2011 Calculus AB free response #4b

WEBVTT
Kind: captions
Language: en

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Part b.

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"Determine the x-coordinate
of the point at which g

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has an absolute maximum
on the interval negative 4

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is less than or equal to x,
is less than or equal to 3.

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And justify your answer."

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So let's just think about
it in general terms.

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If we just think about
a general function

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over an interval, where it
could have an absolute maximum.

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So let me draw some
axes over here.

00:00:22.750 --> 00:00:25.110
And I'm speaking in
general terms first,

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and then we can go back
to our function g, which

00:00:27.110 --> 00:00:29.920
is derived from this
function f right over here.

00:00:29.920 --> 00:00:32.560
So let's say that these
are my coordinate axes,

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and let's say we care
about some interval here.

00:00:34.950 --> 00:00:37.520
So let's say this is the
interval that I care about.

00:00:37.520 --> 00:00:39.480
A function could look
something like this.

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And in this case,
its absolute maximum

00:00:46.190 --> 00:00:49.000
is going to occur at the
beginning of the interval.

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Or a function could look
something like this.

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And then the absolute
maximum could

00:00:55.430 --> 00:00:58.070
occur at the end
point of the interval.

00:00:58.070 --> 00:01:00.550
Or the other possibility
is that the function

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looks something like this.

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At which point,
the maximum would

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be at this critical point.

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And I say critical point, as
opposed to just a point where

00:01:13.750 --> 00:01:16.419
the slope is zero, because
it's possible to the functions

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not differentiable there.

00:01:17.460 --> 00:01:21.819
You could imagine a function
that looks like this,

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and maybe wouldn't be
differentiable there.

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But that point there still
would be the absolute maximum.

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So what we really
just have to do

00:01:31.270 --> 00:01:35.510
is evaluate g at the different
endpoints of this interval,

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to see how high it gets,
or how large of a value

00:01:38.850 --> 00:01:40.800
we get for the g
at the end points.

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And then we have to see if g has
any critical points in between.

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And then evaluate
it there to see

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if that's a candidate
for the global maximum.

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So let's just evaluate g
of the different points.

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So let's start off, let's
evaluate g at negative 4--

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at kind of the lowest
end, or the starting point

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of our interval.

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So g of negative 4 is
equal to 2 times negative 4

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plus the integral from 0
to negative 4 f of t dt.

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The first part is very easy, 2
times negative 4 is negative 8.

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Let me do it over here so
I have some real estate.

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So this is equal to negative 8.

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And instead of leaving this
as 0 to negative 4 f of t dt,

00:02:27.110 --> 00:02:31.280
let's change the bounds
of integration here.

00:02:31.280 --> 00:02:34.910
Especially so that we can get
the lower number as the lower

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bound.

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And that way, it
becomes a little bit

00:02:37.121 --> 00:02:39.190
more natural to think
of it in terms of areas.

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So this expression
right here can

00:02:41.540 --> 00:02:44.460
be rewritten as the
negative of the integral

00:02:44.460 --> 00:02:49.410
between negative
4 and 0 f of t dt.

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And now this expression
right over here

00:02:52.310 --> 00:02:55.860
is the area under f of
t, or in this case, f

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of x-- or the area under f
between negative 4 and 0.

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So it's this area
right over here.

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And we have to be careful,
because this part over here

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is below the x-axis.

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So this we would
consider negative area

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when we think of it
in integration terms.

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And this would be positive area.

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So the total area here is
going to be this positive area

00:03:19.640 --> 00:03:22.050
minus this area right over here.

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So let's think
about what this is.

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So this area-- This section
over here we did this in part a,

00:03:27.570 --> 00:03:30.540
actually, This section--
this is a quarter circle,

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so it's 1/4-- so these
are both quarter circles.

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So we could multiply 1/4 times
the area of this entire circle,

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if we were to draw the entire
thing all the way around.

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It has a radius of 3.

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So the area of the entire circle
would be pi times 3 squared

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or it would be 9 pi.

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And of course we're
going to divide it

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by 4-- multiply it by 1/4 to
just get this quarter circle

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right over there.

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And then this area
right over here,

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the area of the entire
circle, we have a radius of 1.

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So it's going to be
pi times 1 squared.

00:04:08.437 --> 00:04:10.020
So it's going to be
pi, and then we're

00:04:10.020 --> 00:04:13.525
going to divide it
by four, because it's

00:04:13.525 --> 00:04:15.630
only one fourth of that circle.

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And we're going
to subtract that.

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So we have negative pi
and we were multiplying it

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times 1/4 out here,
because it's just

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a quarter circle in either case.

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And we're subtracting
it, because the area

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is below the x-axis.

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And so this simplifies to--
this is equal to 1/4 times

00:04:34.210 --> 00:04:37.767
8 pi, which is the
same thing as 2 pi.

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Did I do that right?

00:04:38.600 --> 00:04:39.850
1/4 times 8 pi.

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So this all simplifies to 2 pi.

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So g of negative 4 is equal
to negative 8 minus 2 pi.

00:04:51.670 --> 00:04:54.010
So clearly it is a
negative number here.

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More negative than negative 8.

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So let's try the other bounds.

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So let's see what
g of positive 3 is.

00:05:02.880 --> 00:05:05.510
I'll do it over here so
I have some more space.

00:05:05.510 --> 00:05:09.290
So g of positive 3,
when x is equal to 3,

00:05:09.290 --> 00:05:12.090
that-- we go back to our
definition-- that is 2 times

00:05:12.090 --> 00:05:18.760
3 plus the integral
from 0 to 3 f of t dt.

00:05:18.760 --> 00:05:21.220
And this is going to be
equal to 2 times 3 is 6.

00:05:21.220 --> 00:05:23.740
And the integral from 0
to 3 f of t dt, that's

00:05:23.740 --> 00:05:24.880
this entire area.

00:05:24.880 --> 00:05:27.010
So we have positive
area over here.

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And then we have an equal
negative area right over here

00:05:30.220 --> 00:05:32.160
because it's below the x-axis.

00:05:32.160 --> 00:05:34.764
So the integral from 0 to
3 is just going to be 0,

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you're going to have
this positive area,

00:05:36.430 --> 00:05:38.629
and then this negative
area right over here

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is going to completely
cancel it out.

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Because it's symmetric
right over here.

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So this thing is going
to evaluate to 0.

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So g of 3 is 6.

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So we already know that
our starting point, g

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of negative 4-- that when
x is equal to negative 4--

00:05:51.930 --> 00:05:54.825
that is not where g
hits a global maximum.

00:05:54.825 --> 00:05:56.200
Because that's a
negative number.

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And we already
found the end point,

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where g hits a positive value.

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So negative 4 is
definitely not a candidate.

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x is equal to 3 is still in the
running for the x-coordinate

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where g has a global
absolute maximum.

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Now what we have
to do is figure out

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any critical points
that g has in between.

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So points in there where
it's either undifferentiable

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or its derivative is equal to 0.

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So let's look at
this derivative.

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So g prime of x-- we just take
the derivative of this business

00:06:24.910 --> 00:06:25.410
up here.

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Derivative of 2x is 2.

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Derivative of this definition
going from 0 to x of f of t

00:06:30.580 --> 00:06:33.640
dt-- we did that in part a, this
is just the fundamental theorem

00:06:33.640 --> 00:06:38.270
of calculus-- this
is just going to be

00:06:38.270 --> 00:06:43.040
plus f of x right over there.

00:06:43.040 --> 00:06:44.860
So it actually
turns out that g is

00:06:44.860 --> 00:06:47.190
differentiable over
the entire interval.

00:06:47.190 --> 00:06:51.430
You give any x value
over this interval,

00:06:51.430 --> 00:06:54.260
we have a value for f of x.
f of x isn't differentiable

00:06:54.260 --> 00:06:55.950
everywhere, but
definitely f of x

00:06:55.950 --> 00:06:57.980
is defined everywhere,
over the interval.

00:06:57.980 --> 00:07:01.120
So you'll get a number here,
and obviously two is just two,

00:07:01.120 --> 00:07:03.501
and you add two to it, and
you get the derivative of g

00:07:03.501 --> 00:07:04.750
at that point in the interval.

00:07:04.750 --> 00:07:07.300
So g is actually differentiable
throughout the interval.

00:07:07.300 --> 00:07:08.810
So the only critical
points would

00:07:08.810 --> 00:07:12.100
be where this derivative
is equal to 0.

00:07:12.100 --> 00:07:14.850
So let's set this
thing equal to 0.

00:07:14.850 --> 00:07:17.230
So we want to solve the
equation-- I'll just rewrite it

00:07:17.230 --> 00:07:20.140
actually-- So we want to solve
the equation g prime of x

00:07:20.140 --> 00:07:25.400
is equal to 0, or 2 plus
f of x is equal to 0.

00:07:25.400 --> 00:07:27.150
You can subtract
2 from both sides,

00:07:27.150 --> 00:07:30.010
and you get f of x is
equal to negative 2.

00:07:30.010 --> 00:07:33.750
So any x that satisfies f
of x is equal to negative 2

00:07:33.750 --> 00:07:38.630
is a point where the
derivative of g is equal to 0.

00:07:38.630 --> 00:07:43.290
And let's see if f of x is equal
to negative 2 at any point.

00:07:43.290 --> 00:07:47.179
So let me draw a line
over here at negative 2.

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We have to look at it visually,
because there's only given us

00:07:49.720 --> 00:07:52.111
this visual
definition of f of x.

00:07:52.111 --> 00:07:54.360
Doesn't equal negative 2,
doesn't equal to negative 2,

00:07:54.360 --> 00:07:59.800
only equals negative
2 right over there.

00:07:59.800 --> 00:08:02.991
And it looks like we're at about
2 and 1/2, but let's get exact.

00:08:02.991 --> 00:08:04.990
Let's actually figure out
the slope of the line,

00:08:04.990 --> 00:08:06.820
and figure out what
x value actually

00:08:06.820 --> 00:08:09.632
gives f of x equal
to negative 2.

00:08:09.632 --> 00:08:11.840
And we could figure out the
slope of this line fairly

00:08:11.840 --> 00:08:15.020
visually-- or figure out the
equation of this line fairly

00:08:15.020 --> 00:08:17.360
visually, we can
figure out its slope.

00:08:17.360 --> 00:08:24.630
If we run 3-- if our
change in x is 3,

00:08:24.630 --> 00:08:29.890
then our change in y,
our rise, is negative 6.

00:08:29.890 --> 00:08:32.450
Change in y is
equal to negative 6.

00:08:32.450 --> 00:08:36.110
Slope is rise over run, or
change in y over change in x.

00:08:36.110 --> 00:08:38.992
So negative 6 divided
by 3 is negative 2.

00:08:38.992 --> 00:08:40.200
It has a slope of negative 2.

00:08:40.200 --> 00:08:42.033
And actually, I could
have done that easier.

00:08:42.033 --> 00:08:45.780
Where if we go forward
one, we go down by 2.

00:08:45.780 --> 00:08:48.760
So we could have seen that
the slope is negative 2.

00:08:48.760 --> 00:08:52.000
So this part of
f of x, we have y

00:08:52.000 --> 00:08:55.260
is equal to negative
2x plus-- and then

00:08:55.260 --> 00:08:57.200
the y-intercept is
pretty straightforward.

00:08:57.200 --> 00:08:59.960
This is at 3-- 1, 2, 3.

00:08:59.960 --> 00:09:01.830
Negative 2x plus 3.

