Question

Find the divergence of the vector field \(\mathrm{F}\). \(\mathbf{F}(x, y, z)=x e^{x} \mathbf{i}+y e^{y} \mathbf{j}\)

Short answer

The divergence of the vector field \( \mathbf{F}(x, y, z) =xe^{x} \mathbf{i}+ye^{y} \mathbf{j} \) is \( \mathrm{div}\ \mathbf{F} = e^{x} + xe^{x} + e^{y} + ye^{y} \)

Step-by-step solution

Understand The Divergence Formula

The formula for the divergence of a vector field in three dimensions \( \mathbf{F}(x, y, z) \) is given by \n\n\[\n\n\mathrm{div}\ \mathbf{F} = \frac{\partial F_i}{\partial x} + \frac{\partial F_j}{\partial y} + \frac{\partial F_k}{\partial z}\n\n\]\n\nwhere \( \mathbf{F_i}, \mathbf{F_j}, \mathbf{F_k} \) are the component functions of the vector field.

Apply Divergence Formula to Vector Field

Looking at the vector field \( \mathbf{F}(x, y, z) = xe^{x} \mathbf{i}+ye^{y} \mathbf{j} \), we notice that there is no dependence on \(z\) and therefore the third term in the divergence formula does not contribute. \n\nSo the divergence of the given vector field is \n\n\[\n\n\mathrm{div}\ \mathbf{F} = \frac{\partial}{\partial x}(x e^{x}) + \frac{\partial}{\partial y}(ye^{y})\n\n\]

Compute the Partial Derivatives

Using the product rule for differentiation, the partial derivatives are computed as follows:\n\n\[\n\n\frac{\partial}{\partial x}(x e^{x}) = e^{x} + xe^{x}\n\n\]\n\nand\n\n\[\n\n\frac{\partial}{\partial y}(ye^{y}) = e^{y} + ye^{y}\n\n\]

Combine the Terms

Combining the computed terms, the divergence of the given vector field denoted as \( \mathrm{div}\ \mathbf{F} \) is \n\n\[\n\n\mathrm{div}\ \mathbf{F} = e^{x} + xe^{x} + e^{y} + ye^{y}\n\n\]

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, offering a way to describe the rate of change of a function with respect to one variable while holding all other variables constant. To compute a partial derivative, one treats all variables except for the one of interest as constants.

For instance, the given vector field \( \mathbf{F}(x, y, z) = xe^{x} \mathbf{i}+ye^{y} \mathbf{j} \) has two components that change with respect to \(x\) and \(y\), respectively. To find the divergence, the partial derivative with respect to each variable is needed. Calculating the partial derivative of \(xe^{x}\) with respect to \(x\) and \(ye^{y}\) with respect to \(y\) involves treating \(e^{x}\) and \(e^{y}\) as functions of their respective variables, while the other variable is considered as a constant. This process isolates the effect each variable has on the vector field, one at a time.

Product Rule for Differentiation

The product rule is an essential differentiation tool used when dealing with products of two functions. It states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

When applied to partial derivatives, the same rule holds. In our exercise, to compute the derivative of \(xe^{x}\), one applies the product rule, treating \(x\) and \(e^{x}\) as separate functions: \(\frac{\partial}{\partial x}(xe^{x}) = e^{x} + xe^{x}\). Similarly, for \(ye^{y}\), the product rule gives: \(\frac{\partial}{\partial y}(ye^{y}) = e^{y} + ye^{y}\). This step is crucial in multivariable calculus for dissecting the interaction between different components of multivariable functions.

Multivariable Calculus

Multivariable calculus extends the concepts of single variable calculus to functions of several variables. It deals with functions whose inputs are vectors and outputs may be either scalar fields or vector fields. Topics include partial derivatives, gradients, divergences, curl, and multiple integrals.

In this context, divergence is a significant operator in multivariable calculus. For a vector field like \( \mathbf{F}(x, y, z) \), divergence measures the extent to which the field spreads out or converges at a given point. It is calculated by taking the dot product of the 'del' operator with the vector field, yielding a scalar function.

To enhance understanding, visualizing a flow represented by the vector field can be helpful. Just as a physical source would increase the flow outward in all directions, a positive divergence at a point indicates 'outflow' from that point. Conversely, negative divergence suggests 'inflow.' Returning to the exercise, following the steps outlined in the solution reveals the divergence without requiring computation with respect to \(z\), as the vector field does not change in that direction. This simplification is an insight multivariable calculus provides, allowing students to focus on the relevant variables contributing to the function's behavior.