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PROFESSOR: Hi.
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Well, I hope you're ready
for second derivatives.
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We don't go higher than that
in many problems, but the
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second derivative is
an important--
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the derivative of the derivative
is an important
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thing to know, especially in
problems with maximum and
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minimum, which is the big
application of derivatives, to
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locate a maximum or a minimum,
and to decide which one it is.
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And I can tell you right away,
locating a maximum, minimum,
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is the first derivative's job.
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The first derivative is 0.
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If I have a maximum or a
minimum, and we'll have
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pictures, somewhere in the
middle of my function I'll
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recognize by derivative
equals 0.
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Slope equals 0, that the
function is leveling off,
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either bending down or bending
up, maximum or minimum.
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OK, and it's the second
derivative that
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tells me which it is.
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The second derivative tells me
the bending of the graph.
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OK, so we now will have
three generations.
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The big picture of calculus
started with two functions:
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the distance and the speed.
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And we discussed in detail the
connection between them.
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How to recover the speed if we
know the distance, take the
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derivative.
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Now comes the derivative of
the speed, which in that
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language, in the
distance-speed-time language,
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the second derivative is the
acceleration, the rate at
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which your speed is changing,
the rate at which you're
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speeding up or slowing down.
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And this is the way I
would write that.
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If the speed is the
first derivative--
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df dt--
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this is the way you write the
second derivative, and you say
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d second f dt squared.
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d second f dt squared.
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OK, so that's you could say the
physics example: distance,
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speed, acceleration.
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And I say physics because, of
course, acceleration is the a
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in Newton's Law f equals ma.
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For a graph, like these graphs
here, I won't especially use
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those physics words.
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I'll use graph words.
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So I would say function
one would be the
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height of the graph.
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And in this case, that height is
y equals x squared, so it's
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a simple parabola.
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Here would be the slope.
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I would use the word "slope"
for the second function.
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And the slope of y equals x
squared we know is 2x, so we
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see the slope increasing.
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And you see on this picture
the slope is increasing.
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As x increases, I'm going
up more steeply.
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Now, it's the second
derivative.
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And what shall I call that?
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Bending.
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Bending is the natural
word for the second
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derivative on a graph.
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And what do I--
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the derivative of 2x is 2, a
constant, a positive constant,
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and that positive constant tells
me that the slope is
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going upwards and that the
curve is bending upwards.
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So in this simple case,
we connect these three
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descriptions of our function.
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It's positive.
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It's slope is positive.
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And its second derivative--
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bending--
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is positive.
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And that gives us a function
that goes like that.
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Now, let me go to a different
function.
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Let me take a second example
now, an example where not
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everything is positive.
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But let's make it familiar.
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Take sine x.
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So sine x starts
out like that.
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So this is a graph of sine x up
to 90 degrees, pi over 2,
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so that's y equals sine x.
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OK, what do you think
about its slope?
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We know the derivative of sine
x, but before we write it
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down, look at the graph.
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The slope is positive, right?
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But the slope actually
starts out at 1.
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Better make it look a little
more realistic.
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That's a slope of 1 there.
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So the slope starts at 1 and the
slope drops to a slope of
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0 up there.
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So a slope of 1.
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I see here is a 1.
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Here I'm graphing y prime.
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dy dx I sometimes write as y
prime, just because it's
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shorter, and particularly, it'll
be shorter for a second
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derivative.
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So y prime, we know the
derivative of sine x is cos x,
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which is pretty neat actually,
that we start with a familiar
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function, and then we get its
twin, its other half.
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And the cosine is the slope of
the sine curve, and it starts
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at 1, a slope of 1, and it comes
down to 0, as we know
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the cosine does.
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So that's a graph
of the cosine.
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And now, of course, we have
three generations.
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I'm going to graph
y double prime.
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Let me put it up here. y double
prime, the second
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derivative, the derivative
of the cosine of x
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is minus sine x.
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OK, let's just--
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from the picture, what
am I seeing here?
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I'm seeing a slope of 0.
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I'm taking now the slope
of the slope.
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So here it starts at 0.
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The slope is downwards, so the
second derivative is going to
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be negative.
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Oh, and it is negative,
minus sign x.
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So the slope starts at 0 and
ends at minus 1 because that
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now comes down at a
negative slope.
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The slope is negative.
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I'm going downhill, and that's
a graph of the second
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derivative.
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And which way is our
function bending?
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It's bending down.
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As I go along, the slope
is dropping.
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And I see that in
the slope curve.
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It's falling.
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And I see it in the
bending curve
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because I'm below 0 here.
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This is bending down, where
that one was bending up.
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I could introduce the word
convex for something that
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bends upwards, and bending down,
I could introduce the
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word concave. But those
are just words.
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The graphs are telling us much
more than the words do.
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OK, so do you see that picture
bending down, but going up?
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So the slope is positive here,
but the second derivative, the
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slope is dropping.
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So the second derivative-- and
you have to pay attention to
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keep them straight.
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The second derivative is telling
us that the original
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one is bending down.
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OK, let me continue these graphs
just a little beyond 90
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degrees, pi over 2, because
you'll see something
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interesting.