00:09:01.830 --> 00:09:05.500
So part of f of x
where clearly it

00:09:05.500 --> 00:09:08.450
equals negative 2 at some point
of that-- this part of f of x

00:09:08.450 --> 00:09:09.890
is defined by this line.

00:09:09.890 --> 00:09:12.604
Obviously this part of f of x
is not defined by that line.

00:09:12.604 --> 00:09:14.020
But to figure out
the exact value,

00:09:14.020 --> 00:09:15.920
we just have to figure
out when this line is

00:09:15.920 --> 00:09:18.100
equal to negative 2.

00:09:18.100 --> 00:09:26.620
So we have negative 2x plus
3 is equal to negative 2.

00:09:26.620 --> 00:09:29.010
And remember, this isn't--
this is what f of x is equal

00:09:29.010 --> 00:09:31.089
to, over the interval
that we care about.

00:09:31.089 --> 00:09:32.880
If we were talking
about f of x over there,

00:09:32.880 --> 00:09:34.879
we wouldn't be able to
put a negative 2x plus 3,

00:09:34.879 --> 00:09:37.930
we would have to have some form
of equation for these circles.

00:09:37.930 --> 00:09:40.910
But right over here, this is
what f of x is, and now we

00:09:40.910 --> 00:09:43.730
can solve this pretty
straightforwardly.

00:09:43.730 --> 00:09:48.470
So we can subtract
3 from both sides,

00:09:48.470 --> 00:09:51.750
and we get negative 2x
is equal to negative 5.

00:09:51.750 --> 00:09:53.670
Divide both sides
by negative 2, you

00:09:53.670 --> 00:09:56.780
get x is equal to negative
5 over negative 2,

00:09:56.780 --> 00:09:58.352
which is equal to 5/2.

00:09:58.352 --> 00:09:59.810
Which is exactly
what we thought it

00:09:59.810 --> 00:10:02.160
was when we looked
at it visually.

00:10:02.160 --> 00:10:04.160
It looked like we were
at about 2 and 1/2, which

00:10:04.160 --> 00:10:06.220
is the same thing as 5/2.

00:10:06.220 --> 00:10:08.376
Now we don't know what this is.

00:10:08.376 --> 00:10:10.250
We don't know if this
is an inflection point.

00:10:10.250 --> 00:10:11.260
Is this a maximum?

00:10:11.260 --> 00:10:12.280
Is this a minimum?

00:10:12.280 --> 00:10:13.970
So really we just
want to evaluate g

00:10:13.970 --> 00:10:17.686
at this point to see if it gets
higher than when we evaluate g

00:10:17.686 --> 00:10:18.600
at 3.

00:10:18.600 --> 00:10:24.140
So let's evaluate g at 5/2.

00:10:24.140 --> 00:10:27.030
So g at 5/2 is going
to be equal to 2 times

00:10:27.030 --> 00:10:42.020
15/2 plus the integral
from 0 to 5/2 of f of t dt.

00:10:42.020 --> 00:10:45.830
So this first part right over
here, the 2's cancel out.

00:10:45.830 --> 00:10:50.020
So this is going
to be equal to 5.

00:10:50.020 --> 00:10:53.402
And then plus the
integral from 0 to 5/2.

00:10:53.402 --> 00:10:55.110
Now, you might be able
to do it visually,

00:10:55.110 --> 00:10:58.210
but we know what the value is
of f of t over this interval,

00:10:58.210 --> 00:11:01.200
we already figured out
the equation for it

00:11:01.200 --> 00:11:02.740
over this interval.

00:11:02.740 --> 00:11:08.780
So it's the integral of
negative 2x plus 3 dt.

00:11:08.780 --> 00:11:10.714
And then let's
just evaluate this.

00:11:10.714 --> 00:11:11.880
Let me get some real estate.

00:11:11.880 --> 00:11:13.463
So this is-- let me
draw a line here.,

00:11:13.463 --> 00:11:14.550
so we don't get confused.

00:11:14.550 --> 00:11:17.390
So this is going to
be equal to 5 plus

00:11:17.390 --> 00:11:18.980
and then I take
the antiderivative.

00:11:18.980 --> 00:11:24.653
The antiderivative of negative
2x is negative x squared.

00:11:27.520 --> 00:11:30.250
So we have negative x squared.

00:11:30.250 --> 00:11:35.020
And the antiderivative of
3 is just going to be 3x.

00:11:35.020 --> 00:11:37.830
So plus 3x.

00:11:37.830 --> 00:11:41.200
And we're going to
evaluate it from 0 to 5/2.

00:11:41.200 --> 00:11:44.060
So this is going to
be equal to 5 plus--

00:11:44.060 --> 00:11:46.450
and I'll do all this
stuff right over here.

00:11:46.450 --> 00:11:48.380
I'll do this stuff in green.

00:11:48.380 --> 00:11:50.200
So when we evaluate
it at 5/2, this

00:11:50.200 --> 00:11:53.560
is going to be
negative 5/2 squared.

00:11:53.560 --> 00:11:57.410
So it's going to be
negative 25 over 4

00:11:57.410 --> 00:12:02.139
plus 3 times 5/2,
which is 15 over 2.

00:12:02.139 --> 00:12:03.930
And then from that,
we're going to subtract

00:12:03.930 --> 00:12:05.170
this evaluated at 0.

00:12:05.170 --> 00:12:10.040
But negative 0 squared
plus 3 times 0 is just 0.

00:12:10.040 --> 00:12:12.140
So this is what
it simplifies to.

00:12:12.140 --> 00:12:15.550
And so what do we
have right over here?

00:12:15.550 --> 00:12:19.320
So let's get our ourselves
a common denominator.

00:12:19.320 --> 00:12:22.730
Looks like a common denominator
right over here could be 4

00:12:22.730 --> 00:12:25.310
So this is equal to--
5 is the same thing

00:12:25.310 --> 00:12:34.220
as 20 over 4 minus 25 over
4 and then plus 30 over 4.

00:12:34.220 --> 00:12:38.680
So 20 plus 30 is
50 minus 25 is 25.

00:12:38.680 --> 00:12:41.850
So this is equal to 25 over 4.

00:12:41.850 --> 00:12:46.270
And 25 over 4 is the
same thing as 6 and 1/4.

00:12:46.270 --> 00:12:49.170
So when we evaluate our
function at this critical point,

00:12:49.170 --> 00:12:51.540
at this thing where the
slope, or the derivative,

00:12:51.540 --> 00:12:54.600
is equal to 0, we
got 6 and 1/4, which

00:12:54.600 --> 00:12:58.480
is higher than six, which is
what g was at this end point.

00:12:58.480 --> 00:13:01.540
And it's definitely higher
than what g was a negative 4.

00:13:01.540 --> 00:13:04.940
So the x-coordinate
of the point at which

00:13:04.940 --> 00:13:08.210
g has an absolute maximum,
on the interval negative 4

00:13:08.210 --> 00:13:14.433
to three is x is equal to 5/2.


2011 Calculus AB free response #4a

WEBVTT
Kind: captions
Language: en

00:00:00.590 --> 00:00:02.850
The continuous
function f is defined

00:00:02.850 --> 00:00:05.000
on the interval
negative 4 is less than

00:00:05.000 --> 00:00:07.870
or equal to x is less
than or equal to 3.

00:00:07.870 --> 00:00:11.480
The graph of f consists of two
quarter circles and one line

00:00:11.480 --> 00:00:12.330
segment.

00:00:12.330 --> 00:00:14.620
So this right here is
one quarter circle,

00:00:14.620 --> 00:00:16.350
then we have another
quarter circle,

00:00:16.350 --> 00:00:18.850
and then it has this
line segment over here,

00:00:18.850 --> 00:00:20.080
as shown in the figure above.

00:00:20.080 --> 00:00:21.746
I put the figure on
the side, so that we

00:00:21.746 --> 00:00:24.170
have some screen real
estate down here.

00:00:24.170 --> 00:00:28.450
Let g of x be equal to 2x plus
the definite integral from 0

00:00:28.450 --> 00:00:31.020
to x of f of t dt.

00:00:31.020 --> 00:00:33.320
All right, that's a
little bit interesting.

00:00:33.320 --> 00:00:34.970
Now, let's do Part a.

00:00:34.970 --> 00:00:35.690
Part a.

00:00:35.690 --> 00:00:38.030
Find g of negative 3.

00:00:38.030 --> 00:00:40.350
So we want to find
g of negative 3.

00:00:40.350 --> 00:00:42.680
So first of all, it
tells us what g of x is.

00:00:42.680 --> 00:00:45.239
So g of negative 3,
every time we see an x,

00:00:45.239 --> 00:00:46.530
we just put a negative 3 there.

00:00:46.530 --> 00:00:48.790
So it's going to
be 2 times negative

00:00:48.790 --> 00:00:56.610
3 plus the definite integral
from 0 to negative 3, f

00:00:56.610 --> 00:00:59.730
of t dt.

00:00:59.730 --> 00:01:01.930
So this first part is
pretty straightforward.

00:01:01.930 --> 00:01:05.604
2 times negative
3 is negative 6.

00:01:05.604 --> 00:01:07.020
And then this part
right over here

00:01:07.020 --> 00:01:10.335
is the definite integral from
0 to negative 3 of f of t dt.

00:01:16.010 --> 00:01:21.470
So this is essentially
the area under f of t--

00:01:21.470 --> 00:01:25.724
or under f of x, if you want to
view it that way-- from 0 to 3.

00:01:25.724 --> 00:01:27.140
And this is pretty
straightforward

00:01:27.140 --> 00:01:29.840
to figure out but we
have to be careful here.

00:01:29.840 --> 00:01:33.380
Because this area
right over here,

00:01:33.380 --> 00:01:35.750
this would be the integral--
let me do this in a color

00:01:35.750 --> 00:01:36.810
that you can see.

00:01:36.810 --> 00:01:38.720
That area right
over there would be

00:01:38.720 --> 00:01:45.590
the integral from
negative 3 to 0 f of t,

00:01:45.590 --> 00:01:50.460
or we could even say f of t
dt or f of x dx, either way.

00:01:50.460 --> 00:01:52.460
That would be this
area right over here.

00:01:52.460 --> 00:01:53.430
They've swapped it.

00:01:53.430 --> 00:01:55.750
They have the larger
number as the lower bound.

00:01:55.750 --> 00:01:57.920
They have 0 as the lower bound.

00:01:57.920 --> 00:02:00.090
So what we could do,
we could rewrite this.

00:02:00.090 --> 00:02:03.250
This is the same thing,
this is equal to negative 6

00:02:03.250 --> 00:02:06.880
minus-- if you swap the
bounds of integration,

00:02:06.880 --> 00:02:09.139
you swap the sign
on the integral.

00:02:09.139 --> 00:02:16.810
So we could say minus from
negative 3 to 0 of f of t dt.

00:02:16.810 --> 00:02:19.920
And now this thing right over
here-- this expression right

00:02:19.920 --> 00:02:25.485
over here is the area
under this quarter circle.

00:02:25.485 --> 00:02:27.110
And the area under
this quarter circle,

00:02:27.110 --> 00:02:29.670
we can just use a little bit
of geometry to figure that out.

00:02:29.670 --> 00:02:31.160
We know its radius.

00:02:31.160 --> 00:02:33.350
The radius here is 3.

00:02:33.350 --> 00:02:34.960
Radius is equal to 3.

