91Ó°ÊÓ

Calculating the gradient of a function is crucial to finding the directional derivative. A gradient is a vector that contains all the partial derivatives of a function, representing the rate of change in each respective direction.
The function given in the exercise is \( f(x, y) = \sqrt{x y} \). To find its gradient, we calculate the partial derivative with respect to \( x \) and \( y \).
For \( x \), the partial derivative \( \frac{\partial f}{\partial x} \) is calculated as \( \frac{1}{2\sqrt{x y}} \times y \), and for \( y \), \( \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x y}} \times x \).

The gradient vector then becomes \( abla f(x, y) = \left( \frac{y}{2\sqrt{x y}}, \frac{x}{2\sqrt{x y}} \right) \). Knowing how to compute this allows us to see how the function changes at different points.

Partial derivatives analyze how a function changes with respect to one variable while keeping others constant. They are the building blocks for gradient calculations.

• For the function \( f(x, y) = \sqrt{x y} \), the partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), indicates how much \( f \) changes with a small change in \( x \), holding \( y \) constant.
• Similarly, \( \frac{\partial f}{\partial y} \) shows the change in \( f \) with a small change in \( y \), keeping \( x \) constant.

These derivatives help construct the gradient, which is a vital element in understanding the function's behavior at any given point.

Normalization of a direction vector turns it into a unit vector, a vector with a length of one. This process is necessary because the directional derivative formula requires a unit vector.
To normalize a vector like \( (3, -4) \), first, compute its magnitude: \( \sqrt{3^2 + (-4)^2} = 5 \).
Then, divide each component of the vector by this magnitude to get the unit vector \( \left( \frac{3}{5}, \frac{-4}{5} \right) \).
This unit vector represents the direction precisely and ensures that the directional derivative is accurate without being influenced by the direction's length.

A unit vector has a magnitude of one, simplifying many vector calculations. Once you've normalized your direction vector, you've got a unit vector.
In this scenario, the unit vector \( \left( \frac{3}{5}, \frac{-4}{5} \right) \) implies that it points in the direction from point \((2,8)\) to point \((5,4)\), but with a consistent magnitude.
This helps standardize directions across different problems, providing a standardized measure of direction that is critical in the calculation of directional derivatives.

Evaluating a function at a specific point helps find exact gradient values at that point. This step forms the foundation for finding the directional derivative.
For \( f(x, y) = \sqrt{x y} \), evaluating at \( P(2,8) \) involves substituting \( x=2 \) and \( y=8 \) into the gradient expression \( abla f(x, y) = \left( \frac{y}{2\sqrt{x y}}, \frac{x}{2\sqrt{x y}} \right) \).
This resulted in \( abla f(2,8) = (1, 0.25) \).
By locating this evaluative point, you can discern how the function behaves in that specific area, laying the groundwork for applying the directional derivative in a clear direction.