[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.