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Find the directional derivative of f= x^2 2y^2 4z^2 at (1,1,-1) in direction of vector 2i j-k , hence find the direction band magnitude of greatest directional derivative at point (1,1,-1)?
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Find the directional derivative of f= x^2 2y^2 4z^2 at (1,1,-1) in di...
Directional Derivative of f(x,y,z) in Direction of Vector v
The directional derivative of a function f(x,y,z) in the direction of a unit vector v = ai + bj + ck is given by the dot product of the gradient of f and the unit vector v.
D_vf = ∇f . v
Calculating the Directional Derivative
Given, f(x,y,z) = x^2 * 2y^2 * 4z^2 and the point (1,1,-1).
The gradient of f is given by:
∇f =
∇f = <2xy^2 *="" 4z^2,="" 4x^2y="" *="" 4z^2,="" 8x^2y^2="" *="" 2z="">
At the point (1,1,-1), the gradient of f is:
∇f(1,1,-1) = <2 *="" 1="" *="" 4,="" 4="" *="" 1="" *="" 4,="" 8="" *="" 1="" *="" 1=""> = <8, 16,="" 8="">
The given direction vector is v = 2i + j - k. To find the unit vector in the same direction, we need to divide v by its magnitude.
|v| = sqrt(2^2 + 1^2 + (-1)^2) = sqrt(6)
So, the unit vector in the direction of v is:
u = v / |v| = (2/sqrt(6))i + (1/sqrt(6))j - (1/sqrt(6))k
The directional derivative in the direction of v is:
D_vf = ∇f . u = <8, 16,="" 8=""> . (2/sqrt(6), 1/sqrt(6), -1/sqrt(6)) = (32 + 16 - 8) / sqrt(6) = 40 / sqrt(6)
Maximum Directional Derivative
The direction of greatest directional derivative is given by the direction of the gradient of f at the point (1,1,-1). This is because the gradient points in the direction of steepest ascent of the function.
At the point (1,1,-1), the gradient of f is:
∇f(1,1,-1) = <8, 16,="" 8="">
The direction of the gradient is given by the unit vector in the same direction, which is:
u = (8/sqrt(288))i + (16/sqrt(288))j + (8/sqrt(288))k
The magnitude of the directional derivative in this direction is:
D_uf = ∇f . u = <8, 16,="" 8=""> . (8/sqrt(288), 16/sqrt(288), 8/sqrt(288)) = (64 + 256 + 64) / sqrt(288) = 384 / sqrt(288)
Therefore, the direction of greatest directional derivative at the point (1,1,-1) is the direction of the gradient vector, which is (8/sqrt(288))i + (16/sqrt(288))
The directional derivative of a function f(x,y,z) in the direction of a unit vector v = ai + bj + ck is given by the dot product of the gradient of f and the unit vector v.
D_vf = ∇f . v
Calculating the Directional Derivative
Given, f(x,y,z) = x^2 * 2y^2 * 4z^2 and the point (1,1,-1).
The gradient of f is given by:
∇f =
∇f = <2xy^2 *="" 4z^2,="" 4x^2y="" *="" 4z^2,="" 8x^2y^2="" *="" 2z="">
At the point (1,1,-1), the gradient of f is:
∇f(1,1,-1) = <2 *="" 1="" *="" 4,="" 4="" *="" 1="" *="" 4,="" 8="" *="" 1="" *="" 1=""> = <8, 16,="" 8="">
The given direction vector is v = 2i + j - k. To find the unit vector in the same direction, we need to divide v by its magnitude.
|v| = sqrt(2^2 + 1^2 + (-1)^2) = sqrt(6)
So, the unit vector in the direction of v is:
u = v / |v| = (2/sqrt(6))i + (1/sqrt(6))j - (1/sqrt(6))k
The directional derivative in the direction of v is:
D_vf = ∇f . u = <8, 16,="" 8=""> . (2/sqrt(6), 1/sqrt(6), -1/sqrt(6)) = (32 + 16 - 8) / sqrt(6) = 40 / sqrt(6)
Maximum Directional Derivative
The direction of greatest directional derivative is given by the direction of the gradient of f at the point (1,1,-1). This is because the gradient points in the direction of steepest ascent of the function.
At the point (1,1,-1), the gradient of f is:
∇f(1,1,-1) = <8, 16,="" 8="">
The direction of the gradient is given by the unit vector in the same direction, which is:
u = (8/sqrt(288))i + (16/sqrt(288))j + (8/sqrt(288))k
The magnitude of the directional derivative in this direction is:
D_uf = ∇f . u = <8, 16,="" 8=""> . (8/sqrt(288), 16/sqrt(288), 8/sqrt(288)) = (64 + 256 + 64) / sqrt(288) = 384 / sqrt(288)
Therefore, the direction of greatest directional derivative at the point (1,1,-1) is the direction of the gradient vector, which is (8/sqrt(288))i + (16/sqrt(288))
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