2015 AP Calculus AB/BC 1ab

WEBVTT
Kind: captions
Language: en

00:00:00.409 --> 00:00:01.638
- [Voiceover] The rate

00:00:01.638 --> 00:00:04.124
at which rainwater flows into a drainpipe

00:00:04.124 --> 00:00:05.666
is modeled by the function R,

00:00:05.666 --> 00:00:09.847
where R of t is equal
to 20sin of t squared

00:00:09.847 --> 00:00:13.723
over 35 cubic feet per hour.

00:00:13.723 --> 00:00:18.042
t is measured in hours and 0
is less than or equal to t,

00:00:18.042 --> 00:00:19.541
which is less than or equal to 8,

00:00:19.541 --> 00:00:21.931
so t is gonna go between 0 and 8.

00:00:21.931 --> 00:00:23.419
The pipe is partially blocked,

00:00:23.419 --> 00:00:26.518
allowing water to drain out the other end

00:00:26.518 --> 00:00:29.710
of the pipe at rate modeled by D of t.

00:00:29.710 --> 00:00:33.124
It's equal to -0.04t to the third power

00:00:33.124 --> 00:00:38.124
plus 0.4t squared plus
0.96t cubic feet per hour.

00:00:38.695 --> 00:00:40.844
For the same interval right over here,

00:00:40.844 --> 00:00:44.304
there are 30 cubic feet
of water in the pipe

00:00:44.304 --> 00:00:47.097
at time t equals 0.

00:00:47.097 --> 00:00:48.205
Alright, Part A.

00:00:48.205 --> 00:00:51.335
How many cubic feet of rainwater

00:00:51.335 --> 00:00:52.629
flow into the pipe

00:00:52.629 --> 00:00:55.288
during the 8 hour time interval

00:00:55.288 --> 00:00:57.471
0 is less than or equal to t

00:00:57.471 --> 00:00:59.395
is less than or equal to 8?

00:00:59.395 --> 00:01:02.064
Alright, so we know the rate,

00:01:02.064 --> 00:01:06.235
the rate that things flow
into the rainwater pipe.

00:01:08.115 --> 00:01:09.544
Let me draw a little rainwater pipe here

00:01:09.544 --> 00:01:11.531
just so that we can
visualize what's going on.

00:01:12.601 --> 00:01:16.377
So if that is the pipe right over there,

00:01:16.377 --> 00:01:18.866
things are flowing in

00:01:18.866 --> 00:01:22.316
at a rate of R of t,

00:01:22.316 --> 00:01:27.180
and things are flowing
out at a rate of D of t.

00:01:27.180 --> 00:01:30.292
And they even tell us that
there is 30 cubic feet

00:01:30.292 --> 00:01:32.370
of water right in the beginning.

00:01:32.370 --> 00:01:34.738
But these are the rates of
entry and the rates of exiting.

00:01:36.098 --> 00:01:39.650
So they're asking how many
cubic feet of water flow into,

00:01:39.650 --> 00:01:44.091
so enter into the pipe, during
the 8-hour time interval.

00:01:44.091 --> 00:01:45.727
So if you have your rate,

00:01:45.727 --> 00:01:49.269
this is the rate at which
things are flowing into it,

00:01:49.269 --> 00:01:51.242
they give it in cubic feet per hour.

00:01:51.242 --> 00:01:55.160
If you multiply times some change in time,

00:01:55.160 --> 00:01:59.374
even an infinitesimally
small change in time, so Dt,

00:01:59.374 --> 00:02:01.243
this is the amount that flows in

00:02:01.243 --> 00:02:03.612
over that very small change in time.

00:02:03.612 --> 00:02:05.121
And so what we wanna do

00:02:05.121 --> 00:02:08.383
is we wanna sum up these amounts

00:02:08.383 --> 00:02:10.787
over very small changes in time

00:02:10.787 --> 00:02:14.200
to go from time is equal to 0,

00:02:14.200 --> 00:02:17.265
all the way to time is equal to 8.

00:02:17.265 --> 00:02:19.229
So this expression right over here,

00:02:19.229 --> 00:02:20.760
this is going to give us

00:02:20.760 --> 00:02:23.129
how many cubic feet of
water flow into the pipe.

00:02:23.129 --> 00:02:24.465
Once again, what am I doing?

00:02:24.465 --> 00:02:27.483
R of t times D of t,
this is how much flows,

00:02:27.483 --> 00:02:31.627
what volume flows in over
a very small interval, dt,

00:02:31.627 --> 00:02:32.927
and then we're gonna sum it up

00:02:32.927 --> 00:02:35.226
from t equals 0 to t equals 8.

00:02:35.226 --> 00:02:37.606
That's the power of the definite integral.

00:02:37.606 --> 00:02:39.963
And so this is going to be equal to

00:02:39.963 --> 00:02:44.963
the integral from 0 to 8 of 20sin

00:02:45.040 --> 00:02:47.997
of t squared

00:02:47.997 --> 00:02:52.948
over 35 dt.

00:02:52.948 --> 00:02:55.627
And lucky for us we can use calculators

00:02:55.627 --> 00:02:57.439
in this section of the AP exam,

00:02:57.439 --> 00:03:00.747
so let's bring out a graphing calculator

00:03:00.747 --> 00:03:03.505
where we can evaluate definite integrals.

00:03:03.505 --> 00:03:04.494
And so let's see.

00:03:04.494 --> 00:03:05.403
We wanna do definite integrals

00:03:05.403 --> 00:03:09.079
so I can click math right over here,

00:03:09.079 --> 00:03:10.484
move down.

00:03:10.484 --> 00:03:12.183
So this function, fn integral,

00:03:12.183 --> 00:03:14.621
this is a integral of a function,

00:03:14.621 --> 00:03:16.641
or a function integral right over here,

00:03:16.641 --> 00:03:18.165
so we press Enter.

00:03:18.165 --> 00:03:19.695
And the way that you do it

00:03:19.695 --> 00:03:21.239
is you first define the function,

00:03:21.239 --> 00:03:22.505
then you put a comma.

00:03:22.505 --> 00:03:24.780
Then you say what variable is the variable

00:03:24.780 --> 00:03:26.324
that you're integrating with respect to.

00:03:26.324 --> 00:03:28.321
And then you put the
bounds of integration.

00:03:29.181 --> 00:03:30.217
So I'm gonna write

00:03:30.217 --> 00:03:33.136
20sin of

00:03:33.136 --> 00:03:36.367
and just cuz it's easier
for me to input x than t,

00:03:36.367 --> 00:03:37.714
I'm gonna use x,

00:03:37.714 --> 00:03:39.188
but if you just do this

00:03:39.188 --> 00:03:40.988
as sin of x squared over 35 dx

00:03:40.988 --> 00:03:42.392
you're gonna get the same value

00:03:42.392 --> 00:03:44.168
so you're going to get

00:03:44.168 --> 00:03:47.522
x squared

00:03:47.522 --> 00:03:49.751
divided by 35.

00:03:51.001 --> 00:03:52.586
Close that parentheses.

00:03:52.586 --> 00:03:55.105
So that is my function there.

00:03:55.105 --> 00:03:57.616
Actually, I don't know if
it's going to understand.

00:04:02.054 --> 00:04:06.303
Let me put the times 2nd, insert, times

00:04:06.303 --> 00:04:08.381
just to make sure it understands that.

00:04:08.381 --> 00:04:10.622
Ok, so that's my function

00:04:10.622 --> 00:04:12.642
and then let me throw a comma here,

00:04:12.642 --> 00:04:15.440
make it clear that I'm
integrating with respect to x.

00:04:15.440 --> 00:04:17.356
I could've put a t here

00:04:17.356 --> 00:04:18.761
and integrated it with respect to t,

00:04:18.761 --> 00:04:20.270
we would get the same value.

00:04:20.270 --> 00:04:22.917
Comma, my lower bound is 0.

00:04:22.917 --> 00:04:25.425
And my upper bound is 8.

00:04:25.425 --> 00:04:27.364
And then close the parentheses

00:04:27.364 --> 00:04:30.213
and let the calculator
munch on it a little bit.

00:04:30.213 --> 00:04:31.378
And we get

00:04:31.378 --> 00:04:34.861
76.570

00:04:34.861 --> 00:04:36.530
so this is

00:04:36.530 --> 00:04:38.565
approximately

00:04:38.565 --> 00:04:40.014
Seventy-six

00:04:40.014 --> 00:04:41.071
point

00:04:41.071 --> 00:04:44.647
five, seven, zero.

00:04:44.647 --> 00:04:46.215
Now let's tackle the next part.

00:04:46.215 --> 00:04:50.174
Is the amount of water in the
pipe increasing or decreasing

00:04:50.174 --> 00:04:53.633
at time t is equal to 3 hours?

00:04:53.633 --> 00:04:56.851
Give a reason for your answer.

00:04:57.861 --> 00:05:00.114
Well, what would make it increasing?

00:05:00.114 --> 00:05:03.421
Well if the rate at
which things are going in

00:05:03.421 --> 00:05:05.719
is larger than the rate
of things going out,

00:05:05.719 --> 00:05:09.165
then the amount of water
would be increasing.

00:05:09.165 --> 00:05:10.178
But if it's the other way around,

00:05:10.178 --> 00:05:12.174
if we're draining faster at t equals 3,

00:05:12.174 --> 00:05:14.578
then things are flowing into the pipe,

00:05:14.578 --> 00:05:17.259
well then the amount of
water would be decreasing.

00:05:17.259 --> 00:05:20.586
Let me be clear, so amount,

00:05:24.143 --> 00:05:25.984
if R of t

00:05:26.994 --> 00:05:28.535
greater than,

00:05:28.535 --> 00:05:29.356
actually let me write it this way,

00:05:29.356 --> 00:05:31.260
if R of 3, t equals 3

00:05:31.260 --> 00:05:33.815
cuz t is given in hour.

00:05:33.815 --> 00:05:35.092
t is measured in hours.

00:05:35.092 --> 00:05:38.017
If R of 3 is greater than D of 3,

00:05:39.353 --> 00:05:41.953
then D of 3,

00:05:41.953 --> 00:05:44.635
If R of 3 is greater than D of 3

00:05:44.635 --> 00:05:46.237
that means water's flowing in

00:05:46.237 --> 00:05:47.375
at a higher rate than leaving.

00:05:47.375 --> 00:05:51.049
So that means that water in pipe,

00:05:51.049 --> 00:05:53.440
let me right then,

00:05:53.440 --> 00:05:54.539
then

00:05:54.539 --> 00:05:56.552
water

00:05:56.552 --> 00:05:59.034
in pipe

00:05:59.034 --> 00:06:00.387
Increasing.

00:06:01.547 --> 00:06:02.577
Increasing.

00:06:03.717 --> 00:06:05.747
And then if it's the other way around,

00:06:05.747 --> 00:06:08.034
if D of 3

00:06:08.034 --> 00:06:10.519
is greater than R of 3,

00:06:10.519 --> 00:06:12.818
then water in pipe decreasing,

00:06:12.818 --> 00:06:14.791
then you're draining faster
than you're putting into it.

00:06:15.871 --> 00:06:20.834
Then water in pipe

00:06:20.834 --> 00:06:22.290
decreasing.

00:06:22.290 --> 00:06:23.290
decreasing.

00:06:23.290 --> 00:06:26.215
So we just have to evaluate
these functions at 3.

00:06:26.215 --> 00:06:27.922
So let's see R.

00:06:27.922 --> 00:06:29.721
Actually I can do it right over here.

00:06:29.721 --> 00:06:32.519
So let me make a little line here.

00:06:32.519 --> 00:06:37.129
R of 3 is equal to,

00:06:37.129 --> 00:06:39.636
well let me get my calculator out.

00:06:39.636 --> 00:06:41.718
This is going to be,

00:06:41.718 --> 00:06:43.860
whoops, not that calculator,

00:06:43.860 --> 00:06:45.889
Let me get this calculator out.

00:06:47.322 --> 00:06:50.924
And I'm assuming that
things are in radians here.

00:06:50.924 --> 00:06:53.258
So I already put my
calculator in radian mode.

00:06:53.258 --> 00:06:54.268
So it's going to be

00:06:54.268 --> 00:06:57.247
20 times sin

00:06:58.097 --> 00:07:01.925
of 3 squared is 9,

00:07:01.925 --> 00:07:04.088
divided by 35,

00:07:04.088 --> 00:07:07.072
and it gives us, this is equal to

00:07:09.012 --> 00:07:11.857
approximately 5.09.

00:07:12.994 --> 00:07:16.740
So this is approximately

00:07:16.740 --> 00:07:19.262
5.09

00:07:19.262 --> 00:07:23.757
and D of 3 is going to be approximately,

00:07:24.827 --> 00:07:26.692
let me get the calculator back out.

00:07:26.692 --> 00:07:27.797
So it is,

00:07:29.231 --> 00:07:34.231
We have -0.04

00:07:34.513 --> 00:07:35.653
times

00:07:36.733 --> 00:07:38.160
3 to the third power,

00:07:38.160 --> 00:07:40.076
so times 27,

00:07:42.597 --> 00:07:46.063
plus 0.4

00:07:46.063 --> 00:07:47.113
times 9,

00:07:48.173 --> 00:07:49.690
times 9, t squared.

00:07:49.690 --> 00:07:51.261
plus

00:07:55.761 --> 00:07:57.284
0.96

00:07:58.917 --> 00:08:00.908
times t,

00:08:00.908 --> 00:08:02.331
times 3.

00:08:02.331 --> 00:08:06.046
And this gives us 5.4.

00:08:09.830 --> 00:08:12.047
So this is equal to 5.4.

00:08:12.047 --> 00:08:16.715
So D of 3 is greater than R of 3,

00:08:17.655 --> 00:08:21.633
so water decreasing.

00:08:21.633 --> 00:08:23.809
We're draining faster than
we're getting water into it

00:08:23.809 --> 00:08:26.817
so water is

00:08:28.689 --> 00:08:31.528
decreasing.


