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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. Now, today we get to move on

from integral formulas and methods of integration back

to some geometry. And this is more or less going

to lead into the kinds of tools you’ll be using in

multivariable calculus. The first thing that we’re going

to do today is discuss arc length. Like all of the cumulative sums

that we’ve worked on, this one has a storyline and the

picture associated to it, which involves dividing

things up. If you have a roadway, if you

like, and you have mileage markers along the road, like

this, all the way up to, say, sn here, then the length along

the road is described by this parameter, s. Which is arc length. And if we look at a graph of

this sort of thing, if this is the last point b, and this is

the first point a, then you can think in terms of having

points above x1, x2, x3, etc. The same as we did with

Riemann sums. And then the way that we’re

going to approximate this is by taking the straight lines

between each of these points. As things get smaller and

smaller, the straight line is going to be fairly close

to the curve. And that’s the main idea. So let me just depict one

little chunk of this. Which is like this. One straight line, and here’s

the curved surface there. And the distance along the

curved surface is what I’m calling delta s, the change in

the length between, so this would be s2 – s1 if I

depicted that one. So this would be delta s is,

say s. si – si – 1, some increment there. And then I can figure out what

the length of the orange segment is. Because the horizontal

distance is delta x. And the vertical distance

is delta y. And so the formula is that the

hypotenuse is delta x ^2 + delta y ^2. Square root. And delta s is approximately

that. So what we’re saying is

that delta s ^2 is approximately this. So this is the hypotenuse. Squared. And it’s very close to the

length of the curve. And the whole idea of calculus

is in the infinitesimal, this is exactly correct. So that’s what’s going to

happen in the limit. And that is the basis for

calculating arc length. I’m going to rewrite that

formula on the next board. But I’m going to write it in

the more customary fashion. We’ve done this before,

a certain amount. But I just want to emphasize

it here because this handwriting is a little

bit peculiar. This ds is really

all one thing. What I really mean is to put

the parenthesis around it. It’s one thing. It’s not d * s, it’s ds. It’s one thing. And we square it. But for whatever reason people

have gotten into the habit of omitting the parentheses. So you’re just going to have

to live with that. And realize that this is not d

of s ^2 or anything like that. And similarly, this is a single

number, and this is a single number. Infinitesimal. So that’s just the way that this

idea here gets written in our notation. And this is the first time we’re

dealing with squares of infinitesimals. So it’s just a little

different. But immediately the first thing

we’re going to do is take the square root. If I take the square root,

that’s the square root of dx ^2 + dy ^2. And this is the form

in which I always remember this formula. Let’s put it in some brightly

decorated form. But there are about 4, 5, 6

other forms that you’ll derive from this, which all mean

the same thing. So this is, as I say, the

way I remember it. But there are other ways

of thinking of it. And let’s just write a

couple of them down. The first one is that you

can factor out the dx. So that looks like this. 1 + (dy / dx)^2. And then I factored

out the dx. So this is a variant. And this is the one which

actually we’ll be using in practice right now

on our examples. So the conclusion is that the

arc length, which if you like is this total sn – s0, if you

like, is going to be equal to the integral from a to b

of the square root of 1 + (dy / dx)^2 dx. In practice, it’s also

very often written informally as this. The integral ds. So the change in this little

variable s, and this is what you’ll see notationally

in many textbooks. So that’s one way of writing it,

and of course the second way of writing it which is

practically the same thing is square root of 1 +

f ‘ ( x^2) dx. Mixing in a little bit

of Newton’s notation. And this is with y=f (x). So this is the formula

for arc length. And as I say, I remember

it this way. But you’re going to have to

derive various variants of it. And you’ll have to use

some arithmetic to get to various formulas. And there will be more later. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: OK, the question is,

is f ‘ (x)^2=f ” (x). And the answer is no. And let’s just see what it is. So, for example, if f ( x)=x

