91Ó°ÊÓ

When we talk about the gradient of a function, particularly in multiple variables, it is an extension of the concept of a derivative. Instead of providing the rate of change in a single direction, it indicates the direction of the steepest ascent for the function at a given point. The gradient is composed of the partial derivatives with respect to each variable of the function.

For example, if we take a function like the one given in the exercise, \( f(x, y, z) = \ln(x^2 + 2y^2 + 3z^2) \), the gradient \( abla f \) will be a vector that includes these partial derivatives. At each point, this vector indicates the proportional rate of change along each axis:

• The derivative with regard to \( x \)
• The derivative with regard to \( y \)
• The derivative with regard to \( z \)

This means for any inputs \( x, y, \) and \( z \), the gradient vector \( abla f \) is \( \left( \frac{2x}{x^2 + 2y^2 + 3z^2}, \frac{4y}{x^2 + 2y^2 + 3z^2}, \frac{6z}{x^2 + 2y^2 + 3z^2} \right) \). Each element represents how the function value changes as each respective variable changes, providing a complete picture of the function's behaviour at that point.

Partial derivatives are a cornerstone of multi-variable calculus, representing the derivative of a multi-variable function with respect to just one of those variables. When computing partial derivatives, we treat all other variables as constants.

In our exercise, to find the gradient \( abla f \), we need partial derivatives of the function \( f \) with respect to each of its variables \( x, y, \) and \( z \). Let's break it down:

• The partial derivative with respect to \( x \) is given by \( \frac{\partial f}{\partial x} = \frac{2x}{x^2 + 2y^2 + 3z^2} \). This measures how \( f \) changes as \( x \) changes, while \( y \) and \( z \) stay constant.
• Similarly, \( \frac{\partial f}{\partial y} = \frac{4y}{x^2 + 2y^2 + 3z^2} \) captures the variability in the \( y \) direction.
• The expression \( \frac{\partial f}{\partial z} = \frac{6z}{x^2 + 2y^2 + 3z^2} \) does the same for changes in \( z \).

These partial derivatives provide the building blocks of the gradient, allowing us to see the direction and rate of the change of the function considering all its variables independently.

The dot product (or scalar product) is an operation that takes two equal-length sequences of numbers, often vectors, and returns a single number. It can be interpreted as the projection of one vector onto another, and it plays a crucial role in finding directional derivatives.

In the context of our exercise, to find the directional derivative \( D_{\mathfrak{u}} f \) at a point \( P \), we take the dot product of the vector given by the gradient \( abla f \) evaluated at that point, and the unit vector \( \mathbf{u} \) which represents the direction in which we're interested in exploring the function's behaviour.

• The gradient \( abla f(P) \) at the point \( P(-1, 2, 4) \) is calculated as \( \left( \frac{-2}{57}, \frac{8}{57}, \frac{24}{57} \right) \).
• The direction vector \( \mathbf{u} = -\frac{3}{13} \mathbf{i}-\frac{4}{13} \mathbf{j}-\frac{12}{13} \mathbf{k} \) is already a unit vector which we use directly in our calculation.

To find the directional derivative, compute \( abla f(P) \cdot \mathbf{u} \). This involves multiplying corresponding components of the vectors and then summing these products, leading to an expression that represents how much the function \( f \) changes at \( P \) in the direction of \( \mathbf{u} \). This is a crucial technique in analyzing how functions behave in specific directions, particularly useful in fields such as physics and engineering.