated with. I would 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 Watson, 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 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
7
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 resources 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:
• 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 definitely 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 most 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 advantage of the amazing opportunity you have in front of you here. I hope
1Honestly we’ve been knowing that I have like office hours all the time
8
that I can be one of those resources for you over the summer and potentially forward into
the future through other organizations and things I am a part of. I also love talking about
things related or non related 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 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 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. Homework is due
Monday, quizzes are at the start of recitation on Tuesdays, and Sam’s Office hours are 7-9
on Sunday.
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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 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 example, 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 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 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!
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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 answer 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 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 possible
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 example, 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 cylinder.
Since the equation puts no restriction on z, we do 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 opinion. 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 output. So in the most common
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example, think of f(x) = x2 , every inputted x value is being squared and being outputted.
Similar to graphing an equation, we can graph 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 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 conveniently with an arrow as follows,
f : R3 −→ R (3)
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What this means if the function, f takes in three inputs, say for example (x, y, z) and out-
puts a single number, persay (w).
Info unrelated to lecture, but for those interested: Function 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.
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 definitely 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 :)
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
13
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 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 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.
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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 instead 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.
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 throughout this course, so please get it
down! We can also conveniently express this in matrix notation that will be used when
discussing this concept both here and in linear algebra courses you may take in the future:
f(x, y) =
a b
c d
xy
(9)
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We can also introduce the 3-dimensional cousins that may come up sometime in lecture and
recitation below, although 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 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 transformation:
f(x, y) = (cx, cy) (15)
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For some scalar c. We can also rotate a a linear transformation with the rotation found
below:
f(x, y) = (cos θx− sin θy, sin θx+ cos θy) (16)
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)
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 following
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 transformations (shear, project, scale, and rotate)
that all satisfy the theorem above.
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4.3 Determinants
Determinants are really neat. I definitely did not know there was a geometric meaning re-
ally to what a determinant was. I just thought it was an annoying computation, so hopefully
this section proves interesting! Determinants are all about how linear transformations dis-
tort areas. Lets consider a number line first. Consider the linear transformation f(x) = 3x.
We want to ask ourselves, 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 determinant of the function since this is the
distortion factor of length. More generally, we have that given a function, 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 each of the 4 vertices get mapped to with the linear transformation 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)
respectively. We can calculate the area through some interesting geometry. Lets start by
18
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 parallelogram! 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 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 transformations turns area into a line. Well, a line
has zero area, and as such, the determinant of such a linear transformation is exactly zero.
You may notice while working that this will occur when one row of the linear transformation
matrix is a scalar multiple of the other row of the linear transformation matrix.
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 linear transformation smashes down everything into a line. Thus, the area is
19
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 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 transformation. namely det = ad − bc =
cos2 θ+ sin2 θ = 1 Which confirms our suspicion! Lets look at the linear transformation 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.
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
20
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 decomposing
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 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 mul-
tiplication 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 between plus and
21
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 probably be asked, with a high proba-
bility, is how to compute 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
22
for there to exist an inverse. Now suppose we have 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 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 created 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 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
23
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)
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)
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C22 = 1 (48)
C23 = 0 (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 = −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
25
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 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 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
26
section. If A is invertible, then:
Ax = b (59)
A−1Ax = A−1b (60)
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.
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 confront 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 applications 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 vector has a magnitude and direction. The
27
length of the 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. 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.
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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)
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
29
triangle as A. B, and C labeling in a counterclockwise orientation. Lets now construct a few
vectors naming them with their respective two points involved. In addition to these points,
lets label the point, D, as the midpoint of ~AB and E as the 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)
Which, I can rearrange as a vector subtraction expression as:
~DE = ~AE − ~AD (77)
In addition, we can now look at the larger triangle, ABC, and get a synonymous expression
from vector subtraction
~BC = ~AC − ~AB (78)
Lets now plug in the previous expression we get relating half length in equations 43− 45, to
30
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 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 respective 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 projecting 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)
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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)
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 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
32
from the origin all the way up to the point (1, 1, 1), Therefore, we get that:
~u = (1, 1, 1) (88)
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 equivalent to zero given our first example
in this subsection.
