A function \( f \) and a point \( P \) are given. Let \( \theta \) correspond to the direction of the directional derivative. Complete parts a. through e. \( f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) \) (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) What angles are associated with the direction of maximum decrease? \( \frac{5 \pi}{6} \) (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) What angles are associated with the direction of zero change? \( \frac{\pi}{3}, \frac{4 \pi}{3} \) (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) c. Write the directional derivative at \( P \) as a function of \( \theta \); call this function \( g(\theta) \). \( \frac{g(\theta)=\square}{(T y p e ~ a n ~ e x a c t ~ a n s w e r, ~ u s i n g ~ r a d i c a l s ~ a s ~ n e e d e d .) ~} \)

To find the directional derivative at point \( P \) as a function of the angle \( \theta \), we start by finding the gradient of the function \( f(x, y) \). The gradient \( \nabla f \) points in the direction of maximum increase, while the negative of the gradient will point in the direction of maximum decrease. Starting with the function, we calculate its partial derivatives: \[ f_x = \frac{\partial f}{\partial x} = \frac{2x}{\sqrt{3 + 2x^2 + 2y^2}}, \quad f_y = \frac{\partial f}{\partial y} = \frac{2y}{\sqrt{3 + 2x^2 + 2y^2}}. \] At the point \( P(\sqrt{3}, -1) \): \[ f_x(\sqrt{3}, -1) = \frac{2\sqrt{3}}{\sqrt{3 + 6 + 2}} = \frac{2\sqrt{3}}{3}. \] \[ f_y(\sqrt{3}, -1) = \frac{-2}{\sqrt{3 + 6 + 2}} = \frac{-2}{3}. \] Thus, the gradient at point \( P \) is: \[ \nabla f(\sqrt{3}, -1) = \left( \frac{2\sqrt{3}}{3}, -\frac{2}{3} \right). \] The directional derivative \( D_{\theta}f \) at angle \( \theta \) can be expressed as: \[ D_{\theta}f = \nabla f \cdot \begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix} = \left( \frac{2\sqrt{3}}{3} \cos(\theta) - \frac{2}{3} \sin(\theta) \right). \] So, the function \( g(\theta) \) representing the directional derivative at \( P \) is: \[ g(\theta) = \frac{2\sqrt{3}}{3} \cos(\theta) - \frac{2}{3} \sin(\theta). \]