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So what happens in the next
part of the graph?
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So this is going--
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the sine curve, of course,
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continues on its way downwards.
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So the slope is going negative,
as I know the cosine
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curve will do, as the cosine
curve will come like that.
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The slope down to minus
1, the slope--
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do you see here?
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The slope is negative, so on
this slope graph, I'm below 0.
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And the slope is 0.
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Let me put a little mark at
these points here, at these
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three points.
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Those are important points.
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In fact, that is a maximum,
of course.
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The sine curve hits
its maximum at 1.
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At that point when it hits its
maximum, what's its slope?
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When you hit a maximum, you're
not going up anymore.
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You haven't started down.
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The slope is 0 right there.
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What's the second derivative?
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What's the bending
at a maximum?
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The bending tells you that the
slope is going down, so the
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bending is negative.
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The bending is negative
at a maximum.
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Good.
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OK, now I'm going to continue
this sine curve for another 90
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degrees, the cosine curve, and
I'll continue the bending
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curve, so I have minus sine
x, which will go back up.
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OK, now what?
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Now what?
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180
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And then, of course, it
would continue along.
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OK, there's something
interesting happening at 180
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degrees, at pi.
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Can I identify that point?
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So there's 180 degrees.
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Something's happening there.
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I don't see--
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I don't quite know how to say
what yet, but something's
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happening there.
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It's got to show up here, and
it has to show up here.
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So whatever is happening is
showing up by a point where y
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double prime, the second
derivative, is 0.
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That's my new little
observation, not as big a deal
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as maximum or minimum.
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This was a max here.
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And we identified it as a max
because the second derivative
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was negative.
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Now I'm interested
in this point.
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Can you see what's happening
at this point as far as
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bending goes?
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This curve is bending down.
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But when I continue, the
bending changes to up.
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This is a point where
the bending changes.
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The second derivative changes
sign, and we see it here.
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Up to this square point,
the bending is below 0.
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The bending is downwards
as I come to here.
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But then there's something
rather special that--
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you see, can I try to
blow that point up?
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Here the bending is down, and
there it turns to up, and
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right in there with the--
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this is called--
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so this is my final word
to introduce--
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inflection point.
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Don't ask me why.
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An inflection point is a
point where the second
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derivative is 0.
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And what does that mean?
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That means at that moment, it
stopped bending down, and it's
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going to start bending up.
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The second derivative is
passing through 0.
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The sign of bending
is changing.
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It's changing from concave
here to convex there.
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That's a significant
point on the graph.
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Not as big a thing as
the max or the min
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that we had over there.
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So let me draw one more example
and identify all these
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different points.
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OK, so here we go.
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I drew it ahead of time because
it's got a few loops,
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and I wanted to get
it in good form.
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OK, here it is.
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This is my function: x cubed
minus x squared.
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Well, before I look at the
picture, what would be the
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first calculus thing I do?
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I take the derivative.
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y prime is the derivative of
x cubed, is three x squared
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minus the derivative of x
squared, which is 2x.
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And now today, I take the
derivative of that.
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I take the second derivative,
y double prime.
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So the second derivative is
the derivative of this.
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x squared is going to give me
2x, and I have a 3, so it's
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all together 6x.
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And minus 2x, the slope of
that is minus 2, right?
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Cubic, quadratic, linear, and
if I cared about y triple
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prime, which I don't,
constant.
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And then the fourth derivatives
and all the rest
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would be 0 for this case.
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OK, now somehow, those
derivatives, those formulas
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for y, y prime, y double prime
should tell me details about
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this graph.
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And the first thing I'm
interested in and the most
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important thing is
max and min.
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So let me set y prime to be--
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which is 3x squared minus 2x.
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I'll set it to be 0 because I
want to look for max, or min.
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And I look for both at the same
time by setting y prime
258
00:16:23,840 --> 00:16:27,950
equals 0, and then I find out
which I've got by looking at y
259
00:16:27,950 --> 00:16:28,760
double prime.
260
00:16:28,760 --> 00:16:31,740
So let me set y prime to be 0.
261
00:16:31,740 --> 00:16:33,150
What are the solutions?
262
00:16:33,150 --> 00:16:39,530
Where are the points on the
curve where it's stationary?
263
00:16:39,530 --> 00:16:44,600
It's not climbing and
it's not dropping?
264
00:16:44,600 --> 00:16:48,020
Well, I see them on
the curve here.
265
00:16:48,020 --> 00:16:53,240
That is a point where
the slope is 0.
266
00:16:53,240 --> 00:16:55,560
And I see one down here.
267
00:16:55,560 --> 00:17:01,220
There is a point where the slope
is 0, but I can find
268
00:17:01,220 --> 00:17:02,580
them with algebra.
269
00:17:02,580 --> 00:17:08,670
I solve 3x squared equals
to 2x, and I see
270
00:17:08,670 --> 00:17:10,359
it's a quadratic equation.
271
00:17:10,359 --> 00:17:12,130
I expect to find two roots.
272
00:17:12,130 --> 00:17:17,480
One of them is x equals 0, and
the other one is what?