00:02:34.960 --> 00:02:37.580
The entire circle
is pi r squared.

00:02:37.580 --> 00:02:39.780
So the entire circle--
the area if this

00:02:39.780 --> 00:02:45.960
was an entire circle right over
here-- that area would be pi r

00:02:45.960 --> 00:02:46.460
squared.

00:02:46.460 --> 00:02:48.280
So pi times 3 squared.

00:02:48.280 --> 00:02:49.680
So 9 pi.

00:02:49.680 --> 00:02:54.580
So it would be 9 pi, but this is
only 1/4 of the entire circle.

00:02:54.580 --> 00:02:56.290
So we want to divide it by 4.

00:02:56.290 --> 00:02:59.290
So this right over here we'll
evaluate to 9 pi over 4.

00:02:59.290 --> 00:03:03.450
So Part a, we get it's
equal to negative 6, that's

00:03:03.450 --> 00:03:11.630
that right over there,
minus 9 pi over 4.

00:03:11.630 --> 00:03:13.710
And that's at least the
first part of Part a.

00:03:13.710 --> 00:03:16.030
Then they say find g prime of x.

00:03:16.030 --> 00:03:19.430
So let's find g prime of x.

00:03:19.430 --> 00:03:22.380
We did the first part of
Part a right over there.

00:03:22.380 --> 00:03:25.842
Then we need to
find g prime of x.

00:03:25.842 --> 00:03:28.050
Well, it's just going to be
the derivative of g of x.

00:03:28.050 --> 00:03:31.439
The derivative of 2x
is just going to be 2.

00:03:31.439 --> 00:03:33.980
And then the derivative-- and
this is the fundamental theorem

00:03:33.980 --> 00:03:36.190
of calculus right over
here-- the derivative

00:03:36.190 --> 00:03:39.350
of the definite integral
from 0 to x of f of t dt,

00:03:39.350 --> 00:03:40.705
that's just going to be f of x.

00:03:43.800 --> 00:03:45.740
So that is g prime of x.

00:03:45.740 --> 00:03:48.260
We did this second
part right over there.

00:03:48.260 --> 00:03:50.840
That is g prime of x.

00:03:50.840 --> 00:03:53.580
And then we need to
evaluate it at negative 3.

00:03:53.580 --> 00:03:58.510
So g prime of
negative 3 is going

00:03:58.510 --> 00:04:02.510
to be equal to 2
plus f of negative 3,

00:04:02.510 --> 00:04:06.300
which is equal to 2 plus-- and
let's look at our-- let's see

00:04:06.300 --> 00:04:08.500
what f of negative 3 is.

00:04:08.500 --> 00:04:10.030
This is our function definition.

00:04:10.030 --> 00:04:14.120
When x is equal to negative
3, our function is at 0.

00:04:14.120 --> 00:04:16.329
So f of negative 3 is just 0.

00:04:16.329 --> 00:04:19.680
So it's just two plus
0, which is equal to 2.

00:04:19.680 --> 00:04:22.300
So g of negative
3 is equal to 2.

00:04:22.300 --> 00:04:24.295
And we're done with Part a.

00:04:24.295 --> 00:04:25.920
And the trickiest
part of this is just

00:04:25.920 --> 00:04:28.580
realizing the bounds
of integration.

00:04:28.580 --> 00:04:31.565
That you might be tempted to
say, oh, the integral between 0

00:04:31.565 --> 00:04:33.690
and negative 3, that's
just this area over here

00:04:33.690 --> 00:04:35.330
because we're going
between 0 and negative 3.

00:04:35.330 --> 00:04:37.060
But you've got to
realize that we actually

00:04:37.060 --> 00:04:39.310
have to swap them and make
them negative if you really

00:04:39.310 --> 00:04:41.780
want to consider this
area right over here.

00:04:41.780 --> 00:04:44.470
That this integral is going to
be the negative of this area.

00:04:44.470 --> 00:04:46.928
And that's what we did when we
went through this step right

00:04:46.928 --> 00:04:47.780
over there.


2011 Calculus AB Free Response #1 parts b c d

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.400
Now let's do Part b).

00:00:00.000 --> 00:00:05.866
Find the average velocity of the particle for the time period between time 0 and time 6.

00:00:00.000 --> 00:00:04.466
So the easy way to think about this, average velocity, is that your

00:00:00.000 --> 00:00:04.000
average, or one way to think about it, the distance that you

00:00:00.000 --> 00:00:04.266
travel, sometimes I use "d" for distance or "s" for displacement

00:00:00.000 --> 00:00:03.533
But your distance in general is equal to your average

00:00:00.000 --> 00:00:01.333
velocity times time.

00:00:00.000 --> 00:00:03.333
Or, your average velocity is equal to the distance

00:00:00.000 --> 00:00:01.866
you travel, divided by time.

00:00:00.000 --> 00:00:03.666
And so how do we figure out the distance that we travel

00:00:00.000 --> 00:00:01.133
over that period?

00:00:00.000 --> 00:00:03.266
Especially because they didn't explicitly give us

00:00:00.000 --> 00:00:01.466
the position function.

00:00:00.000 --> 00:00:01.800
Well if you think about it,

00:00:00.000 --> 00:00:03.266
Let's think about the velocity function over here

00:00:00.000 --> 00:00:01.933
which they did give us,"v(t)"

00:00:00.000 --> 00:00:02.933
and I don't know exactly what it looks like,

00:00:00.000 --> 00:00:04.000
let me just draw a general velocity function right over here

00:00:00.000 --> 00:00:01.466
So this is our t-axis,

00:00:00.000 --> 00:00:03.000
and this is our velocity axis right over here

00:00:00.000 --> 00:00:02.933
and we care about between time 0 and time 6,

00:00:00.000 --> 00:00:03.466
and let's say our function does something like this,

00:00:00.000 --> 00:00:03.733
it doesn't really matter, this is just a general drawing

00:00:00.000 --> 00:00:02.733
over here, How do we figure out the total

00:00:00.000 --> 00:00:01.600
distance over that time?

00:00:00.000 --> 00:00:02.200
Well, if you take any small time,

00:00:00.000 --> 00:00:02.333
right over here,and you multiply it

00:00:00.000 --> 00:00:03.000
times roughly the constant velocity over that

00:00:00.000 --> 00:00:03.200
time period, then that will tell you the area of

00:00:00.000 --> 00:00:02.600
this little column, which will tell you

00:00:00.000 --> 00:00:02.400
approximately, the distance traveled

00:00:00.000 --> 00:00:02.533
over this very small interval of time.

00:00:00.000 --> 00:00:02.933
Because you have time times an approximately

00:00:00.000 --> 00:00:02.666
constant velocity, which will give you a

00:00:00.000 --> 00:00:01.000
distance.

00:00:00.000 --> 00:00:02.533
And so in general if you take the area

00:00:00.000 --> 00:00:02.000
under this curve, from 0 to 6,

00:00:00.000 --> 00:00:02.466
that will give you the total distance

00:00:00.000 --> 00:00:01.000
travelled.

00:00:00.000 --> 00:00:02.666
So "d" is equal to the definite integral

00:00:00.000 --> 00:00:01.400
from 0 to 6 of "v(t)"

00:00:00.000 --> 00:00:02.000
of this thing right over here,

00:00:00.000 --> 00:00:01.000
"v(t)dt"

00:00:00.000 --> 00:00:02.400
So if you want the average velocity,

00:00:00.000 --> 00:00:02.400
you just find the distance, which is

00:00:00.000 --> 00:00:02.133
this definite integral, and then

00:00:00.000 --> 00:00:02.333
you divide that by the time, or the

00:00:00.000 --> 00:00:01.800
elapsed time, in this case,

00:00:00.000 --> 00:00:02.400
I could actually write times delta t

00:00:00.000 --> 00:00:02.333
right over here, our change in time

00:00:00.000 --> 00:00:02.200
right over here, is going to be 6

00:00:00.000 --> 00:00:02.600
we start at 0 and we finish at 6, so we

00:00:00.000 --> 00:00:01.733
just have to evaluate this

00:00:00.000 --> 00:00:03.533
definite integral, which is not that easy to evaluate

00:00:00.000 --> 00:00:02.400
analytically, it's lucky for us that

00:00:00.000 --> 00:00:03.200
they do allow us to use calculators in this part

00:00:00.000 --> 00:00:02.600
so we'll evaluate it using a calculator

00:00:00.000 --> 00:00:02.600
And then we just have to divide it by 6

00:00:00.000 --> 00:00:02.400
So let's first find out teh distance

00:00:00.000 --> 00:00:02.400
travelled between time 0 and time 6.

00:00:00.000 --> 00:00:01.000
And to do that,

00:00:00.000 --> 00:00:01.333
we can use the TI-85

00:00:00.000 --> 00:00:01.333
We'll go to catalog,

00:00:00.000 --> 00:00:02.266
this tells us all of the different

00:00:00.000 --> 00:00:01.333
functions they have.

00:00:00.000 --> 00:00:03.066
And to evaluate definite integrals, we use the


2011 Calculus AB free response #3 (c)

WEBVTT
Kind: captions
Language: en

00:00:00.395 --> 00:00:04.621
Part C, write but do not evaluate an integral expression

00:00:04.621 --> 00:00:07.733
for the volume of the solid generated when R -

00:00:07.733 --> 00:00:09.868
- so that's this region right over here -

00:00:09.868 --> 00:00:14.884
when R is rotated about the horizontal line y=1.

00:00:14.884 --> 00:00:18.228
So y=1 is right over there.

00:00:18.228 --> 00:00:20.933
So, the way I like to think about it - let's think about

00:00:20.933 --> 00:00:24.265
the volume if I were to just take the bottom function - if I were

00:00:24.265 --> 00:00:28.026
to just take f(x) - if I were to just take f(x) - and if I were

00:00:28.026 --> 00:00:32.400
to rotate that function around y=x - if I were to rotate

00:00:32.400 --> 00:00:36.525
this thing around - sorry, around y=1 - what would

00:00:36.525 --> 00:00:39.636
the volume of that be, and then I'm going to subtract

00:00:39.636 --> 00:00:42.516
from that the volume if I were to take the top function - if

00:00:42.516 --> 00:00:44.652
I were to take g(x) and rotate it around.

00:00:44.652 --> 00:00:47.345
So let's first of all think about what volume I would get -

00:00:47.345 --> 00:00:49.342
and this is really the disk method, and I go into it

00:00:49.342 --> 00:00:51.525
in much more detail earlier in the calculus playlist, but

00:00:51.525 --> 00:00:54.265
let's think about the volume if f(x) is rotated around

00:00:54.265 --> 00:00:58.677
that axis. And to do that, let's imagine each of - each

00:00:58.677 --> 00:01:03.599
sliver of that volume. So this is - let me draw a little thing

00:01:03.599 --> 00:01:06.711
right over here - and you could imagine once this little

00:01:06.711 --> 00:01:11.216
silver is rotated it forms the bo- it forms the or you should -

00:01:11.216 --> 00:01:13.863
you could imagine - you could imagine - this length is

00:01:13.863 --> 00:01:17.733
the radius of a disk. And so - and just to imagine that

00:01:17.733 --> 00:01:19.807
let me draw the entire disk.