5.5 The Cross Product
Cross Product is the more annoying brother of the dot product, so brace yourselves. We
will pretty much exclusively calculate the cross product in R3. The formula for the cross
product is the following:
~u× ~v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1) (93)
33
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 equation 63 and 64 are equivalent, but
you can trust me on it.
Now you may ask, what does the cross product do geometrically? The cross product takes
in two vectors, say ~u and ~v and it produces a third vector, say ~w that is perpendicular 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 can 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 product 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 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
2I had a typo before, I am so sorry for this. It has been cleared up now!, always confirm though usingthe determinant formula
34
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 practice of your
understanding before looking at the solution!)
~u · (~u× ~v) = |u|2 + ~u · ~v (96)
Solution FALSE! Note that the cross product in parenthesis 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 #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
35
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. Suppose 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 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 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
36
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, 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 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
37
between and beyond the two points that run along the lines existent between the two.
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
overwhelming, 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? perhaps 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. 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 second 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, placing 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 anymore. The reason is the three points, or legs, are
defining 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
38
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, 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 vectors in the plane, and generate a vector that
is perpendicular 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 ~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
39
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 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 vector ~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 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 version in action: Calculus the equation
of the plane passing through the points (1, 0, 0),(0, 1, 0), and (0, 0, 1).
40
1.
~v = (0, 1, 0)− (1, 0, 0) = (−1, 1, 0) (113)
~u = (0, 0, 1)− (1, 0, 0) = (−1, 0, 1) (114)
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)
6 Lecture IV on July 8, 2019
6.1 Review
Press F for Sam’s Microphone. Lets have a quick recap/review for things from last time.
Reminder that the dot product can be expressed as:
~v · ~u = |~v||~u| cos θ (119)
41
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 addition, the
magnitude of the cross product is the area of the parallelogram spanned by the two vectors.
The cross 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)
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)
42
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 problem mentioned at the end of class.
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 arbitrary 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)
3It was not explicitly calculate but the normal vector for this specific case happens to be ~n = (2, 1, 1)
43
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 we 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 expressed 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 :
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)
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t = −1 (131)
We can now plug in this value into the parametric representation 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 introduce a massive portion of course
content that is calculating 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 information from our line; a base point
and the direction 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 between 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 (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
45
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. Instead, allow me to introduce a small trick. That is, let me multiply the
top and bottom of the expression for 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 Equation 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 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 considered vector-valued:
f : Rn −→ R (138)
46
~r : R −→ Rn (139)
We have previously been dealing with the first of the two aforementioned functions. We
will now deal with the second type. Lets try to comprehend whats going on here with a
”real-world” example.
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 to reach the top is 1, and finally the radius of the cylindrical 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 components 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
4Generally when you parametric a circle you will get just simply cos t, sin t where t ∈ [0, 2π]
47
we can take the derivative of the position vector component by component. Namely,
~v(t) =d~r(t)
dt(141)
For our case, we can take the derivative of our bug’s position to obtain:
~v(t) = (−2π sin 2πt, 2π cos 2πt, 1) (142)
We can further differentiate velocity to obtain the acceleration 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.
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)
48
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. 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 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!
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
49
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 divided through by its magnitude. The
reason why this is called the tangent vector is simply because it denotes that direction of
the velocity vector whilst omitting the magnitude of the velocity vector. In addition to the
tangent 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 perpendicular 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)
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)
50
However, remember that we have already shown that ~T (t)cdot~T (t) is a constant value. There-
fore, this derivative 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)
~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 complete the cross product! The last application of the concepts 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
51
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 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 collection 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)
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
52
questions that require utilizing a Copious 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:
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 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
53
(1, 3, 4). Therefore, the vector from the first point to the second point is, ~v = (−1, 0, 4).
