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So far we've defined
work in one dimension.

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But let's consider
the path of a particle

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in more than one dimension.

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And on this screen it looks
like a two-dimensional path,

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and here's our particle.

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And what we'd like to do is
define the concept of work

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for this type of path.

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Now let's say that
at this instant

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our particle is
experiencing a force F.

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And the particle is being
displaced by a distance, ds,

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which is always
tangent to the path.

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And what we'd like to do
now is talk about the work

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done for this particle.

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Now what we'd like
to first say is

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what we're going to do
is imagine that we're

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going to break the
path down into a bunch

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of small intervals.

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And let's call
this the j-th point

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and the j-th plus 1 point.

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Then our particle
will start at j

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and we'll displace it
delta sj to the j-th

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plus 1 position on the path.

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And here, let's denote
the force by Fj.

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And now what we'd like
to do is define the work

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for this particular section.

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Now our concept of
work is the force

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in the direction of
the displacement.

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So what I'd like to do is I'd
like to define the quantity Fj,

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and I put a little parentheses
upstairs-- parallel lines

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upstairs.

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This is the component
of the force Fj

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in the direction of delta sj.

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Now the way we denote
that is we can then

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say that if I take the dot
product of Fj dot delta sj,

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this is the component
of the force

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in the direction of the motion.

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Because remember,
our dot product,

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if we call this
angle theta j, is

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taking how much of one
vector is in the direction

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of the other times the
magnitude of that displacement.

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And again, when we write that
component in our notation,

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if we call this
angle theta j, we

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know that that's the magnitude
of Fj times sine of-- cosine

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of theta j.

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Now because this is
the amount of force

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in the direction of
the motion, that's

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how we've defined work done.

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And we'll symbolize
the work done

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in taking the particle from j
to j plus 1 by the symbol Wj.

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And now, this is
a scalar quantity.

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And when we want to find
the total work, what

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we have to add up is
from some initial point

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to some final
point, how much work

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is being done in
taking our particle

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from the initial position
to the final position.

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And it's just the sum for all
of these j's, and let's say

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we have n of them of Wj.

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And right now we're
going to call this W,

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it's just that scalar
sum, And, that's

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equal to the sum j goes from
1 to n of Fj dot delta sj.

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So now we'll use the
vector dot product

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because we already know that
this is how much of the force

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is in the direction
of the motion.

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However, this answer depends on
how fine we broke up this path.

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And what we've
seen many times now

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is that the actual work that
we want, let's write this as n

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because it depends on
the number of paths,

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is the limit as n goes to
infinity of this sum of scalar

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amounts of work.

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Now when we take that limit,
as n goes to infinity,

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of this sum, j equals 1
to n of Fj dot delta sj,

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this quantity is formerly what
we mean by a line integral.

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And we denote that
with a new notation.

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We'll denote a line from
our initial to final

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and it's a dot product
of the force dotted

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into the small displacement.

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Now whenever we see the
dot product in an integral

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that distinguishes it
from a normal integral,

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and this is what we
call a line integral.

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Now again, why did we switch
from our delta s's to ds's?

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Because when we take the
limit, as n goes to infinity,

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our grid becomes
finer, and finer,

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and finer until we're
shrinking down to a point

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where we can just talk about the
differential displacement, ds,

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and the force acting on a point.

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And that's what our
line integral is.

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Now, what we'll
learn next is how

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to calculate this line
integral for a number

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of important physical
examples that occur

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in our interest in mechanics.