2011 Calculus BC free response #1a

WEBVTT
Kind: captions
Language: en

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Problem number 1: at time t, a particle moving in the xy-plane

00:00:05.000 --> 00:00:09.800
is at position (x(t), y(t)), where x(t) and y(t)

00:00:09.800 --> 00:00:11.502
are not explicitly given.

00:00:11.764 --> 00:00:14.141
For t is greater than or equal to 0,

00:00:14.141 --> 00:00:17.390
the derivative of x with respect to t is 4t+1

00:00:17.667 --> 00:00:21.797
the derivative of y with respect to t is sin(t^2)

00:00:21.797 --> 00:00:26.240
and at time t=0, x(0)=0 and y(0)=-4

00:00:26.378 --> 00:00:28.800
Fair enough, let's do part a.

00:00:28.800 --> 00:00:33.610
Find the speed of the particle at time t=3,

00:00:33.825 --> 00:00:37.067
and find the acceleration vector of the particle at

00:00:37.067 --> 00:00:40.533
time t is equal to 3 as well.

00:00:40.533 --> 00:00:45.333
So, the speed is really just the magnitude of the velocity vector.

00:00:45.333 --> 00:00:47.137
So what's the velocity vector?

00:00:47.137 --> 00:00:50.979
Our velocity vector as a function of time

00:00:50.979 --> 00:00:55.890
is going to be equal to the derivative of our x position

00:00:55.890 --> 00:00:59.467
as a function of time, or we could say the velocity in the x-direction

00:00:59.467 --> 00:01:06.887
times the i unit vector, plus the velocity in the y-direction times the j unit vector.

00:01:06.887 --> 00:01:08.383
I just wrote this in engineering notation,

00:01:08.383 --> 00:01:12.307
obviously there are many ways that you could specify a vector.

00:01:12.307 --> 00:01:15.800
And, in this case, x'(t), they've already given it to us:

00:01:15.800 --> 00:01:18.000
x'(t) is 4t+1

00:01:18.000 --> 00:01:20.333
they just wrote it as dx/dt.

00:01:20.333 --> 00:01:27.005
So this over here is 4t+1 times the i unit vector,

00:01:27.005 --> 00:01:31.467
and then this here, this y'(t), they gave it to us

00:01:31.467 --> 00:01:33.933
the derivative of y with respect to t is sin(t^2).

00:01:33.933 --> 00:01:41.800
So, plus sin(t^2) times the j unit vector.

00:01:41.800 --> 00:01:46.502
That is our velocity vector. So what is our velocity-

00:01:46.502 --> 00:01:49.134
or this is our velocity vector as a function of time,

00:01:49.134 --> 00:01:53.185
So what is our velocity at time t=3?

00:01:53.400 --> 00:02:01.000
Well, we just have to substitute 3 for t, so 4(3)+1 is 13i.

00:02:01.123 --> 00:02:11.216
Plus 3 squared- 3 squared is 9 -sine of 9. Plus sin9j.

00:02:11.216 --> 00:02:15.000
Our speed, what they want us to figure out is going to be

00:02:15.000 --> 00:02:20.200
equal to- or maybe I should write it as our speed at time 3

00:02:20.200 --> 00:02:26.400
-our speed is going to be equal to the magnitude of our velocity vector at time 3,

00:02:26.400 --> 00:02:34.933
which is equal to the magnitude of 13i+(sin9)j.

00:02:34.933 --> 00:02:40.133
And then this we literally, this is, we use essentially the Pythagorean Theorem

00:02:40.133 --> 00:02:41.133
to find the magnitude.

00:02:41.133 --> 00:02:47.400
This is going to be equal to the square root of 169 (that's 13 squared)

00:02:47.400 --> 00:02:55.746
plus (sin9)^2. So let me get my TI-85 out.

00:02:55.961 --> 00:03:00.048
This part of the exam we are allowed to use calculators.

00:03:00.048 --> 00:03:15.058
So we get the square root of 169+(sin9)^2.

00:03:15.535 --> 00:03:18.608
Alright, that's 13 squared plus sin9 squared.

00:03:18.885 --> 00:03:24.133
And, we get- we get 13.00- I'll just round it to 7,

00:03:24.133 --> 00:03:27.513
just to show that we didn't get just straight up 13,

00:03:27.513 --> 00:03:30.067
that the sin9 squared did do something.

00:03:30.067 --> 00:03:33.255
So 13.007.

00:03:33.255 --> 00:03:38.384
So this is approximately equal to 13.007.

00:03:38.384 --> 00:03:40.381
So we've answered the first part.

00:03:40.381 --> 00:03:45.200
The speed of the particle at time t=3 is 13.007.

00:03:45.200 --> 00:03:49.933
Then they want us to find the acceleration vector of the particle at time t=3.

00:03:49.933 --> 00:03:58.214
So the acceleration vector as a function of time is just going to be the second derivative of x

00:03:58.214 --> 00:04:01.733
with respect to t, or you'd say the acceleration in the x-direction

00:04:01.733 --> 00:04:06.769
as a function as a function of time times i, plus the acceleration in the y-direction.

00:04:06.769 --> 00:04:11.472
So it's the second derivative of y with respect to t in the j-direction.

00:04:11.472 --> 00:04:14.676
And this, of course, is just going to be equal to

00:04:14.676 --> 00:04:19.800
the derivative of x of- the derivative of x- the second derivative of x(t) is just the derivative of

00:04:19.800 --> 00:04:23.667
the first derivative. So the first derivative is 4t+1,

00:04:23.667 --> 00:04:25.892
you take the derivative of that, you just get 4.

00:04:26.215 --> 00:04:34.067
So, we get 4i plus... and then the derivative- the second derivative, or the second derivative of y with respect to t

00:04:34.067 --> 00:04:35.733
is just the derivative of the first derivative.

00:04:35.733 --> 00:04:39.963
The derivative- the first derivative is sin(t^2), so

00:04:39.963 --> 00:04:48.554
derivative of that is going to be 2t*cos(t^2), and of course we have our j.

00:04:48.554 --> 00:04:52.400
We have our j right over there and I just did this from the chain rule.

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The derivative of t^2 with respect to t is 2t,

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derivative of sin(t^2) with respect to t^2 is cos(t^2).

00:05:04.733 --> 00:05:08.440
And so that is the acceleration vector as a function of t

00:05:08.440 --> 00:05:11.400
but they care at time t is equal to 3.

00:05:11.400 --> 00:05:17.795
So, 4 still stays 4, then you have 2(3)+cos(9).

00:05:17.795 --> 00:05:21.733
So let me write this, this is going to be our acceleration vector

00:05:21.733 --> 00:05:24.667
our acceleration vector at time 3 is going to be

00:05:24.667 --> 00:05:32.733
4i plus... and let's just put, if this is 2 times 3, which is 6...

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(6cos9)j.

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Let's get the calculator out so we actually get a value there

00:05:42.575 --> 00:05:45.242
since they allow us to use or almost want us to use a calculator here.

00:05:45.242 --> 00:05:56.133
So, 6cos9 is negative 5.47, or 5.467 if we want.

00:05:56.133 --> 00:06:02.400
So, plus negative- oh, what was I doing, this is j

00:06:02.400 --> 00:06:08.333
-so plus -5.467j.

00:06:08.333 --> 00:06:09.867
So that is our acceleration vector.

00:06:09.867 --> 00:06:12.410
And once again, I wrote it in engineering notation

00:06:12.410 --> 00:06:15.800
(which I'm sure the advanced placement exam people wouldn't mind)

00:06:15.800 --> 00:06:17.733
but you could also write it in kind-of

00:06:20.102 --> 00:06:23.533
the paramiterized vector notation or your ordered-pair notation as a vector

00:06:23.533 --> 00:06:27.733
you could say that the acceleration vector at time 3

00:06:27.733 --> 00:06:29.067
you could write it this way too:

00:06:29.067 --> 00:06:33.369
<4, -5.467>

00:06:33.369 --> 00:06:34.933
These are really specifying the same thing.

00:06:34.933 --> 00:06:36.133
This is the x-component of the vector,

00:06:36.133 --> 00:06:38.000
this is the y-component of the vector.

00:06:38.000 --> 00:06:39.221
This is the x-component of the vector,

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this is the y-component of the vector.


2011 Calculus AB free response #1a

WEBVTT
Kind: captions
Language: en

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Problem 1.

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For 0 is less than or equal to
t is less than or equal to 6,

00:00:04.900 --> 00:00:07.580
a particle is moving
along the x-axis.

00:00:07.580 --> 00:00:11.480
The particle's position x of
t is not explicitly given.

00:00:11.480 --> 00:00:14.260
The velocity of the
particle is given by v of t

00:00:14.260 --> 00:00:17.400
is equal to all of this
business right over here.

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The acceleration of the
particle is given by a of t

00:00:20.310 --> 00:00:22.170
is equal to all of this
business over here.

00:00:22.170 --> 00:00:23.240
They actually didn't
have to give us

00:00:23.240 --> 00:00:25.110
that because the
acceleration is just

00:00:25.110 --> 00:00:26.789
the derivative of the velocity.

00:00:26.789 --> 00:00:29.080
And they also gave us-- they
don't tell us the position

00:00:29.080 --> 00:00:31.640
function, but they tell us
where we start off, x of 0

00:00:31.640 --> 00:00:32.970
is equal to 2.

00:00:32.970 --> 00:00:33.610
Fair enough.

00:00:33.610 --> 00:00:36.630
Now let's do part
A. Is the speed

00:00:36.630 --> 00:00:39.760
of the particle increasing
or decreasing at time t

00:00:39.760 --> 00:00:41.360
equals 5.5?

00:00:41.360 --> 00:00:43.002
Give a reason for your answer.

00:00:43.002 --> 00:00:45.210
It looks like they did
something a little sneaky here

00:00:45.210 --> 00:00:47.000
because they gave us
a velocity function

00:00:47.000 --> 00:00:48.562
and then they ask about a speed.

00:00:48.562 --> 00:00:50.770
And you might say, wait,
aren't those the same thing?

00:00:50.770 --> 00:00:53.250
And I would say, no, they
aren't quite the same thing.

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Velocity is a magnitude
and a direction.

00:00:55.780 --> 00:00:57.200
It is a vector quantity.

00:00:57.200 --> 00:00:58.870
Speed is just a magnitude.

00:00:58.870 --> 00:01:00.000
It is a scalar quantity.

00:01:00.000 --> 00:01:02.452
And to see the difference,
you could have a velocity--

00:01:02.452 --> 00:01:04.410
and this isn't maybe
particular to this problem

00:01:04.410 --> 00:01:05.993
because they don't
give us the units--

00:01:05.993 --> 00:01:09.840
but you could have a velocity
of negative 5 meters per second.

00:01:09.840 --> 00:01:12.159
And maybe if we're talking
about on the x-axis,

00:01:12.159 --> 00:01:13.700
this would mean
we're moving leftward

00:01:13.700 --> 00:01:16.440
at 5 meters per
second on the x-axis.

00:01:16.440 --> 00:01:19.030
So the magnitude is
5 meters per second.

00:01:19.030 --> 00:01:21.980
This is the magnitude.

00:01:21.980 --> 00:01:25.290
And the direction is specified
by the negative number.

00:01:25.290 --> 00:01:27.650
And that is the direction.

00:01:27.650 --> 00:01:30.460
Your velocity could be
negative 5 meters per second,

00:01:30.460 --> 00:01:36.680
but your speed would just
be 5 meters per second.

00:01:36.680 --> 00:01:38.260
So your speed is 5
meters per second

00:01:38.260 --> 00:01:40.602
whether you're going to
the left or to the right.

00:01:40.602 --> 00:01:42.560
Your velocity, you actually
care whether you're

00:01:42.560 --> 00:01:43.640
going to the left or the right.

00:01:43.640 --> 00:01:45.306
So let's just keep
that in mind while we

00:01:45.306 --> 00:01:46.860
try to solve this problem.

00:01:46.860 --> 00:01:50.550
So the best way to figure out
whether our rate of change

00:01:50.550 --> 00:01:53.660
is increasing or decreasing is
to look at the acceleration.

00:01:53.660 --> 00:01:56.040
Because acceleration
is really just

00:01:56.040 --> 00:01:57.722
the rate of change of velocity.

00:01:57.722 --> 00:01:59.180
And then, we can
think a little bit

00:01:59.180 --> 00:02:03.990
about this velocity
versus speed question.

00:02:03.990 --> 00:02:08.222
So what is the
acceleration at time 5.5?

00:02:08.222 --> 00:02:09.180
Get the calculator out.

00:02:09.180 --> 00:02:12.170
We can use calculators for
this part of the AP exam.

00:02:12.170 --> 00:02:14.850
And I assume they intend us to
because this isn't something

00:02:14.850 --> 00:02:17.480
that's easy to
calculate by hand.

00:02:17.480 --> 00:02:21.110
So the acceleration at time 5.5,
we just have to say t is 5.5

00:02:21.110 --> 00:02:22.670
and evaluate this function.

00:02:22.670 --> 00:02:29.350
So 1/2-- I'll just write
0.5-- times e to the t over 4.

00:02:29.350 --> 00:02:31.320
Well t is 5.5.

00:02:31.320 --> 00:02:34.880
5.5 divided by 4.

00:02:34.880 --> 00:02:56.180
And then, times cosine of 5.5
divided by 4 gives us 0.38.

00:02:56.180 --> 00:02:57.070
Did I do that right?

00:02:57.070 --> 00:03:02.320
We have 0.5 times e to
the 5.5 divided by 4,

00:03:02.320 --> 00:03:04.520
times cosine-- oh, sorry.

00:03:04.520 --> 00:03:05.610
I made a mistake.

00:03:05.610 --> 00:03:07.370
That looked a little strange.

00:03:07.370 --> 00:03:09.990
It's not cosine of
5.5 divided by 4,

00:03:09.990 --> 00:03:19.190
it's cosine of e to
the 5.5 divided by 4.

00:03:19.190 --> 00:03:20.350
So let's look at that.

