91Ó°ÊÓ

The gradient of a function is like a compass, guiding you in the direction of the steepest incline.
Imagine you're hiking up a mountain; the gradient shows the path that will take you uphill most quickly. Mathematically, for a function of several variables like \( f(x, y, z) \), the gradient is a vector consisting

• The partial derivative with respect to each variable (\( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \)).

For our function \( f(x, y, z) = x^3 y^2 z^5 - 2xz + yz + 3x \), the gradient is:

• \( abla f = (3x^2 y^2 z^5 - 2z + 3, 2x^3 y z^5 + z, 5x^3 y^2 z^4 - 2x + y) \)

By evaluating at a specific point, say \( P(-1, -2, 1) \), you find the precise direction of steepest ascent at that point, resulting in the vector \( (10, -3, -20) \).
The gradient thus helps you understand how the function behaves in various directions.

Partial derivatives are all about focusing on one variable while treating others as constants.
Think of them as a tool to see how a function changes along only one dimension

• Without interacting with the changes along other axes.

For our function \( f(x, y, z) \), computing the partial derivative with respect to one of the variables, like \( x \), examines the rate at which the function changes as \( x \) changes, while keeping \( y \) and \( z \) constant.
The computed partials were:

• \( \frac{\partial f}{\partial x} = 3x^2 y^2 z^5 - 2z + 3 \)
• \( \frac{\partial f}{\partial y} = 2x^3 y z^5 + z \)
• \( \frac{\partial f}{\partial z} = 5x^3 y^2 z^4 - 2x + y \)

Evaluating these at a specific point makes it possible to find how the function changes locally with respect to each variable.

A unit vector is a vector with a length of one.
It defines a direction but not a magnitude. As a result, it is often used in computations where direction is important but magnitude should not influence the results.

• The process often involves dividing a vector by its magnitude to transform it into a unit vector.

In this exercise, the unit vector \( \mathbf{u} = (0, 0, -1) \) points in the direction of the negative \( z \)-axis.
Its magnitude is 1, making it a perfect candidate to determine the directional derivative without affecting the scale of the gradient.
Using unit vectors, you can precisely calculate how a function like \( f(x, y, z) \) behaves in specific directions, like that of the negative \( z \)-axis, leading to a computation of the directional derivative (in this case, 20).