When an integral defined function is 0

WEBVTT
Kind: captions
Language: en

00:00:00.580 --> 00:00:03.740
So right over here we have the graph of
the function f.

00:00:03.740 --> 00:00:07.250
And we're assuming that f is a function of
t, our horizontal axis.

00:00:07.250 --> 00:00:09.110
Here's the t-axis.

00:00:09.110 --> 00:00:13.680
So that is f of t, lowercase f of t.

00:00:13.680 --> 00:00:16.440
And now let's just, let's define another
function.

00:00:16.440 --> 00:00:20.570
Let's call it capital F of, and it's not
going to be a function of t.

00:00:20.570 --> 00:00:22.480
It's going to be a function of x.

00:00:22.480 --> 00:00:27.050
So capital F of x is equal to.

00:00:27.050 --> 00:00:30.500
We're gonna definite it as the definite
integral between t

00:00:30.500 --> 00:00:35.860
is equal to negative 5 and t is equal to
x.

00:00:35.860 --> 00:00:39.100
Actually let me make those x's in the same
color so they really stand out.

00:00:39.100 --> 00:00:44.310
So f of x is equal to the definite
integral between t is equal to negative 5

00:00:44.310 --> 00:00:49.970
and t is equal to x of f of t, f of t dt.

00:00:49.970 --> 00:00:56.000
And if I had my druthers I'd probably
wouldn't use capital f and lower case f,

00:00:56.000 --> 00:01:00.130
I would use like a g or capital G just so
that when I say F, you don't get confused.

00:01:00.130 --> 00:01:05.640
Well I'm gonna try my best to say
lower-case F of t, capital F of x.

00:01:05.640 --> 00:01:09.790
So this is how we're gonna define the
function capital F of x.

00:01:09.790 --> 00:01:13.770
Definite integral between t is equal to
negative 5, and x of f of t, dt.

00:01:13.770 --> 00:01:16.550
Now given this definition.

00:01:16.550 --> 00:01:22.807
What I wanna think about is at what x
values does

00:01:22.807 --> 00:01:28.820
capital F of x, does capital F of x equal
0?

00:01:28.820 --> 00:01:29.670
So we'll write that down.

00:01:29.670 --> 00:01:35.436
At what x values

00:01:35.436 --> 00:01:40.550
does, this is, is this, I guess you could
say, this equation, at

00:01:40.550 --> 00:01:44.810
what x values is this equation true?

00:01:44.810 --> 00:01:47.710
And I encourage you to pause this video
right

00:01:47.710 --> 00:01:50.480
now and try to think about it on your own.

00:01:50.480 --> 00:01:52.190
And then we can work through it together.

00:01:53.320 --> 00:01:55.590
So I'm assuming you've had a go at it, so
let's just think

00:01:55.590 --> 00:02:01.500
about what this, what this function
capital F of x is really talking about.

00:02:01.500 --> 00:02:03.700
Well it's the one way to think about it

00:02:03.700 --> 00:02:09.919
is it's the area below, between negative
5, between t

00:02:09.919 --> 00:02:13.480
equals negative 5 and t equals x, that is
below

00:02:13.480 --> 00:02:16.930
the function f of t and above the t axis.

00:02:16.930 --> 00:02:19.260
And if the areas is the other way around,
if it's below

00:02:19.260 --> 00:02:23.040
the t axis and above the function, it's
gonna be negative area.

00:02:23.040 --> 00:02:25.570
So we're looking at t equals negative 5,
which

00:02:25.570 --> 00:02:29.860
is this, you could say this boundary right
over here.

00:02:29.860 --> 00:02:32.835
That is t is equal to negative 5.

00:02:33.890 --> 00:02:39.034
And if you put, pick an x value, let's
just say x were negative

00:02:39.034 --> 00:02:44.082
2, so if that is your x value right over
here, capital F would

00:02:44.082 --> 00:02:49.033
describe, would describe this area, this
area it would be a

00:02:49.033 --> 00:02:54.020
negative area because here the function is
below the t axis.

00:02:54.020 --> 00:03:00.450
So, for example capital F of negative 2
would be negative.

00:03:00.450 --> 00:03:03.650
Now, what x values makes this a 0?

00:03:03.650 --> 00:03:04.920
So one might jump out at you.

00:03:04.920 --> 00:03:13.480
Well, if we just, if we made, if we put x,
if we put x right at negative 5.

00:03:13.480 --> 00:03:14.600
Right at negative 5.

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Then there's no width, width here.

00:03:16.480 --> 00:03:18.870
There's not going to be any area.

00:03:18.870 --> 00:03:20.430
So f of, f of negative 5.

00:03:20.430 --> 00:03:27.320
F of negative 5, F of negative capital F
of negative 5, I should be clear here.

00:03:27.320 --> 00:03:32.500
Capital F of negative 5, which is equal to
the definite integral between t is equal

00:03:32.500 --> 00:03:39.570
to negative 5 and t is equal to negative
5, of f of t dt.

00:03:39.570 --> 00:03:41.040
F of t dt.

00:03:41.040 --> 00:03:44.610
Well you have the, you have the, you, you
have the same boundaries here.

00:03:44.610 --> 00:03:48.970
So this is going to be 0, once again this
area you have no width to this area.

00:03:48.970 --> 00:03:51.990
So this is going to be equal to 0.

00:03:51.990 --> 00:03:56.610
So, we can list x equals negative 5 as
being one of the points.

00:03:56.610 --> 00:04:01.360
One of the x values that makes capital F
of x equal 0.

00:04:01.360 --> 00:04:03.560
But let's see if we can find more of them.

00:04:06.160 --> 00:04:09.250
Let's see if we were to go, let me erase
this right over here.

00:04:11.870 --> 00:04:12.470
So, this was.

00:04:12.470 --> 00:04:14.364
We're starting at t equals negative 5.

00:04:15.550 --> 00:04:17.500
Now as x gets larger and larger and
larger,

00:04:17.500 --> 00:04:21.029
when x is equal to negative 3, our area.

00:04:21.029 --> 00:04:27.474
So, let's see this area right over here,
this area right over here is going to be,

00:04:27.474 --> 00:04:34.310
so that is, I see that we have, it's 2,
we're going from negative 5 to negative 3.

00:04:34.310 --> 00:04:36.420
So the distance right over here is two.

00:04:37.430 --> 00:04:40.160
This height, right over here, is 4.

00:04:40.160 --> 00:04:48.940
So, this area, right over here, is 2 times
4 times 1/2, which is going to be 4.

00:04:48.940 --> 00:04:53.200
And, since it's above the function and
below the t-axis, we'll

00:04:53.200 --> 00:04:57.690
write this as a negative, we'll write that
as a negative 4.

00:04:57.690 --> 00:05:01.820
And now, so let's see, so we're starting
when x is equal to 5.

00:05:01.820 --> 00:05:05.440
Uf, capital F of x is equal to 0, that is
you get further and

00:05:05.440 --> 00:05:06.590
further out, you're getting more and more

00:05:06.590 --> 00:05:09.080
negative values, that's more and more
negative values.

00:05:09.080 --> 00:05:11.410
But I just picked out this point right
over here because this

00:05:11.410 --> 00:05:17.150
seems like a point of transition of the
function, and then we have.

00:05:17.150 --> 00:05:22.590
This region, which is just going to add
more negative area to

00:05:22.590 --> 00:05:26.670
capital F of x as x gets larger and larger
and larger.

00:05:26.670 --> 00:05:29.350
And this looks like a really, a quarter
circle.

00:05:29.350 --> 00:05:33.560
This is, it has a quarter circle of radius
4, and now we

00:05:33.560 --> 00:05:39.230
have another quarter circle of radius 4
until we get to x equals 5.

00:05:39.230 --> 00:05:41.800
And this is actually going to be positive
area.

00:05:41.800 --> 00:05:44.660
Because here our function is above the t
axis.

00:05:46.480 --> 00:05:49.710
So when we've gotten this far, so all, the
way I'm thinking about it, my

00:05:49.710 --> 00:05:53.930
x is, my x, when capital F of x equal
negative 5 is, is 0.

00:05:53.930 --> 00:05:56.720
And now our area, as x becomes larger and

00:05:56.720 --> 00:05:59.980
larger, and larger, our area becomes
negative, negative, negative, negative.

00:05:59.980 --> 00:06:01.420
Now it becomes less negative, because
we're

00:06:01.420 --> 00:06:03.510
starting to add, we're starting to add.

00:06:03.510 --> 00:06:06.430
So, for example, if x is equal to 2, if x

00:06:06.430 --> 00:06:09.480
is equal to 2, we would be looking at this
positive.

00:06:09.480 --> 00:06:12.420
But we still have all this huge negative
area to overcome.

00:06:12.420 --> 00:06:14.700
So we're still in negative territory.

00:06:14.700 --> 00:06:17.460
But the more positive we add, we're gonna
become less negative.

00:06:17.460 --> 00:06:19.930
And you go all the way to x

00:06:19.930 --> 00:06:24.760
equals positive 5 and this positive area,
this quarter

00:06:24.760 --> 00:06:27.430
circle right over here of positive area,
is going

00:06:27.430 --> 00:06:31.760
to exactly offset this quarter circle of
negative area.

00:06:31.760 --> 00:06:33.530
You don't even have to think about what
that area is, although

00:06:33.530 --> 00:06:37.340
you can obviously figure it out with pi R
squared and whatever else.

00:06:37.340 --> 00:06:39.100
And so now we just have to keep

00:06:39.100 --> 00:06:43.270
adding more positive area to offset this
negative 4.

00:06:43.270 --> 00:06:45.100
So how do we do that?

00:06:45.100 --> 00:06:50.755
Well the height here, the height right
over here is 4, the height right over

00:06:50.755 --> 00:06:56.630
here is 4 so if we, if we can add our
rectangle that is one wide and 4 high.

00:06:56.630 --> 00:07:01.640
That's positive area of 4, which this has
right over here.

00:07:01.640 --> 00:07:03.280
So this is plus 4.

00:07:03.280 --> 00:07:05.590
Is going to offset this negative 4.

00:07:05.590 --> 00:07:09.140
So we go all the way to x is equal to 6.

00:07:09.140 --> 00:07:13.750
When x is equal to 6, capital F of x is
going to be equal to 0.

00:07:13.750 --> 00:07:15.260
Let's write that down.

00:07:15.260 --> 00:07:17.160
So capital F.

00:07:17.160 --> 00:07:22.912
Capital f of, capital f of positive 6, of
positive

00:07:22.912 --> 00:07:27.879
6, which is going to be equal to the
definite

00:07:27.879 --> 00:07:33.980
integral between negative 5 and positive
6.

00:07:33.980 --> 00:07:37.770
Positive 6 of f of t.

00:07:37.770 --> 00:07:42.150
F of t, dt, well we can break this up.

00:07:42.150 --> 00:07:44.293
We already went through it, I'm just
gonna, going to

00:07:44.293 --> 00:07:46.870
make sure that we really understood what
was going on.

00:07:46.870 --> 00:07:51.130
This is equal to, and I'll just to it all
in one color now.

00:07:51.130 --> 00:07:54.170
This is equal to, actually I'll do it in
these colors that I did here.

00:07:54.170 --> 00:07:55.520
This is equal to the integral.

00:07:56.730 --> 00:08:01.941
The integral between negative 5

00:08:01.941 --> 00:08:07.336
and negative 3 of f of t, dt, plus

00:08:07.336 --> 00:08:12.546
the integral, plus the integral

00:08:12.546 --> 00:08:19.190
between negative 3 and 1 of f of t dt.

00:08:19.190 --> 00:08:27.020
Plus the integral between 1 and 5 of f of
t, dt.

00:08:27.020 --> 00:08:29.430
That describes this right over here.

00:08:29.430 --> 00:08:33.880
And then finally, plus, plus

00:08:34.960 --> 00:08:40.730
the definite integral between 5 and 6 of f
of t, dt.

00:08:40.730 --> 00:08:43.549
f of t, dt.

00:08:43.549 --> 00:08:47.920
And, this describes this negative area.

00:08:47.920 --> 00:08:50.620
This describes this positive area.

00:08:50.620 --> 00:08:52.700
These two net out to be 0.

00:08:52.700 --> 00:08:56.240
And then this area we already figured out,
is.

00:08:56.240 --> 00:08:59.850
Or this definite integral I should say is
negative 4.

00:08:59.850 --> 00:09:04.520
This is negative 4 and this one right over
here is positive 4 and so they

00:09:04.520 --> 00:09:08.520
net out and this of course is equal to 0
which we wanted to figure out.

00:09:08.520 --> 00:09:10.300
Once again how did I do that?

00:09:10.300 --> 00:09:13.770
Well I said okay clearly when x is equal
to negative 5 you have no area and then i

00:09:13.770 --> 00:09:17.530
just kept increasing, kept increasing xs
to larger and

00:09:17.530 --> 00:09:19.840
larger values, I could have gone actually
the other way.

00:09:19.840 --> 00:09:22.520
And, I would have had just more and more
positive values there

00:09:22.520 --> 00:09:25.130
would have been nothing to offset it to
get us back to 0.