^2, which is an example which will come up in a few minutes,

then f ‘ (x)=2x and f ‘ ( x )^2==(2x), which is 4x ^2. Whereas f ” ( x) is equal to,

if I differentiate this another time, it’s equal to 2. So they don’t mean

the same thing. The same thing over here. You can see this dy / dx, this

is the rate of change of y with respect to x. The quantity squared. So in other words, this

thing is supposed to mean the same as that. Yeah. Another question. STUDENT: [INAUDIBLE] PROFESSOR: So the question is,

you got a little nervous because I left out

these limits. And indeed, I did that on

purpose because I didn’t want to specify what was going on. Really, if you wrote it in terms

of ds, you’d have to write it as starting at s0 and

ending at sn to be consistent with the variable s. But of course, if you write it

in terms of another variable, you put that variable in. So this is what we do when we

change variables, right? We have many different choices

for these limits. And this is the clue as to

which variable we use. STUDENT: [INAUDIBLE] PROFESSOR: Correct. s0 and

sn are not the same thing as a and b. In fact, this is xn. And this x0, over here. That’s what a and b are. But s0 and sn are mileage

markers on the road. They’re not the same thing as

keeping track of what’s happening on the x axis. So when we measure arc length,

remember it’s mileage along the curved path. So now, I need to give

you some examples. And my first example is going

to be really basic. But I hope that it helps to give

some perspective here. So I’m going to take the example

y=m x, which is a linear function, a

straight line. And then y ‘ would=m, and so

ds is going to be the square root of 1 + (y ‘) ^2 dx. Which is the square root

of 1 + m ^2 dx. And now, the length, say, if

we go from, I don’t know, let’s say 0 to 10, let’s say. Of the graph is going to be the

integral from 0 to 10 of the square root of

1 + m ^2 dx. Which of course is just 10

square root of 1 + m ^2. Not very surprising. This is a constant. It just factors out and

the integral from 0 to 10 of dx is 10. Let’s just draw a

picture of this. This is something which

has slope m here. And it’s going to 10. So this horizontal is 10. And the vertical is 10 m. Those are the dimensions

of this. And the Pythagorean theorem says

that the hypotenuse, not surprisingly, let’s draw it in

here in orange to remind ourselves that it was the same

type of orange that we had over there, this length

here is the square root of 10 ^2 + (10m)^2. That’s the formula for

the hypotenuse. And that’s exactly

the same as this. Maybe you’re saying duh,

this is obvious. But the point that I’m trying

to make is this. If you can figure out these

formulas for linear functions, calculus tells you how to do

it for every function. The idea of calculus is that

this easy calculation here, which you can do without any

calculus at all, all of the tools, the notations of

differentials and limits and integrals, is going to

make you be able to do it for any curve. Because we can break things

up into these little infinitesimal bits. This is the whole idea of all of

the methods that we had to set up integrals here. This is the main point

of these integrals. Now, so let’s do something

slightly more interesting. Our next example is going to be

the circle, so y=square root of 1 – x ^2. If you like, that’s the

graph of a semicircle. And maybe we’ll set it

up here this way. So that the semicircle goes

around like this. And well start it

here at x=0. And we’ll go over to a. And we’ll take this little

piece of the circle. So down to here. If you like. So here’s the portion of the

circle that I’m going to measure the length of. Now, we know that length. It’s called arc length. And I’m going to give it a name,

