Proving A Gradient Field: A Step-by-Step Guide

Hey there, math enthusiasts! Today, we're diving into the fascinating world of vector fields and gradient fields. Specifically, we'll be tackling an exercise that shows how to prove a given vector field is a gradient field. Understanding gradient fields is super important in various areas of physics and engineering, so let's get started. We'll break down the process into easy-to-follow steps, making sure you grasp the core concepts along the way.

Understanding Gradient Fields

First off, what exactly is a gradient field? Well, a gradient field is a vector field that can be expressed as the gradient of a scalar potential function. In simpler terms, if you have a scalar function (let's call it ${\, \phi }$), the gradient field ${ \vec{V} }$ is obtained by taking the partial derivatives of ${ \phi }$ with respect to each coordinate (x, y, and z in 3D space). Mathematically, ${ \vec{V} = \nabla \phi }$, where ${ \nabla }$ is the del operator (a fancy way of saying gradient).

Now, here's the kicker: gradient fields have a special property. They are conservative. This means that the work done by the field in moving an object between two points doesn't depend on the path taken, but only on the starting and ending points. This concept is fundamental in physics, especially when dealing with potential energy and forces. A key characteristic of gradient fields is that they are irrotational, meaning their curl is zero. This irrotational property is what allows us to verify if a vector field is a gradient field using a set of conditions. So, if we can verify these conditions, we know that our vector field is indeed a gradient field.

The Given Vector Field

Alright, let's get down to the exercise. We're given the vector field:

${ \vec{V} = (2x - y) \vec{e_x} + (2y - x) \vec{e_y} - 4z \vec{e_z} }$

Here, ${ \vec{e_x} }$, ${ \vec{e_y} }$, and ${ \vec{e_z} }$ are the unit vectors along the x, y, and z axes, respectively. Our goal is to prove that this field is a gradient field. To do this, we need to verify a set of conditions that arise from the irrotational nature of gradient fields. Specifically, we must show that:

1. ${ \frac{\partial V_x}{\partial y} = \frac{\partial V_y}{\partial x} }$
2. ${ \frac{\partial V_x}{\partial z} = \frac{\partial V_z}{\partial x} }$
3. ${ \frac{\partial V_y}{\partial z} = \frac{\partial V_z}{\partial y} }$

Let's break down each of these conditions and verify them step by step. It's really just a matter of calculating the partial derivatives and checking if they match up. It may seem intimidating at first, but trust me, it's pretty straightforward once you get the hang of it. We'll methodically go through each step, ensuring that you can follow along with ease. This hands-on approach will not only help you solve the problem but also deepen your understanding of the underlying principles.

Verifying the Conditions

Now, let's get into the nitty-gritty of verifying these conditions. We'll start with the first one, which is ${ \frac{\partial V_x}{\partial y} = \frac{\partial V_y}{\partial x} }$.

Condition 1: ${ \frac{\partial V_x}{\partial y} = \frac{\partial V_y}{\partial x} }$

Here, ${ V_x = 2x - y }$ and ${ V_y = 2y - x }$. Let's find the partial derivatives:

• ${ \frac{\partial V_x}{\partial y} = \frac{\partial}{\partial y}(2x - y) = -1 }$
• ${ \frac{\partial V_y}{\partial x} = \frac{\partial}{\partial x}(2y - x) = -1 }$

Since ${ -1 = -1 }$, the first condition is satisfied! See? Not so bad, right? We just took a couple of partial derivatives and compared the results. This is a crucial step because it confirms that our vector field potentially behaves like a gradient field in the x-y plane. It's like checking the first piece of a puzzle to see if it fits. If this doesn't hold true, we know right away that our field isn't a gradient field, and we can stop. But since it does, we continue to the next set of conditions to provide further evidence.

Condition 2: ${ \frac{\partial V_x}{\partial z} = \frac{\partial V_z}{\partial x} }$

Next up, we need to check if ${ \frac{\partial V_x}{\partial z} = \frac{\partial V_z}{\partial x} }$ holds true. Here, ${ V_x = 2x - y }$ and ${ V_z = -4z }$. Let's calculate the partial derivatives:

• ${ \frac{\partial V_x}{\partial z} = \frac{\partial}{\partial z}(2x - y) = 0 }$
• ${ \frac{\partial V_z}{\partial x} = \frac{\partial}{\partial x}(-4z) = 0 }$

Since ${ 0 = 0 }$, the second condition is also met! This confirms the gradient field in the x-z plane. This is another crucial piece of the puzzle. It shows that the field's behavior in this plane is consistent with what we expect from a gradient field. Again, if this condition were not satisfied, we would know immediately that the field is not a gradient field. So far, so good!

