Question

Given a scalar function \[ V(x, y) = \frac{1}{2} (x^2 + y^2), \] the directional derivative of V in the direction of the vector field \(3yi - 3xj\) at the point (1, 1) is

• A. \( \sqrt{18} \)
• B. 0
• C. \( \dfrac{1}{\sqrt{18}} \)
• D. \( \dfrac{3}{2} \)

Hint:

The directional derivative gives the rate of change of a function in the direction of a vector, and it depends on the gradient and the direction vector.

Answer 1

The Correct Option is B

The directional derivative of a scalar function \(V(x, y)\) in the direction of a vector \( \mathbf{u} \) is given by \[ D_{\mathbf{u}} V = \nabla V \cdot \mathbf{u}, \] where \( \nabla V \) is the gradient of \(V\) and \( \mathbf{u} \) is the unit vector in the direction of the given vector field. Step 1: Compute the Gradient of \( V(x, y) \)
First, compute the gradient of \( V(x, y) = \frac{1}{2} (x^2 + y^2) \): \[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) = (x, y). \] Step 2: Compute the Direction of the Vector Field
The given vector field is \( \mathbf{F} = 3yi - 3xj \), so at the point \( (1, 1) \), \[ \mathbf{F}(1, 1) = 3(1)i - 3(1)j = 3i - 3j. \] The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{F}(1, 1) \) is given by \[ \mathbf{u} = \frac{\mathbf{F}}{|\mathbf{F}|} = \frac{3i - 3j}{\sqrt{3^2 + 3^2}} = \frac{3i - 3j}{3\sqrt{2}} = \frac{i - j}{\sqrt{2}}. \] Step 3: Compute the Directional Derivative
Now, compute the directional derivative: \[ D_{\mathbf{u}} V = \nabla V \cdot \mathbf{u} = (x, y) \cdot \frac{i - j}{\sqrt{2}} = (1, 1) \cdot \frac{1 - 1}{\sqrt{2}} = 0. \] Thus, the correct answer is Option (B). The directional derivative is zero.