Problem 13 - Advanced Engineering Mathematics

Question

In Problems, determine whether the given vector field is a conservative field. If so, find a potential function \(\phi\) for \(\mathbf{F}\). $$ \mathbf{F}(x, y)=y^{2} \cos x y^{2} \mathbf{i}-2 x y \sin x y^{2} \mathbf{j} $$

Step-by-Step Solution

Answer

The vector field is conservative. The potential function is \( \phi(x, y) = \sin(xy^2) + C \).

Step 1: Check the Conservative Condition

For a vector field \( \mathbf{F}(x, y) = M(x, y) \mathbf{i} + N(x, y) \mathbf{j} \) to be conservative, its curl must be zero in two dimensions, which translates to the condition: \[ \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \] Here, \( M(x, y) = y^2 \cos(xy^2) \) and \( N(x, y) = -2xy \sin(xy^2) \). We find the partial derivatives: \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (-2xy \sin(xy^2)) \] \[ = -2y \sin(xy^2) - 2xy^3 \cos(xy^2) \] and \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y^2 \cos(xy^2)) = 2y \cos(xy^2) - 2xy^3 \sin(xy^2) \] Both derivatives are equal, confirming that the field is conservative.

Step 2: Find the Potential Function \(\phi\)

To find the potential function \( \phi(x, y) \) such that \( abla \phi = \mathbf{F} \), integrate \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \). First, integrate \( M(x, y) = y^2 \cos(xy^2) \) with respect to \( x \): \[ \phi_{x}(x, y) = \int y^2 \cos(xy^2) \, dx = \sin(xy^2) + g(y) \] where \( g(y) \) is an arbitrary function of \( y \). Next, integrate \( N(x, y) = -2xy \sin(xy^2) \) with respect to \( y \): \[ \phi_{y}(x, y) = \int -2xy \sin(xy^2) \, dy = \sin(xy^2) + h(x) \] where \( h(x) \) is an arbitrary function of \( x \). Since both partial derivatives match up to a constant function, the potential function is: \[ \phi(x, y) = \sin(xy^2) + C \] where \( C \) is a constant.

Key Concepts

Potential Function

A potential function, often denoted as \( \phi(x, y) \), serves as a scalar function whose gradient results in a given vector field \( \mathbf{F}(x, y) \). This is key when determining if a vector field is conservative. For the vector field to be conservative, a potential function must exist. In simple terms, if you are able to find such a \( \phi \), then the vector field does not "loop" back on itself and has no rotational curl.

Finding a potential function involves integrating the components of the vector field. First, you determine the function of \( \phi \) through integration. If you integrate the vector components with respect to their respective variables, the calculated integrals should equal one another (with potentially differing by a constant or function of the opposite variable). This ensures that the field is conservative and thus a potential function exists.

Vector Calculus

Vector calculus is an essential part of understanding fields and their behaviors in multiple dimensions. It involves differentiating and integrating vector fields, like \( \mathbf{F}(x, y) \) in this exercise.

A few key operations make up vector calculus:

• Gradient: Turning a potential function into a vector field.
• Divergence: Measures the "spread" of a vector field from a point.
• Curl: Indicates how much twisting or curling is present in the field.

In our exercise, the focus lies on determining if a vector field is conservative, by checking if its curl is zero. This is done through partial derivatives, ensuring continuity and consistency across the field.

Partial Derivatives

Partial derivatives are central to understanding and working with multivariable functions, like those encountered in vector calculus. When finding if a vector field is conservative, you need to compute the partial derivatives:

- For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), measures the rate of change while keeping \( y \) constant.- Similarly, \( \frac{\partial f}{\partial y} \) measures the change with respect to \( y \) while \( x \) is constant.

In our vector field \( \mathbf{F}(x, y) \), the step-by-step solution asked us to check:

• \( \frac{\partial N}{\partial x} \)
• \( \frac{\partial M}{\partial y} \)

These derivatives determine if they equate. If they match, thus indicating zero curl, the field is deemed conservative. Each partial derivative explores a specific dimension's influence independently, allowing us to accurately analyze complex fields.