00:02
In this problem, we are given the function f of xy is equal to square root of 4 minus x square minus 2y.
00:13
We are asked to determine the directional derivative of this function at the point 2 negative 2 in the direction of the vector 1 by square root of 10, 3 by square root of 10.
00:26
Now consider v to be the vector 1 by square root of 10, 3 by square root of 10, find the magnitude of this vector which is square root of 1 by square root of 10 the whole square plus 3 by square root of 10 the whole square which is square root of 1 plus 1 by 10 plus 9 by 10 which is square root of 10 by 10 which is 1.
00:55
Therefore, v is a unit vector.
01:00
Now, consider a function f of x ,y, the directional derivative of this function at a point ab in the direction of a unit vector u is evaluated as tu of f of a .b is equal to del f at the point ab.
01:20
Dot u where del f is the gradient vector of the function at the point a b here this vector itself is the unit vector since it has magnitude 1 now we need to find d v of f of 2 negative 2 and by definition it is the gradient vector of the function f at the point 2 negative 2 dot product with the unit vector v.
01:52
Now find the gradient vector del f of xy and it is the vector with components fx and fy where fx is the partial derivative of the function f with respect to x and fy is the partial derivative of the function with respect to y.
02:11
Find fx which is the partial derivative of this function with respect to x keeping y as a constant by chain rule this is 1 by 2 times squared root of 4 minus x square minus 2y times negative 2x and this simplifies to negative x divided by square root of 4 minus x square minus 2y...