00:02
In this problem, we're given a vector field capital f, and we're required to find out the line integral over the curve c, f .dr, where c is a simple closed curve that encloses the origin, but we aren't given any more specifics about c.
00:26
So the only way we can evaluate this is by some property of f, for example, that it may be conservative.
00:33
So let's first check for that.
00:36
So a vector field in its general form is given by a function pfx comma y times the i component plus a function q comma y times the j component.
00:58
And we can compare our given field to find out bf x comma y is equal to 2x y divided by x squared, plus y squared, the whole squared, and q of x comma y is simply equal to y squared minus x squared, the whole divided by x squared plus y squared, the whole squared.
01:35
Now a field is conservative if the partial derivative, so the partial derivative of b, with respect to y is equal to the partial derivative of q with respect to x.
01:51
So let's just find these derivatives, starting with the partial derivative of p, with respect to the partial derivative of y.
02:00
And we can use the quotient rule to find our derivative.
02:11
So first, what i would like to do is actually take the constant out.
02:16
So 2x is a constant with respect to the partial derivative y.
02:21
So we can actually write it down as partial over partial y of y divided by x.
02:30
Squared plus y squared the whole squared and now we can use the quotient rule which gives us 2x times the partial derivative of y with respect to y so partial over partial y of y of of y multiplied by x squared plus y squared the whole squared minus the partial derivative of x squared plus y squared the whole squared times y this entire thing is going to be divided by the denominator squared, so that's going to be x squared plus y squared, the whole squared, squared, squared again.
03:22
And now we can take the derivatives, so that will be 2x times, so the partial derivative of y with respect to y is simply one.
03:35
Now we can take the inner derivative, this expression, which is the derivative of x squared plus y squared with respect to y now x is a constant so the derivative is zero constant with respect to the partial derivative with respect to y and we can find the derivative of y squared using the power rule that gives us 2 y times the and we can use the power rule again to find out the outer derivative that is 2 times x squared plus y squared oh sorry we did not need to take the derivative of the second expression right now, we had to do that later.
04:37
This just comes as it is as being equal to x squared plus y squared, the whole squared.
04:58
Now we'll take the derivative by the same method as we did before.
05:03
So the outer derivative is two times x squared plus y squared, and that is multiplied by the inner derivative.
05:17
That is simply do y.
05:19
That's multiplied by y divided by x squared plus y squared raised with a fourth power and we can actually get rid of one of these by factoring an x squared plus y squared from the denominator from the numerator and canceling it with the denominator so what we'll end up with is 2x times x squared plus y squared divided by x squared plus y squared the whole cubed that will simply give us so x squared minus three y squared divided by x squared plus y squared the whole cubed that means our partial derivative of of b with respect to y turns out to be 2x times x squared minus 3 y squared divided by x squared plus y squared the whole cubed now we can take the partial derivative of q with respect to x partial derivative of q with respect to x with respect to x will be equal to the partial derivative of x of x squared, sorry, y squared minus x squared, divided by x squared plus y squared, the whole squared.
07:57
Now once again, we will be using the quotient rule, so the partial derivative of x of the numerator times the denominator minus the partial derivative of the denominator times the numerator.
08:38
And this entire thing is going to be divided by the square of the denominator that is just x squared plus y squared raised to the fourth power...