00:01
We're getting a function of two variables f.
00:04
And we're told the f has continuous partial derivatives.
00:09
So f is a c -1 function.
00:20
On r2.
00:23
And we're asked to consider the points a, b, c, and d, with coordinates 1 -3 -33, 3, 17, and 615.
00:52
We're told the directional derivative of our function f at the point a in the direction of the vector ab is 3, and the directional derivative at a in the direction of the vector ac is 26.
01:13
We're asked to find the directional derivative of f at a in the direction of the vector ab.
01:19
Well, first of all, let's find what some of these vectors are.
01:25
So let's find vector ab.
01:29
This is the vector with components 3 minus 1 is 2 and 3 minus 3 is 0.
01:35
Therefore, a unit vector, i'll call u sub ab, in the direction of ab, is 1 half, times ad which is 2 -0 likewise vector ac as components 1 minus 1 is 0 and 7 minus 3 is 4 and so a unit vector in the direction of ac u sub ac is the vector 1 4th times ac or 0 4 and the vector ad is the vector with components 6 minus 1 is 5 and 15 minus 3 is 12 and so a unit vector view in a direction of ad well this is going to be well the magnitude of vector ad is a little bit harder to tell this is the square root of 25 plus 144 which is 13 and so this is 1 over 13 times ad which is 512 now he used this book matt 6 like he really now, on the one hand, we know that the directional derivative of our function f at a, the coordinates 1 -3, in the direction of the vector ab, which is also in the direction of the unit vector u sub -a, is given by the gradient of f at 1 -3, dotted with the unit vector u .s .a., is given by the gradient of f at 1 -3, dotted with the unit vector u...