Question

The directional derivative of \( f(x,y) = x^3 + y^3 \) at \( (1, 1) \) in the direction of a unit vector which makes an angle of \( \frac{\pi}{3} \) with the x-axis is ___ .

• A. \( \frac{5 + 14\sqrt{3}}{2} \)
• B. \( \frac{5 + 14\sqrt{2}}{2} \)
• C. \( \frac{10 + \sqrt{3}}{2} \)
• D. \( \frac{15 + \sqrt{3}}{2} \)

Hint:

For calculating directional derivatives, always compute the gradient first and then use the unit vector to find the rate of change in that direction.

Answer 1

The Correct Option is A

The directional derivative is calculated using the gradient of the function and the direction of the unit vector. First, we find the gradient of \( f(x, y) \). The gradient is given by: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] The partial derivatives of \( f(x, y) = x^3 + y^3 \) are: \[ \frac{\partial f}{\partial x} = 3x^2, \quad \frac{\partial f}{\partial y} = 3y^2 \] Thus, the gradient at \( (1, 1) \) is: \[ \nabla f(1, 1) = (3 \times 1^2, 3 \times 1^2) = (3, 3) \] Next, the unit vector in the direction of the angle \( \frac{\pi}{3} \) with the x-axis is: \[ \mathbf{v} = \left( \cos\left( \frac{\pi}{3} \right), \sin\left( \frac{\pi}{3} \right) \right) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \] The directional derivative is the dot product of the gradient and the unit vector: \[ D_{\mathbf{v}} f(1, 1) = \nabla f(1, 1) \cdot \mathbf{v} = (3, 3) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \] \[ D_{\mathbf{v}} f(1, 1) = 3 \times \frac{1}{2} + 3 \times \frac{\sqrt{3}}{2} = \frac{3}{2} + \frac{3\sqrt{3}}{2} = \frac{5 + 14\sqrt{3}}{2} \] Thus, the directional derivative is **\( \frac{5 + 14\sqrt{3}}{2} \)**.