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PROFESSOR: In the twelfth
lecture, we're going to talk

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about maxima and minima.
 

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Let's finish up what
we did last time.

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We really only just started
with maxima and minima.

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And then we're going to
talk about related rates.

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So, right now I want to
give you some examples

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of max-min problems.
 

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And we're going to start
with a fairly basic one.

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So what's the thing
about max-min problems?

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The main thing is that we're
asking you to do a little bit

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more of the interpretation
of word problems.

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So many of the problems are
expressed in terms of words.

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And so, in this case, we have
a wire which is length 1.

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Cut into two pieces.
 

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And then each piece
encloses a square.

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Sorry, encloses a square.
 

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And the problem - so
this is the setup.

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And the problem is to find
the largest area enclosed.

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So here's the problem.
 

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Now, in all of these cases,
in all these cases,

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there's a bunch of words.
 

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And your job is typically
to draw a diagram.

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So the first thing you want
to do is to draw a diagram.

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In this case, it can
be fairly schematic.

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Here's your unit length.
 

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And when you draw the diagram,
you're going to have

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to pick variables.
 

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So those are really
the two main tasks.

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To set up the problem.
 

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So you're drawing a diagram.
 

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This is like word problems
of old, in grade school

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through high school.
 

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Draw a diagram and
name the variables.

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So we'll be doing a
lot of that today.

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So here's my unit length.
 

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And I'm going to choose the
variable x to be the length of

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one of the pieces of wire.
 

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And that makes the
other piece 1 - x.

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And that's pretty much the
whole diagram, except that

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there's something that we
did with the wire after

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we cut it in half.
 

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Namely, we built two
little boxes out of it.

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Like this, these
are our squares.

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And their side lengths are
x / 4 and (1 - x) / 4.

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So, so far, so good.
 

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And now we have to think,
well, we want to find

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the largest area.
 

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So I need a formula for
area in terms of variables

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that I've described.
 

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And so that's the last thing.
 

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I'll give the letter a as
the label for the area.

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And then the area is just
the square of x/4 +

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the square of 1 - x.
 

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Whoops, that strange
2 got in here.

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Over 4.
 

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So far, so good.
 

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Now, The instinct that you'll
have, and I'm going to yield

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to that instinct, is we
should charge ahead and

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just differentiate.
 

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Alright?
 

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That's alright.
 

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We'll find the critical points.
 

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So we know that those
are important points.

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So we're going to find
the critical points.

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In other words, we take the
derivative we set, the

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derivative of a with
respect to x = 0.

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So if I do that
differentiation, I get the,

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well, so the first one, x^2
/ 16, that's 8.

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Sorry.
 

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That's x / 8, right?
 

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That's a derivative of this.
 

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And if I differentiate this, I
get well, the derivative of 1

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- x ^2 is 2 ( 1 - x)( a - 1).
 

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So it's - (1 - x) / 8.
 

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So there are two minus signs in
there, I'll let you ponder that

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differentiation, which I
did by the chain rule.

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Hang on a sec, OK?
 

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Just wait until we're done.
 

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So here's the derivative.
 

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Is there a problem?
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: Right, so
there's a 1/16 here.

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This is x^2 / 16.
 

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And so it's 2x / 8,
over 16, sorry.

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Which has an 8.
 

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That's OK.
 

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Alright, so now, This is equal
to 0 if and only if x = 1 - x.

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That's 2x = = 1, or in
other words x = 1/2.

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Alright?
 

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So there's our critical point.
 

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So x = 1/2 is the
critical point.

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And the critical value, which
is what you get when you

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evaluate a at 1/2, is
(1/2) / 4, that's 1/8.

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So that's (1/8)^2 +
(1/8)^2 which = 1/32.

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So, so far, so good.
 

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But we're not done yet.
 

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We're not done.
 

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So why aren't we done?
 

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Because we haven't
checked the end points.

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So let's check the end points.
 

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Now, in this problem, the
end points are really

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sort of excluded.
 

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The ends are between
0 and 1 here.

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That's the possible
lengths of the cut.

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And so what we should really be
doing is evaluating in the

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limit, so that would be the
right-hand limit as

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x goes to 0 of a.
 

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And if you plug in x = 0, what
you get here is 0 + (1/4)^2.

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Which is 1/16.
 

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And similarly, at the other
end, that's 1 - 1 from

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the left we get (1/4)^2 +
0, which is also 1/16.

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So, what you see is that the
schematic picture of this

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function, and isn't even so
far off from being the right

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picture here, is that it's
level here as 1/16 and then

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it dips down and goes up.
 

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Right?
 

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This is 1/2, this is 1, and
this level here is a half that.

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This is 1/32.
 

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So we did not find, when we
found the critical point

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we did not find the
largest area enclosed.

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We found the least
area enclosed.

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So if you don't pay attention
to what the function looks

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like, not only will you about
half the time get the wrong

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answer, you'll get the
absolute worst answer.

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You'll get the one which
is the polar opposite

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from what you want.
 

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So you have to pay a little
bit of attention to the

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function that you've got.
 

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00:08:53 --> 00:08:55
And in this case it's
just very schematic.

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It dips down and goes up,
and that's true of pretty

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much most functions.
 

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They're fairly simple.
 

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They maybe only have
one critical point.

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They only turn around once.
 

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But then, maybe the critical
point is the maximum or

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maybe it's the minimum.
 

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Or maybe it's neither, in fact.
 

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So we'll be discussing that
maybe some other time.

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So what we find here is that we
have the least area enclosed.

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Enclosed is 1/32.
 

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And this is true when x = 1/2.
 

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So these are equal squares.
 

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And most when there's
only one square.

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Which is more or less
the limiting situation.

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If one of the
pieces disappears.

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Now, so that's the
first kind of example.

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00:10:09 --> 00:10:13
And I just want to make
one more comment about

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terminology before we go on.
 

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And I will introduce it with
the following question.

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What is the minimum?
 

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00:10:31 --> 00:10:39
So, what is the minimum?
 

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Yeah.
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: Right.
 

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The lowest value
of the function.

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00:10:46 --> 00:10:52
So the answer to that
question is 1/32.

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00:10:52 --> 00:10:57
Now, the problem with this
question and you will, so that

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refers to the minimum value.
 

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But then there's this
other question which is

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where is the minimum.
 

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And the answer to
that is x = 1/2.

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So one of them is the minimum
point, and the other one

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is the minimum value.
 

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So they're two separate things.
 

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00:11:35 --> 00:11:39
Now, the problem is that
people are sloppy.

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And especially since you
usually find the critical point

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first, and the value that is
plugging in for a second,

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people will stop short and
they'll give the wrong answer

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00:11:52 --> 00:11:54
to the question, for instance.
 

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Now, both questions are
important to answer.