00:03:20.350 --> 00:03:22.220
So that's one
parentheses I close.

00:03:22.220 --> 00:03:25.540
And now that is the second
parentheses that I've closed.

00:03:25.540 --> 00:03:30.280
And I get negative 1.3-- well, I
just say roughly negative 1.36.

00:03:30.280 --> 00:03:33.270
So this is equal to,
or approximately equal

00:03:33.270 --> 00:03:36.810
to, negative 1.36 if I round it.

00:03:36.810 --> 00:03:39.850
And we don't care about so
much as the actual value.

00:03:39.850 --> 00:03:42.180
What we really care
about is its sign.

00:03:42.180 --> 00:03:46.550
So the acceleration at
time 5.5 is negative,

00:03:46.550 --> 00:03:54.095
which tells us that the
velocity is decreasing.

00:03:54.095 --> 00:03:55.970
Now, you might be tempted
to say, we're done.

00:03:55.970 --> 00:03:57.469
But remember, they're
not asking us,

00:03:57.469 --> 00:04:00.230
is the velocity of the particle
increasing or decreasing?

00:04:00.230 --> 00:04:04.580
They're asking us, is
the speed of the particle

00:04:04.580 --> 00:04:05.760
increasing or decreasing?

00:04:05.760 --> 00:04:07.260
And if you're saying,
hey, how did I

00:04:07.260 --> 00:04:09.134
know that, just remember
acceleration is just

00:04:09.134 --> 00:04:10.680
the rate of change of velocity.

00:04:10.680 --> 00:04:12.400
If acceleration
is negative, that

00:04:12.400 --> 00:04:13.990
means the rate of
change of velocity

00:04:13.990 --> 00:04:15.950
is negative It is going down.

00:04:15.950 --> 00:04:18.279
But anyway, how do we
address this speed issue?

00:04:18.279 --> 00:04:19.850
How do we think about it?

00:04:19.850 --> 00:04:21.519
Well, there's two scenarios.

00:04:21.519 --> 00:04:24.550
If our velocity is
positive at time 5.5,

00:04:24.550 --> 00:04:26.770
so if we have a
positive velocity,

00:04:26.770 --> 00:04:29.436
so let's say our velocity
is 5 meters per second--

00:04:29.436 --> 00:04:31.060
although they don't
give us units here,

00:04:31.060 --> 00:04:32.350
so I won't use units.

00:04:32.350 --> 00:04:34.170
So let's say our velocity is 5.

00:04:34.170 --> 00:04:35.530
And then it's negative.

00:04:35.530 --> 00:04:38.270
So at some point our velocity is
going to be something smaller.

00:04:38.270 --> 00:04:40.960
Then that means that the speed
would also be decreasing.

00:04:40.960 --> 00:04:42.850
So if we have a
positive velocity,

00:04:42.850 --> 00:04:44.940
then the fact that
acceleration is negative

00:04:44.940 --> 00:04:48.950
means that both velocity and
speed would be decreasing.

00:04:48.950 --> 00:04:54.950
On the other hand, if we had
a negative velocity at time t

00:04:54.950 --> 00:04:58.470
equals 5.5, then the
fact that it's decreasing

00:04:58.470 --> 00:05:01.039
means that we're getting
even more negative.

00:05:01.039 --> 00:05:02.705
And if we're getting
even more negative,

00:05:02.705 --> 00:05:04.710
then that means the
speed is increasing.

00:05:04.710 --> 00:05:07.880
The magnitude is increasing
in the leftward direction.

00:05:07.880 --> 00:05:10.380
So what we really need to
do, beyond just evaluating

00:05:10.380 --> 00:05:13.380
the acceleration at
time 5.5, we also

00:05:13.380 --> 00:05:16.280
have to evaluate
the velocity to see

00:05:16.280 --> 00:05:20.160
if it's going in the left
or the rightward direction.

00:05:20.160 --> 00:05:21.770
So let's evaluate the velocity.

00:05:21.770 --> 00:05:25.100
The velocity at 5.5-- and we'll
just get our calculator out

00:05:25.100 --> 00:05:26.190
again.

00:05:26.190 --> 00:05:29.790
Velocity at 5.5-- this is
our velocity function--

00:05:29.790 --> 00:05:33.370
is going to be equal
to 2 times the sine

00:05:33.370 --> 00:05:35.831
of-- let me write it this way.

00:05:35.831 --> 00:05:38.330
Just because I want to make
sure I get my parentheses right.

00:05:38.330 --> 00:05:51.150
2 times the sine-- let me
write it this way-- 2 sine of e

00:05:51.150 --> 00:05:55.940
to the 5.5, that's our
time, divided by 4.

00:05:55.940 --> 00:05:57.650
So I did that part
right over here.

00:05:57.650 --> 00:06:00.210
And I'm going to
close the 2 as well.

00:06:00.210 --> 00:06:01.310
And then plus 1.

00:06:01.310 --> 00:06:03.060
So this is our velocity.

00:06:03.060 --> 00:06:06.490
So our velocity at
time 5.5 is negative.

00:06:06.490 --> 00:06:08.340
So negative 0.45.

00:06:08.340 --> 00:06:10.570
Negative 0.45 roughly.

00:06:10.570 --> 00:06:11.630
So this is negative.

00:06:11.630 --> 00:06:12.875
The velocity is negative.

00:06:17.080 --> 00:06:20.134
So we have this scenario where
the velocity is negative,

00:06:20.134 --> 00:06:22.050
which means we're going
in the left direction.

00:06:22.050 --> 00:06:25.260
And the fact that the
velocity is also decreasing

00:06:25.260 --> 00:06:29.090
means that over time-- at
least at this point in time--

00:06:29.090 --> 00:06:31.960
as we go forward in time, it'll
become even more negative.

00:06:31.960 --> 00:06:33.730
And it'll become
even more negative

00:06:33.730 --> 00:06:35.500
if we wait a little bit longer.

00:06:35.500 --> 00:06:39.830
So that means that the magnitude
of the velocity is increasing.

00:06:39.830 --> 00:06:42.072
It's just going in the
leftward direction.

00:06:42.072 --> 00:06:43.530
So if the magnitude
of the velocity

00:06:43.530 --> 00:06:45.000
is increasing
although it's going

00:06:45.000 --> 00:06:46.541
in the leftward
direction, that means

00:06:46.541 --> 00:06:48.582
that the speed is increasing.

00:06:48.582 --> 00:06:50.040
So the velocity--
so this is one of

00:06:50.040 --> 00:06:53.890
those interesting scenarios--
the velocity is decreasing.

00:06:53.890 --> 00:06:56.980
But the speed, which is
what they're asking us

00:06:56.980 --> 00:07:00.196
in the question,
speed is increasing.

00:07:03.381 --> 00:07:05.380
And if you wanted to do
it really quick with all

00:07:05.380 --> 00:07:07.610
of this explanation I gave
you, you would say, hey, look.

00:07:07.610 --> 00:07:08.480
What's acceleration?

00:07:08.480 --> 00:07:09.660
Is it positive or negative?

00:07:09.660 --> 00:07:10.576
You would evaluate it.

00:07:10.576 --> 00:07:11.910
You'd say, hey, it's negative.

00:07:11.910 --> 00:07:13.724
So you know velocity
is decreasing.

00:07:13.724 --> 00:07:15.640
And then you would say,
hey, what is velocity?

00:07:15.640 --> 00:07:16.750
Is it positive or negative?

00:07:16.750 --> 00:07:17.416
You evaluate it.

00:07:17.416 --> 00:07:18.610
You say it's negative.

00:07:18.610 --> 00:07:21.170
So you have a negative
value that is decreasing.

00:07:21.170 --> 00:07:22.620
So it's becoming more negative.

00:07:22.620 --> 00:07:24.780
So that means its
magnitude is increasing,

00:07:24.780 --> 00:07:27.063
or speed is increasing.


2011 Calculus AB free response #6c

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:04.210
Part C, find the
average value of f

00:00:04.210 --> 00:00:07.100
on the interval negative 1 to 1.

00:00:07.100 --> 00:00:09.020
So the average
value of a function

00:00:09.020 --> 00:00:11.410
over an interval is just
going to be-- so let's

00:00:11.410 --> 00:00:13.030
just write average.

00:00:13.030 --> 00:00:14.800
The average value
of our function

00:00:14.800 --> 00:00:19.525
is just going to be the integral
over the interval negative 1

00:00:19.525 --> 00:00:27.200
to 1 of f of x, d of x,
divided by our change in x.

00:00:27.200 --> 00:00:28.740
Sorry, this is from
negative 1 to 1.

00:00:35.780 --> 00:00:40.120
And our change in x
is 1 minus negative 1.

00:00:40.120 --> 00:00:44.810
So this is going to be equal
to 1/2 times the integral.

00:00:44.810 --> 00:00:47.579
And here, f of x is
piecewise-defined.

00:00:47.579 --> 00:00:49.370
So what we can do is
break up this integral

00:00:49.370 --> 00:00:50.430
into two intervals.

00:00:50.430 --> 00:00:53.655
We can say the integral
of f of x from negative 1

00:00:53.655 --> 00:00:58.900
to 0 of f of x dx plus-- let
me just write it this way

00:00:58.900 --> 00:01:02.540
so we don't have to
keep rewriting the 1/2.

00:01:02.540 --> 00:01:05.810
I'll use brackets--
plus the integral

00:01:05.810 --> 00:01:11.530
from 0 to 1 of f of x dx.

00:01:11.530 --> 00:01:13.380
And the reason why I
broke it up like this

00:01:13.380 --> 00:01:16.770
is because this function
has a different definition,

00:01:16.770 --> 00:01:20.150
or it's piecewise-defined.

00:01:20.150 --> 00:01:23.460
It's different when we're
less than or equal to 0,

00:01:23.460 --> 00:01:25.730
versus when we are
greater than 0.

00:01:25.730 --> 00:01:28.090
So that's why I like to
break it up this way.

00:01:28.090 --> 00:01:31.990
So then we get, this
is equal to 1/2 times--

00:01:31.990 --> 00:01:35.260
and in big brackets like this.

00:01:35.260 --> 00:01:37.180
This first part
right over here, we

00:01:37.180 --> 00:01:40.980
can write as the integral
from negative 1 to 0.

00:01:40.980 --> 00:01:43.300
What is f of x between
negative 1 and 0?

00:01:43.300 --> 00:01:51.010
It's 1 minus 2 sine of x dx.

00:01:51.010 --> 00:01:54.420
And then plus this
thing right over here.

00:01:54.420 --> 00:01:58.550
Plus the integral from 0 to 1.

00:01:58.550 --> 00:02:02.500
And what is our function
when it's between 0 and 1?

00:02:02.500 --> 00:02:08.259
It's e to the negative 4x dx.

00:02:08.259 --> 00:02:11.660
And now we can do each of
these integrals separately.

00:02:11.660 --> 00:02:16.080
And so this is going
to be equal to 1/2--

00:02:16.080 --> 00:02:19.270
and once again, I'll do a big
open bracket right over here.

00:02:19.270 --> 00:02:22.070
My pen is getting loose, let me
tie on the front a little bit

00:02:22.070 --> 00:02:22.570
better.

00:02:22.570 --> 00:02:23.570
There you go.

00:02:23.570 --> 00:02:25.140
All right, back to work.

00:02:25.140 --> 00:02:26.830
So 1/2, open brackets.

00:02:26.830 --> 00:02:28.690
And now, let's take
the antiderivative

00:02:28.690 --> 00:02:31.640
of 1 minus 2 sine
of x, antiderivative

00:02:31.640 --> 00:02:34.870
of 1 with respect
to x is just x.

00:02:34.870 --> 00:02:36.250
Negative 2 sine of x.

00:02:36.250 --> 00:02:40.770
Well, the derivative of cosine
of x is negative sine of x.

00:02:40.770 --> 00:02:44.210
So this is just going
to be 2 cosine of x.

00:02:44.210 --> 00:02:45.670
And you can verify that.

00:02:45.670 --> 00:02:47.670
Derivative of cosine of
x is negative sine of x.

00:02:47.670 --> 00:02:50.640
You have the 2 out front, so
it's negative 2 sine of x.

00:02:50.640 --> 00:02:55.890
So we're going to evaluate
that at 0 and at negative 1.

00:02:55.890 --> 00:02:58.980
And to that, we
are going to add--

00:02:58.980 --> 00:03:01.350
so let's do this definite
integral over here.

00:03:01.350 --> 00:03:05.480
The antiderivative of
e to the negative 4x

00:03:05.480 --> 00:03:13.220
is going to be equal to negative
e to the negative 4x over 4.

00:03:13.220 --> 00:03:17.390
And the way you realize that is,
if this was just an e to the x,

00:03:17.390 --> 00:03:20.070
then it's antiderivative
would just be e to the x.

00:03:20.070 --> 00:03:22.640
If you have an e
to the negative 4x,

00:03:22.640 --> 00:03:25.130
then you know that whatever
its antiderivative is

00:03:25.130 --> 00:03:29.740
is going to be, essentially,
an e to the negative 4x.

00:03:29.740 --> 00:03:31.510
But when you take
its derivative,

00:03:31.510 --> 00:03:33.301
you're going to have
to take the derivative

00:03:33.301 --> 00:03:35.600
of the negative 4x part
because of the chain rule.

00:03:35.600 --> 00:03:38.000
And so you're going to have
a negative 4 that comes out.

00:03:38.000 --> 00:03:41.480
But we don't see a
negative 4 over here,

00:03:41.480 --> 00:03:44.520
so we're obviously going to
have to divide by negative 4

00:03:44.520 --> 00:03:45.710
so it cancels out.

00:03:45.710 --> 00:03:47.900
Another way to
think about it is,

00:03:47.900 --> 00:03:50.930
we could have rewritten
we this thing.

00:03:50.930 --> 00:03:55.720
So this is equal to e
to the negative 4x dx.

00:03:55.720 --> 00:03:58.280
This is exactly
what our problem is,

00:03:58.280 --> 00:04:00.299
what we need to take
the definite integral.

