A function \( f \) and a point \( P \) are given. Let \( \theta \) correspond to the direction of the directional derivative. Complete parts a. through \( e \). \( f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) \) a. Find the gradient and evaluate it at \( P \). The gradient at \( P \) is \( \left\langle\frac{2 \sqrt{3}}{\sqrt{11}},-\frac{2}{\sqrt{11}}\right) \). (Type exact answers, using radicals as needed.) 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 angles 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.)

1. We are given \[ f(x,y)=\sqrt{3+2x^2+2y^2} \] and the point \[ P\bigl(\sqrt{3},\,-1\bigr). \] 2. The gradient is given by \[ \nabla f(x,y)=\left\langle f_x(x,y),\,f_y(x,y)\right\rangle, \] where \[ f_x(x,y)=\frac{2x}{\sqrt{3+2x^2+2y^2}} \quad \text{and} \quad f_y(x,y)=\frac{2y}{\sqrt{3+2x^2+2y^2}}. \] 3. Evaluating at \( P \) we have \[ 3+2x^2+2y^2\big|_{x=\sqrt{3},\,y=-1} = 3+2(3)+2(1)= 3+6+2=11. \] Thus, \[ f_x\bigl(\sqrt{3},-1\bigr)=\frac{2\sqrt{3}}{\sqrt{11}}, \quad f_y\bigl(\sqrt{3},-1\bigr)=\frac{2(-1)}{\sqrt{11}}=-\frac{2}{\sqrt{11}}. \] Therefore, \[ \nabla f\bigl(\sqrt{3},-1\bigr)=\left\langle \frac{2\sqrt{3}}{\sqrt{11}},\,-\frac{2}{\sqrt{11}}\right\rangle. \] 4. The **direction of maximum increase** is in the direction of the gradient. To find its angle \(\theta\) (with respect to the positive \(x\)-axis), we note that the gradient vector components are: \[ \left( \frac{2\sqrt{3}}{\sqrt{11}},\, -\frac{2}{\sqrt{11}} \right). \] The angle \(\theta\) is given by \[ \theta=\arctan\left(\frac{y\text{-component}}{x\text{-component}}\right)=\arctan\left(\frac{-\frac{2}{\sqrt{11}}}{\frac{2\sqrt{3}}{\sqrt{11}}}\right)=\arctan\left(-\frac{1}{\sqrt{3}}\right). \] 5. Since \[ \arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}, \] it follows that \[ \arctan\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}. \] However, we need the angle measured in radians between \(0\) and \(2\pi\). Adding \(2\pi\) we obtain \[ -\frac{\pi}{6}+2\pi=\frac{11\pi}{6}. \] Thus, the angle associated with the **direction of maximum increase** is \[ \boxed{\frac{11\pi}{6}}. \] 6. (For completeness, note that the **direction of maximum decrease** is the direction opposite to the gradient. Its angle is obtained by adding \(\pi\) to \(\frac{11\pi}{6}\) and reducing modulo \(2\pi\): \[ \frac{11\pi}{6}+\pi=\frac{11\pi+6\pi}{6}=\frac{17\pi}{6}. \] Since \(\frac{17\pi}{6}>2\pi\), subtract \(2\pi\) (i.e. \(\frac{12\pi}{6}\)) to obtain \[ \frac{17\pi}{6}-\frac{12\pi}{6}=\frac{5\pi}{6}. \] Also, the directions for **zero change** are those perpendicular to the gradient. These are given by adding and subtracting \(\frac{\pi}{2}\) to \(\frac{11\pi}{6}\). That yields \[ \frac{11\pi}{6}-\frac{\pi}{2}=\frac{11\pi-3\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3}, \] and \[ \frac{11\pi}{6}+\frac{\pi}{2}=\frac{11\pi+3\pi}{6}=\frac{14\pi}{6}=\frac{7\pi}{3}. \] Reducing \(\frac{7\pi}{3}\) modulo \(2\pi\), \[ \frac{7\pi}{3}-2\pi=\frac{7\pi-6\pi}{3}=\frac{\pi}{3}. \] Thus, the zero change directions correspond to \(\theta=\frac{\pi}{3}\) and \(\theta=\frac{4\pi}{3}\).) 7. The answer for part (b) is: The **direction of maximum increase** is given by \[ \boxed{\frac{11\pi}{6}}. \]