273
00:17:17,480 --> 00:17:24,990
If I cancel those x's to find
a non-zero, canceling those
274
00:17:24,990 --> 00:17:31,340
x's leaves me with 3x equals
2 or x equals 2/3.
275
00:17:31,340 --> 00:17:35,570
Yeah, and that's what
our graph shows.
276
00:17:35,570 --> 00:17:43,480
OK, now we can see on the
graph which is a max
277
00:17:43,480 --> 00:17:45,750
and which is a min.
278
00:17:45,750 --> 00:17:48,030
And by the way, let
me just notice, of
279
00:17:48,030 --> 00:17:49,475
course, this is the max.
280
00:17:49,475 --> 00:17:52,530
281
00:17:52,530 --> 00:17:55,790
But let me just notice that
it's what I would
282
00:17:55,790 --> 00:17:59,150
call a local maximum.
283
00:17:59,150 --> 00:18:01,870
It's not the absolute top of
the function because the
284
00:18:01,870 --> 00:18:05,330
function later on is climbing
off to infinity.
285
00:18:05,330 --> 00:18:12,610
This would be way a maximum
in its neighborhood, so a
286
00:18:12,610 --> 00:18:16,910
maximum, and it's only
a local max.
287
00:18:16,910 --> 00:18:22,890
And what do I expect to see at
a maximum at x equals 0?
288
00:18:22,890 --> 00:18:30,220
I expect to see the slope 0 at
x equals 0, which it is.
289
00:18:30,220 --> 00:18:31,510
Check.
290
00:18:31,510 --> 00:18:37,420
And at a maximum, I need to know
the second derivative.
291
00:18:37,420 --> 00:18:39,460
OK, here's my formula.
292
00:18:39,460 --> 00:18:46,720
At x equals 0, I see y double
prime if x is 0 is minus 2.
293
00:18:46,720 --> 00:18:47,970
Good.
294
00:18:47,970 --> 00:18:49,540
295
00:18:49,540 --> 00:18:55,990
Negative second derivative tells
me I'm bending down, as
296
00:18:55,990 --> 00:19:01,850
the graph confirms, and the
place where the slope is 0 is
297
00:19:01,850 --> 00:19:04,630
a maximum and not a minimum.
298
00:19:04,630 --> 00:19:06,990
What about the other one?
299
00:19:06,990 --> 00:19:08,460
What about at x equals 2/3?
300
00:19:08,460 --> 00:19:12,320
301
00:19:12,320 --> 00:19:16,570
At that point, y double prime,
looking at my formula here for
302
00:19:16,570 --> 00:19:20,730
y double prime, is what?
303
00:19:20,730 --> 00:19:24,980
6 times 2/3 is 4 minus
2 is plus 2.
304
00:19:24,980 --> 00:19:26,680
4 minus 2 is 2.
305
00:19:26,680 --> 00:19:28,590
So this will be--
306
00:19:28,590 --> 00:19:33,120
this is positive, so I'm
expecting a min.
307
00:19:33,120 --> 00:19:36,510
At x equals 2/3, I'm
expecting a min.
308
00:19:36,510 --> 00:19:39,070
And, of course, it is.
309
00:19:39,070 --> 00:19:42,530
And again, it's only
a local minimum.
310
00:19:42,530 --> 00:19:45,490
The derivative can only tell
you what's happening very,
311
00:19:45,490 --> 00:19:48,070
very close to that point.
312
00:19:48,070 --> 00:19:51,640
The derivative doesn't know that
over here the function is
313
00:19:51,640 --> 00:19:53,670
going further down.
314
00:19:53,670 --> 00:19:58,360
So this is a min, and
again, a local min.
315
00:19:58,360 --> 00:20:02,370
OK, those are maximum
and minimum
316
00:20:02,370 --> 00:20:04,490
when we know the function.
317
00:20:04,490 --> 00:20:08,255
Oh yeah, I better do the
inflection point.
318
00:20:08,255 --> 00:20:11,140
Do you remember what the
inflection point is?
319
00:20:11,140 --> 00:20:16,310
The inflection point is when
the bending changes from--
320
00:20:16,310 --> 00:20:18,965
up to here I see that
bending down.
321
00:20:18,965 --> 00:20:21,540
322
00:20:21,540 --> 00:20:24,860
From here, I see
it bending up.
323
00:20:24,860 --> 00:20:30,260
So I will not be surprised if
that's the point where the
324
00:20:30,260 --> 00:20:35,550
bending is changing, and 1/3
is the inflection point.
325
00:20:35,550 --> 00:20:37,860
And now how do we find
an inflection point?
326
00:20:37,860 --> 00:20:39,910
How do we identify this point?
327
00:20:39,910 --> 00:20:43,320
Well, y double prime
was negative.
328
00:20:43,320 --> 00:20:45,430
y double prime was positive.
329
00:20:45,430 --> 00:20:51,300
At that point, y double
prime is 0.
330
00:20:51,300 --> 00:20:53,790
This is an inflection point.
331
00:20:53,790 --> 00:20:55,580
And it is.
332
00:20:55,580 --> 00:21:03,800
At x equals to 1/3, I
do have 6 times 1/3.
