Welcome to part one
of the fundamental theorem of line integrals.
The fundamental theorem of line integrals
as stated here, states that a vector field is conservative
then the line integral between any two points
forming curve c is the difference in the values
of the potential function, little f at these two points,
regardless of the path.
So the line integral would be path independent.
This is also sometimes called the gradient theorem.
So now, to evaluate a line integral,
we have a choice of methods.
This first method here, requires us to pre-mor-tahyz
the curve in terms of t, rewrite the vector field
in terms of t, dot it with our prime of t,
integrate with respects to t.
And this method does not require the potential function.
However if c is conservative, and we can easily determine
the potential function, we could take advantage
of the fundamental theorem of line integrals.
The third option is, if f is conservative,
we could use method one but use a much simpler path
because we know that if f is conservative,
the path would be independent.
So in this video, we'll take a look at one example
and evaluate a line integral using all three
of these methods.
And we'll start by using the fundamental theorem
of line integrals.
So first, we'll check to make sure this is
a conservative vector field.
Since this is a vector in our two,
we have to check to make sure that the partial derivative
of g, with respects to x is equal to partial derivative
of f with respects to y, where f would be xy squared
and g would be x squared y.
Well the partial derivative of g with respect to x
would be two xy and the partial derivative of f
with respect to y would also be two xy.
So this a conservative field.
So the next step is to determine the potential function
so we can determine the value of this line integral.
Since f is conservative, we know that the vector field f
is equal to the gradient of little f.
So this tells us the partial derivative of f
with respect to x would be xy squared
and the partial derivative of with respect to y
would be x squared y.
So now we'll integrate xy squared with respects to x
and then integrate x squared y with respects to y
and by comparing these two antiderivatives,
we can determine the potential function.
So here we're going to have 1/2 x squared y squared
plus a function of y and integral to respect to y,
we're going to have the same thing plus a possible function
of x but since these are equal to each other,
we can conclude that the potential function,
f of x, y would be equal to 1/2 x squared y squared
plus the constant c.
So we can use this function to evaluate the line integral.
So let's do that on the next page.
So before we evaluate this, let's take a look at our path.
We start at the origin or the point zero, zero
and we finish at the point three, three
and these are the only two points we need
to determine the value of this line integral.
This is going to be equal to f of three, three
minus f of zero, zero.
Replacing x and y with three, we'll have 1/2
times three squared times three squared
minus and when x and y are zero, this should be zero.
So we have nine times nine, it's 81, times 1/2 or 81/2.
So using the fundamental theorem of line integrals,
we were able to determine the value of this line integral
using the potential function.
So we just used method two.
And now what we're going to do is use method three
where we know F is conservative but now we can define
a simpler path and then use method one
where we pre-mor-tahyz everything in terms of t.
So looking at our path here, the simpler path
would just be the line from the origin
to the point three, three.
So we'll use this path instead of the blue path.
And knowing the equation is line would y is equal x
is going to be helpful.
So if we let x equal t and y equal t,
then t would be on the close interval from zero to three.
Now we can rewrite our line integral in terms of t,
t is going to go from zero to three,
f written in terms of t would be t cubed and t cubed
and our prime of t would be one, one,
so we're going to have t cubed plus t cubed or two t cubed.
Integrating with respect to t, we'd have two
times t the fourth over four
and that will give us 1/2 t to the fourth.
So when t is three, we'll have 1/2
times three to the fourth, which of course, will be 81/2.
Here, we didn't use the fundamental theorem
of line integrals but we did take advantage
of knowing that the fundamental was conservative
and therefore path independent
meaning we could redefine the path.
Now let's take a look at the same problem using
the original technique for integrating line integrals.
So we have two paths to consider, path one and path two.
The equation of curve one would one y equals 3/2 x
and the equation for curve two would be y equals three.
So let's go ahead and pre-mor-tahyz our curve in terms of t.
So for curve one if you let x equal t
and y equal 3/2 t and t would be on the close interval
from zero to two.
And for the second curve, x starts at two and goes to three
so if we let x equal t, an interval for t would be
from two to three, that works perfectly
and the y component would stay constant at three.
So because we have two curves with the set
of two integrals in terms of t,
well first going to have t from zero to t
where xy squared would be t times the quantity 3/2 t squared
that will be nine fourth t to the third
and then x squared y would be t squared times 3/2 t
or 3/2 t cube dotted with our prime of t
that would be one, 3/2, integrated with respect to t.
And our second integral is going to be from two to three
where xy squared would be nine t and x squared y
would be three t squared
and our prime of t would be one, zero.
So as you can see, the original method in this problem
is going to take a lot more work than the previous two methods.
Let's go ahead and finish this on the next page.
You can see here we have 9/4 t cubed
plus another 9/4 t cubed of or 18/4 or 9/2 t cubed.
Here we'll just have nine t
so I'll have 9/2 times t to the fourth over four.
Here we'll have nine times t squared over two.
So replacing t with two, we're going to have 16 divided by four
that's four, four times 9/2, that's going to be 18
and then when t is zero, we'll have zero.
Over here, when t is three, we'll have 81/2
and then we t is two, we'll have 18.
So as the 18 simplify out and the final result is also 81/2.
So hopefully now you see the advantage
of using the fundamental theorem of line integrals
when you know the vector field is conservative.
In part two, we'll take a look at some more examples
just using the fundamental theorem of line integrals.
Thank you for watching.