00:09:25.130 --> 00:09:29.024
But, as we increased x above negative 5,
the

00:09:29.024 --> 00:09:32.040
capital Fof x; the area, is more and more

00:09:32.040 --> 00:09:35.120
and more negative, but then it becomes,
then we

00:09:35.120 --> 00:09:38.920
start adding positive value to it to
offset the negative.

00:09:38.920 --> 00:09:44.666
And, we fully offset the negative at x is
equal to 6.


Functions defined by integrals

WEBVTT
Kind: captions
Language: en

00:00:00.810 --> 00:00:04.080
So this right over here is the graph of
the function g of t.

00:00:04.080 --> 00:00:08.850
It is a function of t, and let's define a
new function.

00:00:08.850 --> 00:00:15.200
Let's call it capital G of x, and it's
equal to the definite

00:00:15.200 --> 00:00:21.790
integral between t is equal to negative 3
and t is equal to x.

00:00:21.790 --> 00:00:27.250
Of our lowercase g of t, g of t, dt.

00:00:27.250 --> 00:00:33.460
So, given how we have just, given how we
have just defined capital G of x,

00:00:33.460 --> 00:00:36.900
let's see if we can evaluate some val,
let's see if we can find some values.

00:00:36.900 --> 00:00:39.610
So, let's evaluate the function capital G.

00:00:39.610 --> 00:00:42.920
At x is equal to 4, and let's also
evaluate

00:00:42.920 --> 00:00:45.930
the function capital G at x is equal to 8.

00:00:45.930 --> 00:00:48.860
And I encourage you to pause the video
right now and try to

00:00:48.860 --> 00:00:53.560
think about these on your own, and then we
can work through them together.

00:00:53.560 --> 00:00:57.340
So let's tackle capital G of 4 first.

00:00:57.340 --> 00:01:02.050
So this is going to be, this is going to
be, well, if x is equal to 4.

00:01:02.050 --> 00:01:05.390
This, this top boundary is going to be 4,
so this is going to be the definite

00:01:05.390 --> 00:01:11.640
integral of g of t between t is equal to
negative 3, and t is equal to 4.

00:01:11.640 --> 00:01:17.210
g of t, g of t, d t.

00:01:17.210 --> 00:01:19.550
Now what's that going to be equal to?

00:01:19.550 --> 00:01:21.240
Well, let's see, let's look at this graph
here.

00:01:21.240 --> 00:01:26.970
So, we have negative 3, this t is equal to
negative 3, which is right over here.

00:01:26.970 --> 00:01:27.990
T is equal to negative 3.

00:01:27.990 --> 00:01:30.920
And we're going to go all the way to is
equal to 4.

00:01:30.920 --> 00:01:34.190
Let me circle that in orange.

00:01:34.190 --> 00:01:37.780
All the way to t is equal to 4 which is
right over here.

00:01:38.820 --> 00:01:39.480
So this is.

00:01:39.480 --> 00:01:41.740
One way to think about this is this is
going to

00:01:41.740 --> 00:01:47.960
be the area above t-axis and below the
graph of G.

00:01:47.960 --> 00:01:51.290
So, it's going to be this area, right over

00:01:51.290 --> 00:01:55.710
here, that is above the t-axis and below
the graph

00:01:55.710 --> 00:02:01.370
of G of T, but, we're not going to add to
it this area because this area over here.

00:02:01.370 --> 00:02:03.050
I'll shade it in yellow.

00:02:03.050 --> 00:02:06.270
This area I'm shading in yellow over here,
this is going to be negative.

00:02:06.270 --> 00:02:07.890
Why's it going to be negative?

00:02:07.890 --> 00:02:10.970
Because we want the area that is above t
and below g.

00:02:10.970 --> 00:02:12.350
This is the other way around.

00:02:12.350 --> 00:02:18.320
This is below the t axis and above the
graph of g.

00:02:18.320 --> 00:02:21.780
So one way to think about it is, we could
split it up.

00:02:21.780 --> 00:02:25.790
So, we could, actually let me just clear
this out so we have more space.

00:02:26.920 --> 00:02:32.930
So, this, right over here, is going to be
equal to the integral.

00:02:32.930 --> 00:02:34.260
I'll do that in this purple color.

00:02:34.260 --> 00:02:43.394
The integral from t equals negative 3 to
0, of g of dt +.

00:02:43.394 --> 00:02:46.500
I'll do this in the other color.

00:02:46.500 --> 00:02:56.450
Plus the interval from 0, t equals 0 to t
is equal to 4 of g of t, dt.

00:02:56.450 --> 00:02:58.470
Now what are each of these going to be?

00:02:58.470 --> 00:03:02.230
Well this is just a triangle where the
base is 3.

00:03:02.230 --> 00:03:03.590
The base has length 3.

00:03:03.590 --> 00:03:05.850
The height is length 3.

00:03:05.850 --> 00:03:09.400
So, it's going to be 3 times 3, which is 9
times one half.

00:03:09.400 --> 00:03:11.870
Because 3 times 3 would give us the area
of this entire square.

00:03:11.870 --> 00:03:13.970
But this triangle is half of that.

00:03:13.970 --> 00:03:16.730
So it's, this right over here, is going to
be 4.5.

00:03:16.730 --> 00:03:19.620
This is going to be 4.5.

00:03:19.620 --> 00:03:21.130
And then, what's this area in yellow?

00:03:21.130 --> 00:03:26.400
Well, let's see, we have a triangle, it's
base is, or it's width here is 4,

00:03:26.400 --> 00:03:31.610
it's height is 4, 4 times 4 is 16, which
would be area of this entire square.

00:03:31.610 --> 00:03:33.420
We take half of that, it's 8.

00:03:33.420 --> 00:03:35.440
Now we don't just add it to it, because,

00:03:35.440 --> 00:03:37.570
once again, this is going to be negative
area.

00:03:37.570 --> 00:03:42.140
The graph over here is below, below the T
axis.

00:03:42.140 --> 00:03:44.710
So this, we would say, we would, this

00:03:44.710 --> 00:03:47.270
integral is going to evaluate to negative
8.

00:03:47.270 --> 00:03:48.980
Once again, why is it negative 8?

00:03:48.980 --> 00:03:52.720
Cause the graph is below, below the T
axis.

00:03:52.720 --> 00:03:53.986
And so what do we get?

00:03:53.986 --> 00:04:00.225
We get g(4) which is this area essentially
minus this area.

00:04:00.225 --> 00:04:05.060
4.5 minus 8 is going to be.

00:04:05.060 --> 00:04:10.510
Let's see 4 minus 8 is negative 4 out of
.5 which is negative 3.5.

00:04:10.510 --> 00:04:12.900
Negative 3.5.

00:04:12.900 --> 00:04:16.510
Now let's try to figure out what g of 8
is.

00:04:16.510 --> 00:04:18.940
So g of 8.

00:04:18.940 --> 00:04:22.800
And if you couldn't figure it out the 1st
time around, try to pause the video again,

00:04:22.800 --> 00:04:25.510
and now that we've figured out g of 4, try
to figure what g of 8 is.

00:04:26.680 --> 00:04:30.620
Well g of 8 is, 1 way to think about it,
it's going to be that.

00:04:30.620 --> 00:04:33.960
Minus that area and then we're going to
add, then we're going to figure

00:04:33.960 --> 00:04:38.870
out the area and then we're going to have
two more areas to think about.

00:04:38.870 --> 00:04:40.090
We're gonna go all the way to 8.

00:04:40.090 --> 00:04:42.450
So actually make, draw the line there.

00:04:42.450 --> 00:04:48.830
So we're gonna have to think about, we're
going to think about this whole area now.

00:04:49.970 --> 00:04:51.790
This whole area now.

00:04:51.790 --> 00:04:54.000
And then we're gonna think about this one.

00:04:55.520 --> 00:04:59.480
So we could write this is going to be
equal to the integral between.

00:04:59.480 --> 00:05:00.770
Actually let me do that purple color.

00:05:01.890 --> 00:05:04.050
So it's going to be this, this integral.

00:05:04.050 --> 00:05:08.880
Which is the integral between negative, t
equals negative 3 and 0.

00:05:08.880 --> 00:05:11.580
G of t, dt.

00:05:11.580 --> 00:05:14.820
Plus this entire yellow region right now.

00:05:14.820 --> 00:05:17.150
The part that we looked at before plus
this part over here.

00:05:17.150 --> 00:05:22.674
So, I'll just write that as, plus the
definated role between t is equal to 0

00:05:22.674 --> 00:05:27.800
and 6 of g of t dt.

00:05:27.800 --> 00:05:29.130
And then finally.

00:05:29.130 --> 00:05:37.150
Plus the definite integral between T is
equal to 6 and T is equal to 8 g(t) dt.

00:05:38.810 --> 00:05:44.120
Now we already know that this first one
evaluates to 4.5, so that one is 4.5.

00:05:44.120 --> 00:05:46.710
Now what's this one going to be?

00:05:46.710 --> 00:05:48.620
We have a triangle.

00:05:48.620 --> 00:05:53.999
Whose base is length 6, height is 4, 6
times

00:05:53.999 --> 00:05:59.640
4 is 24 times one half is going to get us
12.

00:05:59.640 --> 00:06:04.380
So this is going to be, that over there is
twelve and then finally what is this area?

00:06:04.380 --> 00:06:05.660
Oh and we have to be careful.

00:06:05.660 --> 00:06:09.380
This is below the t axis and above the
graph.

00:06:09.380 --> 00:06:10.450
So this is going to be a negative 12.

00:06:10.450 --> 00:06:15.190
And then finally, we have this area, which
is once again going to be a positive area.

00:06:15.190 --> 00:06:18.340
It's below the graph and above the t axis.

00:06:18.340 --> 00:06:23.220
And so let's see, 2 times 4 is 8 times 0.5
is 4.

00:06:23.220 --> 00:06:26.650
So we're going to have.

00:06:26.650 --> 00:06:28.010
Plus 4.

00:06:28.010 --> 00:06:30.410
Plus 4.

00:06:30.410 --> 00:06:33.570
And so what do we have?

00:06:33.570 --> 00:06:39.910
We end up with 4.5 plus 4 is 8.5 minus 12.

00:06:39.910 --> 00:06:42.700
So this is going to be equal to.

00:06:43.710 --> 00:06:50.830
This is going to be equal to 8.8 minus 12
would be negative 4 plus 5.

00:06:50.830 --> 00:06:53.620
It is negative 3.5 again.

00:06:53.620 --> 00:06:56.890
Now why did these two things end up being
the same?

00:06:58.990 --> 00:07:00.490
Well think about what happened here.

00:07:00.490 --> 00:07:04.660
When we went from capital G of 4 to
capital G of 8.

00:07:04.660 --> 00:07:12.130
We subtracted, we subtracted this value
right over here and we added this

00:07:12.130 --> 00:07:17.240
value right over here and you see that
these two triangles have the same areas.

00:07:17.240 --> 00:07:22.670
So we subtracted and added the same amount
to get the same exact value.


Visually determining antiderivative

WEBVTT
Kind: captions
Language: en

00:00:00.790 --> 00:00:05.520
Let's say that this right over here is the
graph of lowercase f of x.

00:00:05.520 --> 00:00:08.050
That's lowercase f of x there.

00:00:08.050 --> 00:00:13.050
And let's say that we have some other
function capital F of x.

00:00:13.050 --> 00:00:18.480
And if you were to take its derivative,
so, capital F prime of x,

00:00:18.480 --> 00:00:23.820
that's equal to lowercase f of x, lower
case f of x.

00:00:23.820 --> 00:00:29.280
So given that, which of these, which of
these, could

00:00:29.280 --> 00:00:33.350
be the graph of capital, of capital F of
x?

00:00:33.350 --> 00:00:36.080
And I encourage you to pause this video,
and try and

00:00:36.080 --> 00:00:38.500
think about it on your own before we work
through it.

00:00:40.790 --> 00:00:44.212
Well if, if this curve is going to be the
derivative of one of them.

00:00:44.212 --> 00:00:46.033
That means that any, for any x value

00:00:46.033 --> 00:00:49.020
it's describing what the instantaneous
rate of change.

00:00:49.020 --> 00:00:52.698
Or what the slope of the tangent line is,
of which

00:00:52.698 --> 00:00:56.880
ever one of these is the possible capital
F of x.

00:00:56.880 --> 00:00:58.740
So, let's just look at a couple of things
right here

00:00:58.740 --> 00:01:01.580
so what, what do we know about lower case
f of x?

00:01:01.580 --> 00:01:04.959
What do we know of which is the derivative
of one of these?

00:01:04.959 --> 00:01:07.370
Well, one thing we know is it's always
positive.

00:01:07.370 --> 00:01:11.550
It, it has as we go to negative infinity,
it asymptotes towards 0.

00:01:11.550 --> 00:01:13.780
But it's always positive.

00:01:13.780 --> 00:01:16.650
So since this is describing the slope of
one of these.

00:01:16.650 --> 00:01:20.060
That means that the slope of one of these
always,

00:01:20.060 --> 00:01:23.860
or out of the candidates, has to always be
positive.