2015 AP Calculus AB/BC 1d

WEBVTT
Kind: captions
Language: en

00:00:00.607 --> 00:00:02.442
- [Voiceover] Part d.

00:00:02.442 --> 00:00:07.341
The pipe can hold 50 cubic feet
of water before overflowing.

00:00:07.341 --> 00:00:10.940
For t greater than eight,
water continue to flow

00:00:10.940 --> 00:00:14.342
into and out of the
pipe at the gives rates

00:00:14.342 --> 00:00:17.128
until the pipe begins to overflow.

00:00:17.128 --> 00:00:20.542
Write, but do not solve, an equation

00:00:20.542 --> 00:00:22.690
involving one or more integrals

00:00:22.690 --> 00:00:27.690
that gives the time w when the
pipe will begin to overflow.

00:00:28.065 --> 00:00:30.433
All right, so the pipe
is going to overflow.

00:00:30.433 --> 00:00:32.593
So we're going to figure
out that gives the time w

00:00:32.593 --> 00:00:34.091
and the pipe begins to overflow.

00:00:34.091 --> 00:00:36.157
So the pipe will begin to overflow

00:00:36.157 --> 00:00:40.116
when it crosses 50 cubic feet of water

00:00:40.116 --> 00:00:42.833
or you could say right when
you hit 50 cubic feet of water

00:00:42.833 --> 00:00:45.642
then it will begin to overflow.

00:00:45.642 --> 00:00:49.066
So we could figure out at
what time does the pipe have

00:00:49.066 --> 00:00:53.677
50 cubic feet of water in it.

00:00:53.677 --> 00:00:58.677
And so we could just say,
well, w of this time.

00:00:58.832 --> 00:01:02.101
So I use uppercase W as my function

00:01:02.101 --> 00:01:03.942
for how much water is in the pipe.

00:01:03.942 --> 00:01:07.392
So capital W of lowercase w is

00:01:07.392 --> 00:01:09.937
going to be equal to 50.

00:01:09.937 --> 00:01:12.909
And so you would just solve for the w.

00:01:12.909 --> 00:01:15.378
And it said write but do
not solve an equation.

00:01:15.378 --> 00:01:17.096
Well, just to make this
a little bit clear,

00:01:17.096 --> 00:01:21.974
if uppercase W of
lowercase w is going to be

00:01:21.974 --> 00:01:26.974
30 plus the integral from zero to w.

00:01:28.012 --> 00:01:30.982
Actually, now since I don't
have t as one of my vars,

00:01:30.982 --> 00:01:34.581
I could just say R of t minus D of t, td.

00:01:34.581 --> 00:01:35.768
So let me just do that.

00:01:35.768 --> 00:01:40.320
R of t minus D of t,

00:01:41.480 --> 00:01:42.502
dt.

00:01:42.502 --> 00:01:44.893
So this is the amount of total water

00:01:44.893 --> 00:01:46.878
in the pipe at time w.

00:01:46.878 --> 00:01:49.955
Well, this is going to be equal to 50.

00:01:49.955 --> 00:01:52.614
So we have just written
an equation involving

00:01:52.614 --> 00:01:55.040
one or more integrals
that gives the time w

00:01:55.040 --> 00:01:56.433
with the pipe will begin to overflow.

00:01:56.433 --> 00:01:57.420
So if you could solve for w,

00:01:57.420 --> 00:02:00.195
that's the time that the
pipe begins to overflow.

00:02:00.195 --> 00:02:01.635
We are assuming that it doesn't just

00:02:01.635 --> 00:02:04.073
right get to 50 and then
somehow come back down

00:02:04.073 --> 00:02:06.720
that it will cross 50 at this time around.

00:02:06.720 --> 00:02:08.241
So you could test a little
bit more if you want.

00:02:08.241 --> 00:02:09.994
You could try that slightly larger w

00:02:09.994 --> 00:02:11.669
or you could see that the rate,

00:02:11.669 --> 00:02:14.336
that you have more flowing in

00:02:14.336 --> 00:02:15.861
than flowing out of that time

00:02:15.861 --> 00:02:19.683
which this R of w is gonna
be greater than D of w

00:02:19.683 --> 00:02:21.250
so it means you're only
gonna be increasing

00:02:21.250 --> 00:02:23.390
so you get across over right at that time.

00:02:23.390 --> 00:02:26.316
So if you want the w
though, you'd solve this.

00:02:26.316 --> 00:02:28.440
Now another option, it's okay, we know w

00:02:28.440 --> 00:02:30.588
is going to be greater than eight.

00:02:30.588 --> 00:02:32.980
So you could say, okay,
how much water do we have

00:02:32.980 --> 00:02:34.280
right at time equals eight?

00:02:34.280 --> 00:02:36.579
We figure that out at the last problem.

00:02:36.579 --> 00:02:40.108
And so you could say we,
at time equals eight,

00:02:40.108 --> 00:02:42.581
we have that much, 48.544,

00:02:42.581 --> 00:02:44.880
and this is approximation
that's pretty close.

00:02:44.880 --> 00:02:48.560
Plus the amount of water
we accumulate between time,

00:02:48.560 --> 00:02:52.740
eight and time w of R of t

00:02:52.740 --> 00:02:56.118
minus D of t,

00:02:56.118 --> 00:02:59.427
dt is equal to 50.

00:02:59.427 --> 00:03:01.667
Either one of these would
get you to the same place.

00:03:01.667 --> 00:03:05.208
So w at right, when do we
hit 50 cubic feet of water?

00:03:05.208 --> 00:03:06.740
And then if you want to test it further,

00:03:06.740 --> 00:03:09.051
you can make sure that your
rate is increasing right at that

00:03:09.051 --> 00:03:12.847
or you have a net positive
inflow of water at that point

00:03:12.847 --> 00:03:16.812
which tells you that you're
just about to start overflowing.


2015 AP Calculus AP 2b

WEBVTT
Kind: captions
Language: en

00:00:00.106 --> 00:00:01.501
- [Voiceover] Part Two, B.

00:00:01.501 --> 00:00:04.845
Region S is the base of a
solid whose cross sections

00:00:04.845 --> 00:00:07.744
perpendicular to the x-axis are squares.

00:00:07.744 --> 00:00:10.186
Find the volume of the solid.

00:00:10.186 --> 00:00:10.847
Alright.

00:00:10.847 --> 00:00:13.681
So, region S, we see it right over here.

00:00:13.681 --> 00:00:16.350
In the last part, we already
said that this function

00:00:16.350 --> 00:00:19.707
is f, and this function is g.

00:00:19.707 --> 00:00:21.502
Now, you could try out
some values and think about

00:00:21.502 --> 00:00:24.520
how'd they behave, or you
could get a sense that, hey,

00:00:24.520 --> 00:00:26.796
like a fourth degree
quadratic's gonna have these ups

00:00:26.796 --> 00:00:29.176
and downs like that
while the, the something,

00:00:29.176 --> 00:00:31.195
well, this is, it has an
x term, but then it has an

00:00:31.195 --> 00:00:32.567
exponential term right here.

00:00:32.567 --> 00:00:35.283
That might have more of a
kind of an upward trend,

00:00:35.283 --> 00:00:38.440
especially as x increases,
and then an upward trend also

00:00:38.440 --> 00:00:40.902
as x becomes more and more negative.

00:00:40.902 --> 00:00:43.258
Because this exponent is
going to become positive.

00:00:43.258 --> 00:00:45.917
So, you can think about 'em
in different ways like that.

00:00:45.917 --> 00:00:48.158
But what's important is region S.

00:00:48.158 --> 00:00:50.005
And there's a couple things
we have to think about.

00:00:50.005 --> 00:00:51.862
We have to think about
the bounds of region S.

00:00:51.862 --> 00:00:55.191
We know that the upper bound
of region S, x is equal to,

00:00:55.191 --> 00:00:57.260
we know x is equal to two here.

00:00:57.260 --> 00:01:00.544
But we don't know this
value right over here.

00:01:00.544 --> 00:01:03.110
We don't know, let's just call that a.

00:01:03.110 --> 00:01:04.946
I'll put a question mark over there.

00:01:04.946 --> 00:01:07.195
We don't know what this x value is.

00:01:07.195 --> 00:01:09.911
But we do know that at that x
value, the functions are going

00:01:09.911 --> 00:01:11.263
to be equal to each other.

00:01:11.263 --> 00:01:15.188
So, we know that f of
a is going to be equal

00:01:15.188 --> 00:01:17.078
to g of a.

00:01:17.078 --> 00:01:19.738
Or, another way, in the last
section, we talked about

00:01:19.738 --> 00:01:21.791
the absolute value of f of
x minus g of x, which we've

00:01:21.791 --> 00:01:24.556
already inputed into our
calculator and saved it.

00:01:24.556 --> 00:01:28.235
We could also say that the
absolute value of f of a

00:01:28.235 --> 00:01:31.743
minus g of a is equal to zero.

00:01:31.743 --> 00:01:34.317
And you don't need the absolute
value there 'cause obviously

00:01:34.317 --> 00:01:36.456
the absolute value of
zero is going to be zero.

00:01:36.456 --> 00:01:38.580
But we've already inputed
this into our calculator.

00:01:38.580 --> 00:01:40.136
So, it's going to be useful.

00:01:40.136 --> 00:01:43.014
So, we can solve that thing we
inputed into our calculator.

00:01:43.014 --> 00:01:44.803
When does that equal zero?

00:01:44.803 --> 00:01:47.545
And actually, let's just do
that right from the get go.

00:01:47.545 --> 00:01:51.641
So, on our calculator, we
already inputed right over,

00:01:51.641 --> 00:01:53.399
let's make sure it's on.

00:01:53.399 --> 00:01:56.344
In our calculator, we already
inputed the absolute value

00:01:56.344 --> 00:01:59.723
of f of x minus g of x.

00:01:59.723 --> 00:02:01.956
And so, we can use our
solver now to figure out when

00:02:01.956 --> 00:02:03.344
does that equal zero.

00:02:03.344 --> 00:02:06.444
It actually equals zero just
into where we see it equals it

00:02:06.444 --> 00:02:08.465
one time, two times, and three times.

00:02:08.465 --> 00:02:10.566
So, we want to find this
place right over here.

00:02:10.566 --> 00:02:13.503
We want to figure out
what x is, or what a is.

00:02:13.503 --> 00:02:16.081
So, let's look at the math functions.

00:02:16.081 --> 00:02:18.020
Actually, let's clear this.

00:02:18.020 --> 00:02:20.039
And actually, let me quit that mode.

00:02:20.039 --> 00:02:22.176
And so, let's go to math.

00:02:22.176 --> 00:02:24.774
Scroll down to the Solver,

00:02:25.754 --> 00:02:27.294
the Solver.

00:02:27.294 --> 00:02:30.639
And let's see, we want to redefine.

00:02:30.639 --> 00:02:32.707
Okay, so we want to see, so
we could go to the orginal

00:02:32.707 --> 00:02:33.739
screen of the Solver.

00:02:33.739 --> 00:02:35.782
And actually, I did this
in the previous problem,

00:02:35.782 --> 00:02:37.703
although we redefined y sub one.

00:02:37.703 --> 00:02:41.240
But you say, okay, when
does zero equal y sub one,

00:02:41.240 --> 00:02:45.094
or when does y sub one,
that's this here, equal zero?

00:02:45.094 --> 00:02:48.042
And so, you press Enter.

00:02:48.042 --> 00:02:49.621
And then, we can put in an initial guess.

00:02:49.621 --> 00:02:51.897
Let's look some place
in-between zero and two.

00:02:51.897 --> 00:02:53.720
So, I'll put in an initial guess of one.

00:02:53.720 --> 00:02:55.616
Now, let's see what it gravitates to.

00:02:55.616 --> 00:02:56.245
Oh, whoops.

00:02:56.245 --> 00:02:59.953
And then, I could do alpha solve.

00:02:59.953 --> 00:03:03.421
Alpha solve, let's let it
munch on it a little bit.

00:03:04.511 --> 00:03:06.981
Let's let it munch.

00:03:08.051 --> 00:03:09.004
It's going.

00:03:09.004 --> 00:03:09.613
And there you go.

00:03:09.613 --> 00:03:11.941
It got to 1.03, which is the one we want.

00:03:11.941 --> 00:03:14.843
It didn't go to x equals
zero or x equals two.

00:03:14.843 --> 00:03:17.770
It went to 1.03.

00:03:17.770 --> 00:03:20.532
So, a is approximately equal to 1.03.

00:03:20.532 --> 00:03:24.910
So, a is approximately equal to 1.03.

00:03:24.910 --> 00:03:26.326
And so, what interval do we take?

00:03:26.326 --> 00:03:28.532
We don't want just the area of s.

00:03:28.532 --> 00:03:30.308
We're saying that s is the base.

00:03:31.508 --> 00:03:32.386
And if we take

00:03:32.386 --> 00:03:34.210
a cross section, we're gonna have squares.

00:03:34.210 --> 00:03:36.345
So, this is going to
be a base of a square.

00:03:36.345 --> 00:03:39.272
So, this length right over here
is going to be the absolute

00:03:39.272 --> 00:03:42.959
value of f of x

00:03:42.959 --> 00:03:45.289
minus g of x.

00:03:45.289 --> 00:03:46.969
That's the length of the base.

00:03:46.969 --> 00:03:49.698
But if we want the area of
that cross section, you can

00:03:49.698 --> 00:03:52.142
imagine, let me draw it at a
little bit of an angle here.

00:03:52.142 --> 00:03:54.128
So, you have a cross
section of a square there.

00:03:54.128 --> 00:03:56.508
You have a cross section of a square here.

00:03:56.508 --> 00:03:59.747
And so, the area of each of
those cross sections is going

00:03:59.747 --> 00:04:02.371
to be the length of the base squared.

00:04:02.371 --> 00:04:05.245
And then, you multiply that
times the dxs, you get that

00:04:05.245 --> 00:04:07.660
volume of each of those little sections.

00:04:07.660 --> 00:04:11.483
And you sum 'em all up, and
then, you are going to get

00:04:11.483 --> 00:04:13.779
the volume of that figure.

