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PROFESSOR: Today we're
going to keep on going
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with related rates.
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And you may recall that last
time we were in the middle of
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a problem with this geometry.
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There was a right triangle.
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There was a road.
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Which was going this way,
from right to left.
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And the police were up here,
monitoring the situation.
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30 feet from the road.
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And you're here.
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And you're heading this way.
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Maybe it's a two lane highway,
but anyway it's only
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going this direction.
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And this distance was 50 feet.
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So, because you're moving,
this distance is varying
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and so we gave it a letter.
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And, similarly, your distance
to the foot of the
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perpendicular with the
road is also varying.
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At this instant it's 40,
because this is a 3,
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4, 5 right triangle.
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So this was the situation
that we were in last time.
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00:01:35 --> 00:01:38
And we're going to pick
up where we left off.
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The question is, are you
speeding if the rate of change
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of D with respect to t
is 80 feet per second.
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Now, technically that would be
- 80, because you're going
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towards the policemen.
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Alright, so D is shrinking at a
rate of - 80 feet per second.
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And I remind you that 95 feet
per second is approximately
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the speed limit.
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Which is 65 miles per hour.
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00:02:21 --> 00:02:24
So, again, this is where
we were last time.
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00:02:24 --> 00:02:30
And, got a little
question mark there.
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00:02:30 --> 00:02:34
And so let's solve
this problem.
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00:02:34 --> 00:02:37
So, this is the setup.
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There's a right triangle.
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So there's a relationship
between these lengths.
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And the relationship is
that x ^2 + 30 ^2 = D ^2.
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So that's the first
relationship that we have.
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And the second relationship
that we have, we've
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already written down.
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Which is dx / dt - oops,
sorry. dD / dt = minus 80.
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00:03:08 --> 00:03:14
Now, the idea here is
relatively straightforward.
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We just want to use
differentiation.
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Now, you could solve for x.
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Alright, x is the square
root of D^2 - 30 squared.
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That's one possibility.
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But this is basically
a waste of time.
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It's a waste of your time.
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So it's easier, or easiest, to
follow this method of implicit
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differentiation, which I want
to encourage you
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to get used to.
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Namely, we just differentiate
this equation with
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respect to time.
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Now, when you do that, you have
to remember that you are not
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allowed to plug in a constant.
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Namely 40, for t.
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You have to keep in mind what's
really going on in this
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problem which is that x
is moving, it's changing.
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And D is also changing.
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So you have to differentiate
first before you
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plug in the values.
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So the easiest thing is
to use, in this case,
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implicit differentiation.
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And if I do that, I get 2x dx /
dt is equal to 2D (dD / dt).
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No more DDT left.
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We hope.
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Except in this blackboard.
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So there's our situation.
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Now, if I just plug in,
now I can plug in.
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In values. so this is after
taking the derivative.
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And, indeed, we have here 2
times the value for x which
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is 40 at this instant.
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And then we have dx / dt.
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And that's equal to
2 (D), which is 50.
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And then dD / dt is - 80.
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So the 80's cancel and
we see that dv / dt =
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- 100 feet per second.
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And so the answer to
the question is yes.
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Although you probably wouldn't
be pulled over for this
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much of a violation.
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So that's, right, it's more
than 65 miles an hour,
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by a little bit.
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So that's the end
of this question.
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And usually in these rate of
change or related rates
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questions, this is considered
to be the answer to the
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question, if you like.
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So that's one example.
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I'm going to give one more
example before we go on to
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some other applications
implicit differentiation.
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So my second example is
going to be, you have a
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it is a conical tank.
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With top of radius
4 feet, let's say.
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And depth 10 feet.
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So that's our situation.
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And then it's being
filled with water.
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So, is being filled with water.
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At 2 cubic feet per minute.
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So there is our situation.
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And then the question is, how
fast is the water rising
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when it is at depth 5 feet?
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So if this thing is half-full
in the sense, well not
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half-full in terms
of total volume.
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00:07:21 --> 00:07:29
But half-full in terms of
height, what's the speed at
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00:07:29 --> 00:07:36
which the water is rising.
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So, how do we set up
problems like this?
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Well, we talked about
this last time.
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00:07:42 --> 00:07:51
The first step is to set up
a diagram and variables.
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And I'm just going to
draw the picture.
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And I'm actually going to
draw the picture twice.
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So here's the conical tank.
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And we have this
radius, which is 4.
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00:08:05 --> 00:08:07
And we have this
height, which is 10.
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So that's just to allow me to
think about this problem.
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00:08:10 --> 00:08:18
Now, it turns out because we
have a varying depth and so
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00:08:18 --> 00:08:21
on, and this is just the top.
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That I'd better depict also
the level at which the water
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00:08:24 --> 00:08:26
actually is at present.
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00:08:26 --> 00:08:30
And furthermore, it's better
to do this schematically,
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as you'll see.
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00:08:31 --> 00:08:38
So the key point here is to
draw this triangle here.
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00:08:38 --> 00:08:45
Which shows me the 10 and
shows me the 4, over here.
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00:08:45 --> 00:08:48
And then imagine that
I'm filling it partway.
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00:08:48 --> 00:08:52
So maybe we'll put that water
level in in another color here.
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00:08:52 --> 00:08:54
So here's the water level.
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00:08:54 --> 00:08:57
And the water level, I'm going
to depict that horizontal
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00:08:57 --> 00:08:59
distance here, as r.
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00:08:59 --> 00:09:00
That's going to be my variable.
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00:09:00 --> 00:09:04
That's the radius of
the top of the water.
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00:09:04 --> 00:09:06
So I'm taking a cross-section
of this, because that
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00:09:06 --> 00:09:10
geometrically the only thing
I have to keep track of.
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00:09:10 --> 00:09:11
At least initially.
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00:09:11 --> 00:09:14
So this is our water level.
141
00:09:14 --> 00:09:18
And it's really rotated around.
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00:09:18 --> 00:09:23
But I'm depicting just this one
half-slice of the picture.
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00:09:23 --> 00:09:26
And then similarly,
I have the height.
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00:09:26 --> 00:09:30
Which is this dimension there.
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00:09:30 --> 00:09:32
Or, if you like, the
depth of the water.
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00:09:32 --> 00:09:42
So the water has filled us
up, up to this point here.
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00:09:42 --> 00:09:46
So I set it up this way so
that it's clear that we
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00:09:46 --> 00:09:47
have two triangles here.
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00:09:47 --> 00:09:50
And that one piece of
information we can get from
150
00:09:50 --> 00:09:53
the geometry is the similar
triangles information.
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00:09:53 --> 00:09:59
Namely, that r / h = 4 / 10.
152
00:09:59 --> 00:10:09
So this is by far the most
typical geometric fact that
153
00:10:09 --> 00:10:13
you'll have to glean
from a picture.
154
00:10:13 --> 00:10:17
So that's one piece of
information that we get
155
00:10:17 --> 00:10:18
from this picture.
156
00:10:18 --> 00:10:23
Now, the second thing we have
to do is set up formulas for
157
00:10:23 --> 00:10:26
the volume of water, and then
figure out what's
158
00:10:26 --> 00:10:27
going on here.
159
00:10:27 --> 00:10:33
So the volume of
water is, of a cone.
160
00:10:33 --> 00:10:37
So again, you have to know
something about geometry to
161
00:10:37 --> 00:10:38
do many of these problems.
162
00:10:38 --> 00:10:41
So you have to know that
the volume of a cone is
163
00:10:41 --> 00:10:44
1/3 base times height.
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00:10:44 --> 00:10:45
Now, this one is upside down.
165
00:10:45 --> 00:10:49
The base is on the top and
it's going down but it
166
00:10:49 --> 00:10:50
works the same way.
167
00:10:50 --> 00:10:52
That doesn't affect volume.
168
00:10:52 --> 00:10:57
So it's 1/3 and the base is
pi r^2, that's the base.
