Problem 7 - Thomas Calculus

Question

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: Understand the Problem

We need to find a potential function \( f(x, y, z) \) such that its gradient \( abla f \) equals the vector field \( \mathbf{F} = 2x \mathbf{i} + 3y \mathbf{j} + 4z \mathbf{k} \). This means that \( \frac{\partial f}{\partial x} = 2x \), \( \frac{\partial f}{\partial y} = 3y \), and \( \frac{\partial f}{\partial z} = 4z \).

Step 2: Integrate with respect to x

Integrate the partial derivative with respect to \( x \):\[ f(x, y, z) = \int 2x \, dx = x^2 + g(y, z) \]Here, \( g(y, z) \) is an arbitrary function of \( y \) and \( z \), since the derivative with respect to \( x \) should still allow dependence on these variables.

Step 3: Integrate with respect to y

Now, differentiate \( f(x, y, z) = x^2 + g(y, z) \) with respect to \( y \) to match \( \frac{\partial f}{\partial y} = 3y \):\[ \frac{\partial f}{\partial y} = \frac{\partial g(y, z)}{\partial y} = 3y \]Integrate with respect to \( y \):\[ g(y, z) = \int 3y \, dy = \frac{3}{2}y^2 + h(z) \]Here, \( h(z) \) is another arbitrary function based on \( z \).

Step 4: Integrate with respect to z

Substitute \( g(y, z) = \frac{3}{2}y^2 + h(z) \) into \( f \) and differentiate with respect to \( z \) to match \( \frac{\partial f}{\partial z} = 4z \):\[ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(x^2 + \frac{3}{2}y^2 + h(z)) = \frac{dh(z)}{dz} = 4z \]Integrate with respect to \( z \):\[ h(z) = \int 4z \, dz = 2z^2 + C \]where \( C \) is a constant.

Step 5: Assemble the Potential Function

Substitute back all parts into \( f(x, y, z) \):\[ f(x, y, z) = x^2 + \frac{3}{2}y^2 + 2z^2 + C \]This is the potential function for the given vector field.

Key Concepts

Gradient Field

In vector calculus, a gradient field is a vector field created as the gradient of a scalar function. This might sound complex, but let’s simplify it. If you imagine a mountain range, the heights at any point define a surface.
The gradient of this surface at any point gives a vector indicating the direction of the steepest climb up the mountain.
For any scalar function, its gradient field is obtained by calculating the partial derivatives of the function with respect to all variables.
These derivatives form components of a vector, creating the gradient vector field.

• The gradient points in the direction of the steepest increase of the function.
• The magnitude of the gradient gives the rate of that increase.

Understanding this helps in solving problems where a scalar function must be determined such that its gradient equals a given vector field.

Potential Function

A potential function is a scalar function whose gradient equals a given vector field. Think of it as a hidden source from which the vector field originates.
To find this function, you need to reverse-engineer the gradient, starting from the given vector field components.
In the exemplary exercise, we search for a potential function for the vector field off provided by \(\mathbf{F} = 2x \mathbf{i} + 3y \mathbf{j} + 4z \mathbf{k}\). The steps involve:

• Matching the vector field with the partial derivatives \( \frac{\partial f}{\partial x} = 2x \), \( \frac{\partial f}{\partial y} = 3y \), and \( \frac{\partial f}{\partial z} = 4z \).
• Integrating each partial derivative separately to find components of the potential function.
• Adding arbitrary functions of the other variables since integration constants can change with unconsidered variables.

Ultimately, this leads to a scalar function that, when the gradient is taken, reproduces the original vector field.

Vector Field

A vector field assigns a vector to each point in space. Imagine wind blowing across a vast plain.
At every point of the plain, there is a vector indicating the wind’s direction and strength.
This representation allows us to describe a range of physical phenomena like electromagnetic fields or fluid flow.
From a calculus perspective, a vector field is expressed in terms of basic unit vectors, like \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) in 3D space.
For instance, in the exercise, the vector field is given by \(\mathbf{F} = 2x \mathbf{i} + 3y \mathbf{j} + 4z \mathbf{k}\).

• The components tell how much the field stretches in the direction of each axis: \(2x\) in the x-direction, \(3y\) in the y-direction, and \(4z\) in the z-direction.
• By using the concepts of gradient and divergence, vector fields can be analyzed to evaluate their behavior over space.

Vector fields are integral in understanding and modeling how systems change and interact across different domains of physics and engineering.