2017 AP Calculus AB/BC 4c

WEBVTT
Kind: captions
Language: en

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- [Instructor] Let's now tackle part C

00:00:03.266 --> 00:00:05.981
which tells us, for T is less than 10,

00:00:05.981 --> 00:00:08.317
an alternate model for
the internal temperature

00:00:08.317 --> 00:00:10.988
of the potato at time T minutes

00:00:10.988 --> 00:00:14.679
is the function G that satisfies
the differential equation.

00:00:14.679 --> 00:00:17.220
The derivative of G with respect to T

00:00:17.220 --> 00:00:19.291
is equal to the negative of

00:00:19.291 --> 00:00:22.086
G minus 27 to the two third power.

00:00:22.086 --> 00:00:24.784
Where G of T is measured
in degrees Celsius

00:00:24.784 --> 00:00:27.489
and G of zero is equal to 91.

00:00:27.489 --> 00:00:30.396
Find an expression for G of T.

00:00:30.396 --> 00:00:33.976
Based on this model, what
is the internal temperature

00:00:33.976 --> 00:00:37.329
of the potato at time T is equal to three?

00:00:37.329 --> 00:00:39.645
So they gave us a differential equation.

00:00:39.645 --> 00:00:42.487
They want us to find an
expression for G of T,

00:00:42.487 --> 00:00:43.869
so essentially find a solution

00:00:43.869 --> 00:00:45.304
to this differential equation.

00:00:45.304 --> 00:00:47.508
And then they want us to
use that solution to find

00:00:47.508 --> 00:00:50.233
the internal temperature
time T equals three.

00:00:50.233 --> 00:00:52.089
So, the first thing to appreciate,

00:00:52.089 --> 00:00:54.608
this is in an AP calculus class.

00:00:54.608 --> 00:00:56.906
And so they're asking us
to solve a differential

00:00:56.906 --> 00:00:58.981
equation, or find a solution
to a differential equation

00:00:58.981 --> 00:01:02.435
it is unlikely to be a really
strange differential equation.

00:01:02.435 --> 00:01:05.464
It's likely to be a separable
differential equation.

00:01:05.464 --> 00:01:06.936
And then once we find that solution

00:01:06.936 --> 00:01:10.308
we just have to evaluate
it at time T equals three.

00:01:10.308 --> 00:01:13.130
So let's rewrite the
differential equation.

00:01:13.130 --> 00:01:15.629
And then, let's try to evaluate.

00:01:15.629 --> 00:01:17.163
Let's see if we can find the solution

00:01:17.163 --> 00:01:18.149
and then we'll evaluate.

00:01:18.149 --> 00:01:19.610
The derivative of G

00:01:19.610 --> 00:01:20.938
with respect to T

00:01:20.938 --> 00:01:24.105
is equal to the negative of G minus 27

00:01:25.170 --> 00:01:27.188
to the two thirds power.

00:01:27.188 --> 00:01:28.784
So,

00:01:28.784 --> 00:01:30.910
if this is going to be a
separable differential equation

00:01:30.910 --> 00:01:32.827
I want to separate the DG

00:01:32.827 --> 00:01:33.744
and the DT.

00:01:34.643 --> 00:01:37.964
So I'm gonna treat my
differential like numbers

00:01:37.964 --> 00:01:39.407
so, or variables.

00:01:39.407 --> 00:01:41.399
So I'm gonna multiply both sides

00:01:41.399 --> 00:01:43.958
times the T differential.

00:01:43.958 --> 00:01:46.632
And so then I'm going to have

00:01:46.632 --> 00:01:48.641
the capital G differential

00:01:48.641 --> 00:01:50.585
DG is equal to

00:01:50.585 --> 00:01:52.039
negative times

00:01:52.039 --> 00:01:52.872
G minus 27

00:01:54.035 --> 00:01:56.690
to the two thirds power DT.

00:01:56.690 --> 00:01:59.341
And the whole notion
here, I'm trying to get

00:01:59.341 --> 00:02:01.108
all of the G things

00:02:01.108 --> 00:02:02.964
on the side with the DG.

00:02:02.964 --> 00:02:04.440
And then all of the things that involve T

00:02:04.440 --> 00:02:06.080
in the side with the DT.

00:02:06.080 --> 00:02:07.556
So we don't see any Ts here

00:02:07.556 --> 00:02:09.161
so really, we just have to get the Gs

00:02:09.161 --> 00:02:11.857
over on this side while
leaving the DT there.

00:02:11.857 --> 00:02:14.201
And then we can integrate both sides.

00:02:14.201 --> 00:02:15.354
So let's see.

00:02:15.354 --> 00:02:16.437
If we were to

00:02:17.753 --> 00:02:19.146
divide

00:02:19.146 --> 00:02:20.432
both sides

00:02:20.432 --> 00:02:21.921
by

00:02:21.921 --> 00:02:22.754
G

00:02:23.707 --> 00:02:25.910
minus 27 to the two thirds.

00:02:25.910 --> 00:02:26.743
G minus 27

00:02:28.535 --> 00:02:30.862
to the two thirds.

00:02:30.862 --> 00:02:31.695
G minus 27

00:02:33.132 --> 00:02:35.151
to the two thirds,

00:02:35.151 --> 00:02:36.111
what do we have?

00:02:36.111 --> 00:02:38.643
Well, we could rewrite the left side as

00:02:38.643 --> 00:02:39.476
G minus 27

00:02:40.945 --> 00:02:42.875
to the negative two thirds

00:02:42.875 --> 00:02:44.302
negative two thirds

00:02:44.302 --> 00:02:45.469
DG is equal to

00:02:47.297 --> 00:02:50.297
I'm left with just this negative DT.

00:02:51.217 --> 00:02:52.134
Is equal to

00:02:53.458 --> 00:02:55.021
negative DT.

00:02:55.021 --> 00:02:55.872
Now let's see.

00:02:55.872 --> 00:02:57.754
Now, we can, or actually just a simplify

00:02:57.754 --> 00:02:59.319
or just to make it a
little bit more obvious

00:02:59.319 --> 00:03:01.461
I could write this as negative one DT

00:03:01.461 --> 00:03:04.790
And now we just have to
integrate both sides.

00:03:04.790 --> 00:03:07.427
So, we could rewrite this as

00:03:07.427 --> 00:03:09.334
let me write it on both sides.

00:03:09.334 --> 00:03:11.447
I'm going to integrate

00:03:11.447 --> 00:03:14.472
I'm going to integrate both sides here.

00:03:14.472 --> 00:03:16.688
And so what does this give me?

00:03:16.688 --> 00:03:18.230
Well, on the left side here

00:03:18.230 --> 00:03:21.259
you could try to do some
U substitutions saying,

00:03:21.259 --> 00:03:23.342
U is equal to G minus 27.

00:03:24.495 --> 00:03:26.425
And then DU would be DG.

00:03:26.425 --> 00:03:28.618
Or, you might recognize that look,

00:03:28.618 --> 00:03:31.131
the derivative of G minus 27

00:03:31.131 --> 00:03:32.048
is just DG.

00:03:35.153 --> 00:03:37.415
The derivative of G minus 27 is

00:03:37.415 --> 00:03:40.118
just going to be equal to one.

00:03:40.118 --> 00:03:42.148
So you could even say, look, I have my

00:03:42.148 --> 00:03:43.294
I have the derivative there.

00:03:43.294 --> 00:03:46.947
So I could really integrate
with respect to G minus 27.

00:03:46.947 --> 00:03:49.414
And so really, I would just
use the reverse power rule.

00:03:49.414 --> 00:03:51.664
I would take my G minus 27.

00:03:52.987 --> 00:03:55.558
I would increment this exponent by one.

00:03:55.558 --> 00:03:57.369
So let's see, negative two thirds plus one

00:03:57.369 --> 00:03:59.249
is positive one third.

00:03:59.249 --> 00:04:01.243
So positive one third power.

00:04:01.243 --> 00:04:03.947
And then I would divide
by this new exponent.

00:04:03.947 --> 00:04:05.836
So dividing by one third
is the same thing as

00:04:05.836 --> 00:04:07.304
multiplying by three.

00:04:07.304 --> 00:04:08.806
So that's the left hand side.

00:04:08.806 --> 00:04:10.541
This is going to be equal to

00:04:10.541 --> 00:04:11.838
the right hand side.

00:04:11.838 --> 00:04:13.479
This is just going to be negative T.

00:04:13.479 --> 00:04:14.876
And for good measure

00:04:14.876 --> 00:04:17.773
I'm going to have A plus C.

00:04:17.773 --> 00:04:20.190
So, how could we solve for C?

00:04:21.683 --> 00:04:24.600
Well they give us some
information, they say G of zero

00:04:24.600 --> 00:04:25.850
is equal to 91.

00:04:27.225 --> 00:04:28.449
So lets write that down.

00:04:28.449 --> 00:04:29.949
So, when T is zero

00:04:31.482 --> 00:04:32.632
G is 91.

00:04:32.632 --> 00:04:34.536
So we can write

00:04:34.536 --> 00:04:36.174
let me write it over here.

00:04:36.174 --> 00:04:37.567
So three times

00:04:37.567 --> 00:04:40.079
so when T is zero G is 91.

00:04:40.079 --> 00:04:41.662
Ninety one minus 27

00:04:42.661 --> 00:04:46.951
to the one third power
is equal to negative T.

00:04:46.951 --> 00:04:49.644
Well now we're saying T is zero.

00:04:49.644 --> 00:04:51.785
So you can write negative zero there.

00:04:51.785 --> 00:04:53.288
Or we could just not write it.

00:04:53.288 --> 00:04:54.712
And then plus C.

00:04:54.712 --> 00:04:57.454
So that's what C is going to be equal to.

00:04:57.454 --> 00:04:59.871
Let's see, 91 minus 27 is 64.

00:05:00.888 --> 00:05:03.534
Sixty four to the one third power

00:05:03.534 --> 00:05:05.671
is positive four.

00:05:05.671 --> 00:05:07.718
And so we have C is equal to 12.

00:05:07.718 --> 00:05:09.135
C is equal to 12.

00:05:10.723 --> 00:05:13.578
And let's see, we want to
write an expression for G of T.

00:05:13.578 --> 00:05:17.373
So now let's just manipulate
this and solve for G.

00:05:17.373 --> 00:05:18.206
So, if we

00:05:19.290 --> 00:05:21.311
let me just take this

00:05:21.311 --> 00:05:23.613
and go right over here.

00:05:23.613 --> 00:05:26.543
So if we divide both sides by three

00:05:26.543 --> 00:05:28.239
we are going to get

00:05:28.239 --> 00:05:29.072
G minus 27

00:05:30.940 --> 00:05:33.773
to the one third power is equal to

00:05:35.001 --> 00:05:36.751
negative T over three

00:05:38.736 --> 00:05:39.569
plus four.

00:05:40.410 --> 00:05:42.333
I just divided both sides by three.

00:05:42.333 --> 00:05:43.924
And I can take the cube of both sides

00:05:43.924 --> 00:05:46.532
and I would get G minus 27

00:05:46.532 --> 00:05:48.238
is equal to

00:05:48.238 --> 00:05:50.284
negative T over three

00:05:50.284 --> 00:05:51.546
plus four

00:05:51.546 --> 00:05:52.901
to the third power.

00:05:52.901 --> 00:05:55.438
And I would just have
to add 27 to both sides.

00:05:55.438 --> 00:05:58.134
And I'll make it clear,
G is a function of T

00:05:58.134 --> 00:06:00.123
is equal to negative T

00:06:00.123 --> 00:06:02.155
over three plus four.

00:06:02.155 --> 00:06:03.997
All of that to the third power.

00:06:03.997 --> 00:06:04.830
Plus 27.

00:06:06.425 --> 00:06:08.217
So I did the first part.

00:06:08.217 --> 00:06:11.427
This is, what do we got cross

00:06:11.427 --> 00:06:13.255
this right over here

00:06:13.255 --> 00:06:14.922
is an expression for G of T.

00:06:14.922 --> 00:06:17.831
It was indeed a separable
differential equation.

00:06:17.831 --> 00:06:19.181
It took a little bit of

00:06:19.181 --> 00:06:21.105
algebraic manipulation to get us there.

00:06:21.105 --> 00:06:22.358
But we were able to do it

00:06:22.358 --> 00:06:23.709
and we were not only able to solve

00:06:23.709 --> 00:06:25.002
for the general solution.

00:06:25.002 --> 00:06:26.583
We were able to find
the particular solution

00:06:26.583 --> 00:06:29.853
using this initial condition right here

00:06:29.853 --> 00:06:32.256
the G of zero is equal to 91.

00:06:32.256 --> 00:06:33.412
Now, we'll do the easy part.

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Based on this model, what
is the internal temperature

00:06:35.557 --> 00:06:37.716
of the potato at time T equals three?

00:06:37.716 --> 00:06:39.507
So G of three

00:06:39.507 --> 00:06:41.294
is equal to negative three

00:06:41.294 --> 00:06:43.989
over three plus, I'll just write it out.

00:06:43.989 --> 00:06:46.382
Negative three over three plus four

00:06:46.382 --> 00:06:49.281
to the third power plus 27.

00:06:49.281 --> 00:06:51.268
Well this is negative one.

00:06:51.268 --> 00:06:53.053
So negative one plus four is three.

00:06:53.053 --> 00:06:55.785
So this is three to the
third power which is 27.

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Plus 27 is equal to 54

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degrees Celsius.

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And, we are done.