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You just need to have
a word to put there.

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So this is a little
bit careless.

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When we say what is the
minimum, some people

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00:12:04 --> 00:12:06
will say 1/2.
 

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And that's literally wrong.
 

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They know what they mean.
 

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But it's just wrong.
 

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And when people ask this
question, they're being sloppy.

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Anyway.
 

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They should maybe be a little
clearer and say what's

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the minimum value.
 

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Or, where is the
value achieved.

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It's achieved at, or where is
the minimum value achieved.

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"Where is min achieved?", would
be a better way of phrasing

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this second question.
 

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So that it has an
unambiguous answer.

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00:12:37 --> 00:12:43
And when people ask you for the
minimum point, they're also -

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so why is it that we call
it the minimum point?

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We have this word, critical
point, which is what x =

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1/2 is in critical value.
 

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And so I'm making those
same distinctions here.

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00:12:52 --> 00:13:01
But there's another notion of
a minimum point, and this is

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00:13:01 --> 00:13:09
an alternative if you like.
 

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00:13:09 --> 00:13:19
The minimum point is
the point (1/2, 1/32).

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00:13:19 --> 00:13:25
Right, that's a
point on the graph.

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00:13:25 --> 00:13:29
It's the point - well, so that
graph is way up there, but

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I'll just put it on there.
 

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That's this point.
 

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And you might say min there.
 

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00:13:36 --> 00:13:38
And you might point to this
point, and you might say max.

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00:13:38 --> 00:13:42
And similarly, this
one might be a max.

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00:13:42 --> 00:13:47
So in other words, what this
means is simply that people

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00:13:47 --> 00:13:49
are a little sloppy.
 

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00:13:49 --> 00:13:51
And sometimes they mean
one thing and sometimes

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00:13:51 --> 00:13:53
they mean another.
 

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00:13:53 --> 00:13:56
And you're just stuck with
this, because there'll be some

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00:13:56 --> 00:13:59
authors who will say one thing
and some people will mean

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00:13:59 --> 00:14:01
another and you just have to
live with this little bit

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00:14:01 --> 00:14:05
of annoying ambiguity.
 

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00:14:05 --> 00:14:05
Yeah?
 

219
00:14:05 --> 00:14:08
STUDENT: [INAUDIBLE]
 

220
00:14:08 --> 00:14:12
PROFESSOR: OK, so that's
a good - very good.

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00:14:12 --> 00:14:16
So here we go, find the
largest area enclosed.

222
00:14:16 --> 00:14:22
So that's sort of a trick
question, isn't it?

223
00:14:22 --> 00:14:28
So there are various -
that's a good thing to ask.

224
00:14:28 --> 00:14:30
That's sort of a
trick question, why?

225
00:14:30 --> 00:14:36
Because according to the rules,
we're trapped between the two

226
00:14:36 --> 00:14:40
maxima at something which
is strictly below.

227
00:14:40 --> 00:14:44
So in other words, one answer
to this question would be, and

228
00:14:44 --> 00:14:48
this is the answer that I
would probably give, is 1/16.

229
00:14:48 --> 00:14:50
But that's not really true.
 

230
00:14:50 --> 00:14:56
Because that's only
in the limit.

231
00:14:56 --> 00:15:01
As x goes to 0, or
as x goes to 1 -.

232
00:15:01 --> 00:15:03
And if you like, the most
is when you've only

233
00:15:03 --> 00:15:04
got one square.
 

234
00:15:04 --> 00:15:07
Which breaks the rules
of the problem.

235
00:15:07 --> 00:15:11
So, essentially, it's
a trick question.

236
00:15:11 --> 00:15:13
But I would answer it this way.
 

237
00:15:13 --> 00:15:16
Because that's the most
interesting part of the answer,

238
00:15:16 --> 00:15:20
which is that it's 1/16 and it
occurs really when one of the

239
00:15:20 --> 00:15:28
squares disappears to nothing.
 

240
00:15:28 --> 00:15:34
So now, let's do
another example here.

241
00:15:34 --> 00:15:41
And I just want to illustrate
the second style, or the

242
00:15:41 --> 00:15:43
second type of question.
 

243
00:15:43 --> 00:15:43
Yeah.
 

244
00:15:43 --> 00:15:51
STUDENT: [INAUDIBLE]
 

245
00:15:51 --> 00:15:56
PROFESSOR: The question is,
since the question was, what

246
00:15:56 --> 00:16:00
was the largest area, why
did we find the least area.

247
00:16:00 --> 00:16:04
The reason is that when we go
about our procedure of looking

248
00:16:04 --> 00:16:10
for the least, or the most,
we'll automatically find both.

249
00:16:10 --> 00:16:12
Because we don't know
which one is which until

250
00:16:12 --> 00:16:14
we compare values.
 

251
00:16:14 --> 00:16:18
And actually, it's much more to
your advantage to figure out

252
00:16:18 --> 00:16:21
both the maximum and minimum
whenever you answer

253
00:16:21 --> 00:16:22
such a question.
 

254
00:16:22 --> 00:16:24
Because otherwise you won't
understand the behavior of

255
00:16:24 --> 00:16:26
the function very well.
 

256
00:16:26 --> 00:16:27
So, the question.
 

257
00:16:27 --> 00:16:30
We started out with one
question, we answered both.

258
00:16:30 --> 00:16:31
We answered two questions.
 

259
00:16:31 --> 00:16:34
We answered the question
of what the largest and

260
00:16:34 --> 00:16:37
the smallest value was.
 

261
00:16:37 --> 00:16:40
STUDENT: Also, I'm wondering
if you can check both

262
00:16:40 --> 00:16:41
the minimum [INAUDIBLE]
 

263
00:16:41 --> 00:16:47
approaches [INAUDIBLE].
 

264
00:16:47 --> 00:16:48
PROFESSOR: Yes.
 

265
00:16:48 --> 00:16:50
One can also use, the
question is, can we use the

266
00:16:50 --> 00:16:51
second derivative test.
 

267
00:16:51 --> 00:16:53
And the answer is, yes we can.
 

268
00:16:53 --> 00:16:56
In fact, you can actually also
stare at this and see that

269
00:16:56 --> 00:16:57
it's a sum of squares.
 

270
00:16:57 --> 00:17:00
So it's always curving up.
 

271
00:17:00 --> 00:17:04
It's a parabola with a
positive second coefficient.

272
00:17:04 --> 00:17:06
So you can differentiate
this twice.

273
00:17:06 --> 00:17:09
If you do you'll get
1/8 + another 1/8

274
00:17:09 --> 00:17:11
and you'll get 1/16.
 

275
00:17:11 --> 00:17:14
So the second
derivative is 1/16.

276
00:17:14 --> 00:17:17
Is 1/4.
 

