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The directional derivative of the function f = x2 - y2 + 2Z2 at the point P(1, 2, 3) in the direction of line PQ where Q is the poin t (5, 0, 4 ) is ________.
Correct answer is '6.11'. Can you explain this answer?
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The directional derivative of the function f = x2 - y2 + 2Z2 at the po...
Here function f = x2 - y2 + 2z2
Now vector
unit vector in direction of
so, directional derivative of f in the direction of
Now vector
unit vector in direction of
so, directional derivative of f in the direction of
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The directional derivative of the function f = x2 - y2 + 2Z2 at the po...
Given function: f = x^2 - y^2 / 2z^2
Point P: (1, 2, 3)
Point Q: (5, 0, 4)
To find the directional derivative of f in the direction of line PQ, we need to follow the below steps:
Step 1: Find the direction vector of line PQ
The direction vector of line PQ is given by:
d = Q - P = (5 - 1, 0 - 2, 4 - 3) = (4, -2, 1)
Step 2: Normalize the direction vector
We need to normalize the direction vector d by dividing it by its magnitude. The magnitude of d is given by:
|d| = sqrt(4^2 + (-2)^2 + 1^2) = sqrt(21)
Therefore, the normalized direction vector is given by:
d' = d / |d| = (4/sqrt(21), -2/sqrt(21), 1/sqrt(21))
Step 3: Find the gradient of the function f at point P
The gradient of f at point P is given by:
grad(f) = (df/dx, df/dy, df/dz)
df/dx = 2x, df/dy = -2y, and df/dz = -y^2/z^3
Therefore, grad(f) at point P is given by:
grad(f)_P = (2(1), -2(2), -2^2/3^3) = (2, -4, -8/27)
Step 4: Find the directional derivative of f at point P in the direction of line PQ
The directional derivative of f at point P in the direction of line PQ is given by:
D_d'f_P = grad(f)_P . d'
where "." represents the dot product.
Substituting the values, we get:
D_d'f_P = (2, -4, -8/27) . (4/sqrt(21), -2/sqrt(21), 1/sqrt(21))
= 2(4/sqrt(21)) - 4(2/sqrt(21)) - 8/27(1/sqrt(21))
= 6.11 (approx)
Therefore, the directional derivative of f at point P in the direction of line PQ is 6.11.
Point P: (1, 2, 3)
Point Q: (5, 0, 4)
To find the directional derivative of f in the direction of line PQ, we need to follow the below steps:
Step 1: Find the direction vector of line PQ
The direction vector of line PQ is given by:
d = Q - P = (5 - 1, 0 - 2, 4 - 3) = (4, -2, 1)
Step 2: Normalize the direction vector
We need to normalize the direction vector d by dividing it by its magnitude. The magnitude of d is given by:
|d| = sqrt(4^2 + (-2)^2 + 1^2) = sqrt(21)
Therefore, the normalized direction vector is given by:
d' = d / |d| = (4/sqrt(21), -2/sqrt(21), 1/sqrt(21))
Step 3: Find the gradient of the function f at point P
The gradient of f at point P is given by:
grad(f) = (df/dx, df/dy, df/dz)
df/dx = 2x, df/dy = -2y, and df/dz = -y^2/z^3
Therefore, grad(f) at point P is given by:
grad(f)_P = (2(1), -2(2), -2^2/3^3) = (2, -4, -8/27)
Step 4: Find the directional derivative of f at point P in the direction of line PQ
The directional derivative of f at point P in the direction of line PQ is given by:
D_d'f_P = grad(f)_P . d'
where "." represents the dot product.
Substituting the values, we get:
D_d'f_P = (2, -4, -8/27) . (4/sqrt(21), -2/sqrt(21), 1/sqrt(21))
= 2(4/sqrt(21)) - 4(2/sqrt(21)) - 8/27(1/sqrt(21))
= 6.11 (approx)
Therefore, the directional derivative of f at point P in the direction of line PQ is 6.11.
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The directional derivative of the function f = x2 - y2 + 2Z2 at the po...
Find the angle between the surfaces x�+y�+z� = 9 and z = x�+y�-3 at the point (2,-1,2)
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The directional derivative of the function f = x2 - y2 + 2Z2 at the point P(1, 2, 3) in the direction of line PQ where Q is the poin t (5, 0, 4 ) is ________.Correct answer is '6.11'. Can you explain this answer?
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