I’m going to call it alpha here. So alpha’s the arc length

along the circle. Now, let’s figure

out what it is. First, in order to do this,

I have to figure out what y ‘ is. Or, if you like, dy / dx. Now, that’s a calculation that

we’ve done a number of times. And I’m going to do it

slightly faster. But you remember it gives

you a square root in the denominator. And then you have the derivative

of what’s inside the square root. Which is – 2x. But then there’s also 1/2,

because in disguise it’s really (1 – x ^2)^2 1/2. So we’ve done this calculation

enough times that I’m not going to carry it

out completely. I want you to think

about what it is. It turns out to – x up

here, because the 1/2 and the 2 cancel. And now the thing that we have

to integrate is this arc length element, as

it’s called. ds. And that’s going to be

the square root of 1 + (y ‘) ^2 dx. And so I’m going to have to

carry out the calculation, some messy calculation here. Which is that this is 1

+ ( – x / square root of 1 – x ^2) ^2. So I have to figure out what’s

under the square root sign over here in order to carry

out this calculation. Now let’s do that. This is 1 + x ^2 / 1 – x ^2. That’s what this

simplifies to. And then that’s equal

to, over a common denominator, (1 – x ^2). 1 – x ^2 + x^2. And there is a little bit

of simplification now. Because the 2x ^2’s cancel. And we get 1 / 1 – x^2. So now I get to finish off the

calculation by actually figuring out what the

arc length is. And what is it? Well, this alpha is equal

to the integral from 0 to a of ds. Well, it’s going to

be the square root of what I have here. This was a square, this is just

what was underneath the square root sign. This is 1 + (y ‘) ^2. Have to take the square

root of that. So what I get here is dx / the

square root of 1 – x ^2. And now, we recognize this. The antiderivative of this is

something that we know. This is the inverse sine. Evaluated at 0 and a. Which is just giving us the

inverse sine of a, because the inverse sine of 0=0. So alpha=the inverse

sine of a. That’s a very fancy way of

saying that sine alpha=a. That’s the equivalent

statement here. And what’s going on here is

something that’s just a little deeper than it looks. Which is this. We’ve just figured out a

geometric interpretation of what’s going on here. That is, that we went a distance

alpha along this arc. And now remember that the

radius here is 1. And this horizontal distance

here is a. This distance here is a. And so the geometric

interpretation of this is that this angle is alpha radians. And sine alpha=a. So this is consistent with our

definition previously, our previous geometric definition

of radians. But this is really your first

true definition of radians. We never actually, people told

you that radians were the arc length along this curve. This is the first time

you’re deriving it. This is the first time you’re

seeing it correctly done. And furthermore, this is the

first time you’re seeing a correct definition of

the sine function. Remember we had this crazy

way, we we defined the exponential function, then we

had another way of defining the ln function as

an integral. Then we defined the exponential

in terms of it. Well, this is the same

sort of thing. What’s really happening here

is that if you want to know what radians are, you have

to calculate this number. If you’ve calculated this number

then by definition if sine is the thing whose alpha

radian amount gives you a, then it must be that this

is sine inverse of a. And so the first thing that gets

defined is the arc sine. And the next thing that gets

defined is the sine afterwards. This is the way the foundational

approach actually works when you start from

first principles. This arc length being one

of the first principles. So now we have a solid

foundation for trig functions. And this is giving

that connection. Of course, it’s consistent with

everything you already knew, so you don’t

have to make any transitional thinking here. It’s just that this is the first

time it’s being done rigorously. Because you only now

have arc length. So these are examples,

as I say, that maybe you already know. And maybe we’ll do one that we

don’t know quite as well. Let’s find the length

of a parabola. This is Example 3. Now, that was what I

was suggesting we were going to do earlier. So this is the function

y=x ^2. y ‘=2x. And so ds=the square root

of 1 + (2x) ^2 dx. And now I can figure out what

a piece of a parabola is. So I’ll draw the piece of

parabola up to a, let’s say, starting from 0. So that’s the chunk. And then its arc length, between

