# Math: Calculus Video #8

## 3Blue1Brown Intro To Calculus Series Summary

--

# Preface ✨

*Hello everyone! My intentions for writing these articles are:*

*Explain technical knowledge about geology in simple terms (to the public)**Document my journey to getting an AD-T Geology degree in one year**Store information and habits for my future self and others (in <7 minutes)*

*Coolio? Sweet. Enjoy the series :-)*

# notes as i watched the video 📹

*p.s. — the video will be linked at the bottom of the article. cheers.*

where something exists, there is always an equal opposite to it.

good vs evil.

sun vs rain.

life vs. death.

and for calculus…

it’s a derivative, and an integral.

well, almost, i should say.

lemme show ya (with da help of grant).

## car example

is there a way to know how quick one has gone if ur car moved from point “a” to point “b” for eight seconds…

but the only data u had was the velocity on the speedometer?

could u?

kinda…

OR…

can i find a distance function (with a name [s(t)]),

that tells me how far i’ve traveled after a given amount of time (t),

between of zero and eight seconds?

well…

let’s say u took note of how quick u were going, and we plotted how quick the car went per second of time that passed (between 0 to 8 seconds).

but now, we need an equation (aka function) that describes how quick we went…

grant has it labeled as “v(t)”, which is code for “velocity as “t” amount of time/seconds passes”… “t” means seconds (between 0 to 8)

velocity is in meters, btw

the reason the equation is “… t(8-t)” is…

the first “t” (outside of parantheses) tells us what second we are on between the interval “0–8” seconds… and yes, it will tell us the velocity at that second.

the second “t” (inside of parantheses) tells us how many seconds are left between the interval “0–8” seconds… and yes, it will tell us the velocity at that second.

ex: v(t) = t(8-t)

v(3) = 3(8–3)

v(3) = 3(5)

v(3) = 15 m/s ||| when 3 seconds have passed, the car’s at 15 meters per second.

u can plug in any number between 0–8 (’cause that’s the time interval as the car started and finished its movement), and u’ll find out the velocity at each second.

but now, for a mind-bender…

in derivatives, we know what our distance is… but velocity is not given.

in integrals, we were given the velocity (as an equation) & the time intervals… but we don’t know how far we traveled IN TOTAL.

the “v(3) = 3(8–3)” part is just for that one point when 3 seconds have passed since the car started moving.

but we need to know the whole trip, not just one point… we’re not figuring out derivatives today lol

the derivative of a change in distance over a change in time can give me the velocity (because that is the literal definition of velocity lol),

ds/dt = v(t)

… but to find the distance, we need a change in velocity over a change in time… yet, we don’t know all of the change…

so the question the mathematician asks is this:

“what function has a derivative of “t(8-t)”… or if we expand it…

we end up getting “8t - t²”…

if **t(8-t) **was the derivative, what was the function that gave us **t(8-t)**?

…’cause they don’t just come from mid-air u know.

and this is where the uses of rectangles to find derivatives can come in handy here.

integrals can approximate the velocity function as if it (“it” → velocity) was constant on many different intervals.

and with these small rectangles, they act as constant, fixed velocity.

and the smaller the changes to the velocity (aka the speed changes faster per new rectangle present),

the rectangles get smaller until the rectangles fit under the line of the curve…

and then we’ll know the distance passed between 0–8 seconds.

each rectangle is a piece of the distance…

**lemme show ya what i mean, look at the photos.**

ok, example time.

chop the time up from 0–8 seconds to small pieces into 0.25 seconds.

now consider a sliver of those times, between 1–1.25 seconds.

the change of time in that blue rectangle (dt) is 0.25 seconds.

when time (t) =1, the velocity is 7 meters /second.

when time (t) = 1.25, the velocity is 8.4 meters/ second.

u find that out by plugging 1 & 1.25 into the equation: v(t) = t(8 — t)

so, how do we find the distance of that specific rectangle?

well, let’s use the moment one second passes.

in that moment, the car’s going 7 m/s.

the time change was 0.25 seconds

multiply that, and we get 1.75 meters, as the distance traveled from 1 to 1.25 seconds.

7 m/s * 0.25 s = 1.75 m

*velocity * change in time = change in distance*

ok, but why multiply?

remember this below: ⏬

the derivative of a change in distance over a change in time can give me the velocity (because that is the literal definition of velocity lol),

ds/dt = v(t)

but to find the distance nudge…

u multiply the velocity (7 m/s) times the change in time (0.25 seconds) from the nudge we made, to find the change in distance.

also, the “seconds portion” in velocity is canceled out, to only leave us with meters.

…but now, do u see the rest of the dark, blurred rectangles?

u might need to zoom in lol

those dark rectangles, we’re going to add all of them up, and we can find the change of those distances, add them up, and find our total distance traveled.

so we’ll use the integral sign: “∫”

but y not “Σ”?

because “∫” is meant to be used as the change in “time stamps” (aka the rectangles) shrink and get closer to a change of zero seconds…

Σ is fixed. ∫ can morph & shrink.

