Problem 1 - Thomas Calculus

Question

Find the \(\mathbf{k}\) -component of \(\operatorname{curl}(\mathbf{F})\) for the following vector fields on the plane. \(\mathbf{F}=(x+y) \mathbf{i}+(2 x y) \mathbf{j}\)

Step-by-Step Solution

Answer

The \( k \)-component of the curl is \( 2y - 1 \).

Step 1: Identify the Vector Field Components

The given vector field is \( \mathbf{F} = (x+y) \mathbf{i} + (2xy) \mathbf{j} \). In component form, this can be expressed as \( \mathbf{F} = (P, Q) \), where \( P = x+y \) and \( Q = 2xy \).

Step 2: Recall the Formula for Curl in 2D

In two dimensions, the curl of a vector field \( \mathbf{F} = (P, Q) \) is given by \( \operatorname{curl}(\mathbf{F}) = (Q_x - P_y) \mathbf{k} \), where \( Q_x \) is the partial derivative of \( Q \) with respect to \( x \) and \( P_y \) is the partial derivative of \( P \) with respect to \( y \).

Step 3: Calculate Partial Derivatives \(Q_x\) and \(P_y\)

Compute the partial derivative \( Q_x \):\[ Q = 2xy \Rightarrow Q_x = \frac{\partial}{\partial x} (2xy) = 2y. \]Compute the partial derivative \( P_y \):\[ P = x+y \Rightarrow P_y = \frac{\partial}{\partial y} (x+y) = 1. \]

Step 4: Compute the \( k \)-component of Curl

Substitute the partial derivatives into the curl formula:\[ \operatorname{curl}(\mathbf{F}) = (Q_x - P_y) \mathbf{k} = (2y - 1) \mathbf{k}. \]

Step 5: Final Expression for the \( k \)-component

The \( k \)-component of \( \operatorname{curl}(\mathbf{F}) \) is \( 2y - 1 \). There is no need for additional components since it's a 2D vector field on the plane.

Key Concepts

Vector Field

In the world of calculus, a vector field is a function that assigns a vector to every point in space. Imagine it as a collection of arrows spread throughout a plane or space, where each arrow has both a direction and a magnitude.
For instance, the vector field mentioned in the original exercise is given by \( \mathbf{F} = (x+y) \mathbf{i} + (2xy) \mathbf{j} \). This means at any point \((x, y)\), this field assigns a vector with:

• a component \((x+y)\) in the \(\mathbf{i}\) direction, which typically represents the x-axis,
• and a component \((2xy)\) in the \(\mathbf{j}\) direction, representing the y-axis.

This conceptualization helps us understand how the vector field behaves across different parts of the plane.

Partial Derivative

A partial derivative is a derivative where we hold some variables constant while we differentiate with respect to one variable. This allows us to understand how a function changes as only one of its input changes.
In the exercise, we had to calculate two partial derivatives:

• \(Q_x = \frac{\partial}{\partial x} (2xy) = 2y\)
• \(P_y = \frac{\partial}{\partial y} (x+y) = 1\)

Here, "partial with respect to \(x\)" means we treat \(y\) as a constant while differentiating, and vice versa for "partial with respect to \(y\)". This technique is essential in multivariable calculus because it helps isolate the effects of single variables in complex functions.

2D Curl

The concept of 2D curl is used to measure the tendency of a vector field to rotate around a point in two-dimensional space. It's represented as a single scalar since in 2D there is essentially only one direction that rotation can occur (out of the plane).
In mathematics, specifically in the exercise, the 2D curl of a vector field \( \mathbf{F} = (P, Q) \) is defined as:

• \(\operatorname{curl}(\mathbf{F}) = (Q_x - P_y) \mathbf{k}\)

Here, \(Q_x\) and \(P_y\) are the partial derivatives of \(Q\) with respect to \(x\) and \(P\) with respect to \(y\) respectively.
This formula provides us the \(k\)-component of the curl, which can indicate the swirling strength of the field.

k-component

In the realm of vector calculus, especially when dealing with a 2D vector field on a plane, the \(k\)-component is crucial. Although we're operating in two dimensions, we're interested in knowing about rotation which "pokes out" in the third dimension, usually represented as the \(k\)-component.
The \(k\)-component of the curl is essentially the only relevant component for 2D fields because it gives the measure of rotation in a plane perpendicular to the x-y plane. In our original exercise, the formula for the 2D curl yielded the \( k \)-component:\[ \operatorname{curl}(\mathbf{F}) = (2y - 1) \mathbf{k} \]
Thus, the expression \((2y - 1)\) represents the intensity of the rotation about a point for this specific vector field. Understanding this component helps in visualizing how much the field "twirls" at each point in the plane.