2017 AP Calculus AB/BC 4b

WEBVTT
Kind: captions
Language: en

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- [Instructor] We're now
going to tackle part (b)

00:00:02.182 --> 00:00:03.731
of the potato problem,

00:00:03.731 --> 00:00:05.678
and it says use the second derivative

00:00:05.678 --> 00:00:08.978
of H with respect to time to
determine whether your answer

00:00:08.978 --> 00:00:12.830
in part (a) is an underestimate
or an overestimate

00:00:12.830 --> 00:00:15.107
of the internal temperature of the potato

00:00:15.107 --> 00:00:17.598
at time t is equal to three.

00:00:17.598 --> 00:00:21.420
So in part (a), we found the
equation of the tangent line

00:00:21.420 --> 00:00:23.767
at time equals zero and we used that

00:00:23.767 --> 00:00:26.969
to estimate what our
internal temperature would be

00:00:26.969 --> 00:00:28.916
at time is equal to three.

00:00:28.916 --> 00:00:31.584
So how is the second derivative
going to help us think

00:00:31.584 --> 00:00:35.170
whether that was an overestimate
or an underestimate?

00:00:35.170 --> 00:00:36.465
Well, the second derivative

00:00:36.465 --> 00:00:39.080
can help us know about concavity.

00:00:39.080 --> 00:00:41.530
It'll let us know, well, is our slope

00:00:41.530 --> 00:00:45.207
increasing over this interval,
or is our slope decreasing?

00:00:45.207 --> 00:00:46.323
And then we can use that

00:00:46.323 --> 00:00:49.559
to estimate whether we
overestimated or not.

00:00:49.559 --> 00:00:52.248
So first, let's just find
the second derivative.

00:00:52.248 --> 00:00:54.110
So we have the first
derivative written up here.

00:00:54.110 --> 00:00:57.179
Let me just rewrite it, and
I'll distribute the negative 1/4

00:00:57.179 --> 00:00:59.453
'cause it'll be a little bit
more straightforward then.

00:00:59.453 --> 00:01:03.934
So if I write the derivative
of H with respect to time

00:01:03.934 --> 00:01:06.851
is equal to negative 1/4

00:01:07.847 --> 00:01:09.550
times our internal temperature,

00:01:09.550 --> 00:01:12.590
which it itself is a function of time,

00:01:12.590 --> 00:01:14.854
and then negative 1/4 times negative 27,

00:01:14.854 --> 00:01:16.604
that would be plus 27

00:01:17.881 --> 00:01:19.128
over four.

00:01:19.128 --> 00:01:22.080
Let me scroll down a little bit.

00:01:22.080 --> 00:01:24.595
So now, let me leave that graph up there

00:01:24.595 --> 00:01:26.267
'cause I think that might be useful.

00:01:26.267 --> 00:01:27.989
What is the second derivative going to be

00:01:27.989 --> 00:01:29.502
with respect to time?

00:01:29.502 --> 00:01:32.030
So I'll write it right over here.

00:01:32.030 --> 00:01:34.447
D, the second derivative of H

00:01:35.546 --> 00:01:39.238
with respect to time is
going to be equal to,

00:01:39.238 --> 00:01:44.230
well, the derivative of this
first term with respect to time

00:01:44.230 --> 00:01:46.620
is going to be the derivative
of this with respect to H

00:01:46.620 --> 00:01:49.115
times the derivative of
H with respect to time.

00:01:49.115 --> 00:01:51.454
So this is equal to negative 1/4.

00:01:51.454 --> 00:01:53.950
That's the derivative of
this term with respect to H,

00:01:53.950 --> 00:01:56.704
and then we want to multiply
that times the derivative of H

00:01:56.704 --> 00:01:57.897
with respect to time.

00:01:57.897 --> 00:02:00.169
This comes just straight
out of the chain rule.

00:02:00.169 --> 00:02:01.504
And then the derivative of a constant,

00:02:01.504 --> 00:02:02.898
how does that change with respect to time?

00:02:02.898 --> 00:02:06.094
It's not gonna change, that
is just going to be zero.

00:02:06.094 --> 00:02:07.512
So just like that, we were able

00:02:07.512 --> 00:02:11.156
to find the second derivative
H with respect to time.

00:02:11.156 --> 00:02:13.382
And now what does this tell us?

00:02:13.382 --> 00:02:17.904
Well, we talked about,
in the previous video,

00:02:17.904 --> 00:02:20.460
that over the interval that we care about,

00:02:20.460 --> 00:02:22.671
for, actually, I can
show you from this graph.

00:02:22.671 --> 00:02:24.192
Over the interval that we care about,

00:02:24.192 --> 00:02:26.192
for t greater than zero,

00:02:27.254 --> 00:02:29.067
it says that our internal temperature

00:02:29.067 --> 00:02:32.598
is always going to be greater than 27.

00:02:32.598 --> 00:02:35.274
And so when you look
at this expression here

00:02:35.274 --> 00:02:39.002
or when you look at this
expression here for dH/dt,

00:02:39.002 --> 00:02:40.689
we talked in the previous video

00:02:40.689 --> 00:02:43.594
how this is always going
to be negative here.

00:02:43.594 --> 00:02:46.592
Because H is always going
to be greater than 27,

00:02:46.592 --> 00:02:48.488
so that part's going to be positive.

00:02:48.488 --> 00:02:50.692
But then we're gonna multiply
it by a negative 1/4,

00:02:50.692 --> 00:02:52.787
so our slope dH/dt,

00:02:52.787 --> 00:02:54.846
our derivative of our
temperature with respect to time,

00:02:54.846 --> 00:02:57.429
is always going to be negative.

00:02:58.397 --> 00:03:01.143
So we could write that this

00:03:01.143 --> 00:03:01.976
or this,

00:03:03.087 --> 00:03:04.934
this is going to be negative,

00:03:04.934 --> 00:03:07.260
let me write it this way.

00:03:07.260 --> 00:03:08.093
Since

00:03:10.548 --> 00:03:12.215
H is greater than 27

00:03:13.655 --> 00:03:15.905
for t is greater than zero,

00:03:16.962 --> 00:03:19.379
we know that dH/dt

00:03:21.368 --> 00:03:23.039
is negative,

00:03:23.039 --> 00:03:24.769
is negative.

00:03:24.769 --> 00:03:27.410
So we could say that this right over here,

00:03:27.410 --> 00:03:29.243
since dH/dt

00:03:31.629 --> 00:03:33.046
is less than zero

00:03:34.000 --> 00:03:37.083
for t is greater than zero,

00:03:37.083 --> 00:03:40.498
the second derivative
of H with respect to t

00:03:40.498 --> 00:03:43.165
is going to be greater than zero

00:03:44.539 --> 00:03:45.886
for

00:03:45.886 --> 00:03:46.886
t is greater

00:03:48.252 --> 00:03:49.616
than zero.

00:03:49.616 --> 00:03:50.919
And so what does that mean?

00:03:50.919 --> 00:03:52.824
It means that we are,

00:03:52.824 --> 00:03:55.273
if your second derivative is positive,

00:03:55.273 --> 00:03:57.946
that means you're concave upward,

00:03:57.946 --> 00:03:59.196
concave upward,

00:04:00.766 --> 00:04:03.433
which means slope is increasing,

00:04:05.030 --> 00:04:05.863
slope

00:04:06.703 --> 00:04:07.620
increasing.

00:04:09.120 --> 00:04:12.599
Or you could say slope
becoming less negative,

00:04:12.599 --> 00:04:13.766
slope becoming

00:04:16.436 --> 00:04:18.026
less

00:04:18.026 --> 00:04:19.532
negative.

00:04:19.532 --> 00:04:20.681
Now what does that mean?

00:04:20.681 --> 00:04:21.759
And you could see it intuitively.

00:04:21.759 --> 00:04:25.021
If the slope is becoming
less and less negative,

00:04:25.021 --> 00:04:27.200
then that means when we approximated

00:04:27.200 --> 00:04:29.432
what our temperature is at t equals three,

00:04:29.432 --> 00:04:32.139
we used a really negative slope.

00:04:32.139 --> 00:04:34.537
When, in reality, our
slope is getting less

00:04:34.537 --> 00:04:36.407
and less and less negative.

00:04:36.407 --> 00:04:39.403
So what we would've done is
we would've over-decreased

00:04:39.403 --> 00:04:43.022
our temperature from t equals
zero to t equals three.

00:04:43.022 --> 00:04:46.669
So that would mean that we
would have underestimated it.

00:04:46.669 --> 00:04:48.023
So let me write that down.

00:04:48.023 --> 00:04:50.176
And I'm running out of
a little bit of space,

00:04:50.176 --> 00:04:52.427
but let me write it right over here.

00:04:52.427 --> 00:04:54.191
So that implies,

00:04:54.191 --> 00:04:57.108
so this implies that underestimate,

00:04:58.391 --> 00:04:59.474
underestimate

00:05:02.081 --> 00:05:04.254
in part (a).

00:05:04.254 --> 00:05:06.319
And if I were taking this on the AP exam,

00:05:06.319 --> 00:05:08.835
I would flesh out my
language a little bit here

00:05:08.835 --> 00:05:11.190
to make it clear, but
hopefully that makes sense.


2017 AP Calculus AB/BC 4a

WEBVTT
Kind: captions
Language: en

00:00:00.558 --> 00:00:02.453
- [Instructor] We are now
going to cover the famous

00:00:02.453 --> 00:00:04.666
or perhaps infamous potato problem

00:00:04.666 --> 00:00:07.249
from the 2017 AP Calculus exam.

00:00:08.218 --> 00:00:09.921
At time t equals zero,

00:00:09.921 --> 00:00:13.348
a boiled potato is taken
from a pot on a stove

00:00:13.348 --> 00:00:15.429
and left to cool in a kitchen.

00:00:15.429 --> 00:00:17.755
The internal temperature of the potato

00:00:17.755 --> 00:00:21.422
is 91 degrees Celsius
at time t equals zero,

00:00:22.332 --> 00:00:25.980
and the internal temperature
of the potato is greater than

00:00:25.980 --> 00:00:29.776
27 degrees for all times
t greater than zero.

00:00:29.776 --> 00:00:31.863
I would guess that the
ambient room temperature

00:00:31.863 --> 00:00:33.673
is 27 degrees Celsius.

00:00:33.673 --> 00:00:36.129
And so that's why the
temperature would approach this,

00:00:36.129 --> 00:00:38.851
but it would always stay a
little bit greater than that,

00:00:38.851 --> 00:00:41.149
as t gets larger and larger.

00:00:41.149 --> 00:00:44.305
The internal temperature of the potato

00:00:44.305 --> 00:00:47.706
at time t minutes can be
modeled by the function H

00:00:47.706 --> 00:00:51.225
that satisfies the
differential equation dH/dt,

00:00:51.225 --> 00:00:53.027
the derivative of our internal temperature

00:00:53.027 --> 00:00:57.339
with respect to time, is equal
to negative 1/4 times our,

00:00:57.339 --> 00:00:59.235
the difference between
our internal temperature

00:00:59.235 --> 00:01:01.076
and the ambient room temperature,

00:01:01.076 --> 00:01:04.193
where H of t is measured
in degrees Celsius

00:01:04.193 --> 00:01:07.132
and H of zero is equal to 91.

00:01:07.132 --> 00:01:11.226
So before I even read part
(a), let's just make sense

00:01:11.226 --> 00:01:13.352
of what this differential
equation is telling us,

00:01:13.352 --> 00:01:16.548
and let's see if it's
consistent with our intuition.

00:01:16.548 --> 00:01:19.565
So let me draw some axes here.

00:01:19.565 --> 00:01:21.315
So this is my y-axis.

00:01:22.961 --> 00:01:25.021
So that is my y-axis.

00:01:25.021 --> 00:01:28.930
And this right over here is my t-axis.

00:01:28.930 --> 00:01:33.780
Now the ambient room temperature
is 27 degrees Celsius,

00:01:33.780 --> 00:01:35.504
so I'll just draw there.

00:01:35.504 --> 00:01:37.709
That's what the room temperature is doing.

00:01:37.709 --> 00:01:40.383
And so we know at t equals zero,

00:01:40.383 --> 00:01:42.311
our potato is at 91 degrees.

00:01:42.311 --> 00:01:46.332
So let's see, that's 27, 91
might be right over there.

00:01:46.332 --> 00:01:49.119
91, this is all in degrees Celsius.

00:01:49.119 --> 00:01:52.019
And what you would expect
intuitively is that it would start

00:01:52.019 --> 00:01:54.477
to cool and when its temperature,

00:01:54.477 --> 00:01:56.260
when there's a big difference between

00:01:56.260 --> 00:01:57.612
the potato and the room,

00:01:57.612 --> 00:01:59.995
maybe its rate of change is steeper

00:01:59.995 --> 00:02:01.574
than when there's a little difference.

00:02:01.574 --> 00:02:06.223
So you would expect the graph
to look something like this.

00:02:06.223 --> 00:02:08.069
So you would expect it

00:02:08.069 --> 00:02:10.651
to look something like
this and then asymptote

00:02:10.651 --> 00:02:14.318
towards a temperature
of 27 degrees Celsius.

00:02:15.584 --> 00:02:17.828
So this is what you would expect to see,

00:02:17.828 --> 00:02:21.619
and this differential equation
is consistent with that,

00:02:21.619 --> 00:02:23.196
where the rate of change,

00:02:23.196 --> 00:02:25.052
notice this is for all
t greater than zero,

00:02:25.052 --> 00:02:26.773
this is going to be a negative value

00:02:26.773 --> 00:02:29.847
because our potato is greater than 27

00:02:29.847 --> 00:02:30.858
for t greater than zero.

00:02:30.858 --> 00:02:33.623
So this part here is going to be positive,

00:02:33.623 --> 00:02:36.132
but then you multiply
positive times negative 1/4.