0 and a of this curve, is the integral from 0 to a of

square root of 1 + 4x ^2 dx. OK, now if you like, this is

the answer to the question. But people hate looking at

answers when they’re integrals if they can be evaluated. So one of the reasons why

we went through all this rigamarole of calculating these

things is to show you that we can actually evaluate

a bunch of these functions here more explicitly. It doesn’t help a lot, but

there is an explicit calculation of this. So remember how you

would do this. So this is just a little

bit of review. What we did in techniques

of integration. The first step is what? A substitution. It’s a trig substitution. And what is it? STUDENT: [INAUDIBLE] PROFESSOR: So x equals

something tan theta. I claim that it’s 1/2 tan, and

I’m going to call it u. Because I’m going to use theta

for something else in a couple of days. OK? So this is the substitution. And then of course dx=

1/2 sec ^2 u du , etc. So what happens if

you do this? I’ll write down the answer,

but I’m not going to carry this out. Because every one of these

is horrendous. But I think I worked it out. Let’s see if I’m lucky. Oh yeah. I think this is what it is. It’s a 1/4 ln 2x + square root

of 1 + 4x ^2 + 1/2 x ( square root of 1 + 4x ^2). Evaluated at a and 0. So yick. I mean, you know. STUDENT: [INAUDIBLE] PROFESSOR: Why I did

I make it 1/2? Because it turns out that

when you differentiate. So the question is, why

there 1/2 there? If you differentiate it without

the 1/2, you get this term and it looks like it’s

going to be just right. But then if you differentiate

this one you get another thing. And it all mixes together. And it turns out that

there’s more. So it turns out that it’s 1/2. Differentiate it and check. So this just an incredibly

long calculation. It would take fifteen minutes

or something like that. But the point is, you

do know in principle how to do these things. STUDENT: [INAUDIBLE] PROFESSOR: Oh, he was talking

about this 1/2, not this crazy 1/2 here. Sorry. STUDENT: [INAUDIBLE] PROFESSOR: Yeah, OK. So sorry about that. Thank you for helping. This factor of 1/2 here comes

about because when you square x, you don’t get tan ^2. When you square 2x, you get 4x

^2 and that matches perfectly with this thing. And that’s why you need

this factor here. Yeah. Another question,

way in the back. STUDENT: [INAUDIBLE] PROFESSOR: The question is, when

you do this substitution, doesn’t the limit from

0 to a change. And the answer is,

absolutely yes. The limits in terms of u are not

the same as the limits in terms of a. But if I then translate back to

the x variables, which I’ve done here in this bottom

formula, of x=0 and x=a, it goes back to those in

the original variables. So if I write things in the

original variables, I have the original limits. If I use the u variables, I

would have to change limits. But I’m not carrying out the

integration, because I don’t want to. So I brought it back

to the x formula. Other questions. OK, so now we’re ready to launch

into three-space a little bit here. We’re going to talk about

surface area. You’re going to be doing a

lot with surface area in multivariable calculus. It’s one of the really

fun things. And just remember, when it gets

complicated, that the simplest things are the

most important. And the simple things are, if

you can handle things for linear functions, you know all

the rest. So there’s going to be some complicated stuff but

it’ll really only involve what’s happening on planes. So let’s start with

surface area. And the example that I’d like to

give, this is the only type of example that we’ll have, is

the surface of rotation. And as long as we have

our parabola there, we’ll use that one. So we have y=x ^2, rotated

around the x axis. So let’s take a look at

what this looks like. It’s the parabola, which

is going like that. And then it’s being spun

around the x axis. So some kind of shape like

this with little circles. It’s some kind of trumpet

shape, right? And that’s the shape

that we’re. Now, again, it’s the surface. It’s just the metal of the

trumpet, not the insides. Now, the principle for figuring

out what the formula for area is, is not that

different from what we did for surfaces of revolution. But it just requires a little

bit of thought and imagination. We have a little chunk of

arc length along here. And we’re going to spin

that around this axis. Now, if this were a horizontal

piece of arc length, then it would spin around just

like a shell. It would just be a surface. But if it’s tilted, if it’s

tilted, then there’s more surface area proportional to the

amount that it’s tilted. So it’s proportional to the

length of the segment that you spin around. So the total is going

to be ds, that’s one of the factors here. Maybe I’ll write that second. That’s one of the dimensions. And then the other dimension