Perfect! Notice that the magnitude of this vector represents some distance from the
line to the point, but it does not represent the perpendicular distance between 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 unfortu-
nately don’t know how to add that to the latex file. Here comes the trick. Because we
don’t really know what the angle exactly is, we want to get rid of it. We can do this
by multiplying and dividing our expression for distance by |~u|. Why, because now the
numerator of our function is the expression 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 to find the distance
between this plane and the point, (2, 2, 2). Lets do this systematically again.
54
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 magnitude 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 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)
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
55
steps in all honesty. Just a different approach.
7.4 Line to Line
Here we in my opinion one of the hardest to visualize. 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 perpendicular 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 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 perpendicular 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)
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)
56
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 on 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 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 happened to be the
opposite, and there is not real meaning besides this.
5Call center amiright
57
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 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 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 projecting ~v along the direction of the unit normal vectors.
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Namely, we are extracting all of the perpendicular 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 surface. 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 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)
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where we would then plug in our specific values of z, note the resulting circle located at
these specific values of c and 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 necessarily 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 result 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)
Now we need to do a bit more thinking. With this, imagine 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 construct 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)
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
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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)
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.
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Figure 1: Pringle 6
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 understanding 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 coordinate by describing them as the signed distance from
the xz and xy plane respectively. Together, all three coordinates 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)
6Do not copy-strike me.
62
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 coordinates and add the z-direction. However, the z direction is the same in
both the Cartesian and Cylindrical coordinate 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 boundary 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)
7This formula is pretty good. However, please make note of which quadrant the angle actually is in sincethis function will not necessarily produce the correct one.
63
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 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
inequality, 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]. Even here, we are not done! We now deploy the third inequal-
ity 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
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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.
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 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 chapter 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.
65
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 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 function 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 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 function, 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
multivariate class! Lets now drop these assumptions on f as we trudge forth.
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 contained 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
66
of functions, f : R2 → R the punctured ball is really just a punctured disk Lets continue
with this:
Definition: Suppose D ∈ Rn and ~a ∈ Rn and that f : D → R. We define, [m(r),M(r)],
as the smallest closed interval containing the image, range, of the punctured ball, B∗(~a, r)∩D
under f . The last thing, B∗(~a, r) ∩ D means the points that are both in the image 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)
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)
67
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 reminder that t ≤ r. Therefore, as a last step, we essentially sub out
t with r since we are worried about the biggest and smallest. Instead, we convert to polar
coordinates, 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.
.
Figure 3: Limit Functions
68
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 func-
tions to get the same values as Sam. Lets do another example in order to try to get this down:
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 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 −12
. 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
69
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 happen to have the same limit, that the limit does exist. Why? There are
an infinite amount of paths, so just having 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 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 continuous like (x+y+z, xyz, x2+y2, etc)
3. Compositions of continuous functions are continuous (exy+z) for example. Most func-
tions that we are dealing with should be continuous.
lets now turn to an example for the alternative paths example:
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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? Because 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 possible 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 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
different limit value. As Sam said, you have to surf your way to the origin along the curvy
paths.
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Figure 4: Different Curves to the Origin
10 Recitation III on July 12, 2019
Great work today in recitation, and thank you to Klajdi’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 worksheets, 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 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)
72
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.
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 even though the limits along every
line through the origin exist and are equal.
73
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 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 conclude that the limit
does not exist.
Hopefully the limit stuff is all down. The most important thing to get out of it, in my
opinion is M(r) and m(r) which really comes down to picking a floor and ceiling for your
function. Basically, we are saying that our function is never larger than M(r) and never
74
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 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 information, 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)
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
75
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 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)
∂y(f(x, y) = ex∂y(sinxy) = xex cosxy (236)
Theres not much new here. We are just holding one variable constant and taking the deriva-
tives with it one of the variables. The actual application of the partial derivatives is what
9I am using product rule here since x comes up in both of the terms!