169
00:10:57 --> 00:11:01
And times h, which
is the height.
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00:11:01 --> 00:11:06
So this is the setup
for this problem.
171
00:11:06 --> 00:11:12
And now, having our
relationship, we have one
172
00:11:12 --> 00:11:14
relationship left that
we have to remember.
173
00:11:14 --> 00:11:17
Because we have one more piece
of information in this problem.
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00:11:17 --> 00:11:20
Namely, how fast the
volume is changing.
175
00:11:20 --> 00:11:22
It's going at 2 cubic
feet per minute.
176
00:11:22 --> 00:11:29
It's increasing, so that
means that dv / dt = 2.
177
00:11:29 --> 00:11:35
So I've now gotten rid of
all the words and I have
178
00:11:35 --> 00:11:38
only formulas left.
179
00:11:38 --> 00:11:43
I started here with the words,
I drew some pictures, and
180
00:11:43 --> 00:11:44
I derived some formulas.
181
00:11:44 --> 00:11:46
Actually, there's
one thing missing.
182
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What's missing?
183
00:11:49 --> 00:11:52
Yeah.
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00:11:52 --> 00:11:57
STUDENT: [INAUDIBLE]
185
00:11:57 --> 00:11:57
PROFESSOR: Exactly.
186
00:11:57 --> 00:11:58
What you want to find.
187
00:11:58 --> 00:12:00
What I left out
is the question.
188
00:12:00 --> 00:12:15
So the question is, what
is dh / dt when h = 5?
189
00:12:15 --> 00:12:20
So that's the question here.
190
00:12:20 --> 00:12:22
Now, we've got the whole
problem and we never have to
191
00:12:22 --> 00:12:25
look at it again if you like.
192
00:12:25 --> 00:12:32
We just have to pay attention
to this piece here.
193
00:12:32 --> 00:12:34
So let's carry it out.
194
00:12:34 --> 00:12:36
So what happens here.
195
00:12:36 --> 00:12:39
So look, you could do this by
implicit differentiation.
196
00:12:39 --> 00:12:42
But it's so easy to express
r as a function of h that
197
00:12:42 --> 00:12:44
that seems kind of foolish.
198
00:12:44 --> 00:12:48
So let's write r as 2/5 h.
199
00:12:48 --> 00:12:51
That's coming from this
first equation here.
200
00:12:51 --> 00:12:53
And then let's
substitute that in.
201
00:12:53 --> 00:13:04
That means that v =
1/3 pi ( 2/5 h^2) h.
202
00:13:04 --> 00:13:06
And now I have to
differentiate that.
203
00:13:06 --> 00:13:08
So now I will use implicit
differentiation.
204
00:13:08 --> 00:13:11
It's very foolish at
this point to take cube
205
00:13:11 --> 00:13:13
roots to solve for h.
206
00:13:13 --> 00:13:16
You just get yourself
into a bunch of junk.
207
00:13:16 --> 00:13:18
So there is a stage at which
we're still using implicit
208
00:13:18 --> 00:13:19
differentiation here.
209
00:13:19 --> 00:13:24
I'm not going to try to solve
for h as a function of v.
210
00:13:24 --> 00:13:26
Instead I'm just
differentiating straight out
211
00:13:26 --> 00:13:28
from this formula, which is
relatively easy to
212
00:13:28 --> 00:13:29
differentiate.
213
00:13:29 --> 00:13:36
So this is dv / dt, which
of course = 2 and if I
214
00:13:36 --> 00:13:39
differentiate it I just get
this constant, pi / 3, this
215
00:13:39 --> 00:13:41
other constant, (2/5)^2.
216
00:13:42 --> 00:13:43
And then I have to
differentiate h^3.
217
00:13:45 --> 00:13:45
(h^2)h.
218
00:13:47 --> 00:13:53
So that's 3h^2 ( dh / dt).
219
00:13:53 --> 00:13:59
That's the chain rule.
220
00:13:59 --> 00:14:02
So now let's plug in
all of our numbers.
221
00:14:02 --> 00:14:07
Again, the other theme is, you
don't plug in numbers, fixed
222
00:14:07 --> 00:14:09
numbers, until everything
has stopped moving.
223
00:14:09 --> 00:14:12
At this point, we've already
calculated our rates of change.
224
00:14:12 --> 00:14:14
So now I can put
in the numbers.
225
00:14:14 --> 00:14:24
So, 2 = pi 3 ( 2/5)^2( 3), and
then h was 5, so this is 5^2.
226
00:14:26 --> 00:14:27
And then I have dh / dt.
227
00:14:27 --> 00:14:30
There's only one unknown
thing left in this
228
00:14:30 --> 00:14:32
problem, which is dh / dt.
229
00:14:32 --> 00:14:33
Everything else is a number.
230
00:14:33 --> 00:14:36
And if you do all the
cancellations, you see
231
00:14:36 --> 00:14:38
that this cancels this.
232
00:14:38 --> 00:14:41
One of the 2's cancels -
well, this cancels this.
233
00:14:41 --> 00:14:42
And then one of the
2's cancels that.
234
00:14:42 --> 00:14:54
So all-told what we have here
is that dh / dt = 1 / 2 pi.
235
00:14:54 --> 00:14:58
And so that happens to
be feet per second.
236
00:14:58 --> 00:15:04
This is the whole story.
237
00:15:04 --> 00:15:05
Questions, way back there.
238
00:15:05 --> 00:15:10
STUDENT: [INAUDIBLE]
239
00:15:10 --> 00:15:11
PROFESSOR: Where did I get --
the question is where did
240
00:15:11 --> 00:15:14
I get r = 2/5 h from.
241
00:15:14 --> 00:15:17
The answer was, it came
from similar triangles,
242
00:15:17 --> 00:15:19
which is over here.
243
00:15:19 --> 00:15:20
I did this similar
triangle thing.
244
00:15:20 --> 00:15:26
And I got this relationship
here. r / h = 4/10, but then
245
00:15:26 --> 00:15:31
I canceled it, got 2/5
and brought the h over.
246
00:15:31 --> 00:15:32
Another question, over here.
247
00:15:32 --> 00:15:45
STUDENT: [INAUDIBLE]
248
00:15:45 --> 00:15:53
PROFESSOR: The question
was the following.
249
00:15:53 --> 00:15:56
Suppose you're at this
stage, can you write
250
00:15:56 --> 00:15:58
from here dv / dh.
251
00:15:58 --> 00:15:59
So, suppose you're here.
252
00:15:59 --> 00:16:01
And work out what this is.
253
00:16:01 --> 00:16:04
It's going to end up being
some constant times h^2.
254
00:16:05 --> 00:16:14
And then also use dv / dt
= dv / dh ( dh / dt).
255
00:16:14 --> 00:16:17
Which the chain rule.
256
00:16:17 --> 00:16:20
And the answer is yes.
257
00:16:20 --> 00:16:26
We can do that, and indeed that
is what my next sentence is.
258
00:16:26 --> 00:16:27
That's exactly what I'm saying.
259
00:16:27 --> 00:16:32
So when I said this - sorry,
when you said this, I did that.
260
00:16:32 --> 00:16:34
That's exactly what I did.
261
00:16:34 --> 00:16:41
This chunk is exactly dv / dh.
262
00:16:41 --> 00:16:43
That's just what I'm doing.
263
00:16:43 --> 00:16:44
OK.
264
00:16:44 --> 00:16:48
So, definitely, that's
what I had in mind.
265
00:16:48 --> 00:16:49
Yeah, another question.
266
00:16:49 --> 00:16:55
STUDENT: [INAUDIBLE]
267
00:16:55 --> 00:16:57
PROFESSOR: You're asking
me whether my arithmetic
268
00:16:57 --> 00:16:58
is right or not?