333
00:21:03,800 --> 00:21:06,640
2 subtract 2, I have 0.
334
00:21:06,640 --> 00:21:10,870
So that is truly an
inflection point.
335
00:21:10,870 --> 00:21:13,900
And now I know all
the essential
336
00:21:13,900 --> 00:21:16,120
points about the curve.
337
00:21:16,120 --> 00:21:19,240
And these are the quantities--
338
00:21:19,240 --> 00:21:20,460
oh!
339
00:21:20,460 --> 00:21:24,860
Say you're an economist. You're
looking now at the
340
00:21:24,860 --> 00:21:30,070
statistics for the US economy
or the world economy.
341
00:21:30,070 --> 00:21:34,300
OK, I suppose we're in a--
342
00:21:34,300 --> 00:21:41,330
we had a local maximum there,
a happy time a little while
343
00:21:41,330 --> 00:21:46,870
ago, but it went downhill,
right?
344
00:21:46,870 --> 00:21:53,400
If y is, say, the gross product
for the world or gross
345
00:21:53,400 --> 00:21:55,500
national product,
it started down.
346
00:21:55,500 --> 00:21:58,120
347
00:21:58,120 --> 00:22:03,110
The slope of that curve
was negative.
348
00:22:03,110 --> 00:22:04,740
The bending was even negative.
349
00:22:04,740 --> 00:22:08,350
It was going down faster
all the time.
350
00:22:08,350 --> 00:22:14,540
Now, at a certain moment, the
economy kept going down, but
351
00:22:14,540 --> 00:22:16,630
you could see some
sign of hope.
352
00:22:16,630 --> 00:22:18,320
And what was the sign of hope?
353
00:22:18,320 --> 00:22:23,290
It was the fact that it
started bending up.
354
00:22:23,290 --> 00:22:28,470
And probably that's where we are
as I'm making this video.
355
00:22:28,470 --> 00:22:33,250
I suspect we're still going
down, but we're bending up.
356
00:22:33,250 --> 00:22:40,090
And at some point, hopefully
tomorrow, we'll hit minimum
357
00:22:40,090 --> 00:22:42,330
and start really up.
358
00:22:42,330 --> 00:22:43,330
So I don't know.
359
00:22:43,330 --> 00:22:45,130
I would guess we're somewhere
in there, and
360
00:22:45,130 --> 00:22:46,880
I don't know where.
361
00:22:46,880 --> 00:22:50,830
If I knew where, mathematics
would be even more useful than
362
00:22:50,830 --> 00:22:53,620
it is, which would
be hard to do.
363
00:22:53,620 --> 00:22:58,690
OK, so that's an example
of how the second
364
00:22:58,690 --> 00:23:00,390
derivative comes in.
365
00:23:00,390 --> 00:23:09,600
Now, I started by giving this
lecture the title Max and Min
366
00:23:09,600 --> 00:23:14,900
and saying those are the biggest
applications of the
367
00:23:14,900 --> 00:23:16,070
derivative.
368
00:23:16,070 --> 00:23:19,880
Set the derivative
to 0 and solve.
369
00:23:19,880 --> 00:23:22,390
Locate maximum points,
minimum points.
370
00:23:22,390 --> 00:23:27,920
That's what calculus is most--
371
00:23:27,920 --> 00:23:33,620
many of the word problems, most
of the ones I see in use,
372
00:23:33,620 --> 00:23:37,790
involve derivative equals 0.
373
00:23:37,790 --> 00:23:41,895
OK, so let me take a
particular example.
374
00:23:41,895 --> 00:23:44,730
375
00:23:44,730 --> 00:23:49,660
So these were graphs, simple
functions which I chose: sine
376
00:23:49,660 --> 00:23:52,920
x, x squared, x cubed
minus x squared.
377
00:23:52,920 --> 00:23:58,520
Now let me tell you the problem
because this is how
378
00:23:58,520 --> 00:24:00,210
math really comes.
379
00:24:00,210 --> 00:24:03,290
Let me tell you the problem, and
let's create the function.
380
00:24:03,290 --> 00:24:05,830
OK, so much it's the problem
I faced this
381
00:24:05,830 --> 00:24:09,080
morning and every morning.
382
00:24:09,080 --> 00:24:09,870
I live here.
383
00:24:09,870 --> 00:24:13,230
So OK, so here's home.
384
00:24:13,230 --> 00:24:20,500
And there is a-- the Mass Pike
is the fast road to MIT.
385
00:24:20,500 --> 00:24:26,360
So let me put in the Mass Pike
here, and let's say that's
386
00:24:26,360 --> 00:24:32,220
MIT, and I'm trying to get there
as fast as possible.
387
00:24:32,220 --> 00:24:40,010
OK, so for part of the time, I'm
going to have to drive on
388
00:24:40,010 --> 00:24:40,720
city streets.
389
00:24:40,720 --> 00:24:43,690
I do have to drive on city
streets, and then I get to go
390
00:24:43,690 --> 00:24:49,440
on the Mass Pike, which is,
let's say, twice as fast. The
391
00:24:49,440 --> 00:24:56,680
question is should I go directly
over to the fast road
392
00:24:56,680 --> 00:24:59,960
and then take off?
