# Lec 31 | MIT 18.01 Single Variable Calculus, Fall 2007 The following content is
OpenCourseWare continue to offer high quality educational
additional materials from hundreds of MIT courses, visit
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
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.

### 31 thoughts on “Lec 31 | MIT 18.01 Single Variable Calculus, Fall 2007”

• July 9, 2010 at 6:25 am

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

• January 14, 2012 at 8:19 am

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

• March 7, 2012 at 8:09 am

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

• April 4, 2012 at 7:30 am

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

• April 11, 2012 at 7:57 pm

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

• April 16, 2012 at 12:38 pm

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.

• June 28, 2012 at 8:39 pm

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

• August 27, 2012 at 12:24 pm

• December 16, 2012 at 6:25 pm

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.

• February 22, 2013 at 8:54 pm

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…

• March 30, 2013 at 5:53 am

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?

• April 3, 2013 at 2:56 pm

7:15 affirmative action and quotas yay!

• April 24, 2013 at 1:05 am

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

• April 24, 2013 at 1:06 am

Fuck you, racist

• April 24, 2013 at 1:08 am

do you mean d2/d^2x?

• April 24, 2013 at 1:36 am

no

• April 24, 2013 at 1:51 am

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."

• May 6, 2013 at 7:37 am

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.

• May 19, 2013 at 5:38 pm

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

• June 4, 2013 at 9:34 pm

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

• December 1, 2013 at 2:39 pm

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

• June 16, 2014 at 4:45 am

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.

• September 7, 2014 at 7:45 am

you would think MIT students would learn this in highschool

• January 10, 2017 at 2:42 pm

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

• October 5, 2017 at 10:13 pm

• March 21, 2018 at 5:41 pm

It's awesome

• April 13, 2019 at 9:05 pm

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).

• July 2, 2019 at 5:19 pm

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

• July 8, 2019 at 3:51 pm

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

• November 29, 2019 at 12:32 am

9:37
19:50
25:18 Surface area
29:43
32:50
36:52
40:40

• 