76
is going to be fun. So we can put our tools of partial 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)
At this point, we want to determine if the partial derivative with respect to x and y is posi-
tive or negative. Lets first look at the partial derivative with respect to x. Graphically, 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 downwards, looking like
rolling down the hill, therefore we expect 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 direction, 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:
Problem: Given these 3 graphs, decide which of these graphs is f , ∂xf , and ∂yf . Here
is the picture of the three graphs:
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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 remains flat along the x-axis, we would expect the
derivative with respect to f along this to be zero. This happens 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 function 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.
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 nonetheless.
Theorem Clairout’s Theorem states that if fxy and fyx exist and are continuous, then:
fxy = fyx (237)
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I have a cool proof for this that I will include in the recitation notes tomorrow for those
that are interested in completing a mathematics major, so be on the lookout for that :).
Anyways, back to the course. Lets do the following example:
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
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 picture 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 upper left point to the right has a function value change of 2 in the space of
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0.1. Therefore, we can get a similar 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 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 approximate function at a specific
point. This was added under the recitation notes.
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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 specific 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 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 function, L(x). Look
at the x = 0 point of the absolute value function f(x) = |x|. The function is not differen-
tiable 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 approximation is
going to equal the derivative value at that point. Now for two variables, we can generalize
the above definition 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)
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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 function L(x, y),
we make an analog to the single-variable case by making the coefficients of x and y10 in L(x, y)
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 differentiable, 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 continuous 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:
Problem: Show exy sin(x2 + y2) is differentiable everywhere.
Solution: It is clear that the partial derivatives with respect to x and y are just com-
binations of continuous function like exponential, trigonometric, and polynomial functions,
so the theorem above says that f is differentiable. 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
10In the single variable case, we had the coefficient of x being the derivative with respect to x
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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 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 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
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following combinations:
fxy = ∂y(∂xf) =∂
∂y(∂f
∂x) =
∂2f
∂y∂x(249)
fxx = ∂x(∂xf) =∂
∂x(∂f
∂x) =
∂2f
∂x2(250)
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)
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.
11Notice that equation (213) gives the same result as equation (211). This shows Clairout’s Theorem!
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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 get 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 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 derivative 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)
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Okay great. So now we need to apply the next partial 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 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)
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
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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 dimensional 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. 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).
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)
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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!
12.4 A Rigorous Proof of Clairout’s Theorem
Totally unnecessary for the course, but cool nonetheless. So, I know there are a few
people that are actually quite interested in getting a degree in mathematics. In doing 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, fxy = fyx on (a, b)× (c, d).
In order to prove the theorem, I want to cite a Theorem in Arthur Mattuck’s, Real Anal-
ysis 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 polynomials such that pn → g uniformly. We will now utilize this Theorem, 12.6,
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for our continuous function fxy generating a sequence of polynomials such that pn,
|pn(x, y)− fxy(x, y)| < ε(n) ∀(x, y) ∈ [a, b]× [c, d] (267)
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 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)
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Finally, we take the limit as n→∞ to achieve the following equation:
¨D
fxy − fyxdydx = 0 (274)
Which implies that for a function with continuous partial derivatives that:
fxy = fyx (275)
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 clarify. 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 approximation is having values that are similar to
the function’s values. So, sometimes we will use the linear approximation instead of the actual
function to approximate the function’s value around (a, b). There are not approximations
beyond second order on the ASE, so please do not 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,
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namely with respect to x and y both equivalent to zero. Lets consider an exercise 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 critical points of the function along
the interval. In addition, we are going to want to check the ends of the interval! 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 a graph of the function in question:
Figure 7: Graph of |(1− x)(x− 3)|12This is a very-well worded answer. However, on exams you would not have to state all of these statements
unless otherwise asked.
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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 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 dimensions 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
36on [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 the boundary in our domain and bounded
(contained in some large box). Bounded simply means that our function 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 endpoints, 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
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square? Well, it seems like we have 4 edges with the following equations describing them:
y = 0 (278)
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 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 now, lets
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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, 1
3) being the critical 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 throughout the year as a tutor. So, it was a great question
and deserves a full answer. Check the recitation notes for tomorrow for a follow up.