00:04:00.299 --> 00:04:01.840
So then, it becomes
completely clear.

00:04:01.840 --> 00:04:04.340
So that the derivative of this
thing is sitting around here,

00:04:04.340 --> 00:04:06.810
we could put a
negative 4 over here.

00:04:06.810 --> 00:04:09.920
But you can't just willy-nilly
throw a negative 4 there.

00:04:09.920 --> 00:04:12.850
You would have to put a
negative 1/4 outside of it

00:04:12.850 --> 00:04:17.029
in order for it to be
negative 1/4 times negative 4.

00:04:17.029 --> 00:04:20.250
You're just multiplying by 1,
which doesn't change the value.

00:04:20.250 --> 00:04:26.370
And then here, you'd clearly
see that this right over here

00:04:26.370 --> 00:04:30.470
is the derivative of
e to the negative 4x.

00:04:30.470 --> 00:04:32.930
So that you'd have e
to the negative 4x.

00:04:32.930 --> 00:04:35.110
The antiderivative of this
is e to the negative 4x.

00:04:35.110 --> 00:04:37.560
And then you have this
negative 1/4 over here.

00:04:37.560 --> 00:04:40.867
So either way, hopefully
that makes sense.

00:04:40.867 --> 00:04:42.450
We go into more
detail of that earlier

00:04:42.450 --> 00:04:43.999
in the calculus playlist.

00:04:43.999 --> 00:04:46.040
So then this is from--
we're going to evaluate it

00:04:46.040 --> 00:04:54.830
at 1 and at 0, and then we
want to close the brackets.

00:04:54.830 --> 00:04:56.340
And so, what do we have here?

00:04:56.340 --> 00:04:57.540
This is equal to 1/2.

00:04:57.540 --> 00:05:00.090
Once again, open the brackets.

00:05:00.090 --> 00:05:02.140
If you evaluate
all of this at 0,

00:05:02.140 --> 00:05:04.830
you get 0 plus 2 cosine of 0.

00:05:04.830 --> 00:05:06.630
Cosine of 0 is 1.

00:05:06.630 --> 00:05:09.580
So you just get a 2 when
you evaluate it at the 0.

00:05:09.580 --> 00:05:11.030
So you get a 2.

00:05:11.030 --> 00:05:13.450
And then from that, we
want to subtract whatever

00:05:13.450 --> 00:05:16.530
we get when we evaluate
it at negative 1.

00:05:16.530 --> 00:05:21.790
We want to subtract negative
1 plus 2 cosine of negative 1.

00:05:26.320 --> 00:05:28.310
So this whole thing
right over here

00:05:28.310 --> 00:05:31.240
evaluates to this whole
thing right over there.

00:05:31.240 --> 00:05:33.670
And then we have plus.

00:05:33.670 --> 00:05:36.400
We want to evaluate this at 1.

00:05:36.400 --> 00:05:41.530
So this gives us negative
e to the negative 4.

00:05:41.530 --> 00:05:46.620
Because negative 4x, when
x is 1, is negative 4.

00:05:46.620 --> 00:05:47.730
Over 4.

00:05:47.730 --> 00:05:49.670
And from that, we
need to subtract

00:05:49.670 --> 00:05:51.090
this thing evaluated at 0.

00:05:54.160 --> 00:05:59.430
So that's going to be
negative e to the 0 over 4.

00:05:59.430 --> 00:06:00.850
e to the 0 is 1.

00:06:00.850 --> 00:06:04.790
So that's just
negative 1 over 4.

00:06:04.790 --> 00:06:09.390
And once again, all of this is
going to be multiplied by 1/2.

00:06:09.390 --> 00:06:11.710
And now we just
have to simplify it.

00:06:11.710 --> 00:06:14.500
So this is equal to-- well, I'll
just still throw out the 1/2

00:06:14.500 --> 00:06:15.000
here.

00:06:15.000 --> 00:06:19.320
This is equal to 1/2 times--
so I'll do this in a new color

00:06:19.320 --> 00:06:19.820
here.

00:06:19.820 --> 00:06:22.070
Let me do it-- well I don't
have too many new colors.

00:06:22.070 --> 00:06:22.830
OK.

00:06:22.830 --> 00:06:25.440
So this is equal to 2 plus 1.

00:06:25.440 --> 00:06:28.820
So this and this together
is going to be equal to 3.

00:06:28.820 --> 00:06:31.390
And then you have
this negative outside.

00:06:31.390 --> 00:06:33.310
So that's why it was a plus 1.

00:06:33.310 --> 00:06:35.930
And then you have a
negative times a positive 2.

00:06:35.930 --> 00:06:39.750
So minus 2 cosine of negative 1.

00:06:39.750 --> 00:06:42.290
And then you have
plus-- or maybe I

00:06:42.290 --> 00:06:47.210
should say minus-- minus e
to the negative 4 over 4.

00:06:47.210 --> 00:06:48.880
And then you have plus 1/4.

00:06:52.140 --> 00:06:54.340
And then we want to
close our brackets.

00:06:54.340 --> 00:06:57.390
And then we could do one last
step of simplification here.

00:06:57.390 --> 00:07:00.270
3, we could add
the 3 to the 1/4.

00:07:00.270 --> 00:07:02.886
3 is the same
thing as 12 over 4.

00:07:02.886 --> 00:07:06.870
12 over 4 plus 1/4 is 13/4.

00:07:06.870 --> 00:07:14.190
So you have 13/4 minus 2
cosine of negative 1 minus

00:07:14.190 --> 00:07:16.680
e to the negative 4 over 4.

00:07:16.680 --> 00:07:19.970
And then, of course, you have
your 1/2 sitting out here.

00:07:19.970 --> 00:07:23.100
And it's not the most
beautiful or simple thing.

00:07:23.100 --> 00:07:24.210
But this is our answer.

00:07:24.210 --> 00:07:28.430
This is the average value of
f of x over that interval.


2011 Calculus AB free response #6b

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:03.510
Part B. For x is not
equal to 0, express

00:00:03.510 --> 00:00:06.900
f prime of x as a
piecewise-defined function.

00:00:06.900 --> 00:00:09.110
Find the value of x
for which f prime of x

00:00:09.110 --> 00:00:10.705
is equal to negative 3.

00:00:10.705 --> 00:00:12.580
So the first thing you
might be wondering is,

00:00:12.580 --> 00:00:16.290
why did we even have to
take out x is equal to 0,

00:00:16.290 --> 00:00:19.100
or why is the derivative not
going to be defined there?

00:00:19.100 --> 00:00:19.770
And that's just
because you're going

00:00:19.770 --> 00:00:22.311
to see that the derivative is
going to be something different

00:00:22.311 --> 00:00:25.410
when we approach x is equal to
0 from the left, versus when we

00:00:25.410 --> 00:00:27.160
approach x is equal
to 0 from the right.

00:00:27.160 --> 00:00:28.980
And that's why they just
took it out of there for us.

00:00:28.980 --> 00:00:31.021
But let's just figure out
what that derivative is

00:00:31.021 --> 00:00:32.870
for all of the
other values of x.

00:00:32.870 --> 00:00:38.480
So f prime of x is equal
to, so for x is less than 0,

00:00:38.480 --> 00:00:41.610
we're going to take the
derivative of this first case.

00:00:41.610 --> 00:00:43.910
So the derivative
of 1 is just 0.

00:00:43.910 --> 00:00:45.870
The derivative of
negative 2 sine of x.

00:00:45.870 --> 00:00:48.130
Well, the derivative of sine
of x is just cosine of x.

00:00:48.130 --> 00:00:55.460
So it's going to be
negative 2 cosine of x for x

00:00:55.460 --> 00:00:57.620
is less than 0.

00:00:57.620 --> 00:00:59.862
And then for x is
greater than 0--

00:00:59.862 --> 00:01:01.320
I'll do this in
another color, I'll

00:01:01.320 --> 00:01:04.910
do it in orange-- we have
this case right over here.

00:01:04.910 --> 00:01:07.220
And we'll just do the
chain rule, derivative

00:01:07.220 --> 00:01:13.230
of negative 4x with
respect to x is negative 4.

00:01:13.230 --> 00:01:20.080
And derivative of e to the
negative 4x with respect

00:01:20.080 --> 00:01:25.327
to negative 4x is just
e to the negative 4x.

00:01:25.327 --> 00:01:27.160
Sometimes you could say
this is a derivative

00:01:27.160 --> 00:01:29.340
of the inside times the
derivative of the outside,

00:01:29.340 --> 00:01:31.400
with respect to the inside.

00:01:31.400 --> 00:01:34.728
So either way, it's negative
4 e to the negative 4x

00:01:34.728 --> 00:01:39.990
for x is greater than 0.

00:01:39.990 --> 00:01:42.060
So we did the first
part, we expressed f

00:01:42.060 --> 00:01:44.910
prime of x as a
piecewise-defined function.

00:01:44.910 --> 00:01:46.410
We didn't define the derivative.

00:01:46.410 --> 00:01:48.600
I actually forgot
a parenthesis here.

00:01:48.600 --> 00:01:52.190
We didn't define the derivative
when x is equal to 0,

00:01:52.190 --> 00:01:54.870
because it's actually not
going to be defined there.

00:01:54.870 --> 00:01:56.470
Now let's do the second part.

00:01:56.470 --> 00:01:59.110
Find the value of x
for which f prime of x

00:01:59.110 --> 00:02:01.370
is equal to negative 3.

00:02:01.370 --> 00:02:03.860
And so if this wasn't
piecewise-defined,

00:02:03.860 --> 00:02:05.960
you'd very simply
just say, look,

00:02:05.960 --> 00:02:08.079
f prime of x is
equal to negative 3.

00:02:08.079 --> 00:02:10.120
You would take whatever
f prime of x is equal to,

00:02:10.120 --> 00:02:12.340
and you'd do some
algebra to solve for it.

00:02:12.340 --> 00:02:14.460
But here you're like,
which case do I use?

00:02:14.460 --> 00:02:16.460
I don't know if the x
that gets us to negative 3

00:02:16.460 --> 00:02:17.980
is going to be less
than 0, or I don't

00:02:17.980 --> 00:02:19.646
know if it's going
to be greater than 0.

00:02:19.646 --> 00:02:21.390
So I don't know
which case to use.

00:02:21.390 --> 00:02:24.130
And one thing that
we realize is to look

00:02:24.130 --> 00:02:26.520
at these functions a
little bit and realize

00:02:26.520 --> 00:02:28.840
that cosine of x is
a bounded function.

00:02:28.840 --> 00:02:33.690
Cosine of x can only go between
positive 1 and negative 1.

00:02:33.690 --> 00:02:35.690
So negative 2
cosine of x can only

00:02:35.690 --> 00:02:39.720
go between positive
2 and negative 2.

00:02:42.890 --> 00:02:45.079
So it can never
get to negative 3.

00:02:45.079 --> 00:02:47.120
So if anything's ever
going to get to negative 3,

00:02:47.120 --> 00:02:49.710
it's going to have to be
this part of the derivative,

00:02:49.710 --> 00:02:51.700
or this part of the
derivative definition.

00:02:51.700 --> 00:02:54.186
So it's going to have to be
this thing right over here.

00:02:54.186 --> 00:02:55.560
And hopefully
there's some values

00:02:55.560 --> 00:02:58.640
of x greater than 0 where
this thing right over here

00:02:58.640 --> 00:03:00.680
is equal to negative 3.

00:03:00.680 --> 00:03:02.410
So let's try it out.

00:03:02.410 --> 00:03:05.870
Negative 4 e to the
negative 4x needs

00:03:05.870 --> 00:03:08.770
to be equal to negative 3.

00:03:08.770 --> 00:03:11.090
We can divide both
sides by negative 4.

00:03:11.090 --> 00:03:14.720
We get e to the negative
4x is equal to negative 3/4

00:03:14.720 --> 00:03:17.730
divided by negative 4 is 3/4.

00:03:17.730 --> 00:03:20.260
We could take the natural
log of both sides,

00:03:20.260 --> 00:03:24.770
and we will get negative 4x
is equal to the natural log

00:03:24.770 --> 00:03:25.635
of 3/4.

00:03:29.054 --> 00:03:31.220
And just to be clear, what
I did here, you literally

00:03:31.220 --> 00:03:34.646
could put the natural log
here, natural log there,

00:03:34.646 --> 00:03:36.270
and you could put
the natural log there

00:03:36.270 --> 00:03:38.720
as well to see that step.

00:03:38.720 --> 00:03:42.660
This is saying, what power do
I have to raise e to, to get e

00:03:42.660 --> 00:03:43.679
to the negative 4x?

00:03:43.679 --> 00:03:45.220
Well, obviously, I
just need to raise

00:03:45.220 --> 00:03:48.110
e to the negative
4x power there.

00:03:48.110 --> 00:03:50.899
So this power is negative 4x.

00:03:50.899 --> 00:03:52.440
And then we just
took the natural log

00:03:52.440 --> 00:03:54.570
of the right-hand side as well.

00:03:54.570 --> 00:03:56.880
And then to solve for x,
we can divide both sides

00:03:56.880 --> 00:03:58.060
by negative 4.

00:03:58.060 --> 00:04:02.950
So you get x is equal to-- or
we could multiply both sides

00:04:02.950 --> 00:04:07.200
by negative 1/4, either way--
negative 1/4 natural log

00:04:07.200 --> 00:04:09.420
of 3/4.

00:04:09.420 --> 00:04:11.650
And what we need to do
is verify that this x.

00:04:11.650 --> 00:04:16.110
So we used this case right over
here, but we have to make sure

00:04:16.110 --> 00:04:21.130
that we can use this case,
that this x is greater than 0.

00:04:21.130 --> 00:04:22.680
And we might be
tempted right when

00:04:22.680 --> 00:04:24.040
we look at this to
say, wait, this looks

00:04:24.040 --> 00:04:25.000
like a negative number.

00:04:25.000 --> 00:04:28.380
But we have to remind ourselves
that the natural log of 3/4,

00:04:28.380 --> 00:04:33.375
since 3/4 is less than
e, the natural log of 3/4

00:04:33.375 --> 00:04:34.750
is going to be a
negative number.

