Question:
Find the directional derivative of $f = xyz$ at the point $(1,-1,-2)$ in the direction of the vector $2i-2j+k$ is
Find the directional derivative of $f = xyz$ at the point $(1,-1,-2)$ in the direction of the vector $2i-2j+k$ is
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Always remember to convert your direction vector into a UNIT vector before taking the dot product.
Updated On: Jul 3, 2026
- 0
- $\frac{7}{3}$
- $\frac{2}{3}$
- 1
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Directional derivative $D_{\hat{u}}f = \nabla f \cdot \hat{u}$, where $\hat{u}$ is the unit vector in the given direction.
Step 2: Meaning
$\nabla f = \frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j + \frac{\partial f}{\partial z}k = (yz)i + (xz)j + (xy)k$. At $(1,-1,-2)$, $\nabla f = 2i - 2j - k$.
Step 3: Analysis
Unit vector $\hat{u} = \frac{2i - 2j + k}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{1}{3}(2i - 2j + k)$.
Step 4: Conclusion
$D_{\hat{u}}f = (2i - 2j - k) \cdot \frac{1}{3}(2i - 2j + k) = \frac{1}{3}(4 + 4 - 1) = \frac{7}{3}$. Final Answer: (B)
Directional derivative $D_{\hat{u}}f = \nabla f \cdot \hat{u}$, where $\hat{u}$ is the unit vector in the given direction.
Step 2: Meaning
$\nabla f = \frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j + \frac{\partial f}{\partial z}k = (yz)i + (xz)j + (xy)k$. At $(1,-1,-2)$, $\nabla f = 2i - 2j - k$.
Step 3: Analysis
Unit vector $\hat{u} = \frac{2i - 2j + k}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{1}{3}(2i - 2j + k)$.
Step 4: Conclusion
$D_{\hat{u}}f = (2i - 2j - k) \cdot \frac{1}{3}(2i - 2j + k) = \frac{1}{3}(4 + 4 - 1) = \frac{7}{3}$. Final Answer: (B)
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