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17.5 Curl and Divergence

gradient: \(\vec{\nabla} f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle\)

Note: In the expression \(\vec{\nabla} f\), \(\vec{\nabla}\) is an operator acting on a scalar \(f\) to produce a vector. It must be vector-like.

The Del Operator

\(\vec{\nabla}\) is called the "del operator" defined as

\[\vec{\nabla} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle = \vec{i} \frac{\partial}{\partial x} + \vec{j} \frac{\partial}{\partial y} + \vec{k} \frac{\partial}{\partial z}\]

in 2D, \(\vec{\nabla} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right\rangle = \vec{i} \frac{\partial}{\partial x} + \vec{j} \frac{\partial}{\partial y}\)

\(\vec{\nabla}\) is a vector-like operator, but it means nothing until applied to a mathematical object (like \(\sin, \cos\))

gradient: \(\vec{\nabla} f = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle\)

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\(\vec{\nabla}\) is like a vector, so we can dot and cross it with another vector.

\(\text{curl } \vec{F} = \vec{\nabla} \times \vec{F}\)

\(\text{curl } \vec{F}\) is defined as \(\vec{\nabla} \times \vec{F}\)

2D Case Verification

last time we defined \(\text{curl } \vec{F}\) when \(\vec{F}\) is 2D: \(\vec{F} = \langle f, g, 0 \rangle\)

from last time: \(\text{curl } \vec{F} = \langle 0, 0, g_x - f_y \rangle\)

verify with the more general formula: \(\vec{\nabla} \times \vec{F}\)

\(\vec{\nabla} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle\)

\(\vec{F} = \langle f, g, 0 \rangle\)

\[\text{curl } \vec{F} = \vec{\nabla} \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & 0 \end{vmatrix}\]

\[= \left\langle \frac{\partial 0}{\partial y} - \frac{\partial g}{\partial z}, -\left( \frac{\partial 0}{\partial x} - \frac{\partial f}{\partial z} \right), \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right\rangle = \langle 0, 0, g_x - f_y \rangle\]

Note: \(\frac{\partial g}{\partial z} = 0\) and \(\frac{\partial f}{\partial z} = 0\) because \(f, g\) do not depend on \(z\). This matches the definition from last time.

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Example: Calculating the Curl of a Vector Field

Given the vector field:

\[ \vec{F} = \langle xy^2z^3, x^3yz^2, x^2y^3z \rangle \]

We want to find the curl of \( \vec{F} \):

\[ \text{curl } \vec{F} = \nabla \times \vec{F} = ? \]

\[ = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy^2z^3 & x^3yz^2 & x^2y^3z \end{vmatrix} \]

\[ = \left\langle \frac{\partial}{\partial y}(x^2y^3z) - \frac{\partial}{\partial z}(x^3yz^2), -\left( \frac{\partial}{\partial x}(x^2y^3z) - \frac{\partial}{\partial z}(xy^2z^3) \right), \frac{\partial}{\partial x}(x^3yz^2) - \frac{\partial}{\partial y}(xy^2z^3) \right\rangle \]

\[ = \langle 3x^2y^2z - 2x^3yz, 3xy^2z^2 - 2xy^3z, 3x^2yz^2 - 2xy^2z^3 \rangle \]

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From a previous lesson → conservative vector field is irrotational

(if \( \vec{F} \) is conservative, then \( \nabla \times \vec{F} = \vec{0} \))

Proof of the Statement

If \( \vec{F} \) is conservative, then:

\[ \vec{F} = \left\langle \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right\rangle = \nabla \phi = \text{grad } \phi \]

Then the curl of the gradient is:

\[ \nabla \times (\nabla \phi) = \text{curl}(\text{grad } \phi) \]

\[ = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{vmatrix} \]

\[ = \left\langle \frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}, -\left( \frac{\partial^2 \phi}{\partial x \partial z} - \frac{\partial^2 \phi}{\partial z \partial x} \right), \frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x} \right\rangle = \vec{0} \]

\( \phi_{zy} - \phi_{yz} = 0 \)

(mixed partials are equal)

So, this proves the statement.

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Divergence of a Vector Field

The dot product of \(\vec{\nabla}\) with \(\vec{F}\) gives us the divergence of \(\vec{F}\).

\(\text{div } \vec{F} = \vec{\nabla} \cdot \vec{F}\)

In 2D, \(\vec{F} = \langle f, g, 0 \rangle\)

\[ \text{div } \vec{F} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \cdot \langle f, g, 0 \rangle \]

\[ = f_x + g_y \quad \text{(same as how we defined last time)} \]

Example

\(\vec{F} = \langle xy^2z^3, x^3yz^2, x^2y^3z \rangle\)

\[ \text{div } \vec{F} = \vec{\nabla} \cdot \vec{F} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \cdot \langle xy^2z^3, x^3yz^2, x^2y^3z \rangle \]

\[ = \frac{\partial}{\partial x}(xy^2z^3) + \frac{\partial}{\partial y}(x^3yz^2) + \frac{\partial}{\partial z}(x^2y^3z) \]

note this is a scalar!

\[ = y^2z^3 + x^3z^2 + x^2y^3 \]

Divergence can be interpreted as the volume change of a small imaginary box that travels w/ the vector field.

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Figure: Vector field plot with arrows pointing away from the origin, showing increasing magnitude in the first quadrant.

greater y-component pull at top, so as it flows, it will look like

the increase in volume indicates a positive divergence in first quadrant

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Vector Field Analysis

Consider the vector field defined by:

\[ \vec{F} = \langle x^2 - y^2 - 4, 2xy \rangle \]

Find curl and divergence.

Figure: A 2D vector field plot on a Cartesian coordinate system from -5 to 5 on both axes.

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Calculating Divergence

The divergence of the vector field \( \vec{F} \) is calculated as follows:

\[ \text{div } \vec{F} = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle \cdot \langle x^2 - y^2 - 4, 2xy, 0 \rangle \]

\[ = 2x + 2x = 4x \]

\( \text{div } \vec{F} = 4x \)

\( \text{div } \vec{F} \) at a point \( (x, y) = 4x \). Divergence can have different values at different locations.

Calculating Curl

The curl of the vector field is:

\[ \text{curl } \vec{F} = \langle 0, 0, 4y \rangle \neq \vec{0} \text{ all the time} \]

So this \( \vec{F} \) is NOT conservative. The rotation strength depends on \( (x, y) \), too.