00:01:23.860 --> 00:01:25.130
And, if we look at this, the slope

00:01:25.130 --> 00:01:27.620
of the tangent line here is, indeed,
always positive.

00:01:27.620 --> 00:01:30.400
The slope of the tangent line here does
look like it's positive.

00:01:30.400 --> 00:01:33.400
Every time we increase an x, we're
increasing by y.

00:01:33.400 --> 00:01:34.580
Here it's positive.

00:01:34.580 --> 00:01:35.510
But here it's negative.

00:01:35.510 --> 00:01:38.110
When we increase by x, we decrease by y.

00:01:38.110 --> 00:01:41.580
So, we can rule, we can rule this one out.

00:01:41.580 --> 00:01:44.510
Now, what else, what else do we know?

00:01:44.510 --> 00:01:46.230
Well, this is the derivative.

00:01:46.230 --> 00:01:48.150
This is telling us the slope of the
tangent line.

00:01:48.150 --> 00:01:53.930
So, for example, when x is equal to, when
x is equal to negative 4,

00:01:53.930 --> 00:01:59.290
f of, f of negative 4 is pretty close to
0.

00:01:59.290 --> 00:02:00.920
It's pretty close to 0.

00:02:00.920 --> 00:02:03.850
So, it's slightly, slightly more than 0.

00:02:03.850 --> 00:02:09.264
So, that tells us that the slope of
tangent line of capital F

00:02:09.264 --> 00:02:14.696
of x has to be pretty close to 0, when x
equals negative 4.

00:02:14.696 --> 00:02:17.996
So, let's see, when x equals negative 4,
the slope of tangent

00:02:17.996 --> 00:02:21.480
line, here, isn't close to 0, this
actually looks closer to 1.

00:02:21.480 --> 00:02:24.280
So, we could rule this one out.

00:02:24.280 --> 00:02:27.850
Over here, when x is equal to negative 4,
the slope of

00:02:27.850 --> 00:02:30.800
the tangent line, yeah, that actually does
look pretty close to 0.

00:02:30.800 --> 00:02:32.170
So I won't rule that one out.

00:02:32.170 --> 00:02:35.550
And over here, the slope of the tangent
line, when x

00:02:35.550 --> 00:02:39.000
is equal to negative 4, that also looks
pretty close to 0.

00:02:39.000 --> 00:02:42.660
So these are still both in the running.

00:02:42.660 --> 00:02:44.760
So let's see how we can think of it
different.

00:02:44.760 --> 00:02:46.290
So let's just pick another point.

00:02:46.290 --> 00:02:53.390
When x is equal to, when x is equal to 0,
f of 0 looks like it's pretty close to 1.

00:02:53.390 --> 00:02:55.120
I don't know if it's exactly to 1.

00:02:55.120 --> 00:02:57.260
Actually it looks almost exactly.

00:02:57.260 --> 00:02:59.450
Almost exactly equal to 1.

00:02:59.450 --> 00:03:04.520
So when capital F of, so at capital F of
0, the

00:03:04.520 --> 00:03:07.770
slope of the tangent line needs to be
pretty close to 1.

00:03:07.770 --> 00:03:10.990
So over here the slope of the tangent
line, when

00:03:10.990 --> 00:03:15.460
x is equal to 0, that looks smaller than
1.

00:03:15.460 --> 00:03:18.780
So this slope is definitely not 1.

00:03:18.780 --> 00:03:22.222
While over here, when x is equal to 0.

00:03:22.222 --> 00:03:27.905
The slope of the tangent line does look,
the slope of the tangent

00:03:27.905 --> 00:03:33.180
line does look pretty, pretty close,
pretty close to 1.

00:03:33.180 --> 00:03:35.730
So this, right over here, looks like the

00:03:35.730 --> 00:03:41.290
best candidate for capital, for capital F
of x.

00:03:41.290 --> 00:03:42.260
So that one right over there.

00:03:42.260 --> 00:03:45.970
Lemme, that is capital F of x.

00:03:45.970 --> 00:03:49.400
And you might say hey these look similar
to each other.

00:03:49.400 --> 00:03:52.700
And fact they look almost or actually they
do look almost identical.

00:03:52.700 --> 00:03:55.190
And you might remember from what you knew
about

00:03:55.190 --> 00:03:58.840
differentiation, that these actually look
like the basic exponential function.

00:03:58.840 --> 00:04:01.130
Were, were I didn't ask you to find out
what the

00:04:01.130 --> 00:04:05.209
actual function was just the possible
anti-derivative of this function would be.

00:04:05.209 --> 00:04:10.307
This is the derivative, lower case f is
the, is the derivative of capital

00:04:10.307 --> 00:04:15.260
f, or you could say that capital f is an
anti derivative of lower case f.

00:04:15.260 --> 00:04:18.180
And when you just inspect this, this looks
like this, the,

00:04:18.180 --> 00:04:21.004
the function, both of these functions is,
are e to the x.

00:04:21.004 --> 00:04:25.440
Because the derivative of e to the x is e
to the x.


Worked example: motion problems (with definite integrals)

WEBVTT
Kind: captions
Language: en

00:00:00.710 --> 00:00:03.860
Let's review a little bit of what we
learned in differential calculus.

00:00:03.860 --> 00:00:08.370
Let's say we have some function S that it
gives us as

00:00:08.370 --> 00:00:12.220
a function of time the position of a
particle in one dimension.

00:00:12.220 --> 00:00:12.903
If we were to take the derivative with
respect to time.

00:00:12.903 --> 00:00:16.500
So if we were to take the derivative with
respect to time

00:00:19.840 --> 00:00:21.080
of this function S.

00:00:21.080 --> 00:00:21.911
What are we going to get?

00:00:21.911 --> 00:00:25.135
Well we're going to get ds, dt or the

00:00:25.135 --> 00:00:29.816
rate in which position changes with
respect to time.

00:00:29.816 --> 00:00:34.159
And what's another word for that, position
changes with respect to time?

00:00:34.159 --> 00:00:35.750
Well, that's just velocity.

00:00:35.750 --> 00:00:39.680
So that we can write as velocity as a
function of time.

00:00:39.680 --> 00:00:43.980
Now, what if we were to take the
derivative of that with respect to time.

00:00:43.980 --> 00:00:47.177
So we could either view this as the second
derivative, we're

00:00:47.177 --> 00:00:50.809
taking the derivative not once, but twice
of our position function.

00:00:50.809 --> 00:00:52.625
Or you could say that we're taking the

00:00:52.625 --> 00:00:55.473
derivative with respect to time of our
velocity function.

00:00:55.473 --> 00:00:59.524
Well this is going to be, we can write
this as, we can write

00:00:59.524 --> 00:01:05.420
this as dv, dt, the rate at which velocity
is changing with respect to time.

00:01:05.420 --> 00:01:07.490
And what's another word for that?

00:01:07.490 --> 00:01:10.000
Well, that's also called acceleration.

00:01:10.000 --> 00:01:12.480
This is going to be our acceleration as a
function of time.

00:01:12.480 --> 00:01:14.560
So, you start with the position function,
take

00:01:14.560 --> 00:01:17.590
it, the position as a function of time.

00:01:17.590 --> 00:01:20.310
Take its derivative with respected time,
you get velocity.

00:01:20.310 --> 00:01:23.800
Take that derivative with respected time,
you get acceleration.

00:01:23.800 --> 00:01:26.210
Well, you could go the other way around.

00:01:26.210 --> 00:01:29.290
If you started with acceleration, if you
started

00:01:29.290 --> 00:01:32.160
with acceleration, and you were to take
the antiderivative.

00:01:32.160 --> 00:01:38.370
If you were to take the antiderivative of
it, the anti, anti, an antiderivative

00:01:38.370 --> 00:01:42.860
of it is going to be, actually let me just
write it this way.

00:01:42.860 --> 00:01:45.230
So an antiderivative, I'll just use the
interval

00:01:45.230 --> 00:01:48.290
symbol to show that I'm taking the
antiderivative.

00:01:48.290 --> 00:01:54.920
Is going to be the integral of the
anti-derivative of a of t.

00:01:54.920 --> 00:01:58.670
And this is going to give you some
expression with a plus c.

00:01:58.670 --> 00:02:02.240
And we could say, well, that's a general
form of our velocity function.

00:02:02.240 --> 00:02:05.380
This is going to be equal to our velocity
function.

00:02:05.380 --> 00:02:08.910
And to find the particular velocity
function, we would have

00:02:08.910 --> 00:02:12.090
to know what the velocity is at a
particular time.

00:02:12.090 --> 00:02:14.390
And then, we could solve for our c.

00:02:14.390 --> 00:02:17.780
Whether then, if we're able to do that and
we were to take the anti-derivative again.

00:02:19.530 --> 00:02:24.173
Then, now we're taking the anti-derivative
of our velocity function,

00:02:24.173 --> 00:02:27.574
which would give us some expression as a
function of t.

00:02:27.574 --> 00:02:29.058
And then, some other constant.

00:02:29.058 --> 00:02:32.576
And, if we could solve for that constant,
then we know, then we know

00:02:32.576 --> 00:02:36.110
what the position is going to be, the
position is a function of time.

00:02:37.960 --> 00:02:39.890
Just like this, it would have some, plus c
here if we

00:02:39.890 --> 00:02:43.920
know our position at a given time we could
solve for that c.

00:02:43.920 --> 00:02:47.430
So now that we've reviewed it a little
bit, but we've rewritten it in.

00:02:47.430 --> 00:02:49.890
I guess you could say, thinking of it not
just from

00:02:49.890 --> 00:02:52.090
the differential point of view from the
derivative point of view.

00:02:52.090 --> 00:02:55.660
But thinking of it from the
anti-derivative point of view.

00:02:55.660 --> 00:02:58.290
Lets see if we can solve an interesting
problem.

00:02:58.290 --> 00:03:00.750
Lets say that we know that the
acceleration of a

00:03:00.750 --> 00:03:05.950
particle is a function of time is equal to
one.

00:03:05.950 --> 00:03:08.938
So it's always accelerating at one unit
per,

00:03:08.938 --> 00:03:11.548
and you know, I'm not giving you time.

00:03:11.548 --> 00:03:14.950
Let, let's just say that we're thinking in
terms of meters and seconds.

00:03:14.950 --> 00:03:20.170
So this is one meter per second, one meter
per second-squared, right over here.

00:03:20.170 --> 00:03:23.380
That's our acceleration as a function of
time.

00:03:23.380 --> 00:03:26.155
And, let's say we don't know the velocity
expressions, but we know

00:03:26.155 --> 00:03:29.318
the velocity at a particular time and we
don't know the position expressions.

00:03:29.318 --> 00:03:32.212
But we know the position at a particular
time.

00:03:32.212 --> 00:03:35.250
So, let's say we know that the velocity,
at time three.

00:03:35.250 --> 00:03:39.420
Let's say three seconds is negative three
meters per second.

00:03:39.420 --> 00:03:42.770
And actually I wanna write the units here,
just to make it a little, a little bit.

00:03:42.770 --> 00:03:45.330
So this is meters per second squared,

00:03:45.330 --> 00:03:46.830
that's going to be our unit for
acceleration.

00:03:46.830 --> 00:03:49.500
This is our unit for velocity.

00:03:49.500 --> 00:03:55.120
And let's say that we know, let's say that
we know that the position at

00:03:55.120 --> 00:03:59.760
time two, at two seconds is equal to
negative ten meters.

00:03:59.760 --> 00:04:02.490
So, if we're thinking in one dimension, of
if this is moving

00:04:02.490 --> 00:04:06.140
along the number line, this is ten to the
left of the origin.

00:04:06.140 --> 00:04:09.508
So, given this information right over
here, and everything that I wrote up here.

00:04:09.508 --> 00:04:14.568
Can we figure out the actual expressions
for velocity as a function of time?

00:04:14.568 --> 00:04:18.890
So not just velocity at time three, but
velocity generally as a function of time.

00:04:18.890 --> 00:04:21.190
And position as a function of time.

00:04:21.190 --> 00:04:23.950
And I encourage you to pause this video
right now.

00:04:23.950 --> 00:04:25.730
And try to figure it out on your own.

00:04:27.130 --> 00:04:28.230
So let's just work through this.

00:04:28.230 --> 00:04:31.410
What is, we know that velocity, as a

00:04:31.410 --> 00:04:36.250
function of time, is going to be the
anti-derivative.

00:04:36.250 --> 00:04:39.910
The anti-derivitive of our acceleration is
a function of time.

00:04:39.910 --> 00:04:42.250
Our acceleration is just one.

00:04:42.250 --> 00:04:45.900
So this is going to be the anti-derivitive
of this right over here

00:04:45.900 --> 00:04:50.890
is going to be t and then we can't forget
our constant plus c.

00:04:50.890 --> 00:04:55.610
And now we can solve for c because we know
v of 3 is negative 3.

00:04:55.610 --> 00:04:57.220
So lets just write that down.

00:04:57.220 --> 00:05:05.385
So v of 3 is going to be equal to 3, 3
plus c.

00:05:05.385 --> 00:05:08.276
I just replaced where I saw the, the t or
every place where

00:05:08.276 --> 00:05:11.580
I have the t I replaced it with this 3
right over here.

