Problem 7 - Thomas Calculus

Question

In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\). $$\mathbf{F}=2 x \mathbf{i}+3 y \mathbf{j}+4 z \mathbf{k}$$

Step-by-Step Solution

Answer

The potential function is \( f(x,y,z) = x^2 + \frac{3}{2}y^2 + 2z^2 + C \).

Step 1: Determine Components of the Field

The given vector field is \( \mathbf{F}=2 x \mathbf{i}+3 y \mathbf{j}+4 z \mathbf{k} \). This can be written in component form as \( \mathbf{F} = (2x, 3y, 4z) \).

Step 2: Define the Potential Function and Compute Partial Derivatives

The potential function \( f(x,y,z) \) must satisfy \( \frac{\partial f}{\partial x} = 2x \), \( \frac{\partial f}{\partial y} = 3y \), and \( \frac{\partial f}{\partial z} = 4z \).

Step 3: Integrate the First Partial Derivative

Integrate \( \frac{\partial f}{\partial x} = 2x \) with respect to \( x \):\[\int 2x \, dx = x^2 + g(y,z),\]where \( g(y,z) \) is an arbitrary function of \( y \) and \( z \).

Step 4: Integrate the Second Partial Derivative

Since \( \frac{\partial f}{\partial y} = 3y \), integrate\[\int 3y \, dy = \frac{3}{2}y^2 + h(x,z),\]where \( h(x,z) \) is an arbitrary function of \( x \) and \( z \).

Step 5: Combine and Resolve Expressions

Combine the integrated results:\( f(x,y,z) = x^2 + \frac{3}{2}y^2 + \) arbitrary functions. Determine consistency between the arbitrary parts using \( \frac{\partial f}{\partial z} = 4z \).

Step 6: Integrate the Third Partial Derivative

Since \( \frac{\partial f}{\partial z} = 4z \), integrate:\[\int 4z \, dz = 2z^2 + c(x,y),\]where \( c(x,y) \) is another arbitrary function.

Step 7: Form the Potential Function

From Steps 3 to 6, the potential function is:\[ f(x,y,z) = x^2 + \frac{3}{2}y^2 + 2z^2 + C, \]where \( C \) is a constant, by ensuring all components are covered.

Key Concepts

Vector Field

A vector field represents a mathematical construct where each point in space is assigned a vector. In three-dimensional space, this vector can be expressed as a combination of its components, typically using the standard unit vectors: \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). For example, the vector field \( \mathbf{F}=2 x \mathbf{i}+3 y \mathbf{j}+4 z \mathbf{k} \) can be broken down into its respective components in the i, j, and k directions.
This representation allows us to visualize and analyze how vectors vary across different positions. Each vector in the field is informed by rules or expressions that may depend on variables like \( x, y, \) and \( z \).

• The components are functions of position, indicating the direction and magnitude of the vectors at every point.
• Vector fields are used to model various physical quantities like electric and magnetic fields, or fluid flows.

Thus, understanding vector fields helps in interpreting a wide range of physical phenomena and mathematical models.

Partial Derivatives

Partial derivatives are used to measure how a function changes in one direction while keeping other variables constant. For multivariable functions such as \( f(x, y, z) \), partial derivatives with respect to \( x \), \( y \), and \( z \) can be written as \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \).
These are essential tools in vector calculus for analyzing multivariable functions and vector fields.

• In our context, partial derivatives help us deduce relationships between the components of a vector field and a potential function.
• For example, if you have \( \frac{\partial f}{\partial x} = 2x \), this tells us specifically how the function \( f \) changes as \( x \) varies, while \( y \) and \( z \) stay fixed.

This makes partial derivatives invaluable in studying fields like physics and engineering, where changes in multidimensional environments are critical.

Integration

Integration, in the context of vector fields, helps us recover a potential function from its derivatives. When you have a partial derivative like \( \frac{\partial f}{\partial x} = 2x \), integrating with respect to \( x \) yields the original function, up to an arbitrary constant or function that might include other variables. This is how we transition from rates of change back to the function itself.

• In our exercise, integration of each partial derivative gave us parts of the potential function \( f(x, y, z) \).
• It is important to treat each variable as independent during integration unless specified otherwise, as in \( \int 2x \, dx = x^2 + g(y,z) \).

This process gradually reconstructs the full potential function by combining all necessary integrations of the partial derivatives.

Arbitrary Function

When integrating partial derivatives to find a potential function, we often encounter arbitrary functions or constants. These arise because integration can only determine the original function up to a constant or additional terms that do not depend on the variable of integration.
For example, when integrating \( \frac{\partial f}{\partial x} = 2x \), you get \( \int 2x \, dx = x^2 + g(y,z) \). Here, \( g(y,z) \) is an arbitrary function that could include any dependencies from \( y \) and \( z \).

• The form of these arbitrary parts needs to be consistent with other parts of the potential function as evaluated from other derivatives.
• These leading terms align the various pieces of the integrated functions ensuring the solution satisfies all the initial conditions or derivative relations.

Recognizing arbitrary functions helps in capturing the full behavior of a multivariable function beyond simple integration results.