Question

The directional derivative of the function \( f(x, y) = 2x^2 + y^2 \) along a line directed from \( (0, 0) \) to \( (1, 1) \), evaluated at the point \( x = 1, y = 1 \), is ___

• A. \( 6\sqrt{2} \)
• B. \( 2\sqrt{2} \)
• C. \( 3\sqrt{2} \)
• D. \( 6 \)

Hint:

Directional derivative = Gradient dot Unit Direction Vector.

Answer 1

The Correct Option is C

The gradient of \( f(x, y) = 2x^2 + y^2 \) is: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (4x, 2y) \] At the point \( (1, 1) \), \[ \nabla f(1,1) = (4, 2) \] The direction vector from \( (0,0) \) to \( (1,1) \) is \( \vec{v} = (1, 1) \). The unit vector in this direction is: \[ \hat{v} = \frac{1}{\sqrt{2}}(1, 1) \] The directional derivative is: \[ \nabla f \cdot \hat{v} = (4, 2) \cdot \frac{1}{\sqrt{2}}(1, 1) = \frac{1}{\sqrt{2}}(4 + 2) = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]