00:04:34.750 --> 00:04:37.540
It's going to be e to
some negative exponent.

00:04:37.540 --> 00:04:42.500
So since this is negative
and this part right over here

00:04:42.500 --> 00:04:47.250
is negative, you have a
negative times a negative,

00:04:47.250 --> 00:04:50.350
so this right over here
is going to be positive.

00:04:50.350 --> 00:04:52.930
So this is a positive
value right over here.

00:04:52.930 --> 00:04:57.660
So you would use this
case right over here.

00:04:57.660 --> 00:05:01.340
So that's our answer-- x
is equal to negative 1/4

00:05:01.340 --> 00:05:02.500
natural log of 3/4.

00:05:02.500 --> 00:05:06.680
Or the derivative, we could
write f prime of negative 1/4

00:05:06.680 --> 00:05:13.180
times the natural log of
3/4 is equal to negative 3.

00:05:13.180 --> 00:05:15.140
And we're done.


2011 Calculus AB free response #6a

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:02.240
Problem number six.

00:00:02.240 --> 00:00:05.430
Let f be defined by
f of x is equal to--

00:00:05.430 --> 00:00:06.360
and we have two cases.

00:00:06.360 --> 00:00:10.790
When x is less than or equal
to 0, f is 1 minus 2 sine of x.

00:00:10.790 --> 00:00:15.920
When x is greater than 0,
f is e to the negative 4x.

00:00:15.920 --> 00:00:19.570
Show that f is
continuous at x equals 0.

00:00:19.570 --> 00:00:21.770
So for something to be
continuous at x equals 0,

00:00:21.770 --> 00:00:23.730
let's think about
what has to happen.

00:00:23.730 --> 00:00:27.570
So if I have a function
here-- So that is my x-axis.

00:00:27.570 --> 00:00:31.710
And let's say this is my y-axis.

00:00:31.710 --> 00:00:33.790
And we care about what
happens at x equals 0.

00:00:33.790 --> 00:00:35.690
So x equals 0 there.

00:00:35.690 --> 00:00:39.220
And let's say that
this is our function.

00:00:39.220 --> 00:00:42.009
So maybe our function maybe
looks something like this.

00:00:42.009 --> 00:00:42.550
I don't know.

00:00:42.550 --> 00:00:44.100
Well, this one in
particular probably

00:00:44.100 --> 00:00:45.183
looks something like this.

00:00:48.770 --> 00:00:51.100
This one in particular
might look-- who knows.

00:00:51.100 --> 00:00:53.075
And then it might look
something like that.

00:00:53.075 --> 00:00:54.700
And the particulars
are that important.

00:00:54.700 --> 00:00:56.700
We just have to think
about what they're asking.

00:00:56.700 --> 00:01:03.780
In order for it to
be continuous here,

00:01:03.780 --> 00:01:07.740
so the limit as we
approach 0 from the left

00:01:07.740 --> 00:01:13.750
should be equal to the
value of the function at 0.

00:01:13.750 --> 00:01:18.170
So the limit of f of x should
be equal to f of 0, which

00:01:18.170 --> 00:01:20.210
should be equal to the
limit as we approach

00:01:20.210 --> 00:01:23.420
from the right, which
should be equal to the limit

00:01:23.420 --> 00:01:25.550
as we approach 0 from the right.

00:01:25.550 --> 00:01:27.650
So that should be
equal to the limit

00:01:27.650 --> 00:01:31.080
as x approaches 0 from
the right of f of x.

00:01:31.080 --> 00:01:35.000
And the reason why this matters
is, if this wasn't the case,

00:01:35.000 --> 00:01:37.670
if f of 0 wasn't the
same as the limit,

00:01:37.670 --> 00:01:40.024
then we might have a
gap right over there.

00:01:40.024 --> 00:01:41.065
So you could have limits.

00:01:41.065 --> 00:01:43.410
So you could have a
situation like this,

00:01:43.410 --> 00:01:45.500
where you have a gap
right over there.

00:01:45.500 --> 00:01:47.480
And then it looks
something like that.

00:01:47.480 --> 00:01:50.680
So the two limits from the
left and the right both exist,

00:01:50.680 --> 00:01:53.070
and the limit at that
point would exist.

00:01:53.070 --> 00:01:56.280
But if the function itself does
not equal that value there,

00:01:56.280 --> 00:01:58.610
if it equals something
else, then the function

00:01:58.610 --> 00:01:59.880
would not be continuous.

00:01:59.880 --> 00:02:01.920
So that's why the
limit has to be

00:02:01.920 --> 00:02:04.780
equal to the value of
the function in order

00:02:04.780 --> 00:02:07.390
for it to be continuous.

00:02:07.390 --> 00:02:09.750
So let's think about
whether all of these things

00:02:09.750 --> 00:02:10.459
equal each other.

00:02:10.459 --> 00:02:13.041
First of all, let's just think
about the value of the function

00:02:13.041 --> 00:02:13.790
there.

00:02:13.790 --> 00:02:15.940
So remember, we're doing part a.

00:02:15.940 --> 00:02:20.240
F of 0 is equal to-- we're
going to use this first case,

00:02:20.240 --> 00:02:23.110
because that's the case for
x is less than or equal to 0.

00:02:23.110 --> 00:02:32.480
So f of 0 is going to be
equal to 1 minus 2 sine of 0.

00:02:32.480 --> 00:02:36.850
Well, sine of 0 is
0, 2 times 0 is 0.

00:02:36.850 --> 00:02:38.310
So this whole thing is 0.

00:02:38.310 --> 00:02:41.170
1 minus 0 is 1.

00:02:41.170 --> 00:02:42.180
Fair enough.

00:02:42.180 --> 00:02:48.050
Now let's think about the
limit as x approaches 0

00:02:48.050 --> 00:02:49.480
from the left-hand side.

00:02:49.480 --> 00:02:54.840
So as we approach 0 from the
left-hand side of f of x.

00:02:54.840 --> 00:02:56.960
So as we approach 0 from
the left-hand hand side,

00:02:56.960 --> 00:02:59.980
we're dealing with values
of x that are less than 0.

00:02:59.980 --> 00:03:03.330
So once again, we're dealing
with this first case up here.

00:03:03.330 --> 00:03:06.700
So this is the limit
as x approaches 0

00:03:06.700 --> 00:03:12.540
from the left-hand side
of 1 minus 2 sine of x.

00:03:12.540 --> 00:03:15.560
Now, sine of x is continuous.

00:03:15.560 --> 00:03:16.690
It's a continuous function.

00:03:16.690 --> 00:03:19.760
So this is going to be the same
thing as 1 minus 2 sine of 0,

00:03:19.760 --> 00:03:23.670
which we've already figured out,
which is exactly equal to 1.

00:03:23.670 --> 00:03:24.611
So this is equal to 1.

00:03:24.611 --> 00:03:26.860
So the value of the limit
as we approach from the left

00:03:26.860 --> 00:03:28.735
is the same as the
value of the function.

00:03:28.735 --> 00:03:30.860
Now let's do it for as we
approach from the right--

00:03:30.860 --> 00:03:34.720
as we approach from values
of x greater than 0.

00:03:34.720 --> 00:03:38.750
So let's think about the
limit as we approach 0

00:03:38.750 --> 00:03:41.540
from the right of f of x.

00:03:41.540 --> 00:03:43.040
So here, we're
dealing with values

00:03:43.040 --> 00:03:44.650
of x that are larger than 0.

00:03:44.650 --> 00:03:47.350
So we're dealing with
this case right over here.

00:03:47.350 --> 00:03:50.600
So that's going to be the
limit as x approaches 0

00:03:50.600 --> 00:03:54.570
from the right of e
to the negative 4x.

00:03:54.570 --> 00:03:57.430
And for the x's that we care
about, or actually in general,

00:03:57.430 --> 00:03:59.630
this is a continuous function.

00:03:59.630 --> 00:04:01.930
This is a continuous function.

00:04:01.930 --> 00:04:03.980
So this is going to
be the same thing as e

00:04:03.980 --> 00:04:06.210
to the negative 4
times 0, which is

00:04:06.210 --> 00:04:08.500
just e to the 0,
which is just 1.

00:04:08.500 --> 00:04:10.880
So this, once again,
is equal to 1.

00:04:10.880 --> 00:04:13.580
So the function is equal to
1 at that point, the limit

00:04:13.580 --> 00:04:15.530
as we approach from
the left is equal to 1,

00:04:15.530 --> 00:04:18.440
and the limit as we approach
from the right is equal to 1.

00:04:18.440 --> 00:04:22.540
So the function is
continuous there.


2011 Calculus AB free response #5c.

WEBVTT
Kind: captions
Language: en

00:00:00.540 --> 00:00:04.150
Part C, find the
particular solution--

00:00:04.150 --> 00:00:06.696
W as a function of t-- to
the differential equations,

00:00:06.696 --> 00:00:08.070
derivative of W
with respect to t

00:00:08.070 --> 00:00:11.670
is equal to 1 over
25 times w minus 300,

00:00:11.670 --> 00:00:15.660
with initial condition, at
time 0, W is equal to 1,400.

00:00:15.660 --> 00:00:21.460
And the units in this problem,
it was 1,400 tons they told us.

00:00:21.460 --> 00:00:24.560
So it might seem strange to
see a differential equations

00:00:24.560 --> 00:00:26.330
problem on an AP exam.

00:00:26.330 --> 00:00:27.970
You say, wait, this isn't fair.

00:00:27.970 --> 00:00:30.926
This isn't a differential
equations class.

00:00:30.926 --> 00:00:33.050
And the thing to keep in
mind is if, on an AP exam,

00:00:33.050 --> 00:00:34.170
you see differential
equations, they're

00:00:34.170 --> 00:00:36.461
really asking you to solve
a differential equation that

00:00:36.461 --> 00:00:38.640
can be solved using
the skills that you

00:00:38.640 --> 00:00:40.730
would learn in an AP class.

00:00:40.730 --> 00:00:43.516
And the main class of
differential equations

00:00:43.516 --> 00:00:44.890
that you could
solve are the ones

00:00:44.890 --> 00:00:46.640
that are separable
differential equations.

00:00:46.640 --> 00:00:48.690
The ones where you can
separate the independent

00:00:48.690 --> 00:00:50.331
and the dependent variables.

00:00:50.331 --> 00:00:52.580
And if you don't have any
idea what I'm talking about,

00:00:52.580 --> 00:00:53.496
I'm about to show you.

00:00:53.496 --> 00:00:56.030
So let me rewrite this
differential equation.

00:00:56.030 --> 00:01:04.069
We have dw, dt is equal to
1 over 25 times W minus 300.

00:01:04.069 --> 00:01:06.300
Now this differential
equation, we have W here.

00:01:06.300 --> 00:01:07.790
And we have this dt.

00:01:07.790 --> 00:01:09.700
But we don't have
t anywhere else.

00:01:09.700 --> 00:01:12.370
So what I want to do is I want
to get the parts that involve W

00:01:12.370 --> 00:01:14.590
onto the left-hand
side, and I want

00:01:14.590 --> 00:01:18.020
to get anything that deals
with dt on the right-hand side.

00:01:18.020 --> 00:01:23.740
So let's divide both sides of
this equation by w minus 300.

00:01:23.740 --> 00:01:29.417
So we would have 1
over W minus 300 dw

00:01:29.417 --> 00:01:31.250
dt-- the derivative of
W with respect to t--

00:01:31.250 --> 00:01:34.880
is equal to-- I divide
this side by W minus 300,

00:01:34.880 --> 00:01:37.710
so it's equal to 1 over 25.

00:01:37.710 --> 00:01:40.050
Now I want to get the
dt's on this side.

00:01:40.050 --> 00:01:42.930
So let's multiply
both sides by dt.

00:01:42.930 --> 00:01:44.430
[INAUDIBLE] the
differential is just

00:01:44.430 --> 00:01:47.210
a very, very small change in t.

00:01:47.210 --> 00:01:51.620
And so we get-- and I'm a little
bit imprecise with my notation

00:01:51.620 --> 00:01:52.120
here.

00:01:52.120 --> 00:01:53.970
But this is one way that
you can think about it when

00:01:53.970 --> 00:01:55.928
you are dealing with the
differential equations

00:01:55.928 --> 00:01:58.040
like this, separable
differential equations.

00:01:58.040 --> 00:02:03.860
You have 1 over W minus
300 dw-- let me make sure

00:02:03.860 --> 00:02:06.600
that the dw doesn't look like
it's in the denominator--

00:02:06.600 --> 00:02:13.730
dw is equal to 1 over 25 dt.

00:02:13.730 --> 00:02:16.830
And now we can integrate both
sides, which is essentially

00:02:16.830 --> 00:02:20.200
just saying that we're taking--
since this is equal to this--

00:02:20.200 --> 00:02:22.950
that each little
increment of W times this

00:02:22.950 --> 00:02:25.960
is equal to each little
increment of dt times 1

00:02:25.960 --> 00:02:27.340
over 25.

00:02:27.340 --> 00:02:31.050
We can then say, well, if we sum
up all of the increments of W

00:02:31.050 --> 00:02:33.360
times this quantity--
if we sum up

00:02:33.360 --> 00:02:35.970
the infinite number of
infinitesimally small

00:02:35.970 --> 00:02:38.320
increments-- that would be
the same thing as summing up

00:02:38.320 --> 00:02:39.480
all of these characters.

00:02:39.480 --> 00:02:41.700
Because each of these
characters is really

00:02:41.700 --> 00:02:44.260
the same as each of
these characters.

00:02:44.260 --> 00:02:45.390
So let's do that sum.

00:02:45.390 --> 00:02:49.360
We're essentially just
integrating both sides of this.

00:02:49.360 --> 00:02:52.260
And here, you might be able
to do this in your head.

00:02:52.260 --> 00:02:54.490
Or we could do the
formal u substitution,

00:02:54.490 --> 00:02:56.320
just to make it clear
what we're doing.

