00:01
In this problem, we are given the shown information, and our first step might be to make sure that our vector field f is conservative.
00:07
To do that, we're going to set this component here equal to some function p, this component equal to some function q, and this component equal to some function r, and we're going to make sure that the partial of p with respect to y is equal to the partial of q with respect to x.
00:28
We're also going to make sure that the partial of p with respect to z is equal to the partial of r with respect to x.
00:38
And lastly, we're going to make sure that the partial of q with respect to z is equal to the partial of r with respect to y.
00:50
Now the partial of p with respect to y is going to be cosine of y, and the partial of q with respect to x is going to be cosine of y, so that's good.
01:02
The partial of p with respect to z is going to be zero.
01:07
Partial of r with respect to x is also going to be zero.
01:12
Now the partial of q with respect to z is going to be negative sign of z.
01:21
Partial of r with respect to y is going to be negative sine of z.
01:26
And so the vector field is conservative.
01:28
We can find our function f.
01:30
To do this, we're going to write p in an integral and solve in terms of x, so sine of y d x.
01:45
And when we solve this, we get negative, or excuse me, we don't have an x in our sign of y, so it's actually just going to be x sine of y, plus some function in y and z.
02:09
That whose derivative will go to zero if we took the partial of this with respect to x.
02:13
Now we're going to take the partial of this with respect to x.
02:14
Now we're going to take the partial of it with respect to y.
02:17
And when we do that, we get the following.
02:29
We're going to get x cosine of y plus g prime of that y z.
02:37
I'm going to set that equal to our function q up above, which was x cosine of y plus cosine of cosine of z.
02:56
From this we can see that g prime of y z is equal to cosine of z and we want to write that in an integral and evaluate it in terms of y and we see that g of some y z is equal to i'm sorry this is just supposed to be equal to g of y z not g prime of y z so the integral of z of y z is equals g of y z and that is equal to y cosine plus some function in z only and now we're going to plug that in into what we got up here.
03:59
And we see that we get x, sine of y, plus y cosine z, plus h of z.
04:11
And we're going to take the partial of this with respect to z.
04:16
And when we do this, we get negative y, sine of z plus h prime of z.
04:29
And we're going to set that equal to our function r up here, which is, negative y sine of z and from this we can see the h prime of z is equal to zero we're to write that an integral with d z and we see that h of z is equal to some constant k we're going to plug that in to what we got right here and when we do that we get that our function of some x y z is equal to x sine of y plus y cosine of z plus our constant k...