00:05:11.580 --> 00:05:12.855
Actually let me make it a little bit
clearer.

00:05:12.855 --> 00:05:18.276
So v of 3, v of 3 is equal to 3 plus c,
and they

00:05:18.276 --> 00:05:23.280
tell us that that's equal to negative 3.

00:05:23.280 --> 00:05:29.040
So that is equal to, that is equal to
negative 3.

00:05:29.040 --> 00:05:30.630
So what's c going to be?

00:05:30.630 --> 00:05:32.260
So if we just look at this part of

00:05:32.260 --> 00:05:35.480
the equation or just this equation right
over here.

00:05:35.480 --> 00:05:43.290
If you subtract 3 from both sides, you
get, you get c is equal to negative 6.

00:05:43.290 --> 00:05:46.660
And so now we know the exact, we know the

00:05:46.660 --> 00:05:51.030
exact expression that defines velocity as
a function of time.

00:05:51.030 --> 00:05:56.349
V of t, v of t is equal to t, t plus

00:05:56.349 --> 00:06:01.410
negative 6 or, t minus 6.

00:06:01.410 --> 00:06:02.380
And we can verify that.

00:06:02.380 --> 00:06:05.980
The derivative of this with respect to
time is just one.

00:06:05.980 --> 00:06:11.040
And when time is equal to 3, time minus 6
is indeed negative 3.

00:06:11.040 --> 00:06:15.730
So we've been able to figure out velocity
a s a function of time.

00:06:15.730 --> 00:06:19.460
So now let's do a similar thing to figure
out position as a function of time.

00:06:19.460 --> 00:06:22.080
We know that position is gonna be an
anti-derivative

00:06:22.080 --> 00:06:25.830
of the velocity function, so let's write
that down.

00:06:25.830 --> 00:06:30.563
So, position, as a function of time, is
going to

00:06:30.563 --> 00:06:35.093
be equal to the anti-derivative of v of t,
dt.

00:06:35.093 --> 00:06:41.178
Which is equal to the anti-derivative of t
minus 6, dt which is

00:06:41.178 --> 00:06:47.615
equal to well the anti-derivative of t, is
t squared over 2.

00:06:47.615 --> 00:06:51.270
So, t squared over 2, we've seen that
before.

00:06:51.270 --> 00:06:56.139
The anti-derivative of negative 6 is
negative 6 t, and of

00:06:56.139 --> 00:07:01.310
course, we can't forget our constant, so,
plus, plus, c.

00:07:01.310 --> 00:07:03.270
So this is what s of t is equal to, s

00:07:03.270 --> 00:07:06.820
of t is equal to all of this business
right over here.

00:07:06.820 --> 00:07:09.480
And now we can try to solve, for our
constant.

00:07:09.480 --> 00:07:12.420
And we do that, using this information
right over here.

00:07:12.420 --> 00:07:16.780
At two seconds were at, our position is
negative two meters.

00:07:16.780 --> 00:07:20.000
So s of 2, or I could just write it this
way.

00:07:20.000 --> 00:07:20.830
Well, let me write it this way.

00:07:20.830 --> 00:07:26.530
S of 2, at 2 seconds, is going to be equal
to 2 squared over 2.

00:07:26.530 --> 00:07:30.700
That is, let's see, that's 4 over 2,
that's going to be 2.

00:07:30.700 --> 00:07:32.770
Minus 6 times 2.

00:07:32.770 --> 00:07:41.510
So, minus 12 + c is equal to negative 10,
is equal to negative 10.

00:07:41.510 --> 00:07:44.817
So, let's see, we get 2 minus 12 is

00:07:44.817 --> 00:07:50.320
negative, is negative 10 plus c equals
negative 10.

00:07:50.320 --> 00:07:54.270
So you add 10 to both sides you get c in
this case is equal to 0.

00:07:54.270 --> 00:07:57.630
So we figured out what our position
function is as well.

00:07:57.630 --> 00:07:59.730
The c right over here is just going to be
0.

00:07:59.730 --> 00:08:08.800
So our position as a function of time is
equal to t squared over 2 minus 6 t.

00:08:08.800 --> 00:08:10.320
And you can verify.

00:08:10.320 --> 00:08:17.250
When t is equal to 2, t squared over 2 is
2, minus 12 is negative 10.

00:08:17.250 --> 00:08:20.680
You take the derivative here, you get t
minus 6.

00:08:20.680 --> 00:08:24.140
And you can see and we already verified
that v of 3 is negative 3.

00:08:24.140 --> 00:08:28.060
And you take the derivative here, you get
a of t, just like that.

00:08:28.060 --> 00:08:30.150
Anyway, hopefully, you found this
enjoyable.


Finding horizontal and vertical asymptotes

WEBVTT
Kind: captions
Language: en

00:00:00.271 --> 00:00:03.428
Voiceover: We have F of X
is equal to three X squared

00:00:03.428 --> 00:00:10.060
minus 18X minus 81, over
six X squared minus 54.

00:00:10.060 --> 00:00:11.878
Now what I want to do in this video

00:00:11.878 --> 00:00:15.127
is find the equations for the horizontal

00:00:15.127 --> 00:00:17.213
and vertical asymptotes

00:00:17.213 --> 00:00:20.111
and I encourage you to
pause the video right now

00:00:20.111 --> 00:00:21.480
and try to work it out on your own

00:00:21.480 --> 00:00:23.801
before I try to work through it.

00:00:23.801 --> 00:00:26.509
I'm assuming you've had a go at it.

00:00:26.509 --> 00:00:28.083
Let's think about each of them.

00:00:28.083 --> 00:00:35.071
Let's first think about
the horizontal asymptote,

00:00:35.071 --> 00:00:37.374
see if there at least is one.

00:00:37.374 --> 00:00:41.822
The horizontal asymptote
is really what is the line,

00:00:41.822 --> 00:00:45.250
the horizontal line that F of X approaches

00:00:45.250 --> 00:00:49.291
as the absolute value of X approaches,

00:00:49.291 --> 00:00:52.981
as the absolute value
of X approaches infinity

00:00:52.981 --> 00:00:55.868
or you could say what does F of X approach

00:00:55.868 --> 00:00:57.912
as X approaches infinity

00:00:57.912 --> 00:01:00.260
and what does F of X approach

00:01:00.260 --> 00:01:03.571
as X approaches negative infinity.

00:01:03.571 --> 00:01:06.172
There's a couple of ways
you could think about it.

00:01:06.172 --> 00:01:08.861
Let me just rewrite the
definition of F of X

00:01:08.861 --> 00:01:09.789
right over here.

00:01:09.789 --> 00:01:15.566
It's three X squared minus 18X minus 81.

00:01:15.566 --> 00:01:21.224
All of that over six X squared minus 54.

00:01:21.224 --> 00:01:23.726
Now there's two ways you
could think about it.

00:01:23.726 --> 00:01:24.759
One you could say, okay,

00:01:24.759 --> 00:01:27.955
as X as the absolute value of X

00:01:27.955 --> 00:01:30.113
becomes larger and larger and larger,

00:01:30.113 --> 00:01:32.854
the highest degree terms in the numerator

00:01:32.854 --> 00:01:39.264
and the denominator are going to dominate.

00:01:39.264 --> 00:01:41.325
What are the highest degree terms?

00:01:41.325 --> 00:01:44.294
Well the numerator you
have three X squared

00:01:44.294 --> 00:01:46.813
and in the denominator
you have six X squared.

00:01:46.813 --> 00:01:53.248
As X approaches, as
the absolute value of X

00:01:53.248 --> 00:01:54.624
approaches infinity,

00:01:54.624 --> 00:01:56.559
these two terms are going to dominate.

00:01:56.559 --> 00:02:02.777
F of X is going to become
approximately three X squared

00:02:02.777 --> 00:02:04.980
over six X squared.

00:02:04.980 --> 00:02:06.896
These other terms are going to matter less

00:02:06.896 --> 00:02:09.304
obviously minus 54 isn't
going to grow at all

00:02:09.304 --> 00:02:11.687
and minus 18X is going to grow much slower

00:02:11.687 --> 00:02:14.027
than the three X squared,

00:02:14.027 --> 00:02:17.318
the highest degree terms are
going to be what dominates.

00:02:17.318 --> 00:02:19.396
If we look at just those terms

00:02:19.396 --> 00:02:22.483
then you could think of
simplifying it in this way.

00:02:22.483 --> 00:02:24.843
F of X is going to get closer and closer

00:02:24.843 --> 00:02:27.893
to 3/6 or 1/2.

00:02:27.893 --> 00:02:30.332
You could say that there's
a horizontal asymptote

00:02:30.332 --> 00:02:35.542
at Y is equal to 1/2.

00:02:35.542 --> 00:02:37.371
Another way we could
have thought about this

00:02:37.371 --> 00:02:39.734
if you don't like this whole little bit

00:02:39.734 --> 00:02:43.206
of hand wavy argument that
these two terms dominate

00:02:43.206 --> 00:02:46.401
is that we can divide the
numerator and the denominator

00:02:46.401 --> 00:02:49.756
by the highest degree or X
raised to the highest power

00:02:49.756 --> 00:02:51.325
in the numerator and the denominator.

00:02:51.325 --> 00:02:54.438
The highest degree term is
X squared in the numerator.

00:02:54.438 --> 00:02:57.838
Let's divide the numerator
and the denominator

00:02:57.838 --> 00:02:59.688
or I should say the highest degree term

00:02:59.688 --> 00:03:01.875
in the numerator and the
denominator is X squared.

00:03:01.875 --> 00:03:03.630
Let's divide both the numerator

00:03:03.630 --> 00:03:05.033
and denominator by that.

00:03:05.033 --> 00:03:10.216
If you multiply the numerator
times one over X squared

00:03:10.216 --> 00:03:13.530
and the denominator
times one over X squared.

00:03:13.530 --> 00:03:15.776
Notice we're not changing the value

00:03:15.776 --> 00:03:19.656
of the entire expression,
we're just multiplying it

00:03:19.656 --> 00:03:22.355
times one if we assume
X is not equal zero.

00:03:23.462 --> 00:03:25.005
We get two.

00:03:25.005 --> 00:03:27.723
In our numerator, let's
see three X squared

00:03:27.723 --> 00:03:29.497
divided by X squared is going to be three

00:03:29.497 --> 00:03:36.073
minus 18 over X minus 81 over X squared

00:03:36.073 --> 00:03:39.080
and then all of that over six X squared

00:03:39.080 --> 00:03:41.412
times one over X squared,
this is going to be six

00:03:41.412 --> 00:03:45.765
and then minus 54 over X squared.

00:03:45.765 --> 00:03:47.638
What's going to happen?

00:03:47.638 --> 00:03:50.622
If you want to think in terms of

00:03:50.622 --> 00:03:52.730
if you want to think of limits

00:03:52.730 --> 00:03:54.417
as something approaches infinity.

00:03:54.417 --> 00:03:57.106
If you want to say the limit as X

00:03:57.106 --> 00:03:58.455
approaches infinity here.

00:03:58.455 --> 00:03:59.408
What's going to happen?

00:03:59.408 --> 00:04:03.297
Well this, this and that
are going to approach zero

00:04:03.297 --> 00:04:08.750
so you're going to approach 3/6 or 1/2.

00:04:08.750 --> 00:04:10.900
Now, if you say this X
approaches negative infinity,

00:04:10.900 --> 00:04:11.955
it would be the same thing.

00:04:11.955 --> 00:04:14.356
This, this and this approach zero

00:04:14.356 --> 00:04:17.391
and once again you approach 1/2.

00:04:17.391 --> 00:04:19.271
That's the horizontal asymptote.

00:04:19.271 --> 00:04:21.744
Y is equal to 1/2.

00:04:21.744 --> 00:04:24.642
Let's think about the vertical asymptotes.

00:04:24.642 --> 00:04:27.556
Let me write that down right over here.

00:04:27.556 --> 00:04:29.429
Let me scroll over a little bit.

00:04:29.429 --> 00:04:37.148
Vertical asymptote or possibly asymptotes.

00:04:37.148 --> 00:04:40.508
Vertical maybe there is more than one.

00:04:40.508 --> 00:04:42.808
Now it might be very tempting to say,

00:04:42.808 --> 00:04:44.623
"Okay, you hit a vertical asymptote"

00:04:44.623 --> 00:04:47.277
"whenever the denominator equals to zero"

00:04:47.277 --> 00:04:49.817
"which would make this
rational expression undefined"

00:04:49.817 --> 00:04:51.894
and as we'll see for this case

00:04:51.894 --> 00:04:55.258
that is not exactly right.

00:04:55.258 --> 00:04:58.461
Just making the denominator
equal to zero by itself

00:04:58.461 --> 00:05:00.700
will not make a vertical asymptote.

00:05:00.700 --> 00:05:01.692
It will definitely be a place

00:05:01.692 --> 00:05:03.003
where the function is undefined

00:05:03.003 --> 00:05:05.785
but by itself it does not
make a vertical asymptote.

00:05:05.785 --> 00:05:08.574
Let's just think about this
denominator right over here

00:05:08.574 --> 00:05:10.004
so we can factor it out.