00:02:56.320 --> 00:03:02.010
If you said u is equal to W
minus 300, then what is du?

00:03:02.010 --> 00:03:03.410
If you wanted the differential.

00:03:03.410 --> 00:03:05.159
So if you want to take
the derivative of u

00:03:05.159 --> 00:03:09.660
with respect to W, you would
say du with respect to W

00:03:09.660 --> 00:03:11.840
is equal to what?

00:03:11.840 --> 00:03:13.150
It is equal to 1.

00:03:13.150 --> 00:03:15.590
The derivative of this
with respect to W is 1.

00:03:15.590 --> 00:03:18.630
Or if you multiply
both sides times dw,

00:03:18.630 --> 00:03:22.440
you have du is equal to dw.

00:03:22.440 --> 00:03:26.290
And so this left-hand side right
over here-- this integral--

00:03:26.290 --> 00:03:32.490
could be rewritten as the
integral of 1 over u du.

00:03:32.490 --> 00:03:34.190
We define this as u.

00:03:34.190 --> 00:03:36.610
And we just saw that du
is the same thing as dw,

00:03:36.610 --> 00:03:38.560
so this could be
rewritten as du.

00:03:38.560 --> 00:03:40.540
And this is one of
the basic integrals,

00:03:40.540 --> 00:03:41.910
so hopefully we've learned.

00:03:41.910 --> 00:03:46.230
And this is equal to the natural
log of the absolute value of u.

00:03:46.230 --> 00:03:48.680
Or if you unwind
the substitution,

00:03:48.680 --> 00:03:51.820
this is equal to the natural
log of the absolute value

00:03:51.820 --> 00:03:55.460
of W minus 300.

00:03:55.460 --> 00:03:56.697
So let me write this down.

00:03:56.697 --> 00:03:58.280
So the left-hand
side right over here.

00:03:58.280 --> 00:03:59.210
And you might be able
do that in your head.

00:03:59.210 --> 00:04:00.001
You say, hey, look.

00:04:00.001 --> 00:04:01.190
You have W minus 300.

00:04:01.190 --> 00:04:03.840
Its derivative is 1, so
it's essentially there.

00:04:03.840 --> 00:04:06.660
So I can just take this
as if this was 1 over-- I

00:04:06.660 --> 00:04:09.320
could just take the
antiderivative here and treat

00:04:09.320 --> 00:04:11.235
this whole thing as a variable.

00:04:11.235 --> 00:04:13.610
So it would just be the natural
log of this entire thing,

00:04:13.610 --> 00:04:15.910
or the absolute value
of that entire thing.

00:04:15.910 --> 00:04:18.480
So the integral of
the left-hand side

00:04:18.480 --> 00:04:24.496
is the natural log of the
absolute value of W minus 300.

00:04:24.496 --> 00:04:26.662
And the derivative of the
right-hand side-- and this

00:04:26.662 --> 00:04:30.360
is a little bit more
straightforward-- or sorry,

00:04:30.360 --> 00:04:33.170
the antiderivative with respect
to t on the right-hand side

00:04:33.170 --> 00:04:39.045
is 1 over 25 t, or we
could say t over 25.

00:04:42.090 --> 00:04:44.160
And then these are both
indefinite integrals.

00:04:44.160 --> 00:04:46.460
You could have a constant
on both sides or one side.

00:04:46.460 --> 00:04:48.501
We could just put it on
one side right over here.

00:04:48.501 --> 00:04:50.440
You could have put
plus some constant.

00:04:50.440 --> 00:04:52.635
You could have added
some C1 over here.

00:04:52.635 --> 00:04:54.260
And then you could've
had C2 over here.

00:04:54.260 --> 00:04:56.426
But then you could've
subtracted C1 from both sides,

00:04:56.426 --> 00:04:59.094
and then just have a
C3 constant over here.

00:04:59.094 --> 00:05:00.510
But to simplify
it, you could just

00:05:00.510 --> 00:05:04.130
say, look, this is going
to be equal to this

00:05:04.130 --> 00:05:05.465
plus some constant.

00:05:05.465 --> 00:05:07.590
You could've put the constant
on that side as well.

00:05:07.590 --> 00:05:08.340
It doesn't matter.

00:05:08.340 --> 00:05:10.860
All the constants
would merge into one.

00:05:10.860 --> 00:05:13.150
Now we've gotten pretty far.

00:05:13.150 --> 00:05:16.950
We want to solve the
particular solution.

00:05:16.950 --> 00:05:18.760
Right now, we only
have this constant.

00:05:18.760 --> 00:05:22.350
And we also haven't written W as
a function of t just right yet.

00:05:22.350 --> 00:05:24.580
They're implicitly expressed.

00:05:24.580 --> 00:05:27.710
It is more of a relationship
than a function right now.

00:05:27.710 --> 00:05:29.480
So the first thing
we can do is try

00:05:29.480 --> 00:05:36.140
to solve for C. We're told
in the problem that W of 0

00:05:36.140 --> 00:05:40.590
is equal to 1400 tons.

00:05:40.590 --> 00:05:41.410
They told us that.

00:05:41.410 --> 00:05:43.610
That's one of our
initial conditions.

00:05:43.610 --> 00:05:49.010
So we know that the natural
log, that when t is equal to 0,

00:05:49.010 --> 00:05:50.990
W is equal to 1400.

00:05:50.990 --> 00:06:00.450
So W is equal to 1,400 when
t is equal to 0 plus C.

00:06:00.450 --> 00:06:02.580
Or another way to
think about it is,

00:06:02.580 --> 00:06:08.314
the natural log of the absolute
value of 1,100 is equal to C.

00:06:08.314 --> 00:06:09.980
And obviously, this
is a positive value.

00:06:09.980 --> 00:06:13.370
So we could just say that C
is equal to the natural log

00:06:13.370 --> 00:06:14.250
of 1,100.

00:06:14.250 --> 00:06:15.830
We can drop the
absolute value sign

00:06:15.830 --> 00:06:18.990
because the absolute
value of 1,100 is 1,100.

00:06:18.990 --> 00:06:22.270
So now we can rewrite
this part right over here.

00:06:22.270 --> 00:06:23.940
Let me do it in yellow.

00:06:23.940 --> 00:06:26.790
Let me start--
that's not yellow.

00:06:26.790 --> 00:06:28.090
Let me start right over here.

00:06:28.090 --> 00:06:31.050
So we could write
the natural log

00:06:31.050 --> 00:06:34.990
of the absolute
value of W minus 300.

00:06:34.990 --> 00:06:38.040
And actually, we know that
W is an increasing function.

00:06:38.040 --> 00:06:41.440
W starts at 1,400 and
only goes up from there

00:06:41.440 --> 00:06:43.450
as t gets larger and larger.

00:06:43.450 --> 00:06:45.500
So if we're starting at
1,400 and only getting

00:06:45.500 --> 00:06:48.420
larger and larger values
for W, this expression right

00:06:48.420 --> 00:06:53.180
here is never going to get
negative for positive t values.

00:06:53.180 --> 00:06:54.534
And that's all we care about.

00:06:54.534 --> 00:06:56.700
And so we can actually drop
the absolute value sign.

00:06:56.700 --> 00:06:58.440
This will always be
a positive value.

00:06:58.440 --> 00:07:04.580
So we could say the absolute
value of W minus 300

00:07:04.580 --> 00:07:07.380
is equal to t over
25-- and we've just

00:07:07.380 --> 00:07:12.590
solved for our constant--
plus the natural log of 1,100.

00:07:12.590 --> 00:07:15.030
And now we can just solve
this explicitly for W.

00:07:15.030 --> 00:07:18.900
We can raise e to both sides.

00:07:18.900 --> 00:07:20.855
This value is equal
to this value.

00:07:20.855 --> 00:07:22.980
The left hand side is equal
to the right hand side.

00:07:22.980 --> 00:07:25.370
So e raised to this power
is the same thing as e

00:07:25.370 --> 00:07:27.150
raised to that power.

00:07:27.150 --> 00:07:32.400
So we have e to the
natural log of W minus 300

00:07:32.400 --> 00:07:36.310
is going to be equal
to e to the t over 25

00:07:36.310 --> 00:07:39.600
plus the natural log of 1,100.

00:07:39.600 --> 00:07:44.050
Let me take this-- actually,
just to get the real estate,

00:07:44.050 --> 00:07:47.020
let me delete that.

00:07:47.020 --> 00:07:47.520
All right.

00:07:47.520 --> 00:07:49.830
Now I can keep working
right below that.

00:07:49.830 --> 00:07:52.060
And so e to the natural
log of something

00:07:52.060 --> 00:07:53.930
is just going to be
equal to that something.

00:07:53.930 --> 00:07:56.013
The natural log is what
power do I have to raise e

00:07:56.013 --> 00:07:57.850
to to get this thing.

00:07:57.850 --> 00:07:59.340
Well I'm raising
e to that power,

00:07:59.340 --> 00:08:00.640
so I'm going to get this thing.

00:08:00.640 --> 00:08:04.510
So the left-hand side
just becomes W minus 300,

00:08:04.510 --> 00:08:07.475
and the right-hand side,
we could rewrite this as e

00:08:07.475 --> 00:08:14.690
to the t over 25 times e to
the natural log of 1,100.

00:08:14.690 --> 00:08:17.352
We have the same base,
different exponents.

00:08:17.352 --> 00:08:18.810
If you wanted to
simplify this, you

00:08:18.810 --> 00:08:20.268
would just add the
exponents, which

00:08:20.268 --> 00:08:21.730
is exactly what we did up here.

00:08:21.730 --> 00:08:24.380
So these two things
are equivalent.

00:08:24.380 --> 00:08:26.820
e to the natural log of
1,100, that is 1,100.

00:08:26.820 --> 00:08:28.780
The natural log of
1,100 is a power

00:08:28.780 --> 00:08:31.280
that you raise e
to to get to 1,100.

00:08:31.280 --> 00:08:34.669
So if you raise e to that
power, you get 1,100.

00:08:34.669 --> 00:08:36.409
So that is 1,100.

00:08:36.409 --> 00:08:46.780
So we get W minus 300 is equal
to 1,100 e to the t over 25.

00:08:46.780 --> 00:08:49.320
And then you just add
300 to both sides.

00:08:49.320 --> 00:08:56.220
W is equal to 1,100 e to
the t over 25 plus 300.

00:08:56.220 --> 00:08:57.430
And we're done.

00:08:57.430 --> 00:08:59.400
We have solved the
differential equation.

00:08:59.400 --> 00:09:01.400
We found the
particular solution.

00:09:01.400 --> 00:09:04.170
We had to solve for C to find
that particular solution.

00:09:04.170 --> 00:09:04.670
We're done.

00:09:04.670 --> 00:09:06.590
And we did it using
just basic integration.

00:09:06.590 --> 00:09:09.330
We didn't have to use any
of the fancy differential

00:09:09.330 --> 00:09:11.690
equation solving
tools that you might

00:09:11.690 --> 00:09:14.690
learn in a more advanced class.


2011 Calculus AB free response #5b

WEBVTT
Kind: captions
Language: en

00:00:00.670 --> 00:00:01.770
Now ready for part b.

00:00:01.770 --> 00:00:04.200
And I've wiped all the
salt off of my fingers,

00:00:04.200 --> 00:00:05.450
and I've had a glass of water.

00:00:05.450 --> 00:00:07.360
So now I'm ready
for some business.

00:00:07.360 --> 00:00:10.640
Find the second derivative
of W with respect

00:00:10.640 --> 00:00:15.290
to t in terms of W. Use
this second derivative

00:00:15.290 --> 00:00:18.880
to determine whether your answer
in part a is an underestimate

00:00:18.880 --> 00:00:21.620
or overestimate for the
amount of solid waste

00:00:21.620 --> 00:00:25.490
that the landfill contains
at time t equals 1/4.

00:00:25.490 --> 00:00:27.100
So let's do the
first things first.

00:00:27.100 --> 00:00:30.440
Let's find the second
derivative in terms of W.

00:00:30.440 --> 00:00:32.627
And we already have the
first derivative over here.

00:00:32.627 --> 00:00:34.960
And in part a, I rewrote it
just with slightly different

00:00:34.960 --> 00:00:35.497
notation.

00:00:35.497 --> 00:00:37.830
But I'll just use this just
because I need to write down

00:00:37.830 --> 00:00:39.996
here and I can still
refer to this thing.

00:00:39.996 --> 00:00:41.370
So let's just take
the derivative

00:00:41.370 --> 00:00:44.350
of both sides of this
differential equation

00:00:44.350 --> 00:00:45.990
with respect to t.

00:00:45.990 --> 00:00:48.160
So you take the derivative
of this left-hand side

00:00:48.160 --> 00:00:50.180
with respect to t, you
get the second derivative

00:00:50.180 --> 00:00:57.070
of W with respect to t as a
function of t is equal to-- now

00:00:57.070 --> 00:00:59.570
you take the derivative
of this with respect to t.

00:00:59.570 --> 00:01:04.220
This is the same thing
as 1/25Wt minus 12.

00:01:04.220 --> 00:01:05.720
And so when you
take the derivative,

00:01:05.720 --> 00:01:07.620
that constant part drops
off, and you're just

00:01:07.620 --> 00:01:15.872
left with 1/25 times
the derivative of Wt.

00:01:15.872 --> 00:01:16.830
Let me make that clear.

00:01:16.830 --> 00:01:26.950
This is the same thing as 1/25W
as a function of t minus 12.

00:01:26.950 --> 00:01:29.370
1/25 of 300 is 12.

00:01:29.370 --> 00:01:30.910
And you take the
derivative of this,

00:01:30.910 --> 00:01:33.810
you get 1/25 times the
first derivative of W

00:01:33.810 --> 00:01:36.710
and then the derivative of this
with respect to t is just 0.

00:01:36.710 --> 00:01:39.710
A constant, obviously, does
not change with respect to t.

00:01:39.710 --> 00:01:41.445
And so we get this
right over here.

00:01:41.445 --> 00:01:42.820
Now this is the
second derivative

00:01:42.820 --> 00:01:45.072
in terms of the
first derivative.

