WEBVTT
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DAVID JORDAN: Hello,
welcome back to recitation.
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The problem I'd like to work
on with you now is a long one.
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So it's going to be practice
computing line integrals.
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So to begin with, we have this
function of two variables.
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f is x to the fifth
plus 3x y cubed.
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And we have this-- C is
the upper semi-circle going
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from (1, 0) to (-1, 0).
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So it's this upper semi-circle
here that we often consider.
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And so the first
thing that we want
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to do is to just compute
the gradient, capital F,
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to be the gradient
of this function f.
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And then parts b
through d, we're
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going to compute this line
integral of this vector field
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f along this curve C.
We're going to compute it
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in three different ways.
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So first of all, we're going
to compute it directly,
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just using the definition.
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And then in Part c, we're going
to compute it using the path
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independence of line
integrals and we're
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going to replace the path
C with a simpler path.
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And then finally
in Part d, we're
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going to use the fundamental
theorem of line integrals.
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Now when you do Part b,
what I want you to do
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is set up the integral.
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You're going to get a very
complicated integral that I
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wouldn't want to try to compute.
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So just set up the integral
completely and then go ahead
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and move on to Part c and d.
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So why don't you pause the
video and work on that,
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and we'll check back
in a few minutes
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and we'll solve it together.
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Welcome back.
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I hope you had some luck
working these problems.
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So let's do the easy one
first, computing the gradient.
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So for the gradient we just take
the two partial derivatives.
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So we get 5 x to the
fourth plus 3 y cubed.
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That's the partial derivative
in the x-direction,
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and in the y-direction,
we just get 9x y squared.
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So now for Part b, we're
asked to compute this integral
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directly.
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So we have to recall
what it means.
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So first of all, if
we go back over here,
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we have this curve
C. And we need
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to give a
parameterization for it,
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and so we're going to
introduce a parameterization
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r of a variable
t, and we're going
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to use that to do
our computations.
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So let's set r of t-- so
this is our usual circle
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that we're used to working
with, so we're just
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going to take the usual
parameterization, cos t and sin
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t.
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And what's important is that
the range is going to be from t
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equals 0 to t equals pi.
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It's t equals pi
because we don't
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want to go all the
way around the circle.
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We just want to
go halfway around
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until we get to negative 1.
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So if that is r of t, then we
can compute the differential
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dr of t.
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And so it's going to be
just taking the derivative.
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So we have negative
sin t and cos t dt.
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And so now we can just write
out this line integral directly.
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So the integral over
C of F dot dr just
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becomes-- so we have the
integral from t equals 0 to pi.
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Those are the ranges
for our curve.
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And now we're going to take the
dot product of F, which was 5 x
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to the fourth plus 3
y cubed, 9x y squared.
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We're just going to dot this
with our dr vector, which
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is minus sin t, cos t.
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Altogether we have dt.
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And so now, notice
that here we've
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got the variables x and y, and
here we've got the variables t,
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but because of our
parameterization,
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we actually know
that, for instance, x
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is cos t and y is sin t.
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So we can write this all out.
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So we have 5 cos to the
fourth t plus 3 sine cubed t.
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So that's this guy
written out in terms of t.
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And then we multiply it
by a negative sine t.
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And then to that we add
the other component.
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So we have plus a 9.
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So we have cos t coming from
the x and another cos t here.
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So we have cos squared t, and
we have a sine squared t dt.
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OK, so that's what it
means to compute this line
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integral directly,
and it's not something
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that I look forward to doing.
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So let's see if we can use
path independence to make
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our lives a little bit simpler.
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So that's going to be c.
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So what we want to do is we want
to replace our original curve
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C with any other curve that
has the same starting point
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and the same ending point.
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And the curve that
I would like to use
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is just a straight
line connecting them.
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There's lots of different
choices that you could do,
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but to me this one
seems the most natural.
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So let's give that a try.
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So let's let r of t be
the curve negative t, 0.
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Negative t because we want
it to run moving to the left.
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And then our range is just
going to be from minus 1 to 1.
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So when t is minus 1, then
we get minus the negative 1
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and it starts at 1.
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And when t is 1 it
goes to negative 1.
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And notice that it goes
along the y equals 0 axis.
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So now we can do
the same computation
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that we did before but
we can use this curve.
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So the thing that
I want to emphasize
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is that if we're
computing a line
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integral of a
gradient function--
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so of a function which
is conservative--
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then we can use any line
and we can use any path that
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connects the two end points.
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We can replace our path.
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And so that's what we did.
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We replaced C_1 with C_2.
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So now this becomes
much easier in two ways.
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So we'll see.
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So our range now is just
t goes from minus 1 to 1.
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And so dr here is
just minus 1, 0.
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That's dr. And there's a dt.
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And let's see.
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So now F, we had
this value for F,
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but notice that the y-coordinate
is always 0 along this curve.
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So the y-coordinate
being 0 means
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that we just have 5 t to
the fourth and then 0 here.
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That's it, because we set
y to be 0 along this curve.
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So altogether this is a
very nice integral to do.
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So just taking this
dot product, all we
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have is minus 5 t
to the fourth dt.
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That's simplified greatly.
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And we just have minus t to
the fifth from 1 to minus 1.
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And so we get simply minus 2.
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So that was a much, much
more straightforward integral
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to do than the one
we started with.
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Now finally, in
d, we're suggested
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to use the fundamental
theorem of line integrals.
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So let's remember
what that says.
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That says that if we have any
curve and the line integral
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that we're taking-- if we know
that we're taking the line
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integral not of
any vector field,
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but of a vector field which is
already the gradient of f, then
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that tells us that
this is simply
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f of the endpoint minus f of
the starting point of our curve.
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So really we don't need
to do any integral at all.
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And so let's see.
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So recall that f was x to
the fifth plus 3x y cubed.
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And so the endpoint-- so we
just need to take f of (-1, 0)
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and subtract f of (1, 0).
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And so plugging this all
in together we get minus 1
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minus a positive 1.
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Altogether we get minus 2.
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And, of course, this
does agree with what
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we did when we computed
using the line integrals.