Question

Alternative construction of potential functions Use the procedure in Exercise 71 to construct potential functions for the following fields. $$\quad \mathbf{F}=\langle x, y\rangle$$

Short answer

Question: Find the potential function for the given vector field $$\mathbf{F} = \langle x, y\rangle$$. Answer: The potential function for the given vector field is $$f(x, y) = \frac{1}{2}x^2 + \frac{1}{2}y^2 + C$$.

Step-by-step solution

Find the partial derivatives of the potential function

From the given vector field $$\mathbf{F} = \langle x, y\rangle$$, we know that $$\nabla f(x, y) = \mathbf{F}$$. This means that the partial derivative of $$f(x, y)$$ with respect to $$x$$ is the first component of the vector field $$\mathbf{F}$$, and the partial derivative of $$f(x, y)$$ with respect to $$y$$ is the second component of the vector field $$\mathbf{F}$$. So: $$\frac{\partial f}{\partial x} = x$$ $$\frac{\partial f}{\partial y} = y$$

Integrate the partial derivatives

Now, we integrate the partial derivatives with respect to their respective variables: $$\int \frac{\partial f}{\partial x} dx = \int x\, dx = \frac{1}{2}x^2 + g(y)$$ $$\int \frac{\partial f}{\partial y} dy = \int y\, dy = \frac{1}{2}y^2 + h(x)$$ where $$g(y)$$ and $$h(x)$$ are functions of integration.

Combine the results and find the potential function

Comparing the two equations above, we can conclude that the potential function $$f(x, y)$$ is of the form: $$f(x, y) = \frac{1}{2}x^2 + \frac{1}{2}y^2 + C$$ where $$C$$ is a constant. This potential function is consistent with the given vector field $$\mathbf{F}$$, as its gradient is equal to $$\mathbf{F}$$: $$\nabla f(x, y) = \langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\rangle = \langle x, y\rangle = \mathbf{F}$$ The potential function for the given vector field $$\mathbf{F} = \langle x, y\rangle$$ is $$f(x, y) = \frac{1}{2}x^2 + \frac{1}{2}y^2 + C$$.

Key concepts

Vector Fields

A vector field is a concept used in mathematics and physics. Think of it as a way to give every point in space a direction and magnitude. In simple terms, it assigns a vector (which has both size and direction) to each point in a region of space. For example, wind across Earth's surface can be described by a vector field, where every point on the map has a vector showing the wind's direction and speed at that location.

In the given exercise, the vector field is \( \mathbf{F} = \langle x, y \rangle \). This means that at any point \((x, y)\), there is a vector pointing in the direction defined by its components \(x\) and \(y\). Understanding vector fields is crucial, especially when dealing with forces like gravity or electromagnetism as they often describe how the field behaves across different locations. Vector fields also help in visualizing and solving problems in multiple dimensions.

Partial Derivatives

Partial derivatives are all about focusing on one variable at a time while treating other variables as constants. They are an essential concept when dealing with functions of more than one variable. For example, if a function depends on two variables, say \(f(x, y)\), calculating \(\frac{\partial f}{\partial x}\) means differentiating with respect to \(x\) while keeping \(y\) constant.

In the context of the exercise, the partial derivatives are calculated from the vector field: \(\frac{\partial f}{\partial x} = x\) and \(\frac{\partial f}{\partial y} = y\). These partial derivatives are key as they hint at how the potential function changes with respect to each variable independently. It is crucial to understand partial derivatives to master fields such as calculus, physics, and engineering, where multi-variable systems are frequently analyzed.

Gradient Fields

Gradient fields play a special role in the study of vector fields. The gradient is a vector operation that acts on scalar functions, and its result is a vector field. When you have a gradient field, this means the vector field can be described as the gradient of some scalar function, termed the potential function.

The exercise illustrates this by creating a gradient field from the vector field \( \mathbf{F} = \langle x, y \rangle \). Here, finding the gradient involves combining the partial derivatives to write the vector field in terms of the potential function \(f(x, y)\). The outcome shows that the potential function is \( f(x, y) = \frac{1}{2}x^2 + \frac{1}{2}y^2 + C \), a key expression linking the vector field and its potential. Understanding gradient fields is important as they are often used to identify force fields and scalar potential landscapes, making them invaluable in physical sciences and optimization tasks.