00:01:45.072 --> 00:01:47.280
But the question asks us,
write the second derivative

00:01:47.280 --> 00:01:50.260
in terms of W. But
lucky for us, we

00:01:50.260 --> 00:01:54.950
know how to express
this as a function of W.

00:01:54.950 --> 00:01:57.036
They gave that to
us in the problem.

00:01:57.036 --> 00:01:57.660
This was given.

00:01:57.660 --> 00:02:00.159
This is just me rewriting
it in a different notation.

00:02:00.159 --> 00:02:01.700
So this is going to
be the same thing

00:02:01.700 --> 00:02:09.710
as 1/25 times the derivative
of W. And the derivative of W

00:02:09.710 --> 00:02:10.462
is this.

00:02:10.462 --> 00:02:11.920
The differential
equation literally

00:02:11.920 --> 00:02:15.030
tell us the derivative
of W is this over here.

00:02:15.030 --> 00:02:22.940
So 1 over 25 times Wt, the
function W as a function of t,

00:02:22.940 --> 00:02:26.510
minus 300.

00:02:26.510 --> 00:02:29.327
And so we can say the
second derivative--

00:02:29.327 --> 00:02:31.660
let me write it a little
smaller so I don't waste space.

00:02:31.660 --> 00:02:34.460
The second derivative
of W as a function of t

00:02:34.460 --> 00:02:42.840
is equal to 1/625 times
W as a function of t,

00:02:42.840 --> 00:02:45.110
which is a function
of t, minus 300.

00:02:45.110 --> 00:02:46.650
So we've done the first part.

00:02:46.650 --> 00:02:50.710
We've found out the
second derivative of W

00:02:50.710 --> 00:02:53.160
in terms of just W.

00:02:53.160 --> 00:02:55.980
Now let's try to address the
second part of their question.

00:02:55.980 --> 00:02:57.660
So we did the first part.

00:02:57.660 --> 00:03:00.220
Now use what we just
figured out to determine

00:03:00.220 --> 00:03:02.505
whether your answer in
part a is an underestimate

00:03:02.505 --> 00:03:05.490
or an overestimate of
the amount of solid waste

00:03:05.490 --> 00:03:08.000
that the landfill
contains at time 1/4.

00:03:08.000 --> 00:03:10.090
So in part a, we
found the slope.

00:03:10.090 --> 00:03:13.360
We found the slope of the
tangent line at time equals 0,

00:03:13.360 --> 00:03:16.010
and we used that slope
to extrapolate out

00:03:16.010 --> 00:03:20.070
to 3 months, or time
equals 1/4, 1/4 of a year.

00:03:20.070 --> 00:03:23.780
Now if the function
W's slope over that 1/4

00:03:23.780 --> 00:03:26.340
was exactly the same as the
slope of the tangent line,

00:03:26.340 --> 00:03:28.780
or if that slope did not
change, then our extrapolation

00:03:28.780 --> 00:03:30.480
would be exactly right.

00:03:30.480 --> 00:03:35.330
If W's slope is
increasing over that time,

00:03:35.330 --> 00:03:38.160
then our estimate would
be an underestimate

00:03:38.160 --> 00:03:39.470
of the correct thing.

00:03:39.470 --> 00:03:41.900
And if the slope is
decreasing over that time,

00:03:41.900 --> 00:03:44.990
then our estimate would
be an overestimate

00:03:44.990 --> 00:03:47.079
of the actual amount
in the landfill.

00:03:47.079 --> 00:03:49.120
And to figure out whether
the slope is increasing

00:03:49.120 --> 00:03:50.690
or decreasing, we
just have to look

00:03:50.690 --> 00:03:53.070
at the value of the
second derivative.

00:03:53.070 --> 00:03:55.300
If the second
derivative is positive,

00:03:55.300 --> 00:03:57.264
that means our
slope is increasing,

00:03:57.264 --> 00:03:58.930
which means that our
extrapolation would

00:03:58.930 --> 00:03:59.997
be an understatement.

00:03:59.997 --> 00:04:01.080
Let me just be clear here.

00:04:01.080 --> 00:04:02.380
Let me draw this.

00:04:02.380 --> 00:04:04.500
Let me make it very clear.

00:04:04.500 --> 00:04:07.510
So let me just draw
a random function.

00:04:07.510 --> 00:04:13.760
So let's say that's W. So this
is a case where W's slope is

00:04:13.760 --> 00:04:16.899
increasing faster,
or W's slope is

00:04:16.899 --> 00:04:18.709
increasing from
that starting point.

00:04:18.709 --> 00:04:20.990
So our starting point--
this was our slope.

00:04:20.990 --> 00:04:24.060
And then W's slope keeps
increasing from there.

00:04:24.060 --> 00:04:25.640
And in this case,
our estimate is

00:04:25.640 --> 00:04:28.900
going to be underestimate
of where W actually

00:04:28.900 --> 00:04:30.840
is after 1/4 of the year.

00:04:30.840 --> 00:04:36.230
If W's slope is exactly the same
as our function over the course

00:04:36.230 --> 00:04:39.470
of the first three months, it
would look something like that.

00:04:39.470 --> 00:04:41.300
Maybe it diverges later on.

00:04:41.300 --> 00:04:43.490
And in this case,
our approximation

00:04:43.490 --> 00:04:44.850
would be really good.

00:04:44.850 --> 00:04:47.380
It would probably
be exact for W.

00:04:47.380 --> 00:04:50.010
And if W's slope,
for whatever reason,

00:04:50.010 --> 00:04:53.160
goes negative after that point,
and they already tells us

00:04:53.160 --> 00:04:57.170
it's an increasing function,
so that is not likely.

00:04:57.170 --> 00:04:59.580
Or it could-- it doesn't
have to even go negative.

00:04:59.580 --> 00:05:03.139
If W's slope decreases, it
could still stay positive.

00:05:03.139 --> 00:05:05.430
And the way I drew it, it
doesn't-- let me draw it like

00:05:05.430 --> 00:05:07.690
this, just to make it clear.

00:05:07.690 --> 00:05:11.670
So let's say we find out that
the slope looks something

00:05:11.670 --> 00:05:13.790
like this.

00:05:13.790 --> 00:05:17.380
This is the slope of the
tangent line at time equals 0.

00:05:17.380 --> 00:05:20.000
If W's slope increases
from that point,

00:05:20.000 --> 00:05:22.440
then w might look
something like that.

00:05:22.440 --> 00:05:27.970
And then our answer to part
a would be an understatement,

00:05:27.970 --> 00:05:30.500
would be an underestimate
of where W actually

00:05:30.500 --> 00:05:31.930
is after a fourth of a year.

00:05:31.930 --> 00:05:35.140
Let me draw it and make it
clear to you what I'm doing.

00:05:35.140 --> 00:05:39.970
So this is my w-axis.

00:05:39.970 --> 00:05:43.320
So this is right at our
initial condition of 1,400.

00:05:43.320 --> 00:05:46.350
And then this right over
here is our time axis,

00:05:46.350 --> 00:05:48.120
and this is that 1/4.

00:05:48.120 --> 00:05:49.960
So in the last
video, we said, hey,

00:05:49.960 --> 00:05:53.200
this is sitting right at 1,411.

00:05:53.200 --> 00:05:55.660
If W's slope increases
from that point,

00:05:55.660 --> 00:05:57.040
then this is an underestimate.

00:05:57.040 --> 00:05:59.540
If W's slope stays
the same, then this

00:05:59.540 --> 00:06:01.880
is actually a very good
estimate because then we're

00:06:01.880 --> 00:06:04.950
going to hit that
point directly with W.

00:06:04.950 --> 00:06:08.160
And if W's slope decreases--
so you can imagine maybe

00:06:08.160 --> 00:06:11.920
W looks something like this.

00:06:11.920 --> 00:06:14.480
It has that slope of the
tangent line when it starts,

00:06:14.480 --> 00:06:16.630
but then the slope decreases.

00:06:16.630 --> 00:06:18.280
And it's still an
increasing function,

00:06:18.280 --> 00:06:19.446
but the slope is decreasing.

00:06:19.446 --> 00:06:23.580
And in that case, we would
have an overestimate.

00:06:23.580 --> 00:06:27.220
And this is a situation--
so this first situation,

00:06:27.220 --> 00:06:28.490
slope is increasing.

00:06:28.490 --> 00:06:34.051
That means W prime
prime is positive.

00:06:34.051 --> 00:06:35.050
The slope is increasing.

00:06:35.050 --> 00:06:38.270
The second derivative
is positive.

00:06:38.270 --> 00:06:39.830
This would be a
byproduct of or this

00:06:39.830 --> 00:06:43.590
would cause the second
derivative to be negative.

00:06:43.590 --> 00:06:45.500
Our slope is decreasing.

00:06:45.500 --> 00:06:47.520
And this would be our
second derivative.

00:06:47.520 --> 00:06:50.130
Our second derivative is 0.

00:06:50.130 --> 00:06:53.089
If your slope isn't changing--
if your slope is constant,

00:06:53.089 --> 00:06:54.630
if your first
derivative is constant,

00:06:54.630 --> 00:06:57.360
your second derivative
is going to be 0.

00:06:57.360 --> 00:06:59.880
So let's just see what
our second derivative is

00:06:59.880 --> 00:07:01.600
at our initial
condition, and then we'll

00:07:01.600 --> 00:07:03.391
have a pretty good
sense of whether we have

00:07:03.391 --> 00:07:06.300
an overestimate or an
underestimate for part a.

00:07:06.300 --> 00:07:08.830
So let's just figure out
what our second derivative

00:07:08.830 --> 00:07:10.790
at time 0.

00:07:10.790 --> 00:07:17.880
It's going to be equal to 1/625
times W of 0, which we know,

00:07:17.880 --> 00:07:19.670
minus 300.

00:07:19.670 --> 00:07:22.690
Well W of 0, the amount of
waste we have at time 0,

00:07:22.690 --> 00:07:26.830
they told us in the
problem, is 1,400 tons.

00:07:26.830 --> 00:07:30.940
1,400 minus 1,300 is 1,100.

00:07:30.940 --> 00:07:33.970
And then 1,100 minus--
or 1,100 divided by 625

00:07:33.970 --> 00:07:35.010
is a small number.

00:07:35.010 --> 00:07:35.580
It's 1.

00:07:35.580 --> 00:07:36.250
something.

00:07:36.250 --> 00:07:37.900
But it is a positive number.

00:07:37.900 --> 00:07:39.910
And that's the
important thing here.

00:07:39.910 --> 00:07:44.770
So this thing, all of this
business, it is positive.

00:07:44.770 --> 00:07:47.130
So the second
derivative is positive,

00:07:47.130 --> 00:07:49.040
which means the
slope is increasing.

00:07:49.040 --> 00:07:53.220
At least right at our starting
point, our slope is increasing.

00:07:53.220 --> 00:07:55.875
Since our slope is increasing
from that point-- and frankly,

00:07:55.875 --> 00:07:57.375
the fact that it's
increasing at all

00:07:57.375 --> 00:07:59.810
and that we know the W is
an increasing function--

00:07:59.810 --> 00:08:04.220
tells us that our estimate in
part a is an underestimate.

00:08:04.220 --> 00:08:07.290
So this is the case
that we're dealing with.

00:08:07.290 --> 00:08:11.910
So our estimate in part
a is an underestimate

00:08:11.910 --> 00:08:15.300
of the actual amount of solid
waste in the landfill at time t

00:08:15.300 --> 00:08:17.009
equals 1/4.


2011 Calculus AB free response #5a

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:01.890
Problem 5.

00:00:01.890 --> 00:00:04.640
"At the beginning
of 2010, a landfill

00:00:04.640 --> 00:00:08.140
contained 1,400
tons of solid waste.

00:00:08.140 --> 00:00:10.750
The increasing
function W--" so I

00:00:10.750 --> 00:00:13.710
guess W constantly increases--
"the increasing function

00:00:13.710 --> 00:00:16.790
W models the total amount
of solid waste stored

00:00:16.790 --> 00:00:18.210
at the landfill.

00:00:18.210 --> 00:00:21.360
Planners estimate that W
will satisfy the differential

00:00:21.360 --> 00:00:24.720
equation." so the derivative
of W with respect to time

00:00:24.720 --> 00:00:29.740
is equal to 1 over 25 times
the quantity W minus 300,

00:00:29.740 --> 00:00:32.250
"for the next 20 years.

00:00:32.250 --> 00:00:33.890
W is measured in tons.

00:00:33.890 --> 00:00:37.580
t is measured in years
from the start of 2010."

00:00:37.580 --> 00:00:39.540
All right, let's get into this.

00:00:39.540 --> 00:00:41.370
So part a.

00:00:41.370 --> 00:00:45.450
"Use the tangent line
to the graph of W at t

00:00:45.450 --> 00:00:48.890
equals 0 to approximate
the amount of solid waste

00:00:48.890 --> 00:00:52.440
that the landfill contains at
the end of the first 3 months

00:00:52.440 --> 00:00:54.050
of 2010."

00:00:54.050 --> 00:00:58.270
So since time is in years, 3
months would be 1/4 of a year.

00:00:58.270 --> 00:01:00.115
So t equals 1/4.

00:01:00.115 --> 00:01:01.990
So it seems at first
this is really daunting.

00:01:01.990 --> 00:01:03.790
They gave us a
differential equation.

00:01:03.790 --> 00:01:07.060
We don't even know what
the actual function W is.

00:01:07.060 --> 00:01:08.920
How do we figure out
its tangent line?

00:01:08.920 --> 00:01:10.294
But we just have
to kind of think

00:01:10.294 --> 00:01:11.410
about what they're asking.

00:01:11.410 --> 00:01:13.370
So regardless of what
W looks like-- so let's

00:01:13.370 --> 00:01:14.540
think about it a little bit.

00:01:14.540 --> 00:01:20.120
So we don't know yet what
W actually does look like.

00:01:20.120 --> 00:01:22.460
But they tell us it's
an increasing function.