00:02:36.132 --> 00:02:38.371
You're constantly going to
have a negative rate of change,

00:02:38.371 --> 00:02:39.361
which makes sense.

00:02:39.361 --> 00:02:41.527
The potato is cooling down.

00:02:41.527 --> 00:02:44.525
And it also makes sense
that your rate of change

00:02:44.525 --> 00:02:47.607
is proportional to the difference
between the temperature

00:02:47.607 --> 00:02:50.177
of the potato and the
ambient room temperature.

00:02:50.177 --> 00:02:51.504
When there's a big difference,

00:02:51.504 --> 00:02:53.835
you expect a steeper rate of change.

00:02:53.835 --> 00:02:55.558
But then when there's
less of a difference,

00:02:55.558 --> 00:02:57.124
the rate of change, you could imagine,

00:02:57.124 --> 00:02:59.256
becomes less and less and less negative,

00:02:59.256 --> 00:03:02.854
as we asymptote towards
the ambient temperature.

00:03:02.854 --> 00:03:06.923
So with this out of the way,
now let's tackle part (a).

00:03:06.923 --> 00:03:11.226
Write an equation for the
line tangent to the graph

00:03:11.226 --> 00:03:13.059
of H at t equals zero.

00:03:13.894 --> 00:03:17.750
Use this equation to approximate
the internal temperature

00:03:17.750 --> 00:03:20.755
of the potato at time t equals three.

00:03:20.755 --> 00:03:22.282
So what are we going to do?

00:03:22.282 --> 00:03:23.764
Well, we're gonna think about

00:03:23.764 --> 00:03:27.462
what's going on at time t equal zero,

00:03:27.462 --> 00:03:28.775
right over here.

00:03:28.775 --> 00:03:32.433
We want the equation of the tangent line,

00:03:32.433 --> 00:03:35.880
which might look something like this,

00:03:35.880 --> 00:03:37.297
at t equals zero.

00:03:38.271 --> 00:03:42.418
So this thing would be
of the form y is equal to

00:03:42.418 --> 00:03:45.297
the slope of the equation
of the tangent line.

00:03:45.297 --> 00:03:48.316
Well, it would be the
derivative of our function

00:03:48.316 --> 00:03:50.857
at that point, so dH/dt,

00:03:50.857 --> 00:03:52.371
times t,

00:03:52.371 --> 00:03:54.585
plus our y-intercept.

00:03:54.585 --> 00:03:57.232
Where does it intersect the y-axis here?

00:03:57.232 --> 00:03:59.755
Well, when t is equal to zero,

00:03:59.755 --> 00:04:02.375
the value of this
equation is going to be 91

00:04:02.375 --> 00:04:04.660
because it intersects our
graph right at that point,

00:04:04.660 --> 00:04:07.058
that point zero comma 91,

00:04:07.058 --> 00:04:07.891
plus 91.

00:04:08.999 --> 00:04:12.261
So what is our derivative
of H with respect to t

00:04:12.261 --> 00:04:16.481
at time t equals zero, right
at this point right over here?

00:04:16.481 --> 00:04:18.349
Well, we just have to look at this.

00:04:18.349 --> 00:04:23.256
You could also write this as
H prime of t right over here.

00:04:23.256 --> 00:04:26.256
So if we want to think about H prime

00:04:27.372 --> 00:04:28.806
of zero,

00:04:28.806 --> 00:04:32.326
that's going to be equal to negative 1/4

00:04:32.326 --> 00:04:33.576
times H of zero

00:04:34.782 --> 00:04:36.128
minus 27.

00:04:36.128 --> 00:04:38.427
What is our initial temperature minus 27?

00:04:38.427 --> 00:04:40.408
This is, of course, 91 degrees.

00:04:40.408 --> 00:04:41.721
They tell us that multiple times.

00:04:41.721 --> 00:04:43.657
We've even drawn it a few times.

00:04:43.657 --> 00:04:44.824
91 minus 27 is

00:04:46.868 --> 00:04:48.241
64.

00:04:48.241 --> 00:04:52.890
64 times negative 1/4
is equal to negative 16.

00:04:52.890 --> 00:04:55.207
So this is negative 16 right over here.

00:04:55.207 --> 00:04:57.195
So just like that, we have the equation

00:04:57.195 --> 00:04:59.246
for the line tangent to the graph

00:04:59.246 --> 00:05:01.031
of H at t is equal to zero.

00:05:01.031 --> 00:05:02.668
I'll write it one more time.

00:05:02.668 --> 00:05:05.128
It is, do a mini drum roll here,

00:05:05.128 --> 00:05:08.128
y is equal to negative 16 t plus 91.

00:05:09.178 --> 00:05:12.089
That's the equation of that
tangent line right over there.

00:05:12.089 --> 00:05:14.506
And then they say we
want to use this equation

00:05:14.506 --> 00:05:16.824
to approximate the internal
temperature of the potato

00:05:16.824 --> 00:05:18.771
at time t equals three.

00:05:18.771 --> 00:05:20.541
So let's say that this is time

00:05:20.541 --> 00:05:23.222
t equals three right over here.

00:05:23.222 --> 00:05:25.264
We want to approximate the temperature

00:05:25.264 --> 00:05:27.400
that this model describes,

00:05:27.400 --> 00:05:29.473
right over here, but we're
gonna do it using the line.

00:05:29.473 --> 00:05:32.609
So we're gonna evaluate
the line at t equals three.

00:05:32.609 --> 00:05:35.581
So then we would get, let's see,

00:05:35.581 --> 00:05:38.172
negative 16 times three

00:05:38.172 --> 00:05:39.005
plus 91

00:05:39.845 --> 00:05:42.512
is equal to, this is negative 48

00:05:43.346 --> 00:05:45.921
plus 91 is equal to, what is that?

00:05:45.921 --> 00:05:46.754
43,

00:05:47.644 --> 00:05:50.894
so this is equal to 43 degrees Celsius.

00:05:51.943 --> 00:05:54.466
So this right over here is the equation

00:05:54.466 --> 00:05:56.527
for the line tangent to the graph of H

00:05:56.527 --> 00:05:58.764
at t is equal to zero,

00:05:58.764 --> 00:06:00.779
and this right over here
is our approximation

00:06:00.779 --> 00:06:03.155
using that equation of the tangent line

00:06:03.155 --> 00:06:06.050
of the internal temperature
of the potato at time

00:06:06.050 --> 00:06:07.429
t is equal to three.


Midpoint sums

WEBVTT
Kind: captions
Language: en

00:00:00.613 --> 00:00:02.308
- [Instructor] What we
wanna do in this video

00:00:02.308 --> 00:00:04.908
is get an understanding
of how we can approximate

00:00:04.908 --> 00:00:06.696
the area under a curve.

00:00:06.696 --> 00:00:07.997
And for the sake of an example,

00:00:07.997 --> 00:00:11.038
we'll use the curve y is
equal to x squared plus one.

00:00:11.038 --> 00:00:13.175
And let's think about the
area under this curve,

00:00:13.175 --> 00:00:15.009
above the x-axis,

00:00:15.009 --> 00:00:18.446
from x equals negative
one to x equals two.

00:00:18.446 --> 00:00:20.257
So that would be

00:00:20.257 --> 00:00:22.370
this area right over here.

00:00:22.370 --> 00:00:25.226
And there's many ways
that I could tackle this,

00:00:25.226 --> 00:00:26.782
but what I'm going to do is

00:00:26.782 --> 00:00:28.918
I'm gonna break up this interval

00:00:28.918 --> 00:00:31.379
into three equal sections

00:00:31.379 --> 00:00:34.491
that are really the bases of rectangles.

00:00:34.491 --> 00:00:36.000
And then we're gonna think
about the different ways

00:00:36.000 --> 00:00:39.181
to define the heights of those rectangles.

00:00:39.181 --> 00:00:40.760
So once again,

00:00:40.760 --> 00:00:44.677
I'm going to approximate
using three rectangles

00:00:47.796 --> 00:00:48.629
of equal

00:00:51.418 --> 00:00:53.043
width.

00:00:53.043 --> 00:00:54.715
And then we'll think
about the different ways

00:00:54.715 --> 00:00:57.618
that we can define the
heights of the rectangles.

00:00:57.618 --> 00:01:00.172
So let's first define the heights

00:01:00.172 --> 00:01:02.308
of each rectangle by

00:01:02.308 --> 00:01:04.607
the value of the function at the midpoint.

00:01:04.607 --> 00:01:06.464
So we see that right over here.

00:01:06.464 --> 00:01:08.090
So let's just make sure

00:01:08.090 --> 00:01:10.505
that it actually makes sense to us.

00:01:10.505 --> 00:01:13.918
So if we look at our first
rectangle right over here,

00:01:13.918 --> 00:01:15.242
actually let's just first appreciate,

00:01:15.242 --> 00:01:17.633
we have split up this x,

00:01:17.633 --> 00:01:20.280
we have split up the interval
from x equals negative one

00:01:20.280 --> 00:01:23.763
to x equals two into three equal sections,

00:01:23.763 --> 00:01:26.155
and then each of them have a width of one.

00:01:26.155 --> 00:01:27.595
If we wanted a better approximation

00:01:27.595 --> 00:01:30.381
we could do more sections
or more rectangles,

00:01:30.381 --> 00:01:32.657
but let's just see how
we would compute this.

00:01:32.657 --> 00:01:35.629
Well the width of each of these is one,

00:01:35.629 --> 00:01:37.974
the height is based on
the value of the function

00:01:37.974 --> 00:01:39.298
at the midpoint.

00:01:39.298 --> 00:01:41.712
The midpoint here is negative 1/2,

00:01:41.712 --> 00:01:43.686
the midpoint here is 1/2,

00:01:43.686 --> 00:01:46.589
the midpoint here is 3/2.

00:01:46.589 --> 00:01:50.513
And so this height is going
to be negative 1/2 squared

00:01:50.513 --> 00:01:51.929
plus one.

00:01:51.929 --> 00:01:55.505
So negative 1/2 squared is 1/4 plus one,

00:01:55.505 --> 00:01:57.316
so that's 5/4.

00:01:57.316 --> 00:02:00.010
So the height here is 5/4.

00:02:00.010 --> 00:02:02.401
So you take 5/4 times one.

00:02:02.401 --> 00:02:05.838
This area is 5/4, let me write that down.

00:02:05.838 --> 00:02:08.206
So if we're doing the midpoint

00:02:08.206 --> 00:02:10.343
to define the height of each rectangle,

00:02:10.343 --> 00:02:12.386
this first one has

00:02:12.386 --> 00:02:14.870
an area of 5/4.

00:02:14.870 --> 00:02:18.121
Do it in a color you
can see, five over four.

00:02:18.121 --> 00:02:19.677
The second one, same idea,

00:02:19.677 --> 00:02:23.206
1/2 squared plus one is 5/4

00:02:23.206 --> 00:02:24.971
times a width of one.

00:02:24.971 --> 00:02:26.759
So 5/4 there.

00:02:26.759 --> 00:02:28.129
So let me add that.

00:02:28.129 --> 00:02:30.381
Plus 5/4.

00:02:30.381 --> 00:02:32.471
And then this third rectangle,

00:02:32.471 --> 00:02:33.911
what's its height?

00:02:33.911 --> 00:02:36.024
Well we're gonna take the
height at the midpoint,

00:02:36.024 --> 00:02:38.357
so 3/2 squared is

00:02:39.577 --> 00:02:42.340
9/4 plus one,

00:02:42.340 --> 00:02:46.519
which is the same thing as 13/4.

00:02:46.519 --> 00:02:48.539
So it has a height of 13/4,

00:02:48.539 --> 00:02:51.070
and then a width of one, so times one,

00:02:51.070 --> 00:02:54.669
which would just give us 13/4.

00:02:54.669 --> 00:02:56.669
So plus 13/4,

00:02:58.176 --> 00:03:00.916
which would give us 23 over four

00:03:00.916 --> 00:03:04.236
which is the same thing as 5 3/4.

00:03:04.236 --> 00:03:07.417
And so this is often known
as a midpoint approximation

00:03:07.417 --> 00:03:10.482
where we're using the
midpoint of each interval

00:03:10.482 --> 00:03:12.642
to define the height of our rectangle.

00:03:12.642 --> 00:03:14.639
But this isn't the only way to do it.

00:03:14.639 --> 00:03:16.821
We could look at the left endpoint

00:03:16.821 --> 00:03:18.284
or the right endpoint,

00:03:18.284 --> 00:03:19.933
and we do that in other videos.

00:03:19.933 --> 00:03:22.162
And if we wanna do it just for kicks here,

00:03:22.162 --> 00:03:24.089
let's just do that really fast.

00:03:24.089 --> 00:03:27.506
So if we wanna look at the left endpoints

00:03:28.408 --> 00:03:30.335
of our interval,

00:03:30.335 --> 00:03:32.054
well here our

00:03:32.054 --> 00:03:34.910
left endpoint is negative one,

00:03:34.910 --> 00:03:37.278
negative one squared plus one is two,

00:03:37.278 --> 00:03:38.361
two times one

00:03:39.763 --> 00:03:40.846
gives us two.

00:03:42.224 --> 00:03:45.126
And then here the left
part of this interval

00:03:45.126 --> 00:03:46.659
is x equals zero,

00:03:46.659 --> 00:03:48.400
zero squared plus one is one,

00:03:48.400 --> 00:03:50.150
one times one is one.

00:03:51.814 --> 00:03:55.413
And now here our left endpoint is one,

00:03:55.413 --> 00:03:57.080
one squared plus one

00:03:57.990 --> 00:04:00.359
is equal to two, times one,

00:04:00.359 --> 00:04:02.797
our base, is equal to two.