393
00:24:59,960 --> 00:25:01,470
Let's take off on
a good morning.
394
00:25:01,470 --> 00:25:04,540
The Mass Pike could be twice
as slow, but let's assume
395
00:25:04,540 --> 00:25:07,070
twice as fast. Should
I go straight over?
396
00:25:07,070 --> 00:25:09,440
Probably not.
397
00:25:09,440 --> 00:25:11,560
That's not the best way.
398
00:25:11,560 --> 00:25:17,050
I should probably pick up the
Mass Pike on some road.
399
00:25:17,050 --> 00:25:23,010
I could go directly to MIT on
the city streets at the slow
400
00:25:23,010 --> 00:25:29,150
rate, say 30 miles an hour or 30
kilometers an hour and 60,
401
00:25:29,150 --> 00:25:37,890
so speeds 30 and 60
as my speeds.
402
00:25:37,890 --> 00:25:44,550
OK, so now I should have
put in some measure.
403
00:25:44,550 --> 00:25:50,350
Let's call that distance
a, whatever it is.
404
00:25:50,350 --> 00:25:52,950
Maybe it's about three miles.
405
00:25:52,950 --> 00:25:54,420
And let me call--
406
00:25:54,420 --> 00:25:56,170
so that's the direct distance.
407
00:25:56,170 --> 00:25:59,650
If I just went direct to the
turnpike, I would go a
408
00:25:59,650 --> 00:26:02,380
distance a at 30 miles
an hour, and then
409
00:26:02,380 --> 00:26:03,700
I would go a distance--
410
00:26:03,700 --> 00:26:06,120
shall I call that b?--
411
00:26:06,120 --> 00:26:07,660
at 60.
412
00:26:07,660 --> 00:26:10,290
So that's one possibility.
413
00:26:10,290 --> 00:26:13,110
But I think it's not the best.
414
00:26:13,110 --> 00:26:15,410
I think better to--
415
00:26:15,410 --> 00:26:17,210
and you know better than me.
416
00:26:17,210 --> 00:26:19,380
I think I should probably
angle over
417
00:26:19,380 --> 00:26:23,170
here and pick up this--
418
00:26:23,170 --> 00:26:26,000
my question is where should
I join the Mass Pike.
419
00:26:26,000 --> 00:26:28,270
And let's--
420
00:26:28,270 --> 00:26:31,480
so we get a calculus problem,
let's model it.
421
00:26:31,480 --> 00:26:35,730
Suppose that I can join it
anywhere I like, not just at a
422
00:26:35,730 --> 00:26:39,930
couple of entrances.
423
00:26:39,930 --> 00:26:41,010
Anywhere.
424
00:26:41,010 --> 00:26:43,790
And the question is where?
425
00:26:43,790 --> 00:26:48,770
So calculus deals with the
continuous choice of x.
426
00:26:48,770 --> 00:26:52,180
So that is the unknown.
427
00:26:52,180 --> 00:26:54,830
I could take that as
the unknown x.
428
00:26:54,830 --> 00:26:57,510
That was a key step, of course,
deciding what should
429
00:26:57,510 --> 00:26:58,510
be the unknown.
430
00:26:58,510 --> 00:27:02,630
I could also have taken this
angle as an unknown, and that
431
00:27:02,630 --> 00:27:05,070
would be quite neat, too.
432
00:27:05,070 --> 00:27:07,530
But let me take that x.
433
00:27:07,530 --> 00:27:12,500
So this distance is
then b minus x.
434
00:27:12,500 --> 00:27:16,590
So that's what I travel
on the Mass Pike,
435
00:27:16,590 --> 00:27:20,340
so my time to minimize.
436
00:27:20,340 --> 00:27:23,930
437
00:27:23,930 --> 00:27:26,480
I'm trying to minimize
my time.
438
00:27:26,480 --> 00:27:35,240
OK, so on this Mass Pike when
I travel at 60, I have
439
00:27:35,240 --> 00:27:40,770
distance divided by 60
is the time, right?
440
00:27:40,770 --> 00:27:42,860
Am I remembering correctly?
441
00:27:42,860 --> 00:27:44,310
Let's just remember.
442
00:27:44,310 --> 00:27:48,390
Distance is speed times time.
443
00:27:48,390 --> 00:27:50,370
That's the one we know.
444
00:27:50,370 --> 00:27:57,180
And then if I divide by the
speed, the time is the
445
00:27:57,180 --> 00:28:00,230
distance divided by the speed,
the distance divided by the
446
00:28:00,230 --> 00:28:02,470
speed on the pike.
447
00:28:02,470 --> 00:28:06,750
And now I have the distance
on the city streets.
448
00:28:06,750 --> 00:28:13,930
OK, so that speed is
going to be 30.
449
00:28:13,930 --> 00:28:17,700
So the time is going to be a bit
longer for the distance,
450
00:28:17,700 --> 00:28:19,310
and what is that distance?
451
00:28:19,310 --> 00:28:21,540
OK, that was a.
