91Ó°ÊÓ

When dealing with multivariable functions, the gradient is an essential tool. It's a vector that points in the direction of greatest increase of a function. The gradient of a function \( f(x, y, z) \), denoted by \( abla f \), can be thought of as a collection of partial derivatives. For the function given in the exercise, \( f(x, y, z) = y \sin^{-1}(xz) \), the gradient \( abla f \) is:

• \( \frac{\partial f}{\partial x} = \frac{y \cdot z}{\sqrt{1 - (xz)^2}} \)
• \( \frac{\partial f}{\partial y} = \sin^{-1}(xz) \)
• \( \frac{\partial f}{\partial z} = \frac{y \cdot x}{\sqrt{1 - (xz)^2}} \)

This gradient vector shows how \( f \) changes as we vary the input slightly in the direction of the x, y, or z axes.
By evaluating the gradient at a specific point, we can find out how the function behaves at that particular location. In our problem, substituting the point \( P \) into the gradient leads to \( abla f(P) = (0, \frac{\pi}{6}, 0) \).
This tells us that at point \( P \), the function remains unchanged if we move slightly in the x or z direction, but increases as \( \frac{\pi}{6} \) as we slightly increase \( y \).

Partial derivatives are the building blocks of the gradient. They show how a multivariable function changes as you vary only one variable, keeping others constant. Let's consider the function \( f(x, y, z) = y \sin^{-1}(xz) \). Each partial derivative represents the rate of change of the function with respect to one of its variables:

• \( \frac{\partial f}{\partial x} \) measures the change in \( f \) as \( x \) alone is varied, with \( y \) and \( z \) held constant.
• \( \frac{\partial f}{\partial y} \) considers the variation in \( f \) when \( y \) changes, but \( x \) and \( z \) are unchanged.
• \( \frac{\partial f}{\partial z} \) captures how altering \( z \) affects \( f \), with other variables fixed.

For our specific function, these partial derivatives help form the gradient vector, which we then use to evaluate shifts at the point \( P \). Given that the partial derivative with respect to \( y \) is \( \sin^{-1}(xz) \), it results in \( \frac{\pi}{6} \) when evaluated at \( P \), indicating how the function \( f \) changes as \( y \) is altered.

A unit vector is critical when calculating the directional derivative as it standardizes the direction, ensuring it has a magnitude of 1. For the vector \( \mathbf{a} = -2\mathbf{i} - 2\mathbf{j} - 2\mathbf{k} \), normalization is necessary to convert \( \mathbf{a} \) into a unit vector \( \mathbf{\hat{u}} \).
Here's a simple explanation of the process:

• Calculate the magnitude of \( \mathbf{a} \): \( \| \mathbf{a} \| = \sqrt{12} = 2\sqrt{3} \).
• Normalize by dividing each component by the magnitude: \( \mathbf{\hat{u}} = \left( \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \right) \).

This unit vector provides the standardized direction necessary for calculating the directional derivative. It ensures consistency, allowing us to measure how rapidly a function \( f \) changes in exactly this chosen direction.