Question

Find the divergence and curl of the given vector field. \(\vec{F}=\nabla f,\) where \(f(x, y, z)=x^{2} y+\sin z\)

Short answer

Divergence is \(2y\); curl is zero vector.

Step-by-step solution

Compute the Gradient of f

First, we'll determine the gradient of the scalar field function, \(f(x, y, z) = x^2 y + \sin z\). The gradient, \(abla f\), is a vector field composed of the partial derivatives of \(f\) with respect to each variable. Compute these as follows:1. \(\frac{\partial f}{\partial x} = 2xy\)2. \(\frac{\partial f}{\partial y} = x^2\)3. \(\frac{\partial f}{\partial z} = \cos z\)So, \(abla f = \langle 2xy, x^2, \cos z \rangle\).

Find the Divergence of \(\vec{F}\)

To find the divergence of \(\vec{F}\), which is \(abla \cdot \vec{F}\), compute the divergence of the gradient vector field \(abla f = \langle 2xy, x^2, \cos z \rangle\):\[abla \cdot \vec{F} = \frac{\partial}{\partial x}(2xy) + \frac{\partial}{\partial y}(x^2) + \frac{\partial}{\partial z}(\cos z)\]Calculate each term:1. \(\frac{\partial}{\partial x}(2xy) = 2y\)2. \(\frac{\partial}{\partial y}(x^2) = 0\)3. \(\frac{\partial}{\partial z}(\cos z) = -\sin z\)Thus, \(abla \cdot \vec{F} = 2y\).

Find the Curl of \(\vec{F}\)

To find the curl of \(\vec{F}\), determine \(abla \times \vec{F}\) for the gradient vector \(abla f = \langle 2xy, x^2, \cos z \rangle\). Remember that the curl of a gradient is always zero:\[abla \times \vec{F} = \mathbf{0}\]This is because the curl of a gradient field results in a zero vector field. Therefore, \(abla \times abla f = \mathbf{0}\).

Key concepts

Divergence

Imagine you are standing in a field of arrows, each representing a vector field at different points in space. The divergence of a vector field helps you understand how much the vectors diverge or spread out from a point. In mathematical terms, divergence is a scalar value calculated from a vector field, expressed as \( abla \cdot \vec{F} \). The divergence operates on a vector by summing the partial derivatives of each component of the vector field with respect to its corresponding variable.
For example, for a vector field \( \vec{F} = \langle F_x, F_y, F_z \rangle \), the divergence is calculated as:

• \( \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \)

In the given problem, we found that the divergence of the gradient field \( \vec{F} = \langle 2xy, x^2, \cos z \rangle \) resolved to \( 2y \). This means that the vector field "spreads out" at a rate proportional to the value of \( 2y \) at any point in the field.

Gradient

The gradient is a crucial concept in vector calculus. It provides a way to measure how a scalar field changes at each point in space. Think of a scalar field as a mountain range, where the height represents different values of the function. The gradient points in the direction where the function increases the most, similar to the steepest path up the mountain.
For a scalar function \( f \), the gradient is a vector field represented by taking partial derivatives of \( f \) with respect to each variable. In our example, we worked with the function \( f(x, y, z) = x^2 y + \sin z \). The gradient, denoted as \( abla f \), is:

• \( \frac{\partial f}{\partial x} = 2xy \)
• \( \frac{\partial f}{\partial y} = x^2 \)
• \( \frac{\partial f}{\partial z} = \cos z \)

Thus, the gradient \( abla f \) forms the vector field \( \langle 2xy, x^2, \cos z \rangle \). This vector field shows the rate and direction of the fastest increase in the function \( f \).

Curl

The curl of a vector field helps to discover rotational effects in the field, like swirls or curls in fluid flow. It measures the tendency to rotate around a point, much like water swirling around a drain.In general, the curl of a vector field \( \vec{F} = \langle F_x, F_y, F_z \rangle \) is given by:

• \( abla \times \vec{F} = \langle \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \rangle \)

However, in our exercise, \( \vec{F} \) was derived as a gradient \( abla f \). An important aspect of vector calculus is that the curl of a gradient is always zero. This means for any gradient field, such as \( \langle 2xy, x^2, \cos z \rangle \), its curl is:

• \( abla \times \vec{F} = \mathbf{0} \)

Thus, there is no rotation or swirling happening in the vector field derived from a gradient, which is corroborated by it being zero.