Question

In Exercises \(3-6,\) determine whether or not the vector field is conservative. $$ \mathbf{F}(x, y)=e^{x}(\sin y \mathbf{i}+\cos y \mathbf{j}) $$

Short answer

The vector field \(\mathbf{F}(x, y)=e^{x}(\sin y \mathbf{i}+\cos y \mathbf{j})\) is conservative.

Step-by-step solution

Compute partial derivatives

Differentiate \( F_1(x,y) = e^{x}\sin y \) with respect to y and \( F_2(x,y) = e^{x}\cos y \) with respect to x. \n\n For \( F_1 \), we will use the chain rule. \n\n \( \partial F_1/\partial y = e^{x}\cos y \) \n\n For \( F_2 \), again the chain rule applies. \n\n \( \partial F_2/\partial x = e^{x}\cos y \)

Compare the results

The definition of a conservative vector field in two dimensions is that the partial derivative of \( F_1 \) with respect to y should equal the partial derivative of \( F_2 \) with respect to x. \n\n Comparing the results of Step 1, we see that \( \partial F_1/\partial y = \partial F_2/\partial x = e^{x}\cos y \). Therefore, the vector field is conservative.

Key concepts

Partial Derivatives

In vector calculus, partial derivatives are vital because they represent how a multivariable function changes with respect to one variable while keeping the others constant. For instance, if we have a vector field \( \mathbf{F}(x, y) = e^{x}(\sin y \mathbf{i}+\cos y \mathbf{j}) \), finding its partial derivatives involves treating each component as a separate function. To check if the vector field \( \mathbf{F} \) is conservative, we focus on the partial derivatives of its components with respect to each variable.

In the given exercise, the partial derivative of the first component \( F_1(x, y) \) with respect to \( y \) and the second component \( F_2(x, y) \) with respect to \( x \) are computed. The concept of partial derivatives is foundational, as a conservative vector field has the property that the mixed second partial derivatives are equal (in domains where they are continuous), known as Clairaut's theorem. So, understanding how to compute them directly impacts one's capacity to determine whether a vector field is conservative or not.

Chain Rule

The chain rule is a powerful derivative rule used to compute the derivative of composite functions. When dealing with multivariable functions and vector fields, such as \( \mathbf{F}(x, y) \), the chain rule helps in finding how quickly a function changes as its input changes. Here, we treated the \( e^{x} \) term as an external function and the \( \sin y \) or \( \cos y \) as an internal function.

In the provided step-by-step solution, the chain rule was applied to differentiate the components of the vector field. When taking the derivative of \( F_1 \) with respect to \( y \), we treated \( e^{x} \) as a constant multiplier, which remains unchanged under differentiation with respect to \( y \). As for the derivative of \( F_2 \) with respect to \( x \), we observed the same rule; \( \cos y \) was treated as a constant. The proper application of the chain rule here is crucial to finding accurate partial derivatives, which is a step towards determining if a vector field is conservative.

Vector Calculus

Vector calculus is the branch of mathematics that covers the differentiation and integration of vector fields, often used in physics and engineering to describe fields like electromagnetism and fluid dynamics. In the context of the given problem, vector calculus is used to analyze the behavior of the vector field \( \mathbf{F} \) and to establish whether or not it is conservative.

A conservative vector field is path-independent, which means the line integral of the field along any path between two points only depends on the endpoints and not the specific path taken. Mathematically, one way to ascertain whether a vector field in two dimensions is conservative is by checking if the partial derivatives mentioned earlier are equal, as seen with \( \partial F_1/\partial y \) and \( \partial F_2/\partial x \). If they agree, that implies the existence of a scalar potential function \( \phi \) such that \( \mathbf{F} = abla\phi \), with \( abla \) being the del operator. This concept connects multivariable calculus with fundamental physics concepts, thus forming a cornerstone of vector calculus.