00:01
Hello everyone, so this is the question that we have.
00:03
We have a function x comma y which is equal to 2 y squared minus x squared we have to compute the formula we have four parts associated with this question a b c and d so a part is that you have to find the gradient of del f x comma y then part b is we have to find the directional directive directional derivative directional derivative at 1 comma 2 so g u f of 1 comma 2 of at the point 1 comma 2 and we have to use the vector which is u equal to i plus j now part c is that we have to give the direction of the fastest rate of increase.
01:10
Direction of fastest rate of increase of f at point e1 power now, last in the final part is i have to give the maximal value, maximum value of the directional derivative, detectional derivative of f at point b1 comma.
01:52
So let's go on to the solution of the question.
01:56
So given to us in part a, if you had fx comma y, the equation is 2y square minus x.
02:06
Let's find a gradient of del of x comma y would be partial differentiation of f with respect to x i cap plus partial differential of f with respect to y cacat.
02:29
So this is going to be if i have fun in values.
02:32
So this will be del f of x comma y is equal to minus 2 x y cap plus my j cap because i know that del f divided by del x is equal to minus 2x and del f divided by del y is equal to minus 2x and del f divided by del y is equal to equal by so this is the answer of my first part.
03:11
Let's move on to the part b that we have to find.
03:15
So direction and derivative.
03:17
Now, tu would be given by gradient of f x comma y when divided by p.
03:27
Multiply the a vector divided by the line 2 of that.
03:34
So i have the gradient to next flagging the formula.
03:39
So my del gradient x comma y is minus 2x i cap plus over y cap.
03:49
Now gradient f, gradient f at 1 comma, negative, 2.
03:55
The values put x is equal to 1 and y is equal to 2 minus 2 1 i cap plus 4 multiplied by 2 j cap which is going to be same f 1 1 by 2 is going to be minus 2 i cap plus 8 j cap so a vector is given to me as lifted a is given to me as i plus j so magnitude of a would be take the coefficients so one square plus one square under root this will be root two now my directional derivative d look is going to be minus two i cap plus eight j cap multiplied by i cap plus j cap and divided by so as we know that vector multiplied by a vector value already is a scalar scalar.
05:02
So this is gonna be this is going to be that's probably the value minus 2 into minus 1.
05:10
So minus 2 multiplied by minus 1 sorry not minus 1 plus 1 plus 8 multiplied by 1 and divided by under the loop of course this is going to be 6 multiplied by little root of 2 this becomes by rationalize or my buy 1 in 2 and divide by root 2 so 6 root 2 divided by 2 this becomes 3 root so i get my direction derivative as 3 root now let's jump on to the part c of this question so part c now tell f x comma y with my is 2x i cap plus over y day cap...