277
00:17:17 --> 00:17:21
And that's an acceptable
way of figuring it out.

278
00:17:21 --> 00:17:23
I'll mention the second
derivative test again,

279
00:17:23 --> 00:17:24
in this second example.
 

280
00:17:24 --> 00:17:32
So let me talk about
a second example.

281
00:17:32 --> 00:17:35
So again, this is going
to be another question.

282
00:17:35 --> 00:17:37
STUDENT: [INAUDIBLE]
 

283
00:17:37 --> 00:17:43
PROFESSOR: The question is,
when I say minimum or maximum

284
00:17:43 --> 00:17:44
point which will I mean.
 

285
00:17:44 --> 00:17:50
STUDENT: [INAUDIBLE]
 

286
00:17:50 --> 00:17:53
PROFESSOR: So I just
repeated the question.

287
00:17:53 --> 00:17:56
So the question is, when
I say minimum point,

288
00:17:56 --> 00:17:58
what will I mean?
 

289
00:17:58 --> 00:18:00
OK?
 

290
00:18:00 --> 00:18:05
And the answer is that for the
purposes of this class I will

291
00:18:05 --> 00:18:09
probably avoid saying that.
 

292
00:18:09 --> 00:18:13
But I will say, probably, where
is the minimum achieved.

293
00:18:13 --> 00:18:14
Just in order to avoid that.
 

294
00:18:14 --> 00:18:17
If I actually sat at I
often am referring to the

295
00:18:17 --> 00:18:19
graph, and I mean this.
 

296
00:18:19 --> 00:18:21
And in fact, when you get your
little review for the second

297
00:18:21 --> 00:18:25
exam, I'll say exactly
that on the review sheet.

298
00:18:25 --> 00:18:28
And I'll make this very clear
when we were doing this.

299
00:18:28 --> 00:18:31
However, I just want to prepare
you for the fact that in real

300
00:18:31 --> 00:18:35
life, and even me when I'm
talking colloquially, when I

301
00:18:35 --> 00:18:38
say what's the minimum point of
something, I might actually be

302
00:18:38 --> 00:18:48
mixing it up with this
other notion here.

303
00:18:48 --> 00:18:54
So let's do another example.
 

304
00:18:54 --> 00:18:57
So this is an example
to get us used to the

305
00:18:57 --> 00:18:59
notion of constraints.
 

306
00:18:59 --> 00:19:08
So we have, so consider
a box without a top.

307
00:19:08 --> 00:19:16
Or, if you like, we're going to
find the box without a top.

308
00:19:16 --> 00:19:35
With least surface area
for a fixed volume.

309
00:19:35 --> 00:19:40
Find the box without a top
with least surface area

310
00:19:40 --> 00:19:42
for a fixed volume.
 

311
00:19:42 --> 00:19:47
The procedure for working
this out is the following.

312
00:19:47 --> 00:19:51
You make this diagram.
 

313
00:19:51 --> 00:19:56
And you set up the variables.
 

314
00:19:56 --> 00:20:00
In this case, we're going to
have four names of variables.

315
00:20:00 --> 00:20:02
We have four letters
that we have to choose.

316
00:20:02 --> 00:20:05
And we'll choose them in a kind
of a standard way, alright?

317
00:20:05 --> 00:20:08
So first I have to tell
you one more thing.

318
00:20:08 --> 00:20:12
Which is something that we
could calculate separately

319
00:20:12 --> 00:20:15
but I'm just going to give
it to you in advance.

320
00:20:15 --> 00:20:16
Which is that it turns
out that the best box

321
00:20:16 --> 00:20:21
has a square bottom.
 

322
00:20:21 --> 00:20:24
And that's going to get rid of
one of our variables for us.

323
00:20:24 --> 00:20:26
So it's got a square
bottom, and so let's

324
00:20:26 --> 00:20:28
draw a picture of it.
 

325
00:20:28 --> 00:20:36
So here's our box.
 

326
00:20:36 --> 00:20:40
Well, that goes down
like this, almost.

327
00:20:40 --> 00:20:49
Maybe I should get it a
little farther down.

328
00:20:49 --> 00:20:52
So here's our box.
 

329
00:20:52 --> 00:20:54
Let's correct that just a bit.
 

330
00:20:54 --> 00:20:57
So now, what about the
dimensions of this box?

331
00:20:57 --> 00:21:02
Well, this is going to be x,
and this is very foreshortened,

332
00:21:02 --> 00:21:03
but it's also x.
 

333
00:21:03 --> 00:21:06
The bottom is x by x, it's
the same dimensions.

334
00:21:06 --> 00:21:12
And then the vertical
dimension is y.

335
00:21:12 --> 00:21:13
So far, so good.
 

336
00:21:13 --> 00:21:16
Now, I promised you two
more letter names.

337
00:21:16 --> 00:21:21
I want to compute the volume.
 

338
00:21:21 --> 00:21:24
The volume is, the base is
x ^2, and the height is y.

339
00:21:24 --> 00:21:26
So there's the volume.
 

340
00:21:26 --> 00:21:33
And then the area, the area is
the area of the bottom, which

341
00:21:33 --> 00:21:37
is x ^2, that's the bottom.
 

342
00:21:37 --> 00:21:40
And then there are
the four sides.

343
00:21:40 --> 00:21:44
And the four sides are
rectangles of dimensions xy.

344
00:21:44 --> 00:21:49
So it's 4xy.
 

345
00:21:49 --> 00:21:53
So these are the sides.
 

346
00:21:53 --> 00:21:56
And remember, there's no top.
 

347
00:21:56 --> 00:21:58
So that's our setup.
 

348
00:21:58 --> 00:22:03
So now, the difference between
this problem and the last

349
00:22:03 --> 00:22:06
problem is that there are two
variables floating around,

350
00:22:06 --> 00:22:09
namely x and y, which
are not determined.

351
00:22:09 --> 00:22:17
But there's what's called
a constraint here.

352
00:22:17 --> 00:22:22
Namely, we've fixed the
relationship between x and y.

353
00:22:22 --> 00:22:30
And so, that means that we can
solve for y in terms of x.

354
00:22:30 --> 00:22:40
So y = v / x ^2.
 

355
00:22:40 --> 00:22:43
And then, we can plug that
into the formula for a.

356
00:22:43 --> 00:23:00
So here we have a which
is x ^2 + 4x ( v / x^2).

357
00:23:00 --> 00:23:01
Question.
 

358
00:23:01 --> 00:23:15
STUDENT: [INAUDIBLE]
 

359
00:23:15 --> 00:23:16
PROFESSOR: The question
is, will you need to

360
00:23:16 --> 00:23:17
know this intuitively?
 

361
00:23:17 --> 00:23:17
No.
 

362
00:23:17 --> 00:23:20
That's something that I
would have to give to you.