is the circumference. Which is 2 pi, in this case y. So that’s the end of

the calculation. This is the area element

of surface area. Now, when you get to 18.02, and

maybe even before that, you’ll also see some people

referring to this area element when it’s a curvy surface like

this with a notation d S. That’s a little confusing

because we have a lower case s here. We’re not going to

use it right now. But the lower case s is

usually arc length. The upper case s is usually

surface area. So. Also used for dA. The area element. Because this is a curved

area element. So let’s figure out

this example. So in the example, is equal to x

^2 then the situation is, we have the surface area is equal

to the integral from, I don’t know, 0 to a if those

are the limits that we wanted to choose. Of 2 pi x ^2, right? Because y=x ^2 ( the square

root of 1 + 4x ^2) dx. Remember we had this from

our previous example. This was ds from previous. And this, of course,

is 2 pi y. Now again, the calculation of

this integral is kind of long. And I’m going to omit it. But let me just point out that

it follows from the same substitution. Namely, x=1/2 tan u. Is going to work for

this integral. It’s kind of a mess. There’s a tan ^2 here

and the sec ^2. There’s another secant

and so on. So it’s one of these trig

integrals that then takes a while to do. So that just is going to

trail off into nothing. And the reason is that

what’s important here is more the method. And the setup of

the integrals. The actual computation, in

fact, you could go to a program and you could plug in

something like this and you would spit out an answer

immediately. So really what we just want is

for you to have enough control to see that it’s an integral

that’s a manageable one. And also to know that if

you plugged it in, you would get an answer. When I actually do carry out a

calculation, though, what I want to do is to do something

that has an answer that you can remember. And that’s a nice answer. So that turns out to be

the example of the surface area of a sphere. So it’s analogous

to this 2 here. And maybe I should remember

this result here. Which was that the arc length

element was given by this. So we’ll save that

for a second. So we’re going to do this

surface area now. So if you like, this

is another example. The surface area of a sphere. This is a good example, and

one, as I say, that has a really nice answer. So it’s worth doing. So first of all, I’m not going

to set it up quite the way I did in Example 2. Instead, I’m going to take the

general sphere, because I’d like to watch the dependence

on the radius. So here this is going

to be the radius. It’s going to be radius a. And now, if I carry out the same

calculations as before, if you think about it for a

second, you’re going to get this result. And then, the rest of the

arithmetic, which is sitting up there in the case, a=1,

will give us that ds=what? Well, maybe I’ll just

carry it out. Because that’s always nice. So we have 1 + x ^2

/ a ^2 – x ^2. That’s 1 + (y ‘) ^2. And now I put this over

a common denominator. And I get a ^2 – x ^2. And I have in the numerator

a ^2 – x ^2 + x^2. So the same cancellation

occurs. But now we get an a ^2

in the numerator. So now I can set up the ds. And so here’s what happens. The area of a section of the

sphere, so let’s see. We’re going to start at some

starting place x1, and end at some place x2. So what does that look like? Here’s the sphere. And we’re starting

at a place x1. And we’re ending

at a place x2. And we’re taking more or less

the slice here, if you like. The section of this sphere. So the area’s going

to equal this. And what is it going to be? Well, so I have here 2 pi y. I’ll write it out, just

leave it as y for now. And then I have ds. So that’s always what the

formula is when you’re revolving around the x axis. And then I’ll plug in

for those things. So 2 pi, the formula for y is

square root a ^2 – x ^2. And the formula for ds, well,

it’s the square root of this times dx. So it’s the square root of

a ^2 / a ^2 – x ^2 dx. So this part is ds. And this part is y. And now, I claim we have a nice cancellation that takes place. Square root of a ^2=a. And then there’s another