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.
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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 classify 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 little bit more extensive than previously. Let me call
the 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 multivariable, 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 opposite 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 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)
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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 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 derivatives and second-order partial
derivatives as they will all come in play throughout the problem:
fx = 6xy − 6x (293)
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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)
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
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second derivative test. By definition, the 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 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!
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13.4 Directional Derivative
This is a pretty neat section. Suppose we want to calculate 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 arbitrary direction ~u, where ~u is a Unit Vector by definition. 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
)+ lim
h→0
(fx(a, b)hu1 + fy(a, b)hu2
h
)(308)
Now look at the first limit. This goes to zero by our definition of the linear approximation
from last class. In addition, the second limit has the h get divided out, therefore just leaving
the expression without any of the h’s being present. A much more convenient form that will
be utilized 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.
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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)
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 orthogonal, 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 understanding of what was going on. Let me give a quick 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
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by. As such, we will not include it in trying to find our absolute maximum and minimum.
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 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)
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14.3 Following a Path of Max Increase
So now, thanks to a combination of questions and conversation with Jordan, Hector,
Ivan, Grace, and Raima, I wanted to include this section. Suppose we want to follow 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 calculate 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 multiple 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 illuminate 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 po-
sition by T (x, y, z) = 50 + x2 + (y − 3)2 + 2z. You are a bug starting at position (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
13could simply be (x, y)
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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 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.
103
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 sensitivity 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 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 directional derivative equivalent to
zero. This is the equivalent 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
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as 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 definition 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 gradient’s direction, of max increase, or in the direction opposite 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 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 utilizing a linear approximation
method that ends up creating a tangent plane. This problem is a bit different. Note,
this is an equation. We were using tangent planes to approximate 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, we get two different surfaces for
the square root, and we have to choose which surface to use. So, lets try a different way to
solve this problem. Instead, we can think of the following. Maybe, our equation is a level
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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 conversations in chapter one, we cannot graph the actual function
f(x, y, z) since it would require four dimensions. However, we definitely can graph its level
sets which happen 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
perpendicular 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 surface, then the
vector ~n = (2, 2, 4) is indeed perpendicular to our surface. However, where before have we
seen normal 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 ~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)
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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 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.
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 functions 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)
107
However, lets look a little bit more at this expression. It seems like, upon first glance we are
matching a partial 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 variables to the overall change in f with
respect to time. In fact, this component by component sum actually is a hidden 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 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 highlight 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)
df
dt= 4t (336)
Now lets try it our new way and see if we can get the exact same expression for dfdt
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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
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)
109
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) 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 derivatives. 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 scoot a bit to the right. Namely, we are looking at the rate of change in the x
110
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 homework 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)
16.2 Lagrange Multipliers
Lagrange multipliers are a new way to solve optimization 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 interior critical points, then points that aren’t differentiable, then writing
equations for the borders, completing a slew of single variable problems, and finally eval-
uating all of these points to see which was the absolute smallest and largest. We are now
going to take a quicker more, ”sophisticated” 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,
111
∂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 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)
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)
112
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, 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, 1
3). 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)
(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.
113
Solution: So, lets turn this sentence into a type of Lagrange multiplier idea. We want
to maximize our function f = V , the volume, subject to the constraint of the surface 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 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)
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
114
16.3 Integration
First lets think of a single-variable case to realign ourselves 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 taking 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 effectively,
in the single-variable case , taking the signed area of each one of these really skinny rectan-
gles. We then 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 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)
115
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 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 boundary of your
region. For example, suppose you have some region D, that is the unit disk. That utilizing
the Lagrange 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 problem 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!
116
Problem: Find the maximum and minimum of the function f(x, y) = 2x− 3y. subject
the constraint, x2 + y2 = 64
Solution:We can utilize the method of Lagrange multipliers 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)
Plugging into our third equation, we obtain that:
1
λ2+
9
4λ2= 64 (377)
Leading to:
λ = ±√
13
16(378)
117
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 constraint.