00:05:10.004 --> 00:05:11.257
Actually let's factor out the numerator

00:05:11.257 --> 00:05:12.298
and the denominator.

00:05:12.298 --> 00:05:16.636
We can rewrite this as F of
X is equal to the numerator

00:05:16.636 --> 00:05:19.155
is clearly every term
is divisible by three

00:05:19.155 --> 00:05:20.253
so let's factor out three.

00:05:20.253 --> 00:05:22.670
It's going to be three times X squared

00:05:22.670 --> 00:05:26.278
minus six X minus 27.

00:05:26.278 --> 00:05:29.680
All of that over the denominator

00:05:29.680 --> 00:05:31.713
each term is divisible by six.

00:05:31.713 --> 00:05:35.495
Six times X squared minus 9

00:05:35.495 --> 00:05:38.578
and let's see if we can
factor the numerators

00:05:38.578 --> 00:05:40.890
and denominators out further.

00:05:40.890 --> 00:05:45.993
This is going to be F of
X is equal to three times

00:05:45.993 --> 00:05:50.106
let's see, two numbers,
their product is negative 27,

00:05:50.106 --> 00:05:51.578
their sum is negative six.

00:05:51.578 --> 00:05:54.355
Negative nine and three seem to work.

00:05:54.355 --> 00:05:58.763
You could have X minus
nine times X plus three.

00:05:58.763 --> 00:06:01.930
Just factor the numerator
over the denominator.

00:06:01.930 --> 00:06:04.000
This is the difference of
squares right over here.

00:06:04.000 --> 00:06:08.507
This would be X minus
three times X plus three.

00:06:08.507 --> 00:06:13.187
When does the denominator equal zero?

00:06:13.187 --> 00:06:18.379
The denominator equals zero

00:06:18.379 --> 00:06:22.420
when X is equal to positive three

00:06:22.420 --> 00:06:26.262
or X is equal to negative three.

00:06:26.262 --> 00:06:28.402
Now I encourage you to pause
this video for a second.

00:06:28.402 --> 00:06:32.202
Think about are both of
these vertical asymptotes?

00:06:32.202 --> 00:06:35.310
Well you might realize that the numerator

00:06:35.310 --> 00:06:40.177
also equals zero when X is
equal to negative three.

00:06:40.177 --> 00:06:43.325
What we can do is actually
simplify this a little bit

00:06:43.325 --> 00:06:44.723
and then it becomes a little bit clear

00:06:44.723 --> 00:06:46.548
where our vertical asymptotes are.

00:06:46.548 --> 00:06:48.483
We could say that F of X,

00:06:48.483 --> 00:06:51.110
we could essentially divide the numerator

00:06:51.110 --> 00:06:53.620
and denominator by X plus three

00:06:53.620 --> 00:06:54.675
and we just have to key,

00:06:54.675 --> 00:06:56.247
if we want the function to be identical,

00:06:56.247 --> 00:06:58.701
we have to keep the [caveat]
that the function itself

00:06:58.701 --> 00:07:02.074
is not defined when X is
equal to negative three.

00:07:02.074 --> 00:07:04.952
That definitely did
make us divide by zero.

00:07:04.952 --> 00:07:06.134
We have to remember that

00:07:06.134 --> 00:07:08.361
but that will simplify the expression.

00:07:08.361 --> 00:07:11.229
This exact same function is going to be

00:07:11.229 --> 00:07:12.544
if we divide the numerator and denominator

00:07:12.544 --> 00:07:13.641
by X plus three,

00:07:13.641 --> 00:07:18.438
it's going to be three times X minus nine

00:07:18.438 --> 00:07:22.377
over six times X minus three

00:07:22.377 --> 00:07:29.361
for X does not equal negative three.

00:07:29.361 --> 00:07:32.037
Notice, this is an identical definition

00:07:32.037 --> 00:07:33.304
to our original function

00:07:33.304 --> 00:07:37.099
and I have to put this
qualifier right over here

00:07:37.099 --> 00:07:38.863
for X does not equal negative three

00:07:38.863 --> 00:07:42.193
because our original function is undefined

00:07:42.193 --> 00:07:44.172
at X equals negative three.

00:07:44.172 --> 00:07:46.891
X equals negative three is
not a part of the domain

00:07:46.891 --> 00:07:48.107
of our original function.

00:07:48.107 --> 00:07:52.514
If we take X plus three
out of the numerator

00:07:52.514 --> 00:07:53.530
and the denominator,

00:07:53.530 --> 00:07:55.132
we have to remember that.

00:07:55.132 --> 00:07:56.743
If we just put this right over here,

00:07:56.743 --> 00:07:58.040
this wouldn't be the same function

00:07:58.040 --> 00:08:00.448
because this without
the qualifier is defined

00:08:00.448 --> 00:08:02.503
for X equals negative three

00:08:02.503 --> 00:08:04.454
but we want to have the
exact same function.

00:08:04.454 --> 00:08:07.415
You'd actually have a
point in discontinuity

00:08:07.415 --> 00:08:08.360
right over here

00:08:08.360 --> 00:08:10.807
and now we could think about
the vertical asymptotes.

00:08:10.807 --> 00:08:13.192
Now the vertical asymptotes
going to be a point

00:08:13.192 --> 00:08:15.111
that makes the denominator equals zero

00:08:15.111 --> 00:08:17.062
but not the numerator equals zero.

00:08:17.062 --> 00:08:20.293
X equals negative three
made both equal zero.

00:08:20.293 --> 00:08:23.264
Our vertical asymptote,
I'll do this in green

00:08:23.264 --> 00:08:25.370
just to switch or blue.

00:08:25.370 --> 00:08:32.704
Our vertical asymptote is going to be

00:08:32.704 --> 00:08:37.122
at X is equal to positive three.

00:08:37.122 --> 00:08:39.246
That's what made the
denominator equal zero

00:08:39.246 --> 00:08:41.826
but not the numerator
so let me write that.

00:08:41.826 --> 00:08:46.146
The vertical asymptote
is X is equal to three.

00:08:46.146 --> 00:08:50.062
Using these two points of information

00:08:50.062 --> 00:08:51.401
or I guess what we just figured out.

00:08:51.401 --> 00:08:54.662
You can start to attempt
to sketch the graph,

00:08:54.662 --> 00:08:56.883
this by itself is not going to be enough.

00:08:56.883 --> 00:08:58.669
You might want to also plot a few points

00:08:58.669 --> 00:09:02.124
to see what happens I
guess around the asymptotes

00:09:02.124 --> 00:09:05.617
as we approach the two
different asymptotes

00:09:05.617 --> 00:09:08.495
but if we were to look at a graph.

00:09:08.495 --> 00:09:13.166
Actually let's just do it for fun here

00:09:13.166 --> 00:09:15.793
just to complete the
picture for ourselves.

00:09:15.793 --> 00:09:21.135
The function is going to
look something like this

00:09:21.135 --> 00:09:22.428
and I'm not doing it at scales.

00:09:22.428 --> 00:09:25.439
That's one and this is
1/2 right over here.

00:09:25.439 --> 00:09:32.595
Y equals 1/2 is the horizontal asymptote.

00:09:32.595 --> 00:09:34.642
Y is equal to 1/2

00:09:34.642 --> 00:09:35.888
and we have a vertical asymptote

00:09:35.888 --> 00:09:37.623
that X is equal to positive three.

00:09:38.823 --> 00:09:42.018
We have one, two ...

00:09:42.018 --> 00:09:43.423
I'm going to do that in blue.

00:09:43.423 --> 00:09:47.778
One, two, three, once again
I didn't draw it to scale

00:09:47.778 --> 00:09:50.396
or the X and Y's aren't on the same scale

00:09:50.396 --> 00:09:53.329
but we have a vertical
asymptote just like that.

00:09:53.329 --> 00:09:55.289
Just looking at this we don't know exactly

00:09:55.289 --> 00:09:56.712
what the function looks like.

00:09:56.712 --> 00:09:58.358
It could like something like this

00:09:58.358 --> 00:10:01.611
and maybe does something like that

00:10:01.611 --> 00:10:04.791
or it could do something like that

00:10:04.791 --> 00:10:10.136
or it could do something
like that and that

00:10:10.136 --> 00:10:13.808
or something like that and that.

00:10:13.808 --> 00:10:15.481
Hopefully you get the idea here

00:10:15.481 --> 00:10:17.064
and to figure out what it does,

00:10:17.064 --> 00:10:18.639
you would actually want
to try out some points.

00:10:18.639 --> 00:10:20.647
The other thing we want
to be clear is that

00:10:20.647 --> 00:10:22.347
the function is also not defined

00:10:22.347 --> 00:10:26.271
at X is equal to negative three.

00:10:27.916 --> 00:10:33.502
Let me make X equals negative three here.

00:10:33.502 --> 00:10:37.632
One, two, three, so
the function might look

00:10:37.632 --> 00:10:39.277
and once again I haven't
tried out the points.

00:10:39.277 --> 00:10:40.824
It could look something like this,

00:10:40.824 --> 00:10:44.848
it could look something
where we're not defined

00:10:44.848 --> 00:10:48.428
at negative three

00:10:48.428 --> 00:10:50.154
and then it goes something like this

00:10:50.154 --> 00:10:52.187
and maybe does something like that

00:10:52.187 --> 00:10:59.262
or maybe it does something like that.

00:10:59.262 --> 00:11:01.411
It's not defined at negative three

00:11:01.411 --> 00:11:02.841
and this would be an asymptote right now

00:11:02.841 --> 00:11:03.832
so we get closer and closer

00:11:03.832 --> 00:11:05.677
and it could go something like that

00:11:05.677 --> 00:11:07.429
or it goes something like that.

00:11:07.429 --> 00:11:09.446
Once again, to decide
which of these it is,

00:11:09.446 --> 00:11:12.318
you would actually want
to try out a few values.

00:11:12.318 --> 00:11:14.181
I encourage you to, after this video,

00:11:14.181 --> 00:11:15.325
try that out on yourself

00:11:15.325 --> 00:11:17.046
and try to figure out
what the actual graph

00:11:17.046 --> 00:11:18.631
of this looks like.


Divergence of telescoping series

WEBVTT
Kind: captions
Language: en

00:00:00.540 --> 00:00:02.650
Lets say that we have the sum

00:00:05.939 --> 00:00:10.660
one minus one plus one minus one plus one
and just keeps going on

00:00:10.660 --> 00:00:14.990
and on and on like that forever and we can
write that with sigma notation.

00:00:14.990 --> 00:00:22.330
This would be the sum from n is equal to
one lower case n equals one to infinity.

00:00:22.330 --> 00:00:25.630
We have an infinite number of terms here
but see this first one, we

00:00:25.630 --> 00:00:30.140
want it to be a positive one, and then we
want to keep switching terms.

00:00:30.140 --> 00:00:36.710
So we could say that this is negative 1 to
the lowercase n minus 1 power.

00:00:36.710 --> 00:00:38.040
Let's just verify that that works.

00:00:38.040 --> 00:00:41.750
When n is equal to 1, it's negative 1 to
the 0 power, which is that.

00:00:41.750 --> 00:00:45.910
When n is equal to 2, it's 2 minus 1; it's
negative

00:00:45.910 --> 00:00:48.820
one to the first power that's equal to
that right over there.

00:00:48.820 --> 00:00:52.480
So this is a way of writing this series.

00:00:52.480 --> 00:01:00.950
Now what I wanna think about is does this
series converge to an actual finite value?

00:01:00.950 --> 00:01:03.800
Or, and this is another way of saying it,
what is the sum?

00:01:03.800 --> 00:01:07.410
Is there a finite sum that is equal to

00:01:07.410 --> 00:01:12.260
this right over here, or does this series
diverge?

00:01:12.260 --> 00:01:14.430
And the way that we can think about that
is

00:01:14.430 --> 00:01:18.990
by thinking about its partial sums let me
write that down.

00:01:18.990 --> 00:01:25.060
The partial, the partial sums of this
series.

00:01:25.060 --> 00:01:28.610
And the way we can define the partial
sums, so we'll give an index here.

00:01:28.610 --> 00:01:33.700
So capital N, so the partial sum is going
to

00:01:33.700 --> 00:01:39.100
be the sum from n equals one but not
infinity but to capital

00:01:39.100 --> 00:01:43.660
N of negative 1 to the n minus 1.

00:01:43.660 --> 00:01:48.820
So just to be clear, what this means, so
the, the partial sum with just

00:01:48.820 --> 00:01:53.830
one term is just gonna be from lowercase n
equals 1 to uppercase N equals 1.

00:01:53.830 --> 00:01:56.770
So it's just going to be this first term
right over here.

00:01:56.770 --> 00:01:58.960
It's just going to be 1.

00:01:58.960 --> 00:02:06.050
The S sub two, s sub two is going to be
equal to one minus one.

00:02:06.050 --> 00:02:08.690
It's gonna be the sum of the first two
terms.

00:02:08.690 --> 00:02:15.560
S sub three, S sub three is going to be
one minus one plus one.

00:02:15.560 --> 00:02:19.950
It's the sum of the first three terms,
which is of course equal to.

00:02:19.950 --> 00:02:22.280
Equal to, let's see this equal to one.