363
00:23:20 --> 00:23:26
I mean, it's actually true that
a lot of things, the correct

364
00:23:26 --> 00:23:28
answer is something symmetric.
 

365
00:23:28 --> 00:23:30
In this last problem, the
minimum turned out to be

366
00:23:30 --> 00:23:33
exactly halfway in between
because there were sort of

367
00:23:33 --> 00:23:35
equal demands from
the two sides.

368
00:23:35 --> 00:23:38
And similarly, here, what
happens is if you elongate

369
00:23:38 --> 00:23:44
one side, you get less - it
actually is involved with

370
00:23:44 --> 00:23:46
a two variable problem.
 

371
00:23:46 --> 00:23:49
Namely, if you have a rectangle
and you have a certain amount

372
00:23:49 --> 00:23:51
of length associated with it.
 

373
00:23:51 --> 00:23:53
What's the optimal thing
you can do with that.

374
00:23:53 --> 00:23:58
But I won't, in other words,
the optimal rectangle, the

375
00:23:58 --> 00:24:01
least perimeter rectangle,
turns out to be a square.

376
00:24:01 --> 00:24:03
That's the little sub-problem
that leads you to

377
00:24:03 --> 00:24:05
this square bottom.
 

378
00:24:05 --> 00:24:09
But so that would have been
a separate max-min problem.

379
00:24:09 --> 00:24:11
Which I'm skipping, because
I what to do this slightly

380
00:24:11 --> 00:24:16
more interesting one.
 

381
00:24:16 --> 00:24:22
So now, here's our formula for
a, and now I want to follow

382
00:24:22 --> 00:24:27
the same procedure as before.
 

383
00:24:27 --> 00:24:29
Namely, we look for
the critical point.

384
00:24:29 --> 00:24:35
Or points.
 

385
00:24:35 --> 00:24:37
So let's take a look.
 

386
00:24:37 --> 00:24:43
So again, a is (x ^2 + 4v) / x.
 

387
00:24:43 --> 00:24:48
And A' = 2x - (4v / x^2).
 

388
00:24:49 --> 00:24:59
So if we set that equal to
0, we get 2x = 2v / x ^2.

389
00:24:59 --> 00:25:01
So 2x^3.
 

390
00:25:01 --> 00:25:04
 


391
00:25:04 --> 00:25:07
How did that happen
to change into 2?

392
00:25:07 --> 00:25:09
Interesting, guess
that's wrong.

393
00:25:09 --> 00:25:12
OK.
 

394
00:25:12 --> 00:25:20
So this is x ^ 3 = 2v.
 

395
00:25:20 --> 00:25:28
And so x = (2 ^ 1/3)( v ^1/3).
 

396
00:25:28 --> 00:25:36
So this is the critical point.
 

397
00:25:36 --> 00:25:38
So we are not done.
 

398
00:25:38 --> 00:25:38
Right?
 

399
00:25:38 --> 00:25:40
We're not done, because we
don't even know whether this is

400
00:25:40 --> 00:25:43
going to give us the worst box
or the best box, from

401
00:25:43 --> 00:25:44
this point of view.
 

402
00:25:44 --> 00:25:48
The one that uses the most
surface area or the least.

403
00:25:48 --> 00:25:51
So let's check the
ends, right away.

404
00:25:51 --> 00:25:54
To see what's happening.
 

405
00:25:54 --> 00:25:56
So can somebody tell me
what the ends, what the

406
00:25:56 --> 00:25:58
end values of x are?
 

407
00:25:58 --> 00:25:59
Where does x range from?
 

408
00:25:59 --> 00:26:05
STUDENT: [INAUDIBLE]
 

409
00:26:05 --> 00:26:07
PROFESSOR: What's the
smallest x can be, yeah.

410
00:26:07 --> 00:26:16
STUDENT: [INAUDIBLE]
 

411
00:26:16 --> 00:26:19
PROFESSOR: OK, the claim was
that the largest x could be

412
00:26:19 --> 00:26:23
root a, because somehow there's
this x ^2 here and you can't

413
00:26:23 --> 00:26:25
get any further past than that.
 

414
00:26:25 --> 00:26:28
But there's a key feature
here of this problem.

415
00:26:28 --> 00:26:32
Which is that a is variable.
 

416
00:26:32 --> 00:26:39
The only thing that's fixed
in the problem is v.

417
00:26:39 --> 00:26:47
So if v is fixed, what
do you know about x?

418
00:26:47 --> 00:26:49
STUDENT: [INAUDIBLE]
 

419
00:26:49 --> 00:26:51
PROFESSOR: x > 0, yeah.
 

420
00:26:51 --> 00:26:53
The lower end point,
that's safe.

421
00:26:53 --> 00:26:55
Because that has to do
geometrically with the fact

422
00:26:55 --> 00:26:59
that we don't have any boxes
with negative dimensions.

423
00:26:59 --> 00:27:02
That would be refused by the
Post Office, definitely.

424
00:27:02 --> 00:27:04
Over and above the
empty top, which they

425
00:27:04 --> 00:27:05
wouldn't accept either.
 

426
00:27:05 --> 00:27:13
STUDENT: [INAUDIBLE]
 

427
00:27:13 --> 00:27:17
PROFESSOR: It's true that
x < square root of v / y.

428
00:27:17 --> 00:27:19
So that's using
this relationship.

429
00:27:19 --> 00:27:26
But notice that y = v / x ^2.
 

430
00:27:26 --> 00:27:30
So 0 to infinity, I just
got a guess there over

431
00:27:30 --> 00:27:32
here, that's right.
 

432
00:27:32 --> 00:27:33
Here's the upper limit.
 

433
00:27:33 --> 00:27:35
So this is really
important to realize.

434
00:27:35 --> 00:27:37
This is most problems.
 

435
00:27:37 --> 00:27:40
Most problems, the variable if
it doesn't have a limitation,

436
00:27:40 --> 00:27:42
usually just goes
out to infinity.

437
00:27:42 --> 00:27:45
And infinity is a very
important end for the problem.

438
00:27:45 --> 00:27:51
It's usually an easy end
to the problem, too.

439
00:27:51 --> 00:27:54
So there's a possibility that
if we push all the way down to

440
00:27:54 --> 00:27:56
x = 0, we'll get a better box.
 

441
00:27:56 --> 00:27:58
It would be very strange box.
 

442
00:27:58 --> 00:28:01
A little bit like our
vanishing enclosure.

443
00:28:01 --> 00:28:05
And maybe an infinitely
long box, also very

444
00:28:05 --> 00:28:06
inconvenient one.
 

445
00:28:06 --> 00:28:07
Might be the best box.
 

446
00:28:07 --> 00:28:10
We'll have to see.
 

447
00:28:10 --> 00:28:12
So let's just take a
look at what happens.