00:04:02.797 --> 00:04:04.724
So here we have a situation

00:04:04.724 --> 00:04:07.325
where we take our left endpoints,

00:04:07.325 --> 00:04:10.877
where it is equal to two plus one

00:04:10.877 --> 00:04:13.060
plus two or five.

00:04:13.060 --> 00:04:15.452
Well we can also look
at the right endpoints

00:04:15.452 --> 00:04:16.869
of our intervals.

00:04:19.608 --> 00:04:21.024
So this first rectangle here,

00:04:21.024 --> 00:04:22.882
clearly under approximating the area

00:04:22.882 --> 00:04:24.368
over this first interval.

00:04:24.368 --> 00:04:26.319
Its right endpoint is zero,

00:04:26.319 --> 00:04:28.292
zero squared plus one is one,

00:04:28.292 --> 00:04:31.845
so a height of one, width
of one, has an area of one.

00:04:31.845 --> 00:04:33.447
Second rectangle here,

00:04:33.447 --> 00:04:35.955
it has a height of, look
at our right endpoint,

00:04:35.955 --> 00:04:38.602
one squared plus one is two,

00:04:38.602 --> 00:04:41.946
times our width of one, well
that's just gonna give us two.

00:04:41.946 --> 00:04:44.384
And then here our right endpoint is two,

00:04:44.384 --> 00:04:48.029
squared plus one is five,
times our width of one,

00:04:48.029 --> 00:04:49.794
gives us five.

00:04:49.794 --> 00:04:51.605
So in this case we get,

00:04:51.605 --> 00:04:54.577
when we look at our right
endpoints of our intervals,

00:04:54.577 --> 00:04:55.410
we get one

00:04:56.388 --> 00:04:57.888
plus two plus five

00:04:58.734 --> 00:05:00.498
is equal to eight.

00:05:00.498 --> 00:05:01.799
And eyeballing this,

00:05:01.799 --> 00:05:03.563
it looks like we're
definitely over counting

00:05:03.563 --> 00:05:05.096
more than under counting,

00:05:05.096 --> 00:05:07.929
and so this looks like
an over approximation.

00:05:07.929 --> 00:05:09.856
So the whole idea here's
just to appreciate

00:05:09.856 --> 00:05:12.642
how we can compute these approximations

00:05:12.642 --> 00:05:14.245
using rectangles.

00:05:14.245 --> 00:05:15.429
And as you can imagine,

00:05:15.429 --> 00:05:18.424
if we added more rectangles
that had skinnier

00:05:18.424 --> 00:05:21.211
and skinnier bases but
still covered the interval

00:05:21.211 --> 00:05:23.951
from x equals negative
one to x equals two,

00:05:23.951 --> 00:05:26.296
we would get better and
better approximations

00:05:26.296 --> 00:05:27.713
of the true area.


Definite integrals intro

WEBVTT
Kind: captions
Language: en

00:00:00.299 --> 00:00:01.207
- [Instructor] What we're
gonna do in this video

00:00:01.207 --> 00:00:05.374
is introduce ourselves to the
notion of a definite integral

00:00:07.199 --> 00:00:11.068
and with indefinite
integrals and derivatives

00:00:11.068 --> 00:00:14.937
this is really one of
the pillars of calculus

00:00:14.937 --> 00:00:17.343
and as we'll see, they're all related

00:00:17.343 --> 00:00:19.491
and we'll see that more
and more in future videos

00:00:19.491 --> 00:00:20.914
and we'll also get a better appreciation

00:00:20.914 --> 00:00:22.093
for even where the notation

00:00:22.093 --> 00:00:24.426
of a definite integral comes from.

00:00:24.426 --> 00:00:27.203
So, let me draw some functions here

00:00:27.203 --> 00:00:28.279
and we're actually gonna
start thinking about

00:00:28.279 --> 00:00:29.862
areas under curves.

00:00:31.032 --> 00:00:33.653
So, let me draw a coordinate axes here,

00:00:33.653 --> 00:00:36.903
so that's my Y axis, this is my X axis,

00:00:38.270 --> 00:00:40.107
actually I'm gonna do two cases.

00:00:40.107 --> 00:00:43.440
So, this is my Y axis, this is my X axis

00:00:46.609 --> 00:00:49.346
and let's say I have some function here,

00:00:49.346 --> 00:00:51.929
so this F of X right over there

00:00:53.223 --> 00:00:56.306
and let's say that this is X equals A

00:00:57.358 --> 00:01:00.021
and let me draw a line
going straight up like that

00:01:00.021 --> 00:01:03.320
and let's say that this is X equals B,

00:01:03.320 --> 00:01:04.487
just like that

00:01:05.605 --> 00:01:08.280
and what we want to do
is concern ourselves

00:01:08.280 --> 00:01:10.555
with the area under the graph,

00:01:10.555 --> 00:01:14.020
under the graph of Y is equal to F of X

00:01:14.020 --> 00:01:17.380
and above the X axis and
between these two bounds,

00:01:17.380 --> 00:01:19.980
between X equals A and X equals B,

00:01:19.980 --> 00:01:22.128
so this area right over here

00:01:22.128 --> 00:01:24.163
and you can already get an appreciation.

00:01:24.163 --> 00:01:25.620
We're not used to finding areas

00:01:25.620 --> 00:01:27.704
where one of the boundaries
or as we'll see in the future,

00:01:27.704 --> 00:01:30.386
many of the boundaries
could actually be curves

00:01:30.386 --> 00:01:33.113
but that's one of the powers
of the definite integral

00:01:33.113 --> 00:01:36.175
and one of the powers
of integral calculus.

00:01:36.175 --> 00:01:39.493
And so, the notation for
this area right over here

00:01:39.493 --> 00:01:41.993
would be the definite integral

00:01:43.262 --> 00:01:46.215
and so, we're gonna have our
lower bound at X equals A,

00:01:46.215 --> 00:01:47.438
so we'll write it there,

00:01:47.438 --> 00:01:50.083
we'll have our upper bound at X equals B,

00:01:50.083 --> 00:01:51.085
right over there.

00:01:51.085 --> 00:01:54.907
We're taking the area
under the curve of F of X,

00:01:54.907 --> 00:01:56.490
F of X and then DX.

00:02:00.666 --> 00:02:01.881
Now, in the future we're going to,

00:02:01.881 --> 00:02:04.780
especially once we start
looking at Riemann sums,

00:02:04.780 --> 00:02:05.747
we'll get a better understanding

00:02:05.747 --> 00:02:07.437
of where this notation comes from.

00:02:07.437 --> 00:02:08.850
This actually comes from Leibniz,

00:02:08.850 --> 00:02:10.647
one of the founders of calculus.

00:02:10.647 --> 00:02:12.725
This is known as the summa symbol

00:02:12.725 --> 00:02:14.021
but for the sake of this video

00:02:14.021 --> 00:02:16.353
you just need to know
what this represents.

00:02:16.353 --> 00:02:17.910
This right over here,

00:02:17.910 --> 00:02:21.448
this represents the area under F of X

00:02:21.448 --> 00:02:24.552
between X equals A and X equals B.

00:02:24.552 --> 00:02:26.727
So, this value and this expression

00:02:26.727 --> 00:02:27.998
should be the same


Khan Academy in the classroom

WEBVTT
Kind: captions
Language: en

00:00:00.396 --> 00:00:04.326
- We have this big
moment and the moment is

00:00:04.326 --> 00:00:05.159
that for

00:00:06.315 --> 00:00:09.315
35 years of my teaching career

00:00:10.472 --> 00:00:14.622
I walked into the classroom having no idea

00:00:14.622 --> 00:00:17.402
if the kids had done the homework

00:00:17.402 --> 00:00:18.735
or

00:00:20.235 --> 00:00:23.997
what their commitment was to the subject.

00:00:23.997 --> 00:00:27.397
And then suddenly there's
this coaching platform

00:00:27.397 --> 00:00:31.531
on Khan Academy that was a
total game-changer for me.

00:00:31.531 --> 00:00:33.698
I wasn't imagining

00:00:34.731 --> 00:00:36.478
that the Khan Academy

00:00:36.478 --> 00:00:38.895
calculus content would become

00:00:42.997 --> 00:00:46.445
a big part of our curriculum.

00:00:46.445 --> 00:00:48.945
I imagined frankly and wrongly

00:00:50.167 --> 00:00:52.946
that we'd use these exercises,

00:00:52.946 --> 00:00:56.244
suggest kids use it for review.

00:00:56.244 --> 00:00:58.917
And when we discovered
the coaching platform

00:00:58.917 --> 00:01:01.782
and how powerful that was,

00:01:01.782 --> 00:01:05.068
a group of us said,

00:01:05.068 --> 00:01:06.051
"Let's give it a go.

00:01:06.051 --> 00:01:08.468
"Let's try using Khan Academy

00:01:10.023 --> 00:01:12.492
"as a major part of our curriculum."

00:01:12.492 --> 00:01:16.159
And my goodness, it
changed the way I teach.

00:01:18.136 --> 00:01:23.009
For instance, five minutes
before I walk into class,

00:01:23.009 --> 00:01:24.077
I can go to the platform

00:01:24.077 --> 00:01:26.613
and I can look through my list of students

00:01:26.613 --> 00:01:29.312
to discover that all but two of them

00:01:29.312 --> 00:01:33.145
had done the homework,
had watched the videos,

00:01:34.471 --> 00:01:36.551
had cleared the hurdle, if you will,

00:01:36.551 --> 00:01:39.373
of the exercises that I had given them.

00:01:39.373 --> 00:01:41.744
So, when I walked in the room,

00:01:41.744 --> 00:01:44.442
I didn't have to go over homework anymore.

00:01:44.442 --> 00:01:46.109
That was liberating.

00:01:48.867 --> 00:01:50.179
So, if there were two students

00:01:50.179 --> 00:01:51.282
who didn't do the homework,

00:01:51.282 --> 00:01:55.048
it gave me the opportunity
to pull them aside

00:01:55.048 --> 00:01:57.669
and say, "Hey, I see
you didn't get to it,"

00:01:57.669 --> 00:02:00.372
or, "I see you struggled with it.

00:02:00.372 --> 00:02:03.563
"Is there a way that you
and I can meet later today?

00:02:03.563 --> 00:02:07.494
"It's because I don't want
you to get behind in this."

00:02:07.494 --> 00:02:09.418
The first 15 minutes of class now

00:02:09.418 --> 00:02:11.096
all the sudden we're breaking new ground.

00:02:11.096 --> 00:02:13.429
We're doing harder problems.

00:02:16.268 --> 00:02:18.734
The kids responded so well to it

00:02:18.734 --> 00:02:21.244
because I think they had
years and years and years

00:02:21.244 --> 00:02:23.229
of math teachers going over the homework

00:02:23.229 --> 00:02:26.270
for the first 15 minutes of class.

00:02:26.270 --> 00:02:28.671
The poor kids must've been bored to death,

00:02:28.671 --> 00:02:31.122
or, "Why bother doing the homework

00:02:31.122 --> 00:02:34.004
because he'll do it on the board anyway?"

00:02:34.004 --> 00:02:37.224
That was just totally liberating

00:02:37.224 --> 00:02:40.882
and gave me an opportunity to really

00:02:40.882 --> 00:02:43.382
think hard about teaching.

00:02:44.305 --> 00:02:46.473
Since we started using Khan Academy,

00:02:46.473 --> 00:02:49.623
the one thing that we can't help

00:02:49.623 --> 00:02:53.705
but notice is that we're having more kids

00:02:53.705 --> 00:02:55.750
make it to the end of BC Calculus,

00:02:55.750 --> 00:02:59.105
and it's clear to me that we're having

00:02:59.105 --> 00:03:02.997
more girls and more underrepresented kids

00:03:02.997 --> 00:03:07.164
finish our BC Calculus class
than we ever did before.

00:03:08.154 --> 00:03:09.821
I gotta believe it's

00:03:11.389 --> 00:03:13.638
our new way about thinking about teaching.

00:03:13.638 --> 00:03:16.997
And using Khan Academy in the classroom

00:03:16.997 --> 00:03:18.738
and for homework assignments

00:03:18.738 --> 00:03:21.405
has gotta be a big part of that.


Meet Jeff, a creator of AP Statistics on Khan Academy

WEBVTT
Kind: captions
Language: en

00:00:04.328 --> 00:00:07.509
- I was a teacher for 10 years
in Kalamazoo Public Schools.

00:00:07.509 --> 00:00:10.319
They're a mid-size urban
district in southwest Michigan,

00:00:10.319 --> 00:00:12.966
and my first three years
I taught algebra I,

00:00:12.966 --> 00:00:15.218
geometry, algebra II,
the core math classes,

00:00:15.218 --> 00:00:17.958
but I also taught an
introductory statistics course.

00:00:17.958 --> 00:00:19.165
And then in my fourth year teaching,

00:00:19.165 --> 00:00:21.325
I started teaching AP Statistics

00:00:21.325 --> 00:00:23.531
and taught it for my last
seven years teaching.

00:00:23.531 --> 00:00:25.830
I love how AP courses
really stretch students

00:00:25.830 --> 00:00:28.152
and give them a chance to

00:00:28.152 --> 00:00:30.056
really challenge themselves

00:00:30.056 --> 00:00:33.446
and take on a subject
and a curriculum that

00:00:33.446 --> 00:00:35.327
is just a little bit harder than

00:00:35.327 --> 00:00:38.577
a standard high school
curriculum might be for them.