00:01:22.460 --> 00:01:25.760
So at time 0 it has
1,400 tons in it.

00:01:25.760 --> 00:01:27.910
They tell us that right up here.

00:01:27.910 --> 00:01:29.227
And that it increases.

00:01:29.227 --> 00:01:31.560
We don't know what the function
actually does look like.

00:01:31.560 --> 00:01:33.590
But let's say that
that is W. So what

00:01:33.590 --> 00:01:37.105
they're saying in
problem a is use

00:01:37.105 --> 00:01:40.240
a tangent line to the
graph at t equals 0.

00:01:40.240 --> 00:01:42.650
Let me draw a W a
little bit differently.

00:01:42.650 --> 00:01:46.069
So W might look like this.

00:01:46.069 --> 00:01:48.110
So what they're saying
is, find the tangent line,

00:01:48.110 --> 00:01:50.860
find the slope of that
tangent line at t equals 0.

00:01:50.860 --> 00:01:53.370
So there's some slope
right over there.

00:01:53.370 --> 00:01:57.750
And then we can use
this slope to create

00:01:57.750 --> 00:01:59.310
a linear approximation
of where we're

00:01:59.310 --> 00:02:03.080
going to be a quarter
of a year from then.

00:02:03.080 --> 00:02:05.910
So although we don't
know what W is just yet,

00:02:05.910 --> 00:02:09.220
we could take the slope
of this line out--

00:02:09.220 --> 00:02:12.370
so this is our t axis--
to t is equal to 1/4.

00:02:12.370 --> 00:02:16.330
And wherever that line takes us
out after a fourth of a year,

00:02:16.330 --> 00:02:18.380
that will be at least
a decent approximation.

00:02:18.380 --> 00:02:22.290
We're extrapolating forward
from that first point

00:02:22.290 --> 00:02:23.791
and the current slope.

00:02:23.791 --> 00:02:26.290
So we really just have to figure
out the slope of this line,

00:02:26.290 --> 00:02:29.110
and just see where that line
is at t is equal to 1/4.

00:02:29.110 --> 00:02:30.680
And you say, wait.

00:02:30.680 --> 00:02:33.370
How do we know what
the derivative of W

00:02:33.370 --> 00:02:34.772
is at 0 without knowing W?

00:02:34.772 --> 00:02:36.230
And that's where
we can go straight

00:02:36.230 --> 00:02:37.540
to this differential equation.

00:02:37.540 --> 00:02:39.706
We could actually rewrite
this differential equation

00:02:39.706 --> 00:02:41.540
using slightly
different notation.

00:02:41.540 --> 00:02:43.570
This is the derivative
of W with respect to t.

00:02:43.570 --> 00:02:48.860
We could write that as W
prime of t is equal to 1

00:02:48.860 --> 00:02:54.960
over 25 times the function
W, which is a function of t,

00:02:54.960 --> 00:02:56.635
minus 300.

00:02:56.635 --> 00:02:58.010
And when you look
at it this way,

00:02:58.010 --> 00:02:59.593
it becomes a little
bit clearer on how

00:02:59.593 --> 00:03:02.400
to figure out what the
derivative of W is at 0.

00:03:05.479 --> 00:03:07.770
The slope of that line
literally is just the derivative

00:03:07.770 --> 00:03:10.130
of W evaluated at 0.

00:03:10.130 --> 00:03:12.785
So let's literally take
the derivative of W

00:03:12.785 --> 00:03:15.050
and evaluate it at 0.

00:03:15.050 --> 00:03:18.920
So we have W prime
of 0 is equal to 1

00:03:18.920 --> 00:03:23.090
over 25 times W of
0, which we know.

00:03:23.090 --> 00:03:25.720
We know this is 1,400
tons of solid waste.

00:03:25.720 --> 00:03:29.400
This is how much waste
there is at time 0.

00:03:29.400 --> 00:03:30.640
Minus 300.

00:03:30.640 --> 00:03:33.450
So this right over
here is 1,400.

00:03:33.450 --> 00:03:36.550
And so we have W prime
at time equals 0.

00:03:36.550 --> 00:03:39.820
So our slope at time equals
0, or our derivative at time

00:03:39.820 --> 00:03:48.778
equals 0, is equal to 1 over 25
times 1,400 minus 300 is 1,100.

00:03:48.778 --> 00:03:57.050
And 25 goes into
1,100 44 times, right?

00:03:57.050 --> 00:03:58.780
It goes four times
into each 100.

00:03:58.780 --> 00:03:59.970
We have 11 hundreds here.

00:03:59.970 --> 00:04:01.780
So it'll go 44 times.

00:04:01.780 --> 00:04:03.820
So it goes 44.

00:04:03.820 --> 00:04:10.040
So the slope of that line is 44.

00:04:10.040 --> 00:04:11.200
We could say m for slope.

00:04:11.200 --> 00:04:13.300
Or actually, let me just
write down the word.

00:04:13.300 --> 00:04:16.575
The slope of this line is 44.

00:04:16.575 --> 00:04:18.644
And I just ate some
peanuts or something.

00:04:18.644 --> 00:04:19.810
So my voice is a little dry.

00:04:19.810 --> 00:04:20.660
So bear with me.

00:04:20.660 --> 00:04:23.530
But the slope of
this line is 24.

00:04:23.530 --> 00:04:25.847
So how do we use that
to find an approximation

00:04:25.847 --> 00:04:27.930
for the amount of waste
that the landfill contains

00:04:27.930 --> 00:04:29.860
at the end of the
first three months?

00:04:29.860 --> 00:04:31.430
Well, let me zoom
in a little bit.

00:04:31.430 --> 00:04:33.040
And actually, my fingers
are all salty, too.

00:04:33.040 --> 00:04:34.956
So maybe I don't have a
proper grip on my pen.

00:04:34.956 --> 00:04:37.460
But I'll try my best.

00:04:37.460 --> 00:04:39.450
Let me just zoom in
a little bit more.

00:04:39.450 --> 00:04:41.220
So we're starting at 1,400 tons.

00:04:41.220 --> 00:04:42.840
So this is my W-axis.

00:04:42.840 --> 00:04:44.530
This is my time axis.

00:04:44.530 --> 00:04:46.950
We're starting at 1,400 tons.

00:04:46.950 --> 00:04:48.400
And we are increasing
from there.

00:04:48.400 --> 00:04:51.120
They're telling us that it
is an increasing function.

00:04:51.120 --> 00:04:53.720
So maybe it looks
something like that.

00:04:53.720 --> 00:04:57.190
And then our slope
right over here.

00:04:57.190 --> 00:04:58.790
And I don't know,
I haven't drawn W

00:04:58.790 --> 00:05:00.940
exactly that accurately.

00:05:00.940 --> 00:05:03.480
I'm just guessing what it
might look like at this point.

00:05:03.480 --> 00:05:07.410
The tangent line
has a slope of 44.

00:05:07.410 --> 00:05:09.320
So that is our tangent line.

00:05:09.320 --> 00:05:13.890
And what that's
saying is if we go out

00:05:13.890 --> 00:05:16.920
one unit of time,
which is one year,

00:05:16.920 --> 00:05:19.770
then we would have
gone up 44 in tonnage.

00:05:19.770 --> 00:05:24.610
So if we use this line as an
approximation, after one year,

00:05:24.610 --> 00:05:27.482
at this point, would
have 1,444 tons.

00:05:27.482 --> 00:05:29.440
But we're not trying to
approximate a year out.

00:05:29.440 --> 00:05:32.710
We're trying to approximate
1/4 of a year out.

00:05:32.710 --> 00:05:34.366
So we're trying to approximate.

00:05:34.366 --> 00:05:35.740
So this would be
half a year out.

00:05:35.740 --> 00:05:40.500
This is 1/4 of a year out.

00:05:40.500 --> 00:05:43.580
We're trying to approximate
that point right over there.

00:05:43.580 --> 00:05:46.397
So it's going to be 1,400,
this point right over here.

00:05:46.397 --> 00:05:48.480
And we could write down
the equation of this line,

00:05:48.480 --> 00:05:49.190
if we like.

00:05:49.190 --> 00:05:52.400
We could say this
line-- so we'll

00:05:52.400 --> 00:05:56.500
call this the W approximation,
because this isn't exactly

00:05:56.500 --> 00:06:01.010
our W function-- this is equal
to the slope of our line, 44,

00:06:01.010 --> 00:06:04.110
times time, plus
our W intercept,

00:06:04.110 --> 00:06:08.410
we could say, or plus our
initial condition, plus 1,400.

00:06:08.410 --> 00:06:11.220
So if you put time is
equal to 1/4 in there,

00:06:11.220 --> 00:06:13.290
you get it equal to 44.

00:06:13.290 --> 00:06:16.030
So let me write it this way.

00:06:16.030 --> 00:06:20.530
So our approximate W, at time
is equal to 1/4 of a year,

00:06:20.530 --> 00:06:25.840
is equal to 44 times
1/4 plus 1,400.

00:06:25.840 --> 00:06:27.200
I'm running out of space.

00:06:27.200 --> 00:06:30.410
And my salty hands are having
trouble writing this properly.

00:06:30.410 --> 00:06:36.510
And so 44 times 1/4, or 44
divided by 4, is 11 plus 1,400.

00:06:36.510 --> 00:06:37.740
And so you add them together.

00:06:37.740 --> 00:06:41.890
You get 1,411 or 1,411 tons.

00:06:41.890 --> 00:06:43.050
This is our approximation.

00:06:43.050 --> 00:06:45.440
We just took the slope
from our starting point

00:06:45.440 --> 00:06:47.650
and used that slope
as an approximation.

00:06:47.650 --> 00:06:49.670
It probably is not
the exact amount

00:06:49.670 --> 00:06:51.850
of tonnage based on
the actual function W.

00:06:51.850 --> 00:06:53.850
But it's an OK approximation.

00:06:53.850 --> 00:06:59.360
But that's our first answer
for Part A. 1,411 tons.


2011 Calculus AB free response #4d

WEBVTT
Kind: captions
Language: en

00:00:00.700 --> 00:00:07.180
Part d, find the average rate of change of
f on the interval from x is greater than

00:00:07.180 --> 00:00:09.590
or equal, negative four is less than or
equal

00:00:09.590 --> 00:00:11.100
to x, which is less than or equal to
three.

00:00:11.100 --> 00:00:13.920
So really the interval that they've
depicted right over here.

00:00:13.920 --> 00:00:17.360
And then they say there's no point c
between in that interval for

00:00:17.360 --> 00:00:20.350
which f prime of c is equal to the average
rate of change.

00:00:20.350 --> 00:00:24.490
Explain why the statement does not
contradict the Mean Value theorem.

00:00:24.490 --> 00:00:25.620
Fascinating.

00:00:25.620 --> 00:00:25.840
Alright.

00:00:25.840 --> 00:00:27.520
Let's do this first part first.

00:00:27.520 --> 00:00:30.730
The average rate of change of f on the
interval.

00:00:30.730 --> 00:00:34.270
Sounds like a very fancy thing, but the
average rate of change on the interval

00:00:34.270 --> 00:00:38.492
is really just a slope of the line that
connects the end points of the interval.

00:00:38.492 --> 00:00:41.060
So these, this right over here, those are
the end

00:00:41.060 --> 00:00:43.990
points, and let's figure out the slope of
that line.

00:00:43.990 --> 00:00:48.630
So, going we, going from this point to
that point, our change

00:00:48.630 --> 00:00:51.820
in x, we do this in a color that you're
likely to see.

00:00:51.820 --> 00:00:55.430
Our change in x over here, our change in
x, change in x,

00:00:57.490 --> 00:00:58.820
is equal to 7.

00:00:58.820 --> 00:01:02.920
And you can get that by taking 3 minus
negative 4 or

00:01:02.920 --> 00:01:07.670
you could literally just count, 1, 2, 3,
4, 5, 6, 7.

00:01:07.670 --> 00:01:09.060
That's our change in x.

00:01:09.060 --> 00:01:14.350
And our change in y, when we, when we run
over 7, when we go 7

00:01:14.350 --> 00:01:19.540
to the right, our change in y, our change
in y is equal to negative 2.

00:01:19.540 --> 00:01:23.780
We went from negative 1 to negative 3 is
equal to negative 2.

00:01:23.780 --> 00:01:30.580
So, slope, slope, which is change in y
over change in x, rise over run.

00:01:30.580 --> 00:01:37.080
Is equal to 7 over negative 2, or well
actually let me, the other way around.

00:01:37.080 --> 00:01:40.230
The change in y is negative 2 over 7.

00:01:40.230 --> 00:01:44.040
So negative two 7s and that's all it is.

00:01:44.040 --> 00:01:46.200
Now you could, you could kind of.

00:01:46.200 --> 00:01:47.620
Think of it in, in, kind of a more

00:01:47.620 --> 00:01:50.050
fancy sense, but you're gonna get the
exact same answer

00:01:50.050 --> 00:01:53.110
if you said, oh well look, you know, the
average,

00:01:53.110 --> 00:01:55.170
the average rate of change of f on the
interval.

00:01:55.170 --> 00:01:59.560
Well, the rate of change of f is f prime,
f prime of x.

00:01:59.560 --> 00:02:02.600
That is the rate of change of f at any
point x.

00:02:02.600 --> 00:02:04.190
And so if you want to find the average
value

00:02:04.190 --> 00:02:07.090
of this over the interval, you would
integrate it from

00:02:07.090 --> 00:02:11.680
our starting point, from negative 4 to 3
dx, and

00:02:11.680 --> 00:02:13.900
then you would divide it by your change in
x.

00:02:13.900 --> 00:02:17.930
So then you would divide it by, you would
have it over 1 over 7.

00:02:17.930 --> 00:02:19.350
Over 1 over 7.

00:02:19.350 --> 00:02:22.450
But then this, this part right over here
is just going to

00:02:22.450 --> 00:02:26.130
be the same thing as f of 3 minus f of 4.

00:02:26.130 --> 00:02:29.700
And then this over here you have a 7 in
the denominator.

00:02:29.700 --> 00:02:32.290
So this is really just your change in x.