good cancellation. As you can see. Now, what we get here is the

integral from x1 to x2, of 2 pi a dx, which is about

the easiest integral you can imagine. It’s just the integral

of a constant. So it’s 2 pi a ( x2 – x1). Let’s check this in a

couple of examples. And then see what it’s saying

geometrically, a little bit. So what this is saying, so

special cases that you should always check when you have a

nice formula like this, at least. But really with anything

in order to make sure that you’ve got the

right answer. If you take, for example,

the hemisphere. So you take 1/2 of

this sphere. So that would be starting

at 0, sorry. And ending at a. So that’s the integral

from 0 to a. So this is the case

x1=0. x2=a. And what you’re going to

get is a hemisphere. And the area is (2 pi a ) a. Or in other words, 2 pi a ^2. And if you take the whole

sphere, that’s starting at x1=- a, and x2=a, you’re

getting (2 pi a) ( a – (- a)). Which is 4 pi a ^2. That’s the whole thing. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: The question is,

would it be possible to rotate around the y axis? And the answer is yes. It’s legal to rotate

around the y axis. And there is, if you use

vertical slices as we did here, that is, well they’re sort

of tips of slices, it’s a different idea. But anyway, it’s using dx as the

integral of the variable of integration. So we’re checking each little

piece, each little strip of that type. If we use dx here,

we get this. If you did the same thing

rotated the other way, and use dy as the variable, you get

exactly the same answer. And it would be the

same calculation. Because they’re parallel. So you’re, yep. STUDENT: [INAUDIBLE] PROFESSOR: Can you do service

area with shells? Well, ah shell shape. The short answer is not quite. The shell shape is a vertical

shell which is itself already three-dimensional, and

it has a thickness. So this is just a matter

of terminology, though. This thickness is this dx, when

we do this rotation here. And then there are two

other dimensions. If we have a curved surface,

there’s no other dimension left to form a shell. But basically, you can chop

things up into any bits that you can actually measure. That you can figure out

what the area is. That’s the main point. Now, I said we were going to,

we’ve just launched into three-dimensional space. And I want to now move on to

other space-like phenomena. But we’re going to do this. So this is also a preparation