448
00:28:12 --> 00:28:18
So we're looking at a, at 0 +.
 

449
00:28:18 --> 00:28:25
And that's x^2 + 4v
/ x with x at 0 +.

450
00:28:25 --> 00:28:26
So what happens to that?
 

451
00:28:26 --> 00:28:35
Notice right here, this
is going to infinity.

452
00:28:35 --> 00:28:38
So this is infinite.
 

453
00:28:38 --> 00:28:42
So that turns out
to be a bad box.

454
00:28:42 --> 00:28:45
Let's take a look
at the other end.

455
00:28:45 --> 00:28:51
So this is x ^2 + 4v /
x, x going to infinity.

456
00:28:51 --> 00:28:59
And again, this term here means
that this thing is infinite.

457
00:28:59 --> 00:29:02
So the shape of this thing,
I'll draw this tiny little

458
00:29:02 --> 00:29:05
schematic diagram over here.
 

459
00:29:05 --> 00:29:11
The shape of this thing
is like this, right?

460
00:29:11 --> 00:29:14
And so, when we find that one
turnaround point, which

461
00:29:14 --> 00:29:20
happened to be at this strange
point 2/3 , (2 ^ 1/3)( v ^

462
00:29:20 --> 00:29:24
1/3), that is going
to be the minimum.

463
00:29:24 --> 00:29:30
So we've just discovered
that it's the minimum.

464
00:29:30 --> 00:29:31
Which is just what
we were hoping for.

465
00:29:31 --> 00:29:38
This is going to be
the optimal box.

466
00:29:38 --> 00:29:45
Now, since you asked earlier
and since it's worth checking

467
00:29:45 --> 00:29:51
this as well, let's also check
an alternative justification.

468
00:29:51 --> 00:30:03
So an alternative to
checking ends is the

469
00:30:03 --> 00:30:11
second derivative test.
 

470
00:30:11 --> 00:30:14
I do not recommend the
second derivative test.

471
00:30:14 --> 00:30:17
I try my best, when I give you
problems, to make it really

472
00:30:17 --> 00:30:19
hard to apply the second
derivative test.

473
00:30:19 --> 00:30:22
But in this example, the
function is simple enough

474
00:30:22 --> 00:30:24
so that it's perfectly OK.
 

475
00:30:24 --> 00:30:28
If you take the derivative
here, remember, this

476
00:30:28 --> 00:30:34
was whatever it was,
2x - (4v / x ^2).

477
00:30:34 --> 00:30:38
If I take the second
derivative, it's

478
00:30:38 --> 00:30:43
2 + (8v / x ^3).
 

479
00:30:43 --> 00:30:45
And that's positive.
 

480
00:30:45 --> 00:30:49
So this thing is concave up.
 

481
00:30:49 --> 00:30:52
And that's consistent with
its being, the critical

482
00:30:52 --> 00:30:59
point is a min.
 

483
00:30:59 --> 00:31:00
Is a minimum point.
 

484
00:31:00 --> 00:31:04
See how I almost said,
is a min, as opposed

485
00:31:04 --> 00:31:05
to minimum point.
 

486
00:31:05 --> 00:31:05
So watch out.
 

487
00:31:05 --> 00:31:06
Yes.
 

488
00:31:06 --> 00:31:12
STUDENT: [INAUDIBLE]
 

489
00:31:12 --> 00:31:13
PROFESSOR: You're one
step ahead of me.

490
00:31:13 --> 00:31:16
The question is, is this the
answer to the question or

491
00:31:16 --> 00:31:20
would we have to give y and
a and so on and so forth.

492
00:31:20 --> 00:31:24
So, again, this is something
that I want to emphasize and

493
00:31:24 --> 00:31:26
take my time with right now.
 

494
00:31:26 --> 00:31:30
Because it depends, what kind
of real life problem you're

495
00:31:30 --> 00:31:33
answering, what kind of
answer is appropriate.

496
00:31:33 --> 00:31:36
So, so far we've found
the critical point.

497
00:31:36 --> 00:31:38
We haven't found the
critical value.

498
00:31:38 --> 00:31:42
We haven't found the
dimensions of the box.

499
00:31:42 --> 00:31:44
So we're going to spend a
little bit more time on

500
00:31:44 --> 00:31:48
this, exactly in order to
address these questions.

501
00:31:48 --> 00:31:50
So, first of all.
 

502
00:31:50 --> 00:31:51
The value of y.
 

503
00:31:51 --> 00:31:55
So, so far we have x =
(2 ^ 1/3)( v ^ 1/3).

504
00:31:55 --> 00:31:57
And certainly if you're going
to build the box, you also want

505
00:31:57 --> 00:32:00
to know what the y value is.
 

506
00:32:00 --> 00:32:04
The y value is going
to be, let's see.

507
00:32:04 --> 00:32:11
Well, it's v / x ^2, so that's
v / ((2 ^ 1/3)( v ^ 1/3)

508
00:32:11 --> 00:32:19
^2, which comes out to be
(2 ^ - 2/3)( v ^ 1/3).

509
00:32:19 --> 00:32:22
So there's the y value.
 

510
00:32:22 --> 00:32:29
On top of that, we could
figure out the value of a.

511
00:32:29 --> 00:32:31
So that's also a perfectly
reasonable part of the answer.

512
00:32:31 --> 00:32:34
Depending on what one is
interested in, you might care

513
00:32:34 --> 00:32:38
how much money it's going to
cost you to build this box.

514
00:32:38 --> 00:32:39
This optimal box.
 

515
00:32:39 --> 00:32:41
And so you plug in
the value of a.

516
00:32:41 --> 00:32:43
So a, let's see, is up here.
 

517
00:32:43 --> 00:32:47
It's x ^2 + 4v / x.
 

518
00:32:47 --> 00:32:56
So that's going to be ((2
^ 1/3)( v ^ 1/3) ^2 + (4v

519
00:32:56 --> 00:33:02
/ (2 ^ 1/3)( v ^ 1/3)).
 

520
00:33:02 --> 00:33:06
And if you work that all out,
what you get turns out to

521
00:33:06 --> 00:33:13
be 3 ( 2 ^ 1/3)( v ^ 2/3).
 

522
00:33:13 --> 00:33:17
So if you like, one way of
answering this question

523
00:33:17 --> 00:33:23
is these three things.
 

524
00:33:23 --> 00:33:26
That would be the minimum point
corresponding to the graph.

525
00:33:26 --> 00:33:28
That would be the answer
to this question.

526
00:33:28 --> 00:33:32
But the reason why I'm carrying
it out in such detail is I want

527
00:33:32 --> 00:33:35
to show you that there are much
more meaningful ways of

528
00:33:35 --> 00:33:37
answering this question.
 