00:02:32.290 --> 00:02:35.268
This is your change in x which is, which
was by definition

00:02:35.268 --> 00:02:38.940
or how we, how we actually set up this
average right over here.

00:02:38.940 --> 00:02:41.020
And this over here is really just your
change in y.

00:02:41.020 --> 00:02:44.110
So it really is just a slope between the
end points.

00:02:44.110 --> 00:02:45.680
So, we did the first part.

00:02:45.680 --> 00:02:50.840
Our average rate of change of f on the
interval is negative two 7s.

00:02:50.840 --> 00:02:52.910
And then let's think about the second
part.

00:02:52.910 --> 00:02:55.290
You see there is no point c in that
interval.

00:02:55.290 --> 00:02:58.440
For which f prime of c is equal to the
average rate of change.

00:02:58.440 --> 00:03:02.980
Explain why this statement does not
contradict the Mean Value Theorem.

00:03:02.980 --> 00:03:06.520
So the Mean Value Theorem, just as a
little bit of a review, as a

00:03:06.520 --> 00:03:10.050
little bit of review, it says that if we
have some type of an interval.

00:03:10.050 --> 00:03:11.650
If you have an interval.

00:03:11.650 --> 00:03:15.590
So, let me draw some axes right over here.

00:03:15.590 --> 00:03:17.550
If you have an interval.

00:03:17.550 --> 00:03:19.230
Let me draw an interval like this.

00:03:21.160 --> 00:03:24.720
And over that interval you have a
differentiable function.

00:03:24.720 --> 00:03:28.800
You have a differentiable function, so
maybe my function looks like this.

00:03:28.800 --> 00:03:33.470
It says at least, there is at least one
point c on that interval

00:03:33.470 --> 00:03:37.830
where the derivative at that point c is
equal to the average rate of change.

00:03:37.830 --> 00:03:40.500
So the way I've drawn it right over here,
the average rate of

00:03:40.500 --> 00:03:44.440
change of this function, I'll do it in
magenta, is this right over here.

00:03:45.570 --> 00:03:48.100
And the Mean Value Theorem says, is it, if
this

00:03:48.100 --> 00:03:51.210
is deferentiable, there's at least one
point c in this interval.

00:03:51.210 --> 00:03:52.630
Where I have the same slope.

00:03:52.630 --> 00:03:54.390
Where the tangent line has the same slope.

00:03:54.390 --> 00:03:58.800
Where the derivative is the same as the
slope, as that average slope.

00:03:58.800 --> 00:04:01.450
And you can see here, is probably right
over here you have

00:04:01.450 --> 00:04:04.330
one of those points and actually we
probably have multiple of them.

00:04:04.330 --> 00:04:06.210
We probably have another point right over
here that's

00:04:06.210 --> 00:04:09.600
like that, and then another point there
that's like that.

00:04:09.600 --> 00:04:11.680
And if you think about it, it's kind of.

00:04:11.680 --> 00:04:13.620
It's kind of intuitive that at some point,

00:04:13.620 --> 00:04:15.700
you know, here we're, we obviously have a
larger

00:04:15.700 --> 00:04:17.800
slope and over we have a smaller slope and

00:04:17.800 --> 00:04:21.269
since our, since it's differentialable,
our derivative is continuous.

00:04:21.269 --> 00:04:23.410
So at some point, at some point the slope

00:04:23.410 --> 00:04:27.150
has to get to exactly what the average
slope is.

00:04:27.150 --> 00:04:28.320
Now, let's think about our little

00:04:28.320 --> 00:04:30.540
conundrum with this question right over
here.

00:04:30.540 --> 00:04:33.510
Why is there not a point c for which f

00:04:33.510 --> 00:04:36.020
prime of c is equal to the average rate of
change?

00:04:36.020 --> 00:04:38.020
You can even verify that for yourself
because

00:04:38.020 --> 00:04:41.470
from negative 4 to 0 our slope is
positive.

00:04:41.470 --> 00:04:46.460
We have a positive slope here, and then
the slope just jumps down to negative 2.

00:04:46.460 --> 00:04:49.690
It just jumps down to negative 2 which is
a much more negative slope than this.

00:04:49.690 --> 00:04:53.970
So, it never goes to negative 2 7s, and
the reason is that.

00:04:53.970 --> 00:04:57.440
This is not a differentiable function at x
is equal to 0.

00:04:57.440 --> 00:04:59.600
It is not differentiable at x equals 0.

00:04:59.600 --> 00:05:01.920
Our slope jumps here and because it's

00:05:01.920 --> 00:05:05.140
not differentiable the Mean Value theorem
doesn't apply.

00:05:05.140 --> 00:05:07.010
Now you can imagine if this was
differentiable, if this

00:05:07.010 --> 00:05:10.400
did have a continuous derivative, then you
would find a point.

00:05:10.400 --> 00:05:14.680
So if this looks something like this
instead, I'll continue it over here.

00:05:14.680 --> 00:05:17.540
If it looks something like this instead

00:05:17.540 --> 00:05:20.020
where it was, it had a continuous
derivative

00:05:20.020 --> 00:05:23.970
then there would be a point where the
slope was the same as the average slope.

00:05:23.970 --> 00:05:26.310
Maybe it would have been right over there.

00:05:26.310 --> 00:05:28.880
So it's really because, it's really
because

00:05:28.880 --> 00:05:32.040
f is not differentiable over the entire
interval.

00:05:32.040 --> 00:05:34.380
It does not have a continuous derivative.

00:05:34.380 --> 00:05:38.630
The derivative jumps from a, from a
positive value approaching 0 here

00:05:38.630 --> 00:05:42.400
and it jumps just straight down to
negative 2 right over here.

00:05:42.400 --> 00:05:44.839
It doesn't go continuously through all the
values in between.


2011 Calculus AB free response #4c

WEBVTT
Kind: captions
Language: en

00:00:00.640 --> 00:00:04.420
Part C. Find all values of
x on the interval negative 4

00:00:04.420 --> 00:00:07.670
is less than x is less than
3 for which the graph of g

00:00:07.670 --> 00:00:09.580
has a point of inflection.

00:00:09.580 --> 00:00:11.460
Give a reason for your answer.

00:00:11.460 --> 00:00:15.660
So an inflection
point is a point

00:00:15.660 --> 00:00:24.220
where the sign of the
second derivative changes.

00:00:24.220 --> 00:00:26.560
So if you take the second
derivative at that point,

00:00:26.560 --> 00:00:29.345
or as we go close to that point,
or as we cross that point,

00:00:29.345 --> 00:00:30.720
it goes from
positive to negative

00:00:30.720 --> 00:00:32.390
or negative to positive.

00:00:32.390 --> 00:00:34.110
And to think about
that visually,

00:00:34.110 --> 00:00:35.920
you could think
of some examples.

00:00:35.920 --> 00:00:42.030
So if you have a curve that
looks something like this,

00:00:42.030 --> 00:00:44.820
you'll notice that over
here the slope is negative,

00:00:44.820 --> 00:00:46.040
but it's increasing.

00:00:46.040 --> 00:00:47.980
It's getting less
negative, less negative.

00:00:47.980 --> 00:00:48.740
Then it goes to 0.

00:00:48.740 --> 00:00:49.781
Then it keeps increasing.

00:00:49.781 --> 00:00:53.110
Slope is increasing, increasing,
all the way to there,

00:00:53.110 --> 00:00:55.690
and then it starts
getting less positive.

00:00:55.690 --> 00:00:56.720
So it starts decreasing.

00:00:56.720 --> 00:00:58.100
So it's increasing.

00:00:58.100 --> 00:01:00.739
The slope is increasing over
at this point right over here.

00:01:00.739 --> 00:01:02.280
So even though the
slope is negative,

00:01:02.280 --> 00:01:03.821
it's getting less
negative over here.

00:01:03.821 --> 00:01:04.989
So it's increasing.

00:01:04.989 --> 00:01:06.730
And then the slope
keeps increasing.

00:01:06.730 --> 00:01:08.480
It keeps on getting
more and more positive

00:01:08.480 --> 00:01:09.470
up to about this point.

00:01:09.470 --> 00:01:11.261
And then the slope is
positive, but then it

00:01:11.261 --> 00:01:12.340
becomes less positive.

00:01:12.340 --> 00:01:14.840
So the slope begins
decreasing after that.

00:01:18.130 --> 00:01:21.320
So this right over here
is a point of inflection.

00:01:21.320 --> 00:01:25.427
The slope has gone from
increasing to decreasing.

00:01:25.427 --> 00:01:27.510
And if the other thing
happened, if the slope went

00:01:27.510 --> 00:01:28.680
from decreasing to
increasing, that

00:01:28.680 --> 00:01:30.180
would also be a
point of inflection.

00:01:30.180 --> 00:01:32.710
So if this was maybe some
type of a trigonometric curve,

00:01:32.710 --> 00:01:34.450
then you might see
something like this.

00:01:34.450 --> 00:01:39.380
And so this also would it
be a point of inflection.

00:01:39.380 --> 00:01:43.030
But for this, our g of x is kind
of hard to visualize the way

00:01:43.030 --> 00:01:44.702
they've defined it
right over here.

00:01:44.702 --> 00:01:46.410
So the best way to
think about it is just

00:01:46.410 --> 00:01:49.790
figure out where its second
derivative has a sign change.

00:01:49.790 --> 00:01:53.450
And to think about that, we have
to find its second derivative.

00:01:53.450 --> 00:01:55.490
So let's write g of x over here.

00:01:55.490 --> 00:01:59.070
We know g of x is equal to
2x plus the definite integral

00:01:59.070 --> 00:02:02.572
from 0 to x of f of t dt.

00:02:02.572 --> 00:02:04.030
We've already taken
its derivative,

00:02:04.030 --> 00:02:05.010
but we'll do it again.

00:02:05.010 --> 00:02:10.430
g prime of x is equal to 2
plus-- fundamental theorem

00:02:10.430 --> 00:02:11.130
of calculus.

00:02:11.130 --> 00:02:14.920
The derivative of this right
over here is just f of x.

00:02:14.920 --> 00:02:17.950
And if we have the second
derivative of g-- g prime prime

00:02:17.950 --> 00:02:21.010
of x-- this is equal to--
derivative of 2 is just 0.

00:02:21.010 --> 00:02:25.090
And then derivative of
f of x is f prime of x.

00:02:25.090 --> 00:02:28.790
So asking where this has a
sign change, asking where

00:02:28.790 --> 00:02:30.820
our second derivative
has a sign change,

00:02:30.820 --> 00:02:35.180
is equivalent to asking where
does the first derivative of f

00:02:35.180 --> 00:02:38.610
have a sign change?

00:02:38.610 --> 00:02:41.230
And asking where the
first derivative of f

00:02:41.230 --> 00:02:44.540
has a sign change is
equivalent to saying

00:02:44.540 --> 00:02:47.980
where does the slope of
f have a sign change?

00:02:47.980 --> 00:02:52.900
You can view this as the slope
or the instantaneous slope

00:02:52.900 --> 00:02:53.490
of f.

00:02:53.490 --> 00:02:57.210
So we want to know when the
slope of f has a sign change.

00:02:57.210 --> 00:02:58.330
So let's think about.

00:02:58.330 --> 00:03:00.550
Over here, the
slope is positive.

00:03:00.550 --> 00:03:01.213
It's going up.

00:03:01.213 --> 00:03:02.432
It's up.

00:03:02.432 --> 00:03:03.890
It's increasing,
but it's positive.

00:03:03.890 --> 00:03:05.050
And that's what we care about.

00:03:05.050 --> 00:03:05.600
So let's write it.

00:03:05.600 --> 00:03:06.433
I'll do it in green.

00:03:06.433 --> 00:03:10.040
So the slope is positive
this entire time.

00:03:10.040 --> 00:03:10.780
It's increasing.

00:03:10.780 --> 00:03:11.260
It's increasing.

00:03:11.260 --> 00:03:11.843
It's positive.

00:03:11.843 --> 00:03:13.570
It's getting less positive now.

00:03:13.570 --> 00:03:16.820
It's starting to decrease, but
the slope is still positive.

00:03:16.820 --> 00:03:19.510
The slope is still
positive all the way

00:03:19.510 --> 00:03:20.882
until we get right over there.

00:03:20.882 --> 00:03:22.590
You can see it gets
pretty close to zero.

00:03:22.590 --> 00:03:24.220
And then the slope
gets a negative.

00:03:24.220 --> 00:03:26.320
And then right over here,
the slope is negative.

00:03:29.520 --> 00:03:31.530
The slope is negative
right over here.

00:03:31.530 --> 00:03:32.530
So this is interesting.

00:03:32.530 --> 00:03:36.140
Because even though f is
actually not differentiable

00:03:36.140 --> 00:03:38.960
right here-- so f is not
differentiable at that point

00:03:38.960 --> 00:03:39.669
right over there.

00:03:39.669 --> 00:03:42.251
And you could see it because the
slope goes pretty close to 0,

00:03:42.251 --> 00:03:43.850
and then it just
jumps to negative 3.

00:03:43.850 --> 00:03:47.700
So you have a discontinuity
of the derivative right

00:03:47.700 --> 00:03:50.430
over there, but we do
have a sign change.

00:03:50.430 --> 00:03:54.600
We go from having
a positive slope

00:03:54.600 --> 00:03:56.220
on this part of
the curve to having

00:03:56.220 --> 00:03:58.660
a negative slope over
this part of the curve.

00:03:58.660 --> 00:04:00.990
So we experience a sign
change right over here

00:04:00.990 --> 00:04:04.640
at x is equal to 0, a sign
change in the first derivative

00:04:04.640 --> 00:04:07.610
of f, which is the same
thing as saying a sign

00:04:07.610 --> 00:04:10.060
change in the second
derivative of g.

00:04:10.060 --> 00:04:12.650
And a sign change in the
second derivative of g

00:04:12.650 --> 00:04:19.970
tells us that, when x is
equal to 0, the graph of g

00:04:19.970 --> 00:04:22.860
has a point of inflection.