for 18.02, where you’ll be doing this a tremendous

amount. We’re going to talk now about

parametric equations. Really just parametric curves. So you’re going to see this

now and we’re going to interpret it a couple of times,

and we’re going to think about polar coordinates. These are all preparation for

thinking in more variables, and thinking in a different

way than you’ve been thinking before. So I want you to prepare

your brain to make a transition here. This is the beginning

of the transition to multivariable thinking. We’re going to consider

curves like this. Which are described with x being

a function of t and y being a function of t. And this letter t is called

the parameter. In this case you should think

of it, the easiest way to think of it is as time. And what you have is what’s

called a trajectory. So this is also called

a trajectory. And its location, let’s

say, at time 0, is this location here. Of (0, y ( 0)), that’s

a point in the plane. And then over here, for

instance, maybe it’s (x ( 1), y (1)). And I drew arrows along here to

indicate that we’re going from this place over

to that place. These are later times. t=1

is a later time than t=0. So that’s just a very casual,

it’s just the way we use these notations. Now let me give you the first

example, which is x=a cos t, y=a sin t. And the first thing to figure

out is what kind of curve this is. And to do that, we want to

figure out what equation it satisfies in rectangular

coordinates. So to figure out what curve this

is, we recognize that if we square and add. So we add x ^2 to y^2, we’re

going to get something very nice and clean. We’re going to get a^2 cos

^2 t + a ^2 sin ^2 t. Yeah that’s right, OK. Which is just a ^2 (cos^2

+ sin ^2), or in other words a^2. So lo and behold, what

we have is a circle. And then we know what

shape this is now. And the other thing I’d like

to keep track of is which direction we’re going

on the circle. Because there’s more to this

parameter then just the shape. There’s also where we

are at what time. This would be, think

of it like the trajectory of a planet. So here, I have to do this by

plotting the picture and figuring out what happens. So at t=0, we have (x, y) is

equal to, plug in here (a cos 0, a sin 0). Which is just a * 1

+ a * 0, so a0. And that’s here. That’s the point (a, 0). We know that it’s the

circle of radius a. So we know that the curve

is going to go around like this somehow. So let’s see what happens

at t=pi / 2. So at that point, we have

(x,y)=( a cos pi / 2, a sin pi / 2). Which is (0, a), because

sine of pi / 2=1. So that’s up here. So this is what happens

at t=0. This is what happens

at t=pi / 2. And the trajectory clearly

goes this way. In fact, this turns out

to be t=pi, etc. And it repeats at t=2 pi. So the other feature that we

have, which is qualitative feature, is that it’s

counterclockwise. No the last little bit is going

to be the arc length. Keeping track of

the arc length. And we’ll do that next time.

Thank you. I've just learned that a few days ago, and you helped me to understand it better.

I Really Like The Video From Your Parametric equations, arclength, surface area

So Smiley You Make Available This Video Arc Length Surface Area and My Favourite the Parametric Equation

"Is f ' (x) squared the same as f "(x)?" asks an MIT student.

the confusion arises from leibniz notation where d^2/dx^2 would represent the second derivative. still pretty dumb question though aha

Yeah theres an equivalent notation in the function notation two but it contains parentheses f^(2) (x) (the 2's little), yeah nothing much, just thought it was funny, but it does happen- smart people can derp moment sometimes XD.

MIT, thank you so much. This is truly a gift.

Awesome..Its much helpful.

Am I the only one who doesn't seem think mr jerison isn't a good teacher? I mean compared to the brilliant herbert gross and christine breiner he seems to not be able to connect with us students and get what might be difficult or need more to students. In one of the lectures he couldn't even understand that the student who asked a question had a problem with negative numbers n that's the freakin basics .. PS I totally get this a purely subjective thing.

I think he meant is f'(x)^2 the same thing as f''(x). lol i had the same question the same time i saw this 😛 leibniz notation is too pro for me…

Could anyone tell me why we usually take the doubly infinitesimal number to be 0 but here it is valid to write ds^2 =/= 0?

7:15 affirmative action and quotas yay!

It made me realize that I'm smarter than 1 MIT student – for a minute, anyway.

Fuck you, racist

do you mean d2/d^2x?

no

cry about it. When you do something that stupid in MIT while tons of capable white and asian students get closed out due to "fairness" and "equality" I won't shut my mouth. Fuck you.

"…Thus, the libertarian opposes compulsory segregation and police brutality, but also opposes compulsory integration and such absurdities as ethnic quota systems in jobs."

I think he's actually not very sympathetic with students but he masters the subject and I find his explanations very enlightening, and hear him increases my knowledge and understanding a lot.

don't judge man, it was a legit question..

I agree. Some people don't need to be an ass.

I was wondering if a semicircle is differentiable at x=+1. I would say no, since it becomes minus infinity.

Great video, but misleading title. "Parametric Equations" in the name, and 40/45 minutes are about arc length and surface area? I kept waiting for the Parametric Equations topic to kick in I didn't realize the video was almost over.

you would think MIT students would learn this in highschool

i must say some guys ask really stupid questions. i felt like the professor having hard time not to say that.

I didn't get it about the deeper understanding about radians! 🙁

It's awesome

To everyone being rude to the student, he's actually not completely off like you wrongly assume. Because if you have parentheses around the number in the superscript/exponent it can be used to denote a higher order derivative, like f⁽⁵⁾(x) can be used instead of the much less readable f'''''(x).

min 25:04 it would have been worth mentioning that the resulting integral is a mix of real + imaginary components

How can I calculate volumen of a solid of revolution generated by a curve turned aroun an inclined line?

9:37

19:50

25:18 Surface area

29:43

32:50

36:52

40:40

36:03