00:02:22.280 --> 00:02:27.070
This one over here is equal to zero.

00:02:27.070 --> 00:02:30.640
S sub 4, we could keep going, S sub four
is going to

00:02:30.640 --> 00:02:35.650
be one minus one plus one minus one which
is equal to zero again.

00:02:35.650 --> 00:02:43.440
So once again, the question is, does this
sum converge to some finite value?

00:02:43.440 --> 00:02:45.530
And I encourage you to pause this video
and think about

00:02:45.530 --> 00:02:48.180
it, given what we see about the partial
sums right over here.

00:02:49.450 --> 00:02:54.230
So in order for a series to converge, that
means that the limit, an

00:02:54.230 --> 00:02:59.400
infinite series to converge, that means
that the limit, the limit, so if you're a

00:02:59.400 --> 00:03:04.930
convergence, convergence is the same
thing, is the same

00:03:04.930 --> 00:03:09.980
thing as saying that the limit as capital,

00:03:09.980 --> 00:03:14.470
the limit as capital N approaches infinity
of our partial

00:03:14.470 --> 00:03:18.400
sums is equal to some finite.

00:03:18.400 --> 00:03:24.310
Let me just write like this, is equal to
some Finite, so Finite Value.

00:03:24.310 --> 00:03:26.760
So, what is this limit going to be?

00:03:26.760 --> 00:03:27.690
Well, let's see if we can write this.

00:03:27.690 --> 00:03:30.015
So, this is going to be, let's see s sub

00:03:30.015 --> 00:03:32.020
n, if we want to write this in general
terms.

00:03:32.020 --> 00:03:38.200
We already see if s, if capital N is odd,
it's equal to 1.

00:03:38.200 --> 00:03:40.790
If capital N is even, it's equal to 0.

00:03:40.790 --> 00:03:47.140
So, we can write, lets write this down so
s sub n I could write it like this

00:03:47.140 --> 00:03:51.960
is going to be one if n odd

00:03:51.960 --> 00:03:56.620
it's equal to zero if n even.

00:03:56.620 --> 00:04:01.900
So what's the limit as s sub n approaches
infinity so what's the limit.

00:04:01.900 --> 00:04:06.250
What's the limit, as N approaches infinity
of S sub N.

00:04:07.340 --> 00:04:08.810
Well, this limit doesn't exist.

00:04:08.810 --> 00:04:11.630
It keeps oscillating between these points.

00:04:11.630 --> 00:04:14.470
You give me, you, you go one more, it goes
from 1 to 0.

00:04:14.470 --> 00:04:16.660
You give me one more, it goes from 0 to 1.

00:04:16.660 --> 00:04:20.450
So it actually is not approaching a finite
value.

00:04:20.450 --> 00:04:24.790
So this right over here does not exist.

00:04:29.280 --> 00:04:31.120
It's tempting, because it's bounded.

00:04:31.120 --> 00:04:35.160
It's only, it keeps oscillating between 1
and 0.

00:04:35.160 --> 00:04:41.100
But it does not go to one particular value
as n approaches infinity.

00:04:41.100 --> 00:04:46.090
So here we would say that our series s
diverges.

00:04:46.090 --> 00:04:51.810
Our series S diverges.


Local linearization example

WEBVTT
Kind: captions
Language: en

00:00:00.630 --> 00:00:04.430
Consider the function F of X is equal to
the natural log of X squared.

00:00:04.430 --> 00:00:10.100
Let L of X be the local linearization of F
at A is equal to E.

00:00:10.100 --> 00:00:12.720
What is a rule for L of X?

00:00:12.720 --> 00:00:16.922
Let's just make sure we understand what
they're asking for, and I've copied and

00:00:16.922 --> 00:00:21.450
pasted this exact exercise on my little
scratchpad so we can work through it.

00:00:21.450 --> 00:00:25.330
So, let's just visualize what f of x looks
like.

00:00:25.330 --> 00:00:27.090
And what l of x is.

00:00:27.090 --> 00:00:33.630
What the local linearization, if we are
centering it at x equals e.

00:00:33.630 --> 00:00:36.603
If we're saying that there's the set, when
they say

00:00:36.603 --> 00:00:40.320
a that's just a convention for what are we
approximating around.

00:00:40.320 --> 00:00:44.030
What the locality, what the x value that
we are approximating around.

00:00:44.030 --> 00:00:49.950
So lets just, let me grab some points here
just to visualize f of x.

00:00:49.950 --> 00:00:52.662
So let's say this is 1, this is 2, this

00:00:52.662 --> 00:00:56.230
is, actually let me spread them out a
little bit more.

00:00:57.290 --> 00:01:02.810
So this is 1, this is 2, this is 3.

00:01:02.810 --> 00:01:04.420
F of 1 is gonna be 0.

00:01:04.420 --> 00:01:07.820
The natural log of 1 squared is just 0.

00:01:07.820 --> 00:01:09.910
And let's see, we could think of F of

00:01:09.910 --> 00:01:12.910
E, and that's actually where we're
centering our linearization.

00:01:12.910 --> 00:01:16.380
We're gonna think about values in that
locality.

00:01:16.380 --> 00:01:20.030
So E is roughly right over here.

00:01:20.030 --> 00:01:21.890
E is right, roughly right over here.

00:01:21.890 --> 00:01:24.176
And F of E is equal to natural log of

00:01:24.176 --> 00:01:28.070
E squared is, well, that's just going to
be 2.

00:01:28.070 --> 00:01:33.027
So this is going to be, you're gonna have
1 and 2 and so, this is also

00:01:33.027 --> 00:01:39.220
going to be a point on the graph and it's
going to look something like this.

00:01:39.220 --> 00:01:43.350
Gonna look something, something like that.

00:01:43.350 --> 00:01:47.960
That's what, that's the graph of Y is
equal to F of X.

00:01:47.960 --> 00:01:51.030
y is equal to f of x.

00:01:51.030 --> 00:01:53.300
Now, what do they mean when they say the
local,

00:01:53.300 --> 00:01:57.390
the local linearizaton of f at a, is equal
to e.

00:01:57.390 --> 00:01:59.480
Well the convention here, is they use the
letter a

00:01:59.480 --> 00:02:02.220
when they're talking about, when they're
talking about local linearization.

00:02:02.220 --> 00:02:04.050
Say okay, this is what we're approximating
around.

00:02:04.050 --> 00:02:06.640
So this our a value right over here.

00:02:06.640 --> 00:02:09.400
So we're essentially just gonna think
about the slope of the tangent

00:02:09.400 --> 00:02:13.650
line at that point, or this point E comma
F of E.

00:02:13.650 --> 00:02:18.160
So the L of X is going to describe, L

00:02:18.160 --> 00:02:22.560
of X is going to describe this line right
over here.

00:02:22.560 --> 00:02:28.294
So this is, this right over here, is Y is
equal to L of x, but we write it in a

00:02:28.294 --> 00:02:33.661
way that is well suited for local
linearization, for approximating

00:02:33.661 --> 00:02:38.850
the value of the function around, well,
for x values around e.

00:02:38.850 --> 00:02:41.190
So what do we mean by that?

00:02:41.190 --> 00:02:44.700
Well, the general way of describing L of x
say

00:02:44.700 --> 00:02:49.820
that L of x is equal to the function
evaluated.

00:02:49.820 --> 00:02:53.187
At that point that you're centering around
at a, at a, so

00:02:53.187 --> 00:02:56.920
it's going this, it's going to be this
value right over here.

00:03:01.420 --> 00:03:05.184
Plus the slope of the tangent line today,
plus the slope of the tangent line

00:03:05.184 --> 00:03:07.755
today times how far, times the difference
between

00:03:07.755 --> 00:03:10.790
the x value that you're approximating and
a.

00:03:10.790 --> 00:03:13.050
And once again why does this make sense?

00:03:13.050 --> 00:03:15.162
Well let's say that you're trying to

00:03:15.162 --> 00:03:17.970
approximate half of this value right over
here.

00:03:17.970 --> 00:03:22.260
So the function evaluated there would be
right over there, but we want to use

00:03:22.260 --> 00:03:27.810
the local linearization, so we essentially
want to say, what is l of this x value?

00:03:27.810 --> 00:03:30.550
So we want to figure out what is this
right over here.

00:03:30.550 --> 00:03:34.180
Well you start at f of a, which is this.

00:03:34.180 --> 00:03:37.410
And then you multiply, or to that, you add

00:03:37.410 --> 00:03:40.260
the slope times the difference between
these two values.

00:03:40.260 --> 00:03:42.270
X minus a is this distance.

00:03:42.270 --> 00:03:46.330
You multiply that times the slope, that's
gonna give you your change in y.

00:03:46.330 --> 00:03:48.840
So your change in y, plus this y right
over

00:03:48.840 --> 00:03:51.930
here, is going to give you the y value of
that.

00:03:51.930 --> 00:03:52.470
Right over there.

00:03:52.470 --> 00:03:55.120
So this could be useful for approximating,
let's

00:03:55.120 --> 00:03:57.720
say you want to approximate what, what
the, what

00:03:57.720 --> 00:04:02.330
the natural log of e plus 0.2 squared is,

00:04:02.330 --> 00:04:05.100
and so this could, could be, could be
valuable.

00:04:05.100 --> 00:04:08.450
But let's just work through this for this
particular case.

00:04:08.450 --> 00:04:12.960
We know that, so we're saying that a is
equal to e.

00:04:12.960 --> 00:04:18.410
So let's replace, this should be e, this
should be e, and this should be e.

00:04:18.410 --> 00:04:20.399
And now what is f of e?

00:04:20.399 --> 00:04:23.910
Well this is this right over here is the
natural

00:04:23.910 --> 00:04:28.680
log of e squared and now whats f-prime of
e?

00:04:28.680 --> 00:04:31.320
Well let's just calculate that down here.

00:04:31.320 --> 00:04:35.740
So f-prime of x we can just use the chains
rules.

00:04:35.740 --> 00:04:37.550
It's the derivative of the natural log of

00:04:37.550 --> 00:04:40.190
x squared with respect to x squared, so
that's

00:04:40.190 --> 00:04:48.120
going to be 1 over x squared, times the
derivative of x squared with respect to x.

00:04:48.120 --> 00:04:54.290
So times 2x and so this is going to be
equal to 2 over x.

00:04:54.290 --> 00:04:57.722
So f prime of e is going to be two over e,
and

00:04:57.722 --> 00:05:03.608
then times x minus e so that's, that's our
rule, our definition for l

00:05:03.608 --> 00:05:08.021
of x and lets see which one of these lets
see which one of

00:05:08.021 --> 00:05:13.780
these oh and of course we can simplify the
natural log of e squared.

00:05:13.780 --> 00:05:15.405
The power that have to raise e to get to

00:05:15.405 --> 00:05:18.270
e squared, well this is of course equal to
2.

00:05:18.270 --> 00:05:21.230
So 2 plus 2 over e times x minus e.

00:05:21.230 --> 00:05:25.000
So 2 plus 2 over e times x minus e.

00:05:25.000 --> 00:05:27.080
It's that choice right over there.

00:05:27.080 --> 00:05:31.730
But just to make it clear, how this could
be useful, let's say that we're trying to

00:05:31.730 --> 00:05:38.603
approximate, we're trying to approximate f
of e plus 0.1.

00:05:38.603 --> 00:05:41.850
So, we're trying to approximate that.

00:05:41.850 --> 00:05:44.663
Now, like if I just put e plus 0.1 and I
squared and I have

00:05:44.663 --> 00:05:46.386
to figure out the natural log of that

00:05:46.386 --> 00:05:49.450
without a calculator, that seems hard to
me.

00:05:49.450 --> 00:05:52.210
But now we can use a local linearization.

00:05:52.210 --> 00:05:58.020
This is going to be approximately equal to
2 plus,

00:05:58.020 --> 00:06:02.840
2 over e, times, well what's e plus 0.1,
minus e.

00:06:02.840 --> 00:06:05.300
Well it's just gonna be 0, 0.1.

00:06:05.300 --> 00:06:09.261
And I guess you probably still need a
calculator to figure this out,

00:06:09.261 --> 00:06:14.150
or you could just leave it here, anyway
this is a pretty good approximation.

00:06:14.150 --> 00:06:22.250
For what the natural log of e plus 0.1
squared is going to be equal to.

00:06:22.250 --> 00:06:25.960
And let's just, let's just check our
answer.

00:06:25.960 --> 00:06:28.680
So, it is that one right there.

00:06:28.680 --> 00:06:29.980
And we got it right.


Local linearization

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:03.330
Let's say that we're interested in
approximating what

00:00:03.330 --> 00:00:07.939
the square root of 4.36 is equal to.

00:00:07.939 --> 00:00:09.859
So we want to figure apro, we want to
figure out

00:00:09.859 --> 00:00:13.440
an approximation of this, and we don't
have a calculator at hand.

00:00:13.440 --> 00:00:15.030
Well, one way to think about it is we know

00:00:15.030 --> 00:00:18.780
what, we know what the square root of 4
is.

00:00:18.780 --> 00:00:20.520
We know that this is positive 2.