529
00:33:37 --> 00:34:02
So let me explain that.
 

530
00:34:02 --> 00:34:07
So let me go through some more
meaningful answers here.

531
00:34:07 --> 00:34:14
The first more meaningful
answer is the following idea

532
00:34:14 --> 00:34:29
simply, what are known as
dimensionless variables.

533
00:34:29 --> 00:34:33
So the first thing that you
notice is the scaling law.

534
00:34:33 --> 00:34:36
That a / v ^ 2/3 is the thing
that's a dimensionless

535
00:34:36 --> 00:34:37
quantity.
 

536
00:34:37 --> 00:34:42
That happens to
be 3 ( 2 ^ 1/3).

537
00:34:42 --> 00:34:43
So that's one thing.
 

538
00:34:43 --> 00:34:45
If you want to expand the
volume, you'll have to

539
00:34:45 --> 00:34:49
expand the area by the
2/3 power of the volume.

540
00:34:49 --> 00:34:55
And if you think of the area as
being in, say, square inches,

541
00:34:55 --> 00:35:00
and the volume of the box as
being in cubic inches, then you

542
00:35:00 --> 00:35:02
can see that this is a
dimensionless quantity and you

543
00:35:02 --> 00:35:05
have a dimensionless number
here, which is a characteristic

544
00:35:05 --> 00:35:09
independent of what
a and v were.

545
00:35:09 --> 00:35:13
The other dimensionless
quantity is the y:x.

546
00:35:15 --> 00:35:19
So x : y.
 

547
00:35:19 --> 00:35:23
So, again, that's inches
divided by inches.

548
00:35:23 --> 00:35:32
And it's (2 ^ 1/3)( v ^ 1/3)
/ ( 2 ^ - 2/3)( v ^ 1/3),

549
00:35:32 --> 00:35:36
which happens to be 2.
 

550
00:35:36 --> 00:35:41
So this is actually the best
answer to the question.

551
00:35:41 --> 00:35:46
And it shows you that
the box is a 2:1 box.

552
00:35:46 --> 00:35:50
If this is 2 and this is
1, that's the good box.

553
00:35:50 --> 00:35:57
And that is just the shape,
if you like, and it's

554
00:35:57 --> 00:36:02
the optimal shape.
 

555
00:36:02 --> 00:36:04
And certainly that,
aesthetically, that's

556
00:36:04 --> 00:36:12
the cleanest answer
to the question.

557
00:36:12 --> 00:36:13
There was a question
right here.

558
00:36:13 --> 00:36:13
Yes.
 

559
00:36:13 --> 00:36:20
STUDENT: [INAUDIBLE]
 

560
00:36:20 --> 00:36:22
PROFESSOR: Could you repeat
that, I couldn't hear.

561
00:36:22 --> 00:36:24
STUDENT: I'm wondering if
you'd be able to get that

562
00:36:24 --> 00:36:26
answer if you [INAUDIBLE]
 

563
00:36:26 --> 00:36:30
square.
 

564
00:36:30 --> 00:36:32
PROFESSOR: The question is,
could we have gotten the answer

565
00:36:32 --> 00:36:34
if we weren't told that
the bottom was square.

566
00:36:34 --> 00:36:39
The answer is, yes in
18.02 with multivariable.

567
00:36:39 --> 00:36:41
You would have to have three
letters here, an x, a y,

568
00:36:41 --> 00:36:43
and a z, if you like.
 

569
00:36:43 --> 00:36:48
And then you'd have to work
with all three of them.

570
00:36:48 --> 00:36:53
So I separated out into one,
there's a separate one

571
00:36:53 --> 00:36:55
variable problem that
you can do for the base.

572
00:36:55 --> 00:36:57
And then this is a second
one variable problem

573
00:36:57 --> 00:36:58
for this other thing.
 

574
00:36:58 --> 00:37:01
And it's just two consecutive
one variable problems that

575
00:37:01 --> 00:37:03
solve the multivariable
problem.

576
00:37:03 --> 00:37:06
Or, as I say in multivariable
calculus, you can just

577
00:37:06 --> 00:37:08
do it all in one step.
 

578
00:37:08 --> 00:37:09
Yeah?
 

579
00:37:09 --> 00:37:11
STUDENT: [INAUDIBLE]
 

580
00:37:11 --> 00:37:15
PROFESSOR: Why did I divide
x by y, rather than

581
00:37:15 --> 00:37:17
y by x, or in any?
 

582
00:37:17 --> 00:37:20
So, again, what I was aiming
for was dimensionless

583
00:37:20 --> 00:37:22
quantities.
 

584
00:37:22 --> 00:37:26
So x and y are measured
in the same units.

585
00:37:26 --> 00:37:29
And also the proportions
of the box.

586
00:37:29 --> 00:37:34
So that's another word
for this is proportions.

587
00:37:34 --> 00:37:38
Are something that's universal,
independent of the volume v.

588
00:37:38 --> 00:37:42
It's something you can say
about any box, at any scale.

589
00:37:42 --> 00:37:46
Whether it be, you know,
something by Cristo

590
00:37:46 --> 00:37:48
in the Common.
 

591
00:37:48 --> 00:37:50
Maybe we'll get in here
to do some fancy --

592
00:37:50 --> 00:37:55
STUDENT: [INAUDIBLE]
 

593
00:37:55 --> 00:37:58
PROFESSOR: The proportions
is with geometric problems

594
00:37:58 --> 00:38:01
typically, when there's a
scaling to the problem.

595
00:38:01 --> 00:38:04
Where the answer is the
same at small scales

596
00:38:04 --> 00:38:05
and at large scales.
 

597
00:38:05 --> 00:38:07
This is capturing that.
 

598
00:38:07 --> 00:38:10
So that's why, the ratios
are what's capturing that.

599
00:38:10 --> 00:38:11
And that's why it's
aesthetically the

600
00:38:11 --> 00:38:13
nicest thing to ask.
 

601
00:38:13 --> 00:38:18
STUDENT: So, what exactly does
the ratio of the area to the

602
00:38:18 --> 00:38:20
volume ratio [INAUDIBLE]
 

603
00:38:20 --> 00:38:21
tell us?
 

604
00:38:21 --> 00:38:23
PROFESSOR: Unfortunately,
this number is a

605
00:38:23 --> 00:38:25
really obscure number.
 

606
00:38:25 --> 00:38:28
So the question is what
does this tell us.

607
00:38:28 --> 00:38:30
The only thing that I want
to emphasize is what's on

608
00:38:30 --> 00:38:31
the left-hand side here.
 

609
00:38:31 --> 00:38:35
Which is, it's the area to the
2/3 power of the volume, so

610
00:38:35 --> 00:38:38
it's a dimensionless quantity
that happens to be this.

