[INTRO MUSIC]
Hey everyone, Grant here. This is the first video in a series on the essence of calculus.
and I'll be publishing the following videos once per day for the next 10 days.
The goal here, as the name suggests, is to really get the heart of the subject out
in one binge watchable set. But with a topic that's as broad as calculus.
there's a lot of things that can mean. So, here's what I've in my mind specifically.
Calculus has a lot of rules and formulas which are often presented as
things to be memorised.
Lots of derivative formulas, the product rule, the chain rule, implicit differentiation,
the fact that integrals and derivatives are opposite, Taylor
series; just a lot of things like that.
And my goal is for you to come away
feeling like you could have invented
calculus yourself. That is, cover all
those core ideas, but in a way that makes
clear where they actually come from and
what they really mean using an
all-around visual approach. Inventing
math is no joke, and there is a
difference between being told why
something's true and actually generating
it from scratch. But at all points I want
you to think to yourself if you were an
early mathematician, pondering these
ideas and drawing out the right diagrams,
does it feel reasonable that you could
have stumbled across these truths yourself?
In this initial video, I want to
show how you might stumble into the core
ideas of calculus by thinking very
deeply about one specific bit of
geometry: the area of a circle. Maybe you
know that this is pi times its radius
squared, but why? Is there a nice way to
think about where this formula comes from?
Well, contemplating this problem and
leaving yourself open to exploring the
interesting thoughts that come about can
actually lead you to a glimpse of three
big ideas in calculus: integrals,
derivatives, and the fact that they're opposites.
But the story starts more simply—just
you and a circle; let's say with radius three.
You're trying to figure out its
area, and after going through a lot of
paper trying different ways to chop up
and rearrange the pieces of that area,
many of which might lead to their own
interesting observations, maybe you try
out the idea of slicing up the circle
into many concentric rings. This should
seem promising because it respects the
symmetry of the circle, and math has a
tendency to reward you when you respect
its symmetries. Let's take one of those
rings which has some inner radius R
that's between 0 and 3.
If we can find a nice
expression for the area of each
ring like this one, and if we have a nice
way to add them all up, it might lead us
to an understanding of the full circle's area.
Maybe you start by imagining
straightening out this ring.
And you could try thinking through
exactly what this new shape is and what
its area should be, but for simplicity
let's just approximate it as a rectangle.
The width of that rectangle is the
circumference of the original ring, which
is two pi times R. Right? I mean that's
essentially the definition of pi; and its
thickness? Well that depends on how
finely you chopped up the circle in the
first place, which was kind of arbitrary.
In the spirit of using what will come to
be standard calculus notation, let's call
that thickness dr, for a tiny difference
in the radius from one ring to the next.
Maybe you think of it as something like 0.1.
So, approximating this unwrapped ring
as a thin rectangle, its area is 2 pi
times R, the radius, times dr, the
little thickness. And even though that's
not perfect, for smaller and smaller
choices of dr, this is actually going to
be a better and better approximation for
that area, since the top and the bottom
sides of this shape are going to get
closer and closer to being exactly the
same length. So let's just move forward
with this approximation, keeping in the
back of our minds that it's slightly
wrong, but it's going to become more
accurate for smaller and smaller choices
of dr. That is, if we slice up the circle
into thinner and thinner rings. So just
to sum up where we are, you've broken up
the area of the circle into all of these
rings, and you're approximating the area
of each one of those as two pi times its
radius times dr, where the specific
value for that inner radius ranges from
zero, for the smallest ring, up to just
under three, for the biggest ring, spaced
out by whatever the thickness is that you
choose for dr—something like 0.1.
And notice that the spacing
between the values here corresponds to
the thickness dr of each ring, the
difference in radius from one ring to
the next. In fact, a nice way to think
about the rectangles approximating each
ring's area is to fit them all up-right
side by side along this axis. Each one
has a thickness dr, which is why they
fit so snugly right there together, and
the height of any one of these
rectangles sitting above some specific
value of R—like 0.6—is
exactly 2 pi times
that value. That's the circumference of the
corresponding ring that this rectangle approximates. Pictures like this two pi R
can actually get kind of tall for the
screen. I mean 2 times pi times 3
is around 19, so let's just throw
up a y-axis that's scaled a little
differently, so that we can actually fit
all of these rectangles on the screen.
