Find the directional derivative of the function at \( P \) in the direction of \( \mathbf{v} \). \[ f(x, y)=\frac{x}{y}, \quad P(2,2), \mathbf{v}=-\mathbf{j} \]

To find the directional derivative of the function \( f(x, y) = \frac{x}{y} \) at the point \( P(2,2) \) in the direction of the vector \( \mathbf{v} = -\mathbf{j} \), we first need to find the gradient \( \nabla f \). 1. **Calculate the gradient \( \nabla f \)**: - The partial derivative with respect to \( x \): \[ f_x = \frac{\partial}{\partial x} \left(\frac{x}{y}\right) = \frac{1}{y} \] - The partial derivative with respect to \( y \): \[ f_y = \frac{\partial}{\partial y} \left(\frac{x}{y}\right) = -\frac{x}{y^2} \] - Therefore, the gradient is: \[ \nabla f = \left( \frac{1}{y}, -\frac{x}{y^2} \right) \] 2. **Evaluate the gradient at point \( P(2, 2) \)**: \[ \nabla f(2, 2) = \left( \frac{1}{2}, -\frac{2}{2^2} \right) = \left( \frac{1}{2}, -\frac{2}{4} \right) = \left( \frac{1}{2}, -\frac{1}{2} \right) \] 3. **Normalize the direction vector \( \mathbf{v} \)**: The direction vector is \( \mathbf{v} = (0, -1) \). The magnitude of \( \mathbf{v} \) is: \[ ||\mathbf{v}|| = \sqrt{0^2 + (-1)^2} = 1 \] The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is: \[ \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} = (0, -1) \] 4. **Calculate the directional derivative**: The directional derivative \( D_{\mathbf{u}} f \) is given by: \[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} \] Now compute the dot product: \[ D_{\mathbf{u}} f = \left( \frac{1}{2}, -\frac{1}{2} \right) \cdot (0, -1) = \frac{1}{2} \cdot 0 + \left(-\frac{1}{2}\right) \cdot (-1) = 0 + \frac{1}{2} = \frac{1}{2} \] Thus, the directional derivative of the function \( f \) at the point \( P(2,2) \) in the direction of \( \mathbf{v} = -\mathbf{j} \) is \( \boxed{\frac{1}{2}} \).