Problem 106 - Calculus Volume 3

Question

For the following exercises, determine whether the vector field is conservative and, if it is, find the potential function. $$ \mathbf{F}(x, y)=2 x y^{3} \mathbf{i}+3 y^{2} x^{2} \mathbf{j} $$

Step-by-Step Solution

Answer

The vector field is conservative with the potential function \( f(x, y) = x^2y^3 + C \).

Step 1: Check if the vector field is conservative

A vector field \( \mathbf{F} = P(x,y) \mathbf{i} + Q(x,y) \mathbf{j} \) is conservative if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). Here, \( P(x,y) = 2xy^3 \) and \( Q(x,y) = 3y^2x^2 \). Let's compute these partial derivatives.

Step 2: Compute \( \frac{\partial P}{\partial y} \)

To find \( \frac{\partial P}{\partial y} \) where \( P(x,y) = 2xy^3 \), we use the derivative rule for polynomials.\[ \frac{\partial}{\partial y} (2xy^3) = 6xy^2 \]

Step 3: Compute \( \frac{\partial Q}{\partial x} \)

Now, let's find \( \frac{\partial Q}{\partial x} \) where \( Q(x,y) = 3y^2x^2 \).\[ \frac{\partial}{\partial x} (3y^2x^2) = 6xy^2 \]

Step 4: Verify equality of partial derivatives

We have obtained \( \frac{\partial P}{\partial y} = 6xy^2 \) and \( \frac{\partial Q}{\partial x} = 6xy^2 \). Since the partial derivatives are equal, the vector field is conservative.

Step 5: Find the potential function

For a conservative vector field, the potential function \( f(x,y) \) satisfies \( abla f = \mathbf{F} \). Start by integrating \( P(x,y) = 2xy^3 \) with respect to \( x \). \[ \int 2xy^3 \, dx = x^2y^3 + g(y) \] where \( g(y) \) is a function of \( y \) alone.

Step 6: Differentiate and match to \( Q(x,y) \)

Differentiate the expression \( x^2y^3 + g(y) \) with respect to \( y \) and match it to \( Q(x,y) \). \[ \frac{\partial}{\partial y}(x^2y^3 + g(y)) = 3x^2y^2 + g'(y) \]Set it equal to \( Q(x,y) = 3y^2x^2 \):\[ 3x^2y^2 + g'(y) = 3y^2x^2 \] implying \( g'(y) = 0 \), so \( g(y) \) is a constant.

Step 7: Conclude the potential function

Since \( g(y) \) is constant, the potential function is \[ f(x, y) = x^2y^3 + C \] where \( C \) is an arbitrary constant. This function satisfies the conditions for the potential of the vector field.

Key Concepts

Partial Derivatives

In the context of vector calculus, partial derivatives are essential for understanding the behavior of multivariable functions. They describe how a function changes as one of its variables is varied while keeping others constant. This is fundamental in determining the nature of a vector field.
For a vector field \( \mathbf{F} = P(x,y) \mathbf{i} + Q(x,y) \mathbf{j} \), checking if \( \mathbf{F} \) is conservative involves computing the partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \).

• For \( P(x,y) = 2xy^3 \), applying the derivative rule gives \( \frac{\partial}{\partial y} (2xy^3) = 6xy^2 \).
• For \( Q(x,y) = 3y^2x^2 \), differentiating gives \( \frac{\partial}{\partial x} (3y^2x^2) = 6xy^2 \).

The equality of these partial derivatives \( 6xy^2 = 6xy^2 \) confirms that the field is conservative.

Potential Function

A potential function, denoted usually as \( f(x, y) \), is strongly associated with the concept of a conservative vector field. If a vector field is conservative, there exists a potential function such that the gradient of this function equals the vector field itself.
For the vector field \( \mathbf{F}(x, y)=2 x y^{3} \mathbf{i}+3 y^{2} x^{2} \mathbf{j} \), the potential function \( f(x, y) \) can be found by integrating \( P(x,y) \) and ensuring it satisfies \( abla f = \mathbf{F} \).

• Integrate \( P(x, y) = 2xy^3 \) with respect to \( x \), resulting in \( x^2y^3 + g(y) \).
• Differentiating \( x^2y^3 + g(y) \) with respect to \( y \) equates to \( Q(x,y) = 3y^2x^2 \), leading to \( g'(y) = 0 \) and indicating \( g(y) \) is a constant.

Thus, the potential function is \( f(x, y) = x^2y^3 + C \), where \( C \) is an arbitrary constant.

Vector Calculus

Vector calculus is a powerful branch of mathematics used to study vector fields and their properties. It combines principles from calculus and linear algebra to address multivariable functions and understand fields that have directions and magnitudes.
In this exercise, vector calculus helps us determine if a vector field is conservative. A conservative vector field implies the existence of a potential function, making it easier to analyze the field's behavior.

• To identify if the field \( \mathbf{F}(x, y) = 2xy^3 \mathbf{i} + 3y^2x^2 \mathbf{j} \) is conservative, we utilize partial derivatives.
• Ensure \( abla \times \mathbf{F} = 0 \), which for two dimensions translates to the equality of the computed partial derivatives.

The concepts of vector calculus allow us to assess the relations of components within vector fields and their potential functions, offering comprehensive insights into their mathematical and physical implications.