00:04:13.779 --> 00:04:17.540
So, the volume of that figure
is going to be equal to,

00:04:17.540 --> 00:04:22.387
so the volume is going to
be equal to the interval

00:04:22.387 --> 00:04:27.281
from x equals a all the way,
so that's approximately,

00:04:28.581 --> 00:04:32.548
well, I'll just write a
over here to x equals two.

00:04:32.548 --> 00:04:35.996
Of this business, this is
the length of one side of

00:04:35.996 --> 00:04:36.860
the square.

00:04:36.860 --> 00:04:38.423
But we want to square it
'cause that's gonna give us

00:04:38.423 --> 00:04:40.862
the area of that cross section.

00:04:40.862 --> 00:04:43.020
And then, so let's do that.

00:04:43.020 --> 00:04:47.560
So, times, I could say the
absolute value of f of x

00:04:47.560 --> 00:04:49.555
minus g of x.

00:04:49.555 --> 00:04:51.333
I could just square that.

00:04:51.333 --> 00:04:54.004
Or, if I want, I could
just put parentheses there.

00:04:54.004 --> 00:04:55.838
I'll actually leave it as
absolute value since that's

00:04:55.838 --> 00:04:59.066
what I've already inputed
into my calculator.

00:04:59.066 --> 00:05:02.316
And, so, that's gonna give
the area of each of those

00:05:02.316 --> 00:05:05.079
cross sections
perpendicular to the x-axis.

00:05:05.079 --> 00:05:07.472
And you multiply it times d
x, you get the volume of that

00:05:07.472 --> 00:05:08.450
little section.

00:05:08.450 --> 00:05:09.721
And then, you add 'em all up from

00:05:10.781 --> 00:05:13.225
x equals a to x equals two.

00:05:13.225 --> 00:05:15.319
And luckily, we've already
done a lot of the work here

00:05:15.319 --> 00:05:16.506
on our calculator.

00:05:17.506 --> 00:05:20.668
Our value of a is stored
in the variable x already.

00:05:20.668 --> 00:05:22.468
So, let me quit this.

00:05:22.468 --> 00:05:26.147
And then, so, let me go
back to my math functions.

00:05:26.147 --> 00:05:30.270
Let me go to definite interval
or a function interval.

00:05:30.270 --> 00:05:33.412
Select that.

00:05:33.412 --> 00:05:35.529
And now, I'm not gonna do y sub one.

00:05:35.529 --> 00:05:37.580
I'm gonna do y sub one squared.

00:05:37.580 --> 00:05:41.573
So, let me go variables, y variables.

00:05:41.573 --> 00:05:43.690
It's a function, y sub one.

00:05:43.690 --> 00:05:46.673
And I am going to, let me make sure.

00:05:46.673 --> 00:05:48.404
I'm going to square it.

00:05:48.404 --> 00:05:50.111
That's what I do right over here.

00:05:50.111 --> 00:05:52.293
My variable of integration is x.

00:05:52.293 --> 00:05:55.764
I'm going to do it from x.

00:05:55.764 --> 00:05:58.221
The value of a is stored in the variable x

00:05:58.221 --> 00:05:59.350
in my calculator right now.

00:05:59.350 --> 00:06:00.044
It's a little confusing.

00:06:00.044 --> 00:06:01.890
This first x is a variable of integration.

00:06:01.890 --> 00:06:04.614
The next x is my lower
bound of integration.

00:06:04.614 --> 00:06:07.430
And then, my upper bound is two.

00:06:07.430 --> 00:06:09.815
And so, the most complex part
here is some of the short cuts

00:06:09.815 --> 00:06:11.650
I'm doing with the calculator.

00:06:11.650 --> 00:06:14.076
And we let the calculator munch on it.

00:06:14.076 --> 00:06:17.037
And it's approximately 1.283.

00:06:17.037 --> 00:06:19.541
So, it's approximately

00:06:19.541 --> 00:06:23.671
1.283 cubic units.

00:06:23.671 --> 00:06:25.588
And we are done.


2015 AP Calculus AB/BC 3cd

WEBVTT
Kind: captions
Language: en

00:00:00.422 --> 00:00:03.635
- [Voiceover] Bob is riding his
bicycle along the same path.

00:00:03.635 --> 00:00:05.348
For zero is less than or equal to t

00:00:05.348 --> 00:00:06.658
is less than or equal to 10.

00:00:06.658 --> 00:00:10.176
Bob's velocity is modeled
by B of t is equal to

00:00:10.176 --> 00:00:12.834
t to the third minus
six t squared plus 300,

00:00:12.834 --> 00:00:16.143
where t is measured in minutes
and B of t is measured in

00:00:16.143 --> 00:00:17.733
meters per minute.

00:00:17.733 --> 00:00:22.191
Find Bob's acceleration
at time t equals five.

00:00:22.191 --> 00:00:23.515
Well acceleration,

00:00:23.515 --> 00:00:25.724
this is a velocity
function right over here,

00:00:25.724 --> 00:00:27.407
so the acceleration is
going to be derivative

00:00:27.407 --> 00:00:29.781
of the velocity function
with respect to time.

00:00:29.781 --> 00:00:32.199
What is the rate of change of
velocity with respect to time?

00:00:32.199 --> 00:00:33.465
That's acceleration.

00:00:33.465 --> 00:00:38.465
So we really just want to evaluate this,

00:00:38.516 --> 00:00:41.594
Bob's acceleration at t equals five,

00:00:41.594 --> 00:00:45.112
that's going to be B prime of five.

00:00:45.112 --> 00:00:47.685
So let's first figure
out what B prime of t is.

00:00:47.685 --> 00:00:50.239
B prime of t is equal to,

00:00:50.239 --> 00:00:51.203
we'll take the derivative here.

00:00:51.203 --> 00:00:52.201
It's pretty straightforward.

00:00:52.201 --> 00:00:53.559
Just use the Power Rule.

00:00:53.559 --> 00:00:58.559
So it's going to be three
t squared minus 12t,

00:00:58.850 --> 00:01:02.507
two times negative six
is 12 or negative 12,

00:01:02.507 --> 00:01:05.444
and then the derivative
of 300 doesn't change

00:01:05.444 --> 00:01:06.182
with respect to time,

00:01:06.182 --> 00:01:07.634
so it's just a zero.

00:01:07.634 --> 00:01:12.124
And so B prime of five is
going to be equal to three

00:01:12.124 --> 00:01:17.124
times five squared minus 12 times five

00:01:17.258 --> 00:01:21.319
which is equal to 75 minus 60

00:01:21.319 --> 00:01:23.340
which is equal to 15,

00:01:23.340 --> 00:01:25.371
and the units here, this is acceleration.

00:01:25.371 --> 00:01:26.370
So this is going to be,

00:01:26.370 --> 00:01:28.750
his velocity was in meters per minute,

00:01:28.750 --> 00:01:30.358
and so this is going meters per minute,

00:01:30.358 --> 00:01:31.503
per minute because remember,

00:01:31.503 --> 00:01:33.058
time is in terms of minutes.

00:01:33.058 --> 00:01:35.845
So we could write as meters,

00:01:35.845 --> 00:01:38.632
and we write it out meters per minute,

00:01:38.632 --> 00:01:43.053
per minute which is the same
thing as meters per minute,

00:01:43.053 --> 00:01:45.897
meters per minute squared.

00:01:45.897 --> 00:01:48.518
All right. Let's do the next part.

00:01:48.518 --> 00:01:50.603
Based on the model B from part c,

00:01:50.603 --> 00:01:54.260
find Bob's average velocity
during the interval

00:01:54.260 --> 00:01:56.184
from zero is less than or equal to t

00:01:56.184 --> 00:01:57.669
is less than or equal to 10.

00:01:57.669 --> 00:01:59.981
And if the notion of average velocity

00:01:59.981 --> 00:02:03.149
or average value of a function
is completely foreign to you,

00:02:03.149 --> 00:02:04.533
I encourage you to watch the videos

00:02:04.533 --> 00:02:08.191
on Khan Academy on find
the average of a function.

00:02:08.191 --> 00:02:12.114
But just to kind of cut to the chase,

00:02:12.114 --> 00:02:14.368
the average velocity,

00:02:14.368 --> 00:02:15.955
the average velocity,

00:02:15.955 --> 00:02:19.217
is going to be the area
under the velocity curve

00:02:19.217 --> 00:02:22.433
divided by our change in time.

00:02:22.433 --> 00:02:26.474
So the area under the velocity
curve from t equals zero

00:02:26.474 --> 00:02:30.351
to t equals 10 of B of t

00:02:30.351 --> 00:02:33.579
of B of t, dt,

00:02:33.579 --> 00:02:36.714
divided by our change in time.

00:02:36.714 --> 00:02:38.216
So it's going to be divided by,

00:02:38.216 --> 00:02:39.946
well you're going from zero to 10.

00:02:39.946 --> 00:02:43.800
So 10 minus zero is
going to be equal to 10.

00:02:43.800 --> 00:02:46.362
And if you wanted the
intuition here it's like well,

00:02:46.362 --> 00:02:48.589
if you know the area of
something and if you wanted

00:02:48.589 --> 00:02:50.524
to find its average height,

00:02:50.524 --> 00:02:52.917
you could just divide by its width.

00:02:52.917 --> 00:02:54.195
And that's what we're doing here.

00:02:54.195 --> 00:02:56.203
If we know the area of something,

00:02:56.203 --> 00:02:58.450
we want to figure out its average height,

00:02:58.450 --> 00:03:00.354
and so you divide by its width.

00:03:00.354 --> 00:03:03.071
That's I guess a very
high-level intuition for

00:03:03.071 --> 00:03:05.637
where this expression came from.

00:03:05.637 --> 00:03:08.795
And so this is going to be
equal to one tenth times

00:03:08.795 --> 00:03:11.314
the integral from zero to 10,

00:03:11.314 --> 00:03:16.314
and B of t is t to the
third power minus 6t squared

00:03:18.710 --> 00:03:22.390
plus 300 dt,

00:03:22.390 --> 00:03:25.792
and so this is going to
be equal to one over 10.

00:03:25.792 --> 00:03:27.208
Take the anti-derivative here,

00:03:27.208 --> 00:03:29.832
so this is going to be t
to the fourth over four,

00:03:29.832 --> 00:03:32.618
t to the fourth over four,

00:03:32.618 --> 00:03:34.453
and then this is going to be,

00:03:34.453 --> 00:03:36.295
if we increase the exponent here by

00:03:36.295 --> 00:03:37.503
what is t to the third,

00:03:37.503 --> 00:03:39.140
and then you divide by three.

00:03:39.140 --> 00:03:41.420
So it's negative six divided by three is

00:03:41.420 --> 00:03:44.578
negative two, t to the third,

00:03:44.578 --> 00:03:47.886
and then plus 300t,

00:03:47.886 --> 00:03:49.581
300t,

00:03:49.581 --> 00:03:51.938
And I'm going to evaluate it.

00:03:51.938 --> 00:03:56.938
I am going to evaluate it
at 10 and subtract from that

00:03:57.186 --> 00:03:58.927
and evaluate it at zero.

00:03:58.927 --> 00:04:01.482
And so this is going to be equal to,

00:04:01.482 --> 00:04:03.950
this is going to be equal to one tenth,

00:04:03.950 --> 00:04:05.552
that same one tenth there,

00:04:05.552 --> 00:04:07.828
and when you evaluate all this at 10

00:04:07.828 --> 00:04:08.971
what'd everybody get?

00:04:08.971 --> 00:04:12.708
Let's see 10 to the fourth power is,

00:04:12.708 --> 00:04:17.708
10 to the fourth power is
10,000 divided by four is 2,500,

00:04:18.491 --> 00:04:20.388
2,500,

00:04:20.388 --> 00:04:23.209
and then minus two times 10 to the third.

00:04:23.209 --> 00:04:25.420
So it's two times a thousand.

00:04:25.420 --> 00:04:28.206
So minus 2,000.

00:04:28.206 --> 00:04:30.308
And then 300 times 10,

00:04:30.308 --> 00:04:33.059
well that's plus 3,000,

00:04:33.059 --> 00:04:35.764
and then you subtract all
of this evaluated at zero

00:04:35.764 --> 00:04:37.877
which is just going to be zero.

00:04:37.877 --> 00:04:39.898
So this going to be equal to,

00:04:39.898 --> 00:04:43.741
2,500 minus 2,000 is 500 plus 3,000.

00:04:43.741 --> 00:04:44.786
This is all,

00:04:44.786 --> 00:04:47.398
simplifies to 3,500,

00:04:47.398 --> 00:04:49.220
or 3,500,

00:04:49.220 --> 00:04:50.846
and then you divide it by 10.

00:04:50.846 --> 00:04:54.590
This is going to be 350,

00:04:54.590 --> 00:04:55.771
350,

00:04:55.771 --> 00:05:00.771
and it's an average
velocity meters per 350,

00:05:01.634 --> 00:05:03.631
meters per minute.

00:05:03.631 --> 00:05:05.041
And we are done.


2015 AP Calculus AB/BC 3a

WEBVTT
Kind: captions
Language: en

00:00:00.503 --> 00:00:02.870
- [Voiceover] Johanna jogs
along a straight path.

00:00:02.870 --> 00:00:05.248
For zero is less than or equal
to t is less than or equal

00:00:05.248 --> 00:00:07.825
to 40, Johanna's velocity
is given by a differentiable

00:00:07.825 --> 00:00:09.347
function v.

00:00:09.347 --> 00:00:12.295
Selected values of v of t,
where t is measured in minutes

00:00:12.295 --> 00:00:15.592
and v of t is measured in
meters per minute, are given

00:00:15.592 --> 00:00:16.707
in the table above.

00:00:16.707 --> 00:00:18.147
We see that right over there.

00:00:18.147 --> 00:00:21.328
Use the data in the table
to estimate the value of

00:00:21.328 --> 00:00:24.533
not v of 16 but v prime of 16.

00:00:24.533 --> 00:00:27.446
And we see here, they don't
even give us v of 16, so how

00:00:27.446 --> 00:00:29.873
do we think about v prime of 16.