00:00:38.577 --> 00:00:41.340
And it prepares them for a
level of rigor in college

00:00:41.340 --> 00:00:43.895
that can be intimidating
for a lot of students

00:00:43.895 --> 00:00:45.659
when they first enter a college course,

00:00:45.659 --> 00:00:47.354
but students who take AP classes,

00:00:47.354 --> 00:00:49.700
in my experience, tend
to feel more comfortable

00:00:49.700 --> 00:00:51.372
once they get to college.

00:00:51.372 --> 00:00:53.554
I discovered Khan Academy when

00:00:53.554 --> 00:00:55.923
a coworker of mine showed it to me,

00:00:55.923 --> 00:00:58.964
and this was back when
Khan Academy first started,

00:00:58.964 --> 00:01:01.542
and a coworker showed it
to me and I could actually

00:01:01.542 --> 00:01:04.003
get students the practice they needed,

00:01:04.003 --> 00:01:06.348
often they needed at home as well.

00:01:06.348 --> 00:01:08.554
Because our students
would for various reasons

00:01:08.554 --> 00:01:10.180
miss a lot of days of school,

00:01:10.180 --> 00:01:12.037
and so getting students
the help they needed

00:01:12.037 --> 00:01:15.033
in the class or outside of the classroom,

00:01:15.033 --> 00:01:16.913
Khan Academy was a great help.

00:01:16.913 --> 00:01:18.632
When I was an AP teacher,

00:01:18.632 --> 00:01:20.931
I liked to use a variety of methods

00:01:20.931 --> 00:01:23.810
for engaging my students with the concepts

00:01:23.810 --> 00:01:24.808
we were learning,

00:01:24.808 --> 00:01:27.409
so I would some days like to

00:01:27.409 --> 00:01:29.475
give a lecture on the material

00:01:29.475 --> 00:01:32.448
or other days maybe play
a game or do a simulation

00:01:32.448 --> 00:01:35.350
or maybe other days let them be self-paced

00:01:35.350 --> 00:01:37.719
in some review problems.

00:01:37.719 --> 00:01:40.760
And Khan Academy offers a
way for teachers to do that

00:01:40.760 --> 00:01:42.154
in their classrooms.

00:01:42.154 --> 00:01:43.965
You have content creators like myself

00:01:43.965 --> 00:01:47.541
working at Khan Academy
who have been teachers,

00:01:47.541 --> 00:01:50.118
teaching AP for many years,

00:01:50.118 --> 00:01:53.555
and we also have external
reviewers who work closely

00:01:53.555 --> 00:01:55.969
with The College Board
and who are table readers

00:01:55.969 --> 00:01:57.688
and grade the AP Test,

00:01:57.688 --> 00:01:59.778
all working together to
make sure that the content

00:01:59.778 --> 00:02:01.844
is high quality for students.

00:02:01.844 --> 00:02:03.911
My entire career as a
teacher I felt like I was

00:02:03.911 --> 00:02:06.233
doing the world good by teaching

00:02:06.233 --> 00:02:09.507
multiple groups of students a subject

00:02:09.507 --> 00:02:11.480
that I thought would be incredibly useful

00:02:11.480 --> 00:02:13.988
for their professional, academic,

00:02:13.988 --> 00:02:16.937
and everyday lives because
teaching statistics,

00:02:16.937 --> 00:02:18.423
I feel like I'm equipping students

00:02:18.423 --> 00:02:22.590
with how to make informed
decisions in their lives.


Inflection points from graphs of function & derivatives

WEBVTT
Kind: captions
Language: en

00:00:00.212 --> 00:00:01.829
- [Instructor] What we're going
to do in this video is try

00:00:01.829 --> 00:00:05.597
to get a graphical appreciation
for inflection points,

00:00:05.597 --> 00:00:08.752
which we also cover in some
detail in other videos.

00:00:08.752 --> 00:00:10.638
So, the first thing to appreciate

00:00:10.638 --> 00:00:14.177
is an inflection point
is a point on our graph

00:00:14.177 --> 00:00:17.414
where our slope goes from
decreasing to increasing

00:00:17.414 --> 00:00:20.234
or from increasing to decreasing.

00:00:20.234 --> 00:00:23.602
So, right over here I have
the graph of some function,

00:00:23.602 --> 00:00:24.837
and let me draw the slope

00:00:24.837 --> 00:00:26.954
of a tangent line at different points.

00:00:26.954 --> 00:00:29.341
So, when x is equal to negative two,

00:00:29.341 --> 00:00:31.524
that is what the tangent line looks like.

00:00:31.524 --> 00:00:32.970
And you can see its slope.

00:00:32.970 --> 00:00:34.938
And then as we increase x,

00:00:34.938 --> 00:00:39.510
we can see that the slope is
positive, but it is decreasing.

00:00:39.510 --> 00:00:42.038
Then it goes to zero, and
then it goes negative,

00:00:42.038 --> 00:00:44.612
and the slope keeps decreasing,

00:00:44.612 --> 00:00:47.813
all the way until we
get, it looks like we get

00:00:47.813 --> 00:00:50.044
to about x equals negative one,

00:00:50.044 --> 00:00:54.028
and then our slope
begins to increase again.

00:00:54.028 --> 00:00:55.745
So, something interesting happened

00:00:55.745 --> 00:00:58.292
right at x equals negative one,

00:00:58.292 --> 00:00:59.807
and so that's a pretty good indication.

00:00:59.807 --> 00:01:01.179
We're just doing it graphically here.

00:01:01.179 --> 00:01:02.556
We're not proving it.

00:01:02.556 --> 00:01:05.382
But that at this point right over here,

00:01:05.382 --> 00:01:08.710
we have an inflection point,
so let me write that down.

00:01:08.710 --> 00:01:10.466
So, let me show you that again

00:01:10.466 --> 00:01:12.336
now that the point is labeled.

00:01:12.336 --> 00:01:15.114
For x at negative two,
we have a positive slope.

00:01:15.114 --> 00:01:17.553
It decreases, decreases, decreases.

00:01:17.553 --> 00:01:22.330
It's negative, it still
decreases, x equals negative one,

00:01:22.330 --> 00:01:26.143
and then our slope
begins increasing again.

00:01:26.143 --> 00:01:27.139
So, that's how you could tell it

00:01:27.139 --> 00:01:29.142
just from the function itself.

00:01:29.142 --> 00:01:31.927
But you could also tell inflection points

00:01:31.927 --> 00:01:34.261
by looking at your first derivative.

00:01:34.261 --> 00:01:36.523
Remember, an inflection
point is when our slope goes

00:01:36.523 --> 00:01:38.231
from increasing to decreasing

00:01:38.231 --> 00:01:40.322
or from decreasing to increasing.

00:01:40.322 --> 00:01:43.224
The derivative is just the
slope of the tangent line.

00:01:43.224 --> 00:01:45.736
So, this right over here,
this is the derivative

00:01:45.736 --> 00:01:48.236
of our original blue function.

00:01:49.663 --> 00:01:51.875
So, here we can see the interesting parts.

00:01:51.875 --> 00:01:53.322
And so notice what's happening.

00:01:53.322 --> 00:01:56.644
On the derivative, the
derivative is decreasing,

00:01:56.644 --> 00:01:58.873
which means the slope of our tangent line

00:01:58.873 --> 00:02:02.491
of our original function is
decreasing, and we saw that.

00:02:02.491 --> 00:02:06.158
Notice, while the derivative
is decreasing right over here,

00:02:06.158 --> 00:02:08.499
our slope will be decreasing.

00:02:08.499 --> 00:02:10.490
Our slope is positive.

00:02:10.490 --> 00:02:13.651
Our slope is positive, but decreasing.

00:02:13.651 --> 00:02:16.322
Then it becomes negative, but decreasing,

00:02:16.322 --> 00:02:17.938
all the way until this point,

00:02:17.938 --> 00:02:20.554
which is at x equals negative one.

00:02:20.554 --> 00:02:22.191
So, let's do that again.

00:02:22.191 --> 00:02:25.489
So, our slope is positive and decreasing,

00:02:25.489 --> 00:02:30.110
and then right over about
there, right over here,

00:02:30.110 --> 00:02:31.571
our slope keeps decreasing,

00:02:31.571 --> 00:02:33.654
but then it actually turns negative.

00:02:33.654 --> 00:02:35.422
And it keeps decreasing all the way

00:02:35.422 --> 00:02:38.036
until x equals negative one,

00:02:38.036 --> 00:02:40.959
and then our slope
begins increasing again.

00:02:40.959 --> 00:02:42.578
So, the derivative begins increasing,

00:02:42.578 --> 00:02:43.977
which means the slope of our tangent line

00:02:43.977 --> 00:02:46.485
of our original function
begins increasing.

00:02:46.485 --> 00:02:48.825
So, that point is interesting.

00:02:48.825 --> 00:02:52.640
An inflection point, one way
to identify an inflection point

00:02:52.640 --> 00:02:56.504
from the first derivative is
to look at a minimum point

00:02:56.504 --> 00:03:01.430
or to look at a maximum point,
because that shows a place

00:03:01.430 --> 00:03:04.485
where your derivative
is changing direction.

00:03:04.485 --> 00:03:06.193
It's going from increasing to decreasing,

00:03:06.193 --> 00:03:08.892
or in this case from
decreasing to increasing,

00:03:08.892 --> 00:03:12.780
which tells you that this is
likely an inflection point.

00:03:12.780 --> 00:03:15.739
Now, let's think about
the second derivative.

00:03:15.739 --> 00:03:16.850
So, right over here,

00:03:16.850 --> 00:03:19.921
this is the derivative of the derivative.

00:03:19.921 --> 00:03:21.773
And I could zoom out to
look at the whole thing.

00:03:21.773 --> 00:03:23.861
You actually can't see the
whole thing right over here.

00:03:23.861 --> 00:03:25.286
Actually, I can zoom out a little bit more

00:03:25.286 --> 00:03:28.982
so that you can really
see what's going on.

00:03:28.982 --> 00:03:30.984
And so what's interesting here?

00:03:30.984 --> 00:03:34.840
Well, it looks like right
at x equals negative one,

00:03:34.840 --> 00:03:39.092
we cross, our second
derivative crosses the x axis,

00:03:39.092 --> 00:03:40.633
so let me label that.

00:03:40.633 --> 00:03:43.103
So, right over there, we cross the x axis,

00:03:43.103 --> 00:03:45.487
which is exactly where we
have the inflection point.

00:03:45.487 --> 00:03:46.904
And that makes sense,

00:03:46.904 --> 00:03:49.091
because if our second derivative goes

00:03:49.091 --> 00:03:51.119
from being negative to positive,

00:03:51.119 --> 00:03:52.604
that means our first derivative goes

00:03:52.604 --> 00:03:54.965
from being decreasing to increasing,

00:03:54.965 --> 00:03:57.463
which means the slope of our tangent line

00:03:57.463 --> 00:04:01.696
of our function goes from
decreasing to increasing.

00:04:01.696 --> 00:04:03.290
We've seen that over and over,

00:04:03.290 --> 00:04:06.696
decreasing to increasing right over here.

00:04:06.696 --> 00:04:07.529
Now, it's important to realize

00:04:07.529 --> 00:04:10.100
the second derivative doesn't
need to just touch the x axis.

00:04:10.100 --> 00:04:11.641
It needs to cross it.

00:04:11.641 --> 00:04:12.474
So, you might say,

00:04:12.474 --> 00:04:16.313
"Well, what about this point
right over here, two, zero?"

00:04:16.313 --> 00:04:20.484
The second derivative
touches the x axis there,

00:04:20.484 --> 00:04:22.137
but it doesn't cross it,

00:04:22.137 --> 00:04:26.313
so we never go from our
derivative increasing

00:04:26.313 --> 00:04:28.593
to our derivative decreasing.

00:04:28.593 --> 00:04:30.289
So, big takeaways,

00:04:30.289 --> 00:04:31.729
you could figure out the inflection point

00:04:31.729 --> 00:04:33.385
from either the graph of the function,

00:04:33.385 --> 00:04:34.817
from the graph of the derivative,

00:04:34.817 --> 00:04:36.752
or the graph of the second derivative.

00:04:36.752 --> 00:04:38.085
On the function itself,

00:04:38.085 --> 00:04:40.499
you just wanna inspect the
slopes of the tangent line

00:04:40.499 --> 00:04:41.696
and think about where does it go

00:04:41.696 --> 00:04:44.901
from decreasing to increasing?

00:04:44.901 --> 00:04:47.919
Or the other way around, from
increasing to decreasing.

00:04:47.919 --> 00:04:51.051
If you're looking at the first derivative,

00:04:51.051 --> 00:04:55.174
you really just wanna look
at minimum or maximum points.

00:04:55.174 --> 00:04:56.827
And if you're looking at
the second derivative,

00:04:56.827 --> 00:04:58.617
which have we in orange, you wanna look at

00:04:58.617 --> 00:05:02.093
at what x value are we
crossing the x axis?

00:05:02.093 --> 00:05:05.710
Not just touching it,
but crossing the x axis.


Worked example: Inflection points from second derivative

WEBVTT
Kind: captions
Language: en

00:00:00.950 --> 00:00:03.713
- [Instructor] Let g be a
twice differentiable function

00:00:03.713 --> 00:00:05.241
defined over the closed interval

00:00:05.241 --> 00:00:06.685
from negative seven to seven,

00:00:06.685 --> 00:00:09.704
so it includes those
endpoints of the interval.

00:00:09.704 --> 00:00:13.521
This is the graph of its second
derivative, g prime prime,

00:00:13.521 --> 00:00:14.780
so that's the graph right over there,

00:00:14.780 --> 00:00:17.540
y is equal to g prime prime of x.

00:00:17.540 --> 00:00:20.579
And they ask us how many inflection points

00:00:20.579 --> 00:00:22.662
does the graph of g have?

