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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

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is that some of the
stuff that we've

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been doing with
derivatives can also

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be phrased in terms
of differentials.

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That for many purposes they
form a just a different language

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to compute the same things.

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So in particular one example of
that is linear approximation.

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

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but I'd like you to do it using
the method of differentials.

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So in particular I'd like you
to approximate the square root

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of 21.

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And the way I'd like
you to do that is

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by using a linear
approximation to the function

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f of x equals the square root
of 10x minus x squared based

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at the point 2, x_0 equals 2.

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Note that, you know,
why are we using

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this function to 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

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21 means approximate
the function value at 3.

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And so, you know, this function
has a nice value at x equals 2,

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so we can use linear
approximation at 2

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to try and compute
an approximation

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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

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to 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,

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we have this formula,
which is that f

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of x plus dx-- 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

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in terms of differentials.

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So in our case we have that
our-- that the point we want

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is x equals 2 and y equals 4.

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And so we have, we
know what dx is also,

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because we want the
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
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,

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we just use our straight-forward
rules for doing this.

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So we take a derivative and then
we 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

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squared to 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

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of the inside function,
which is 10 minus 2x dx.

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So really it's 10dx minus
2x*dx but I just, you know,

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pulled 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

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at this particular
point in question.

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So at this particular
point in question

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we have x equals 2
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, that's 16.

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To the minus 1/2, times--
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

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is 4, so 16 to the
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

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that we're after is just
f of 3, f of x plus dx

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is approximately
equal to our value y

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that we, at our original point,
which was 4, plus the dy.

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So plus 3/4.

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So the linear
approximation that we

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get from this method
of differentials

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is exactly the
same approximation

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that we would get if we
did this the other way.

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

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in this differential form.

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Somehow it seems a little more
straightforward sometimes.

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And OK, and so the answer that
we get out of it is 4 plus 3/4.

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All right, so that's that.