269
00:16:58 --> 00:17:03
STUDENT: [INAUDIBLE]
270
00:17:03 --> 00:17:06
PROFESSOR: Pi - per second.
271
00:17:06 --> 00:17:07
Oh.
272
00:17:07 --> 00:17:09
This should - no, OK, right.
273
00:17:09 --> 00:17:11
I guess it's per minute.
274
00:17:11 --> 00:17:12
Since the other one
is per minute.
275
00:17:12 --> 00:17:14
Thank you.
276
00:17:14 --> 00:17:16
Yes.
277
00:17:16 --> 00:17:17
Was there another question?
278
00:17:17 --> 00:17:21
Probably also fixing my
seconds to minutes.
279
00:17:21 --> 00:17:23
Way back there.
280
00:17:23 --> 00:17:26
STUDENT: I don't understand,
why did you have
281
00:17:26 --> 00:17:27
to do all that.
282
00:17:27 --> 00:17:30
Isn't the speed 2 cubic
feet per minute?
283
00:17:30 --> 00:17:34
PROFESSOR: The speed at which
you're filling it is 2 cubic
284
00:17:34 --> 00:17:37
feet, but the water level is
rising at a different rate,
285
00:17:37 --> 00:17:40
depending on whether you're
low down or high up.
286
00:17:40 --> 00:17:44
It depends on how wide the
pond, the surface, is.
287
00:17:44 --> 00:17:45
So in fact it's not.
288
00:17:45 --> 00:17:48
In fact, the answer wasn't
2 cubic - it wasn't.
289
00:17:48 --> 00:17:49
There's a rate there.
290
00:17:49 --> 00:17:52
That is, that's how much
volume is being added.
291
00:17:52 --> 00:17:54
But then there's another number
that we're keeping track
292
00:17:54 --> 00:17:55
of, which is the height.
293
00:17:55 --> 00:17:58
Or, if you like, the
depth of the water.
294
00:17:58 --> 00:18:00
OK.
295
00:18:00 --> 00:18:01
So this is the whole point
about related rates.
296
00:18:01 --> 00:18:03
Is you have one variable,
which is v, which is
297
00:18:03 --> 00:18:05
volume of something.
298
00:18:05 --> 00:18:07
You have another variable,
which is h, which is the height
299
00:18:07 --> 00:18:10
of the cone of water there.
300
00:18:10 --> 00:18:13
And you're keeping track of one
variable in terms of the other.
301
00:18:13 --> 00:18:15
And the relationship will
always be a chain rule
302
00:18:15 --> 00:18:17
type of relationship.
303
00:18:17 --> 00:18:19
So, therefore, you'll be able
to - if you know one you'll be
304
00:18:19 --> 00:18:20
able to figure out the other.
305
00:18:20 --> 00:18:22
Provided you get all
of the related rates.
306
00:18:22 --> 00:18:24
These are what are
called related rates.
307
00:18:24 --> 00:18:25
This is a rate of
something with respect
308
00:18:25 --> 00:18:28
to something, etc. etc.
309
00:18:28 --> 00:18:30
So it's really all
about the chain rule.
310
00:18:30 --> 00:18:38
And just fitting it
to word problems.
311
00:18:38 --> 00:18:41
So there's a couple
of examples.
312
00:18:41 --> 00:18:44
And I'll let you
work out some more.
313
00:18:44 --> 00:18:50
So now, the next thing that
I want to do is to give you
314
00:18:50 --> 00:18:54
one more max-min problem.
315
00:18:54 --> 00:18:58
Which has to do with
this device, which
316
00:18:58 --> 00:19:00
I brought with me.
317
00:19:00 --> 00:19:02
So this is a ring.
318
00:19:02 --> 00:19:05
Happens to be a napkin ring,
and this is some parachute
319
00:19:05 --> 00:19:09
string, which I use
when I go backpacking.
320
00:19:09 --> 00:19:15
And the question is if you
have a - you think of this
321
00:19:15 --> 00:19:17
is a weight, it's flexible.
322
00:19:17 --> 00:19:19
It's allowed to
move along here.
323
00:19:19 --> 00:19:23
And the question is, if I fix
these two ends where my two
324
00:19:23 --> 00:19:25
hands are, where my right
hand is here and where my
325
00:19:25 --> 00:19:28
left hand is over there.
326
00:19:28 --> 00:19:32
And the question is, where
does this ring settle down.
327
00:19:32 --> 00:19:36
Now, it's obvious, or should be
maybe obvious, is that if my
328
00:19:36 --> 00:19:41
two hands are at equal heights,
it should settle in the middle.
329
00:19:41 --> 00:19:44
The question that we're trying
to resolve is what if one hand
330
00:19:44 --> 00:19:47
is a little higher
than the other.
331
00:19:47 --> 00:19:49
What happens, or
if the other way.
332
00:19:49 --> 00:19:52
Where does it settle down?
333
00:19:52 --> 00:19:55
So in order to depict this
problem properly, I need two
334
00:19:55 --> 00:19:57
volunteers to help me out.
335
00:19:57 --> 00:20:00
Can I have some help?
336
00:20:00 --> 00:20:03
OK.
337
00:20:03 --> 00:20:06
So I need one of you to hold
the right side, and one of
338
00:20:06 --> 00:20:08
you to hold the left side.
339
00:20:08 --> 00:20:09
OK.
340
00:20:09 --> 00:20:11
And just hold it against
the blackboard.
341
00:20:11 --> 00:20:12
We're going to trace.
342
00:20:12 --> 00:20:15
So stick it about here,
in the middle somewhere.
343
00:20:15 --> 00:20:17
And now we want to make
sure that this one is a
344
00:20:17 --> 00:20:18
little higher, alright?
345
00:20:18 --> 00:20:22
So we'll have to - yeah, let's
get a little higher up.
346
00:20:22 --> 00:20:25
That's probably good enough.
347
00:20:25 --> 00:20:29
So the experiment
has been done.
348
00:20:29 --> 00:20:31
We now see where this thing is.
349
00:20:31 --> 00:20:33
But, so now hold on tight.
350
00:20:33 --> 00:20:35
This thing stretches.
351
00:20:35 --> 00:20:38
So we want to get it
stretched so that we can
352
00:20:38 --> 00:20:40
see what it is properly.
353
00:20:40 --> 00:20:43
So this thing isn't heavy
enough for this demonstration.
354
00:20:43 --> 00:20:46
I should've had a ten-ton
brick attached there.
355
00:20:46 --> 00:20:50
But if I did that, than I
would tax my, right, I would
356
00:20:50 --> 00:20:52
tax your abilities to.
357
00:20:52 --> 00:20:54
Right, so we're going to
try to trace out what the
358
00:20:54 --> 00:21:03
possibilities are here.
359
00:21:03 --> 00:21:06
So this is, roughly speaking,
where this thing - and so
360
00:21:06 --> 00:21:07
now where does it settle.
361
00:21:07 --> 00:21:10
Well, it settles about here.
362
00:21:10 --> 00:21:12
So we're going to put that
as x marks the spot.
363
00:21:12 --> 00:21:15
OK, thank you very much.
364
00:21:15 --> 00:21:20
Got to remember where those
dots -- OK, all right.
365
00:21:20 --> 00:21:24
Sorry, I forgot to
mark the spots.
366
00:21:24 --> 00:21:27
OK, so here's the situation.
367
00:21:27 --> 00:21:29
We have a problem.
368
00:21:29 --> 00:21:33
And we've hung a string.
369
00:21:33 --> 00:21:39
And it went down like this
and then it went like that.
370
00:21:39 --> 00:21:43
And we discovered
where it settled.
371
00:21:43 --> 00:21:44
Physically.
372
00:21:44 --> 00:21:47
So we want to explain this
mathematically, and see what's
373
00:21:47 --> 00:21:49
going on with this problem.
374
00:21:49 --> 00:21:50
So this is a real-life problem.
