00:01
Hi there, so for this problem, we are told that a flat circular plate has the shape of the region x square plus y square, and this is less or equal than one.
00:16
The plate including the boundary, where x square plus y square is equal to 1, is heated so that the temperature at a point, xy is given by the following expression.
00:31
So the temperature that depends on x and y is equal to x square, this plus three times y square, and this plus three divided by two times x.
00:51
So we need to find the temperatures at the hottest and coldest point on the plane.
00:59
So, okay, so we need to find that the hottest temperature and the coldest, and the coldest, and the coldest, the temperature in this point, on the point of some this plate.
01:18
So with that said, what we need to do, first of all, is to take the derivatives of this temperature expression that we are given with respect to x and y, set those equal to.
01:32
To zero and solve, we will obtain some critical point.
01:37
So let's do that.
01:38
So first of all, the imperative of the temperature function with respect to x is just simply two times x, because that is the derivative of x squared.
01:55
This plus three divided by two.
02:02
Okay.
02:04
So, and then we set this equal to 0.
02:11
And if we solve for x in this case, we will have that x it's the value of minus 3 divided by 4.
02:20
Now, the derivative of the temperature with respect to y is just 6 times y, and then we set this equal to 0.
02:34
So from this, we obtained that y is equal to 0.
02:39
So as you can see, one critical point is, the point minus 3 divided by 4 and 0.
02:49
So what we can do is to evaluate these points in the temperature so that we obtain some temperature that later we can determine if it is a minimum or a maximum.
03:02
So the temperature evaluated at this point is x squared so that will be minus 3 divided by 4 that to the square.
03:16
This plus 3 divided by 2 times minus 3 divided by 4.
03:25
And let me see, well, let me use my calculator to give you the result for this.
03:32
So the value of the web time from this is minus 9 divided by 16.
03:37
So once we set that, we need to determine another other points.
03:47
And, and then what we can do is to solve for y from here using the equality, of course.
03:57
So that will be that y is equal to the square root of 1 minus x square.
04:03
So we can substitute that into the temperate function.
04:09
So when we do that, we obtain the following.
04:15
So now that the emperor only in terms of x is equal to x squared this plus 3 times 1 minus x square, yes.
04:41
And this plus 3 divided by 2 times x...