# areas under graphs 📈

see this?

the capital “T” is the current distance, that has passed according to how much we are trying to sum up.

∫ decides the starting point and the final point.

distance = velocity * each small nudge (aka rectangle)

WHAT IS THE DERIVATIVE OF THE ORIGINAL EQUATION/FUNCTION?*

* aka, if v(t) = t(8-t), what function made that derivative happen?

s(T) is the distance traveled according to the maximum value of “T”.

and s(T) was part of what made velocity possible.

v(T) = velocity btw

and to find the distance,

we can switch the equation up by doing this:

ds = v(T) * dT

change in distance = velocity * change in time

and the approximation of how much distance changed from two specific points in time will become more exact as dT get thinner.

…and maybe so thin, the thickness of dT nearly, nearly, becomes zero.

and this “dT” adds extra area that we didn’t originally account for, because of the limit of the “T” line*

* check first photo on “areas under graphs”

ok, now we know what the “dx” does:

gives us the nudge, or change to distance at a specific point on the area we want to consider.

but now, we must find its integral, or anti-derivative.

the anti-derivative is used to find out what the entire function is, if we were to graph it out.

if the function “t(8-t)” expanded, is “8t-t²”…

we reverse it.

**what function makes “8t” if we were to take that derivative?**

4t².

take the derivative 4t²… we get 8t. ||| **(4t² → 2*4(t)¹ → 8(t)¹ → 8t**

ok, what about -t²? ||| p.s. - ignore the negative symbol for now, add it later.

(1/3) t³.

take the derivative (1/3) t³… we get t². ||| **(1/3) t³ → 3 * (1/3) t² → 1 * t² → t²**

ok, now add the negative sign to get -(1/3)t³.

**the anti-derivative is 4t² — (1/3)t³.**

good.

BUT!

how do we know if the graph can move up or down?

that’s y u may have seen some integrals have an answer like this:

4t² — (1/3)t³ + Cthe “C” means a constant number, (like 3,4,5, 6/5, 0.332,π, … whole numbers, integers, fractions, real numbers,)

and it can change the positioning (or where the graph is located) of the function on the x-y axis.

HOWEVER!

because we know where the start & end point of the integral (0 & 8 seconds)…

we know where our potential constants lie.

when u know the start & end points, it’s called a definite integral.

definite → starts & ends somewhere

if u don’t know the start & end points, it’s called an indefinite integral.

indefinite → idk where tf to plot it exactly lol

… and that’s when u use “+ C”, to indicate many possibilities of its positioning.

p.s. — u can see it in action, when u don’t know the start + end points.

notice how the slope of a specific point stays the same, no matter where it moves vertically.

# integral 📊

when we do know where it starts & ends, we get something like this.

“T” is the terminal point, or the end section.

looking at the photo, if “T” was 5, the number 5 would replace the letter “T” wherever a “T” is present on the integral and the equation.

but in our problem, the start was 0, and our “T” is 8 seconds.

and if we solve it, we get this.

**the anti-derivative is used to find out what the entire function is, if we were to graph it out.**

and it’s being used to identify the distance traveled by the imaginary car from 0 to 8 seconds (if velocity was the only thing we could measure), and that’s 85.33 meters.

ok, but what about if we didn’t have 0 as our starting point?

in fact, what if we didn’t want to measure all of the distance?

maybe just a sliver of it?

what if just from between one to seven seconds (instead of zero to eight seconds):

u would just do plug & play.

the higher end variable goes first, then the smaller variable.

and even if we had constants within the integral, they cancel themselves out thanks to the negative sign…

anytime i want to integrate some function, ask yourself:

WHAT DOES THE SUM OF ALL THESE RECTANGLES APPROACH AS “dx” GETS CLOSER TO ZERO?

dx → change in distance from one rectangle to another.

# the fundamental theorem of calculus 👑

we start off with finding the anti-derivative.

the derivative of dF is the colored part of the photo, what’s under the whole integral.

then we divide by the width of each nudge, or dx.

this gives us the function.

when we multiply the inputs of the function (f(x)), times the nudge in all the little rectangles, (as the dx goes to 0), and summing them up, gives us the integral we have for the top & bottom numbers.

this is the fundamental theorem of calculus.

it takes all the derivatives, adds them up, and integrates them all to create an area.

and, to actually find it, all u need is the area from where u wanna start calculating to where u wanna stop.

F(a) & F(b), btw…

but that’s crazy how simple it is!!

the area under the curve = the distance traveled by the car.

and when one understands that each dx means a change in distance, velocity, and time…

they are all bound to the same derivative.

so when we find out its anti-derivative, we know the graph, now we need to know what part of the graph we will look at and find its area under the curve.

yep, calculus actually now makes sense…

’cause for one to exist, its inverse must be burning alive.

**CJ**

*© 2023–2100 by Carlos Manuel Jarquín Sánchez. All Rights Reserved.*