Part 1 of 8 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 \( \langle\square, \square \). (Type exact answers, using radicals as needed.)

To find the gradient of the function \( f(x, y) = \sqrt{3 + 2x^2 + 2y^2} \), we first compute the partial derivatives of \( f \) with respect to \( x \) and \( y \). 1. **Partial derivative with respect to \( x \)**: Using the chain rule: \[ \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{3 + 2x^2 + 2y^2}} \cdot (4x) = \frac{2x}{\sqrt{3 + 2x^2 + 2y^2}} \] 2. **Partial derivative with respect to \( y \)**: Similarly, \[ \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{3 + 2x^2 + 2y^2}} \cdot (4y) = \frac{2y}{\sqrt{3 + 2x^2 + 2y^2}} \] Now, we evaluate these partial derivatives at the point \( P(\sqrt{3}, -1) \). First, we find \( 3 + 2(\sqrt{3})^2 + 2(-1)^2 \): \[ 3 + 2(3) + 2(1) = 3 + 6 + 2 = 11 \] So, \( \sqrt{3 + 2x^2 + 2y^2} \) at \( P \) is \( \sqrt{11} \). Now, substituting \( x = \sqrt{3} \) and \( y = -1 \): - For \( \frac{\partial f}{\partial x} \): \[ \frac{\partial f}{\partial x} \bigg|_P = \frac{2\sqrt{3}}{\sqrt{11}} \] - For \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial y} \bigg|_P = \frac{2(-1)}{\sqrt{11}} = \frac{-2}{\sqrt{11}} \] Thus, the gradient at \( P \) is: \[ \nabla f(P) = \left\langle \frac{2\sqrt{3}}{\sqrt{11}}, \frac{-2}{\sqrt{11}} \right\rangle \] So the gradient at \( P \) is \( \langle \frac{2\sqrt{3}}{\sqrt{11}}, \frac{-2}{\sqrt{11}} \rangle \).