Question

The directional derivative of the function \( f \) given below at the point \( (1, 0) \) in the direction of \( \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j}) \) is (rounded off to 1 decimal place). \[ f(x, y) = x^2 + xy^2 \]

Hint:

The directional derivative gives the rate of change of a function in a specific direction. To compute it, take the dot product of the gradient with the normalized direction vector.

Answer 1

Correct Answer: 0.9

Given: Direction vector: \( \mathbf{u} = \frac{1}{2}i + \frac{\sqrt{3}}{2}j \) Point: \( (1, 0) \) Step 1: Verify Unit Vector \[ \|\mathbf{u}\| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = 1. \] The direction vector is already a unit vector. Step 2: Compute Gradient (Assuming \( f(x, y) = x^2 + y^2 \)) \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2x, 2y). \] At \( (1, 0) \): \[ \nabla f(1, 0) = (2, 0). \] Step 3: Directional Derivative \[ D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = (2, 0) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) = 1. \] Final Answer For \( f(x, y) = x^2 + y^2 \), the directional derivative is \(\boxed{1.0}\). Note The problem is incomplete without the explicit form of \( f \). The answer depends on the function's gradient at \( (1, 0) \).