611
00:38:38 --> 00:38:43
If you do this, for example, in
general with planar diagrams,

612
00:38:43 --> 00:38:47
circumferenced area is
a bad ratio to take.

613
00:38:47 --> 00:38:49
What you want to take
is the square of

614
00:38:49 --> 00:38:51
circumference to area.
 

615
00:38:51 --> 00:38:52
Because the square of
circumference has the same

616
00:38:52 --> 00:38:56
dimensions; that is, say,
inches squared to area.

617
00:38:56 --> 00:38:58
Which is in square inches.
 

618
00:38:58 --> 00:39:01
So, again, it's these
dimensionless quantities

619
00:39:01 --> 00:39:03
that you want to cook up.
 

620
00:39:03 --> 00:39:06
And those are the ones that
will have universal properties.

621
00:39:06 --> 00:39:11
The most famous of these is the
circle that encloses the most

622
00:39:11 --> 00:39:13
area for its circumference.
 

623
00:39:13 --> 00:39:17
And, again, that's only true
if you take the square

624
00:39:17 --> 00:39:18
of the circumference.
 

625
00:39:18 --> 00:39:25
You do the units correctly.
 

626
00:39:25 --> 00:39:26
Anyway.
 

627
00:39:26 --> 00:39:29
So we're here,
we've got a shape.

628
00:39:29 --> 00:39:31
We've got an answer
to this question.

629
00:39:31 --> 00:39:36
And I now want to
do this problem.

630
00:39:36 --> 00:39:38
Well, let's put it this way.
 

631
00:39:38 --> 00:39:40
I wanted to do this problem
by a different method.

632
00:39:40 --> 00:39:43
I think I'll take
the time to do it.

633
00:39:43 --> 00:39:46
So I want to do this
problem by a slightly

634
00:39:46 --> 00:39:48
different method here.
 

635
00:39:48 --> 00:39:59
So, here's Example 2 by
implicit differentiation.

636
00:39:59 --> 00:40:02
So the same example, but
now I'm going to do it by

637
00:40:02 --> 00:40:03
implicit differentiation.
 

638
00:40:03 --> 00:40:07
Well, I'll tell you the
advantages and disadvantages

639
00:40:07 --> 00:40:08
to this method here.
 

640
00:40:08 --> 00:40:20
So the situation is, you
have to start the same way.

641
00:40:20 --> 00:40:24
So here is the starting
place of the problem.

642
00:40:24 --> 00:40:34
And the goal was the minimum
of a with v constant.

643
00:40:34 --> 00:40:38
So this was the situation
that we were in.

644
00:40:38 --> 00:40:45
And now, what I want to do
is just differentiate.

645
00:40:45 --> 00:40:47
The function y is implicitly
a function of x, so I can

646
00:40:47 --> 00:40:54
differentiate the
first expression.

647
00:40:54 --> 00:41:00
And that yields 0 =
2xy + (x ^2 )( y').

648
00:41:00 --> 00:41:03
 


649
00:41:03 --> 00:41:07
So this is giving me my
implicit formula for y',

650
00:41:07 --> 00:41:13
So y' = - 2xy / x ^2.
 

651
00:41:13 --> 00:41:19
Or in other words, - 2y / x.
 

652
00:41:19 --> 00:41:22
And then I also have the dA/dx.
 

653
00:41:24 --> 00:41:28
Now, you may notice I'm not
using primes quite as much.

654
00:41:28 --> 00:41:32
Because all of the variables
are varying, and so here I'm

655
00:41:32 --> 00:41:34
emphasizing that it's a
differentiation with

656
00:41:34 --> 00:41:36
respect to the variable x.
 

657
00:41:36 --> 00:41:46
And this becomes
2x + 4y + 4xy'.

658
00:41:48 --> 00:41:53
So again, this is using
the product rule.

659
00:41:53 --> 00:41:57
And now I can plug in for what
y' is, which is right above it.

660
00:41:57 --> 00:42:09
So this is 2x + 4y
+ 4x ( - 2y / x).

661
00:42:09 --> 00:42:14
And that's equal to 0.
 

662
00:42:14 --> 00:42:24
And so let's gather
that together.

663
00:42:24 --> 00:42:25
So what do we have?
 

664
00:42:25 --> 00:42:36
We have 2x + 4y, and then,
altogether, this is 8 - 8y = 0.

665
00:42:36 --> 00:42:41
So that's the same
thing as 2x = 4y.

666
00:42:41 --> 00:42:44
The - 4y goes to
the other side.

667
00:42:44 --> 00:42:54
And so, x / y = 2.
 

668
00:42:54 --> 00:42:59
So this, I claim, so you have
to decide for yourself.

669
00:42:59 --> 00:43:03
But I claim that
this is faster.

670
00:43:03 --> 00:43:08
It's faster, and also it gets
to the heart of the matter,

671
00:43:08 --> 00:43:10
which is this scale in
variant proportions.

672
00:43:10 --> 00:43:13
Which is basically also nicer.
 

673
00:43:13 --> 00:43:16
So it gets to the
nicer answer, also.

674
00:43:16 --> 00:43:19
So those are the
advantages that this has.

675
00:43:19 --> 00:43:23
So it's faster, and it
gets to this, I'm going

676
00:43:23 --> 00:43:26
to call it nicer.
 

677
00:43:26 --> 00:43:41
And the disadvantage
is it did not check.

678
00:43:41 --> 00:43:59
Whether this critical point
is a max, min, or neither.

679
00:43:59 --> 00:44:02
So we didn't quite
finish the problem.

680
00:44:02 --> 00:44:10
But we got to the
answer very fast.

681
00:44:10 --> 00:44:11
Yeah, question.
 

682
00:44:11 --> 00:44:13
STUDENT: [INAUDIBLE]
 

683
00:44:13 --> 00:44:17
PROFESSOR: How would
you check it?

684
00:44:17 --> 00:44:18
STUDENT: [INAUDIBLE]
 

685
00:44:18 --> 00:44:20
PROFESSOR: Well, so it
gives you a candidate.

686
00:44:20 --> 00:44:23
The answer is - so the question
is, how would you check it?

687
00:44:23 --> 00:44:27
The answer is that for this
particular problem, the only

688
00:44:27 --> 00:44:31
way to do it is to do
something like this.

689
00:44:31 --> 00:44:34
So in other words, it doesn't
save you that much time.

690
00:44:34 --> 00:44:38
But with many, many, examples,
you actually can tell

691
00:44:38 --> 00:44:42
immediately that if the two
ends, the thing is, say, 0, and

692
00:44:42 --> 00:44:43
inside it's positive.
 

693
00:44:43 --> 00:44:44
Things like that.
 

694
00:44:44 --> 00:44:53
So in many, many, cases
this is just as good.