00:00:23.672 --> 00:00:27.430
So, let's just remind ourselves
what an inflection point is.

00:00:27.430 --> 00:00:31.638
So, that is when we go from
being concave downwards

00:00:31.638 --> 00:00:34.812
to concave upwards, so
something like this.

00:00:34.812 --> 00:00:36.241
And another way to think about it,

00:00:36.241 --> 00:00:40.227
a point where our slope goes
from decreasing to increasing.

00:00:40.227 --> 00:00:43.134
So, here our slope is that,
then it's a little lower,

00:00:43.134 --> 00:00:45.371
then it's a little lower,
then it's a little lower,

00:00:45.371 --> 00:00:46.834
but then all of a sudden
it starts increasing again.

00:00:46.834 --> 00:00:49.662
It starts increasing, getting
higher, higher, and higher.

00:00:49.662 --> 00:00:51.596
So, that would be an inflection point,

00:00:51.596 --> 00:00:55.190
whatever x value where
that would actually happen.

00:00:55.190 --> 00:00:56.632
That would be the inflection point.

00:00:56.632 --> 00:00:58.229
You could go the other way around.

00:00:58.229 --> 00:01:01.856
You could have a function
that looks something like this

00:01:01.856 --> 00:01:04.115
where we have a negative slope,

00:01:04.115 --> 00:01:05.631
but then our slope is increasing,

00:01:05.631 --> 00:01:07.342
slope is increasing, slope is increasing,

00:01:07.342 --> 00:01:09.359
but then our slope
begins decreasing again.

00:01:09.359 --> 00:01:12.526
This too would be an inflection point.

00:01:13.401 --> 00:01:15.876
So, in other videos we go
into more of the intuition

00:01:15.876 --> 00:01:18.627
of how do you think about the
first and second derivatives

00:01:18.627 --> 00:01:21.027
of a function at an inflection point?

00:01:21.027 --> 00:01:22.057
But the big picture,

00:01:22.057 --> 00:01:24.333
at least for the purposes
of this worked example,

00:01:24.333 --> 00:01:27.218
is to realize when you're
looking at the second derivative,

00:01:27.218 --> 00:01:28.952
you have an inflection point

00:01:28.952 --> 00:01:32.637
where the second derivative
crosses the x axis.

00:01:32.637 --> 00:01:35.908
It's not enough to just touch the x axis.

00:01:35.908 --> 00:01:38.942
You must cross the x axis.

00:01:38.942 --> 00:01:42.865
And so right over here we
are crossing the x axis,

00:01:42.865 --> 00:01:44.957
so that is an inflection point.

00:01:44.957 --> 00:01:47.315
Right over here, we are
crossing the x axis,

00:01:47.315 --> 00:01:48.974
so that is an inflection point.

00:01:48.974 --> 00:01:51.702
Here and here we touch the x axis.

00:01:51.702 --> 00:01:54.402
Our second derivative is equal to zero.

00:01:54.402 --> 00:01:56.342
But we don't cross.

00:01:56.342 --> 00:01:59.563
We don't cross the actual x axis.

00:01:59.563 --> 00:02:01.623
We don't go from being
positive to negative.

00:02:01.623 --> 00:02:05.314
We stay non-negative this entire time.

00:02:05.314 --> 00:02:07.375
Similarly right over here,

00:02:07.375 --> 00:02:09.896
maybe something interesting
happens past this point,

00:02:09.896 --> 00:02:10.997
but they're telling us that the function

00:02:10.997 --> 00:02:12.863
is only defined over this interval,

00:02:12.863 --> 00:02:15.499
so actually nothing happens
beyond getting that point,

00:02:15.499 --> 00:02:18.004
so we're not going to cross the x axis.

00:02:18.004 --> 00:02:19.154
So, to answer the question,

00:02:19.154 --> 00:02:21.941
how many inflection points
does the graph g have,

00:02:21.941 --> 00:02:23.948
well, it has two inflection points

00:02:23.948 --> 00:02:26.570
looking at the second derivative here.

00:02:26.570 --> 00:02:28.285
Now, we know the answer.

00:02:28.285 --> 00:02:29.285
Why does that make sense?

00:02:29.285 --> 00:02:32.535
Why do you have to cross the x axis?

00:02:33.393 --> 00:02:34.837
Well, let's just imagine.

00:02:34.837 --> 00:02:37.410
Let's say that this is the
graph of a second derivative.

00:02:37.410 --> 00:02:39.577
So, this is f prime prime.

00:02:41.170 --> 00:02:45.337
So, the first derivative, for
example, could look like this.

00:02:47.026 --> 00:02:50.526
The first derivative might look like this,

00:02:51.572 --> 00:02:53.468
where over here we have a negative slope,

00:02:53.468 --> 00:02:56.349
negative slope, negative
slope, negative slope,

00:02:56.349 --> 00:02:57.982
but it's getting closer and closer,

00:02:57.982 --> 00:02:59.885
and then right over here,

00:02:59.885 --> 00:03:03.970
all of a sudden the slope
becomes positive and increasing.

00:03:03.970 --> 00:03:06.553
So, that would be f prime of x.

00:03:08.476 --> 00:03:10.235
And then you could think about, well,

00:03:10.235 --> 00:03:13.833
if this is describing the
derivative of our function,

00:03:13.833 --> 00:03:16.137
then what's our function
going to look like?

00:03:16.137 --> 00:03:19.256
Well, our function over here

00:03:19.256 --> 00:03:21.760
would have a very positive slope,

00:03:21.760 --> 00:03:24.558
but then the slope would keep decreasing

00:03:24.558 --> 00:03:27.438
all the way up until this point,

00:03:27.438 --> 00:03:29.487
and then it increases again.

00:03:29.487 --> 00:03:32.470
So, we have a positive
slope right over here.

00:03:32.470 --> 00:03:34.242
So, for example, our function
might look like this.

00:03:34.242 --> 00:03:36.718
It might have a very positive slope,

00:03:36.718 --> 00:03:39.182
but then the slope keeps decreasing,

00:03:39.182 --> 00:03:40.500
and then right over here all of a sudden,

00:03:40.500 --> 00:03:43.462
the slope begins increasing.

00:03:43.462 --> 00:03:46.108
The slope begins increasing again.

00:03:46.108 --> 00:03:50.955
And so here we were concave
downward over this first part,

00:03:50.955 --> 00:03:52.508
over this first part.

00:03:52.508 --> 00:03:54.778
We have a positive slope,
but it's decreasing.

00:03:54.778 --> 00:03:56.852
Positive slope, but it's decreasing.

00:03:56.852 --> 00:04:00.747
And then we go to having a positive slope,

00:04:00.747 --> 00:04:03.471
but now we are increasing again.

00:04:03.471 --> 00:04:05.680
And so this should give you
a good sense for why you need

00:04:05.680 --> 00:04:09.680
to cross the x axis in
the second derivative.


Existence theorems intro

WEBVTT
Kind: captions
Language: en

00:00:00.591 --> 00:00:02.976
- [Instructor] What we're going
to talk about in this video

00:00:02.976 --> 00:00:06.505
are three theorems that are
sometimes collectively known as

00:00:06.505 --> 00:00:08.088
existence theorems.

00:00:08.972 --> 00:00:11.046
So the first that we're
going to talk about

00:00:11.046 --> 00:00:13.885
is the intermediate value theorem.

00:00:13.885 --> 00:00:15.327
And the common thread here,

00:00:15.327 --> 00:00:17.272
all of the existence theorems, say,

00:00:17.272 --> 00:00:19.942
hey, we're looking for
something over an interval.

00:00:19.942 --> 00:00:23.465
There exists an x value, between a and b,

00:00:23.465 --> 00:00:25.681
where something interesting happens.

00:00:25.681 --> 00:00:27.583
In the intermediate value theorem,

00:00:27.583 --> 00:00:31.588
we assume that if we're continuous
over the closed interval

00:00:31.588 --> 00:00:34.342
from a to b, and in fact all
of these existence theorems

00:00:34.342 --> 00:00:36.471
assume that our function is continuous

00:00:36.471 --> 00:00:38.847
over the closed interval from a to b,

00:00:38.847 --> 00:00:43.014
then we take on every value
between f of a, and f of b,

00:00:44.129 --> 00:00:45.753
or another way to think about it is,

00:00:45.753 --> 00:00:48.355
pick a value from f,
between f of a and f of b,

00:00:48.355 --> 00:00:51.074
including f of a, or f of b.

00:00:51.074 --> 00:00:53.074
Let's call that value l.

00:00:53.987 --> 00:00:57.540
The intermediate value
theorem tells us that there,

00:00:57.540 --> 00:01:00.362
if we make the assumption
that f is continuous

00:01:00.362 --> 00:01:01.951
over this interval,

00:01:01.951 --> 00:01:04.995
then there must be a value between a and b

00:01:04.995 --> 00:01:07.586
that takes on the value l.

00:01:07.586 --> 00:01:10.699
And I challenge you, try to
draw a continuous function

00:01:10.699 --> 00:01:13.492
that goes from a, comma f of a,

00:01:13.492 --> 00:01:15.008
to b, comma f of b,

00:01:15.008 --> 00:01:17.294
that does not go through l.

00:01:17.294 --> 00:01:19.900
If you're continuous
you've got to go through l.

00:01:19.900 --> 00:01:21.651
The only way that you can't go through,

00:01:21.651 --> 00:01:23.809
the only you could avoid going through l

00:01:23.809 --> 00:01:25.421
is if you were discontinuous,

00:01:25.421 --> 00:01:27.106
if you had a discontinuity right there,

00:01:27.106 --> 00:01:28.549
then you could avoid going through l,

00:01:28.549 --> 00:01:30.827
but we're assuming that we're
continuous over the interval.

00:01:30.827 --> 00:01:33.186
And so here, it's pretty intuitive,

00:01:33.186 --> 00:01:34.962
that if you're continuous,

00:01:34.962 --> 00:01:36.295
there exists a c

00:01:37.215 --> 00:01:38.764
between

00:01:38.764 --> 00:01:41.508
a and b, including it could be a or b,

00:01:41.508 --> 00:01:43.904
that takes on any of those values,

00:01:43.904 --> 00:01:46.404
and in this case, this particular value

00:01:46.404 --> 00:01:49.589
that sits between f of a, and f of b.

00:01:49.589 --> 00:01:51.752
So that's the first existence theorem.

00:01:51.752 --> 00:01:54.252
The second is what's often known as

00:01:54.252 --> 00:01:56.799
the extreme value theorem.

00:01:56.799 --> 00:01:59.502
And this one is similarly
intuitive, I think.

00:01:59.502 --> 00:02:01.745
Once again, it assumes
that f is continuous

00:02:01.745 --> 00:02:05.260
over a comma, over this closed interval.

00:02:05.260 --> 00:02:07.316
But here, we say look,

00:02:07.316 --> 00:02:09.831
if it is continuous over
that closed interval,

00:02:09.831 --> 00:02:12.735
then there exists,
that's why they're called

00:02:12.735 --> 00:02:14.324
existence theorems,

00:02:14.324 --> 00:02:16.501
there exists values between a and b,

00:02:16.501 --> 00:02:18.474
and it could happen at a or b,

00:02:18.474 --> 00:02:21.244
where the function takes on a maximum

00:02:21.244 --> 00:02:24.533
and there's a value between a and b,

00:02:24.533 --> 00:02:26.629
where the function
takes on a minimum value

00:02:26.629 --> 00:02:28.364
over that interval.

00:02:28.364 --> 00:02:30.375
So once again, try to draw a function

00:02:30.375 --> 00:02:33.370
that does not take on a
minimum and maximum value

00:02:33.370 --> 00:02:34.711
over that interval.

00:02:34.711 --> 00:02:37.387
If I just draw a straight line here,

00:02:37.387 --> 00:02:40.551
well then the maximum
value happens when x is b,

00:02:40.551 --> 00:02:42.956
the minimum value happens when x is a.

00:02:42.956 --> 00:02:44.373
If I do something

00:02:45.872 --> 00:02:46.705
like that,

00:02:47.854 --> 00:02:49.479
the maximum value

00:02:49.479 --> 00:02:52.724
occurs, once again, it is
occurring in this interval,

00:02:52.724 --> 00:02:54.908
it's occurring right here at c,

00:02:54.908 --> 00:02:57.441
and then the minimum
value is occurring at a.

00:02:57.441 --> 00:03:00.073
If I draw something like,

00:03:00.073 --> 00:03:01.571
let me draw it like this,

00:03:01.571 --> 00:03:04.005
if I draw something like this,

00:03:04.005 --> 00:03:07.565
then the minimum value is
occurring at this x value,

00:03:07.565 --> 00:03:11.029
and then the maximum value
is occurring at that x value.

00:03:11.029 --> 00:03:12.881
This is just saying that they exist,

00:03:12.881 --> 00:03:14.388
they exist over that interval,

00:03:14.388 --> 00:03:16.222
and it might happen at
a, it might happen at b.

00:03:16.222 --> 00:03:18.669
And once again, the only way
I can construct something

00:03:18.669 --> 00:03:21.410
where these maximum or minimum values,

00:03:21.410 --> 00:03:23.651
these extreme values don't exist,

00:03:23.651 --> 00:03:25.242
is if I make it discontinuous.

00:03:25.242 --> 00:03:27.294
So, for example, what if I had a graph

00:03:27.294 --> 00:03:28.627
that looked like

00:03:30.044 --> 00:03:32.726
right at where we thought
we were gonna have

00:03:32.726 --> 00:03:35.342
a maximum value, we're discontinuous?

00:03:35.342 --> 00:03:37.779
Well now, you don't have
a clear maximum value.