00:01:19.807 --> 00:01:22.361
So if this is rotated around - if this is rotated around

00:01:22.361 --> 00:01:27.655
it will become a disk - it will become a disk - it will

00:01:27.655 --> 00:01:31.278
become a disk that looks something like that, and

00:01:31.278 --> 00:01:36.340
I'll just call the depth of the disk - so the disk if you

00:01:36.340 --> 00:01:38.337
imagine it a coin this is kind of the side of the coin,

00:01:38.337 --> 00:01:42.330
the depth of the coin - the depth of the coin right over there.

00:01:42.330 --> 00:01:47.021
I know I can draw that better than that. So the depth

00:01:47.021 --> 00:01:51.990
of the coin is just like that. It's a fixed depth, and I'm going

00:01:51.990 --> 00:01:55.008
to call that dx - so it's just this distance right over here

00:01:55.008 --> 00:01:59.095
it is the dx - and what is going to be the area of that coin?

00:01:59.095 --> 00:02:02.346
Well the area of the surface of this is coin - so let me -

00:02:02.346 --> 00:02:07.083
let me do it like this... in a different color. So I want to

00:02:07.083 --> 00:02:11.355
do it in blue - the area of this coin is just pi times the radius

00:02:11.355 --> 00:02:15.024
of that coin, squared. And what is the radius of the coin?

00:02:15.024 --> 00:02:18.089
Well, the radius of the coin is this height - is this

00:02:18.089 --> 00:02:20.467
height right over here. And what is that height?

00:02:20.467 --> 00:02:26.123
Well it is 1-f(x). So that is equal to the radius.

00:02:26.123 --> 00:02:30.495
So the area - the surf- the area - the kind of the face

00:02:30.495 --> 00:02:32.795
of this coin is going to be pi - the area of the face of this

00:02:35.086 --> 00:02:39.033
coin is going to be pi times the radius squared,

00:02:39.033 --> 00:02:46.092
which is equal to pi times one minus f of x - one minus f of x squared -

00:02:46.092 --> 00:02:49.111
that's this blue area right over here, and if I want to

00:02:49.111 --> 00:02:52.269
find the volume of this coin I would multiply it by -

00:02:52.269 --> 00:02:56.913
by the depth of the coin. So times - so times dx.

00:02:56.913 --> 00:03:00.767
And if I wanted to find the volume of this entire solid -

00:03:00.767 --> 00:03:04.761
this entire rotated, i would want to find the sum of

00:03:04.761 --> 00:03:06.467
all of these volumes. So this is just the disk right

00:03:06.467 --> 00:03:09.359
over here but I can have another disk, a similar

00:03:09.359 --> 00:03:12.749
disk that I do right over here, I could have another

00:03:12.749 --> 00:03:15.133
disk right over here, and I want to take the sum

00:03:15.133 --> 00:03:19.668
of ALL of those disks. So I want to take - I want to

00:03:19.668 --> 00:03:22.594
take - so the volume is going to be the sum over all

00:03:22.594 --> 00:03:26.495
of those disks - so x goes from 0 - which is this bounding

00:03:26.495 --> 00:03:35.272
point - to x is equal to 1/2, times pi(1-f(x))^2 - this is

00:03:35.272 --> 00:03:39.266
the area of e- the fa- the area of the face of each of those disks

00:03:39.266 --> 00:03:41.774
and then I multiply time the depth of each of those disks -

00:03:41.774 --> 00:03:44.282
no this gives the volume of each of those disks, and

00:03:44.282 --> 00:03:46.418
I'm taking the sum of all of them.

00:03:46.418 --> 00:03:49.436
So this is the volume - this is the volume if I were to just

00:03:49.436 --> 00:03:54.034
rotate f(x) around y=1 - actually I should just write

00:03:54.034 --> 00:03:57.424
dx here - and so this right over here - this expression -

00:03:57.424 --> 00:04:00.721
I just did that so they really are equal - this is

00:04:00.721 --> 00:04:02.733
obviously just the volume of each of those disks.

00:04:02.733 --> 00:04:07.037
So this is the volume if I were to take f(x) around y=1.

00:04:07.037 --> 00:04:10.752
Let's figure out - so let me call this volume of f(x) -

00:04:10.752 --> 00:04:14.096
and by the same logic - the same exact logic - we can

00:04:14.096 --> 00:04:18.200
figure out the volume if we take g(x) - if we rotate g(x)

00:04:18.200 --> 00:04:23.268
around - if we rotate g(x) - if we construct disks like this

00:04:23.268 --> 00:04:27.819
and rotate them around y=1. And so the volume -

00:04:27.819 --> 00:04:33.438
if I take g(x) around y=1 would be 0 to 1/2 times

00:04:33.438 --> 00:04:39.011
pi(1-g(x) - cause 1-g(x) is each of these radius' right

00:04:39.011 --> 00:04:44.166
over here - each of these radius' - that squared, dx.

00:04:44.166 --> 00:04:47.138
And so the volume of what they're asking us - the

00:04:47.138 --> 00:04:50.389
volume of the solid generated when R is rotated - well

00:04:50.389 --> 00:04:53.872
R is kind of the space in between f(x) and g(x) - so

00:04:53.872 --> 00:04:56.449
so what's going to be - is g- the volume is going

00:04:56.449 --> 00:04:58.771
to be the difference between these volumes - it's

00:04:58.771 --> 00:05:01.093
going to be this volume - this is kind of the outer

00:05:01.093 --> 00:05:03.183
volume, and we're going to take out its hollow core -

00:05:03.183 --> 00:05:05.459
we're going to hollow it out be subtracting this volume.

00:05:05.459 --> 00:05:08.524
So the volume of that region is going to be the integral -

00:05:08.524 --> 00:05:13.028
I'll do this in a new color - integral from 0 to 1/2 of

00:05:13.028 --> 00:05:25.149
pi(1-f(x))^2 dx MINUS the integral from 0 to 1/2 of pi(1-g(x))^2 dx

00:05:25.149 --> 00:05:28.200
and this is completely valid answer but you might want to simplify it

00:05:28.200 --> 00:05:32.022
we have the same bounds of integration - we have the same

00:05:32.022 --> 00:05:34.530
variable of integration - and actually we have this pi over here

00:05:34.530 --> 00:05:36.805
so we could just factor that out - and so this is

00:05:36.805 --> 00:05:40.196
the same thing as pi times the integral from 0 to 1/2

00:05:40.196 --> 00:05:49.994
of (1-f(x))^2 MINUS (1-g(x))^2 -

00:05:49.994 --> 00:05:55.428
and then all of that dx. And then - and actually you probably

00:05:55.428 --> 00:05:57.146
would want to do the - you probably would want to do

00:05:57.146 --> 00:06:00.258
this while taking the AP Exam, not just leaving it in

00:06:00.258 --> 00:06:02.200
terms of f(x) and g(x) - you would actually want to write

00:06:02.200 --> 00:06:06.388
the expression for what f(x)... f(x) and g(x) are.

00:06:06.388 --> 00:06:08.803
So really the best answer would probably be:

00:06:08.803 --> 00:06:12.843
pi times the integral from 0 to 1/2, times

00:06:12.843 --> 00:06:18.694
[1(1-... well f(x) is 8x to the third power - 8x to the

00:06:18.694 --> 00:06:23.152
third power, squared. Minus, (1-g(x)... and g(x) is

00:06:23.152 --> 00:06:30.165
sin of pi x. That squared - that squared times dx.

00:06:30.165 --> 00:06:33.137
And that would be our answer and you could see why

00:06:33.137 --> 00:06:34.670
they didn't want us to go through the trouble

00:06:34.670 --> 00:06:36.434
of evaluating it - they just wanted us to set up

00:06:36.434 --> 99:59:59.999
this integral.


2011 Calculus AB free response #3 (a & b)

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:03.680
Let R be the region in the
first quadrant enclosed

00:00:03.680 --> 00:00:05.670
by the graphs of f
of x is equal to 8x

00:00:05.670 --> 00:00:08.830
to the third and g of x
is equal to sine of pi x,

00:00:08.830 --> 00:00:10.052
as shown in the figure above.

00:00:10.052 --> 00:00:11.760
And they drew the
figure right over here.

00:00:11.760 --> 00:00:14.060
Part a, write the equation
for the line tangent

00:00:14.060 --> 00:00:16.574
to the graph of f at
x is equal to 1/2.

00:00:16.574 --> 00:00:17.990
So let me just
redraw it here just

00:00:17.990 --> 00:00:20.530
so that-- I like drawing on
the black background, I guess,

00:00:20.530 --> 00:00:22.730
is the main reason
why I'm redrawing it.

00:00:22.730 --> 00:00:26.860
So the function f of x is
equal to 8x to the third,

00:00:26.860 --> 00:00:28.870
looks like this.

00:00:28.870 --> 00:00:29.920
It looks like that.

00:00:29.920 --> 00:00:34.530
This is our f of x-axis,
and this is our x-axis.

00:00:34.530 --> 00:00:36.590
And we want the equation
for the line tangent

00:00:36.590 --> 00:00:38.440
at x is equal to 1/2.

00:00:38.440 --> 00:00:40.975
So this is x is equal to 1/2.

00:00:40.975 --> 00:00:44.060
If you go up here, if
you evaluate f of 1/2,

00:00:44.060 --> 00:00:47.240
you get 8 times
1/2 to the third,

00:00:47.240 --> 00:00:49.562
which is 8 times
1/8, which is 1.

00:00:49.562 --> 00:00:51.770
And they actually gave us
that already on this point.

00:00:51.770 --> 00:00:54.730
This is the point 1/2 comma 1.

00:00:54.730 --> 00:00:57.690
And we need to find the
equation for the tangent line.

00:00:57.690 --> 00:01:00.664
So the tangent line will
look something like that.

00:01:00.664 --> 00:01:02.330
And to figure out
this equation, we just

00:01:02.330 --> 00:01:04.580
need to figure out
its slope, and then

00:01:04.580 --> 00:01:08.220
we know a point that it's on--
and you could use point-slope

00:01:08.220 --> 00:01:11.230
or you could just use your kind
of standard slope-intercept

00:01:11.230 --> 00:01:13.404
form to give an
equation for that line.

00:01:13.404 --> 00:01:15.320
So the first part, let's
figure out its slope.

00:01:15.320 --> 00:01:16.694
And the slope of
the tangent line

00:01:16.694 --> 00:01:18.670
is going to be the
same slope as the slope

00:01:18.670 --> 00:01:20.816
of our function at that point.

00:01:20.816 --> 00:01:22.190
Or another way to
think about it,

00:01:22.190 --> 00:01:27.170
it is going to be
f prime of 1/2,

00:01:27.170 --> 00:01:29.400
or the derivative
evaluated at 1/2.

00:01:29.400 --> 00:01:30.860
The derivative
gives us the slope

00:01:30.860 --> 00:01:32.750
of that line at any point.

00:01:32.750 --> 00:01:35.110
So what is f prime
of x? f prime of x

00:01:35.110 --> 00:01:36.840
is just the derivative of this.

00:01:36.840 --> 00:01:41.320
So 3 times 8 is 24
times x squared.

00:01:41.320 --> 00:01:43.050
24 x squared.

00:01:43.050 --> 00:01:50.260
f prime of 1/2 is equal
to 24 times 1/2 squared,

00:01:50.260 --> 00:01:54.610
which is equal to 24 times
1/4, which is equal to 6.

00:01:54.610 --> 00:01:58.780
So the slope of this
line is equal to 6.

00:01:58.780 --> 00:02:00.000
I'll use m for slope.

00:02:00.000 --> 00:02:02.660
That's the convention
that we used

00:02:02.660 --> 00:02:04.900
when we first learned
it in algebra.

