Question

find direction derivative of f= (xz)/(x^2+y^2) at (1,-1,1) in the direction of (A vector) = i -2j +k​

Answer 1

Answer:

Therefore, the directional derivative of f = (xz)/(x^2+y^2) at (1,-1,1) in the direction of A = i - 2j + k is 5/9.

Step-by-step explanation:

To find the directional derivative of f = (xz)/(x^2+y^2) at (1,-1,1) in the direction of the vector A = i - 2j + k, we need to first find the gradient of f at the given point:

grad(f) = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

where ∂f/∂x, ∂f/∂y, and ∂f/∂z are the partial derivatives of f with respect to x, y, and z, respectively.

Computing the partial derivatives, we have:

∂f/∂x = z(x^2 - y^2)/(x^2 + y^2)^2

∂f/∂y = -2xyz/(x^2 + y^2)^2

∂f/∂z = x/(x^2 + y^2)

Substituting the point (1,-1,1) into these expressions, we get:

∂f/∂x(1,-1,1) = 1/2

∂f/∂y(1,-1,1) = 2/9

∂f/∂z(1,-1,1) = 1/2

Therefore, the gradient of f at (1,-1,1) is:

grad(f)(1,-1,1) = (1/2)i + (2/9)j + (1/2)k

To find the directional derivative of f in the direction of A, we take the dot product of A and the gradient of f:

D_A(f) = grad(f)(1,-1,1) ⋅ A

= (1/2)i ⋅ i - (2/9)j ⋅ 2j + (1/2)k ⋅ k

= 1/2 - 4/9 + 1/2

= 5/9

Therefore, the directional derivative of f = (xz)/(x^2+y^2) at (1,-1,1) in the direction of A = i - 2j + k is 5/9.

To know more about direction derivative refer :

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