Problem 8 - Calculus: Early Transcendental Functions

Question

Determine whether \(F\) is conservative. If it is, find a potential function \(f.\) $$\mathbf{F}(x, y)=\left\langle y \cos x y-2 x y, x \cos x y-x^{2}\right\rangle$$

Step-by-Step Solution

Answer

The vector field \(F\) is conservative and its potential function \(f\) is \(f(x, y) = y\sin(xy) - x^{2}y\).

Step 1: Compute the partial derivatives

First, the partial derivatives need to be calculated: \(\frac{\partial M}{\partial y} = \cos(xy) - xy\sin(xy) - 2x\) and \(\frac{\partial N}{\partial x} = \cos(xy) - xy\sin(xy) - x\) These partial derivatives will check if the vector field is conservative.

Step 2: Compare the partial derivatives

Now, compare \(\frac{\partial M}{\partial y} \) and \(\frac{\partial N}{\partial x}\). If they are the same, then the vector field is conservative. Comparing the computed derivatives, they are indeed the same. Hence, the field \(\mathbf{F}(x, y)\) is conservative.

Step 3: Calculate potential function

Since we confirmed the vector field is conservative, a potential function \(f\) can be found. To find \(f\), integrate \(M(x, y)\) with respect to \(x\) and \(N(x, y)\) with respect to \(y\) which gives \(f_1(x, y) = y\sin(xy) - x^{2}y + g_1(y)\) and \(f_2(x, y) = x\sin(xy) - \frac{1}{2}x^{2} + g_2(x)\). A common function for both \(f_1\) and \(f_2\) will be our potential function. Hence, after comparing the result of these integrations, the potential function is found to be \(f(x, y) = y\sin(xy) - x^{2}y\).

Key Concepts

Potential Function

In vector calculus, a potential function is a scalar function whose gradient yields a given vector field. If a vector field \(\mathbf{F}(x, y)\) is conservative, it implies there exists a potential function \(f(x, y)\) such that \(abla f = \mathbf{F}\). This means that if you take the gradient of the potential function, you get back the original vector field. It's a fundamental concept because it can often simplify understanding complex vector fields by reducing them to a simpler scalar field.

In the original exercise, once we confirmed that the vector field \(\mathbf{F}(x, y)\) is conservative, we were tasked with finding this potential function. This typically involves integrating the components of the vector field. The process ensures compatibility, meaning the mixed partial derivatives must be equal. In essence, a potential function provides a kind of 'shortcut' to understand the behavior of vectors in a field.

Partial Derivatives

Partial derivatives are crucial in finding a potential function. They measure how a function changes as only one variable changes, keeping the others constant. For a vector field \(\mathbf{F}(x, y) = \langle M(x, y), N(x, y) \rangle\), ensuring it is conservative requires the partial derivatives to satisfy certain conditions.
To check if a vector field is conservative, you calculate the partial derivative of \(M\) with respect to \(y\), noted as \(\frac{\partial M}{\partial y}\), and the partial derivative of \(N\) with respect to \(x\), noted as \(\frac{\partial N}{\partial x}\).

If these derivatives are equal, then the vector field is conservative, which means a potential function \(f\) exists. In our exercise, after computing these partial derivatives, we found they matched, confirming that the vector field is conservative. Partial derivatives, thus, not only aid in finding the existence of potential functions but also suggest the particular form they might take.

Vector Calculus

Vector calculus is a branch of mathematics focusing on vector fields and operations like gradient, divergence, and curl. It's used extensively in physics and engineering to solve problems involving fields and flows.
In our context, identifying whether a vector field is conservative relates to understanding its fundamental properties and behaviors through calculus. Conservative vector fields are those where the line integral is path-independent, and this can be verified through calculating partial derivatives.

The step-by-step solution of the given exercise heavily employs tools from vector calculus: it checks the equality of mixed partial derivatives of the vector field components to ascertain if a potential function exists. Once confirmed, integration—a tool from calculus—was used to derive the potential function. Vector calculus, hence, provides a robust framework that simplifies analyzing and understanding complex vector fields.