00:02:04.900 --> 00:02:06.480
So the slope is going to be 6.

00:02:06.480 --> 00:02:08.240
So the general
equation for this line

00:02:08.240 --> 00:02:12.551
is y is equal to mx plus b.

00:02:12.551 --> 00:02:13.300
This is the slope.

00:02:13.300 --> 00:02:14.700
This is the y-intercept.

00:02:14.700 --> 00:02:17.010
We already know
that the slope is 6,

00:02:17.010 --> 00:02:19.240
and then we can use the
fact that the line goes

00:02:19.240 --> 00:02:22.810
through the point 1/2, 1
to figure out what b is.

00:02:22.810 --> 00:02:29.440
So when y is 1, 1 is equal to
our slope times x. x is 1/2.

00:02:29.440 --> 00:02:31.200
Or another way to say
it, when x is 1/2,

00:02:31.200 --> 00:02:34.950
y is 1 plus some y-intercept.

00:02:34.950 --> 00:02:36.810
Or if I take x is
1/2, I multiply

00:02:36.810 --> 00:02:40.210
it times the slope plus the
y-intercept, I should get 1.

00:02:40.210 --> 00:02:43.600
And so I get 1 is
equal to 3 plus b.

00:02:43.600 --> 00:02:46.610
I can subtract 3 from both
sides, and I get negative 2

00:02:46.610 --> 00:02:47.870
is equal to b.

00:02:47.870 --> 00:02:51.120
So the equation of the
line is going to be y

00:02:51.120 --> 00:02:54.670
is equal to 6x minus 2.

00:02:54.670 --> 00:02:58.580
That is the equation
of the tangent line.

00:02:58.580 --> 00:03:03.190
Now part b, find
the area of R. So R

00:03:03.190 --> 00:03:06.810
is this region right over here.

00:03:06.810 --> 00:03:09.630
It's bounded above
by g of x, which

00:03:09.630 --> 00:03:13.160
they've defined as sine of pi x.

00:03:13.160 --> 00:03:18.350
It's bounded below by f
of x, or 8x to the third.

00:03:18.350 --> 00:03:21.680
So the area is going
to be-- actually,

00:03:21.680 --> 00:03:23.870
let me just do it-- I'll
scroll down a little bit.

00:03:23.870 --> 00:03:27.580
I still want to be able to see
this graph right over here.

00:03:27.580 --> 00:03:29.440
Part b.

00:03:29.440 --> 00:03:37.650
The area of R is going to be
equal to the integral from 0,

00:03:37.650 --> 00:03:40.390
that's this point of
intersection right over here,

00:03:40.390 --> 00:03:42.100
to 1/2.

00:03:42.100 --> 00:03:43.470
So let me make it clear.

00:03:43.470 --> 00:03:45.260
This is 0 to 1/2.

00:03:47.950 --> 00:03:50.660
And then the
function on the top--

00:03:50.660 --> 00:03:52.307
so we could just take
the area of that,

00:03:52.307 --> 00:03:54.390
but then we're going to
have to subtract from that

00:03:54.390 --> 00:03:57.010
the area underneath
the function below.

00:03:57.010 --> 00:03:59.760
Or one way to think about
it is the integral from 0

00:03:59.760 --> 00:04:04.495
to 1/2 of the top function is
g of x, which is sine of pi x.

00:04:09.360 --> 00:04:11.920
But if we just evaluated
this integral--

00:04:11.920 --> 00:04:14.630
let me just put a dx over here.

00:04:14.630 --> 00:04:16.730
If we just evaluated
this, we would

00:04:16.730 --> 00:04:20.557
get the area of
this entire region.

00:04:20.557 --> 00:04:22.140
But what we need to
do is subtract out

00:04:22.140 --> 00:04:26.880
the area underneath the
second, underneath f of x.

00:04:26.880 --> 00:04:28.660
We just subtract out
the area under that.

00:04:28.660 --> 00:04:31.710
So we just subtract
from that f of x.

00:04:31.710 --> 00:04:35.430
And f of x, we already
saw, is 8x to the third.

00:04:35.430 --> 00:04:38.920
Is 8x to the third power.

00:04:38.920 --> 00:04:40.574
And now we can
just evaluate this.

00:04:40.574 --> 00:04:41.990
So let me draw a
little line here.

00:04:41.990 --> 00:04:45.290
It's getting a little bit messy.

00:04:45.290 --> 00:04:46.850
I'll just do it down here.

00:04:46.850 --> 00:04:50.390
So we need to take the
antiderivative of sine of pi x.

00:04:50.390 --> 00:04:56.170
Well, the derivative of cosine
of x is negative sine x.

00:04:56.170 --> 00:05:02.650
The derivative of cosine of pi
x is negative pi cosine of pi x.

00:05:02.650 --> 00:05:05.200
So the antiderivative
of sine of pi

00:05:05.200 --> 00:05:13.060
x is going to be negative
1 over pi cosine of pi x.

00:05:13.060 --> 00:05:14.560
And you can verify
it for yourself.

00:05:14.560 --> 00:05:15.160
And you might say, wait.

00:05:15.160 --> 00:05:16.770
Sal, how did you know
it was a negative.

00:05:16.770 --> 00:05:18.145
Well, I put the
negative there so

00:05:18.145 --> 00:05:21.390
that when I take the derivative
of the cosine of pi x,

00:05:21.390 --> 00:05:24.100
I would get a negative sign, but
that negative will cancel out

00:05:24.100 --> 00:05:25.766
the negative to give
me a positive here.

00:05:25.766 --> 00:05:27.850
And you say, why do you
put a 1 over pi here.

00:05:27.850 --> 00:05:30.099
Well, when you take the
derivative of this thing using

00:05:30.099 --> 00:05:32.760
the chain rule, you take
the derivative of the pi x,

00:05:32.760 --> 00:05:35.560
you'll get pi, that you
would multiply everything by,

00:05:35.560 --> 00:05:38.380
and then you would get
negative sine of pi x.

00:05:38.380 --> 00:05:40.464
And that pi doesn't
show up here.

00:05:40.464 --> 00:05:42.380
So I need something for
it to cancel out with.

00:05:42.380 --> 00:05:46.070
And that's what this 1 over pi
is going to cancel out with.

00:05:46.070 --> 00:05:49.640
And you could do use
substitution and all the rest,

00:05:49.640 --> 00:05:51.920
if you found something
like that useful.

00:05:51.920 --> 00:05:54.700
But it's in general,
a good habit

00:05:54.700 --> 00:06:00.580
or I guess it's good to be able
to do this almost by sight.

00:06:00.580 --> 00:06:03.090
So, and you can verify
that this derivative is

00:06:03.090 --> 00:06:04.410
equal to sine of pi x.

00:06:04.410 --> 00:06:07.010
So the antiderivative
of sine of pi x is this.

00:06:07.010 --> 00:06:10.440
The antiderivative of
negative 8x to the third power

00:06:10.440 --> 00:06:16.260
is negative 8-- I'm going to
divide it by 4-- so negative 2x

00:06:16.260 --> 00:06:18.030
to the fourth power.

00:06:18.030 --> 00:06:21.560
And all I did is I
incremented the 3 to a 4,

00:06:21.560 --> 00:06:23.362
and then I divided
by 8 by the 4.

00:06:23.362 --> 00:06:25.070
And you could take
the derivative of this

00:06:25.070 --> 00:06:27.920
to verify this it is the
same thing as negative 8x

00:06:27.920 --> 00:06:29.730
to the third power.

00:06:29.730 --> 00:06:33.650
And we're going to have to
evaluate that from 0 to 1/2.

00:06:33.650 --> 00:06:39.190
When you evaluated 1/2,
I'm going to get negative 1

00:06:39.190 --> 00:06:48.770
over pi cosine of pi
over 2 minus 2 times 1/2

00:06:48.770 --> 00:06:51.630
to the fourth power is 1/16.

00:06:51.630 --> 00:06:54.160
So that's evaluated at 1/2.

00:06:54.160 --> 00:06:57.450
And then from that, I'm
going to subtract negative 1

00:06:57.450 --> 00:06:59.609
over pi cosine of 0.

00:06:59.609 --> 00:07:00.650
Let me just write it out.

00:07:00.650 --> 00:07:10.590
So minus negative 1 over
pi cosine of pi times 0.

00:07:10.590 --> 00:07:13.780
Let me just write
cosine of 0 pi,

00:07:13.780 --> 00:07:18.100
I could write, or pi times 0,
minus 2 times 0 to the fourth.

00:07:18.100 --> 00:07:19.750
So that's just
going to be minus 0.

00:07:19.750 --> 00:07:21.660
So let's evaluate this.

00:07:21.660 --> 00:07:26.250
So to simplify it, we have
a cosine of pi over 2.

00:07:26.250 --> 00:07:29.350
Cosine of pi over 2
is just going to be 0,

00:07:29.350 --> 00:07:31.680
so this whole thing
just becomes zero.

00:07:31.680 --> 00:07:33.790
And then you have a
negative 2 divided by 16.

00:07:33.790 --> 00:07:36.830
That's negative 1/8.

00:07:36.830 --> 00:07:40.760
And then from that, I'm going
to subtract this business here.

00:07:40.760 --> 00:07:43.250
Cosine of 0, this is 1.

00:07:43.250 --> 00:07:47.677
So this is just a
negative 1 over pi.

00:07:47.677 --> 00:07:48.760
And then I have a 0 there.

00:07:48.760 --> 00:07:49.635
so I can ignore that.

00:07:49.635 --> 00:07:56.090
So this is equal to negative
1 over 8 plus 1 over pi.

00:07:56.090 --> 00:07:56.767
And we are done.

00:07:56.767 --> 00:07:59.100
This part of it, you're not
allowed to use a calculator.

00:07:59.100 --> 00:08:03.880
So this is about
as far as I would

00:08:03.880 --> 00:08:06.474
expect them to
expect you to get.

00:08:06.474 --> 00:08:07.640
And so I'll leave you there.

00:08:07.640 --> 00:08:12.528
In the next video,
we will do part c.


2011 Calculus AB free response #2 (c & d)

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:03.050
We're now on part c. "Evaluate
the definite integral

00:00:03.050 --> 00:00:05.850
from 0 to 10 of
H prime of (t)dt.

00:00:05.850 --> 00:00:08.060
Using correct units,
explain the meaning

00:00:08.060 --> 00:00:11.150
of the expression in the
context of this problem."

00:00:11.150 --> 00:00:13.140
So we know that if we
just want to evaluate

00:00:13.140 --> 00:00:18.070
this definite integral from
0 to 10 of H prime of (t)dt,

00:00:18.070 --> 00:00:21.640
this is just the same thing as
evaluating the anti-derivative

00:00:21.640 --> 00:00:25.120
of this thing right over here,
which is just H of t at 10.

00:00:25.120 --> 00:00:27.430
And subtract from that the
anti-derivative of this

00:00:27.430 --> 00:00:28.550
evaluated at 0.

00:00:28.550 --> 00:00:30.930
This is the second fundamental
theorem of calculus.

00:00:30.930 --> 00:00:33.984
This is exactly how we
evaluate definite integrals.

00:00:33.984 --> 00:00:35.400
And when you look
at it from this,

00:00:35.400 --> 00:00:37.590
you just see that
when you evaluate it,

00:00:37.590 --> 00:00:40.110
it just gives us the
difference in temperature

00:00:40.110 --> 00:00:41.560
from 0 minutes to 10 minutes.

