A function \( f \) and a point \( P \) are given. Let \( \theta \) correspond to the direction of the directional derivative. Complete parts (a) through (e). \( f(x, y)=16-3 x^{2}-2 y^{2}, P(2,3) \) a. Find the gradient and evaluate it at \( P \). The gradient at \( P \) is \( \langle-12,-12\rangle \). b. Find the angles \( \theta \) (with respect to the positive \( x \)-axis) associated with the directions of maximum increase, maximum decrease, and zero change. What angle(s) is/are associated with the direction of maximum increase? \( \square \) (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.)

To find the angles associated with maximum increase, maximum decrease, and zero change, we start with the gradient evaluated at point \( P(2, 3) \), which is \( \nabla f(2, 3) = \langle -12, -12 \rangle \). The direction of maximum increase is along the gradient vector itself. To find the angle \( \theta \) of this vector with respect to the positive \( x \)-axis, we can use the tangent function: \[ \tan(\theta) = \frac{y}{x} = \frac{-12}{-12} = 1. \] The angle \( \theta \) corresponding to this tangent value is \( \frac{3\pi}{4} + k\pi \) for any integer \( k \). Within the range \( [0, 2\pi) \), the specific angle for maximum increase is \( \frac{5\pi}{4} \). The direction of maximum decrease is in the opposite direction of the gradient, which gives us \( \theta = \frac{5\pi}{4} + \pi = \frac{9\pi}{4} \). However, we can simplify this to \( \theta = \frac{\pi}{4} \) when we place it within the range. For zero change, the angles are \( \theta = \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), which are orthogonal to the gradient direction. So, we have: - Maximum increase direction angle: \( \frac{5\pi}{4} \) - Maximum decrease direction angle: \( \frac{3\pi}{4} \) - Zero change angles: \( \frac{\pi}{2}, \frac{3\pi}{2} \) The angles associated with maximum increase are \( \frac{5\pi}{4} \).