00:00:29.873 --> 00:00:34.145
Well, just remind ourselves,
this is the rate of change

00:00:34.145 --> 00:00:38.755
of v with respect to time
when time is equal to 16.

00:00:38.755 --> 00:00:41.250
So, we can estimate it, and
that's the key word here,

00:00:41.250 --> 00:00:42.331
estimate.

00:00:42.331 --> 00:00:45.975
We can estimate v prime of 16
by thinking about what is our

00:00:45.975 --> 00:00:49.429
change in velocity over our change in time

00:00:49.429 --> 00:00:51.319
around 16.

00:00:51.319 --> 00:00:53.665
And when we look at it over
here, they don't give us

00:00:53.665 --> 00:00:55.704
v of 16, but they give us v of 12.

00:00:55.704 --> 00:00:57.842
They give us v of 20.

00:00:57.842 --> 00:01:01.035
So, let's figure out our
rate of change between 12,

00:01:01.035 --> 00:01:03.444
t equals 12, and t equals 20.

00:01:03.444 --> 00:01:07.567
So, our change in velocity,
that's going to be v of 20,

00:01:10.627 --> 00:01:12.992
minus v of 12.

00:01:12.992 --> 00:01:15.880
And then our change in time is going to be

00:01:15.880 --> 00:01:18.290
20 minus 12.

00:01:18.290 --> 00:01:21.452
And so, this is going
to be equal to v of 20

00:01:21.452 --> 00:01:23.632
is 240.

00:01:23.632 --> 00:01:25.316
240 v of 12.

00:01:25.316 --> 00:01:27.355
We see right there is 200.

00:01:28.585 --> 00:01:32.385
And so, this is going to be 40 over eight,

00:01:32.385 --> 00:01:34.496
which is equal to five.

00:01:34.496 --> 00:01:36.758
And we would be done.

00:01:36.758 --> 00:01:38.986
For good measure, it's good
to put the units there.

00:01:38.986 --> 00:01:40.555
So, this is our rate.

00:01:40.555 --> 00:01:44.189
This is how fast the velocity
is changing with respect

00:01:44.189 --> 00:01:45.268
to time.

00:01:45.268 --> 00:01:48.833
So, the units are gonna be
meters per minute per minute.

00:01:48.833 --> 00:01:50.906
So, we could write this as meters

00:01:52.326 --> 00:01:54.626
per minute squared,

00:01:54.626 --> 00:01:59.605
per minute, meters per minute squared.

00:01:59.605 --> 00:02:01.046
And we're done.

00:02:01.046 --> 00:02:02.743
Now, if you want to get a
little bit more of a visual

00:02:02.743 --> 00:02:04.719
understanding of this, and
what I'm about to do, you would

00:02:04.719 --> 00:02:08.036
not actually have to
do on the actual exam.

00:02:08.036 --> 00:02:09.923
Well, let's just try to graph.

00:02:09.923 --> 00:02:12.505
Let's graph these points here.

00:02:12.505 --> 00:02:16.216
Let me give myself some space to do it.

00:02:16.216 --> 00:02:20.464
So, if we were, if we tried
to graph it, so I'll just do

00:02:20.464 --> 00:02:23.348
a very rough graph here.

00:02:23.348 --> 00:02:26.669
So, let's say this is
y is equal to v of t.

00:02:26.669 --> 00:02:29.909
And we see that v of
t goes as low as -220.

00:02:29.909 --> 00:02:32.160
It goes as high as 240.

00:02:32.160 --> 00:02:35.037
So, let me give, so I want
to draw the horizontal axis

00:02:35.037 --> 00:02:37.038
some place around here.

00:02:37.038 --> 00:02:39.149
T is positive.

00:02:39.149 --> 00:02:41.324
It would look something like that.

00:02:41.324 --> 00:02:44.226
And so, let's just make, let's make this,

00:02:45.406 --> 00:02:48.196
let's make that 200 and,

00:02:48.196 --> 00:02:50.667
let's make that 300.

00:02:51.737 --> 00:02:55.771
And so, then this would be 200 and 100.

00:02:57.761 --> 00:03:00.500
200 and 100 there.

00:03:00.500 --> 00:03:02.488
And then that would be -100.

00:03:02.488 --> 00:03:04.541
And then that would be -200.

00:03:04.541 --> 00:03:08.372
And then that would be -300.

00:03:08.372 --> 00:03:11.994
And we see on the t axis,
our highest value is 40.

00:03:11.994 --> 00:03:13.874
We go between zero and 40.

00:03:13.874 --> 00:03:16.719
And so, these obviously
aren't at the same scale.

00:03:16.719 --> 00:03:18.113
But this is going to be zero.

00:03:18.113 --> 00:03:22.044
If we put 40 here, and then
if we put 20 in-between.

00:03:23.264 --> 00:03:25.299
And so, this would be 10.

00:03:25.299 --> 00:03:27.458
Let me do a little bit to the right.

00:03:27.458 --> 00:03:28.933
That would be 10.

00:03:28.933 --> 00:03:32.058
And then, that would be 30.

00:03:32.058 --> 00:03:34.141
And so, what points do they give us?

00:03:35.171 --> 00:03:38.140
So, they give us, I'll do these in orange.

00:03:38.140 --> 00:03:40.218
Zero, comma, zero.

00:03:40.218 --> 00:03:43.678
So, that is right over there.

00:03:43.678 --> 00:03:47.545
They give us when time is
12, our velocity is 200.

00:03:47.545 --> 00:03:50.858
So, when the time is 12,
which is right over there,

00:03:50.858 --> 00:03:53.860
our velocity is going to be 200.

00:03:53.860 --> 00:03:55.555
So, that's that point.

00:03:55.555 --> 00:04:00.071
When our time is 20, our
velocity is going to be 240.

00:04:00.071 --> 00:04:03.434
So, when our time is
20, our velocity is 240,

00:04:03.434 --> 00:04:05.852
which is gonna be right over there.

00:04:05.852 --> 00:04:10.443
And then, when our time is
24, our velocity is -220.

00:04:10.443 --> 00:04:12.555
So, she switched directions.

00:04:12.555 --> 00:04:14.198
So, when it's 24.

00:04:14.198 --> 00:04:18.114
So, 24 is gonna be roughly over here.

00:04:18.114 --> 00:04:20.757
It's gonna be -220.

00:04:20.757 --> 00:04:25.648
So, -220 might be right over there.

00:04:25.648 --> 00:04:29.944
And then, finally, when time
is 40, her velocity is 150,

00:04:29.944 --> 00:04:31.465
positive 150.

00:04:31.465 --> 00:04:32.945
So, at 40,

00:04:32.945 --> 00:04:36.745
it's positive 150.

00:04:36.745 --> 00:04:39.489
And so, these are just
sample points from her

00:04:39.489 --> 00:04:40.800
velocity function.

00:04:40.800 --> 00:04:43.348
But what we wanted to do is
we wanted to find in this

00:04:43.348 --> 00:04:48.038
problem, we want to say,
okay, when t is equal to 16,

00:04:48.038 --> 00:04:52.979
when t is equal to 16,
what is the rate of change?

00:04:52.979 --> 00:04:56.947
And we don't know much about,
we don't know what v of 16 is.

00:04:56.947 --> 00:04:59.038
But what we could do is, and
this is essentially what we did

00:04:59.038 --> 00:04:59.828
in this problem.

00:04:59.828 --> 00:05:02.382
We could say, alright, well,
we can approximate with

00:05:02.382 --> 00:05:05.518
the function might do by
roughly drawing a line here.

00:05:06.538 --> 00:05:09.220
So, if you draw a line
there, and you say, alright,

00:05:09.220 --> 00:05:13.748
well, v of 16, or v prime
of 16, I should say.

00:05:13.748 --> 00:05:18.098
So, v prime of 16 is going
to be approximately the slope

00:05:18.098 --> 00:05:23.062
is going to be approximately
the slope of this line.

00:05:23.062 --> 00:05:25.232
That's going to be our best
job based on the data that

00:05:25.232 --> 00:05:28.729
they have given us of estimating
the value of v prime of 16.

00:05:28.729 --> 00:05:32.959
So, we literally just did
change in v, which is that one,

00:05:32.959 --> 00:05:36.831
delta v over change in t over

00:05:38.231 --> 00:05:40.081
delta t to get the slope

00:05:40.081 --> 00:05:42.959
of this line, which was
our best approximation

00:05:42.959 --> 00:05:47.019
for the derivative when t is equal to 16.


2015 AP Calculus 2c

WEBVTT
Kind: captions
Language: en

00:00:00.384 --> 00:00:01.917
- [Voiceover] Let h be
the vertical distance

00:00:01.917 --> 00:00:04.901
between the graphs of
f and g and region s.

00:00:04.901 --> 00:00:08.442
Find the rate at which h
changes with respect to x

00:00:08.442 --> 00:00:12.226
when x is equal to 1.8.

00:00:13.841 --> 00:00:15.687
We have region s right over here.

00:00:15.687 --> 00:00:17.706
You can't see it that
well since I drew over it.

00:00:17.706 --> 00:00:20.749
And what you see when we're in region s,

00:00:20.749 --> 00:00:24.472
the function f is greater
than the function g.

00:00:24.472 --> 00:00:26.297
It's above the function g.

00:00:27.267 --> 00:00:29.282
We can write h of x,

00:00:29.282 --> 00:00:30.812
we could write h of x,

00:00:30.812 --> 00:00:33.673
as being equal to f of x

00:00:33.673 --> 00:00:37.307
minus g of x and what we wanna do

00:00:37.307 --> 00:00:39.350
is we wanna find the
rate at which h changes

00:00:39.350 --> 00:00:40.986
with respect to x.

00:00:40.986 --> 00:00:44.310
We could write that as h prime of,

00:00:44.310 --> 00:00:45.690
we could say h prime of x

00:00:45.690 --> 00:00:47.930
but we want the rate
when x is equal to 1.8.

00:00:47.930 --> 00:00:52.794
So h prime of 1.8 is
what we wanna figure out.

00:00:52.794 --> 00:00:54.733
We could evaluate f prime of 1.8

00:00:54.733 --> 00:00:56.311
and g prime of 1.8 and to do that

00:00:56.311 --> 00:00:58.251
we would take the derivatives of each

00:00:58.251 --> 00:01:00.492
of these things and we
know how to do that.

00:01:00.492 --> 00:01:03.185
It's within our capabilities,
but it's important to realize

00:01:03.185 --> 00:01:04.253
when you're taking the AP test

00:01:04.253 --> 00:01:06.668
that you have a calculator
at your disposal.

00:01:06.668 --> 00:01:09.211
And a calculator can numerically integrate

00:01:09.211 --> 00:01:12.333
and can numerically evaluate derivatives.

00:01:12.333 --> 00:01:14.528
And so when they actually want us

00:01:14.528 --> 00:01:17.559
to find the area or evaluate an integral,

00:01:17.559 --> 00:01:19.005
where they give the endpoints,

00:01:19.005 --> 00:01:20.296
or evaluate a derivative at a point,

00:01:20.296 --> 00:01:21.667
well that's a pretty good
sign that you could probably

00:01:21.667 --> 00:01:23.526
use your calculator here.

00:01:23.526 --> 00:01:25.621
And what's extra good about this

00:01:25.621 --> 00:01:28.913
is we have already
essentially inputted h of x

00:01:28.913 --> 00:01:31.107
in the previous steps
and really in part A.

00:01:32.207 --> 00:01:33.755
I defined this function here,

00:01:33.755 --> 00:01:35.345
and this function is essentially h of x.

00:01:35.345 --> 00:01:37.446
I took the absolute value of it

00:01:37.446 --> 00:01:39.385
so it's always positive
over either region,

00:01:40.265 --> 00:01:42.415
but I could delete the
absolute value if we want.

00:01:43.305 --> 00:01:45.015
Delete that absolute value,

00:01:45.015 --> 00:01:47.268
then I have to get rid of
that parentheses at the end.

00:01:48.348 --> 00:01:50.917
Let me delete that.

00:01:50.917 --> 00:01:52.655
And so notice this is h of x.

00:01:52.655 --> 00:01:53.607
We have our f of x,

00:01:53.607 --> 00:01:58.607
which was 1 plus x plus e
to the x squared minus 2x,

00:01:58.665 --> 00:02:02.275
and then from that we subtract g of x.

00:02:02.275 --> 00:02:06.885
G of x was a positive x to the
fourth but we're subtracting

00:02:06.885 --> 00:02:07.729
so negative x to the fourth,

00:02:07.729 --> 00:02:09.900
let me show you g of x right over here.

00:02:09.900 --> 00:02:11.596
G of x is right over here, and notice

00:02:11.596 --> 00:02:12.815
we are subtracting it.

00:02:12.815 --> 00:02:15.171
So the y1, as I've defined
it in my calculator,

00:02:15.171 --> 00:02:17.297
and I just pressed this y
equals button right over here,

00:02:17.297 --> 00:02:18.702
that is my h of x.

00:02:18.702 --> 00:02:21.806
Now I can go back to the other screen,

00:02:21.806 --> 00:02:25.233
and I can evaluate its derivative
when x is equal to 1.8.

00:02:25.233 --> 00:02:26.614
I go to math.

00:02:26.614 --> 00:02:28.519
I scroll down.

00:02:29.449 --> 00:02:32.423
We have n derivative right here

00:02:32.423 --> 00:02:35.750
and so click Enter there.

00:02:35.750 --> 00:02:37.602
And then what are we gonna
take the derivative of?

00:02:37.602 --> 00:02:39.773
Well the function y sub 1
that I've already defined

00:02:39.773 --> 00:02:40.981
in my calculator.

00:02:40.981 --> 00:02:44.383
I can go to variables, y variables,

00:02:44.383 --> 00:02:47.239
it's already selected function
so I'll just press Enter.

00:02:47.239 --> 00:02:49.480
And I'll select the function y sub 1

00:02:49.480 --> 00:02:50.419
that I've already defined.

00:02:50.419 --> 00:02:52.324
So I'm taking the derivative of y sub 1.