00:00:20.520 --> 00:00:22.740
The principle root of 4 is positive 2.

00:00:22.740 --> 00:00:25.464
So, okay, this is going to be a little bit
more than two.

00:00:25.464 --> 00:00:28.400
Well, let's say that we wanna get a little
bit more

00:00:28.400 --> 00:00:31.677
accurate, and so what I'm going to show
you in this video

00:00:31.677 --> 00:00:35.025
is a method for doing that, for
approximating the value of

00:00:35.025 --> 00:00:38.910
a function near, near a value where we
already know the value.

00:00:38.910 --> 00:00:40.570
So what am I talking about?

00:00:40.570 --> 00:00:44.000
So let's just imagine that we had the
function.

00:00:44.000 --> 00:00:47.766
We have the function f of x is equal to
the square root of

00:00:47.766 --> 00:00:53.340
x, which is, of course, the same thing as
x to the one-half power.

00:00:53.340 --> 00:00:55.570
So we know what f of 2 is.

00:00:55.570 --> 00:01:01.058
We know, we know that f of 2, I'm sorry,
we know that f of 4 is.

00:01:01.058 --> 00:01:04.519
We know that f of 4 is the square root of
4, which is

00:01:04.519 --> 00:01:08.465
going to be equal to 2 or the principle
root of 4, which is

00:01:08.465 --> 00:01:13.378
equal to positive 2 and what we want to
approximate, we wanna figure out

00:01:13.378 --> 00:01:17.513
what f of, we wanna figure out what f of
4.36 is equal to.

00:01:17.513 --> 00:01:19.631
This is just another way of framing the

00:01:19.631 --> 00:01:22.553
exact same question that we started off
this video.

00:01:22.553 --> 00:01:25.740
So let's just imagine our function.

00:01:25.740 --> 00:01:28.070
Let's just imagine it for a second.

00:01:28.070 --> 00:01:32.397
So, let me draw some axes.

00:01:32.397 --> 00:01:35.436
This is my y-axis.

00:01:35.436 --> 00:01:44.284
This is my, this is my x-axis, and let's
graph y is equal to f of x.

00:01:44.284 --> 00:01:46.389
So let's say it looks something like this.

00:01:47.750 --> 00:01:50.360
Y equals f of x looks something like that.

00:01:51.530 --> 00:01:53.200
That's pretty decent.

00:01:53.200 --> 00:01:57.671
All right, so that's, that right there is
y is equal to

00:01:57.671 --> 00:02:00.820
f of x, and we know f of 4 is equal to 2.

00:02:00.820 --> 00:02:05.260
F of 4 is equal to 2, so this is when x is
equal to 4.

00:02:05.260 --> 00:02:09.340
I haven't drawn it really to scale, but
hopefully, this is clear enough.

00:02:09.340 --> 00:02:11.852
So that right over here is going to be 2.

00:02:11.852 --> 00:02:12.710
That's f of 4.

00:02:12.710 --> 00:02:20.280
And what we wanna approximate is f of
4.36, so 4.36 might be right around, right

00:02:20.280 --> 00:02:23.620
around there, and so we want to
approximate,

00:02:23.620 --> 00:02:27.870
we wanna approximate this y value right
over here.

00:02:27.870 --> 00:02:29.718
We want to approximate that.

00:02:29.718 --> 00:02:32.371
Right over here is f of 4.36, and, once

00:02:32.371 --> 00:02:36.120
again, we're assuming we don't have a
calculator at hand.

00:02:37.160 --> 00:02:41.690
So, how can we do that using what we know
about derivatives?

00:02:41.690 --> 00:02:45.769
Well, what if we were to figure out an
equation for the line

00:02:45.769 --> 00:02:51.460
that is tangent to the point, to tangent
to this point right over here.

00:02:51.460 --> 00:02:56.262
So the equation of the tangent line at x
is equal to 4, and then we use

00:02:56.262 --> 00:03:01.916
that linearization, that linearization
defined to approximate values local

00:03:01.916 --> 00:03:06.750
to it, and this technique is called local
linearization.

00:03:06.750 --> 00:03:11.010
So what I'm saying is, let's figure out
what this, the equation of this line is.

00:03:11.010 --> 00:03:13.800
Let's call that l of x.

00:03:13.800 --> 00:03:15.500
And then we can use that to appro, and

00:03:15.500 --> 00:03:17.910
then we can evaluate that at 4.36, and
hopefully

00:03:17.910 --> 00:03:20.620
that will be a little bit easier to do

00:03:20.620 --> 00:03:23.360
than to try and figure out this right over
here.

00:03:23.360 --> 00:03:25.130
So how would we do that?

00:03:25.130 --> 00:03:30.395
Well, one way to think about it, and
obviously, there are

00:03:30.395 --> 00:03:35.763
many ways to express a line, but one way
to think about it is,

00:03:35.763 --> 00:03:41.165
okay, it's going to l of x is going to be
f of 4, which is 2.

00:03:41.165 --> 00:03:46.403
It's going to be f of 4 plus the slope,
the slope at, at x equals

00:03:46.403 --> 00:03:51.253
4, which is, of course, the derivative f
prime of 4, so

00:03:51.253 --> 00:03:56.500
that's going to be the slope of this line
of l of x is f prime of 4.

00:03:56.500 --> 00:03:58.130
Let me make that clear.

00:03:58.130 --> 00:04:00.833
So this right over here is the slope.

00:04:00.833 --> 00:04:03.500
The slope when x is at, at x equals 4, so

00:04:03.500 --> 00:04:06.481
this is a slope of this entire line, and
so any

00:04:06.481 --> 00:04:09.305
other point on this is gonna be f' of 4
plus

00:04:09.305 --> 00:04:12.770
the slope times how far you are away from
x equals 4.

00:04:12.770 --> 00:04:16.920
So it's going to be times x minus 4.

00:04:16.920 --> 00:04:19.370
Let's just, let's just validate that this
makes sense.

00:04:19.370 --> 00:04:23.485
When we put 4.36 here, when we put 4.36
here, actually let me

00:04:23.485 --> 00:04:27.810
zoom in on this graph just to make things
a little bit clearer.

00:04:27.810 --> 00:04:31.603
So, if this is, so I'm gonna do a zoom in.

00:04:31.603 --> 00:04:33.760
I'm gonna do a zoom in.

00:04:33.760 --> 00:04:39.960
I'm gonna try to zoom in into this region
right over here.

00:04:39.960 --> 00:04:42.630
So, this is the point.

00:04:42.630 --> 00:04:49.640
This is the point (4, f of 4), and we are
going to graph l of x.

00:04:49.640 --> 00:04:50.670
So let me do that.

00:04:50.670 --> 00:04:56.560
So this right over here is l of x.

00:04:56.560 --> 00:04:56.850
That's l of x.

00:04:56.850 --> 00:05:01.313
And let's say this, right over here, this
right over

00:05:01.313 --> 00:05:06.359
here, is the point (4.36, f of 4.36), and
the way

00:05:06.359 --> 00:05:11.505
that we're, we're gonna approximate this
value is to figure

00:05:11.505 --> 00:05:16.580
out what, to figure out, what this value
is right over here.

00:05:16.580 --> 00:05:17.706
And what is this one going to be?

00:05:17.706 --> 00:05:27.193
This right over here is going to be, this
is going to be (4.36, L of 4.36).

00:05:27.193 --> 00:05:31.946
This line evaluated when x is equal to
4.36, and what is that going to be?

00:05:31.946 --> 00:05:34.910
What is that going to be equal to?

00:05:34.910 --> 00:05:35.320
Well, let's see.

00:05:35.320 --> 00:05:36.340
Let's just evaluate it.

00:05:36.340 --> 00:05:38.470
L of 4.36 is going to be f of 4.

00:05:38.470 --> 00:05:44.641
So, it's going to be 2 plus the
derivative, so the

00:05:44.641 --> 00:05:50.830
slope of this line plus f prime of 4 times
x minus 4.

00:05:50.830 --> 00:05:56.740
So, 4.36 minus 4 is going to be times
0.36, and that makes sense.

00:05:56.740 --> 00:06:00.621
You're starting at 2 and you're saying,
okay, my

00:06:00.621 --> 00:06:03.590
change in x, my change in x is 4.36.

00:06:03.590 --> 00:06:09.100
So, my change in y is going to be my slope
times that change

00:06:09.100 --> 00:06:13.580
in x to get me that value, to get me that
value right over there.

00:06:13.580 --> 00:06:17.765
So, let's figure out, let's figure out
what this, let's

00:06:17.765 --> 00:06:22.170
figure out what this thing, what this
thing actually is.

00:06:22.170 --> 00:06:27.350
So to do that, we need to figure out f
prime of 4, so let's go back up here.

00:06:27.350 --> 00:06:30.800
I'll try to leave actually, I'll leave
this little visualization here.

00:06:30.800 --> 00:06:34.663
So let's see, f prime, f prime of x is
going to be

00:06:34.663 --> 00:06:40.622
one-half x to the negative one-half just
using the power rule over here.

00:06:40.622 --> 00:06:46.142
So f prime of 4, f prime of 4 is equal to
one-half

00:06:46.142 --> 00:06:51.663
times 4 to the negative one-half, which
is, of

00:06:51.663 --> 00:06:56.932
course, equal to one-half times one-half.

00:06:56.932 --> 00:06:58.119
4 to the one-half would be 2.

00:06:58.119 --> 00:07:00.264
4 to the negative one-half is going to be
one-half.

00:07:00.264 --> 00:07:03.080
So, this is equal to 1 4th.

00:07:03.080 --> 00:07:10.048
So L of, we deserve a little bit of a drum
roll now, L of 4.36 is equal

00:07:10.048 --> 00:07:16.680
to f of 4, is equal to f of 4, which is
eq, let me just rewrite it.

00:07:16.680 --> 00:07:23.860
It's f of 4 plus f prime of 4 plus, gee,
why am I switching to that color.

00:07:23.860 --> 00:07:25.520
Let me do the yellow.

00:07:25.520 --> 00:07:31.571
Plus f prime of 4, times, times,

00:07:31.571 --> 00:07:36.001
times, 4.36.

00:07:36.001 --> 00:07:39.383
4.36.

00:07:39.383 --> 00:07:42.802
Let me make this actually a new color,
just so we see it.

00:07:42.802 --> 00:07:48.467
So, 4.36, so times 4.36 minus 4, minus

00:07:48.467 --> 00:07:54.130
4, and actually let me make all the 4s one
color,

00:07:54.130 --> 00:07:59.816
too, so you see it's the same, so just
like that.

00:07:59.816 --> 00:08:01.464
So what is this going to be?

00:08:01.464 --> 00:08:05.670
Well, this we already established is
positive 2.

00:08:05.670 --> 00:08:10.000
This we already established, and we do
this in a yellow color.

00:08:10.000 --> 00:08:19.390
This, we already established is 1 4th, and
this part right over here is 0.36.

00:08:19.390 --> 00:08:24.507
So, this is going to be equal to 2 plus 1
4th times

00:08:24.507 --> 00:08:29.862
0.36, is 0.9 or is 0.09, so this is going

00:08:29.862 --> 00:08:35.250
to be equal to, this is going to be equal
to 2.09.

00:08:35.250 --> 00:08:37.645
So that is our approximation and should be
at

00:08:37.645 --> 00:08:39.974
least the way, based on how I graphed it,
a

00:08:39.974 --> 00:08:42.630
little higher than the actual value of the
square

00:08:42.630 --> 00:08:45.447
root of 4.36, but we could write that up
here.

00:08:45.447 --> 00:08:49.010
This is going to be approximately, let me
just write it this way.

00:08:50.450 --> 00:08:51.942
The square root, I'll just write it down
here.

00:08:51.942 --> 00:09:00.002
So we could say the square root of 4.36,
which is the same thing as f of 4.36.

00:09:00.002 --> 00:09:04.198
This is approximately equal to 2.09.

00:09:04.198 --> 00:09:07.549
Now let's just say we happen to find a
calculator, and just

00:09:07.549 --> 00:09:11.247
out of curiosity let's see how good of an
approximation that is.

00:09:11.247 --> 00:09:13.150
Let's get a calculator out.

00:09:13.150 --> 00:09:22.220
And so, we wanna do the square root of
4.36, and we get 2.088.

00:09:22.220 --> 00:09:24.660
So, we actually, if we round to the
nearest hundredths,

00:09:24.660 --> 00:09:28.170
we got a pretty good approximation just
like we saw.

00:09:28.170 --> 00:09:31.832
In this, in this indicative graph right
over here, it is,

00:09:31.832 --> 00:09:36.320
our approximation was indeed a little bit
higher than the actual value.


Derivative of natural logarithm

WEBVTT
Kind: captions
Language: en

00:00:00.589 --> 00:00:02.267
- [Voiceover] We know that the derivative

00:00:02.267 --> 00:00:06.583
with respect to x,of e to the x, is equal

00:00:06.583 --> 00:00:08.606
to e to the x, which I think is one

00:00:08.606 --> 00:00:11.403
of the neatest things in mathematics.

00:00:11.403 --> 00:00:14.587
It just builds my
respect for the number e.

00:00:14.587 --> 00:00:18.008
But that's not what we're
here to do, to just praise e.