00:00:41.560 --> 00:00:43.309
We're taking the
temperature at 10 minutes

00:00:43.309 --> 00:00:47.140
and from that we're subtracting
from that the temperature at 0

00:00:47.140 --> 00:00:47.670
minutes.

00:00:47.670 --> 00:00:53.810
So this is really our
change in temperature

00:00:53.810 --> 00:00:55.357
over those 10 minutes.

00:00:55.357 --> 00:00:56.690
And we can actually evaluate it.

00:00:56.690 --> 00:00:59.080
We know what our temperature
was after 10 minutes.

00:00:59.080 --> 00:01:02.400
H of 10 is 43 degrees Celsius.

00:01:02.400 --> 00:01:04.769
So this is 43 right over here.

00:01:04.769 --> 00:01:06.440
And from that, we're
going to subtract

00:01:06.440 --> 00:01:08.280
our initial temperature,
our temperature

00:01:08.280 --> 00:01:13.140
at 0 minutes, which
is 66 degrees Celsius.

00:01:13.140 --> 00:01:15.220
We're going to subtract
66 degrees Celsius.

00:01:15.220 --> 00:01:19.150
This gives us negative
23 degrees Celsius.

00:01:19.150 --> 00:01:21.560
So our change in temperature
is negative 23 degrees.

00:01:21.560 --> 00:01:24.960
Or our temperature has gone
down 23 degrees Celsius

00:01:24.960 --> 00:01:27.600
over the course of
the first 10 minutes.

00:01:27.600 --> 00:01:31.290
So that is part c
right over there.

00:01:31.290 --> 00:01:32.500
Now, let's do part d.

00:01:36.040 --> 00:01:39.475
"At time t equals 0, biscuits
with temperature 100 degrees

00:01:39.475 --> 00:01:40.600
were removed from an oven."

00:01:40.600 --> 00:01:42.099
So now we're talking
about biscuits.

00:01:42.099 --> 00:01:43.710
We started with
tea, now biscuits.

00:01:43.710 --> 00:01:45.500
"The temperature of
the biscuits at time t

00:01:45.500 --> 00:01:47.320
is modeled by a
differentiable function

00:01:47.320 --> 00:01:50.270
B, for which it is
known that B prime of t"

00:01:50.270 --> 00:01:52.450
is equal to this
business right over here.

00:01:52.450 --> 00:01:54.410
"Using the given
models at time t

00:01:54.410 --> 00:01:57.800
equals 10, how much cooler are
the biscuits than the tea?"

00:01:57.800 --> 00:02:00.050
Well, we know what the
temperature of the tea is.

00:02:00.050 --> 00:02:02.591
So we just have to figure out
the temperature of the biscuits

00:02:02.591 --> 00:02:04.930
to figure out how much
cooler they are than the tea.

00:02:04.930 --> 00:02:07.013
And to figure out the
temperature of the biscuits,

00:02:07.013 --> 00:02:09.860
we can essentially just
use the exact same idea.

00:02:09.860 --> 00:02:12.250
We can say, we know that
the biscuits started off

00:02:12.250 --> 00:02:14.040
at 100 degrees Celsius.

00:02:14.040 --> 00:02:16.570
And we could say, well, how
much did they cool down?

00:02:16.570 --> 00:02:18.210
Or what was their
change in temperature

00:02:18.210 --> 00:02:19.359
over the 10 minutes?

00:02:19.359 --> 00:02:20.900
If we know the change
in temperature,

00:02:20.900 --> 00:02:22.691
and we know it started
at 100 degrees, then

00:02:22.691 --> 00:02:24.590
we can use that
information to get

00:02:24.590 --> 00:02:26.390
what its temperature
is at 10 minutes.

00:02:26.390 --> 00:02:28.200
And then we can
answer their question.

00:02:28.200 --> 00:02:30.400
So the change in temperature
over those 10 minutes

00:02:30.400 --> 00:02:33.190
is just the definite
integral from 0 to 10

00:02:33.190 --> 00:02:35.835
of this business
right over here.

00:02:35.835 --> 00:02:36.710
Let me just write it.

00:02:36.710 --> 00:02:39.370
So you see it's the
exact same pattern

00:02:39.370 --> 00:02:40.749
as what we saw over here.

00:02:40.749 --> 00:02:41.790
This was for the cookies.

00:02:41.790 --> 00:02:43.840
Now we're talking
about the biscuits.

00:02:43.840 --> 00:02:45.830
And this right
over here is going

00:02:45.830 --> 00:02:48.680
to be the definite
integral from 0 to 10.

00:02:48.680 --> 00:02:51.400
B prime of t-- they give
it to us right over here--

00:02:51.400 --> 00:03:00.490
is negative 13.84e to
the negative 0.173t, dt.

00:03:00.490 --> 00:03:04.110
And now we just evaluate this.

00:03:04.110 --> 00:03:05.840
Well, we can evaluate this.

00:03:05.840 --> 00:03:08.620
What we'd have to
is-- because we

00:03:08.620 --> 00:03:11.791
know the derivative of e
to the a x is a e to the x.

00:03:11.791 --> 00:03:12.540
Let me write that.

00:03:12.540 --> 00:03:18.049
The derivative of e to the
ax is equal to ae to the ax,

00:03:18.049 --> 00:03:19.090
just from the chain rule.

00:03:19.090 --> 00:03:21.234
Derivative of the
inside is just a.

00:03:21.234 --> 00:03:23.150
And then we multiply
that times the derivative

00:03:23.150 --> 00:03:24.050
of the entire thing.

00:03:24.050 --> 00:03:26.650
And derivative of e to
the x is just e to the x.

00:03:26.650 --> 00:03:31.110
Or we could say that the
integral of e to the ax dx

00:03:31.110 --> 00:03:34.742
is equal to 1 over a
e to the ax plus c.

00:03:34.742 --> 00:03:36.450
And you could take
the derivative of this

00:03:36.450 --> 00:03:40.230
to see that you would
get this right over here.

00:03:40.230 --> 00:03:43.100
So using that same idea,
the anti-derivative

00:03:43.100 --> 00:03:46.880
of this over here is just
going to be the negative 13.84.

00:03:46.880 --> 00:03:51.130
And we're going to divide
by this coefficient on the t

00:03:51.130 --> 00:03:52.210
right over here.

00:03:52.210 --> 00:03:54.814
So negative 0.173.

00:03:54.814 --> 00:03:56.230
And you could use
your calculator.

00:03:56.230 --> 00:03:58.438
Calculators are allowed for
this part of the problem.

00:03:58.438 --> 00:03:59.900
But we could do
this analytically.

00:03:59.900 --> 00:04:08.340
Times e to the
negative 0.173 times t.

00:04:08.340 --> 00:04:11.662
We're going to evaluate
that from 0 to 10.

00:04:11.662 --> 00:04:13.120
So we're going to
evaluate it at 10

00:04:13.120 --> 00:04:15.819
and subtract from that
this thing evaluated at 0.

00:04:18.510 --> 00:04:21.019
Let me just factor out
this part right over here.

00:04:21.019 --> 00:04:27.800
So it's going to be negative
13.84 over negative 0.173

00:04:27.800 --> 00:04:31.720
times this evaluated
at when t is 10.

00:04:31.720 --> 00:04:34.720
So e to the-- if we
multiply this times 10--

00:04:34.720 --> 00:04:37.460
this is negative 1.73.

00:04:37.460 --> 00:04:39.110
That's when I
evaluated it at 10.

00:04:39.110 --> 00:04:41.040
And from that we
want to subtract

00:04:41.040 --> 00:04:42.440
when this is evaluated at 0.

00:04:42.440 --> 00:04:46.180
If the exponent here is 0, if t
is 0, the whole exponent is 0.

00:04:46.180 --> 00:04:49.980
So e to the 0 power
is just equal to 1.

00:04:49.980 --> 00:04:51.590
And now we can get
our calculator out

00:04:51.590 --> 00:04:53.512
to evaluate this.

00:04:53.512 --> 00:04:57.212
So I'll get my TI-85 out.

00:04:57.212 --> 00:04:59.670
I'm just going to evaluate this
inner expression right over

00:04:59.670 --> 00:05:00.169
here.

00:05:00.169 --> 00:05:07.150
So e to the negative 1.73.

00:05:07.150 --> 00:05:13.180
And from that I
want to subtract 1.

00:05:13.180 --> 00:05:16.900
So that gives me negative 0.822.

00:05:16.900 --> 00:05:18.570
So that's this part
in parentheses.

00:05:18.570 --> 00:05:21.410
And then I want to multiply it
times what I have out front.

00:05:21.410 --> 00:05:24.222
So my previous
answer times-- and I

00:05:24.222 --> 00:05:26.680
won't write the negatives just
because the negative divided

00:05:26.680 --> 00:05:28.263
by negative is going
to be a positive.

00:05:28.263 --> 00:05:38.455
So it's just 13.84
divided by 0.173.

00:05:38.455 --> 00:05:39.830
And I didn't write
the negatives,

00:05:39.830 --> 00:05:41.270
because they just cancel out.

00:05:41.270 --> 00:05:44.860
And that gives me
negative 65.817.

00:05:44.860 --> 00:05:48.170
So this is our change in
temperature for the biscuits.

00:05:48.170 --> 00:05:49.915
It's our change in temperature.

00:05:49.915 --> 00:05:50.790
Let me write it down.

00:05:50.790 --> 00:05:54.130
Negative 65.82 degrees.

00:05:54.130 --> 00:05:59.290
So this is negative
65.82 degrees Celsius.

00:05:59.290 --> 00:06:03.080
This is our change in
temp for the biscuits.

00:06:03.080 --> 00:06:05.460
Now we know that they
started off at 100 degrees,

00:06:05.460 --> 00:06:10.450
and they went down by 65.82
degrees over the 10 minutes.

00:06:10.450 --> 00:06:11.980
They started at
100, and then they

00:06:11.980 --> 00:06:14.550
went down by this
previous amount.

00:06:14.550 --> 00:06:15.770
So, whoops.

00:06:15.770 --> 00:06:17.920
I want to subtract the answer.

00:06:17.920 --> 00:06:20.450
So minus the answer.

00:06:20.450 --> 00:06:21.124
Oh, sorry.

00:06:21.124 --> 00:06:23.165
I want to actually add
the change in temperature.

00:06:23.165 --> 00:06:24.490
I want to be very careful here.

00:06:24.490 --> 00:06:26.100
It went down by 65 degrees.

00:06:26.100 --> 00:06:36.210
So I really should just
say 100 minus 65.817.

00:06:36.210 --> 00:06:37.830
I could keep adding
digits if I want.

00:06:37.830 --> 00:06:39.750
247, that's enough.

00:06:39.750 --> 00:06:44.140
So after 10 minutes the
biscuits are 34.18 degrees.

00:06:44.140 --> 00:06:50.420
And we already knew what
the temperature of the tea

00:06:50.420 --> 00:06:51.570
is after 10 minutes.

00:06:51.570 --> 00:06:53.805
We already knew
it was 43 degrees.

00:06:53.805 --> 00:07:00.100
So the biscuits
are at 34 degrees.

00:07:00.100 --> 00:07:02.440
The tea is at 43 degrees.

00:07:02.440 --> 00:07:04.370
So if we do 43 is
where the tea is, then

00:07:04.370 --> 00:07:07.230
we subtract the temperature
of the biscuits,

00:07:07.230 --> 00:07:11.360
we see that the biscuits
are 8.82 degrees cooler.

