Problem 218 - Calculus Volume 3

Question

Find the curl of \(\mathbf{F}\). $$ \mathbf{F}(x, y, z)=x y z \mathbf{i}+x^{2} y^{2} z^{2} \mathbf{j}+y^{2} z^{3} \mathbf{k} $$

Step-by-Step Solution

Answer

The curl of \( \mathbf{F} \) is \( (2yz^3 - 2x^2y^2z)\mathbf{i} + xy\mathbf{j} + (2xy^2z^2 - xz)\mathbf{k} \).

Step 1: Understand the Curl Formula

The curl of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k} \). Identify \( P = xyz \), \( Q = x^2y^2z^2 \), and \( R = y^2z^3 \).

Step 2: Calculate Partial Derivatives for \(\mathbf{i}\)-Component

Find \( \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(y^2z^3) = 2yz^3 \) and \( \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(x^2y^2z^2) = 2x^2y^2z \). The \( \mathbf{i} \)-component is \( 2yz^3 - 2x^2y^2z \).

Step 3: Calculate Partial Derivatives for \(\mathbf{j}\)-Component

Find \( \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(xyz) = xy \) and \( \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(y^2z^3) = 0 \). The \( \mathbf{j} \)-component is \( xy - 0 = xy \).

Step 4: Calculate Partial Derivatives for \(\mathbf{k}\)-Component

Find \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2y^2z^2) = 2xy^2z^2 \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xyz) = xz \). The \( \mathbf{k} \)-component is \( 2xy^2z^2 - xz \).

Step 5: Combine Components to Form Curl

Combine the components: \( abla \times \mathbf{F} = (2yz^3 - 2x^2y^2z)\mathbf{i} + xy\mathbf{j} + (2xy^2z^2 - xz)\mathbf{k} \).

Key Concepts

Multivariable Calculus

Multivariable calculus extends concepts from single variable calculus to multiple variables. This enables handling more complex problems involving functions of several variables, like the vector field in our exercise.

In multivariable calculus, we often deal with functions that take input as points in two or more dimensions. For example, the vector field \( \mathbf{F}(x, y, z) \) described in the exercise is a function of three variables. This branch explores how these functions change concerning each input variable, leading to development of concepts like gradients and curls.

• The **gradient** generalizes derivatives to vector fields, representing directional changes at every point.
• The **curl** informs about the rotation of vector fields, which is what we calculate in this exercise.

The curl is a pivotal concept in areas such as electromagnetism where fields are vector quantities defined by multiple variables. Through methods like partial differentiation, we decompose and understand such vector fields more completely.

Vector Calculus

Vector calculus, a field essential for physics and engineering, builds upon concepts of multivariable calculus. It focuses on functions that have both magnitude and direction, otherwise known as vectors. In vector calculus, understanding the dynamics of vector fields is crucial.

We often encounter real-world phenomena represented as vector fields, such as the flow of fluid in space. The curl function calculates the degree of rotation of a point within these fields. It reveals where and how much a field rotates around a point—much like water swirling around in a whirlpool.

• **Vector operations** such as the cross product help derive details like the curl.
• **Curl** can identify rotational characteristics, indicating how field lines twist or rotate.

These operations have practical applications, helping us understand properties like the rotational velocity in fluid dynamics. By mastering vector calculus, the behavior of physical systems described by vector fields can be modeled with higher precision.

Partial Derivatives

Partial derivatives are a fundamental element of multivariable calculus. They measure how a function changes as one of the several input variables is varied, keeping others constant. This is essential for operations like calculating the curl, as seen in our exercise.

The process involves differentiating a multivariable function with respect to one variable at a time. Here's how it works:

• Differentiate \( y^2z^3 \) with respect to \( y \) to obtain \( 2yz^3 \)—this is a partial derivative.
• Differentiate \( x^2y^2z^2 \) with respect to \( z \) to yield \( 2x^2y^2z \).

Partial derivatives were crucial in determining each component of the curl in the solution.

By evaluating these derivatives, we can understand the contribution of each variable to the overall change in the function, allowing deeper insight into the behavior of complex systems. Mastering partial derivatives empowers you to tackle advanced calculus problems by systematically breaking them down into simpler, more manageable parts.