\( 1 \leftarrow \quad \) First, compute the gradient of the function \( p(x, y)=\sqrt{10-3 x^{2}-y^{2}} \). Then evaluate it at the point \( (-1,-1) \). The gradient is \( \nabla p(x, y)=\langle\square, \square \).

To compute the gradient of the function \( p(x, y) = \sqrt{10 - 3x^2 - y^2} \), we first need to find the partial derivatives with respect to \( x \) and \( y \). 1. Find \( p_x \): \[ p_x = \frac{\partial}{\partial x} (\sqrt{10 - 3x^2 - y^2}) = \frac{1}{2\sqrt{10 - 3x^2 - y^2}} \cdot (-6x) = \frac{-3x}{\sqrt{10 - 3x^2 - y^2}} \] 2. Find \( p_y \): \[ p_y = \frac{\partial}{\partial y} (\sqrt{10 - 3x^2 - y^2}) = \frac{1}{2\sqrt{10 - 3x^2 - y^2}} \cdot (-2y) = \frac{-y}{\sqrt{10 - 3x^2 - y^2}} \] Therefore, the gradient is given by: \[ \nabla p(x, y) = \left\langle -\frac{3x}{\sqrt{10 - 3x^2 - y^2}}, -\frac{y}{\sqrt{10 - 3x^2 - y^2}} \right\rangle \] Next, we evaluate the gradient at the point \( (-1, -1) \): 1. Plugging in \( x = -1 \) and \( y = -1 \): \[ \nabla p(-1, -1) = \left\langle -\frac{3(-1)}{\sqrt{10 - 3(-1)^2 - (-1)^2}}, -\frac{-1}{\sqrt{10 - 3(-1)^2 - (-1)^2}} \right\rangle \] Calculate the denominator: \[ 10 - 3(-1)^2 - (-1)^2 = 10 - 3 - 1 = 6 \] Therefore, \( \sqrt{6} \) is the result of that calculation. 2. Thus, \[ \nabla p(-1, -1) = \left\langle \frac{3}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle \] So the gradient evaluated at the point \( (-1, -1) \) is: \[ \nabla p(-1, -1) = \left\langle \frac{3}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle \] Finally, filling in the boxes gives: \[ \nabla p(x, y) = \left\langle \frac{3}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle \]