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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like us to
work on the following problem

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that has to do with tangent
planes and approximations.

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So we know that a rectangle--
we'll say a rectangle has

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sides x and y, and we know that
to find the area of the

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rectangle then, we just take x
times y, and I would like us

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to approximate the area
for x equal to 2.1

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and y equal to 2.8.

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And obviously, with this type of
equation, it's not hard to

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just compute this, but I'd like
us to use the tangent

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plane approximation to determine
the value, and then

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we'll compare it to the actual
value, just to give us an idea

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of how we can use the tangent
plane to approximate things.

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And obviously, I'd like to
do this near x equal

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2 and y equal 3.

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So when you're doing your
tangent plane approximation,

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do the approximation at x
equal 2 and y equal 3.

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And then when you're done with
that, I would like you to

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answer this question.

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So near x equal 2 and
y equal 3, which

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has the greater effect?

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A change in x or an
equal change in y?

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So if I change by x in the same
value as I change by y,

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which will have the
greater impact?

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So why don't you pause the tape,
work on the problems,

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and when you're ready to see how
I do them, bring it back

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up, and I'll come back
and show you.

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OK, welcome back.

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So we're going to work on this
problem, and as I mentioned,

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the first thing we want to
do is do a tangent plane

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approximation near x equal
2 and y equal 3.

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So let me write down what we're
going to need in order

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to do that.

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First, we'll remind ourselves
that in this case, what I'm

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going to approximate is the area
function for a rectangle,

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which is A of x, y is
equal to x times y.

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So that's the function I'm going
to be approximating.

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And now the actual
approximation, the tangent

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plane approximation, has
the following form.

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So we know A of x, y
is approximately--

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well, it's going to be the area
evaluated at the point

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I'm interested in, which we
said was 2 comma 3, right,

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plus the x-derivative of area
evaluated at 2 comma 3 times

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the change in x.

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And the change in x is where
x is now versus where we

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started, which is
at x equal 2.

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And then the same thing
in terms of y.

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So you do the y-derivative
evaluated at 2 comma 3, and

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then you do the change in y.

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And since we started at y equals
3, the change in y is y

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minus 3, OK?

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So now we obviously need to fill
in three quantities here.

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The three things we need to
evaluate: area evaluated at 2,

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3, its x-derivative evaluated at
2, 3, and its y-derivative

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evaluated at 2, 3.

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So let me just point
out what we have.

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Area evaluated at 2, 3, well,
that's just 2 times 3, which

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we can do, so that's 6.

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That's pretty easy.

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Now A sub x, so the derivative
of the area function with

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respect to x.

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The derivative of the area
function with respect

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to x is just y.

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y in that case we treat
as a constant.

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We take this derivative.

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We just get y back.

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And so, A sub x evaluated at 2,
3 is going to be y, which y

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here is 3, so we get A sub
x is equal to 3 at 2, 3.

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In a similar vein, we can
immediately look and see that

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A sub y evaluated at 2,
3 is going to be 2.

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And the reason for that, of
course, is if we look back

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here, the derivative of A with
respect to y is x, so

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evaluating it at 2,
3 gives us 2.

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OK.

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So now we just have to
fill everything in.

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So my tangent plane
approximation now says I get 6

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plus 3 times x minus 2 plus
2 times y minus 3.

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And what I wanted was the
area at 2.1 comma 2.8.

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So if I fill in those values for
x and y-- so 2.1 is x and

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2.8 is y-- if I fill those in,
I get that this is equal to--

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well, I'll keep writing

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approximately just to be safe--

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is 6 plus 3 times--

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well, 2.1 minus 2 gives me a
0.1 there and 2 times 2.8

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minus 3 gives me a negative 0.2
there, so I get a negative

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0.4, I get a positive 0.3,
so together this

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is a negative 0.1.

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6 minus 0.1 gives me 5.9.

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So the area based on
the approximation

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is 5.9 square units.

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And, in actuality, if you
multiply it out, I think you

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get something like 5.88, so
the approximation is very

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good, right?

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We're close to 2, 3 and that is
one of the reasons we can

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know it's going to be very,
it should be very good.

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It should be pretty good, OK?

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Now, I just have to answer the
second part of the question.

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So let me remind us what
the second part was.

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It was near x equal 2
and y equal 3, which

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has a greater effect?

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A change in x or an
equal change in y?

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And to do that, it's really
easiest to come back and look

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at maybe this line.

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I'll underline this
line right here.

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Actually, I'll box it, OK?

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This value will represent
the change in x.

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So we started at 2, and we
go somewhere, and that'll

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represent the change in x.

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This value represents
the change in y.

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So if we look at which has a
greater impact near 2, 3, if

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my change in x and my change in
y are equal, then obviously

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this term has a bigger impact,
because there's a 3.

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The coefficient here is 3 and
the coefficient here is 2.

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And so the point is, changes
in x will have a greater

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effect than changes in y.

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And what this corresponds to
pictorially is if we make a

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slice where we keep y constant
and we look at the curve in

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the xz-plane, that corresponds
to the fact that the

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derivative in the xz-plane of
the curve there is more

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significant-- the curve there
has a steeper derivative--

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than if I kept the x-value
fixed and I looked in the

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yz-plane and I looked
at the curve there.

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So the sort of one-dimensional
picture is that the derivative

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of a curve in the x-direction,
keeping y fixed, is steeper

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than the derivative of the
curve in the y-direction,

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keeping x fixed.

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So maybe hopefully that wasn't
more confusing than

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it should have been.

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From here, you can see it right
away, but that's sort of

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the picture of it as well.

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So I think that's
where I'll stop.

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