375
00:21:50 --> 00:21:53
It honestly is the problem
you have to solve if you
376
00:21:53 --> 00:21:54
want to build a bridge.
377
00:21:54 --> 00:21:55
Like, a suspension bridge.
378
00:21:55 --> 00:21:57
This, among many problems.
379
00:21:57 --> 00:21:59
It's a very serious and
important problem.
380
00:21:59 --> 00:22:02
And this is the simplest
one of this type.
381
00:22:02 --> 00:22:04
So we've got our shape here.
382
00:22:04 --> 00:22:06
This should be a straight
line, maybe not quite
383
00:22:06 --> 00:22:10
as angled as that.
384
00:22:10 --> 00:22:13
The first, we've already drawn
the diagram and we've more or
385
00:22:13 --> 00:22:16
less visualized what's
going on here.
386
00:22:16 --> 00:22:21
But the first step after the
diagram is to give letters.
387
00:22:21 --> 00:22:24
Is to label things
appropriately.
388
00:22:24 --> 00:22:29
And I do not expect you to be
able to do this, at this stage.
389
00:22:29 --> 00:22:31
Because it requires a
lot of experience.
390
00:22:31 --> 00:22:33
But I'm going to do it for you.
391
00:22:33 --> 00:22:34
We're going to just do that.
392
00:22:34 --> 00:22:37
So the simplest thing to do
is to use the coordinates
393
00:22:37 --> 00:22:39
of the plane.
394
00:22:39 --> 00:22:43
And if you do that, it's also
easiest to use the origin.
395
00:22:43 --> 00:22:46
My favorite number is 0 and
it should be yours, too.
396
00:22:46 --> 00:22:51
So we're going to make
this point be (0, 0).
397
00:22:51 --> 00:22:54
Now, there's another fixed
point in this problem.
398
00:22:54 --> 00:22:57
And it's this point over here.
399
00:22:57 --> 00:22:58
And we don't know what
its coordinates are.
400
00:22:58 --> 00:23:00
So we're just going to give
them letters, a and b.
401
00:23:00 --> 00:23:02
But those letters are
going to be fixed
402
00:23:02 --> 00:23:06
numbers in this problem.
403
00:23:06 --> 00:23:08
And we want to solve it for
all possible a's and b's.
404
00:23:08 --> 00:23:10
Now, the interesting
thing, remember, is what
405
00:23:10 --> 00:23:12
happens when b Is not 0.
406
00:23:12 --> 00:23:16
If b = 0, we already have a
clue that the point should
407
00:23:16 --> 00:23:16
be the center point.
408
00:23:16 --> 00:23:19
It should be exactly that x,
the middle point, which I'm
409
00:23:19 --> 00:23:22
going to label in a second,
is halfway in between.
410
00:23:22 --> 00:23:25
So now the variable point that
I'm going to use is down here.
411
00:23:25 --> 00:23:30
I'm going to call
this point (x, y).
412
00:23:30 --> 00:23:31
So here's my setup.
413
00:23:31 --> 00:23:34
I've now given labels
to all the things on
414
00:23:34 --> 00:23:37
the diagram so far.
415
00:23:37 --> 00:23:42
Most of the things
on the diagram.
416
00:23:42 --> 00:23:47
So now, what else
do I have to do?
417
00:23:47 --> 00:23:54
Well, I have to explain
to you that this is a
418
00:23:54 --> 00:23:56
minimization problem.
419
00:23:56 --> 00:24:01
What happens, actually,
physically is that the weight
420
00:24:01 --> 00:24:03
settles to the lowest point.
421
00:24:03 --> 00:24:05
That's the thing that has the
lowest potential energy.
422
00:24:05 --> 00:24:09
So we're minimizing a function.
423
00:24:09 --> 00:24:13
And it's this curve here.
424
00:24:13 --> 00:24:17
The constraint is that we're
restricted to this curve.
425
00:24:17 --> 00:24:18
So this is a constraint curve.
426
00:24:18 --> 00:24:24
And we want the lowest
point of this curve.
427
00:24:24 --> 00:24:29
So now, we need a little bit
more language in order to
428
00:24:29 --> 00:24:31
describe what it is
that we've got.
429
00:24:31 --> 00:24:36
And the constraint curve, we
got it in a particular way.
430
00:24:36 --> 00:24:39
Namely, we strung some
string from here to there.
431
00:24:39 --> 00:24:42
And what happens at all these
points is that the total length
432
00:24:42 --> 00:24:45
of the string is the same.
433
00:24:45 --> 00:24:49
So one way of expressing the
constraint is that the length
434
00:24:49 --> 00:24:52
of the string is constant.
435
00:24:52 --> 00:24:54
And so in order to figure out
what the constraint is, what
436
00:24:54 --> 00:24:58
this curve is, I have to
describe that analytically.
437
00:24:58 --> 00:25:01
And I'm going to do that by
drawing in some helping lines.
438
00:25:01 --> 00:25:05
Namely, some right triangles to
figure out what this length is.
439
00:25:05 --> 00:25:07
And what the other length is.
440
00:25:07 --> 00:25:09
So this length is pretty
easy if I draw a
441
00:25:09 --> 00:25:11
right triangle here.
442
00:25:11 --> 00:25:14
Because we go over x
and we go down y.
443
00:25:14 --> 00:25:18
So this length is the
square root of x^2 + y^2.
444
00:25:19 --> 00:25:22
That's the Pythagorean theorem.
445
00:25:22 --> 00:25:26
Similarly, over here, I'm
going to get another length.
446
00:25:26 --> 00:25:28
Which is a little
bit of a mess.
447
00:25:28 --> 00:25:30
It's the vertical.
448
00:25:30 --> 00:25:35
So I'm just going to label one
half of it, so that you see.
449
00:25:35 --> 00:25:38
So this horizontal
distance is x.
450
00:25:38 --> 00:25:41
And this horizontal distance
from this top point with this
451
00:25:41 --> 00:25:44
right angle, over there.
452
00:25:44 --> 00:25:47
It starts at x and ends at a.
453
00:25:47 --> 00:25:50
The right-most point is
a in the x coordinate.
454
00:25:50 --> 00:25:56
So the whole distance is a - x.
455
00:25:56 --> 00:26:00
So that's this leg of
this right triangle.
456
00:26:00 --> 00:26:05
And, similarly, the vertical
distance will be b - y.
457
00:26:05 --> 00:26:08
And so, the formula here, which
is a little complicated for
458
00:26:08 --> 00:26:15
this length, is the square root
of (a - x)^2 + (b - y)^2.
459
00:26:17 --> 00:26:21
So here are the two formulas
that are going to allow me
460
00:26:21 --> 00:26:22
to set up my problem now.
461
00:26:22 --> 00:26:24
So, my goal is to set it
up the way I did here,
462
00:26:24 --> 00:26:26
just with formulas.
463
00:26:26 --> 00:26:40
And not with diagrams
and not with names.
464
00:26:40 --> 00:26:42
So here's what I'd like to do.
465
00:26:42 --> 00:26:46
I claim that what's
constrained, if I'm along that
466
00:26:46 --> 00:26:48
curve, is that the total
length is constant.
467
00:26:48 --> 00:26:50
So that's this statement here.
468
00:26:50 --> 00:26:55
The square root of x^2 + y^2
+ this other square root.
469
00:26:55 --> 00:27:02
These are the two
lengths of string.
470
00:27:02 --> 00:27:08
Is equal to some number,
L, which is constant.
471
00:27:08 --> 00:27:16
And this, as I said, is what
I'm calling my constraint.
472
00:27:16 --> 00:27:16
Yeah.
473
00:27:16 --> 00:27:20
STUDENT: [INAUDIBLE]
474
00:27:20 --> 00:27:24
PROFESSOR: So the question is,
shouldn't it be b plus y.
475
00:27:24 --> 00:27:28
No, and the reason is that
y is a negative number.