A nice way to think about this setup is to
draw the graph of two pi r which is a
straight line that has a slope two pi.
Each of these rectangles extends up to
the point where it just barely touches that graph.
Again we're being approximate here.
Each of these rectangles only
approximates the area of the
corresponding ring from the circle, but
remember, that approximation, 2 pi R
times dr, gets less and less wrong as
the size of dr gets smaller and smaller.
And this has a very beautiful meaning
when we're looking at the sum of the
areas of all those rectangles.
For smaller and smaller choices of dr, you
might at first think that that turns the
problem into a monstrously large sum.
I mean there's many many rectangles to
consider and the decimal precision of
each one of their areas is going to be
an absolute nightmare! But notice; all of
their areas in aggregate just looks like
the area under a graph, and that portion
under the graph is just a triangle.
A triangle with a base of 3 and a height
that's 2 pi times 3. So its area, 1/2
base times height, works out to be
exactly pi times 3 squared; or if the
radius of our original circle was some
other value capital R that area comes
out to be pi times R squared, and that's
the formula for the area of a circle!
It doesn't matter who you are or what you
typically think of math that right there
is a beautiful argument.
But if you want to think like a
mathematician here,
you don't just care about finding the
answer; you care about developing general
problem-solving tools and techniques.
So take a moment to meditate on
what exactly just happened and why it worked,
cause the way that we transitioned
from something approximate to something
precise is actually pretty subtle, and it
cuts deep to what calculus is all about.
You had this problem that can be
approximated with the sum of many small
numbers, each of which looked like 2 pi R
times dr for values of R ranging
between 0 and 3.
Remember, the small number dr here
represents our choice for the thickness
of each ring—for example 0.1. And there
are two important things to note here.
First of all, not only is dr a factor in
the quantities we're adding up—2 pi R
times dr—it also gives the spacing
between the different values of R.
And secondly, the smaller our choice for dr
the better the approximation.
Adding all of those numbers could be seen in a
different pretty clever way as adding
the areas of many thin rectangles
sitting underneath a graph. The graph of
the function 2 pi R in this case.
Then—and this is key—by considering smaller
and smaller choices for dr corresponding
to better and better approximations of the original problem, this sum, thought
of as the aggregate area of those
rectangles, approaches the area under the
graph; and because of that, you can
conclude that the answer to the original
question in full un-approximated
precision is exactly the same as the
area underneath this graph.
A lot of other hard problems in math and
science can be broken down and
approximated as the sum of many small
quantities. Things like figuring out how
far a car has traveled based on its
velocity at each point in time. In a case
like that you might range through many
different points in time and at each one
multiply the velocity at that time times
a tiny change in time, dt, which would
give the corresponding little bit of
distance traveled during that little
time. I'll talk through the details of
examples like this later in the series,
but at a high level many of these types
of problems turn out to be equivalent to
finding the area under some graph.
In much the same way that our circle
problem did. This happens whenever the
quantities that you're adding up,
the one whose sum approximates the
original problem, can be thought of as
the areas of many thin rectangles
sitting side-by-side like this.
If finer and finer approximations of the
original problem correspond to thinner
and thinner rings, then the original
problem is going to be equivalent to
finding the area under some graph.
Again, this is an idea we'll see in more detail
later in the series, so don't worry if
it's not 100% clear right now.
The point now is that you, as the
mathematician having just solved a
problem by reframing it as the area
under a graph, might start thinking about
how to find the areas under other graphs.
I mean we were lucky in the circle
problem that the relevant area turned
out to be a triangle. But imagine instead
something like a parabola, the graph of x
squared. What's the area underneath that
curve say between the values of x equals
zero and x equals 3? Well, it's hard
to think about, right? And let me reframe
that question in a slightly different way:
we'll fix that left endpoint in place at
zero and let the right endpoint vary.
