91Ó°ÊÓ

Understanding the gradient vector is vital in multivariable calculus. The gradient of a scalar function \( \phi(x, y, z) \) is a vector field denoted by \(\abla \phi \). Each component of this vector is a partial derivative of the function with respect to one of its variables. This means that for \( \phi = x e^{y} + y z^{2} + x y z \), the gradient vector will look like this:

\(\frac{\partial \phi}{\partial x} = e^{y} + y z \) \(\frac{\partial \phi}{\partial y} = x e^{y} + z^{2} + x z \) \(\frac{\partial \phi}{\partial z} = 2 y z + x y \)

The gradient vector, \( \abla \phi \), captures the direction and rate of the fastest increase of the function. For example, at the point \( (2, 0, 3) \), the gradient vector evaluates to \((1, 11, 0)\). This vector tells us in which direction the function \( \phi(x, y, z) \) changes most rapidly and at what rate.

Partial derivatives are a cornerstone in understanding how functions of multiple variables change when altering one variable at a time. To find a partial derivative, we differentiate the function with respect to one variable while keeping the other variables constant. For instance, given the function \( \phi = x e^{y} + y z^{2} + x y z \):

The partial derivative with respect to \( x \) is \(\frac{\partial \phi}{\partial x} = e^{y} + y z \).
The partial derivative with respect to \( y \) is \(\frac{\partial \phi}{\partial y} = x e^{y} + z^{2} + x z \).
The partial derivative with respect to \( z \) is \(\frac{\partial \phi}{\partial z} = 2 y z + x y \).

These derivatives indicate how \( \phi \) changes as each individual variable is incremented slightly. Evaluating these at a specific point, like \( (2, 0, 3) \), gives immediate rates of change in those directions: \(\frac{\partial \phi}{\partial x}(2, 0, 3) = 1 \), \(\frac{\partial \phi}{\partial y}(2, 0, 3) = 11 \), and \(\frac{\partial \phi}{\partial z}(2, 0, 3) = 0\).

The dot product is a way to multiply two vectors to get a scalar value. It's essential in computing the directional derivative. Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), their dot product is calculated as:

\( \mathbf{a} \cdot \mathbf{b} = a_{x} \cdot b_{x} + a_{y} \cdot b_{y} + a_{z} b_{z} \)

In the context of the directional derivative, the dot product between the gradient vector \(\abla \phi \), and the unit direction vector \( \mathbf{u}_{\footnotesize \mathbf{A}} \), signifies how much of the gradient points in the specified direction. For our example:

\( \abla \phi (2, 0, 3) = (1, 11, 0) \)
\( \mathbf{u}_{\footnotesize \mathbf{A}} = \left( \frac{3}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right) \)

The dot product calculation is, therefore: \( \abla \phi (2, 0, 3) \cdot \mathbf{u}_{\footnotesize \mathbf{A}} = 1 \cdot \frac{3}{\sqrt{14}} + 11 \cdot \frac{-2}{\sqrt{14}} + 0 \cdot \frac{1}{\sqrt{14}} = \frac{-19}{\sqrt{14}} \).

Normalization is the process of converting a vector into a unit vector, i.e., a vector with a magnitude (length) of 1, while maintaining its direction. To normalize a vector \( \mathbf{A} = 3 \mathbf{i} - 2 \mathbf{j} + \mathbf{k} \), we divide each component by its magnitude. The magnitude of \( \mathbf{A} \) is calculated as:

\( \| \mathbf{A} \| = \sqrt{3^{2} + (-2)^{2} + 1^{2}} = \sqrt{9 + 4 + 1} = 14\)

Then, the unit vector \( \mathbf{u}_{\footnotesize \mathbf{A}} \) is:

\( \mathbf{u}_{\footnotesize \mathbf{A}} = \frac{1}{\sqrt{14}} (3, -2, 1)=\left(\frac{3}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{1}{\sqrt{14}}\right) \)

By using this normalized vector, we can effectively project onto the specified direction, ensuring that our directional derivative calculation is accurate.