Welcome to a lesson on line integrals
of vector fields.
More specifically, we'll determine the work done
by a vector field.
To determine work in a straight line,
we can dot the force and displacement vectors.
However, if the path is not in a straight line,
we could break up the path into smaller and smaller pieces,
and the accumulation of these increments of work
give us an infinite limit of the change in F
and the change in D, which leads to the line integral
of the force field dotted with differential r
along the curve c.
And to integrate this in terms of t,
we can rewrite differential r as r prime of t dt,
then we can rewrite the vector field f
as a vector field in terms of t
using the x, y, and z components of
our vector value function r of t for the path
of the line integral.
And we can also express work as a line integral of
the force field dotted with the unit tangent vector
integrated with respects to s.
Where F dotted with T represents increment of force,
and ds represents a small change in arch length.
But for this video, we're going to use this form here.
Let's take a look at our first example.
We're going to determine the work done by the force field
represented by the vector field F of xy equals negative y
comma x along y equals x squared minus two
form the point zero negative two to the point two, two.
I've sketched our path of integration here.
Here's the point zero negative two,
and here's the point two, two
along the curve y equals x squared minus two.
So, our first step is to write our path of integration
as a vectored valued function r of t.
So, if we let x equal t
and since the equation of this path is
y equals x squared minus two.
We could let the y component equal t square minus two.
And the interval for t would be from zero to two.
Let's go and set up our def integral in terms of t.
The [mumbles] be from zero to two.
We'll rewrite our vector field as a function of t
using x of t and y of t.
So, the X component will be the opposite of y of t.
Well, here's y of t.
So instead t squared minus two it'll be two minus t squared.
For the y component is equal to x,
x of t is equal to t.
We'll dot this with our prime of t.
So, we'll have one and two t.
Let's go ahead and evaluate this on the next page.
So, we'll have two minus t squared times one.
That's two minus t squared
plus t times two t.
That'll be two t squared.
Let's go and combine our like terms.
We'll have two plus t squared.
So, we'll have two t plus 1/3 t cubed.
So, when t is two, we'll have four plus
this'll be 8/3.
And, then when t is zero these are both zero.
Well, four is equal to 12/3.
So, that'll be 20/3, so the work is 20/3.
Let's take a look at the given path in this force field.
So, here we see the force field
and the path we're taking is from this point here
to the point two, two.
Now, notice how the vector field is moving counter-clock
wise and the path is also moving counter-clock wise.
And, that's the reason why the work is positive.
The motion is in the same direction as the force field,
the work will be positive.
If it's in the opposite direction of the force field,
the work should be negative.
Now notice for this example our force was two-dimensional
and now we'll take a look at one that's three-dimensional.
So, it's the same question but now we have
a three-dimensional force field,
and our path is now a space curve rather than a plane curve.
The process will be exactly the same.
And, for this example we already have r of t
as well as the interval for t.
So, we can go ahead and set up our def integral
in terms of t.
[mumbles] integrals will be from zero to pi.
We rewrite the given force field as a function of t,
using x of t, y of t,
and z of t.
So, the x component is equal to x.
That'll be two cosine t.
The y component will be y, so that's still sine t,
and the z component is the opposite of z.
So that'll be negative t.
We'll dot this with our prime of t.
So, we'll have negative two sine t.
Cosine t and one.
So, I have negative four cosine t sine t.
Plus cosine t sine t
minus t.
Let's go and combine our like terms.
We have negative three cosine t sine t
minus t.
Let's go ahead and finish this on the next page.
And integrate this let's go ahead and let u equals cosine t.
So, differential u is going to be negative sine t dt.
If we do have a factor of negative sine t dt.
So, this just simplifies to three u.
So, we'll have three times u squared over two
where 3/2 cosine t squared.
Then we'll have minus t squared over two.
First, evaluate this at pi.
Well, cosine and pi is equal to negative one,
negative one squared is one.
So, we'll have 3/2 then minus pi squared over two.
Then when t is zero, cosine zero is one,
one squared it one times 3/2 and then minus zero.
So, the 3/2 simplify out,
and we're left with negative pi squared over two
as the work performed by this vector field.
Let's take a look at a graph of this.
So, here's our three-dimensional vector field in blue
our force field.
And then in red you can see the path through space.
So, as a particle moves along this path from here to here,
it's moving against the force field,
and that's why the work has a negative value this time.
That'll do it for this video.
Thank you for watching.