00:07:11.360 --> 00:07:16.530
8.82 degrees Celsius cooler.

00:07:16.530 --> 00:07:21.660
So we could say the biscuits
are 8.82 degrees Celsius cooler

00:07:21.660 --> 00:07:24.190
than the tea.

00:07:24.190 --> 00:07:27.180
And we're done with
part d as well.


2011 Calculus AB free response #2 (a & b)

WEBVTT
Kind: captions
Language: en

00:00:00.770 --> 00:00:02.760
As a pot of tea cools, the temperature

00:00:02.760 --> 00:00:06.260
the tea is modeled by a differentiable
function h.

00:00:06.260 --> 00:00:09.780
For 0 is less than or equal to, t is less
than or equal to 10.

00:00:09.780 --> 00:00:12.600
Where time t is measured in minutes.

00:00:12.600 --> 00:00:16.830
And temperature h of t is measured in
degrees Celsius.

00:00:16.830 --> 00:00:21.090
Values of h of t, at selected time,
selected values of time

00:00:21.090 --> 00:00:24.380
t are shown in the table above, so that's
right over there.

00:00:24.380 --> 00:00:27.420
Use the data in the table to approximate
the rate at which

00:00:27.420 --> 00:00:32.110
the temperature of the t is changing at
time t equals 3.5.

00:00:32.110 --> 00:00:36.130
Show the computations that lead to your
answer.

00:00:36.130 --> 00:00:38.232
So, they give us a bunch of points, so
let's just graph this,

00:00:38.232 --> 00:00:40.730
just so we can visualize what this data's
telling us a little bit.

00:00:40.730 --> 00:00:43.810
And, I suspect that this might be useful,

00:00:43.810 --> 00:00:48.280
so, this, right over here, is our
temperature axis.

00:00:48.280 --> 00:00:52.370
This is our temperature at any moment, in
degrees Celsius.

00:00:52.370 --> 00:00:57.560
And then, this is our time axis, and they
gave us the time of 0 2 5 9 and 10.

00:00:57.560 --> 00:01:00.186
They give us a temperature of those times.

00:01:00.186 --> 00:01:01.590
So, this is 0.

00:01:01.590 --> 00:01:03.300
2, 5's A little bit further.

00:01:03.300 --> 00:01:07.320
5, 9, And then 10.

00:01:07.320 --> 00:01:11.890
And at time zero, the temperature's 66
degrees, Celsius.

00:01:11.890 --> 00:01:14.640
66, So it's right over there.

00:01:14.640 --> 00:01:18.080
At time 2, it is 60 degrees Celsius.

00:01:18.080 --> 00:01:21.330
At time two, it is 60 degrees celsius.

00:01:21.330 --> 00:01:24.780
At time 5, it is 52 degrees Celsius.

00:01:24.780 --> 00:01:30.610
So time 5, maybe it's right over here,
this is 52 degrees Celsius.

00:01:30.610 --> 00:01:36.520
At time 9, or after 9 minutes, I should
say, it's at 44 degrees Celsius.

00:01:36.520 --> 00:01:42.050
So this 44 degrees Celsius, and after 10
minutes it's 43 degrees Celsius.

00:01:42.050 --> 00:01:44.600
So this is really just a graph showing how
it

00:01:44.600 --> 00:01:48.430
cools off over those ten minutes, over
those ten minutes.

00:01:48.430 --> 00:01:50.130
That's what the data is telling us.

00:01:50.130 --> 00:01:53.270
It's kind of a curve like this, and we've
sampled it at these points.

00:01:53.270 --> 00:01:55.340
So going back to Part A.

00:01:55.340 --> 00:01:58.010
Use the data in the table to approximate
the rate at which

00:01:58.010 --> 00:02:02.190
the temperature of the tea is changing at
time t equals 3.5.

00:02:02.190 --> 00:02:04.420
So the rate of change is really just a
slope

00:02:04.420 --> 00:02:08.289
of this curve at time t is equal to 3.5.

00:02:08.289 --> 00:02:13.450
So really, we just wanna find the slope at
that point right over there.

00:02:13.450 --> 00:02:18.610
We wanna find the slope, and we don't know
the actual function.

00:02:18.610 --> 00:02:21.220
So the best way to approximate the slope
at that point, is

00:02:21.220 --> 00:02:27.050
really just find the slope, between, time,
between minute five, and minute two.

00:02:27.050 --> 00:02:31.840
So the rate of change, so we could say, we
could say that the rate, the

00:02:31.840 --> 00:02:35.160
rate, is going to be approximately, our
change

00:02:35.160 --> 00:02:38.340
in temperature over that, over those three
minutes.

00:02:38.340 --> 00:02:42.930
So, h of 5 minus h of 2, and this is
obviously going

00:02:42.930 --> 00:02:47.675
to be in degrees Celsius, over the number
of minutes that changed, so we

00:02:47.675 --> 00:02:51.870
went from 5 minutes or from, we end at 5
minutes, and we started

00:02:51.870 --> 00:02:58.090
at 2 minutes, and so the rate, the rate is
going to be approximately.

00:02:58.090 --> 00:03:02.990
At five minutes, our temperature is 52
degrees, 52 degrees.

00:03:02.990 --> 00:03:07.350
At two minutes, our temperature is 60
degrees, 60 degrees.

00:03:07.350 --> 00:03:10.020
And then, this is over change in three
minutes.

00:03:10.020 --> 00:03:11.490
5 Minus 2 is 3.

00:03:11.490 --> 00:03:14.410
So, this gets us, let me just pull to the
right a little bit,.

00:03:14.410 --> 00:03:16.820
This gives us negative 8.

00:03:16.820 --> 00:03:21.350
Negative 8 degrees Celsius over three
minutes.

00:03:21.350 --> 00:03:25.390
Or the rate at, the rate at, at three and
a half

00:03:25.390 --> 00:03:27.260
minutes is going to be approximately

00:03:27.260 --> 00:03:31.990
negative 8 3rds, degrees Celsius per
minute.

00:03:31.990 --> 00:03:35.840
So that is part a, part a.

00:03:35.840 --> 00:03:39.132
And then part b, using correct units,
explain the

00:03:39.132 --> 00:03:42.040
meaning of 1 over 10, 1 over 10, times

00:03:42.040 --> 00:03:47.510
the definite integral from 0 to 10 of h of
t dt in the context of this problem.

00:03:47.510 --> 00:03:50.400
Use a trapezoidal sum with the 4
sub-intervals

00:03:50.400 --> 00:03:53.090
indicated by the table to estimate this
thing.

00:03:53.090 --> 00:03:55.390
So, the integral from 0 to 10 of h of t is

00:03:55.390 --> 00:04:00.050
really the area of this entire, under this
curve, right over here.

00:04:00.050 --> 00:04:02.360
That's the integral, that's this part
right over here.

00:04:02.360 --> 00:04:04.340
And then, we're going to divide that.

00:04:04.340 --> 00:04:06.280
We're going to divide that by.

00:04:06.280 --> 00:04:07.780
We're going to divide that by 10.

00:04:07.780 --> 00:04:09.950
So, what this is really giving us.

00:04:09.950 --> 00:04:10.510
This is.

00:04:10.510 --> 00:04:12.410
This expression, right over here.

00:04:12.410 --> 00:04:13.830
Is the average,

00:04:17.079 --> 00:04:18.107
this is the average temperature.

00:04:18.107 --> 00:04:20.410
This is the average temperature over that
time period.

00:04:20.410 --> 00:04:23.340
The integral of the temperature function
divided by the

00:04:23.340 --> 00:04:27.390
total amount of time that has, that has
elapsed.

00:04:27.390 --> 00:04:30.900
And then, in using correct units, well
it's the average temperature

00:04:30.900 --> 00:04:34.000
and the average temperature is once again
going to be in degrees.

00:04:34.000 --> 00:04:34.820
Degrees Celsius.

00:04:34.820 --> 00:04:37.340
And it makes sense, cuz this right over
here is in degrees Celsius.

00:04:37.340 --> 00:04:41.330
You're multiplying it by time right over
here, but then you could divide by.

00:04:41.330 --> 00:04:43.850
You're dividing, I'm assuming, by time
right over here

00:04:43.850 --> 00:04:46.630
as, so this is times minutes, divided by
minutes.

00:04:46.630 --> 00:04:49.180
So then you just get degrees Celsius
again.

00:04:49.180 --> 00:04:51.310
So this is, this expression right over
here

00:04:51.310 --> 00:04:55.030
is just the average temperature over those
ten minutes.

00:04:55.030 --> 00:04:57.570
They they say use a trapezoidal sum with
the

00:04:57.570 --> 00:05:01.390
four sub intervals indicated by the table
to estimate this.

00:05:01.390 --> 00:05:03.440
So, a trapezoidal sum is, we don't know
the

00:05:03.440 --> 00:05:05.050
exact function here, so we won't be able
to

00:05:05.050 --> 00:05:08.790
analytically evaluate this definite
integral, but what we can

00:05:08.790 --> 00:05:13.040
do is divide, is divide this area into
four sections.

00:05:13.040 --> 00:05:15.620
They tell us to use this, four
subintervals.

00:05:15.620 --> 00:05:18.280
And, we'll essentially divide it into four
trapezoids.

00:05:18.280 --> 00:05:20.510
So, this is one trapezoid right over here.

00:05:21.860 --> 00:05:23.980
So this is one trapezoid right over there.

00:05:25.080 --> 00:05:28.550
That's my first trapezoid from zero to two
on the base.

00:05:28.550 --> 00:05:30.220
And you see on the left end of that

00:05:30.220 --> 00:05:32.880
trapezoid is height of 66 the right end is
60.

00:05:32.880 --> 00:05:37.680
And the the next trapezoid will go from
two to five.

00:05:37.680 --> 00:05:39.160
Let me do that in a more.

00:05:39.160 --> 00:05:41.300
Color that contrast a little bit better.

00:05:41.300 --> 00:05:43.270
The next trapezoid will go from 2 to 5.

00:05:43.270 --> 00:05:46.880
The next trapezoid goes from 2 to 5.

00:05:46.880 --> 00:05:50.160
And then the third trapezoid goes from,

00:05:53.650 --> 00:05:54.662
goes from 5 to 9.

00:05:54.662 --> 00:05:55.423
The third trapezoid goes from 5 to 9.

00:05:55.423 --> 00:05:58.480
And then, the fourth trapezoid goes from 9
to 10.

00:05:58.480 --> 00:05:59.850
Goes from 9 to 10.

00:05:59.850 --> 00:06:01.320
And so if I want to.

00:06:01.320 --> 00:06:05.268
If I want to approximate the definite
integral part right here, before we

00:06:05.268 --> 00:06:08.790
divide by 10, I just need to find the area
of these four trapezoids.

00:06:08.790 --> 00:06:10.980
So, let's do that.

00:06:10.980 --> 00:06:13.450
The area of this first trapezoid is going
to

00:06:13.450 --> 00:06:17.950
be the base, which is 2 times the average
height.

00:06:17.950 --> 00:06:20.710
The height if the left side for this
trapped is 66.

00:06:20.710 --> 00:06:22.920
The height on the right side of this
trapezoid is 60.

00:06:22.920 --> 00:06:25.250
The average height right over here is
going to

00:06:25.250 --> 00:06:28.280
be 63, just the average of 60 and 66.

