Question

A function is defined in Cartesian coordinate system as \[ f(x, y) = x e^y. \] The value of the directional derivative of the function (in integer) at the point \( (2, 0) \) along the direction of the straight line segment from point \( (2, 0) \) to point \( \left( \frac{1}{2}, 2 \right) \) is \(\underline{\hspace{1cm}}\).

Hint:

The directional derivative of a function at a point can be calculated using the gradient and the unit vector in the direction of the line.

Answer 1

Correct Answer: 1

The formula for the directional derivative of a function \( f(x, y) \) at a point \( (x_0, y_0) \) along a unit vector \( \mathbf{v} = (v_x, v_y) \) is: \[ D_{\mathbf{v}} f(x_0, y_0) = \nabla f(x_0, y_0) \cdot \mathbf{v} \] where \( \nabla f(x_0, y_0) \) is the gradient of \( f(x, y) \) at \( (x_0, y_0) \), and \( \mathbf{v} \) is the unit vector in the direction of the line. The gradient of \( f(x, y) = x e^y \) is: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( e^y, x e^y \right) \] At \( (x_0, y_0) = (2, 0) \), the gradient is: \[ \nabla f(2, 0) = \left( e^0, 2 e^0 \right) = (1, 2) \] The direction vector from \( (2, 0) \) to \( \left( \frac{1}{2}, 2 \right) \) is: \[ \mathbf{v} = \left( \frac{1}{2} - 2, 2 - 0 \right) = \left( -\frac{3}{2}, 2 \right) \] To make it a unit vector, we divide by its magnitude: \[ |\mathbf{v}| = \sqrt{\left( -\frac{3}{2} \right)^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} \] Thus, the unit vector is: \[ \mathbf{v} = \left( \frac{-3}{5}, \frac{4}{5} \right) \] Now, the directional derivative is: \[ D_{\mathbf{v}} f(2, 0) = \nabla f(2, 0) \cdot \mathbf{v} = (1, 2) \cdot \left( \frac{-3}{5}, \frac{4}{5} \right) \] \[ D_{\mathbf{v}} f(2, 0) = 1 \times \frac{-3}{5} + 2 \times \frac{4}{5} = \frac{-3}{5} + \frac{8}{5} = \frac{5}{5} = 1 \] Thus, the directional derivative at the point \( (2, 0) \) is \( \boxed{1} \).