To calculate the gradient of the function $p(x, y) = \sqrt{16 - 3x^2 - y^2}$, we need to compute the partial derivatives with respect to both $x$ and $y$.

Step 1: Partial derivative with respect to $x$

The gradient of $p(x, y)$ with respect to $x$ involves applying the chain rule.

$p(x, y) = (16 - 3x^2 - y^2)^{1/2}$

The derivative of the square root function is:

$\frac{d}{dx} \left[ (16 - 3x^2 - y^2)^{1/2} \right] = \frac{1}{2} (16 - 3x^2 - y^2)^{-1/2} \cdot \frac{d}{dx}(16 - 3x^2 - y^2)$

Now, take the derivative of the inside expression with respect to $x$:

$\frac{d}{dx} (16 - 3x^2 - y^2) = -6x$

So, the partial derivative of $p(x, y)$ with respect to $x$ is:

$\frac{\partial p}{\partial x} = \frac{1}{2} (16 - 3x^2 - y^2)^{-1/2} \cdot (-6x)$

Simplifying:

$\frac{\partial p}{\partial x} = \frac{-3x}{\sqrt{16 - 3x^2 - y^2}}$

Step 2: Partial derivative with respect to $y$

Now, take the partial derivative of $p(x, y)$ with respect to $y$, again applying the chain rule:

$\frac{d}{dy} \left[ (16 - 3x^2 - y^2)^{1/2} \right] = \frac{1}{2} (16 - 3x^2 - y^2)^{-1/2} \cdot \frac{d}{dy}(16 - 3x^2 - y^2)$

Taking the derivative of the inside expression with respect to $y$:

$\frac{d}{dy} (16 - 3x^2 - y^2) = -2y$

So, the partial derivative of $p(x, y)$ with respect to $y$ is:

$\frac{\partial p}{\partial y} = \frac{1}{2} (16 - 3x^2 - y^2)^{-1/2} \cdot (-2y)$

Simplifying:

$\frac{\partial p}{\partial y} = \frac{-y}{\sqrt{16 - 3x^2 - y^2}}$

Step 3: The gradient of $p(x, y)$

The gradient of $p(x, y)$, denoted as $\nabla p(x, y)$, is the vector formed by the partial derivatives with respect to $x$ and $y$:

$\nabla p(x, y) = \left( \frac{\partial p}{\partial x}, \frac{\partial p}{\partial y} \right)$

Substitute the expressions for the partial derivatives:

$\nabla p(x, y) = \left( \frac{-3x}{\sqrt{16 - 3x^2 - y^2}}, \frac{-y}{\sqrt{16 - 3x^2 - y^2}} \right)$

Thus, the gradient of $p(x, y)$ is:

$\nabla p(x, y) = \frac{-1}{\sqrt{16 - 3x^2 - y^2}} \left( 3x, y \right)$

Would you like more details on the application of this result or its meaning?

Related Questions:

1. How does the gradient of a function relate to its directional derivative?
2. What is the physical interpretation of the gradient in this context?
3. Can the gradient be used to find the critical points of the function?
4. How does the gradient change for functions defined in three dimensions?
5. How do gradients interact in multivariable optimization problems?

Tip: The gradient vector points in the direction of the steepest ascent of the function at any given point.