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PROFESSOR: Hi.

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Welcome back to recitation.

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You've been talking about
differentials.

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And one of the things that
you've shown is that some of

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the stuff that we've been doing
with derivatives can

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also be phrased in terms
of differentials.

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That for many purposes they
form a just a different

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language to compute
the same things.

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So in particular one example
of that is linear

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approximation.

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So I have a problem here that's
a linear approximation

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problem, but I'd like you to
do it using the method of

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differentials.

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So in particular I'd like
you to approximate the

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square root of 21.

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And the way I'd like you to do
that is by using a linear

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approximation to the function f
of x equals the square root

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of 10x minus x squared based at
the point 2, x0 equals 2.

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Note that, you know, why are
we using this function to

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approximate square root of 21?

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Well because at 3, the function
value at 3, f of 3,

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is exactly equal to the
square root of 21.

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So to approximate the square
root 21 means approximate the

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function value at 3.

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And so, you know, this function
has a nice value at x

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equals 2, so we can use linear
approximation at 2 to try and

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compute an approximation to the
function at x equals 3.

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So why don't you take a couple
of minutes, work this out,

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come back and we can work
it out together.

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All right, welcome back.

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So we're doing linear
approximation here.

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And we're doing it with
differentials.

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So, in general, when we use the
method of differentials to

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compute a linear approximation,
what we have is

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that if y equals f of x, so if y
is given as a function of x,

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then we have that so if we want
to change x a little bit

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and figure out what the change
in y is, we have this formula,

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which is that f of x plus dx--

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so the function value
at a nearby point--

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is approximately equal
to y plus dy.

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So this is our formula for
linear approximation in terms

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of differentials.

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So in our case we have that our,
that the point we want is

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x equals 2 and y equals 4.

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And so we have, we know was dx
is also, because we want the

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approximation at the point 3.

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So we want dx to be equal to
1 for this approximation.

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And so the question is, what
is dy going to be?

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So that's the value
that we're after.

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That tells us how much the
function value changes.

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So to compute dy,
well dy is the

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differential of the function.

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So this is d of the
square root of 10x

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minus x squared, quantity.

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OK, so to compute this
differential now, we just use

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our straight-forward rules
for doing this.

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So we take a derivative
and then we

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have dx's where necessary.

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So this is d of square root
of 10x minus x squared.

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So the outermost function is
the square root function.

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So this is going to be 1/2 times
10x minus x squared to

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the minus 1/2.

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So that's 1 over the square root
of 10x minus x squared.

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And now I need to multiply by
the derivative of the inside

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function, which is
10 minus 2x dx.

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So really it's 10dx minus 2xdx
but I just, you know, pulled

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that dx out in front.

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So we have a differential
equal to a differential.

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OK, and so, but we want this
value at this particular point

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in question.

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So at this particular point in
question we have x equals 2

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and dx is equal to 1.

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So the value of dy that we want
is equal to 1/2 times--

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well this is again, OK, 10
times 2 minus 2 squared.

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That's 16 to the minus
1/2 times--

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so here x equals 2.

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10 minus 4 is 6.

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Times dx is just this 1 that
we're interested in.

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OK so 16 to the minus 1/2, 16 to
the 1/2 is 4, so 16 to the

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minus 1/2 is 1/4.

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So this is 1/2 times
1/4 times 6.

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So that's 6/8, which is 3/4.

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So our dy is 3/4.

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So that means our linear
approximation that we're after

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is just f of 3, f of x plus dx
is approximately equal to our

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value y that we, at our original
point, which was 4

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plus the dy.

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So plus 3/4.

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So the linear approximation that
we get from this method

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of differentials is exactly the
same approximation that we

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would get if we did this
the other way.

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But I think it's kind of
nice to do it in this

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differential form.

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Somehow it seems a little more
straightforward sometimes.

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And OK, andso the answer
that we get out of

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it is 4 plus 3/4.

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All right, so that's that.

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