Gradient of a Square Root Function in 3D

Find the gradient of the function $f(x, y) = \sqrt{36 - x^2 - y^2}$ and evaluate it at the point $(-3, 2)$.

When dealing with gradients in multivariable calculus, it's important to understand that the gradient of a function gives you a vector that points in the direction of the greatest rate of increase of that function. For a function of two variables like $f(x, y)$, the gradient is a vector composed of the partial derivatives of the function with respect to each variable. This vector essentially maps the steepness and the direction of the incline at any given point in the function's domain.

In this problem, you are asked to find the gradient of the function $f(x, y) = \sqrt{36 - x^2 - y^2}$. The function represents a portion of the sphere with a radius of 6 centered at the origin, but it only includes the upper hemisphere since it is defined as the square root of a potentially non-positive expression. The constraint within the square root indicates that $(x, y)$ points must lie within or on this circle in the xy-plane to ensure the output is real.

After computing the expression for the gradient, evaluating it at a specific point, in this case $(-3, 2)$, involves substituting these values into the gradient. This allows you to determine the direction and rate of change of the function at that particular point. Such evaluations are useful in fields that require optimization or in understanding how a function behaves near critical points. Familiarizing yourself with gradient calculations and interpretations is a key competency in multivariable calculus, which will support further exploration into topics such as linearization and optimization.

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