00:06:28.280 --> 00:06:32.770
So, this area right over here is, this is
126,

00:06:32.770 --> 00:06:37.210
that's the area of the green part right
over there.

00:06:37.210 --> 00:06:39.970
By the same logic, the area of this orange
part.

00:06:40.990 --> 00:06:42.530
I want to do that in the orange color.

00:06:43.630 --> 00:06:48.250
The area of this orange part, the base
right over here is 3.

00:06:48.250 --> 00:06:49.000
Base is 3.

00:06:49.000 --> 00:06:50.520
And then the average height.

00:06:50.520 --> 00:06:51.890
The height over here is 60.

00:06:51.890 --> 00:06:54.060
The height over here is 52.

00:06:54.060 --> 00:06:56.790
This is, they're seated 8 apart, so the
average is going

00:06:56.790 --> 00:06:59.010
to be 4 away from each is going to be 56.

00:06:59.010 --> 00:07:04.800
So this is, the area here is going to be 3
times, 3 times 56.

00:07:04.800 --> 00:07:07.540
Which is 150 plus 18.

00:07:07.540 --> 00:07:11.740
150 Plus 18 is 168.

00:07:11.740 --> 00:07:18.110
That's the area of this orange trapezoid,
and then the area of this blue trapezoid.

00:07:18.110 --> 00:07:20.640
The base right over here is 4, and then

00:07:20.640 --> 00:07:24.397
then average height, the height here is
52, the height

00:07:24.397 --> 00:07:27.464
here is 44, the average of 52 and 44, let

00:07:27.464 --> 00:07:31.240
see they're 10, they are, sorry, they are
8 apart.

00:07:31.240 --> 00:07:34.020
So then it's just gonna be 4 from each of
these so it's 48, so

00:07:34.020 --> 00:07:40.200
the average height here is 48, and so 4
times 48 is 160 plus 32.

00:07:40.200 --> 00:07:47.690
So it's 192, and then finally this last
trapezoid, it's base is only 1.

00:07:47.690 --> 00:07:50.720
Its base right over here is only 1, and
then its

00:07:50.720 --> 00:07:56.820
height is 43 at the left side, 43 at the
right side.

00:07:56.820 --> 00:08:00.303
So its average height is 43.5, 43.5.

00:08:00.303 --> 00:08:03.200
So it's 43.5 times 1.

00:08:03.200 --> 00:08:09.600
So that's just going to be 43.5, and so if
we add up the areas under,

00:08:09.600 --> 00:08:11.150
or the areas of all these trapezoids, we

00:08:11.150 --> 00:08:13.710
have a pretty good approximation for the
definite integral.

00:08:13.710 --> 00:08:14.590
I'll just use, do the.

00:08:14.590 --> 00:08:16.748
Use the calculator for this part right
over here.

00:08:16.748 --> 00:08:25.951
So we have 126, plus 168, plus 192, plus
43.5, 43.5.

00:08:25.951 --> 00:08:27.630
Gives us 529.5.

00:08:27.630 --> 00:08:31.470
So this is equal to 529.5.

00:08:31.470 --> 00:08:34.740
And so that's our

00:08:36.780 --> 00:08:40.000
approximation for the definite integral
part.

00:08:40.000 --> 00:08:45.090
But then we also have to divide it by 10,
or we have to multiply time 1 10th.

00:08:45.090 --> 00:08:47.630
So we've evaluated, so this part right

00:08:47.630 --> 00:08:51.780
over here, we got 529.5 as our
approximation.

00:08:51.780 --> 00:08:54.030
It's not going to be exact, but using the,
using

00:08:54.030 --> 00:08:56.980
the trapezoidal sum, the sum of the areas
of these trapezoids.

00:08:56.980 --> 00:09:00.975
And now we have to multiply by 1 10th, so
lets do that.

00:09:00.975 --> 00:09:03.606
So, or we could just divide by 10, which

00:09:03.606 --> 00:09:07.417
actually we don't need a calculator for
that, 52.95.

00:09:07.417 --> 00:09:11.615
So this whole thing, this whole thing
evaluates to 52.95,

00:09:11.615 --> 00:09:14.760
and I'll do the next two parts in the next
video.


Early Show

WEBVTT
Kind: captions
Language: en

00:00:00.060 --> 00:00:02.310
SPEAKER 1: We continue our
Education in America series

00:00:02.310 --> 00:00:04.640
this morning with one man
who is turning the learning

00:00:04.640 --> 00:00:06.280
process upside down.

00:00:06.280 --> 00:00:08.520
SPEAKER 2: His series
of really simple videos

00:00:08.520 --> 00:00:11.800
is being called revolutionary,
as CBS News correspondent John

00:00:11.800 --> 00:00:12.810
Blackstone reports.

00:00:16.994 --> 00:00:18.410
JOHN BLACKSTONE:
It would be music

00:00:18.410 --> 00:00:21.980
to the ears of any educator,
what 10-year-old Megan

00:00:21.980 --> 00:00:25.330
Tsern has to say about
her favorite teacher.

00:00:25.330 --> 00:00:28.520
MEGAN TSERN: He makes
math really easy

00:00:28.520 --> 00:00:30.317
in a really kind of
conversational way.

00:00:30.317 --> 00:00:32.150
SALMAN KHAN: It would
be on the number line.

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JOHN BLACKSTONE:
He is Salman Khan.

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SALMAN KHAN: It's negative $7.

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It's like saying that I owe $7.

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JOHN BLACKSTONE:
Khan is the voice

00:00:39.200 --> 00:00:42.030
and the brains behind
the Khan Academy,

00:00:42.030 --> 00:00:44.770
the free educational
website that some believe

00:00:44.770 --> 00:00:46.735
is leading a
revolution in teaching.

00:00:46.735 --> 00:00:48.110
SALMAN KHAN:
There's no nonsense.

00:00:48.110 --> 00:00:50.310
There's no defocusing
on the material.

00:00:50.310 --> 00:00:51.720
It really is what it is.

00:00:51.720 --> 00:00:54.804
JOHN BLACKSTONE: Khan's lessons
are unscripted and unadorned.

00:00:54.804 --> 00:00:57.220
SALMAN KHAN: Let's divide both
sides of this equation by--

00:00:57.220 --> 00:00:59.230
JOHN BLACKSTONE: His
voice over a black screen

00:00:59.230 --> 00:01:01.520
as he writes or doodles.

00:01:01.520 --> 00:01:04.620
His style has changed little
since this all began back

00:01:04.620 --> 00:01:09.270
in 2004 as YouTube videos
to help his young cousins.

00:01:09.270 --> 00:01:10.814
SALMAN KHAN: 9 times the 6.

00:01:10.814 --> 00:01:12.230
JOHN BLACKSTONE:
He's now recorded

00:01:12.230 --> 00:01:14.940
more than 2,400
lessons on everything

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from simple arithmetic
to college-level biology.

00:01:18.960 --> 00:01:20.614
Anyone can watch for free.

00:01:20.614 --> 00:01:22.780
SALMAN KHAN: People are
more likely to say thank you

00:01:22.780 --> 00:01:24.630
when it's free.

00:01:24.630 --> 00:01:27.290
And you know, thank
you is worth a lot.

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JOHN BLACKSTONE: But what
are Khan's lessons worth?

00:01:29.560 --> 00:01:33.840
The Los Altos School District
is trying to figure that out.

00:01:33.840 --> 00:01:37.470
Last year, the district used
Khan's lessons in five classes.

00:01:37.470 --> 00:01:40.570
The students did well enough
that this year the Khan method

00:01:40.570 --> 00:01:45.200
is in 40 classes, to be used
by every fifth and sixth grader

00:01:45.200 --> 00:01:46.130
in the district.

00:01:46.130 --> 00:01:48.816
ALYSSA GALLAGHER: We found
that our students are really

00:01:48.816 --> 00:01:50.190
engaged in this
type of learning,

00:01:50.190 --> 00:01:52.874
that it was a great addition
to what we were already doing.

00:01:52.874 --> 00:01:54.790
JOHN BLACKSTONE: You
might say this California

00:01:54.790 --> 00:01:55.880
School has flipped.

00:01:55.880 --> 00:01:57.600
Students get their
lectures at home

00:01:57.600 --> 00:01:59.780
and come to school
to do their homework,

00:01:59.780 --> 00:02:02.407
fundamentally changing
the role of the teacher.

00:02:02.407 --> 00:02:04.490
COURTNEY CADWELL: The kids
can really move through

00:02:04.490 --> 00:02:07.533
at their own pace, and then
I can see who's struggling.

00:02:07.533 --> 00:02:08.949
JOHN BLACKSTONE:
At home, students

00:02:08.949 --> 00:02:11.009
listen to Khan's online lessons.

00:02:11.009 --> 00:02:13.290
In the classroom,
they practice what

00:02:13.290 --> 00:02:16.100
they learned while teacher
Courtney Cadwell works

00:02:16.100 --> 00:02:17.142
with those who need help.

00:02:17.142 --> 00:02:18.849
COURTNEY CADWELL: When
they're having fun

00:02:18.849 --> 00:02:20.490
and they're wanting
to do math, that

00:02:20.490 --> 00:02:22.431
tells me that this
is a good thing.

00:02:22.431 --> 00:02:24.180
JOHN BLACKSTONE: This
educational overhaul

00:02:24.180 --> 00:02:27.310
comes from a man who was
working as a hedge fund analyst

00:02:27.310 --> 00:02:29.220
until 2009.

00:02:29.220 --> 00:02:34.150
Khan has a Harvard MBA and
three degrees from MIT.

00:02:34.150 --> 00:02:36.550
You're kind of a math
geek, I guess, are you?

00:02:36.550 --> 00:02:38.570
SALMAN KHAN: Yes, in the
best form of the word.

00:02:38.570 --> 00:02:39.986
JOHN BLACKSTONE:
In the beginning,

00:02:39.986 --> 00:02:42.930
he financed the Khan Academy
from his own savings,

00:02:42.930 --> 00:02:45.599
until he was discovered
by some people with money.

00:02:45.599 --> 00:02:46.890
BILL GATES: Well, it's amazing.

00:02:46.890 --> 00:02:48.500
I think you've
just got a glimpse

00:02:48.500 --> 00:02:50.300
of the future of education.

00:02:50.300 --> 00:02:50.890
Thank you.

00:02:50.890 --> 00:02:53.140
JOHN BLACKSTONE: The Bill
and Melinda Gates Foundation

00:02:53.140 --> 00:02:55.135
and Google are among
the big donors that

00:02:55.135 --> 00:02:59.320
have allowed the Khan Academy
to add nearly a dozen software

00:02:59.320 --> 00:03:00.640
engineers.

00:03:00.640 --> 00:03:02.980
MEGAN TSERN: It's really
smartly done on the computer,

00:03:02.980 --> 00:03:04.880
and I love being
on the computer.

00:03:04.880 --> 00:03:07.440
JOHN BLACKSTONE: With more
than two million users a month,

00:03:07.440 --> 00:03:10.410
many others seem to love
Khan's lessons, too.

00:03:10.410 --> 00:03:12.930
SALMAN KHAN: I've already
taught 70 million lessons.

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JOHN BLACKSTONE: And he will be
happy to teach millions more.

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SALMAN KHAN: Smallest
to the largest.

00:03:18.549 --> 00:03:20.340
JOHN BLACKSTONE: John
Blackstone, CBS News.

00:03:20.340 --> 00:03:22.925
Mountain View, California.