Question

Help to gradient and directional derivative .

Answer 1

What is the definition of gradient?

Answer 2

You have a function \( f : \mathbb{R}^2 \rightarrow \mathbb{R} \), grad(f) is defined as ...

Answer 3

The partiel derivatives of the function respect to x and y .

Answer 4

Right. Hence in this case grad(f) = (..., ...) what?

And then the directional derivative in a (unit direction) is given by
\( D_u f = grad(f).u \), the inner product between the gradient and the direction \( u \).

Answer 5

\[ f(x,y) = x^3 + y^2 - xy \]

Answer 6

grad(f) = <3x^2, 2y> = (3x^2)i + (2y)j

fx(1,2) = 3
fy(1,2) = 4

Duf(x,y) = 3u1+4u2 .

Answer 7

No.
\[ grad(f) = \nabla f = (\partial f / \partial x, \partial f / \partial y) = (3x^2 - y, 2y - x) \]

Answer 8

Thus \( \nabla f(1,2) = (3-2,4-1)=(1,3) \)

Answer 9

Okay. So we get:

fx(1,2) = 3*(1)^2-2 = 1
fy(1,2) = 2*2-1 = 3

Duf(x,y) = u1+3u2 .

Is that correct?

Answer 10

Yes.

Answer 11

Thank you, James .. Could you help me with the second questions. I dont get how to find the vector, when the directional derivative is equal to zero.