00:03:37.779 --> 00:03:40.267
Similarly I could do something like this,

00:03:40.267 --> 00:03:43.396
where now we do not have
a clear minimum value,

00:03:43.396 --> 00:03:45.675
and so there does not exist an x

00:03:45.675 --> 00:03:48.154
where the function
takes on a minimum value

00:03:48.154 --> 00:03:49.959
in that interval.

00:03:49.959 --> 00:03:52.370
But once again, we are assuming
that we are continuous,

00:03:52.370 --> 00:03:55.137
and so we will find these extreme values,

00:03:55.137 --> 00:03:57.556
and we have other videos where
we go into much more depth

00:03:57.556 --> 00:03:59.209
on it.

00:03:59.209 --> 00:04:00.323
Last, but not least,

00:04:00.323 --> 00:04:01.448
to complete the trifecta,

00:04:01.448 --> 00:04:03.458
we have the mean value theorem,

00:04:03.458 --> 00:04:05.200
and this one is also intuitive.

00:04:05.200 --> 00:04:07.415
It starts going into differentiatability

00:04:07.415 --> 00:04:08.902
in the derivative,

00:04:08.902 --> 00:04:10.819
and it adds an extra constraint for us.

00:04:10.819 --> 00:04:12.978
Above and beyond saying
that we're continuous

00:04:12.978 --> 00:04:14.575
over the closed interval,

00:04:14.575 --> 00:04:16.438
we also assume that we're differentiable

00:04:16.438 --> 00:04:17.913
over the open interval.

00:04:17.913 --> 00:04:19.604
If you're differentiable over an interval,

00:04:19.604 --> 00:04:21.609
it does mean that you're continuous,

00:04:21.609 --> 00:04:22.897
but if you're continuous,

00:04:22.897 --> 00:04:24.809
it does not necessarily mean
that you're differentiable.

00:04:24.809 --> 00:04:26.293
And all this tells us is,

00:04:26.293 --> 00:04:28.855
if I have a continuous and
differentiable function

00:04:28.855 --> 00:04:30.815
over this interval,

00:04:30.815 --> 00:04:32.738
differentiable over the open interval,

00:04:32.738 --> 00:04:34.765
continuous over the closed interval,

00:04:34.765 --> 00:04:36.293
so let me draw that,

00:04:36.293 --> 00:04:38.640
so let me draw something like this,

00:04:38.640 --> 00:04:40.554
and if I were to compute
the average rate of change

00:04:40.554 --> 00:04:42.247
from a comma f of a

00:04:42.247 --> 00:04:43.962
to b comma f of b,

00:04:43.962 --> 00:04:47.871
so the average rate of change
I will do in this pink color,

00:04:47.871 --> 00:04:51.531
so that would be the slope
of this line right over here,

00:04:51.531 --> 00:04:53.222
that would be the average rate of change,

00:04:53.222 --> 00:04:54.943
the slope of that line.

00:04:54.943 --> 00:04:56.597
The mean value theorem tells us

00:04:56.597 --> 00:04:58.698
that there exists a point c

00:04:58.698 --> 00:05:01.381
where the derivative of
our function at that point,

00:05:01.381 --> 00:05:03.751
the slope of the tangent
line, at that point,

00:05:03.751 --> 00:05:06.624
is the same as the average rate of change.

00:05:06.624 --> 00:05:07.940
And we could eyeball that here,

00:05:07.940 --> 00:05:09.531
it looks like for this curve,

00:05:09.531 --> 00:05:11.255
there's actually several points

00:05:11.255 --> 00:05:13.105
where the derivative looks the same

00:05:13.105 --> 00:05:14.678
as the average rate of change.

00:05:14.678 --> 00:05:16.218
Maybe right over here the
slope of the tangent line

00:05:16.218 --> 00:05:18.625
looks like it has the exact same slope

00:05:18.625 --> 00:05:20.280
as the average rate of change.

00:05:20.280 --> 00:05:22.385
So that could be our c, it exists!

00:05:22.385 --> 00:05:23.676
We feel good it exists.

00:05:23.676 --> 00:05:25.764
But there could more than one value.

00:05:25.764 --> 00:05:28.358
Right over there, it looks like the slope

00:05:28.358 --> 00:05:30.330
is the same as the average rate of change,

00:05:30.330 --> 00:05:32.998
that could also be our c right over there.

00:05:32.998 --> 00:05:35.200
So how could we construct something

00:05:35.200 --> 00:05:36.500
where that isn't true?

00:05:36.500 --> 00:05:38.345
Well, if we don't assume differentiability

00:05:38.345 --> 00:05:39.658
over the interval,

00:05:39.658 --> 00:05:41.251
we can actually find a continuous function

00:05:41.251 --> 00:05:42.683
where it isn't true,

00:05:42.683 --> 00:05:44.866
where you can find a
point whose derivative

00:05:44.866 --> 00:05:47.105
is the same as the average rate of change.

00:05:47.105 --> 00:05:49.356
So, for example, here's my counter case.

00:05:49.356 --> 00:05:52.567
Maybe something like an
absolute value function,

00:05:52.567 --> 00:05:54.883
where the average rate,

00:05:54.883 --> 00:05:58.261
where the average rate of change,

00:05:58.261 --> 00:06:01.092
let me draw it a little bit better.

00:06:01.092 --> 00:06:03.486
So this was some type of
an absolute value function,

00:06:03.486 --> 00:06:05.328
where this is linear up to this point,

00:06:05.328 --> 00:06:06.768
and then linear up to this point.

00:06:06.768 --> 00:06:08.788
We're not going to be
differentiable over here,

00:06:08.788 --> 00:06:11.344
so not, the derivative
isn't defined at this point

00:06:11.344 --> 00:06:13.219
right over here.

00:06:13.219 --> 00:06:15.959
Well now, you can't, at no
point over this interval

00:06:15.959 --> 00:06:18.747
is the slope of the tangent line

00:06:18.747 --> 00:06:20.497
the same as the slope conn--,

00:06:20.497 --> 00:06:23.455
or is the same as the
average rate of change.

00:06:23.455 --> 00:06:25.230
You might try to make an argument that

00:06:25.230 --> 00:06:26.691
oh, maybe right over there,

00:06:26.691 --> 00:06:28.339
but we're not differentiable over there.

00:06:28.339 --> 00:06:29.864
There isn't a well-defined tangent line,

00:06:29.864 --> 00:06:31.712
or there isn't a well-defined tangent line

00:06:31.712 --> 00:06:33.469
and a well-defined derivative or slope

00:06:33.469 --> 00:06:34.969
of a tangent line.

00:06:35.890 --> 00:06:37.811
So in other videos we will go into depth.

00:06:37.811 --> 00:06:39.675
But it's nice to look
at them all together.

00:06:39.675 --> 00:06:41.356
See what they are all talking about.

00:06:41.356 --> 00:06:43.130
They're all talking about
the existence of an x value

00:06:43.130 --> 00:06:45.774
in the interval where
something interesting happens,

00:06:45.774 --> 00:06:48.523
where we take on a value
between f of a and f of b,

00:06:48.523 --> 00:06:50.692
where we take on extreme values,

00:06:50.692 --> 00:06:53.368
or where the derivative at that point

00:06:53.368 --> 00:06:55.507
is the same as the average rate of change

00:06:55.507 --> 00:06:57.007
over the interval.


Identifying f, f', and f'' based on graphs

WEBVTT
Kind: captions
Language: en

00:00:00.575 --> 00:00:04.343
- [Instructor] Let F be a
twice differentiable function.

00:00:04.343 --> 00:00:07.022
One of these graphs is the graph of F.

00:00:07.022 --> 00:00:08.707
One is of F prime.

00:00:08.707 --> 00:00:11.571
And one is of the second derivative of F

00:00:11.571 --> 00:00:13.236
F prime prime.

00:00:13.236 --> 00:00:16.596
Match each function with
its appropriate graph.

00:00:16.596 --> 00:00:17.980
So I encourage you to pause the video

00:00:17.980 --> 00:00:20.277
and try to figure out which of these is F,

00:00:20.277 --> 00:00:21.684
which of these is F prime,

00:00:21.684 --> 00:00:24.450
and which of these is F prime prime.

00:00:24.450 --> 00:00:27.000
All right, now let's work
through this together.

00:00:27.000 --> 00:00:29.353
So there's a couple of
ways to think about it.

00:00:29.353 --> 00:00:31.155
One way to think about it is

00:00:31.155 --> 00:00:34.083
if these were graphs of polynomials

00:00:34.083 --> 00:00:37.623
the then F would be a
higher degree polynomial.

00:00:37.623 --> 00:00:41.459
So you have F, if you were to
take the derivative of that

00:00:41.459 --> 00:00:43.192
you'd get to F prime.

00:00:43.192 --> 00:00:44.930
So this is derivative with respect to X.

00:00:44.930 --> 00:00:47.315
And if you take the derivative
with respect to X of that

00:00:47.315 --> 00:00:49.500
you get F prime prime.

00:00:49.500 --> 00:00:52.106
So, whatever degree polynomial this is

00:00:52.106 --> 00:00:53.513
if it is a polynomial,

00:00:53.513 --> 00:00:54.752
this will be a lower degree

00:00:54.752 --> 00:00:57.481
and then this would be an
even lower degree polynomial.

00:00:57.481 --> 00:00:59.911
So let's see if we can
make sense of these graphs

00:00:59.911 --> 00:01:01.236
using that lens.

00:01:01.236 --> 00:01:03.455
Well, one way to think about it is

00:01:03.455 --> 00:01:05.302
higher degree polynomials are gonna have

00:01:05.302 --> 00:01:08.321
more minima or more potential minimum

00:01:08.321 --> 00:01:09.598
or maximum points.

00:01:09.598 --> 00:01:11.277
And it looks like B here

00:01:11.277 --> 00:01:13.412
has the most minima and maxima.

00:01:13.412 --> 00:01:15.012
A has the second most.

00:01:15.012 --> 00:01:17.000
And then C has the third most.

00:01:17.000 --> 00:01:18.932
So my first hypothesis

00:01:18.932 --> 00:01:20.573
is that

00:01:20.573 --> 00:01:22.751
what we have right over here B

00:01:22.751 --> 00:01:25.106
that this is the graph of F

00:01:25.106 --> 00:01:27.689
that A is the graph of F prime.

00:01:28.617 --> 00:01:31.986
And that C is the graph
of the second derivative

00:01:31.986 --> 00:01:33.460
F prime prime.

00:01:33.460 --> 00:01:34.995
So that's just my hypothesis.

00:01:34.995 --> 00:01:37.530
Now let's see if it actually makes sense.

00:01:37.530 --> 00:01:38.828
And what I can do is

00:01:38.828 --> 00:01:40.882
I can look at points where I know

00:01:40.882 --> 00:01:42.719
that the derivative would be zero.

00:01:42.719 --> 00:01:43.969
And then look at the derivative

00:01:43.969 --> 00:01:45.600
to confirm what I think is the derivative

00:01:45.600 --> 00:01:46.854
to confirm that they are.

00:01:46.854 --> 00:01:48.786
And then I could also look at trends.

00:01:48.786 --> 00:01:51.241
So let's say, if this is F, so let's say

00:01:51.241 --> 00:01:52.750
and that was just my initial guess

00:01:52.750 --> 00:01:54.236
that this is gonna be the function F.

00:01:54.236 --> 00:01:56.979
So let's see, I have a slope of zero,

00:01:56.979 --> 00:02:00.351
a slope of a tangent line
of zero right over there.

00:02:00.351 --> 00:02:03.672
So slope of zero right over there.

00:02:03.672 --> 00:02:06.235
I have a slope of zero

00:02:06.235 --> 00:02:07.630
right over there.

00:02:07.630 --> 00:02:09.629
I have a slope of zero

00:02:09.629 --> 00:02:10.992
right over there.

00:02:10.992 --> 00:02:14.623
I have a slope of zero,
a horizontal tangent line

00:02:14.623 --> 00:02:16.069
right over there.

00:02:16.069 --> 00:02:17.275
So let's see.

00:02:17.275 --> 00:02:19.342
If this is the derivative

00:02:19.342 --> 00:02:22.197
then this function should be equal to zero

00:02:22.197 --> 00:02:23.678
at this X value,

00:02:23.678 --> 00:02:24.876
this X value,

00:02:24.876 --> 00:02:25.920
this X value,

00:02:25.920 --> 00:02:27.096
and this X value.

00:02:27.096 --> 00:02:28.592
Because the slope of the tangent line

00:02:28.592 --> 00:02:29.871
at those X values

00:02:29.871 --> 00:02:31.449
looks like zero.

00:02:31.449 --> 00:02:34.540
So this function is equal to zero here.

00:02:34.540 --> 00:02:35.419
Here.

00:02:35.419 --> 00:02:36.297
Here.

00:02:36.297 --> 00:02:37.130
And, here.

00:02:38.202 --> 00:02:40.944
Well these seem to coincide.

00:02:40.944 --> 00:02:43.428
This coincides with this.

00:02:43.428 --> 00:02:45.245
This coincides with that.

00:02:45.245 --> 00:02:47.435
And that coincides with that.

00:02:47.435 --> 00:02:49.644
And then that coincides with that.

00:02:49.644 --> 00:02:52.125
So I'm already feeling
pretty good about it.

00:02:52.125 --> 00:02:53.490
So now let's see if

00:02:53.490 --> 00:02:56.547
this F prime prime, what
I think is F prime prime

00:02:56.547 --> 00:02:58.905
is indeed the derivative of F prime.

00:02:58.905 --> 00:03:01.676
So, it's got a minimum
point right over here

00:03:01.676 --> 00:03:04.498
it looks like the slope of
the tangent line here is zero

00:03:04.498 --> 00:03:06.324
let me do this in a different color.

