Directional derivative in direction of vector v
The directional derivative in the direction of a vector v measures how a function f(x,y,z) changes in the direction of the vector v at a given point
The directional derivative in the direction of a vector v measures how a function f(x,y,z) changes in the direction of the vector v at a given point.
The directional derivative in the direction of v is denoted as D_vf(x,y,z) or ∇vf(x,y,z), where ∇v represents the gradient operator acting on the function f.
To calculate the directional derivative, we use the dot product of the gradient of f with the unit vector u in the direction of v.
D_vf(x,y,z) = ∇f(x,y,z) · u
To find the unit vector u in the direction of v, we divide the vector v by its magnitude.
u = v / ||v||
The magnitude of v is given by:
||v|| = √(v1^2 + v2^2 + v3^2)
Let’s work through an example to illustrate how to calculate the directional derivative in the direction of v.
Example:
Consider the function f(x,y,z) = x^2 + y^2 – z^2 and the vector v = ⟨1, 2, 3⟩. We want to find the directional derivative of f in the direction of v at the point (2, -1, 3).
1. Calculate the unit vector u:
||v|| = √(1^2 + 2^2 + 3^2) = √(1 + 4 + 9) = √14
u = v / ||v|| = ⟨1, 2, 3⟩ / √14 = ⟨1/√14, 2/√14, 3/√14⟩
2. Calculate the gradient of f:
∇f(x,y,z) = ⟨∂f/∂x, ∂f/∂y, ∂f/∂z⟩
∂f/∂x = 2x, ∂f/∂y = 2y, ∂f/∂z = -2z
∇f(x,y,z) = ⟨2x, 2y, -2z⟩
3. Calculate the directional derivative:
D_vf(x,y,z) = ∇f(x,y,z) · u
D_vf(2,-1,3) = ⟨2(2), 2(-1), -2(3)⟩ · ⟨1/√14, 2/√14, 3/√14⟩
D_vf(2,-1,3) = ⟨4, -2, -6⟩ · ⟨1/√14, 2/√14, 3/√14⟩
D_vf(2,-1,3) = (4)(1/√14) + (-2)(2/√14) + (-6)(3/√14)
D_vf(2,-1,3) = 4/√14 – 4/√14 – 18/√14
D_vf(2,-1,3) = -18/√14
Therefore, the directional derivative of f in the direction of vector v at (2,-1,3) is -18/√14.
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