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The value of the directional derivative of the function θ (x, y, z) = xy2 + yz2 + zx2 at the point (2, -1, 1) in the direction of the vector p = i + 2j + 2k is
- a)1
- b)0.95
- c)0.93
- d)0.9
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The value of the directional derivative of the function θ (x, y,...
Understanding the Directional Derivative
The directional derivative of a function measures how the function changes as you move in a specific direction. It is computed using the gradient of the function and the direction vector.
Step 1: Calculate the Gradient
For the function θ(x, y, z) = xy² + yz² + zx², we first need to find the gradient (∇θ):
- ∂θ/∂x = y² + 2zx
- ∂θ/∂y = 2xy + z²
- ∂θ/∂z = 2yz + x²
Now, substitute the point (2, -1, 1):
- ∂θ/∂x at (2, -1, 1) = (-1)² + 2(1)(2) = 1 + 4 = 5
- ∂θ/∂y at (2, -1, 1) = 2(2)(-1) + 1² = -4 + 1 = -3
- ∂θ/∂z at (2, -1, 1) = 2(-1)(1) + 2² = -2 + 4 = 2
Thus, ∇θ(2, -1, 1) = (5, -3, 2).
Step 2: Normalize the Direction Vector
The direction vector p = i + 2j + 2k needs to be normalized:
- Magnitude of p = sqrt(1² + 2² + 2²) = sqrt(1 + 4 + 4) = sqrt(9) = 3
- Normalized vector u = (1/3, 2/3, 2/3)
Step 3: Calculate the Directional Derivative
The directional derivative D_uθ at the point is given by:
D_uθ = ∇θ · u
Calculating this:
D_uθ = (5, -3, 2) · (1/3, 2/3, 2/3) = (5/3) + (-6/3) + (4/3) = 5/3 - 6/3 + 4/3 = 3/3 = 1.
Conclusion
Thus, the value of the directional derivative of the function at the specified point in the given direction is indeed 1, confirming the correct answer is option 'A'.
The directional derivative of a function measures how the function changes as you move in a specific direction. It is computed using the gradient of the function and the direction vector.
Step 1: Calculate the Gradient
For the function θ(x, y, z) = xy² + yz² + zx², we first need to find the gradient (∇θ):
- ∂θ/∂x = y² + 2zx
- ∂θ/∂y = 2xy + z²
- ∂θ/∂z = 2yz + x²
Now, substitute the point (2, -1, 1):
- ∂θ/∂x at (2, -1, 1) = (-1)² + 2(1)(2) = 1 + 4 = 5
- ∂θ/∂y at (2, -1, 1) = 2(2)(-1) + 1² = -4 + 1 = -3
- ∂θ/∂z at (2, -1, 1) = 2(-1)(1) + 2² = -2 + 4 = 2
Thus, ∇θ(2, -1, 1) = (5, -3, 2).
Step 2: Normalize the Direction Vector
The direction vector p = i + 2j + 2k needs to be normalized:
- Magnitude of p = sqrt(1² + 2² + 2²) = sqrt(1 + 4 + 4) = sqrt(9) = 3
- Normalized vector u = (1/3, 2/3, 2/3)
Step 3: Calculate the Directional Derivative
The directional derivative D_uθ at the point is given by:
D_uθ = ∇θ · u
Calculating this:
D_uθ = (5, -3, 2) · (1/3, 2/3, 2/3) = (5/3) + (-6/3) + (4/3) = 5/3 - 6/3 + 4/3 = 3/3 = 1.
Conclusion
Thus, the value of the directional derivative of the function at the specified point in the given direction is indeed 1, confirming the correct answer is option 'A'.
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Community Answer
The value of the directional derivative of the function θ (x, y,...
Given that,
ϕ = xy2 + yz2 + zx2
directional vector (p) = I + 2j + 2K
Directional derivative = 
∇ϕ at the point (2, -1, 1) is
∇ϕ = ((-1)2 + 2(2)(1)) î + (2(2)(-1) + (1)2) ĵ + (2(-1)(1) + (2)2)k̂
= 5î - 3ĵ + 2k̂
Directional derivative = 
= 5 - 6 + 4 / 3
= 1
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