00:02:52.324 --> 00:02:55.692
I'm taking the derivative
with respect to x.

00:02:55.692 --> 00:02:57.665
And I'm going to evaluate that derivative,

00:02:57.665 --> 00:03:00.705
when x is equal to 1.8.

00:03:00.705 --> 00:03:01.925
That simple.

00:03:01.925 --> 00:03:04.108
And then I click Enter.

00:03:04.108 --> 00:03:05.048
And there you have it.

00:03:05.048 --> 00:03:09.484
It's approximately -3.812.

00:03:10.664 --> 00:03:15.664
So approximately -3.812.

00:03:17.002 --> 00:03:18.399
And we're done.

00:03:18.399 --> 00:03:22.104
One thing you might appreciate
from this entire question,

00:03:22.104 --> 00:03:23.937
and even the question 1,

00:03:23.937 --> 00:03:26.945
is that they really wanted to make sure

00:03:26.945 --> 00:03:28.919
that you understand the
underlying conceptual ideas

00:03:28.919 --> 00:03:31.682
behind derivatives and integrals.

00:03:31.682 --> 00:03:33.968
And if you understand
the conceptual ideas,

00:03:33.968 --> 00:03:36.868
of how do you use them to solve problems,

00:03:36.868 --> 00:03:38.300
and you have your calculator at disposal,

00:03:38.300 --> 00:03:39.924
then these are not too hairy,

00:03:39.924 --> 00:03:42.479
that these can be done fairly quickly.


2015 AP Calculus AB/BC 3b

WEBVTT
Kind: captions
Language: en

00:00:00.338 --> 00:00:03.145
- [Voiceover] Part B, using correct units,

00:00:03.145 --> 00:00:06.186
explain the meaning of
the definite integral,

00:00:06.186 --> 00:00:07.603
so it's definite integral from zero

00:00:07.603 --> 00:00:09.518
from T equals zero to T equals 40

00:00:09.518 --> 00:00:12.609
of the absolute value of V of T DT

00:00:12.609 --> 00:00:14.664
in the context of the problem.

00:00:14.664 --> 00:00:17.149
Approximate the value of that integral

00:00:17.149 --> 00:00:21.108
using a right Riemann sum
with the four subintervals

00:00:21.108 --> 00:00:23.348
indicated in the table.

00:00:23.348 --> 00:00:25.264
All right so let's first
tackle the first part.

00:00:25.264 --> 00:00:28.225
So what's the meaning of
this definite integral?

00:00:28.225 --> 00:00:32.648
So the velocity, velocity is
a vector, it has direction,

00:00:32.648 --> 00:00:34.923
and you can see when you look up here

00:00:34.923 --> 00:00:37.049
that sometimes the velocity is positive

00:00:37.049 --> 00:00:39.405
and sometimes it is negative,

00:00:39.405 --> 00:00:42.365
and a reasonable interpretation of that

00:00:42.365 --> 00:00:44.722
is well this is when she's
walking in one direction

00:00:44.722 --> 00:00:47.810
and then that's where
whatever direction we assume

00:00:47.810 --> 00:00:49.831
is the positive direction
and this is when she's

00:00:49.831 --> 00:00:51.386
walking the other way,

00:00:51.386 --> 00:00:54.730
and so she has a positive velocity here

00:00:54.730 --> 00:00:56.088
and then she has a negative velocity.

00:00:56.088 --> 00:00:58.352
She has some speed in the other direction

00:00:58.352 --> 00:01:01.603
and then she has a
positive velocity again.

00:01:01.603 --> 00:01:03.704
So if you were to take the
absolute value of this,

00:01:03.704 --> 00:01:06.607
if you just cared about the
magnitude of the velocity,

00:01:06.607 --> 00:01:08.081
well we're talking about,

00:01:08.081 --> 00:01:10.323
we're talking about speed,

00:01:10.323 --> 00:01:11.402
but we're not just talking about

00:01:11.402 --> 00:01:13.271
the absolute value of velocity,

00:01:13.271 --> 00:01:16.208
we're integrating the
absolute value of velocity

00:01:16.208 --> 00:01:18.426
over time from time equals zero

00:01:18.426 --> 00:01:20.167
to time equals 40.

00:01:20.167 --> 00:01:21.885
So if you integrate speed,

00:01:21.885 --> 00:01:23.441
if you're saying okay for every

00:01:23.441 --> 00:01:27.029
for every little amount of time DT

00:01:27.029 --> 00:01:28.991
she has some speed,

00:01:28.991 --> 00:01:31.684
well that's going to give you
the distance that she travels.

00:01:31.684 --> 00:01:33.971
So the distance remember,
is the difference

00:01:33.971 --> 00:01:35.806
between distance and displacement

00:01:35.806 --> 00:01:38.824
is displacement you can
think of as the net distance

00:01:38.824 --> 00:01:41.483
it takes if you go back
and forth a bunch of times

00:01:41.483 --> 00:01:43.062
it's going to add up to the distance,

00:01:43.062 --> 00:01:45.465
but they're going to net out
with respect to each other

00:01:45.465 --> 00:01:46.591
in displacement.

00:01:46.591 --> 00:01:49.389
So this right over is going
to give us the total distance,

00:01:49.389 --> 00:01:54.389
not displacement, that she
travels over those 40 minutes.

00:01:54.893 --> 00:01:57.168
If there was no absolute value sign here

00:01:57.168 --> 00:01:59.548
then we'd be talking
about the displacement

00:01:59.548 --> 00:02:01.522
and so let me write that.

00:02:01.522 --> 00:02:03.867
So this integral from zero to 40

00:02:03.867 --> 00:02:07.350
absolute value V of T DT

00:02:07.350 --> 00:02:12.350
is the total distance

00:02:13.666 --> 00:02:16.963
distance she travels

00:02:16.963 --> 00:02:18.328
she travels

00:02:19.483 --> 00:02:21.062
over the 40 minutes

00:02:21.062 --> 00:02:25.649
over the 40, over the 40 minutes,

00:02:25.649 --> 00:02:29.144
remember our unit of
time here is in minutes.

00:02:29.144 --> 00:02:30.466
All right.

00:02:30.466 --> 00:02:34.007
So using correct units and so this is

00:02:34.007 --> 00:02:35.528
and so the units,

00:02:35.528 --> 00:02:37.143
let me write it this way.

00:02:37.143 --> 00:02:38.663
Is a total distance she travels

00:02:38.663 --> 00:02:40.021
over the 40 minutes.

00:02:40.021 --> 00:02:42.841
The units are meters.

00:02:42.841 --> 00:02:46.710
The units are meters,

00:02:46.710 --> 00:02:49.810
are meters.

00:02:49.810 --> 00:02:50.949
All right,

00:02:50.949 --> 00:02:53.329
now this says approximate
the value of that integral

00:02:53.329 --> 00:02:55.044
using a right Riemann sum

00:02:55.044 --> 00:02:57.865
with the four subintervals
indicated in the table.

00:02:57.865 --> 00:03:01.011
So what do they mean
by a right Riemann sum?

00:03:01.011 --> 00:03:03.431
Well actually, let me
just, let me just draw,

00:03:03.431 --> 00:03:06.236
let me draw the absolute
value of the V of T

00:03:06.236 --> 00:03:08.291
and I already took the
trouble of the last part

00:03:08.291 --> 00:03:09.963
of drawing the axis and
you wouldn't actually

00:03:09.963 --> 00:03:10.926
have to do this on the test,

00:03:10.926 --> 00:03:12.703
but I'm doing this to just help you

00:03:12.703 --> 00:03:15.466
understand what we're actually going to do

00:03:15.466 --> 00:03:17.927
when we talk about a right Riemann sum.

00:03:17.927 --> 00:03:21.909
So if we, so in orange I have V of T,

00:03:21.909 --> 00:03:24.397
but if I want to do the
absolute value of V of T,

00:03:24.397 --> 00:03:25.720
I'll do that in magenta,

00:03:25.720 --> 00:03:27.125
so the absolute value of that is that,

00:03:27.125 --> 00:03:28.588
absolute value of that is that,

00:03:28.588 --> 00:03:30.028
absolute value of that is that,

00:03:30.028 --> 00:03:35.028
absolute value of negative 220
is going to be positive 220.

00:03:35.217 --> 00:03:37.319
So we saw here when time,

00:03:37.319 --> 00:03:41.266
when T is equal to 24
V of T is negative 220,

00:03:41.266 --> 00:03:42.938
she's running in the opposite direction

00:03:42.938 --> 00:03:44.598
or jogging on the other direction,

00:03:44.598 --> 00:03:46.885
but we're going to take
the absolute value of that.

00:03:46.885 --> 00:03:51.657
So when T is equal to 24,

00:03:51.657 --> 00:03:56.354
her speed, we could
say, is going to be 220,

00:03:56.354 --> 00:03:58.375
we're going to take the
absolute value of it.

00:03:58.375 --> 00:04:01.186
So it's going to be right around,

00:04:01.186 --> 00:04:02.730
right around there

00:04:02.730 --> 00:04:03.763
and then the absolute value of this

00:04:03.763 --> 00:04:05.563
is going to be that point again

00:04:05.563 --> 00:04:08.442
and so we don't know the
actual velocity function.

00:04:08.442 --> 00:04:10.126
The actual velocity
function, I don't know,

00:04:10.126 --> 00:04:11.867
it might look something like,

00:04:11.867 --> 00:04:15.803
it might look something like, like this.

00:04:15.803 --> 00:04:18.392
It goes like that, maybe it dips down.

00:04:18.392 --> 00:04:20.133
I should say the speed function.

00:04:20.133 --> 00:04:22.409
We don't know the actual speed function.

00:04:22.409 --> 00:04:26.333
So this is the absolute value of V of T,

00:04:26.333 --> 00:04:28.969
which is the speed function you could say.

00:04:28.969 --> 00:04:33.427
Speed, speed function,

00:04:33.427 --> 00:04:35.505
and so if we wanted to
actually integrate it,

00:04:35.505 --> 00:04:36.945
so if we wanted to integrate it,

00:04:36.945 --> 00:04:38.265
well we don't know how to do it

00:04:38.265 --> 00:04:39.426
if we don't know,

00:04:39.426 --> 00:04:40.762
we don't know how to do it exactly

00:04:40.762 --> 00:04:42.770
without knowing the exact function,

00:04:42.770 --> 00:04:46.926
but we can approximate what
the integral is going to be

00:04:46.926 --> 00:04:48.726
with a right Riemann sum

00:04:48.726 --> 00:04:51.269
and a Riemann sum is just
breaking up this area

00:04:51.269 --> 00:04:54.067
into rectangles and then
finding the total area

00:04:54.067 --> 00:04:55.448
of those rectangles.

00:04:55.448 --> 00:04:57.329
And then there's a couple
of ways you could do it,

00:04:57.329 --> 00:04:59.187
you could use, you could use

00:04:59.187 --> 00:05:02.368
for each rectangle you
could use the left boundary

00:05:02.368 --> 00:05:04.249
as the height or you could
use the right boundary

00:05:04.249 --> 00:05:05.061
as the height

00:05:05.061 --> 00:05:08.881
and so they tell us to
use a right Riemann sum.

00:05:08.881 --> 00:05:10.286
A right Riemann sum,

00:05:10.286 --> 00:05:12.329
so for each of the rectangles,

00:05:12.329 --> 00:05:13.641
for each of the intervals
we're going to use

00:05:13.641 --> 00:05:16.404
the value of our function
on the right side

00:05:16.404 --> 00:05:18.749
to approximate the area.

00:05:18.749 --> 00:05:19.983
So let's do that.

00:05:19.983 --> 00:05:23.060
So let me divide this into four rectangles

00:05:23.060 --> 00:05:25.080
and they give us the four intervals.

00:05:25.080 --> 00:05:29.503
So one interval goes
from zero to T equals 12,

00:05:29.503 --> 00:05:30.943
we see that there.

00:05:30.943 --> 00:05:32.157
The other integral goes from 12 to 20,

00:05:32.157 --> 00:05:34.270
the next integral goes from 20 to 24,

00:05:34.270 --> 00:05:36.615
the next integral goes from 24 to 40,

00:05:36.615 --> 00:05:38.217
and if the whole notation of a Riemann sum

00:05:38.217 --> 00:05:39.471
is completely foreign to you,

00:05:39.471 --> 00:05:41.677
I encourage you to watch
the videos on Khan Academy

00:05:41.677 --> 00:05:43.513
on Riemann sums.

00:05:43.513 --> 00:05:45.777
And so here we are.

00:05:45.777 --> 00:05:48.435
So the first interval's from zero to 12

00:05:48.435 --> 00:05:50.538
and so we want to use
the value of our function

00:05:50.538 --> 00:05:53.892
on the right side to define
the height of the rectangle.

00:05:53.892 --> 00:05:55.553
So let me define.

00:05:55.553 --> 00:05:57.455
So that rectangle is going to look,

00:05:57.455 --> 00:06:01.356
is going to look like this,

00:06:01.356 --> 00:06:03.597
so that's a rectangle right over there.

00:06:03.597 --> 00:06:07.231
Then the next interval goes from 12 to 20.

00:06:07.231 --> 00:06:10.509
The next rectangle goes from 12 to 20,

00:06:10.509 --> 00:06:11.775
we see that there.

00:06:11.775 --> 00:06:13.354
From 12 to 20.

00:06:13.354 --> 00:06:15.595
So let's use a value of our function at 20

00:06:15.595 --> 00:06:17.928
to define the height of this rectangle.

00:06:17.928 --> 00:06:20.031
So just like that.

00:06:21.198 --> 00:06:23.763
Look like, let me do it,

00:06:23.763 --> 00:06:24.843
I'll just hand draw that.

00:06:24.843 --> 00:06:27.780
It would look like that

00:06:27.780 --> 00:06:30.276
and then the next rectangle goes from

00:06:30.276 --> 00:06:32.737
T equals 20 to T equals 24

00:06:32.737 --> 00:06:35.025
and we want to use the right hand side

00:06:35.025 --> 00:06:37.440
to define the height of that rectangle.