00:03:06.324 --> 00:03:08.564
So this looks like we have a slope

00:03:08.564 --> 00:03:09.732
of zero there.

00:03:09.732 --> 00:03:12.117
It looks like we have a slope of zero

00:03:12.117 --> 00:03:13.505
right over there.

00:03:13.505 --> 00:03:15.450
And it looks like we have a slope of zero

00:03:15.450 --> 00:03:16.368
right

00:03:16.368 --> 00:03:17.616
over there.

00:03:17.616 --> 00:03:20.071
So at that point for that X value.

00:03:20.071 --> 00:03:20.943
That X value.

00:03:20.943 --> 00:03:22.028
And that X value.

00:03:22.028 --> 00:03:24.028
And we see that this function is indeed

00:03:24.028 --> 00:03:26.465
equal to zero, right over here.

00:03:26.465 --> 00:03:27.649
Right over here.

00:03:27.649 --> 00:03:28.792
And right over here.

00:03:28.792 --> 00:03:30.972
At the exact same X values.

00:03:30.972 --> 00:03:33.017
So, based on what I just said,

00:03:33.017 --> 00:03:35.084
I would feel pretty good if
I was under time pressure

00:03:35.084 --> 00:03:36.899
I'd already feel pretty good about

00:03:36.899 --> 00:03:38.196
how I'd match things up.

00:03:38.196 --> 00:03:41.149
But you could also look
at trends between points.

00:03:41.149 --> 00:03:42.316
So for example

00:03:43.539 --> 00:03:45.434
let me pick out a trend.

00:03:45.434 --> 00:03:46.767
So, between this

00:03:48.607 --> 00:03:50.580
between this maximum point

00:03:50.580 --> 00:03:52.135
and this minimum point

00:03:52.135 --> 00:03:53.292
at first,

00:03:53.292 --> 00:03:55.292
the slope is decreasing.

00:03:56.646 --> 00:03:58.552
So the slope is getting more

00:03:58.552 --> 00:03:59.820
and more negative.

00:03:59.820 --> 00:04:02.095
And you see from here

00:04:02.095 --> 00:04:05.059
to here is indeed getting
more and more negative.

00:04:05.059 --> 00:04:07.008
And then it starts getting
more and more positive

00:04:07.008 --> 00:04:08.737
starting right about here.

00:04:08.737 --> 00:04:11.372
So that trend seems to be consistent.

00:04:11.372 --> 00:04:13.674
And so you could look for
these trends, as well.

00:04:13.674 --> 00:04:15.739
But I find looking at
the minima and maxima

00:04:15.739 --> 00:04:17.122
and where you see

00:04:17.122 --> 00:04:18.972
horizontal tangent lines,

00:04:18.972 --> 00:04:21.405
or where the slope of
the tangent line is zero,

00:04:21.405 --> 00:04:23.534
those are the easiest to
test for your derivatives

00:04:23.534 --> 00:04:26.201
or for your second derivatives.


Differentiating using multiple rules: strategy

WEBVTT
Kind: captions
Language: en

00:00:00.682 --> 00:00:02.540
- [Instructor] So I have two
different expressions here

00:00:02.540 --> 00:00:04.607
that I wanna take the derivative of.

00:00:04.607 --> 00:00:07.091
And what I want you to
do is pause the video

00:00:07.091 --> 00:00:09.715
and think about how you
would first approach

00:00:09.715 --> 00:00:11.874
taking the derivative of this expression

00:00:11.874 --> 00:00:14.011
and how that might be
the same or different

00:00:14.011 --> 00:00:15.938
as your approach in taking the derivative

00:00:15.938 --> 00:00:17.563
of this expression.

00:00:17.563 --> 00:00:20.350
The goal here isn't to compute
the derivatives all the way,

00:00:20.350 --> 00:00:23.345
but really to just think
about how we identify

00:00:23.345 --> 00:00:25.262
what strategies to use.

00:00:26.921 --> 00:00:29.057
Okay so let's first tackle this one.

00:00:29.057 --> 00:00:32.053
And the key when looking
at a complex expression

00:00:32.053 --> 00:00:35.025
like either of these is
to look at the big picture

00:00:35.025 --> 00:00:37.300
structure of the expression.

00:00:37.300 --> 00:00:39.065
So one way to think about it is,

00:00:39.065 --> 00:00:43.477
let's look at the outside
rather than the inside details.

00:00:43.477 --> 00:00:45.149
So if we look at the outside here,

00:00:45.149 --> 00:00:46.482
we have the sine

00:00:48.701 --> 00:00:49.784
of something.

00:00:50.930 --> 00:00:53.926
So there's a sine of
something going on here

00:00:53.926 --> 00:00:56.689
that I'm going to circle in red

00:00:56.689 --> 00:00:58.477
or in this pink color.

00:00:58.477 --> 00:01:00.358
So that's how my brain thinks about it.

00:01:00.358 --> 00:01:02.006
From the outside I'm like okay,

00:01:02.006 --> 00:01:04.026
big picture I'm taking the sine

00:01:04.026 --> 00:01:05.745
of some stuff.

00:01:05.745 --> 00:01:07.881
I might be taking some
stuff to some exponent.

00:01:07.881 --> 00:01:10.783
In this case, I'm inputting
in a trigonometric expression.

00:01:10.783 --> 00:01:12.804
But if you have a situation like that,

00:01:12.804 --> 00:01:14.777
it's a good sign that

00:01:14.777 --> 00:01:17.633
the chain rule is in order.

00:01:17.633 --> 00:01:19.375
So let me write that down.

00:01:19.375 --> 00:01:21.766
So we would wanna use in this case

00:01:21.766 --> 00:01:24.622
the chain rule, C.R. for chain rule.

00:01:24.622 --> 00:01:26.201
And how would we apply it?

00:01:26.201 --> 00:01:29.568
Well we would take the
derivative of the outside

00:01:29.568 --> 00:01:32.424
with respect to this inside

00:01:32.424 --> 00:01:34.491
times the derivative of this inside

00:01:34.491 --> 00:01:36.047
with respect to x.

00:01:36.047 --> 00:01:37.556
And I'm gonna write it
the way that my brain

00:01:37.556 --> 00:01:39.506
sometimes thinks about it.

00:01:39.506 --> 00:01:41.689
So we can write this as

00:01:41.689 --> 00:01:43.268
the derivative

00:01:43.268 --> 00:01:45.010
with respect to

00:01:45.010 --> 00:01:46.472
that something,

00:01:46.472 --> 00:01:48.307
and I'm just gonna make that pink circle

00:01:48.307 --> 00:01:51.604
for the something rather
than writing it all again,

00:01:51.604 --> 00:01:53.771
of sine of that something,

00:01:57.409 --> 00:01:59.290
sine of that something,

00:01:59.290 --> 00:02:01.891
not even thinking about what
that something is just yet,

00:02:01.891 --> 00:02:04.141
times, times the derivative

00:02:05.048 --> 00:02:07.278
with respect to x

00:02:07.278 --> 00:02:08.778
of that something.

00:02:09.669 --> 00:02:11.202
This is just an application

00:02:11.202 --> 00:02:13.036
of the chain rule.

00:02:13.036 --> 00:02:15.288
No matter what was here in this

00:02:15.288 --> 00:02:16.589
pink colored circle,

00:02:16.589 --> 00:02:18.191
it might have been
something with square roots

00:02:18.191 --> 00:02:20.234
and logarithms and whatever else,

00:02:20.234 --> 00:02:22.928
as long as it's being
contained within the sine,

00:02:22.928 --> 00:02:24.669
I would move to this step.

00:02:24.669 --> 00:02:26.782
The derivative with
respect to that something

00:02:26.782 --> 00:02:28.129
of sine of that something

00:02:28.129 --> 00:02:31.380
times the derivative with
respect to x of the something.

00:02:31.380 --> 00:02:34.398
Now what would that be
tangibly in this case?

00:02:34.398 --> 00:02:36.148
Well this first part,

00:02:37.092 --> 00:02:39.251
I will do it in orange,

00:02:39.251 --> 00:02:42.595
this first part would just be cosine

00:02:42.595 --> 00:02:44.678
of x squared plus five

00:02:46.659 --> 00:02:48.159
times cosine of x.

00:02:50.095 --> 00:02:52.417
So that's that circle right over there.

00:02:52.417 --> 00:02:55.947
Let me close the cosine right over there.

00:02:55.947 --> 00:02:58.988
And then times the
derivative with respect to x,

00:02:58.988 --> 00:03:02.170
times the derivative with respect to x,

00:03:02.170 --> 00:03:04.515
of all of this again,

00:03:04.515 --> 00:03:07.255
of x squared plus five

00:03:07.255 --> 00:03:08.755
times cosine of x.

00:03:10.831 --> 00:03:13.385
And then I would close my brackets.

00:03:13.385 --> 00:03:14.894
And of course I wouldn't be done yet,

00:03:14.894 --> 00:03:17.541
I have more derivative taking to do.

00:03:17.541 --> 00:03:20.629
Here now I would look at the big structure

00:03:20.629 --> 00:03:22.139
of what's going on,

00:03:22.139 --> 00:03:24.461
and I have two expressions
being multiplied.

00:03:24.461 --> 00:03:26.597
I don't have just one

00:03:26.597 --> 00:03:29.592
big expression that's being an input

00:03:29.592 --> 00:03:31.775
into like a sine function
or cosine function

00:03:31.775 --> 00:03:34.770
or one big expression that's
taken to some exponent.

00:03:34.770 --> 00:03:37.487
I have two expressions being multiplied.

00:03:37.487 --> 00:03:38.404
I have this

00:03:39.693 --> 00:03:41.776
being multiplied by this.

00:03:43.455 --> 00:03:45.893
And so if I'm just
multiplying two expressions,

00:03:45.893 --> 00:03:48.958
that's a pretty good clue
that to compute this part,

00:03:48.958 --> 00:03:52.418
I would then use the product rule.

00:03:52.418 --> 00:03:54.368
And I could keep doing
that and compute it,

00:03:54.368 --> 00:03:55.738
and I encourage you to do so,

00:03:55.738 --> 00:03:57.479
but this is more about the strategies

00:03:57.479 --> 00:03:59.267
and how do you recognize them.

00:03:59.267 --> 00:04:01.148
But now let's go to the other example.

00:04:01.148 --> 00:04:04.376
Well this looks a lot more like this step

00:04:04.376 --> 00:04:06.349
of the first problem than the beginning

00:04:06.349 --> 00:04:08.161
of the original problem.

00:04:08.161 --> 00:04:12.038
Here I don't have a
sine of a bunch of stuff

00:04:12.038 --> 00:04:14.918
or a bunch of stuff being
raised to one exponent.

00:04:14.918 --> 00:04:18.145
Here I have the product of two expressions

00:04:18.145 --> 00:04:20.072
just like we saw over here.

00:04:20.072 --> 00:04:21.989
We have this expression

00:04:23.602 --> 00:04:26.899
being multiplied by this expression.

00:04:26.899 --> 00:04:28.734
So my brain just says okay
I have two expressions,

00:04:28.734 --> 00:04:30.730
then I'm going to use the product rule.

00:04:30.730 --> 00:04:32.333
Two expressions being multiplied,

00:04:32.333 --> 00:04:33.703
I'm going to use the product rule.

00:04:33.703 --> 00:04:35.491
If it was one expression being divided

00:04:35.491 --> 00:04:37.813
by another expression, then I
would use the quotient rule.

00:04:37.813 --> 00:04:39.600
But in this case it's going
to be the product rule.

00:04:39.600 --> 00:04:41.110
And so that tells me

00:04:41.110 --> 00:04:44.152
that this is going to be the
derivative with respect to x

00:04:44.152 --> 00:04:46.152
of the first expression,

00:04:47.333 --> 00:04:49.260
just gonna do that with the orange circle,

00:04:49.260 --> 00:04:51.628
times the second expression,

00:04:51.628 --> 00:04:53.323
I'm gonna do that with the blue circle,

00:04:53.323 --> 00:04:56.156
plus the first expression,
not taking its derivative,

00:04:56.156 --> 00:04:58.406
the first expression, times

00:05:00.220 --> 00:05:03.470
the derivative with respect to x of,

00:05:05.259 --> 00:05:07.342
of the second expression.

00:05:10.947 --> 00:05:11.946
Once again here,

00:05:11.946 --> 00:05:13.757
this is just the product rule.

00:05:13.757 --> 00:05:16.125
You can substitute sine
of x squared plus five

00:05:16.125 --> 00:05:17.635
where you see this orange circle,

00:05:17.635 --> 00:05:19.121
and you can substitute cosine of x

00:05:19.121 --> 00:05:20.282
where you see this blue circle.

00:05:20.282 --> 00:05:22.673
But the whole point here
isn't to actually solve this

00:05:22.673 --> 00:05:23.974
or compute this,

00:05:23.974 --> 00:05:25.599
but really to just show

00:05:25.599 --> 00:05:28.177
how you identify the
structures in these expressions

00:05:28.177 --> 00:05:30.406
to think about well do I
use the chain rule first

00:05:30.406 --> 00:05:32.333
and then use the product rule here?

00:05:32.333 --> 00:05:34.423
Or in this case do I use
the product rule first?

00:05:34.423 --> 00:05:35.607
And even once you do this,

00:05:35.607 --> 00:05:36.791
you're not going to be done.

00:05:36.791 --> 00:05:38.811
Then to compute this derivative,

00:05:38.811 --> 00:05:40.622
you're going to have
to use the chain rule,

00:05:40.622 --> 00:05:43.084
and you'll keep going until
you don't have any more

00:05:43.084 --> 00:05:44.751
derivatives to take.