Compute the directional derivative of the function \( g(x, y)=\sin (\pi(x-3 y)) \) at the point \( P(-3,1) \) in the direction \( \left\langle\frac{3}{5}, \frac{4}{5}\right) \). Be sure to use a unit vector for the direction vector. The directional derivative is (Type an exact answer, using \( \pi \).

To compute the directional derivative of the function \( g(x, y) = \sin (\pi(x - 3y)) \) at the point \( P(-3, 1) \) in the direction \( \mathbf{u} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \), we first need to find the gradient \( \nabla g(x, y) \). 1. **Calculate the gradient**: We find the partial derivatives of \( g \): \[ g_x = \frac{\partial g}{\partial x} = \pi \cos(\pi(x - 3y)) \] \[ g_y = \frac{\partial g}{\partial y} = -3\pi \cos(\pi(x - 3y)) \] The gradient is given by: \[ \nabla g(x, y) = \left\langle g_x, g_y \right\rangle = \left\langle \pi \cos(\pi(x - 3y)), -3\pi \cos(\pi(x - 3y)) \right\rangle \] 2. **Evaluate the gradient at the point \( P(-3, 1) \)**: First, we compute \( x - 3y \) at the point \( P(-3, 1) \): \[ -3 - 3(1) = -6 \] Now plug this into the gradient: \[ g_x(-3, 1) = \pi \cos(-6\pi) = \pi \cdot 1 = \pi \] \[ g_y(-3, 1) = -3\pi \cos(-6\pi) = -3\pi \cdot 1 = -3\pi \] Thus, \[ \nabla g(-3, 1) = \left\langle \pi, -3\pi \right\rangle \] 3. **Compute the directional derivative**: The directional derivative \( D_{\mathbf{u}} g = \nabla g \cdot \mathbf{u} \): \[ D_{\mathbf{u}} g = \left\langle \pi, -3\pi \right\rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \] Calculating this dot product: \[ D_{\mathbf{u}} g = \pi \cdot \frac{3}{5} + (-3\pi) \cdot \frac{4}{5} = \frac{3\pi}{5} - \frac{12\pi}{5} = \frac{-9\pi}{5} \] Thus, the directional derivative is \[ \boxed{\frac{-9\pi}{5}}. \]