00:06:37.440 --> 00:06:38.775
You can see it's a little irregular

00:06:38.775 --> 00:06:39.680
and that's okay,

00:06:39.680 --> 00:06:41.945
they all don't have to
have the same width.

00:06:41.945 --> 00:06:45.677
So the height is something like this,

00:06:45.677 --> 00:06:50.677
and then the last interval
is T equals 24 to T equals 40

00:06:51.145 --> 00:06:54.768
and so we use the right
side as the height.

00:06:54.768 --> 00:06:55.986
So the interval,

00:06:55.986 --> 00:06:57.472
so this was the last interval we did,

00:06:57.472 --> 00:06:59.991
the next interval is going to be that

00:06:59.991 --> 00:07:02.487
and then the height we
use the right hand side

00:07:02.487 --> 00:07:05.727
to define the height that's
why it's a right Riemann sum

00:07:05.727 --> 00:07:09.546
and so our approximation of the integral

00:07:09.546 --> 00:07:12.321
of the absolute value of V
of T is going to be the area

00:07:12.321 --> 00:07:15.165
of these rectangles over here,

00:07:15.165 --> 00:07:17.996
and so we could write,

00:07:17.996 --> 00:07:22.931
we could write the
integral from zero to 40

00:07:22.931 --> 00:07:27.737
of the absolute value of V of T DT

00:07:27.737 --> 00:07:30.419
is approximately,

00:07:30.419 --> 00:07:32.834
well it's going to be
this first change in time

00:07:32.834 --> 00:07:35.692
times this height, so
this first change in time

00:07:35.692 --> 00:07:38.571
this is a change in time of 12 times,

00:07:38.571 --> 00:07:42.101
so it's going to be 12 times,

00:07:42.101 --> 00:07:43.099
what's the height?

00:07:43.099 --> 00:07:47.132
Well V of 12 we knew was 200,

00:07:47.132 --> 00:07:49.558
V of 12 is 200.

00:07:49.558 --> 00:07:51.876
So it's going to be 12 times,

00:07:51.876 --> 00:07:56.555
12 times 200 plus,

00:07:56.555 --> 00:07:57.995
okay the height here

00:07:57.995 --> 00:08:00.131
or the next interval goes from 12 to 20,

00:08:00.131 --> 00:08:02.209
so it's going to be eight times

00:08:02.209 --> 00:08:03.881
and then the height here

00:08:03.881 --> 00:08:07.248
is going to be what is V of 20,

00:08:07.248 --> 00:08:11.904
I think that was V of 20 was 240.

00:08:11.904 --> 00:08:16.501
V of 20 was 240,

00:08:16.501 --> 00:08:19.845
and then the next interval
goes from 20 to 24

00:08:19.845 --> 00:08:22.922
so that only has a change in time of four

00:08:22.922 --> 00:08:25.627
and then what is the height at 24?

00:08:25.627 --> 00:08:27.891
The height at 24,

00:08:27.891 --> 00:08:29.864
and remember we took the
absolute value of V of T

00:08:29.864 --> 00:08:31.989
so it's 220.

00:08:31.989 --> 00:08:35.867
So four times 220

00:08:35.867 --> 00:08:37.586
and then our last rectangle,

00:08:37.586 --> 00:08:41.068
our change in time we go from 24 to 40,

00:08:41.068 --> 00:08:43.181
so our change in time is 16

00:08:43.181 --> 00:08:44.601
times the height at 40.

00:08:44.601 --> 00:08:45.263
What was that?

00:08:45.263 --> 00:08:46.683
That was 150 I think.

00:08:46.683 --> 00:08:49.040
Yup, that was a 150.

00:08:49.040 --> 00:08:51.501
So times 150.

00:08:51.501 --> 00:08:53.776
150,

00:08:53.776 --> 00:08:55.843
and so, just to make
clear what I just did,

00:08:55.843 --> 00:08:58.095
I'm just adding up the sums of the areas

00:08:58.095 --> 00:08:59.338
of these rectangles.

00:08:59.338 --> 00:09:02.902
The area of this
rectangle is 12 times 200,

00:09:02.902 --> 00:09:04.794
area of this rectangle's eight times 240,

00:09:04.794 --> 00:09:07.116
the area of this
rectangle's four times 220,

00:09:07.116 --> 00:09:10.054
area of this rectangle's 16 times 150,

00:09:10.054 --> 00:09:12.017
and then let's see if
we could calculate that,

00:09:12.017 --> 00:09:13.735
we're not allowed to use
our calculator for this part

00:09:13.735 --> 00:09:15.116
so we're going to do it by hand.

00:09:15.116 --> 00:09:18.123
12 times 200 is going to be 2400,

00:09:18.123 --> 00:09:22.302
plus eight times 240,

00:09:22.302 --> 00:09:24.543
that's going to be 1600,

00:09:24.543 --> 00:09:29.543
1600 plus 3200 that is 1920.

00:09:29.721 --> 00:09:30.499
Did I do that right?

00:09:30.499 --> 00:09:32.693
Eight times 40 is 3200,

00:09:32.693 --> 00:09:36.095
or sorry eight times 40 is 320

00:09:36.095 --> 00:09:37.999
plus 1600, yup that's right.

00:09:37.999 --> 00:09:42.004
Plus, this is going to be 880

00:09:42.004 --> 00:09:45.465
plus, let's see 15 times 15

00:09:45.465 --> 00:09:48.715
is going to be, would be 2250

00:09:48.715 --> 00:09:50.793
that would be 15 times 150,

00:09:50.793 --> 00:09:53.138
and so we're going to have another 150

00:09:53.138 --> 00:09:55.623
and so this is going to be 2400.

00:09:55.623 --> 00:10:00.623
2400, whoops, it's 2400.

00:10:01.393 --> 00:10:02.775
Then we want to add all these together,

00:10:02.775 --> 00:10:04.075
let me just do it the way that,

00:10:04.075 --> 00:10:05.619
the easiest way.

00:10:05.619 --> 00:10:10.619
2400, plus 1920, plus 880, plus,

00:10:12.573 --> 00:10:15.244
plus 2400 again,

00:10:15.244 --> 00:10:20.062
plus 2400 again will give us,

00:10:20.062 --> 00:10:23.719
will give us, we get a zero, we get a 10,

00:10:23.719 --> 00:10:27.005
so let's see we have nine plus nine

00:10:27.005 --> 00:10:32.005
is 18, plus eight is 26
and then we have seven.

00:10:34.005 --> 00:10:36.548
So we get 7,600,

00:10:36.548 --> 00:10:41.548
so is equal to 7,600

00:10:42.005 --> 00:10:45.987
and just to be clear,
this would be in meters.

00:10:45.987 --> 00:10:49.226
In, in meters.

00:10:49.226 --> 00:10:50.701
In meters.

00:10:50.701 --> 00:10:51.838
Now you might be saying wait, wait, wait

00:10:51.838 --> 00:10:53.638
didn't we just find an area,

00:10:53.638 --> 00:10:55.960
wouldn't areas be in meters squared?

00:10:55.960 --> 00:10:59.768
But remember, we're finding
the area under the velocity,

00:10:59.768 --> 00:11:01.800
under the velocity graph,

00:11:01.800 --> 00:11:03.146
and so if you look at our units,

00:11:03.146 --> 00:11:04.749
say for this rectangle right here

00:11:04.749 --> 00:11:08.499
you're multiplying 12 minutes,

00:11:08.499 --> 00:11:13.003
you're multiplying 12 minutes times 200,

00:11:13.003 --> 00:11:15.488
times 200 meters per minute,

00:11:15.488 --> 00:11:19.087
times 200 meters per minute.

00:11:19.087 --> 00:11:20.364
So when you multiply them out,

00:11:20.364 --> 00:11:21.885
the minutes the cancel with the minutes

00:11:21.885 --> 00:11:25.542
and you're left with 2400 meters.

00:11:25.542 --> 00:11:26.819
So even though it's area,

00:11:26.819 --> 00:11:28.468
if you look at the units,

00:11:28.468 --> 00:11:32.073
it's going to be in
meters not meters squared.


2015 AP Calculus AB 2a

WEBVTT
Kind: captions
Language: en

00:00:00.547 --> 00:00:03.439
- [Voiceover] Let F and G
be the functions defined by

00:00:03.439 --> 00:00:06.967
F of X is equal to one plus
X plus E to the X squared

00:00:06.967 --> 00:00:10.799
minus two X, and G of
X equal X to the fourth

00:00:10.799 --> 00:00:15.799
minus 6.5X squared plus six X plus two.

00:00:16.046 --> 00:00:19.657
Let R and S be the two
regions enclosed by the graphs

00:00:19.657 --> 00:00:23.442
of F and G shown in the figure above.

00:00:23.442 --> 00:00:25.938
So here I have the graphs
of the two functions

00:00:25.938 --> 00:00:28.759
and they enclose regions R and S.

00:00:28.759 --> 00:00:30.245
So the first thing they
want us to figure out

00:00:30.245 --> 00:00:35.245
is find the sum of the
areas of region R and S.

00:00:35.761 --> 00:00:37.757
So the sum of those areas,

00:00:37.757 --> 00:00:38.628
you can think about it.

00:00:38.628 --> 00:00:41.960
We're going to go from X
equals zero right over here

00:00:41.960 --> 00:00:45.907
to X equals two, so we're
going to take the integral

00:00:45.907 --> 00:00:47.928
from X equals, I'm
going to write this down

00:00:47.928 --> 00:00:52.246
area of R plus S is equal to,

00:00:52.246 --> 00:00:54.557
let me just write that
a little bit neater.

00:00:54.557 --> 00:00:59.557
So the area of R plus S
is going to be equal to,

00:01:01.546 --> 00:01:04.797
let's see we can take the
integral from X equals zero

00:01:04.797 --> 00:01:07.165
to X equals two,

00:01:07.165 --> 00:01:09.418
X equals zero to X equals two,

00:01:09.418 --> 00:01:11.147
and what are we going to integrate?

00:01:11.147 --> 00:01:13.123
Well we're going to
integrate the difference

00:01:13.123 --> 00:01:14.317
between the two functions

00:01:14.317 --> 00:01:16.116
and really the absolute
value of the difference

00:01:16.116 --> 00:01:17.138
where we want the sums,

00:01:17.138 --> 00:01:18.740
we don't want to have negative area here,

00:01:18.740 --> 00:01:22.397
we want the sum of the
areas of regions R and S

00:01:22.397 --> 00:01:25.800
and at some point you have
G of X is above F of X

00:01:25.800 --> 00:01:27.994
and at other points F
of X is above G of X,

00:01:27.994 --> 00:01:29.317
but if we take the absolute value,

00:01:29.317 --> 00:01:31.523
it doesn't matter which one
we're subtracting from the other

00:01:31.523 --> 00:01:33.636
we're just getting the absolute
value of the difference.

00:01:33.636 --> 00:01:36.480
So let's take the absolute value of,

00:01:36.480 --> 00:01:40.834
the absolute value of F of X minus,

00:01:40.834 --> 00:01:44.735
minus G of X DX.

00:01:44.735 --> 00:01:46.441
So that's going to be
the sum of the areas,

00:01:46.441 --> 00:01:47.858
or we could say this is
going to be the integral

00:01:47.858 --> 00:01:50.900
from zero to two of the absolute value

00:01:50.900 --> 00:01:55.900
F of X is one plus X
plus E to the X squared

00:01:56.415 --> 00:02:00.060
minus two X, minus G of X,

00:02:00.060 --> 00:02:05.060
so minus X to the fourth,
plus 6.5X squared,

00:02:05.621 --> 00:02:08.454
minus six X minus two.

00:02:08.454 --> 00:02:10.308
Do the absolute value.

00:02:10.308 --> 00:02:11.620
DX.

00:02:11.620 --> 00:02:13.105
Now this would be pretty hairy to solve

00:02:13.105 --> 00:02:14.547
if we did not have access to a calculator

00:02:14.547 --> 00:02:16.790
but lucky for us on
this part of the AP exam

00:02:16.790 --> 00:02:19.054
we can use a graphing calculator,

00:02:19.054 --> 00:02:22.050
so let's do that to evaluate
this definite integral here.

00:02:22.050 --> 00:02:23.396
If you're wondering why
I'm seeing minus two

00:02:23.396 --> 00:02:25.254
instead of plus two
remember we're substracting,

00:02:25.254 --> 00:02:27.008
we're subtracting G of X, we're finding

00:02:27.008 --> 00:02:28.377
the difference between them.

00:02:28.377 --> 00:02:32.196
So let's input this
function into my calculator

00:02:32.196 --> 00:02:33.288
and I'm going to do it with the same thing

00:02:33.288 --> 00:02:34.589
that I did in part one.

00:02:34.589 --> 00:02:35.830
Where I'm just going to define,

00:02:35.830 --> 00:02:37.409
let me turn it on.

00:02:37.409 --> 00:02:38.454
All right.

00:02:38.454 --> 00:02:41.171
So I'm going to actually clear that out

00:02:41.171 --> 00:02:44.410
and I'm going to define this
whole expression as Y one.

00:02:44.410 --> 00:02:49.410
So I am going to take the absolute value,

00:02:49.948 --> 00:02:52.689
so let me see where the absolute value

00:02:52.689 --> 00:02:56.868
it's been a little while since
I last used one of these.

00:02:56.868 --> 00:02:59.689
So should give me some math,

00:02:59.689 --> 00:03:01.396
math number, oh there you go.

00:03:01.396 --> 00:03:04.658
Absolute value, so it's the absolute value

00:03:04.658 --> 00:03:09.658
of one plus X, plus 2nd E to the

00:03:13.889 --> 00:03:17.115
and then we have X squared

00:03:17.115 --> 00:03:22.115
minus two X and then
close that parenthesis

00:03:23.211 --> 00:03:28.211
and then you have minus
X to the fourth power

00:03:28.772 --> 00:03:33.772
plus 6.5 times X squared minus six times X

00:03:38.373 --> 00:03:41.218
minus two and then we have
to close the parenthesis

00:03:41.218 --> 00:03:42.693
around the absolute value.