452
00:28:21,540 --> 00:28:22,790
This was x.
453
00:28:22,790 --> 00:28:25,330
454
00:28:25,330 --> 00:28:30,390
Pythagoras is the great leveler
of mathematics.
455
00:28:30,390 --> 00:28:33,580
456
00:28:33,580 --> 00:28:37,400
That's the distance on
the city streets.
457
00:28:37,400 --> 00:28:43,430
And now what do I do?
458
00:28:43,430 --> 00:28:45,440
I've got an expression
for the time.
459
00:28:45,440 --> 00:28:49,770
This is the quantity I'm
trying to minimize.
460
00:28:49,770 --> 00:28:53,820
I minimize it by taking its
derivative and set the
461
00:28:53,820 --> 00:28:55,670
derivative to 0.
462
00:28:55,670 --> 00:28:58,150
Take the derivative and set
the derivative to 0.
463
00:28:58,150 --> 00:29:01,980
So now this is where I use
the formulas of calculus.
464
00:29:01,980 --> 00:29:05,220
So the derivative, now I'm ready
to write the derivative,
465
00:29:05,220 --> 00:29:06,730
and I'll set it to 0.
466
00:29:06,730 --> 00:29:10,140
So the derivative of that, b is
a constant, so I have minus
467
00:29:10,140 --> 00:29:13,870
1/60; is that OK?
468
00:29:13,870 --> 00:29:17,220
Plus whatever the derivative
of this is.
469
00:29:17,220 --> 00:29:20,390
Well, I have 1/30.
470
00:29:20,390 --> 00:29:22,080
I always take the
constant first.
471
00:29:22,080 --> 00:29:24,980
Now I have to deal with
that expression.
472
00:29:24,980 --> 00:29:28,720
That is some quantity
square root.
473
00:29:28,720 --> 00:29:34,420
The square root is the 1/2
power, so I have 1/2 times
474
00:29:34,420 --> 00:29:37,600
this quantity to one
lower power.
475
00:29:37,600 --> 00:29:39,790
That's the minus 1/2 power.
476
00:29:39,790 --> 00:29:44,770
That means that I still have a
square root, but now it's a
477
00:29:44,770 --> 00:29:46,700
minus 1/2 power.
478
00:29:46,700 --> 00:29:48,560
It's down here.
479
00:29:48,560 --> 00:29:58,490
And then the chain rule says
don't forget the derivative of
480
00:29:58,490 --> 00:30:00,180
what's inside, which is 2x.
481
00:30:00,180 --> 00:30:03,430
482
00:30:03,430 --> 00:30:08,050
OK, depending on what order
you've seen these videos and
483
00:30:08,050 --> 00:30:13,360
read text, you know the chain
rule, or you see it now.
484
00:30:13,360 --> 00:30:17,940
It's a very, very valuable rule
to find derivatives as
485
00:30:17,940 --> 00:30:19,910
the function gets complicated.
486
00:30:19,910 --> 00:30:23,460
And the thing to remember,
there will be a proper
487
00:30:23,460 --> 00:30:24,840
discussion of the chain rule.
488
00:30:24,840 --> 00:30:26,120
It's so important.
489
00:30:26,120 --> 00:30:29,800
But you're seeing it here that
the thing to remember is take
490
00:30:29,800 --> 00:30:33,780
also the derivative of what's
inside the a squared plus x
491
00:30:33,780 --> 00:30:36,850
squared, and the derivative of
the x squared is the 2x.
492
00:30:36,850 --> 00:30:39,120
OK, and that I have
to set to 0.
493
00:30:39,120 --> 00:30:43,180
And, of course, I'm going
to cancel the 2's, and
494
00:30:43,180 --> 00:30:44,250
I'll set it to 0.
495
00:30:44,250 --> 00:30:45,990
What does that mean
"set to zero"?
496
00:30:45,990 --> 00:30:47,300
Here's something minus.
497
00:30:47,300 --> 00:30:48,920
Here's something plus.
498
00:30:48,920 --> 00:30:51,450
I guess what I really want
is to make them equal.
499
00:30:51,450 --> 00:30:55,120
500
00:30:55,120 --> 00:31:03,570
When the 1/60 equals this
messier expression, at that
501
00:31:03,570 --> 00:31:07,850
point the minus term cancels
the plus term.
502
00:31:07,850 --> 00:31:11,640
I get 0 for the derivative,
so I'm looking for
503
00:31:11,640 --> 00:31:14,450
derivative equals 0.
504
00:31:14,450 --> 00:31:17,000
That's my equation now.
505
00:31:17,000 --> 00:31:19,250
OK, now I just have
to solve it.
506
00:31:19,250 --> 00:31:22,260
All right, let's see.
507
00:31:22,260 --> 00:31:24,810
If I wanted to solve that,
I would probably multiply
508
00:31:24,810 --> 00:31:27,820
through by 60.
509
00:31:27,820 --> 00:31:29,750
Can I do this?
510
00:31:29,750 --> 00:31:32,600
I'll multiply both
sides by 60.
511
00:31:32,600 --> 00:31:35,670
That will cancel the 30 and
leave an extra 2, so
512
00:31:35,670 --> 00:31:38,910
I'll have a 2x here.
