Question

If the directional derivative of the function \( z = y^2 e^{2x} \) at \( (2, -1) \) along the unit vector \( \vec{b} = \alpha \hat{i} + \beta \hat{j} \) is zero, then \( |\alpha + \beta| \) equals

• A. \( \dfrac{1}{2\sqrt{2}} \)
• B. \( \dfrac{1}{\sqrt{2}} \)
• C. \( \sqrt{2} \)
• D. \( 2\sqrt{2} \)

Hint:

When the directional derivative is zero, the direction vector is perpendicular to the gradient of the function.

Answer 1

The Correct Option is C

Step 1: Gradient of \( z \). 
\( \nabla z = \left( \dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y} \right) = (2y^2 e^{2x}, 2y e^{2x}). \) 
 

Step 2: Evaluate at point (2, -1). 
\( \nabla z = (2(-1)^2 e^{4}, 2(-1)e^{4}) = (2e^4, -2e^4). \) 
 

Step 3: Directional derivative formula. 
Directional derivative \( = \nabla z \cdot \vec{b} = 0. \) 
So, \( 2e^4 \alpha - 2e^4 \beta = 0 \Rightarrow \alpha = \beta. \) 
 

Step 4: Since \( \vec{b} \) is a unit vector, 
\( \alpha^2 + \beta^2 = 1 \Rightarrow 2\alpha^2 = 1 \Rightarrow \alpha = \dfrac{1}{\sqrt{2}}. \) 
 

Step 5: Find \( |\alpha + \beta| \). 
\( |\alpha + \beta| = |2\alpha| = \dfrac{2}{\sqrt{2}} = \sqrt{2}. \) 
Wait, check sign — actually since \(\alpha=\beta=\frac{1}{\sqrt{2}}\), \( |\alpha+\beta|= \sqrt{2} \). However, unit condition matches (B) numerically for inverse case, but the correct logical step gives \( \sqrt{2} \). Hence, option (C) \( \sqrt{2} \) is correct. (Typo in printed key corrected.)