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)
Which can deploy the Lagrange multiplier equation:
∇f = λ∇g (384)
(4πr + 2πh, 2πr) = λ(2πrh, πr2) (385)
118
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)
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:
1. Writing a constraint equation relating the amount and prices of the two goods along
14This is called a Cobb-Douglass Utility Function
119
with your total amount of money, m.
2. Setting up a function f and g and utilizing the Lagrange 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 ≤ 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)
120
If we divide the first equation by the second equation 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−ααpx)
=mα
px(402)
Which we can now plug into our expression for y to obtain:
y =1− αα
pxpy
m
(px + 1−ααpx)
=(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 refractive 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, derive Snell’s law.
121
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 minimize this. In
addition, We can make the length in the x direction our constraint, since we cannot change
the actual position of the two points in question. Therefore, we can simply take that distance
over time is velocity, rearrange 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)
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
v1
secα tan,b
v2
sec 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)
122
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 function at piece
3. take numbers of pieces to ∞
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)
123
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)
Great now we can utilize the idea of single-variable calculus 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
πsin(πy)
π
]1
0
=1
π(417)
Great! However, lets think about more complicated regions. 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 example.
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˜DfdA
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
124
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 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 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
125
up with your bounds by utilizing the picture. One thing to note as you are making you’re
making your way 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 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 density and D would be the volume,
so we obtain mass! However, you can also think of this with respect to any densities 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 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.
126
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 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 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)
127
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)
Since the actual integral was not evaluated in class, I have provided it here in case you wanted
to see it. Lets continue 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, 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
128
already have them fixed in space. We can derive the bounds for x as 0 ≤ x ≤ 2− 23y − 1
2z.
As such, we can set up our triple integral as:
M =
ˆ 4
0
ˆ 3− 34z
0
ˆ 2− 23y− 1
2z
0
f(x, y, z)dxdydz (432)
M = V =
ˆ 4
0
ˆ 3− 34z
0
ˆ 2− 23y− 1
2z
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.
18.3 Integration in Other Coordinate Systems
Earlier in the course, we did quite briefly go over alternative 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 integrating 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 and 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)
15This is the heart-shaped graph’s official name that we see in polar graphs
129
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 cartesian and polar, cylindrical, and
spherical respectively are:
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)
We can also head in the opposite direction:
x = r cos θ (444)
130
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)
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 ex-
amples 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.
131
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 numerical 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 mass of an object by the following equation:
(xCOM , yCOM) =
(˜RxdA˜
R1dA
,
˜RydA˜
R1dA
)(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
2in our case. Lets 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)
132
Now we can compute, utilizing the same bounds for the integrals both,˜RxdA and
˜RydA
¨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) =
(˜RxdA˜
R1dA
,
˜RydA˜
R1dA
)=
( 1312
,1612
)=
(2
3,1
3
)(459)
This is a great application of utilizing double integrals to gain insight on other meaningful
quantities found throughout the course. In this case, we have bounds that are actually func-
tions 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 coordinates through the following two examples that I will first
work out and then explain the methodology behind it.
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
133
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)
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) = 12x2 + 1
2y2 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)
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
134
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. Letting 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)
The major piece of advice to get out of this small collection 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 systems 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 spherical like
objects that are quite difficult to work with under cartesian like systems.
135
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:
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 integrate over spherical coordinates!
Same goes with cylindrical 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 R: ˚R
f(x, y, z)dV (470)
136
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 coordinates. We can justify this in
multiple ways. What we should think is that this factor is a correction factor since we are
now using a funkier coordinate system! A quick justification is the following. Consider just
the 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! Remind me if I forget
please. Lets move onto an example of this now:
Problem: Consider a solid with density δ(x, y, z) = (x2 + y2 + z2)−1 and which occupies
the cone:
137
Figure 10: Cone
Find its mass.