476
00:27:28 --> 00:27:30
It's below 0.
477
00:27:30 --> 00:27:39
So it's actually the sum,
- y is a positive number.
478
00:27:39 --> 00:27:43
Alright, so here's the formula.
479
00:27:43 --> 00:27:53
And then, we want to find
the minimum of something.
480
00:27:53 --> 00:27:56
So what is it that we're
finding the minimum of?
481
00:27:56 --> 00:27:58
This is actually the hardest
part of the problem,
482
00:27:58 --> 00:27:59
conceptually.
483
00:27:59 --> 00:28:02
I tried to prepare it, but it's
very hard to figure this out.
484
00:28:02 --> 00:28:07
We're finding the least what?
485
00:28:07 --> 00:28:10
It's y.
486
00:28:10 --> 00:28:11
We got a name for that.
487
00:28:11 --> 00:28:17
So we want to find
the lowest y.
488
00:28:17 --> 00:28:20
Now, the reason why it seems a
little weird is you might think
489
00:28:20 --> 00:28:21
of y as just being a variable.
490
00:28:21 --> 00:28:25
But really, y is a function.
491
00:28:25 --> 00:28:32
It's really y = y ( x )
is defined implicitly by
492
00:28:32 --> 00:28:34
the constraint equation.
493
00:28:34 --> 00:28:36
That's what that curve is.
494
00:28:36 --> 00:28:42
And notice the bottom point is
exactly the place where the
495
00:28:42 --> 00:28:46
tangent line will
be horizontal.
496
00:28:46 --> 00:28:48
Which is just what we want.
497
00:28:48 --> 00:29:03
So from the diagram, the
bottom point is where y' = 0.
498
00:29:03 --> 00:29:04
So this is the critical point.
499
00:29:04 --> 00:29:25
Yeah?
500
00:29:25 --> 00:29:27
STUDENT: [INAUDIBLE]
501
00:29:27 --> 00:29:28
PROFESSOR: Exactly.
502
00:29:28 --> 00:29:32
So I'm deriving for you, so the
question is, could I have just
503
00:29:32 --> 00:29:35
tried to find y' =
0 to begin with.
504
00:29:35 --> 00:29:37
The answer is yes, absolutely.
505
00:29:37 --> 00:29:39
And in fact I'm leading
in that direction.
506
00:29:39 --> 00:29:42
I'm just showing you, so I'm
trying to make the following,
507
00:29:42 --> 00:29:43
very subtle, point.
508
00:29:43 --> 00:29:49
Which is in maximum-minimum
problems, we always have to
509
00:29:49 --> 00:29:51
keep track of two things.
510
00:29:51 --> 00:29:55
Often the interesting point
is the critical point.
511
00:29:55 --> 00:29:56
And that indeed turns out
to be the case here.
512
00:29:56 --> 00:29:59
But we always have
to check the ends.
513
00:29:59 --> 00:30:02
And so there are several
ways of checking the ends.
514
00:30:02 --> 00:30:03
One is, we did this
physical problem.
515
00:30:03 --> 00:30:05
We can see that it's
coming up here.
516
00:30:05 --> 00:30:06
We can see that it's
coming up here.
517
00:30:06 --> 00:30:10
Therefore the bottom must
be at this critical point.
518
00:30:10 --> 00:30:14
So that's OK, so that's
one way of checking it.
519
00:30:14 --> 00:30:18
Another way of checking it is
the reasoning that I just gave.
520
00:30:18 --> 00:30:19
But it's really the
same reasoning.
521
00:30:19 --> 00:30:22
I'm pointing to this thing and
I'm showing you that the bottom
522
00:30:22 --> 00:30:23
is somewhere in the middle.
523
00:30:23 --> 00:30:26
So, therefore, it is a place
of horizontal tangency.
524
00:30:26 --> 00:30:28
That's the reasoning
that I'm using.
525
00:30:28 --> 00:30:32
So, again, this is to avoid
having to evaluate a limit of
526
00:30:32 --> 00:30:35
an end or to use the second
derivative test, which is a
527
00:30:35 --> 00:30:42
total catastrophe in this case.
528
00:30:42 --> 00:30:45
OK, now.
529
00:30:45 --> 00:30:48
There's one other thing that
you might know about this if
530
00:30:48 --> 00:30:51
you've seen this geometric
construction before.
531
00:30:51 --> 00:30:54
With a string and chalk.
532
00:30:54 --> 00:30:58
Which is that this
curve is an eclipse.
533
00:30:58 --> 00:30:59
It turns out, this is a
piece of an eclipse.
534
00:30:59 --> 00:31:00
It's a huge ellipse.
535
00:31:00 --> 00:31:03
These two points turn
out to be the so-called
536
00:31:03 --> 00:31:05
foci of the ellipse.
537
00:31:05 --> 00:31:09
However, that geometric fact
is totally useless for
538
00:31:09 --> 00:31:10
solving this problem.
539
00:31:10 --> 00:31:12
Completely useless.
540
00:31:12 --> 00:31:14
If you actually write out the
formulas for the ellipse,
541
00:31:14 --> 00:31:17
you'll discover that you have
a much harder problem
542
00:31:17 --> 00:31:18
on your hands.
543
00:31:18 --> 00:31:22
And it will take you twice or
ten times as long, so, it's
544
00:31:22 --> 00:31:25
true that it's an ellipse,
but it doesn't help.
545
00:31:25 --> 00:31:29
OK, so what we're going to
do instead is much simpler.
546
00:31:29 --> 00:31:32
We're going to leave this
expression along and
547
00:31:32 --> 00:31:35
we're just going to
differentiate implicitly.
548
00:31:35 --> 00:31:42
So again, we use implicit
differentiation on the
549
00:31:42 --> 00:31:49
constraint equation.
550
00:31:49 --> 00:31:51
So that's the equation
which is directly above
551
00:31:51 --> 00:31:54
me there, at the top.
552
00:31:54 --> 00:31:57
And I have to differentiate
it with respect to x.
553
00:31:57 --> 00:31:59
So that's a little ugly,
but we've done this
554
00:31:59 --> 00:32:01
a few times before.
555
00:32:01 --> 00:32:03
When you differentiate a square
root, the square root goes
556
00:32:03 --> 00:32:08
into the denominator.
557
00:32:08 --> 00:32:11
And there's a factor of 1/2, so
there's a 2x which cancels.
558
00:32:11 --> 00:32:12
So I claim it's this.
559
00:32:12 --> 00:32:17
Now, because y depends on x,
there's also a y y' here.
560
00:32:17 --> 00:32:21
So technically speaking, it's
twice this with a half.
561
00:32:21 --> 00:32:24
2 / 2 times that.
562
00:32:24 --> 00:32:30
So that's the differentiation
of the first piece of this guy.
563
00:32:30 --> 00:32:32
Now I'm going to do this
second one, and it's
564
00:32:32 --> 00:32:34
also the chain rule.
565
00:32:34 --> 00:32:39
But you're just going to have
to let me do it for you.
566
00:32:39 --> 00:32:41
Because it's just a little
bit too long for you
567
00:32:41 --> 00:32:43
to pay attention to.
568
00:32:43 --> 00:32:46
It turns out there's a minus
sign that comes out, because
569
00:32:46 --> 00:32:49
there's a - x and a - y there.
570
00:32:49 --> 00:32:52
And then the numerator, the
denominator is the same
571
00:32:52 --> 00:32:55
massive square root.
572
00:32:55 --> 00:32:58
So it's (a - x)^2 ( b - y)^2.
573
00:32:59 --> 00:33:06
And the numerator is a -
x, which is what replaces
574
00:33:06 --> 00:33:07
the x over here.
575
00:33:07 --> 00:33:11
And then another term,
which is (b - y)y'.
576
00:33:13 --> 00:33:15
I claim that that's
analogous to what I
577
00:33:15 --> 00:33:17
did in the first term.
