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C Find the directional derivative of f(x,y,2) =xly+x+z at the point P(,2,3), the direction of V= <,1,2>...

Question

C Find the directional derivative of f(x,y,2) =xly+x+z at the point P(,2,3), the direction of V= <,1,2>

C Find the directional derivative of f(x,y,2) =xly+x+z at the point P(,2,3), the direction of V= <,1,2>



Answers

Calculate the directional derivative in the direction of $\mathbf{v}$ at the given point. Remember to use a unit vector in your directional derivative computation.
$$ f(x, y)=e^{x y-y^{2}}, \quad \mathbf{v}=\langle 12,-5\rangle, \quad P=(2,2) $$

Be this P. Find the direction of find the directional derivative of F. In the direction of you at the point P where you is the unit factor in the v direction. So magnitude of V is one square plus one square square roots squared it to. So you whenever squared it to. Whenever squared it to. Let's go ahead and make that Squared up to over two. All right. And then Glf it's the sign of something. So the derivative is co sign of something Times the derivative of X -Y, which is one and then kind of x minus Y times the drift, even the x minus y minus one. So then tell F dot you Squirt of two or 2 times the coastline of x minus y minus the cosine of x minus Y. We have to choose equal to zero. No matter what point you put in there. It's like directional derivative In the direction of U. of F. at any point is zero. This function and this unit vector.

We need to calculate the ingredient of F, which is equal to the coastline of X minus y common the negative co sign of X minus y. Evaluating this Grady int of f at the point p which is equal to pi over to calm a pyre or sex, we can simplify this down to 1/2 on the native 1/2 now plaguing this into the formula to find the directional derivative off the creating the F at the point p, we can simplify this down to zero.

For this problem, we are given the function F of x. Y equals X, divided by Y, the 0.11. And the vector negative J. And we are asked to find the directional derivative of F in the direction of V. So our first step would be to normalize V. But we can note that because we are only going in one direction or one dimension for V, it's going to have a magnitude of one. It's already a unit factor. The next step is to find the gradient of F at some generic point X. Y. Well, in the I direction we are in the X direction I direction, no big difference. We'll have I over why. And in the J direction we'll have minus X over Y squared J. Then we want to evaluate our gradient at are given point. So we'll have just I then minus 1/1 squared. So we'll have I minus J as our gradient. And I should specify that that's not at X. Y, that is at the 0.11. Now that we have that we can find the directional derivative in the direction of the by taking the dot product of this with v. Well, we just have I time zero plus negative or negative one, times negative one. So we'd have their directional derivative is going to equal one

News for the Grady into another. The radiant of F is equal to two acts. Why Cube Comma three X squared y squared. Now evaluating the great end of F at the point p which is equal to make tea to calm. You know what we end up with later for common 12. Planning this into the formula and simplifying little bit We end up with dese of you off f off The point is equal to eight divided by the square.


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