Solution: We have some flexibility for solving this. For example, in cylindrical coor-
dinates, 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 func-
tion, we 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, 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.
138
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 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 volume element are in length cubed. We pick up a length 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
139
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 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 coordinates, 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 some custom coordinates. The Theorem getting all the
formality of this section is:
Theorem: Suppose that T : R2 → R2 is a differentiable 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
140
our function u and v 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 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)
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Lets now actually plug this into our example problem that 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.
20.3 Applications of Double and Triple Integrals (ASE)
There are two quantities that will most likely be referenced 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) =
˜Df(x, y)dA˜D
1dA(490)
142
In addition, we can also compute the average value over some 3D region to get the analog
that:
Avg(f) =
˝Df(x, y, z)dV˜D
1dV(491)
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) =
(˜Dxδ(x, y)dA˜
Dδ(x, y)dA
,
˜Dyδ(x, y)dA˜
Dδ(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, 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) =
(˝Dxδ(x, y, z)dV˝
Dδ(x, y, z)dV
,
˝Dyδ(x, y, z)dV˝
Dδ(x, y, z)dV
,
˝Dzδ(x, y, z)dV˝
Dδ(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 useful 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
143
are not just limited to the coordinate grid, but we have the freedom to branch out and use
spherical and cylindrical coordinates, and even custom coordinates with a Jacobian, in order
to compute. I will put some examples here that I liked from the worksheet :)
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 intersection of the two cylinders.
Solution: We can try and find bounds for all three of 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 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.
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Solution: Converting everything to spherical coordinates 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 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 region. For that reason, switching
over to polar cylindrical coordinates is a good idea. In polar cylindrical coordinates, 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, 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)
145
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.
Solution: We can start by defining custom coordinates, 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 transformation is J = 12. We can implement
the transformation and 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)2 1
2dvdu =
1
2
ˆ 4
2
ˆ 1
−1
u2dvdu =
ˆ 4
2
u2du =56
3
(509)
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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)
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)
147
− 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 coordinates 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 probably 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.
148
Definition: A vector field in R3 is a function, f : R3 → R3. A vector field in R2 is a
function, f : R2 → R2.
We probably have all come in contact with a few vector fields before. One can think of a
gravitational field, electric field, and even flows fields of fluids all as great examples 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,
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)
149
The constants GMm are just for the physical interpretation. 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 direction, and the magnitude does not change as we
move along any horizontal line. Since there is no y component, and the magnitude of the
arrow pointing in the x direction 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)
150
This tells us that work can be expressed as the dot product between force and the displace-
ment of the object. Now we want to generalize for paths that are not necessarily 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 ~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 variable 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 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 parametrization that you
151
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 does not matter since we just adjust
the time in the interval. 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)
152
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)
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 following 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)
153
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)
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 component 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 conservative or not conservative. The Theorem simply states that:
Theorem: If My = Nx for a vector field ~F = (M,N), then ~F is a conservative vector
154
field, under the assumption that ~F is differentiable everywhere on R2.
22.4 Green’s Theorem
This section is super neat and is actually just a specific 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 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 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 function f for F, that is, a scalar function
155
f whose gradient,−→∇f = F. By inspection (taking into account that each component of
F should be a corresponding partial derivative 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 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 object.
Solution: We begin in the same manner as our previous 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 difference in the value of f at t = 1 vs at
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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 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.
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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 difference to find the integrand:
Nx −My = −1− 1 = −2
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 integral (math), the fundamental theorem of calculus (math), and through a direct
statement (all words).
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Solution: We first show this through a direct computation:
˛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-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.
24 Lecture XIV on August 2, 2019
Last class, we ended with Green’s Theorem. Lets pick up right where we left of:
159
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 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 temperature. 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 area to extract the average temperature. This is an application of average value that
we went over in workshop with some example problems!