578
00:33:17 --> 00:33:20
And you'll just have to
check this on your own.
579
00:33:20 --> 00:33:21
Because I did it too
fast for you to be able
580
00:33:21 --> 00:33:23
to check yourself.
581
00:33:23 --> 00:33:26
Now, that's going to be equal
to what on the right-hand side.
582
00:33:26 --> 00:33:29
What's the derivative of
L with respect to x?
583
00:33:29 --> 00:33:30
It's 0.
584
00:33:30 --> 00:33:32
It's not changing
in the problem.
585
00:33:32 --> 00:33:37
Although my parachute
stuff stretches.
586
00:33:37 --> 00:33:40
We tried to stretch it
to its fullest extent.
587
00:33:40 --> 00:33:45
So that we kept it fixed,
that was the goal here.
588
00:33:45 --> 00:33:50
So now, this looks
like a total mess.
589
00:33:50 --> 00:33:52
But, it's not.
590
00:33:52 --> 00:33:54
And let me show you why.
591
00:33:54 --> 00:33:57
It simplifies a great deal.
592
00:33:57 --> 00:34:01
And let me show
you exactly how.
593
00:34:01 --> 00:34:04
So, first of all, the whole
point is we're looking for
594
00:34:04 --> 00:34:06
the place where y' = 0.
595
00:34:06 --> 00:34:14
So that means that these
terms go away. y' = 0.
596
00:34:14 --> 00:34:16
So they're gone.
597
00:34:16 --> 00:34:21
And now what we have is
the following equation.
598
00:34:21 --> 00:34:27
It's x / square root of x^2 +
y^2 squared is equal to, if I
599
00:34:27 --> 00:34:30
put it on the other side the
minus sign is changed to a
600
00:34:30 --> 00:34:36
plus sign. a - x divided by
this other massive object,
601
00:34:36 --> 00:34:43
(a - x)^2 + (b - y)^2.
602
00:34:43 --> 00:34:45
So this is what it's
simplifies to.
603
00:34:45 --> 00:34:51
Now again, that also might
look complicated to you.
604
00:34:51 --> 00:34:57
But I claim that this is
something, this is a kind of
605
00:34:57 --> 00:35:01
equilibrium equation that can
be interpreted geometrically,
606
00:35:01 --> 00:35:05
in a way that is very
meaningful and important.
607
00:35:05 --> 00:35:09
So first of all, let me observe
for you that this x is
608
00:35:09 --> 00:35:10
something on our picture.
609
00:35:10 --> 00:35:13
And the square root of
x^2 + y^2 is something
610
00:35:13 --> 00:35:14
on our picture.
611
00:35:14 --> 00:35:19
Namely, if I go over to the
picture, here was x and
612
00:35:19 --> 00:35:20
this was a right triangle.
613
00:35:20 --> 00:35:25
And this hypotenuse was
square root of x^2 + y^2.
614
00:35:25 --> 00:35:34
So, if I call this angle alpha,
then this is the sine of alpha.
615
00:35:34 --> 00:35:34
Right?
616
00:35:34 --> 00:35:36
It's a right triangle,
that's the opposite leg.
617
00:35:36 --> 00:35:44
So this guy is the
sine of alpha.
618
00:35:44 --> 00:35:48
Similarly, the other side has
an interpretation for the
619
00:35:48 --> 00:35:50
other right triangle.
620
00:35:50 --> 00:35:56
If this angle is beta, then the
opposite side is a - x, and
621
00:35:56 --> 00:36:00
the hypotenuse is what was in
the denominator over there.
622
00:36:00 --> 00:36:10
So this side is sine of beta.
623
00:36:10 --> 00:36:14
And so what this condition
is telling us is
624
00:36:14 --> 00:36:19
that alpha = beta.
625
00:36:19 --> 00:36:22
Which is the hidden
symmetry in the problem.
626
00:36:22 --> 00:36:25
I don't know if you can
actually see it when I
627
00:36:25 --> 00:36:28
show you this thing.
628
00:36:28 --> 00:36:33
But, no matter how I tilt it,
actually the two angles from
629
00:36:33 --> 00:36:39
the horizontal are the same.
630
00:36:39 --> 00:36:41
In the middle it's kind
of obvious that that
631
00:36:41 --> 00:36:42
should be the case.
632
00:36:42 --> 00:36:44
But on the sides it's
not obvious that that's
633
00:36:44 --> 00:36:46
what's happening.
634
00:36:46 --> 00:36:49
Now, this has even, so
that's a symmetry, if you
635
00:36:49 --> 00:36:50
like, of the situation.
636
00:36:50 --> 00:36:52
These two angles are equal.
637
00:36:52 --> 00:36:55
But there's something
more to be said.
638
00:36:55 --> 00:36:58
If you do a forced diagram for
this, what you'll discover is
639
00:36:58 --> 00:37:03
that the tension on the
two lines is the same.
640
00:37:03 --> 00:37:06
Which means that when you build
something which is hanging like
641
00:37:06 --> 00:37:11
this, it will involve
the least stress.
642
00:37:11 --> 00:37:14
If you hang something very
heavy, and one side carries
643
00:37:14 --> 00:37:16
twice as much load as the
other, then you have twice
644
00:37:16 --> 00:37:19
as much chance of its
falling and breaking.
645
00:37:19 --> 00:37:23
If they each hold the same
strength, then you've
646
00:37:23 --> 00:37:26
distributed the load in a
much more balanced way.
647
00:37:26 --> 00:37:29
So this is a kind of a balance
condition, and it's very
648
00:37:29 --> 00:37:31
typical of minimization
problems.
649
00:37:31 --> 00:37:34
And fortunately, there are nice
solutions which distribute
650
00:37:34 --> 00:37:37
the weight reasonably well.
651
00:37:37 --> 00:37:41
That's certainly the principal
of suspension bridges.
652
00:37:41 --> 00:37:42
So yeah, one more question.
653
00:37:42 --> 00:37:44
STUDENT: [INAUDIBLE]
654
00:37:44 --> 00:37:49
PROFESSOR: OK, so the question
where it hangs, that is, what
655
00:37:49 --> 00:37:53
the formula for x is and
what the formula for y is.
656
00:37:53 --> 00:37:56
And other things, like the
equation for the ellipse and
657
00:37:56 --> 00:37:58
lots of other stuff like that.
658
00:37:58 --> 00:38:02
Those are things that I will
answer for you in a set of
659
00:38:02 --> 00:38:03
notes which I will hand out.
660
00:38:03 --> 00:38:06
And they're just a mess.
661
00:38:06 --> 00:38:08
You see they're, just as in
that other problem that we
662
00:38:08 --> 00:38:12
did last time, there's some
illuminating things you can say
663
00:38:12 --> 00:38:15
about the problem and then
there's some messy formulas.
664
00:38:15 --> 00:38:19
You know, you want to try
to pick out the simple
665
00:38:19 --> 00:38:19
things that you can say.
666
00:38:19 --> 00:38:23
In fact, that's a property of
math, you want to focus on the
667
00:38:23 --> 00:38:24
more comprehensible things.
668
00:38:24 --> 00:38:26
On the other hand,
it can be done.
669
00:38:26 --> 00:38:30
It just takes a little
bit of computation.
670
00:38:30 --> 00:38:33
So I didn't answer the question
of what the lowest y was.
671
00:38:33 --> 00:38:39
But I'll do that for you.
672
00:38:39 --> 00:38:42
Maybe I'll just mention one
more thing about this problem.
673
00:38:42 --> 00:38:44
I'm using problems from
a completely different
674
00:38:44 --> 00:38:45
point of view.
675
00:38:45 --> 00:38:50
If you sort of roll the
ellipse around, you get
676
00:38:50 --> 00:38:55
the same phenomenon
from each place here.
677
00:38:55 --> 00:38:58
So it doesn't matter
where a and 0 are.