Condition 3: ${ \frac{\partial V_y}{\partial z} = \frac{\partial V_z}{\partial y} }$

Finally, we need to check the last condition: ${ \frac{\partial V_y}{\partial z} = \frac{\partial V_z}{\partial y} }$. Here, ${ V_y = 2y - x }$ and ${ V_z = -4z }$. Let's calculate the partial derivatives:

• ${ \frac{\partial V_y}{\partial z} = \frac{\partial}{\partial z}(2y - x) = 0 }$
• ${ \frac{\partial V_z}{\partial y} = \frac{\partial}{\partial y}(-4z) = 0 }$

Since ${ 0 = 0 }$, the third condition is also satisfied! This confirms the gradient field in the y-z plane. Now that we've checked all three conditions and found that they are true, we can confidently conclude that the given vector field is indeed a gradient field.

Conclusion: It's a Gradient Field!

Alright, folks, we've done it! By systematically checking the conditions ${ \frac{\partial V_x}{\partial y} = \frac{\partial V_y}{\partial x} }$, ${ \frac{\partial V_x}{\partial z} = \frac{\partial V_z}{\partial x} }$, and ${ \frac{\partial V_y}{\partial z} = \frac{\partial V_z}{\partial y} }$, we've proven that the given vector field ${ \vec{V} = (2x - y) \vec{e_x} + (2y - x) \vec{e_y} - 4z \vec{e_z} }$ is a gradient field. This means there exists a scalar potential function ${ \phi }$ such that ${ \vec{V} = \nabla \phi }$. The fact that all three conditions hold true is a direct consequence of the irrotational nature of gradient fields. This exercise highlights the importance of understanding the properties of gradient fields and the conditions that define them.

So, what does this mean in the grand scheme of things? Well, because our field is a gradient field, it's conservative. This is a big deal in physics, as it simplifies many calculations. It means that the work done by the field is independent of the path taken, which makes it easier to predict and understand the behavior of systems governed by this field. Think about it: if you had to calculate the work done for every possible path, it would be a nightmare! But because it's a gradient field, you only need to know the starting and ending points.

Beyond the Basics: Finding the Potential Function

Now, here's a bonus for those who want to take it a step further. Since we've confirmed that the field is a gradient field, we can actually find the scalar potential function ${ \phi }$ itself! Remember, ${ \vec{V} = \nabla \phi }$, which means:

• ${ \frac{\partial \phi}{\partial x} = 2x - y }$
• ${ \frac{\partial \phi}{\partial y} = 2y - x }$
• ${ \frac{\partial \phi}{\partial z} = -4z }$

To find ${ \phi }$, we can integrate these equations. Let's start with the first one:

${ \phi(x, y, z) = \int (2x - y) dx = x^2 - xy + f(y, z) }$

Here, ${ f(y, z) }$ is an arbitrary function of y and z (since the derivative with respect to x would be zero). Now, let's take the partial derivative of this with respect to y:

${ \frac{\partial \phi}{\partial y} = -x + \frac{\partial f}{\partial y} }$

But we know ${ \frac{\partial \phi}{\partial y} = 2y - x }$. Therefore:

${ 2y - x = -x + \frac{\partial f}{\partial y} }$

Which means:

${ \frac{\partial f}{\partial y} = 2y }$

Integrating this with respect to y, we get:

${ f(y, z) = y^2 + g(z) }$

where ${ g(z) }$ is an arbitrary function of z. So, now we have:

${ \phi(x, y, z) = x^2 - xy + y^2 + g(z) }$

Finally, let's take the partial derivative with respect to z:

${ \frac{\partial \phi}{\partial z} = \frac{\partial g}{\partial z} }$

But we know ${ \frac{\partial \phi}{\partial z} = -4z }$. Therefore:

${ \frac{\partial g}{\partial z} = -4z }$

Integrating this with respect to z, we get:

${ g(z) = -2z^2 + C }$

where C is a constant. Putting it all together, the potential function is:

${ \phi(x, y, z) = x^2 - xy + y^2 - 2z^2 + C }$

So there you have it! We've not only shown that the vector field is a gradient field but we've also found the potential function. This shows the entire process for this exercise. This function is extremely useful because it can be used to describe the potential energy of a particle moving through this field. The constant C doesn't affect the field itself, but it can be adjusted based on the specific context of the problem. What a journey!

Conclusion and Next Steps

We successfully proved that the given vector field is a gradient field. We did this by verifying the necessary conditions based on the irrotational nature of gradient fields. The satisfaction of all the condition proves that the field acts like a potential function. If you found this helpful, why not try similar exercises? Practice makes perfect, and working through more examples will solidify your understanding of gradient fields. The concepts of gradient fields are essential for understanding many physical phenomena, such as electromagnetism and fluid dynamics. They are also widely used in computer graphics for generating realistic lighting and shading. The ability to identify and work with gradient fields is a valuable skill for any aspiring scientist or engineer. Keep exploring, keep learning, and keep the math adventures going! Thanks for joining me on this mathematical journey. Until next time, keep calculating!