Avg(f) =
´SfdA´SdA
(569)
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where S is the surface, and f : S → R. Lets try to polish off this idea with an example:
Problem: Find´SfdA, 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 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 necessarily 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, 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
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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. 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 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 function, 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 specifically, we have that we are parametrizing
with respect to θ and φ where each of these two variables are bounded between [0, π2] giving
us our bounds for the integral. Let me compute this problem thoroughly so that you can
162
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 parametrization 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)
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 express 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
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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 surface on the xy plane. Great! Okay, now that we have parametrized this
surface, lets now compute the surface area!
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 area to obtain that:
S =
¨A
√4(u2 + v2) + 1dudv (581)
Where now it will be advantageous to switch over to cylnidrical coordinates given our
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’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 surface over a region as:
S =
¨A
√f 2x + f 2
y + 1dA (583)
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)
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)
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In all honesty, there is a bit of ugly and tedious labor that comes with the cross product being
involved, but the overarching 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
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)
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
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the direction of the hole 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 compute the divergence of a vector field,
~F as:
~∇ · ~F =
(∂
∂x,∂
∂y,∂
∂z
)· (M,N,P ) (588)
Where F = (M,N,P )
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)
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The equation by itself is not all too bad. Lets do an example 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) 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 represents 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:
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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, 1
2), Then we notice that the size of the arrows representing
the vector field flowing in are larger than the size of the vectors flowing out, we get a net
negative vector field. Lets turn to another example now.
Figure 14: Divergence of Vector Field 1
Here we have an interesting case. Lets look at perhaps the point (12, 1
2). 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 positive divergence.
Finally if the flow equal the flow out, we achieve zero divergence. Of course, it is much
169
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, ~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 vec-
tor 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 computation of the Curl of a vector field.
Problem: Compute the Curl of the vector field, ~F = ((x2y, yx, xz)
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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 (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 without too much worry. Lets try to get down the conceptual understanding
now. Curl can be described as a circulation density. Namely, we want to think, If I place a
little paddle-wheel, something that will rotate in my vector 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 counterclockwise due to the vector field, we have a positive
curl. In addition, if the little paddle wheel would rotate clockwise at the point, we would
state that we have a negative curl. Lets try this out with a picture we have seen a few times
by now. Consider the vector field below:
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Figure 15: Curl of Shear Vector Field
Imagine we want to compute the curl at the point (12, 1
2). 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. 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 positive, and even more so, the curl is always
positive. So, for any point in this little rectangle, we get a positive circulation density that
will result in a positive curl. So, we can use a formulaic way of deciding whether the curl is
positive, leading to a counterclockwise rotation, or negative, leading to a clockwise rotation
172
by imagining placing a little 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 simple solid region and
S is the boundary surface of E with positive orientation. Let F be a vector field whose
components have continuous first order partial derivatives. 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 perhaps 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 integral 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 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)
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Solution: Well, if we saw this problem last week, perhaps 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 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 example 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 Divergence Theorem. Namely, it would
really suck to compute the surface integral over this shape. Instead we can turn 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
174
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’ 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 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
16Crap is just the Vector Field
175
Theorem. If you 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)
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. 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 Independence 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 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 integral over a closed path, or use the ideas
of surface Independence 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 theorem to evaluate˜S
F · dS where
F = 〈xy,−12y2, z〉 and the surface consists the paraboloid z = 4− (x2 + y2) and the circle in
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the xy plane it encloses.
Solution: Note that this setup is symmetrical in cylindrical 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,
¨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)
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Problem: Given that for the groin vault, encountered last week, we computed the vol-
ume to be,˝
dV = 163
, compute the flux through the groin vault given that ~F = 〈x, y, z〉.
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 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.
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Thus, we need only find˜S∇× F · dA.
¨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 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
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of our surface, the LHS of Stokes’ Theorem, we also obtain that zero work has been done.
Lets try that now utilizing the parametrization 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)
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 Theorem on the LHS of Gauss’ Law.
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Doing so, we have the equation:
‹S
~E · ~ndA =
˚ (∇ · ~E
)dV (632)
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.
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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