695
00:44:53 --> 00:44:58
So now I'm going to
change subjects here.

696
00:44:58 --> 00:45:03
But the subject that I'm going
to talk about next is almost,

697
00:45:03 --> 00:45:07
is very, very closely linked.
 

698
00:45:07 --> 00:45:10
Namely, I talked about
implicit differentiation.

699
00:45:10 --> 00:45:12
Now, we're going to just
talk about dealing with

700
00:45:12 --> 00:45:13
lots of variables.
 

701
00:45:13 --> 00:45:15
And rates of change.
 

702
00:45:15 --> 00:45:17
So, essentially, we're
going to talk about the

703
00:45:17 --> 00:45:19
same type of thing.
 

704
00:45:19 --> 00:45:23
So, I'm going to tell you
about a subject which is

705
00:45:23 --> 00:45:25
called related rates.
 

706
00:45:25 --> 00:45:28
Which is really just another
excuse for getting used to

707
00:45:28 --> 00:45:32
setting up variables
and equations.

708
00:45:32 --> 00:45:34
So, here we go.
 

709
00:45:34 --> 00:45:36
Related rates.
 

710
00:45:36 --> 00:45:40
And i'm going to illustrate
this with one example

711
00:45:40 --> 00:45:44
today, one tomorrow.
 

712
00:45:44 --> 00:45:47
So here's my example for today.
 

713
00:45:47 --> 00:45:50
So, again, this is going
to be a police problem.

714
00:45:50 --> 00:45:54
But this is going to be a word
problem and - sorry, I'm don't

715
00:45:54 --> 00:45:56
want to scare you, no police.
 

716
00:45:56 --> 00:45:59
Well, there are police in the
story but they're not present.

717
00:45:59 --> 00:46:05
So, but I'm going to draw it
immediately with the diagram

718
00:46:05 --> 00:46:08
because I'm going to
save us the trouble.

719
00:46:08 --> 00:46:11
Although, you know, the point
here is to get from the

720
00:46:11 --> 00:46:15
words to the diagram.
 

721
00:46:15 --> 00:46:21
So you have the police, and
they're 30 feet from the road.

722
00:46:21 --> 00:46:25
And here's the road.
 

723
00:46:25 --> 00:46:37
And you're coming along, here,
in your, let's see, in your car

724
00:46:37 --> 00:46:39
going in this direction here.
 

725
00:46:39 --> 00:46:43
And the police have radar.
 

726
00:46:43 --> 00:46:46
Which is bouncing
off of your car.

727
00:46:46 --> 00:46:53
And what they read off is
that you're 50 feet away.

728
00:46:53 --> 00:46:57
They also know that you're
approaching along the line

729
00:46:57 --> 00:47:12
of the radar at a rate
of 80 feet per second.

730
00:47:12 --> 00:47:19
Now, the question is,
are you speeding.

731
00:47:19 --> 00:47:20
That's the question.
 

732
00:47:20 --> 00:47:29
So when you're speeding, by the
way, up 95 feet per second

733
00:47:29 --> 00:47:32
is about 65 miles per hour.
 

734
00:47:32 --> 00:47:35
So that's the threshold here.
 

735
00:47:35 --> 00:47:41
So what I want to do now is
show you how you set up

736
00:47:41 --> 00:47:43
a problem like this.
 

737
00:47:43 --> 00:47:46
This distance is 50.
 

738
00:47:46 --> 00:47:50
This is 30, and because it's
the distance to a straight

739
00:47:50 --> 00:47:52
line you know that
this is a right angle.

740
00:47:52 --> 00:47:54
So we know that this
is a right triangle.

741
00:47:54 --> 00:47:58
And this is set out to be a
right triangle, which is an

742
00:47:58 --> 00:48:00
easy one, a 3, 4, 5 right
triangle just so that we can

743
00:48:00 --> 00:48:05
do the computations easily.
 

744
00:48:05 --> 00:48:09
So now, the question is, how do
we put the letters in to make

745
00:48:09 --> 00:48:12
his problem work, to figure out
what the rate of change is.

746
00:48:12 --> 00:48:15
So now, let me explain
that right now.

747
00:48:15 --> 00:48:18
And we will actually do the
computation next time.

748
00:48:18 --> 00:48:22
So the first thing is, you
have to understand what's

749
00:48:22 --> 00:48:24
changing and what's not.
 

750
00:48:24 --> 00:48:30
And we're going to use t
for time, in seconds.

751
00:48:30 --> 00:48:36
And now, an important distance
here is the distance to this

752
00:48:36 --> 00:48:37
foot of this perpendicular.
 

753
00:48:37 --> 00:48:41
So I'm going to name that x.
 

754
00:48:41 --> 00:48:42
I'm going to give
that letter x.

755
00:48:42 --> 00:48:44
Now, x is varying.
 

756
00:48:44 --> 00:48:47
The reason why I need a letter
for it as opposed to this 40 is

757
00:48:47 --> 00:48:50
that it's going to have a rate
of change with respect to t.

758
00:48:50 --> 00:48:55
And, in fact, it's related to,
the question is whether dx / dt

759
00:48:55 --> 00:49:00
is faster or slower than 95.
 

760
00:49:00 --> 00:49:01
So that's the thing
that's varying.

761
00:49:01 --> 00:49:04
Now, there's something
else that's varying.

762
00:49:04 --> 00:49:06
This distance here
is also varying.

763
00:49:06 --> 00:49:08
So we need a letter for that.
 

764
00:49:08 --> 00:49:11
We do not need a
letter for this.

765
00:49:11 --> 00:49:12
Because it's never changing.
 

766
00:49:12 --> 00:49:15
We're assuming the
police are parked.

767
00:49:15 --> 00:49:17
They're not ready to roar out
and catch you just yet, and

768
00:49:17 --> 00:49:19
they're certainly not in
motion when they've got the

769
00:49:19 --> 00:49:20
radar guns aimed at you.
 

770
00:49:20 --> 00:49:24
So you need to know something
about the sociology

771
00:49:24 --> 00:49:28
and style of police.
 

772
00:49:28 --> 00:49:30
So you need to know things
about the real world.

773
00:49:30 --> 00:49:36
Now, the last bit is,
what about this 80 here.

774
00:49:36 --> 00:49:37
So this is how fast
you're approaching.

775
00:49:37 --> 00:49:40
Now, that's measured
along the radar gun.

776
00:49:40 --> 00:49:44
I claim that that's d by
dt of this quantity here.

777
00:49:44 --> 00:49:46
So this is d is also changing.
 

778
00:49:46 --> 00:49:49
That's why we needed a
letter for it, too.

779
00:49:49 --> 00:49:51
So, next time, we'll just
put that all together

780
00:49:51 --> 00:49:55
and compute dx / dt.
 

781
00:49:55 --> 00:49:55