00:00:18.008 --> 00:00:19.281
What we really want to think about is,

00:00:19.281 --> 00:00:21.282
what's the derivative
of the inverse function?

00:00:21.282 --> 00:00:22.878
What's the derivative, with respect

00:00:22.878 --> 00:00:27.673
to x, of the natural log of x?

00:00:27.673 --> 00:00:29.588
We've done this several
times, where we know

00:00:29.588 --> 00:00:31.196
the derivative of a
function and we now want

00:00:31.196 --> 00:00:33.499
to find the derivative
of the inverse function.

00:00:33.499 --> 00:00:36.267
What we can do, let's say y is

00:00:36.267 --> 00:00:39.538
equal to the natural log of x.

00:00:39.538 --> 00:00:41.255
This is another way of saying that y

00:00:41.255 --> 00:00:45.162
is the power that we
raise e to, to get to x.

00:00:45.162 --> 00:00:47.991
So, this is equivalent to saying that e

00:00:47.991 --> 00:00:52.130
to the y power is equal to x.

00:00:52.130 --> 00:00:54.890
Now we can take the derivative of both

00:00:54.890 --> 00:00:58.195
sides of this equation with respect to x.

00:00:58.195 --> 00:00:59.713
So let's do that.

00:00:59.713 --> 00:01:01.883
The derivative of both sides with respect

00:01:01.883 --> 00:01:04.748
to x, do a little bit of
implicit differentiation.

00:01:04.748 --> 00:01:07.595
Really just an application
of the chain rule.

00:01:07.595 --> 00:01:11.634
So, on the left-hand side right over here,

00:01:11.634 --> 00:01:13.640
this is going to be the
derivative of e to the y

00:01:13.640 --> 00:01:17.656
with respect to y, which is
just going to be e to the y

00:01:17.656 --> 00:01:21.819
times the derivative
of y with respect to x.

00:01:21.819 --> 00:01:23.524
On the right hand side
here, the derivative

00:01:23.524 --> 00:01:26.797
of x with respect to x is one.

00:01:26.797 --> 00:01:29.015
Now, to solve for the derivative,

00:01:29.015 --> 00:01:32.253
we just divide both sides by e to the y.

00:01:32.253 --> 00:01:35.615
We get the derivative of y with respect

00:01:35.615 --> 00:01:40.615
to x is equal to one
over e to the y power.

00:01:47.370 --> 00:01:49.738
Now, what is y equal to?

00:01:49.738 --> 00:01:54.617
Well, we know that y is equal
to the natural log of x.

00:01:54.617 --> 00:01:55.748
So let's write that down.

00:01:55.748 --> 00:02:00.748
This is equal to one over
e to the natural log of x.

00:02:07.339 --> 00:02:11.347
What is e to the natural
log of x going to be?

00:02:12.541 --> 00:02:14.300
The natural log of x is the power

00:02:14.300 --> 00:02:16.584
I need to raise e to, to get to x.

00:02:16.584 --> 00:02:18.617
So if I actually raise e to that power,

00:02:18.617 --> 00:02:21.063
to that exponent, I'm going to get x.

00:02:21.063 --> 00:02:26.063
This is going to be equal to one over x.

00:02:29.701 --> 00:02:32.552
So this thing simplifies to x.

00:02:32.552 --> 00:02:33.464
We are done.

00:02:33.464 --> 00:02:35.060
We just figured out if y is equal

00:02:35.060 --> 00:02:36.774
to the natural log of x, the derivative

00:02:36.774 --> 00:02:39.362
of y with respect to x is one over x.

00:02:39.362 --> 00:02:41.308
Or you could say the
derivative of the natural log

00:02:41.308 --> 00:02:46.308
of x with respect to x
is equal to one over x.

00:02:46.817 --> 00:02:49.417
So this is equal to, this right

00:02:49.417 --> 00:02:54.417
over here, is equal to one over x.

00:02:56.138 --> 00:02:59.058
One over x, which is also a pretty

00:02:59.058 --> 00:03:02.886
neat result in mathematics.

00:03:02.886 --> 00:03:06.760
Not quite as exciting as this
one, but still pretty neat.


Derivative of inverse tangent

WEBVTT
Kind: captions
Language: en

00:00:00.820 --> 00:00:06.446
We already know that the derivative with
respect to x of

00:00:06.446 --> 00:00:12.193
tangent of x is equal to the secant of x
squared, which is

00:00:12.193 --> 00:00:17.967
of course the same thing of one over
cosine of x squared.

00:00:17.967 --> 00:00:21.615
Now what we wanna do in this video, like
we've done in the last

00:00:21.615 --> 00:00:26.103
few videos, is figure out what the
derivative of the inverse function of the

00:00:26.103 --> 00:00:29.958
tangent of x is, or in particular, let's
see if we can figure out

00:00:29.958 --> 00:00:34.390
what the derivative with respect to x of
the inverse tangent of x is.

00:00:34.390 --> 00:00:36.411
And I encourage you to pause this video
and

00:00:36.411 --> 00:00:38.489
use a technique similar to the one, or
very

00:00:38.489 --> 00:00:40.336
close to the one that we've used in the

00:00:40.336 --> 00:00:42.560
last two videos, to figure out what this
is.

00:00:43.820 --> 00:00:48.133
Well let's set y equal to the inverse
tangent

00:00:48.133 --> 00:00:52.920
of x, y is equal to inverse tangent of x.

00:00:52.920 --> 00:00:57.290
That is the same thing as saying that the
tangent

00:00:57.290 --> 00:01:03.110
of y, the tangent of y is equal to x.

00:01:03.110 --> 00:01:05.884
All I've done, now you can kind of think
of it

00:01:05.884 --> 00:01:09.243
as I've just taken the tangent of both
sides right over

00:01:09.243 --> 00:01:12.529
here, and now we can take the derivative
of both sides

00:01:12.529 --> 00:01:17.050
with respect, the derivative of both sides
with respect to x.

00:01:17.050 --> 00:01:20.110
And on the left hand side, we can just
apply the chain rule.

00:01:20.110 --> 00:01:26.130
Derivative of tangent of y with respect to
y, is going to be secant

00:01:26.130 --> 00:01:32.490
squared of y, which is the same thing as
one over cosine of y squared.

00:01:32.490 --> 00:01:33.420
I like to write it this way.

00:01:33.420 --> 00:01:36.590
It's it'll, keeps it a little bit simpler
at

00:01:36.590 --> 00:01:39.613
least in, in my brain, but when we're
applying

00:01:39.613 --> 00:01:43.155
the [INAUDIBLE], it's gonna be the
derivative of tangent

00:01:43.155 --> 00:01:45.883
of y with respect to y times the
derivative of

00:01:45.883 --> 00:01:50.821
y with respect to x, times the derivative
of y with respect to x, and on the right

00:01:50.821 --> 00:01:53.697
hand side, the derivative of x with
respect to

00:01:53.697 --> 00:01:56.633
x, well that's just going to be equal to
one.

00:01:56.633 --> 00:02:00.601
And so we get, if we wanna solve for the
derivative y with

00:02:00.601 --> 00:02:05.767
respect to x, we just multiply both sides
times the cosine of y squared.

00:02:05.767 --> 00:02:10.286
And we get the derivative of y with
respect

00:02:10.286 --> 00:02:14.680
to x is equal to cosine of y squared.

00:02:14.680 --> 00:02:16.781
And like we've seen in previous videos,

00:02:16.781 --> 00:02:19.603
this isn't that satisfying because, you
know, I've

00:02:19.603 --> 00:02:23.100
written the derivative of y with respect
to x as a, as a function of y.

00:02:23.100 --> 00:02:26.855
But what we're really interested in is
writing it as a function of x,

00:02:26.855 --> 00:02:30.768
and to do that, we'd express this somehow
in terms of the tangent of y.

00:02:30.768 --> 00:02:33.884
And the reason why the tangent of y is
interesting is

00:02:33.884 --> 00:02:37.460
because we already know that tangent of y
is equal to x.

00:02:37.460 --> 00:02:41.500
So we can rewrite this using a little bit
of trigonometric identities.

00:02:41.500 --> 00:02:43.587
Then we can, with tangent of y, we can

00:02:43.587 --> 00:02:46.090
substitute all the tangent of y's with an
x.

00:02:46.090 --> 00:02:47.270
So let's see if we can do that.

00:02:47.270 --> 00:02:48.910
And this seems a little bit tricky.

00:02:48.910 --> 00:02:53.072
They introduce a tangent of y, we'd wanna
have a sine, divided by a

00:02:53.072 --> 00:02:58.210
cosine, that's what tangent is, and this
is just a straight up cosine squared y.

00:02:58.210 --> 00:03:01.957
So this is really gonna take a little bit
more experimentation than at

00:03:01.957 --> 00:03:06.410
least some of the other, the, than the
last two examples we've done.

00:03:06.410 --> 00:03:09.220
So one thing we could do is we could say
hey, let's just divide by one.

00:03:09.220 --> 00:03:11.360
Dividing by one never hurt anyone.

00:03:11.360 --> 00:03:13.960
So we could say this is the same thing,

00:03:13.960 --> 00:03:16.900
this is the same thing as cosine squared
y.

00:03:16.900 --> 00:03:18.876
And I'm really doing this to, to see if I

00:03:18.876 --> 00:03:20.969
can in, start to express it as some type
of a

00:03:20.969 --> 00:03:24.519
rational, a rational expression which
might involve, at some point,

00:03:24.519 --> 00:03:27.650
a sine divided by cosine and it could have
a tangent.

00:03:27.650 --> 00:03:28.900
So, let's divide by one.

00:03:28.900 --> 00:03:34.436
But we know from the Pythagorean, the
Pythagorean identity that one

00:03:34.436 --> 00:03:39.781
is equal to sine squared of y plus cosine
squared of y, and so let's

00:03:39.781 --> 00:03:45.233
try that or we could write cosine squared
of y plus sine squared of y.

00:03:45.233 --> 00:03:47.967
Once again, why was I able to divide by
this expression?

00:03:47.967 --> 00:03:50.724
Well, this expression by the Pythagorean
identity, which

00:03:50.724 --> 00:03:52.939
really comes out of the unit circle
definition of

00:03:52.939 --> 00:03:54.830
trig functions, this is equal to one, so

00:03:54.830 --> 00:03:58.020
I have not changed the value of this
expression.

00:03:58.020 --> 00:04:00.461
Now what makes this interesting is if I

00:04:00.461 --> 00:04:03.511
wanted to introduce a sine divided by a
cosine,

00:04:03.511 --> 00:04:06.106
I could just divide the numerator and the

00:04:06.106 --> 00:04:09.537
denominator by cosine square, so lets do
that, lets

00:04:09.537 --> 00:04:12.284
multiply one over cosine or else divide
the

00:04:12.284 --> 00:04:15.183
numerator by cosine square of y and divide
the

00:04:15.183 --> 00:04:18.690
denominator, by cosine square of y, or
multiply each

00:04:18.690 --> 00:04:21.560
of them by one over cosine square of y.

00:04:21.560 --> 00:04:23.050
What's that going to give us?

00:04:23.050 --> 00:04:25.372
Well the numerator, these characters are
going to

00:04:25.372 --> 00:04:27.880
cancel you're just going to have a one.

00:04:27.880 --> 00:04:32.130
And the denominator this time this this,
that's going to be equal to one.

00:04:32.130 --> 00:04:37.716
And then you're gonna have sine squared,
sine squared y over,

00:04:37.716 --> 00:04:43.095
over cosine squared y, and this is, this
is the goal that I was

00:04:43.095 --> 00:04:48.760
trying to achieve, I have a sine divided
by cosine squared.

00:04:48.760 --> 00:04:51.950
So this right over here, this is the same
thing, actually let me

00:04:51.950 --> 00:04:55.220
just write it this way, let me, let me
write it this way.

00:04:55.220 --> 00:05:04.470
This is the same thing as sine of y over
cosine of y, cosine of y.

00:05:04.470 --> 00:05:09.427
Whole thing squared, which is of course
the same thing

00:05:09.427 --> 00:05:14.970
as one over one plus tangent of y, tangent
of y squared.

00:05:14.970 --> 00:05:17.310
This is equal to this.

00:05:17.310 --> 00:05:18.460
Now why is that useful?

00:05:18.460 --> 00:05:21.670
Well we know that x is equal to tangent of
y.

00:05:21.670 --> 00:05:27.472
So this is going to be equal to, this is
going to be equal

00:05:27.472 --> 00:05:33.150
to one, one over, one plus tangent of y is
equal to x,

00:05:33.150 --> 00:05:38.600
x squared, x squared, which is pretty
exciting.

00:05:38.600 --> 00:05:43.010
We just figured out the derivative of y
with respect to x.

00:05:43.010 --> 00:05:45.366
So the derivative of this thing with
respect

00:05:45.366 --> 00:05:47.680
to x is one over one plus x squared.

00:05:47.680 --> 00:05:49.640
So we could write that right up here.

00:05:49.640 --> 00:05:55.654
So this is going to be equal to one over

00:05:55.654 --> 00:06:01.671
one plus x squared, and we are done.