Find the directional derivative of the function in the direction of the unit vector \( \mathbf{u}=\cos (\theta) \mathbf{i}+\sin (\theta) \mathbf{j} \). \[ f(x, y)=\sin (16 x+19 y), \quad \theta=\frac{\pi}{3} \]

To find the directional derivative of the function \( f(x, y) = \sin(16x + 19y) \) in the direction of the unit vector \( \mathbf{u} = \cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j} \) where \( \theta = \frac{\pi}{3} \), we first need to compute the gradient of the function \( f \). The gradient \( \nabla f \) is given by: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Calculating the partial derivatives: 1. \( \frac{\partial f}{\partial x} = 16 \cos(16x + 19y) \) 2. \( \frac{\partial f}{\partial y} = 19 \cos(16x + 19y) \) Thus, \[ \nabla f = \left( 16 \cos(16x + 19y), 19 \cos(16x + 19y) \right) \] Now we find \( \mathbf{u} \): \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] So, \[ \mathbf{u} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \] Now, we compute the directional derivative \( D_{\mathbf{u}} f \) using the formula: \[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} \] Calculating the dot product: \[ D_{\mathbf{u}} f = \left( 16 \cos(16x + 19y), 19 \cos(16x + 19y) \right) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \] \[ D_{\mathbf{u}} f = 16 \cos(16x + 19y) \cdot \frac{1}{2} + 19 \cos(16x + 19y) \cdot \frac{\sqrt{3}}{2} \] \[ D_{\mathbf{u}} f = \frac{16}{2} \cos(16x + 19y) + \frac{19\sqrt{3}}{2} \cos(16x + 19y) \] \[ D_{\mathbf{u}} f = \frac{1}{2} \cos(16x + 19y) (16 + 19\sqrt{3}) \] Therefore, the directional derivative of the function \( f \) in the direction of the unit vector \( \mathbf{u} \) is: \[ D_{\mathbf{u}} f = \frac{1}{2} \cos(16x + 19y) (16 + 19\sqrt{3}) \]