WEBVTT
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PROFESSOR: Hi.
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Welcome back to recitation.
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Last time in lecture
you started learning
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about implicit differentiation.
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And you saw some examples of
how implicit differentiation can
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be used to compute
derivatives of functions
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defined implicitly.
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So let's do another
example today.
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So here I have a curve that's
defined by the implicit
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equation y cubed plus
x cubed equals 3x*y.
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And I'd like to
know what the slope
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of the tangent
line to that curve
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is at the point (4/3, 2/3).
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So before we start
doing anything,
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let me just make a
couple of observations.
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If you don't believe me,
that the point (4/3, 2/3)
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is on this curve, you can always
check by plugging the values in
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and confirm that really, yes,
4/3 cubed plus 2/3 cubed is
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equal to 3 times 2/3 times 4/3.
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So it's-- how I found this
point is maybe a little magical.
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Because as you can
see, this equation
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is really a tough
one to solve for y.
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What you sort of--
natural thing to want
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to do when asked this
question is to solve for y
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and get an equation for
y in terms of x and then
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take the derivative using the
various differentiation rules
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that you've learned.
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But here, this is-- I'll
let you in a secret.
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There is a way to do this.
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But it's really hard.
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It's really ugly and it's
beyond the scope of this course.
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So really we're much
better off treating
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this as an implicit
differentiation problem
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than as an explicit
differentiation problem.
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So having said that, why don't
you take a minute or two.
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Try and have a go
at this yourself.
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And then we'll come back and
work through it together.
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All right, so welcome back.
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We were in the middle-- we were
just about to start, actually--
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solving this problem, computing
the slope of the tangent line
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to the curve y cubed plus
x cubed equals 3x*y at this
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point, (4/3, 2/3).
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So the slope of the tangent
line is the value y prime
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of x at that point.
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So we need to answer
this question.
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What we need to do is we need
to find the derivative of y.
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But as I said earlier, this
is tough to do explicitly,
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to find y in terms of
x, so we're going to use
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implicit differentiation.
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So, so we start
with this equation,
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y cubed plus x cubed equals 3x*y
and we can take a derivative
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with respect to x.
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So some parts--
all right, so let's
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start with it in the
order it's given.
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So you have y cubed.
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If you take a derivative of y
cubed with respect to x, what
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you need to use the chain
rule because y is implicitly
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a function of x and so y
cubed is the chain rule.
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It's the cubed function
applied to the y function.
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And this is true of implicit
differentiation in general.
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That the reason
that we can do this,
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the really fundamental
reason this works
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is that we have the chain rule
and that it lets us evaluate
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derivatives of compositions.
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So in our case we have,
so we take a derivative
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of the whole thing, of this
whole I'm going to write,
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this is a little
sloppy notation,
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but I hope you know what I mean.
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We have this
identity and so we're
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going to take a derivative
of the whole thing.
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And so the first
part on the left, we
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get the derivative of y cubed.
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So by the chain
rule, so we first
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take the derivative of
the cube function at y
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and then multiply by
the derivative of y.
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So this is the
derivative of y cubed.
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It just gives us 3 y squared.
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So that's what happens when you
just deal with the cubed part.
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But then we need to multiply
by-- in the chain rule--
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by the derivative of the inside.
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Which in this
context is dy by dx.
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OK.
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Plus the derivative of x cubed.
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That's more straightforward.
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Nothing really
complicated going on here.
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We've seen this for
a little while now.
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It's just 3x squared equal--
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OK.
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So on the right now, we don't
actually have a chain rule,
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we have a product
rule situation here.
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We have 3 times x times y.
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So 3 is just a constant.
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And so we could just
pull it out in front.
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So we take the
derivative of x*y.
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So we take the derivative of
the first times the second
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plus the derivative of the
second times the first.
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So the derivative of
the first is just--
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ah sorry, x is the first,
so its derivative is 1.
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So we got 3 times
the second is y.
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Plus-- OK, so we
take the first times
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the derivative of the
second, which is dy by dx.
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So because this is
an identity it holds
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for all values of x and y.
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This equality follows
just by taking
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the derivative of both sides.
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Good.
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So now the thing
we want is that we
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want the slope of the tangent
line at a particular point.
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So we want to isolate dy/dx.
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That's the thing
we're trying to find.
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So here, if you're only
interested in dy/dx,
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this is actually a linear
equation in some sense, right?
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We have dy/dx, a
constant, something--
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or, it's not a constant--
something times dy/dx
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plus something equals something
plus something times dy/dx.
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There are no squares of
dy/dx is what I really mean.
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So OK.
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So that that's nice.
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It makes it relatively
easier to solve,
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so we can just combine all
the terms with dy/dx in them.
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Let's say we'll combine maybe,
put them over here and put
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everything else over there.
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So over here we
get, so dy/dx, so we
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have a 3 y squared minus a 3x.
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And on the other side we have
a 3y minus a 3 x squared.
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And so this is times,
multiplication there.
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And so we want dy by
dx just by itself.
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So we can just divide through
by 3 y squared minus 3x.
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So then we have dy/dx
is equal to-- well,
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all right, so there
are a lot of 3's here.
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There's a constant
multiple of 3 on this side,
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a constant multiple
of 3 on this side.
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Those are going to cancel.
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So this is y minus x squared
over y squared minus x.
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OK, so at any point
(x, y) on this curve,
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the slope of the tangent line is
given by this expression here.
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And we're interested
in a particular point
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in this problem.
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We're interested in the
point 4/3 comma 2/3.
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So at let me take that back up
here so at the point 4/3 comma
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2/3 we have dy by dx.
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So OK, we just we just
plug that value of y
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and that value of x into
this formula that we've got.
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So that's 2/3 minus 4/3
squared is 16/9 over-- well,
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let's see, 2/3 squared
is 4/9 minus 4/3.
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All right, so we have a
little bit of rational number
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arithmetic here.
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Maybe I'll multiply top
and bottom through by 9
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to get 6 minus 16
over 4 minus 12.
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So this is negative 10 over
negative 8, which is 5 over 4.
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And if we go back to
the picture that I drew,
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it actually looks pretty
reasonable over here, right?
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This slope of this
tangent line is actually
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a little bit bigger than 1.
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Great.
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So that's that.