678
00:38:58 --> 00:38:59
You'll get the same phenomenon.
679
00:38:59 --> 00:39:02
That is, the tangent line.
680
00:39:02 --> 00:39:06
So, this angle and that
angle will be equal.
681
00:39:06 --> 00:39:10
So you can also read that as
being the angle over here and
682
00:39:10 --> 00:39:11
the angle over here are equal.
683
00:39:11 --> 00:39:14
That is, beta, that
is the complementary
684
00:39:14 --> 00:39:17
angles are also equal.
685
00:39:17 --> 00:39:21
And if you interpret this as a
ray of light, and this as a
686
00:39:21 --> 00:39:24
mirror, then this would be
saying that if you start at one
687
00:39:24 --> 00:39:27
focus, every ray of light will
bounce and go to
688
00:39:27 --> 00:39:29
the other focus.
689
00:39:29 --> 00:39:32
So that's a property
that an ellipse has.
690
00:39:32 --> 00:39:36
More precisely, a property
that this kind of curve has.
691
00:39:36 --> 00:39:41
And in fact, a few years ago
there was this great piece of
692
00:39:41 --> 00:39:45
art at something called the
DeCordova Museum, which I
693
00:39:45 --> 00:39:50
recommended very highly to you
go sometime to visit in
694
00:39:50 --> 00:39:52
your four years here.
695
00:39:52 --> 00:39:59
There was a collection of
miniature golf holes.
696
00:39:59 --> 00:40:02
So they had a bunch of
mini golf pieces of art.
697
00:40:02 --> 00:40:04
And every one was completely
different from the other.
698
00:40:04 --> 00:40:07
And one of them was
called hole in one.
699
00:40:07 --> 00:40:11
And the tee was at one
focus of the ellipse.
700
00:40:11 --> 00:40:15
And the hole was at the
other focus of the ellipse.
701
00:40:15 --> 00:40:19
So, no matter how you hit
the golf ball, it always
702
00:40:19 --> 00:40:21
goes into the hole.
703
00:40:21 --> 00:40:24
No matter where it bounces,
it just, one bounce
704
00:40:24 --> 00:40:26
and it's in the hole.
705
00:40:26 --> 00:40:30
So that's actually a
consequence of the computation
706
00:40:30 --> 00:40:40
that we just did.
707
00:40:40 --> 00:40:42
Time to go on.
708
00:40:42 --> 00:41:06
We're going to now talk
about something else.
709
00:41:06 --> 00:41:10
So our next topic is
Newton's method.
710
00:41:10 --> 00:41:24
Which is one of the greatest
applications of calculus.
711
00:41:24 --> 00:41:27
And I'm going to describe
it here for you.
712
00:41:27 --> 00:41:36
And we'll illustrate it on an
example, which is solving
713
00:41:36 --> 00:41:39
the equation x^2 = 5.
714
00:41:39 --> 00:41:42
We're going to find
the square root of 5.
715
00:41:42 --> 00:41:45
Now, you can actually solve
any equation this way.
716
00:41:45 --> 00:41:47
Any equation that you
understand, you can solve
717
00:41:47 --> 00:41:49
this way, essentially.
718
00:41:49 --> 00:41:52
So even though I'm doing it for
the square root of 5, which is
719
00:41:52 --> 00:41:57
something that you can figure
out on your calculator, in fact
720
00:41:57 --> 00:42:00
this is really at the heart
of many of the ways in
721
00:42:00 --> 00:42:02
which calculators work.
722
00:42:02 --> 00:42:04
So, the first thing is
to make this problem a
723
00:42:04 --> 00:42:06
little bit more abstract.
724
00:42:06 --> 00:42:13
We're going to set
f ( x ) = x^2 - 5.
725
00:42:13 --> 00:42:19
And then we're going
to solve f (x) = 0.
726
00:42:19 --> 00:42:24
So this is the sort of standard
form for solving such a.
727
00:42:24 --> 00:42:27
So you take some either
complicated or simple function
728
00:42:27 --> 00:42:30
of x, linear functions are
boring, quadratic functions
729
00:42:30 --> 00:42:32
are already interesting.
730
00:42:32 --> 00:42:36
And cubic functions, as I've
said a few times, you don't
731
00:42:36 --> 00:42:37
even have formulas for solving.
732
00:42:37 --> 00:42:39
So this would be the only
method you have for
733
00:42:39 --> 00:42:43
solving them numerically.
734
00:42:43 --> 00:42:45
So here's how it works.
735
00:42:45 --> 00:42:50
So the idea, I'll
plot this function.
736
00:42:50 --> 00:42:55
Here's the function, it's
a parabola, y = x^2 - 5.
737
00:42:55 --> 00:43:01
It dips below 0, sorry, it
should be centered, but anyway.
738
00:43:01 --> 00:43:06
And now the idea here is
to start with a guess.
739
00:43:06 --> 00:43:10
And square root of 5 is
pretty close to the square
740
00:43:10 --> 00:43:12
root of 4, which is 2.
741
00:43:12 --> 00:43:17
So my first guess is
going to be 2, here.
742
00:43:17 --> 00:43:30
So start with initial guess.
743
00:43:30 --> 00:43:32
So that's our first guess.
744
00:43:32 --> 00:43:37
And now, what we're going to
do is we're going to pretend
745
00:43:37 --> 00:43:40
that the function is linear.
746
00:43:40 --> 00:43:41
That's all we're going to do.
747
00:43:41 --> 00:43:45
And then if the function were
linear, we're going to try
748
00:43:45 --> 00:43:47
to find where the 0 is.
749
00:43:47 --> 00:43:50
So if the function is linear,
what we'll use is we'll plot
750
00:43:50 --> 00:43:58
the point where 2 is on, that
is, the point (2, f ( 2 )),
751
00:43:58 --> 00:44:06
and then we're going to
draw the tangent line here.
752
00:44:06 --> 00:44:15
And this is going to be
our new guess. x = x,
753
00:44:15 --> 00:44:22
which I'll call x1.
754
00:44:22 --> 00:44:26
So the idea here is that this
point may be somewhat far from
755
00:44:26 --> 00:44:31
where it crosses, but this
point will be a little closer.
756
00:44:31 --> 00:44:33
And now we're going to do
this over and over again.
757
00:44:33 --> 00:44:38
And see how fast it gets to
the place we're aiming for.
758
00:44:38 --> 00:44:42
So we have to work out
what the formulas are.
759
00:44:42 --> 00:44:53
And that's the strategy.
760
00:44:53 --> 00:45:01
So now, the first step here
is, we have our guess, we
761
00:45:01 --> 00:45:06
have our tangent line.
762
00:45:06 --> 00:45:13
Which has the formula
y - y0 = m(x - x0).
763
00:45:13 --> 00:45:16
So that's the general
form for a tangent line.
764
00:45:16 --> 00:45:20
And now, I have to
tell you what x1 is.
765
00:45:20 --> 00:45:29
In terms of this tangent
line. x1 is the x intercept.
766
00:45:29 --> 00:45:32
The tangent line, if you look
over here at the diagram,
767
00:45:32 --> 00:45:36
the tangent line is
the orange line.
768
00:45:36 --> 00:45:41
Where that crosses the axis,
that's where I want to put x1.
769
00:45:41 --> 00:45:43
So that's the x intercept.
770
00:45:43 --> 00:45:46
Now, how do you find
the x intercept?
771
00:45:46 --> 00:45:49
You find it by setting y = 0.
772
00:45:49 --> 00:45:52
That horizontal line is y = 0.
773
00:45:52 --> 00:45:59
So I set y = 0, and I get
0 - y0 = m (x1 - x0).
774
00:45:59 --> 00:46:02
So I changed two things
in this equation.
775
00:46:02 --> 00:46:04
I plugged in 0 here, for y.
