Compute The Directional Derivative Of The Following Function At The Given Point Pin The Direction Of

The directional derivative of the function at the given point P in the direction of the given vector is:

(8/21)√(2).

Directional derivative

The directional derivative of a function in the direction of a unit vector is the rate at which the function changes in that direction.

To compute the directional derivative of f(x, y) = ln(5 + 3x^2 + 2y^2) at the point P(2, -1) in the direction of the vector ⟨1, 1⟩, we need to:

Compute the gradient of f(x, y) at P(2, -1).Normalize the direction vector ⟨1, 1⟩ to obtain a unit vector.Compute the dot product of the gradient of f at P with the unit direction vector.The gradient of f(x, y) is given by:

1) ∇f(x, y) = (6x / (5 + 3x^2 + 2y^2), 4y / (5 + 3x^2 + 2y^2))

Therefore, the gradient of f at P(2, -1) is:

∇f(2, -1) = (24/21, -4/21)

2) To obtain a unit vector in the direction of ⟨1, 1⟩, we need to divide it by its length:

||⟨1, 1⟩|| = √(1^2 + 1^2) = sqrt(2)

Therefore, a unit vector in the direction of ⟨1, 1⟩ is given by:

u = ⟨1, 1⟩ / √2) = ⟨√(2)/2, √(2)/2⟩

3) The directional derivative of f at P in the direction of u is given by:

D_uf(2, -1) = ∇f(2, -1) · u

where “·” denotes the dot product. Substituting the values for ∇f(2, -1) and u, we get:

D_uf(2, -1) = (24/21, -4/21) · (√(2)/2, √(2)/2)

= (24/21)(√(2)/2) + (-4/21)(√(2)/2)

= (8/21)√(2)

Therefore, the directional derivative of f(x, y) at P(2, -1) in the direction of ⟨1, 1⟩ is (8/21)√(2).

To know more about the “directional derivative“: brainly.com/question/12873145

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