Are you able to find a function A(x)
that gives you the area under this
parabola between 0 and x? A function
A(x) like this is called an integral of
x-squared. Calculus holds within it the
tools to figure out what an integral
like this is, but right now it's just a
mystery function to us. We know it gives
the area under the graph of x squared
between some fixed left point and some
variable right point, but we don't know
what it is. And again, the reason we care
about this kind of question is not just
for the sake of asking hard geometry
questions; it's because many practical
problems that can be approximated by
adding up a large number of small things
can be reframed as a question about an
area under a certain graph. And I'll tell
you right now that finding this area
this integral function, is genuinely hard
and whenever you come across a genuinely
hard question in math a good policy is
to not try too hard to get at the answer
directly, since usually you just end up
banging your head against a wall.
Instead, play around with the idea, with no
particular goal in mind. Spend some time
building up familiarity with the
interplay between the function defining
the graph, in this case x squared, and the
function giving the area.
In that playful spirit if you're lucky
here's something that you might notice
When you slightly increase x by some
tiny nudge dx look at the resulting
change in area represented with this
sliver that I'm going to call dA for a
tiny difference in area. That sliver can
be pretty well approximated with a
rectangle one whose height is x squared
and whose width is dx, and the smaller
the size of that nudge dx the more that
sliver actually looks like a rectangle.
Now this gives us an interesting way to
think about how A(x) is related to
x-squared. A change to the output of A,
this little dA, is about equal to x
squared, where X is whatever input you
started at, times dx,
the little nudge to the input that
caused A to change. Or rearranged dA
divided by dx, the ratio of a tiny change
in A to the tiny change in x that caused it, is approximately whatever x squared
is at that point, and that's an
approximation that should get better and
better for smaller and smaller choices
of dx. In other words, we don't know what
A(x) is; that remains a mystery, but we
do know a property that this mystery
function must have. When you look at two
nearby points for example 3 & 3.001
consider the change to the output of A
between those two points—the difference
between the mystery function evaluated
at 3.001 and evaluated at 3. That change
divided by the difference in the input
values, which in this case is 0.001,
should be about equal to the value of x
squared for the starting input—in this
case 3 squared.
And this relationship between tiny
changes to the mystery function and the
values of x-squared itself is true at
all inputs not just 3. That doesn't
immediately tell us how to find A(x),
but it provides a very strong clue that
we can work with.
and there's nothing special about the
graph x squared here. Any function
defined as the area under some graph has
this property that dA divided by
dx—a slight nudge to the output of A divided
by a slight nudge to the input that
caused it—is about equal to the height
of the graph at that point.
Again, that's an approximation that gets better and
better for smaller choices of dx.
And here, we're stumbling into another big
idea from calculus: "Derivatives". This
ratio dA divided by dx is called the
derivative of A, or more technically the
derivative is whatever this ratio
approaches as dx gets smaller and
smaller. Although, I dive much more deeply
into the idea of a derivative in the
next video, but loosely speaking it's a
measure of how sensitive a function is
to small changes in its input. You'll see
as the series goes on that there are
many many ways that you can visualize a
derivative depending on what function
you're looking at and how you think
about tiny nudges to its output.
And we care about derivatives because
they help us solve problems, and in our
little exploration here, we already have
a slight glimpse of one way that they're
used. They are the key to solving
integral questions, problems that require
finding the area under a curve.
Once you gain enough familiarity with computing
derivatives, you'll be able to look at a
situation like this one where you don't
know what a function is but you do know
that its derivative should be x squared
and from that reverse engineer what the
function must be. And this back and forth
between integrals and derivatives where
the derivative of a function for the
area under a graph gives you back the
function defining the graph itself is
called the "Fundamental theorem of
calculus". It ties together the two big
ideas of integrals and derivatives, and
it shows how, in some sense, each one is
an inverse of the other.
All of this is only a high-level view:
just a peek at some of the core ideas
that emerge in calculus, and what follows
in the series are the details for
derivatives and integrals and more.
At all points I want you to feel that you
could have invented calculus yourself.
That if you drew the right pictures and
played with each idea in just the right
way, these formulas and rules and
constructs that are presented could have
just as easily popped out naturally from
your own explorations, and before you go
it would feel wrong not to give the
people who supported this series on
Patreon a well-deserved thanks both for
their financial backing as well as for
the suggestions they gave while the
series was being developed.
You see supporters got early access to
the videos as I made them, and they'll
continue to get early access for future
essence of type series and as a thanks
to the community
I keep ads off of new videos for their
first month. I'm still astounded that I
can spend time working on videos like
these, and in a very direct way you are
the one to thank for that.