WEBVTT

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- [Voiceover] Part c, let y
equals f of x be the particular

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solution to the differential
equation with the

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initial condition f of
two is equal to three.

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Does f have a relative
minimum, a relative maximum,

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or neither at x equals two?

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Justify your answer.

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Well, to think about
whether we have a realtive

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minimum or relative maximum, we can say,

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well, what's the derivative at that point?

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If it's zero, then it's a good candidate

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that we're dealing with a,

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that it could be a relative
minimum or maximum,

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if it's not zero, then it's neither.

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And then if it is zero,
if we wanna figure out

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relative minimum or relative maximum,

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we can evaluate the sign
of the second derivative.

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So let's just think about this.

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So, we want to evaluate f prime,

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we wanna figure out what f prime of,

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f prime of two is equal to.

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So, we know that f prime of x,

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f prime of x, which is
the same thing as dy dx,

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is equal to two times x minus y.

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We saw that in the last problem.

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And so f prime of two,
I'll write it this way,

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f prime of two is going to be equal to

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two times two,

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two times two minus

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whatever y is when x is equal to two.

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Well do we know what y
is when x equals to two?

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Sure, they tell us right over here.

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Y is equal to f of x,

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so when x is equal to two,

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when x is equal to two,

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y is equal to three,

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so two times two minus three.

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And so this is going to be
equal to four minus three

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is equal to one.

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And so since the derivative
at two is not zero,

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this is not going to be a
minimum, a relative minimum

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or a relative maximum, so you could say

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since, since f prime of two,

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f prime of two does not equal zero,

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this, we have a, f has,
let me write it this way

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f has neither, neither minimum,

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or relative minimum I guess I could say,

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relative min or

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relative max

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at x equals two.

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Alright, let's do the next one.

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Find the values of the constants m and b,

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for which y equals m
x plus b is a solution

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to the differential equation.

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Alright, this one is interesting.

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So let's, actually let's
just write down everything we

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know before we even think
that y equals m x plus b

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could be a solution to
the differential equation.

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So, we know that, we know that dy

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over dx is equal to two x minus y,

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they told us that.

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We also know that the second derivative,

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the second derivative of y,
with respect to x, is equal to

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two minus dy dx. We
figured that out in part b

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of this problem.

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And then, we could also express this

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we saw that we could also write that

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as two minus two x plus
y, if you just substitute,

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if you substitute this in for that.

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So it's two minus two x plus y.

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So let me write it that way.

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This is also equal to
two minus two x plus y.

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So that's everything that we know

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before we even thought that
maybe there's a solution

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of y equals m x plus b.

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So now let's start with
y equals m x plus b.

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So if y is equal to m x
plus b, y equals m x plus b,

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so this is the equation of a line,

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then dy dx is going to be equal to,

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well the derivative of this
with respect to x is just m,

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the derivative of this with
respect to x, this is constant

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this is not going to
change with respect to x,

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it's just zero.

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And that makes sense,

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the rate of change of y with
respect to x is the slope,

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is the slope of our line.

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So, can we use, and this
is really all that we know,

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we could keep, actually
we could go even further,

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we could take the second derivative here,

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the second derivative
of y with respect to x.

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Well, that's going to be zero.

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The second derivative of
a, of a linear function,

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well it's going to be
zero, you see that here.

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So this is, this is all of
the information that we have.

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We get this from the previous
parts of the problem,

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and we get this just taking the
first and second derivatives

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of, of y equals m x plus b.

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So given this, can we figure out,

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can we figure out what m and b are?

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Alright, so we could,
if we said m is equal to

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two x minus y, that doesn't seem right.

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This one is a tricky one.

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Well, let's see, we know
that the second derivative

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is going to be equal to zero.

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We know that this is
going to be equal to zero

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for this particular solution.

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And we know dy dx is equal to m.

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We know this is m.

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And so there you have it,
we have enough information

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to solve for m.

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We know that zero is equal to two minus m.

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So zero is equal to two minus m.

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And so we can add m to
both sides and we get

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m is equal to, m is equal to two.

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So, that by itself, was quite useful.

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And then, what we could say, let's see,

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can we solve this further?

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Well we know that this,
right over here, dy dx,

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this is m, this is m.

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And it's equal to two.

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So, we could say that two
is equal to 2 x minus y.

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Two is equal to 2 x minus y.

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And then let's see, if we solve for y,

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add y to both sides,
subtract two from both sides,

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we get y is equal to two x minus two.

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And there we have our whole solution.

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And so you have your m, right over there.

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That is m.

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And then we also have our b.

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This one was a tricky one.

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Anytime that you, you have to, ya know,

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do something like this and it
doesn't just jump out at you,

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and if it wasn't obvious, it
didn't jump out at me at first,

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when I looked at this problem,
I said, well let me just

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write down everything that they told us,

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so they wrote this before,

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and then we say okay this
is going to be a solution.

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And so let me see if I
can somehow solve, so,

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let's see what I didn't use.

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I didn't use, I didn't use that.

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I did use this.

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I absolutely used that.

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I did use that, I did use
that, and I did use that.

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So this was a little bit
of a fun little puzzle

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where I just wrote down all
the information they gave us

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and I tried to figure out, based on that,

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whether I could figure out m and m and b.

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And this is pretty neat,
that this a solution

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two x minus two.

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If we go to our slope
field above, it's not,

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it wouldn't have jumped out at me.

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But if you think about, if you think about

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so two x minus two, it's
y intercept would be

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negative two like that, let me
do this in a different color,

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and so the line would look something,

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would like something like this.

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The line would look something like this.

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And you can verify that
any one of these points,

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at any one of these points, the slope,

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the slope is equal, the
slope is equal to two.

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If we're at the point two comma two,

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well it's gonna be two
times two minus two is two.

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One comma zero, two times
one minus zero is two.

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Negative two comma, or sorry
zero comma negative two

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well, zero minus negative two, that's two.

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So you see this is
pretty neat, the slope is

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changing all around it, but this is

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this is a linear solution

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to that original differential
equation, that was,

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that was pretty cool.