513
00:31:38,910 --> 00:31:42,600
And let me multiply also by
this miserable square root
514
00:31:42,600 --> 00:31:45,565
that's in the denominator
to get it up there.
515
00:31:45,565 --> 00:31:51,180
516
00:31:51,180 --> 00:31:54,160
I think that's what I've got.
517
00:31:54,160 --> 00:31:57,020
That's the same equation as
this one, just simplified.
518
00:31:57,020 --> 00:31:59,010
Multiply through by 60.
519
00:31:59,010 --> 00:32:01,370
Multiply through by square
root of a squared plus x
520
00:32:01,370 --> 00:32:04,200
squared, and it's
looking good.
521
00:32:04,200 --> 00:32:07,070
All right, how am I going
to solve that?
522
00:32:07,070 --> 00:32:11,240
Well, the only mess up
is the square root.
523
00:32:11,240 --> 00:32:14,480
Get rid of that by squaring
both sides.
524
00:32:14,480 --> 00:32:18,690
So now I square both sides,
and I get a squared plus x
525
00:32:18,690 --> 00:32:23,440
squared, and the square
of 2x is 4x squared.
526
00:32:23,440 --> 00:32:28,855
All right, now I have an
equation that's way better.
527
00:32:28,855 --> 00:32:32,550
528
00:32:32,550 --> 00:32:36,000
In fact, even better
if I subtract x
529
00:32:36,000 --> 00:32:37,080
squared from both sides.
530
00:32:37,080 --> 00:32:39,850
My equation is telling
me that a squared
531
00:32:39,850 --> 00:32:42,710
should be 3x squared.
532
00:32:42,710 --> 00:32:46,060
In other words, this
good x is--
533
00:32:46,060 --> 00:32:51,540
now I'm ready to take the square
root and find x itself.
534
00:32:51,540 --> 00:32:53,310
So put the 3 here.
535
00:32:53,310 --> 00:32:54,830
Take the square root.
536
00:32:54,830 --> 00:32:58,735
I'm getting a over the
square root of 3.
537
00:32:58,735 --> 00:33:02,470
538
00:33:02,470 --> 00:33:08,920
So there is a word problem, a
minimum problem, where we had
539
00:33:08,920 --> 00:33:14,170
to create the function to
minimize, which was the time,
540
00:33:14,170 --> 00:33:18,810
trying to get to work as
quickly as possible.
541
00:33:18,810 --> 00:33:24,190
After naming the key quantity x,
then taking the derivative,
542
00:33:24,190 --> 00:33:29,150
then simplifying, that's where
the little work of calculus
543
00:33:29,150 --> 00:33:32,660
comes in, in the end getting
something nice, solving it,
544
00:33:32,660 --> 00:33:34,830
and getting the answer a
over square root of 3.
545
00:33:34,830 --> 00:33:40,120
So we now know what to do
driving in if there's an
546
00:33:40,120 --> 00:33:42,000
entrance where we
want to get it.
547
00:33:42,000 --> 00:33:47,100
And actually, it is a
beautiful answer.
548
00:33:47,100 --> 00:33:50,390
If this is a over the square
root of 3, this will turn out
549
00:33:50,390 --> 00:33:55,980
to be 30 degrees, pi over 6--
550
00:33:55,980 --> 00:33:57,640
I think.
551
00:33:57,640 --> 00:34:00,140
Yeah, I think that's right.
552
00:34:00,140 --> 00:34:03,260
So that's the conclusion
from calculus.
553
00:34:03,260 --> 00:34:04,990
Drive at a 30-degree angle.
554
00:34:04,990 --> 00:34:07,290
Hope that there's a road
going that way--
555
00:34:07,290 --> 00:34:09,000
sorry about that point--
556
00:34:09,000 --> 00:34:12,159
and join the turnpike.
557
00:34:12,159 --> 00:34:15,830
And probably the reason
for that nice
558
00:34:15,830 --> 00:34:19,239
answer, 30 degrees, came--
559
00:34:19,239 --> 00:34:23,860
I can't help but imagine that
because I chose 30 and 60
560
00:34:23,860 --> 00:34:30,060
here, a ratio of 1:2, and then
somehow the fact that the sine
561
00:34:30,060 --> 00:34:34,110
of 30 degrees is 1/2,
those two facts
562
00:34:34,110 --> 00:34:35,560
have got to be connected.
563
00:34:35,560 --> 00:34:39,860
So I change these 30 and 60
numbers, I'll change my
564
00:34:39,860 --> 00:34:43,260
answer, but basically, the
picture won't change much.
565
00:34:43,260 --> 00:34:51,949
And there's another little
point to make to really
566
00:34:51,949 --> 00:34:54,960
complete this problem.
567
00:34:54,960 --> 00:34:58,830
It could have happened that the
distance on the turnpike
568
00:34:58,830 --> 00:35:04,140
was very small and that
this was a dumb move.
569
00:35:04,140 --> 00:35:06,550
That 30-degree angle could
be overshooting
570
00:35:06,550 --> 00:35:11,070
MIT if MIT was there.