776
00:46:04 --> 00:46:07
And I said that the place
where that happens is
777
00:46:07 --> 00:46:13
going to be where x is x1.
778
00:46:13 --> 00:46:15
So now let's solve.
779
00:46:15 --> 00:46:19
And what we get here is
- y divided by - sorry,
780
00:46:19 --> 00:46:26
-y0 / m = x1 - x0.
781
00:46:26 --> 00:46:30
And now I can get
a formula for x1.
782
00:46:30 --> 00:46:46
So x1 = x0 - (y0 / m).
783
00:46:46 --> 00:46:53
I now need to tell you what
this formula means, in
784
00:46:53 --> 00:46:56
terms of the function f.
785
00:46:56 --> 00:46:59
So first of all, x0 is
x0, whatever x0 is.
786
00:46:59 --> 00:47:06
And y0, I claim, is f (x0).
787
00:47:06 --> 00:47:11
And m is the slope
at that same place.
788
00:47:11 --> 00:47:17
So it's f' (x0).
789
00:47:17 --> 00:47:19
And this is the whole story.
790
00:47:19 --> 00:47:25
This is the formula which
will enable us to calculate
791
00:47:25 --> 00:47:34
basically any root.
792
00:47:34 --> 00:47:36
I'm going to repeat this
formula, so I'm going to tell
793
00:47:36 --> 00:47:40
you again what Newton's method
is, and put a little more
794
00:47:40 --> 00:47:43
colorful box around it.
795
00:47:43 --> 00:47:53
So Newton's method involves
the following formula.
796
00:47:53 --> 00:47:58
In order to get the n + 1st
point, that's our better and
797
00:47:58 --> 00:48:03
better guess, I'm going to take
the nth one and then I'm going
798
00:48:03 --> 00:48:04
to plug in this formula.
799
00:48:04 --> 00:48:09
So f ( xn ) / f' ( xn ).
800
00:48:09 --> 00:48:14
So this is the basic formula,
and if you like, this is the
801
00:48:14 --> 00:48:18
idea of just repeating
what I had before.
802
00:48:18 --> 00:48:22
Now, we've gone from geometry,
from just pictures, to an
803
00:48:22 --> 00:48:24
honest to goodness formula
which is completely
804
00:48:24 --> 00:48:29
implementable and very easy to
implement in any case.
805
00:48:29 --> 00:48:33
So let's see how it works in
the case that we've got.
806
00:48:33 --> 00:48:41
Which is x0 = 2.
f(x) = x^2 - 5.
807
00:48:41 --> 00:48:44
Let's see how to
implement this formula.
808
00:48:44 --> 00:48:49
So first of all, I have to
calculate for you, f' (x).
809
00:48:49 --> 00:48:55
That's 2x.
810
00:48:55 --> 00:49:05
And so, x1 = x0 -, so
f' , sorry, f ( x )
811
00:49:05 --> 00:49:08
would be x0^2 - 5.
812
00:49:08 --> 00:49:10
That's what's in the numerator.
813
00:49:10 --> 00:49:18
And in the denominator I have
the derivative, so that's 2 x0.
814
00:49:18 --> 00:49:23
And so all told, well, let's
work it out in two steps here.
815
00:49:23 --> 00:49:29
This is - 1/2 x0 for the first
term, and then + (5/2 /
816
00:49:29 --> 00:49:31
x0) for the second term.
817
00:49:31 --> 00:49:40
And these two terms combine, so
we have here 1/2 x0 + 5/2 with
818
00:49:40 --> 00:49:43
an x0 in the denominator.
819
00:49:43 --> 00:49:53
So here's the formula
for x1. in this case.
820
00:49:53 --> 00:50:03
Now I'd like to show you
how well this works.
821
00:50:03 --> 00:50:09
So first of all, we have x1
= 1/2 ( 2), if I plug in
822
00:50:09 --> 00:50:16
x1, + 5/4, which is 9/4.
823
00:50:16 --> 00:50:28
And x2, I have 1/2 (
9/4) + 5/2 ( 4/9).
824
00:50:28 --> 00:50:30
That's the next one.
825
00:50:30 --> 00:50:37
And if you work this
out, it's 161/72.
826
00:50:37 --> 00:50:41
And then x3 is kind of long.
827
00:50:41 --> 00:50:43
But I will just write down what
it is, so that you can see it's
828
00:50:43 --> 00:50:50
1/2 ( 161/72) + 5/2 and then
I do the reciprocal of that.
829
00:50:50 --> 00:50:56
So 72/161.
830
00:50:56 --> 00:51:00
So let's see how
good these are.
831
00:51:00 --> 00:51:05
I carefully calculated
how far off they are.
832
00:51:05 --> 00:51:07
Somewhere on my notes.
833
00:51:07 --> 00:51:11
So I'll just take a look
and see what I said.
834
00:51:11 --> 00:51:13
Oh yeah, I did do it.
835
00:51:13 --> 00:51:20
So, what's the square root of 5
minus - so here's n, here's the
836
00:51:20 --> 00:51:24
square root of 5 -
xn, if you like.
837
00:51:24 --> 00:51:26
Or the other way around.
838
00:51:26 --> 00:51:28
The size of this.
839
00:51:28 --> 00:51:32
You'll have to decide on your
homework whether it comes out
840
00:51:32 --> 00:51:34
positive or negative, to the
right or to the left
841
00:51:34 --> 00:51:35
of the answer.
842
00:51:35 --> 00:51:37
But let's do this.
843
00:51:37 --> 00:51:41
So when n = 0, the guess was 2.
844
00:51:41 --> 00:51:45
And we're off by
about 2 ( 10 ^ - 1).
845
00:51:45 --> 00:51:49
And if n = 1, so that's
this 9/4, that's off
846
00:51:49 --> 00:51:53
by about 10 ^ - 2.
847
00:51:53 --> 00:51:58
And then n = 2, that's
this number here, right?
848
00:51:58 --> 00:52:03
And that's off by
about 4 ( 10 ^ - 5).
849
00:52:03 --> 00:52:06
That's already as good an
approximation to the square
850
00:52:06 --> 00:52:10
root of 5 as you'll
ever need in your life.
851
00:52:10 --> 00:52:17
If you do 3, this number here
turns out to be accurate
852
00:52:17 --> 00:52:20
to 10 ^ - 10 or so.
853
00:52:20 --> 00:52:23
This goes right off to the
edge of my calculator,
854
00:52:23 --> 00:52:23
this one here.
855
00:52:23 --> 00:52:29
So already, with the third
iterate. you're of way past
856
00:52:29 --> 00:52:31
the accuracy that you
need for most things.
857
00:52:31 --> 00:52:32
Yep, question.
858
00:52:32 --> 00:52:32
STUDENT: [INAUDIBLE]
859
00:52:32 --> 00:52:38
PROFESSOR: How come
the x0 disappears?
860
00:52:38 --> 00:52:39
STUDENT: [INAUDIBLE]
861
00:52:39 --> 00:52:56
PROFESSOR: So, from here
to here [INAUDIBLE]
862
00:52:56 --> 00:52:56
STUDENT: [INAUDIBLE]
863
00:52:56 --> 00:52:58
PROFESSOR: Hang on, folks,
we have a question.
864
00:52:58 --> 00:52:59
Let's just answer it.
865
00:52:59 --> 00:53:03
So here we have an x0, and here
we have - 1/2, there's an
866
00:53:03 --> 00:53:08
x0^2 and an x which cancel.
867
00:53:08 --> 00:53:11
And here we have a
minus, and a - 5/2 x0.
868
00:53:11 --> 00:53:16
So I combine the 1 - 1/2,
I got + 1/2, that's all.
869
00:53:16 --> 00:53:17
OK?
870
00:53:17 --> 00:53:18
Alright, thanks.
871
00:53:18 --> 00:53:21
We'll have to ask other
questions after class.
872
00:53:21 --> 00:53:22