00:03:42.693 --> 00:03:44.770
All right, so we inputted Y one

00:03:44.770 --> 00:03:46.257
and so now let's go over here

00:03:46.257 --> 00:03:48.509
and also evaluate this definite integral.

00:03:48.509 --> 00:03:51.214
So we go to math, and then we scroll down

00:03:51.214 --> 00:03:56.214
to definite for function integral.

00:03:56.229 --> 00:03:58.996
So click on that and
we're going to use Y one,

00:03:58.996 --> 00:04:00.398
so we go to variables,

00:04:00.398 --> 00:04:02.235
we go to the right to go to Y variables,

00:04:02.235 --> 00:04:04.417
which it's a function
variable that we just defined,

00:04:04.417 --> 00:04:07.018
and so we select Y one,

00:04:07.018 --> 00:04:08.401
that's what we just inputted.

00:04:08.401 --> 00:04:11.941
Our variable of integration is X

00:04:11.941 --> 00:04:14.077
and our bounds of integration,

00:04:14.077 --> 00:04:15.633
we'll we're going to go from X equals zero

00:04:15.633 --> 00:04:16.817
to X equals two.

00:04:16.817 --> 00:04:19.940
So we go from zero to two

00:04:19.940 --> 00:04:24.038
and then we let the calculator
munch on it a little bit

00:04:24.038 --> 00:04:29.038
and we get, it's taking
some time to calculate,

00:04:31.538 --> 00:04:34.417
it's still, it's still munching on it.

00:04:34.417 --> 00:04:39.227
Let's see this is taking
a good bit of time.

00:04:41.160 --> 00:04:45.444
There you ago, all right,
so it's approximately 2.00,

00:04:45.444 --> 00:04:46.617
if you want to get a
little bit more precise,

00:04:46.617 --> 00:04:48.126
2.004.

00:04:48.126 --> 00:04:53.126
So this is approximately 2.004.


2015 AP Calculus AB 5c

WEBVTT
Kind: captions
Language: en

00:00:00.964 --> 00:00:02.207
- [Voiceover] So part c.

00:00:02.207 --> 00:00:05.039
Find the x-coordinates of
all points of inflection

00:00:05.039 --> 00:00:06.827
for the graph of f.

00:00:06.827 --> 00:00:09.138
Give a reason for your answer.

00:00:09.138 --> 00:00:11.204
So points of inflection,

00:00:11.204 --> 00:00:13.585
points of inflection

00:00:14.745 --> 00:00:17.716
happen

00:00:18.844 --> 00:00:22.474
when we go

00:00:22.474 --> 00:00:25.949
from concave upwards,

00:00:25.949 --> 00:00:29.552
concave upwards

00:00:29.552 --> 00:00:34.552
to downwards

00:00:35.205 --> 00:00:39.415
or vice versa.

00:00:40.325 --> 00:00:43.396
So this is true if and only if

00:00:43.396 --> 00:00:46.116
f prime prime of x

00:00:46.116 --> 00:00:50.954
goes from

00:00:50.954 --> 00:00:54.870
positive

00:00:54.870 --> 00:00:57.680
to negative

00:00:57.680 --> 00:01:02.577
or vice versa.

00:01:02.577 --> 00:01:05.271
So where do we see f prime prime of x

00:01:05.271 --> 00:01:07.676
going from positive to negative?

00:01:07.676 --> 00:01:11.531
Well that's going to
be true if and only if

00:01:11.531 --> 00:01:14.677
f prime of x goes from

00:01:14.677 --> 00:01:17.870
being increasing to
decreasing or vice versa.

00:01:17.870 --> 00:01:20.517
F prime goes

00:01:20.517 --> 00:01:25.517
from increasing

00:01:26.287 --> 00:01:30.513
to decreasing

00:01:30.513 --> 00:01:32.394
or vice versa.

00:01:32.394 --> 00:01:34.646
We're seeing a lot of vice versa here.

00:01:34.646 --> 00:01:36.393
So now let's, and I
want you to think of it

00:01:36.393 --> 00:01:39.319
in terms of f prime because
we have the graph of f prime.

00:01:39.319 --> 00:01:42.442
So f prime goes from
increasing to decreasing

00:01:42.442 --> 00:01:45.786
or vice versa or we could go
from decreasing to increasing.

00:01:45.786 --> 00:01:46.796
Let's think about it.

00:01:46.796 --> 00:01:50.836
Let's see, over here, f prime is

00:01:50.836 --> 00:01:54.598
decreasing, decreasing,
decreasing, decreasing

00:01:54.598 --> 00:01:56.223
and then it increases.

00:01:56.223 --> 00:01:59.102
So we have a point of inflection
right over here, right?

00:01:59.102 --> 00:02:01.634
When f prime of x is zero.

00:02:02.844 --> 00:02:04.366
That's because a prime is differentiable

00:02:04.366 --> 00:02:06.432
so the derivative is definitely,

00:02:06.432 --> 00:02:08.929
the derivative is zero right at that point

00:02:08.929 --> 00:02:10.171
of inflection right over here.

00:02:10.171 --> 00:02:12.872
So with that happens at
x equals negative one.

00:02:12.872 --> 00:02:16.583
And over here, then f
prime starts increasing

00:02:16.583 --> 00:02:18.057
but then right at x equals one

00:02:18.057 --> 00:02:19.915
then it starts decreasing.

00:02:19.915 --> 00:02:23.212
So at x equals one, we have
another point of inflection.

00:02:23.212 --> 00:02:26.359
That's where we have that zero,

00:02:26.359 --> 00:02:30.979
that zero of tangent line with slope zero.

00:02:30.979 --> 00:02:32.419
And then we're decreasing, decreasing,

00:02:32.419 --> 00:02:35.577
decreasing, decreasing,
decreasing, increasing.

00:02:35.577 --> 00:02:36.995
All right, so this is going to be another

00:02:36.995 --> 00:02:39.896
point of inflection, x equals three.

00:02:39.896 --> 00:02:43.006
So these are at three
points of inflection.

00:02:43.006 --> 00:02:47.248
So this happens,

00:02:47.248 --> 00:02:52.117
happens at x equals negative one,

00:02:52.117 --> 00:02:56.401
x equals one and x equals three.

00:02:56.401 --> 00:03:01.401
So these three points
on our graph of f prime

00:03:01.521 --> 00:03:03.634
where we see f prime goes
from decreasing to increasing

00:03:03.634 --> 00:03:07.210
or increasing to decreasing
or decreasing to increasing.

00:03:07.210 --> 00:03:10.533
All right, all right.

00:03:10.533 --> 00:03:14.403
Now, well, I'll do the last
part on the next video.


2015 AP Calculus AB/BC 4cd

WEBVTT
Kind: captions
Language: en

00:00:00.239 --> 00:00:04.424
- [Voiceover] Part c, let y
equals f of x be the particular

00:00:04.424 --> 00:00:07.040
solution to the differential
equation with the

00:00:07.040 --> 00:00:11.486
initial condition f of
two is equal to three.

00:00:11.486 --> 00:00:15.714
Does f have a relative
minimum, a relative maximum,

00:00:15.714 --> 00:00:18.767
or neither at x equals two?

00:00:18.767 --> 00:00:20.889
Justify your answer.

00:00:20.889 --> 00:00:22.199
Well, to think about
whether we have a realtive

00:00:22.199 --> 00:00:24.258
minimum or relative maximum, we can say,

00:00:24.258 --> 00:00:26.706
well, what's the derivative at that point?

00:00:26.706 --> 00:00:28.633
If it's zero, then it's a good candidate

00:00:28.633 --> 00:00:30.063
that we're dealing with a,

00:00:30.063 --> 00:00:32.549
that it could be a relative
minimum or maximum,

00:00:32.549 --> 00:00:34.751
if it's not zero, then it's neither.

00:00:34.751 --> 00:00:36.114
And then if it is zero,
if we wanna figure out

00:00:36.114 --> 00:00:37.705
relative minimum or relative maximum,

00:00:37.705 --> 00:00:40.845
we can evaluate the sign
of the second derivative.

00:00:40.845 --> 00:00:42.360
So let's just think about this.

00:00:42.360 --> 00:00:45.675
So, we want to evaluate f prime,

00:00:45.675 --> 00:00:49.314
we wanna figure out what f prime of,

00:00:49.314 --> 00:00:53.142
f prime of two is equal to.

00:00:53.142 --> 00:00:56.712
So, we know that f prime of x,

00:00:56.712 --> 00:01:01.576
f prime of x, which is
the same thing as dy dx,

00:01:01.576 --> 00:01:05.001
is equal to two times x minus y.

00:01:05.001 --> 00:01:06.982
We saw that in the last problem.

00:01:06.982 --> 00:01:10.028
And so f prime of two,
I'll write it this way,

00:01:10.028 --> 00:01:14.184
f prime of two is going to be equal to

00:01:14.184 --> 00:01:16.064
two times two,

00:01:16.064 --> 00:01:19.121
two times two minus

00:01:19.121 --> 00:01:21.761
whatever y is when x is equal to two.

00:01:21.761 --> 00:01:24.668
Well do we know what y
is when x equals to two?

00:01:24.668 --> 00:01:26.873
Sure, they tell us right over here.

00:01:26.873 --> 00:01:29.056
Y is equal to f of x,

00:01:29.056 --> 00:01:31.029
so when x is equal to two,

00:01:31.029 --> 00:01:32.664
when x is equal to two,

00:01:32.664 --> 00:01:35.197
y is equal to three,

00:01:35.197 --> 00:01:37.937
so two times two minus three.

00:01:37.937 --> 00:01:40.653
And so this is going to be
equal to four minus three

00:01:40.653 --> 00:01:42.128
is equal to one.

00:01:42.128 --> 00:01:47.128
And so since the derivative
at two is not zero,

00:01:47.190 --> 00:01:50.081
this is not going to be a
minimum, a relative minimum

00:01:50.081 --> 00:01:53.112
or a relative maximum, so you could say

00:01:53.112 --> 00:01:58.112
since, since f prime of two,

00:01:59.240 --> 00:02:02.514
f prime of two does not equal zero,

00:02:02.514 --> 00:02:07.411
this, we have a, f has,
let me write it this way

00:02:07.411 --> 00:02:12.411
f has neither, neither minimum,

00:02:14.114 --> 00:02:16.274
or relative minimum I guess I could say,

00:02:16.274 --> 00:02:19.535
relative min or

00:02:19.535 --> 00:02:23.045
relative max

00:02:23.045 --> 00:02:26.324
at x equals two.

00:02:26.324 --> 00:02:29.064
Alright, let's do the next one.

00:02:29.064 --> 00:02:32.942
Find the values of the constants m and b,

00:02:32.942 --> 00:02:37.487
for which y equals m
x plus b is a solution

00:02:37.487 --> 00:02:39.583
to the differential equation.

00:02:39.583 --> 00:02:41.905
Alright, this one is interesting.

00:02:41.905 --> 00:02:44.078
So let's, actually let's
just write down everything we

00:02:44.078 --> 00:02:47.257
know before we even think
that y equals m x plus b

00:02:47.257 --> 00:02:49.903
could be a solution to
the differential equation.

00:02:49.903 --> 00:02:54.507
So, we know that, we know that dy

00:02:54.507 --> 00:02:58.939
over dx is equal to two x minus y,

00:02:58.939 --> 00:03:00.273
they told us that.

00:03:00.273 --> 00:03:02.281
We also know that the second derivative,

00:03:02.281 --> 00:03:07.281
the second derivative of y,
with respect to x, is equal to

00:03:07.519 --> 00:03:12.519
two minus dy dx. We
figured that out in part b

00:03:12.824 --> 00:03:14.333
of this problem.

00:03:14.333 --> 00:03:17.269
And then, we could also express this

00:03:17.269 --> 00:03:18.911
we saw that we could also write that

00:03:18.911 --> 00:03:22.599
as two minus two x plus
y, if you just substitute,

00:03:22.599 --> 00:03:25.375
if you substitute this in for that.

00:03:25.375 --> 00:03:27.826
So it's two minus two x plus y.

00:03:27.826 --> 00:03:29.282
So let me write it that way.

00:03:29.282 --> 00:03:33.603
This is also equal to
two minus two x plus y.

00:03:33.603 --> 00:03:35.785
So that's everything that we know

00:03:35.785 --> 00:03:39.081
before we even thought that
maybe there's a solution

00:03:39.081 --> 00:03:42.189
of y equals m x plus b.

00:03:42.189 --> 00:03:44.513
So now let's start with
y equals m x plus b.

00:03:44.513 --> 00:03:48.924
So if y is equal to m x
plus b, y equals m x plus b,

00:03:48.924 --> 00:03:50.547
so this is the equation of a line,

00:03:50.547 --> 00:03:54.996
then dy dx is going to be equal to,

00:03:54.996 --> 00:03:57.362
well the derivative of this
with respect to x is just m,

00:03:57.362 --> 00:04:00.007
the derivative of this with
respect to x, this is constant

00:04:00.007 --> 00:04:01.576
this is not going to
change with respect to x,

00:04:01.576 --> 00:04:02.842
it's just zero.

00:04:02.842 --> 00:04:03.654
And that makes sense,

00:04:03.654 --> 00:04:06.864
the rate of change of y with
respect to x is the slope,

00:04:06.864 --> 00:04:09.460
is the slope of our line.

00:04:09.460 --> 00:04:12.163
So, can we use, and this
is really all that we know,

00:04:12.163 --> 00:04:14.233
we could keep, actually
we could go even further,

00:04:14.233 --> 00:04:16.074
we could take the second derivative here,

00:04:16.074 --> 00:04:19.177
the second derivative
of y with respect to x.

00:04:19.177 --> 00:04:21.009
Well, that's going to be zero.

00:04:21.009 --> 00:04:23.549
The second derivative of
a, of a linear function,

00:04:23.549 --> 00:04:25.582
well it's going to be
zero, you see that here.

00:04:25.582 --> 00:04:28.592
So this is, this is all of
the information that we have.