571
00:35:11,070 --> 00:35:16,310
So that's a case in which the
minimum time didn't happen
572
00:35:16,310 --> 00:35:21,210
where the derivative
bottomed out.
573
00:35:21,210 --> 00:35:27,701
If MIT was here, the good idea
would be go straight for it.
574
00:35:27,701 --> 00:35:31,580
575
00:35:31,580 --> 00:35:36,180
Yeah, the extra part on the
turn-- you wouldn't drive on
576
00:35:36,180 --> 00:35:38,090
the turnpike at all.
577
00:35:38,090 --> 00:35:41,990
And that's a signal that somehow
in the graph, which I
578
00:35:41,990 --> 00:35:48,910
didn't graph this function, but
if I did, then this stuff
579
00:35:48,910 --> 00:35:52,580
would be locating the minimum
of the graph.
580
00:35:52,580 --> 00:35:57,000
But this extra example where you
go straight for MIT would
581
00:35:57,000 --> 00:36:00,110
be a case in which the minimum
is at the end.
582
00:36:00,110 --> 00:36:02,650
And, of course, that
could happen.
583
00:36:02,650 --> 00:36:06,520
You could have a graph that just
goes down, and then it
584
00:36:06,520 --> 00:36:09,380
ends, so the minimum is there.
585
00:36:09,380 --> 00:36:11,660
Even though the graph looks like
it's still going down,
586
00:36:11,660 --> 00:36:13,290
the graph ended.
587
00:36:13,290 --> 00:36:14,300
What can you do?
588
00:36:14,300 --> 00:36:16,500
That's the best point
there is.
589
00:36:16,500 --> 00:36:21,840
OK, so that is a--
590
00:36:21,840 --> 00:36:26,910
can I recap this lecture
coming first over here?
591
00:36:26,910 --> 00:36:30,870
So the lecture is about maximum
and minimum, and we
592
00:36:30,870 --> 00:36:35,310
learned which it is by the
second derivative.
593
00:36:35,310 --> 00:36:37,490
So then we had examples.
594
00:36:37,490 --> 00:36:40,980
There was an example of a
minimum when the second
595
00:36:40,980 --> 00:36:43,030
derivative was positive.
596
00:36:43,030 --> 00:36:47,110
Here was an example of a local
maximum when the second
597
00:36:47,110 --> 00:36:50,230
derivative was negative.
598
00:36:50,230 --> 00:36:55,240
Here with the sine and cosine,
those are nice examples.
599
00:36:55,240 --> 00:36:58,810
And it takes some patience
to go through them.
600
00:36:58,810 --> 00:37:03,750
I suggest you take another
simple function, like start
601
00:37:03,750 --> 00:37:06,100
with cosine x.
602
00:37:06,100 --> 00:37:07,290
Find its maximum.
603
00:37:07,290 --> 00:37:08,330
Find its minimum.
604
00:37:08,330 --> 00:37:11,700
Find its inflection points so
the inflection points are
605
00:37:11,700 --> 00:37:17,750
where the bending is 0 because
it's changing from bending one
606
00:37:17,750 --> 00:37:19,380
way to bending the other way.
607
00:37:19,380 --> 00:37:23,080
We didn't need an inflection
test--
608
00:37:23,080 --> 00:37:27,070
so actually, I didn't complete
the lecture, because I didn't
609
00:37:27,070 --> 00:37:30,140
compute the second derivative
and show that
610
00:37:30,140 --> 00:37:33,190
this was truly a minimum.
611
00:37:33,190 --> 00:37:34,270
I could have done that.
612
00:37:34,270 --> 00:37:37,040
I would have had to take the
derivative of this, which
613
00:37:37,040 --> 00:37:41,900
would be one level messier,
and look at its sign.
614
00:37:41,900 --> 00:37:43,610
I wouldn't have to
set it to 0.
615
00:37:43,610 --> 00:37:47,310
I would be looking at the sign
of the second derivative.
616
00:37:47,310 --> 00:37:52,910
And in this problem, it would
be safely come out positive
617
00:37:52,910 --> 00:37:57,210
sign, meaning bending upwards,
meaning that this point I've
618
00:37:57,210 --> 00:38:02,540
identified by all these steps
was truly the minimum time,
619
00:38:02,540 --> 00:38:04,230
not a maximum.
620
00:38:04,230 --> 00:38:10,120
OK, that's a big part of
important calculus
621
00:38:10,120 --> 00:38:11,670
applications.
622
00:38:11,670 --> 00:38:12,870
Thanks.
623
00:38:12,870 --> 00:38:14,650
NARRATOR: This has been
a production of MIT
624
00:38:14,650 --> 00:38:17,040
OpenCourseWare and
Gilbert Strang.
625
00:38:17,040 --> 00:38:19,310
Funding for this video was
provided by the Lord
626
00:38:19,310 --> 00:38:20,530
Foundation.
627
00:38:20,530 --> 00:38:23,660
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628
00:38:23,660 --> 00:38:26,740
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629
00:38:26,740 --> 00:38:28,300
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630
00:38:28,300 --> 00:38:30,378