00:04:28.592 --> 00:04:31.350
We get this from the previous
parts of the problem,

00:04:31.350 --> 00:04:33.890
and we get this just taking the
first and second derivatives

00:04:33.890 --> 00:04:37.317
of, of y equals m x plus b.

00:04:37.317 --> 00:04:41.230
So given this, can we figure out,

00:04:41.230 --> 00:04:45.391
can we figure out what m and b are?

00:04:45.391 --> 00:04:50.204
Alright, so we could,
if we said m is equal to

00:04:50.204 --> 00:04:55.204
two x minus y, that doesn't seem right.

00:04:55.546 --> 00:04:57.950
This one is a tricky one.

00:04:57.950 --> 00:05:00.806
Well, let's see, we know
that the second derivative

00:05:00.806 --> 00:05:03.453
is going to be equal to zero.

00:05:03.453 --> 00:05:05.542
We know that this is
going to be equal to zero

00:05:05.542 --> 00:05:08.317
for this particular solution.

00:05:08.317 --> 00:05:12.149
And we know dy dx is equal to m.

00:05:12.149 --> 00:05:13.717
We know this is m.

00:05:13.717 --> 00:05:15.220
And so there you have it,
we have enough information

00:05:15.220 --> 00:05:16.032
to solve for m.

00:05:16.032 --> 00:05:19.011
We know that zero is equal to two minus m.

00:05:19.011 --> 00:05:22.299
So zero is equal to two minus m.

00:05:22.299 --> 00:05:25.120
And so we can add m to
both sides and we get

00:05:25.120 --> 00:05:30.120
m is equal to, m is equal to two.

00:05:30.832 --> 00:05:34.942
So, that by itself, was quite useful.

00:05:34.942 --> 00:05:38.158
And then, what we could say, let's see,

00:05:38.158 --> 00:05:40.933
can we solve this further?

00:05:40.933 --> 00:05:43.497
Well we know that this,
right over here, dy dx,

00:05:43.497 --> 00:05:46.285
this is m, this is m.

00:05:46.285 --> 00:05:48.688
And it's equal to two.

00:05:48.688 --> 00:05:52.531
So, we could say that two
is equal to 2 x minus y.

00:05:52.531 --> 00:05:56.455
Two is equal to 2 x minus y.

00:05:56.455 --> 00:05:58.320
And then let's see, if we solve for y,

00:05:58.320 --> 00:06:01.342
add y to both sides,
subtract two from both sides,

00:06:01.342 --> 00:06:05.152
we get y is equal to two x minus two.

00:06:05.152 --> 00:06:07.328
And there we have our whole solution.

00:06:07.328 --> 00:06:09.973
And so you have your m, right over there.

00:06:09.973 --> 00:06:10.952
That is m.

00:06:10.952 --> 00:06:14.572
And then we also have our b.

00:06:14.572 --> 00:06:16.819
This one was a tricky one.

00:06:16.819 --> 00:06:19.717
Anytime that you, you have to, ya know,

00:06:19.717 --> 00:06:21.761
do something like this and it
doesn't just jump out at you,

00:06:21.761 --> 00:06:24.005
and if it wasn't obvious, it
didn't jump out at me at first,

00:06:24.005 --> 00:06:26.007
when I looked at this problem,
I said, well let me just

00:06:26.007 --> 00:06:28.260
write down everything that they told us,

00:06:28.260 --> 00:06:29.650
so they wrote this before,

00:06:29.650 --> 00:06:32.339
and then we say okay this
is going to be a solution.

00:06:32.339 --> 00:06:35.701
And so let me see if I
can somehow solve, so,

00:06:35.701 --> 00:06:37.057
let's see what I didn't use.

00:06:37.057 --> 00:06:39.867
I didn't use, I didn't use that.

00:06:39.867 --> 00:06:42.119
I did use this.

00:06:42.119 --> 00:06:44.569
I absolutely used that.

00:06:44.569 --> 00:06:47.669
I did use that, I did use
that, and I did use that.

00:06:47.669 --> 00:06:49.666
So this was a little bit
of a fun little puzzle

00:06:49.666 --> 00:06:52.162
where I just wrote down all
the information they gave us

00:06:52.162 --> 00:06:54.484
and I tried to figure out, based on that,

00:06:54.484 --> 00:06:59.255
whether I could figure out m and m and b.

00:06:59.255 --> 00:07:01.143
And this is pretty neat,
that this a solution

00:07:01.143 --> 00:07:02.725
two x minus two.

00:07:02.725 --> 00:07:05.632
If we go to our slope
field above, it's not,

00:07:05.632 --> 00:07:07.027
it wouldn't have jumped out at me.

00:07:07.027 --> 00:07:09.263
But if you think about, if you think about

00:07:09.263 --> 00:07:11.300
so two x minus two, it's
y intercept would be

00:07:11.300 --> 00:07:14.356
negative two like that, let me
do this in a different color,

00:07:14.356 --> 00:07:17.717
and so the line would look something,

00:07:17.717 --> 00:07:20.405
would like something like this.

00:07:20.405 --> 00:07:22.727
The line would look something like this.

00:07:22.727 --> 00:07:25.548
And you can verify that
any one of these points,

00:07:25.548 --> 00:07:28.202
at any one of these points, the slope,

00:07:28.202 --> 00:07:33.202
the slope is equal, the
slope is equal to two.

00:07:33.267 --> 00:07:36.666
If we're at the point two comma two,

00:07:36.666 --> 00:07:39.174
well it's gonna be two
times two minus two is two.

00:07:39.174 --> 00:07:42.680
One comma zero, two times
one minus zero is two.

00:07:42.680 --> 00:07:46.268
Negative two comma, or sorry
zero comma negative two

00:07:46.268 --> 00:07:49.240
well, zero minus negative two, that's two.

00:07:49.240 --> 00:07:51.202
So you see this is
pretty neat, the slope is

00:07:51.202 --> 00:07:53.570
changing all around it, but this is

00:07:53.570 --> 00:07:55.356
this is a linear solution

00:07:55.356 --> 00:07:57.402
to that original differential
equation, that was,

00:07:57.402 --> 00:07:59.730
that was pretty cool.


2015 AP Calculus AB 5a

WEBVTT
Kind: captions
Language: en

00:00:00.256 --> 00:00:03.652
- [Voiceover] The figure above
shows the graph of f prime,

00:00:03.652 --> 00:00:06.612
the derivative of a
twice-differentiable function f,

00:00:06.612 --> 00:00:08.047
on the interval, that's a closed interval,

00:00:08.047 --> 00:00:09.580
from negative three to four.

00:00:09.580 --> 00:00:12.262
The graph of f prime
has horizontal tangents

00:00:12.262 --> 00:00:14.526
at x equals negative one, x equals one,

00:00:14.526 --> 00:00:15.860
and x equals three.

00:00:15.860 --> 00:00:19.192
So you have a horizontal
tangent right over,

00:00:19.192 --> 00:00:21.921
a horizontal tangent right over there.

00:00:23.041 --> 00:00:24.524
And let me draw that a little bit neater,

00:00:24.524 --> 00:00:25.458
right over there,

00:00:25.458 --> 00:00:27.320
a horizontal tangent right over there,

00:00:27.320 --> 00:00:29.362
and a horizontal tangent right over there.

00:00:30.976 --> 00:00:31.557
Alright.

00:00:31.557 --> 00:00:33.345
The areas of the regions bounded

00:00:33.345 --> 00:00:36.475
by the x-axis and the graph of f prime

00:00:36.475 --> 00:00:38.731
on the intervals negative two to one,

00:00:38.731 --> 00:00:40.346
closed intervals from negative two to one,

00:00:40.346 --> 00:00:42.694
so this region right over here,

00:00:42.694 --> 00:00:45.142
and the region from one to four,

00:00:45.142 --> 00:00:46.591
so this region right over there,

00:00:46.591 --> 00:00:48.494
they tell us have,

00:00:48.494 --> 00:00:50.911
have the areas are 9 and 12, respectively.

00:00:50.911 --> 00:00:54.203
So that area's 9 and that area is 12.

00:00:54.203 --> 00:00:55.729
So Part A,

00:00:55.729 --> 00:00:59.955
Find all x coordinates at
which f has a relative maximum.

00:00:59.955 --> 00:01:03.473
Give a reason for your answer.

00:01:03.473 --> 00:01:05.527
All x-coordinates at which f

00:01:05.527 --> 00:01:06.955
has a relative maximum.

00:01:06.955 --> 00:01:07.629
So you might say,

00:01:07.629 --> 00:01:09.196
"Oh wait, wait this looks
like a relative maximum

00:01:09.196 --> 00:01:10.509
over here, but this isn't f.

00:01:10.509 --> 00:01:12.413
This is the graph of f prime."

00:01:12.413 --> 00:01:13.793
So let's think about when we used to

00:01:13.793 --> 00:01:15.490
we don't have a graph of f in front of us.

00:01:15.490 --> 00:01:17.427
So let's think about
what it needs to be true

00:01:17.427 --> 00:01:20.943
for f to have a relative
maximum at a point.

00:01:20.943 --> 00:01:23.210
We are probably familiar with

00:01:23.210 --> 00:01:25.415
what relative maximum will look like.

00:01:25.415 --> 00:01:28.214
It'd look like a little lump, like that.

00:01:28.214 --> 00:01:30.072
It could also actually look like that

00:01:30.072 --> 00:01:32.173
but since this is differentiable function

00:01:32.173 --> 00:01:33.671
over the interval, we're
probably not dealing

00:01:33.671 --> 00:01:36.213
with a relative maximum
that looks like that.

00:01:37.571 --> 00:01:41.803
And so what do we know about
a relative maximum point?

00:01:41.803 --> 00:01:44.754
So let's say that's our relative maximum.

00:01:44.754 --> 00:01:46.883
As we approach our relative maximum

00:01:46.883 --> 00:01:50.202
from values, as we have x
values that are approaching

00:01:50.202 --> 00:01:54.043
the x value of our relative maximum point,

00:01:54.043 --> 00:01:56.659
as we approach it from
values below that x value,

00:01:56.659 --> 00:01:59.699
we see that we have a positive slope.

00:01:59.699 --> 00:02:02.839
Our function needs to be increasing.

00:02:02.839 --> 00:02:04.751
So over here.

00:02:05.784 --> 00:02:08.686
Over here we see f is increasing,

00:02:08.686 --> 00:02:11.827
going into the relative maximum point,

00:02:11.827 --> 00:02:12.726
f is increasing,

00:02:12.726 --> 00:02:15.227
which means that the derivative of f

00:02:15.227 --> 00:02:17.312
the derivative of f must
be greater than zero.

00:02:18.342 --> 00:02:20.874
And then after we past that maximum point,

00:02:20.874 --> 00:02:22.491
after we past that maximum point,

00:02:22.491 --> 00:02:26.247
we see that our function
needs to be decreasing.

00:02:26.247 --> 00:02:27.569
Do this in another color.

00:02:27.569 --> 00:02:29.711
We see that our funciton is decreasing

00:02:29.711 --> 00:02:33.090
right over here, so f decreasing,

00:02:34.100 --> 00:02:36.764
decreasing, which means

00:02:36.764 --> 00:02:39.684
that f prime of x needs
to be less than zero.

00:02:40.524 --> 00:02:43.745
So our relative maximum point

00:02:43.745 --> 00:02:46.247
should happen at an x value.

00:02:46.247 --> 00:02:47.312
It should happen at an x value

00:02:47.312 --> 00:02:50.284
where our first derivative transitions

00:02:50.284 --> 00:02:52.478
from being greater than zero

00:02:52.478 --> 00:02:55.244
to being less than zero.

00:02:55.244 --> 00:02:56.437
So what x value,

00:02:56.437 --> 00:02:57.262
so let me say this,

00:02:57.262 --> 00:03:00.737
so we have

00:03:00.737 --> 00:03:05.407
f has relative, let me
just write it shorthand,

00:03:05.407 --> 00:03:10.351
relative maximum at x values

00:03:11.505 --> 00:03:13.526
where

00:03:13.526 --> 00:03:16.561
f prime transitions,

00:03:16.561 --> 00:03:17.631
transitions,

00:03:19.022 --> 00:03:21.703
transitions from

00:03:21.703 --> 00:03:26.272
from positive, positive,

00:03:27.871 --> 00:03:32.713
to negative, to, let me write
this a little bit neater,

00:03:32.713 --> 00:03:34.493
to negative.

00:03:35.653 --> 00:03:37.713
to negative.

00:03:37.713 --> 00:03:39.923
And where do we see f prime transitioning

00:03:39.923 --> 00:03:41.902
from positive to negative?

00:03:41.902 --> 00:03:43.933
Well over here we only
see that happening once.

00:03:43.933 --> 00:03:47.138
We see right here f prime is
positive, positive, positive,

00:03:47.138 --> 00:03:49.077
and then it goes negative,
negative, negative.

00:03:49.077 --> 00:03:50.797
So we see,

00:03:50.797 --> 00:03:54.143
we see f prime is positive over here,

00:03:55.273 --> 00:03:58.020
And then, right when we
hit x equals negative two,

00:03:58.020 --> 00:04:00.702
f prime becomes negative.

00:04:00.702 --> 00:04:03.849
F prime becomes negative.

00:04:03.849 --> 00:04:05.706
So we know that the function itself,

00:04:05.706 --> 00:04:06.426
not f prime,

00:04:06.426 --> 00:04:07.877
f must be increasing here

00:04:07.877 --> 00:04:10.768
because f prime is positive,

00:04:10.768 --> 00:04:13.369
and then our function at f

00:04:13.369 --> 00:04:14.518
is decreasing here

00:04:14.518 --> 00:04:17.084
because f prime is negative.

00:04:18.054 --> 00:04:20.969
And so this happens at x equals two.

00:04:20.969 --> 00:04:22.773
So let me write that down.

00:04:22.773 --> 00:04:24.870
This happens at x equals two.

00:04:24.870 --> 00:04:27.550
This happens,

00:04:27.550 --> 00:04:31.969
happens at x equals two.

00:04:31.969 --> 00:04:33.301
And we're done.