Problem 136 - Calculus Volume 3

Question

For the following exercises, show that the following vector fields are conservative by using a computer. Calculate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) for the given curve. \(\mathbf{F}=\left(x y^{2}+3 x^{2} y\right) \mathbf{i}+(x+y) x^{2} \mathbf{j} ; \quad C\) is the curve consisting of line segments from (1,1) to (0,2) to (3,0)

Step-by-Step Solution

Answer

\( \int_{C} \mathbf{F} \cdot d\mathbf{r} = -1 \).

Step 1: Checking for Conservativeness

A vector field \( \mathbf{F} \) is conservative if it can be written as the gradient of a scalar potential function \( f \). To check this, we need to verify if \( abla \times \mathbf{F} = 0 \). The vector field is \( \mathbf{F} = (x y^2 + 3x^2 y)\mathbf{i} + ((x+y)x^2)\mathbf{j} \). Compute \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \), where \( M = xy^2 + 3x^2y \) and \( N = (x+y)x^2 \).

Step 2: Compute Partial Derivatives

Compute \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} ((x+y)x^2) = 3x^2 + 2xy \) and \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (xy^2 + 3x^2y) = 2xy + 3x^2 \). Both partial derivatives simplify to \( 3x^2 + 2xy \), confirming that \( \mathbf{F} \) is conservative \( (abla \times \mathbf{F} = 0) \).

Step 3: Finding the Potential Function

Since the field is conservative, there exists a potential function \( f(x,y) \) such that \( abla f = \mathbf{F} \). Therefore, \( \frac{\partial f}{\partial x} = xy^2 + 3x^2y \) and \( \frac{\partial f}{\partial y} = (x+y)x^2 \). Integrate \( \frac{\partial f}{\partial x} \) with respect to \( x \) to find \( f(x,y) = \frac{x^2y^2}{2} + x^3y + g(y) \).

Step 4: Solve for \( g(y) \)

Differentiate the potential function \( f(x,y) = \frac{x^2y^2}{2} + x^3y + g(y) \) with respect to \( y \) and equate to \( \frac{\partial f}{\partial y} = (x+y)x^2 \). Find \( g'(y) \): \( 2xy + x^3 + g'(y) = x^3 + 2xy + yx^2 \) implies \( g'(y) = yx^2 \). Integrate \( g'(y) \) to get \( g(y) = \frac{yx^2}{2} + C \). Combine terms in \( f(x,y) \) to get the potential function: \( f(x,y) = \frac{x^2y^2}{2} + x^3y + \frac{yx^2}{2} + C \).

Step 5: Evaluate the Line Integral

Since \( \mathbf{F} \) is conservative, the integral over a curve depends only on the endpoints. Evaluate \( f \) at the endpoints of the curve \( C \).\( f(3,0) - f(1,1) = (0) - (\frac{1^2 \cdot 1^2}{2} + 0 + 0.5) = -\frac{1}{2} - 0.5 = -1 \). Thus, \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = f(3,0) - f(1,1) = -1 \).

Key Concepts

Potential Functions

In vector calculus, potential functions are closely associated with conservative vector fields. A vector field \( \mathbf{F} \) is conservative if there exists a scalar function \( f(x, y) \), known as the potential function, such that the gradient of \( f \) is equal to \( \mathbf{F} \); that is, \( abla f = \mathbf{F} \). This means that the vector field can be represented as the gradient of some scalar function, providing deep insight into the nature of the field.

To find a potential function for a given conservative vector field, we solve for \( f \) such that \( \frac{\partial f}{\partial x} = M \) and \( \frac{\partial f}{\partial y} = N \), where \( M \) and \( N \) are the components of \( \mathbf{F} \). By integrating these partial derivatives, we determine the corresponding expression for \( f(x, y) \).

Potential functions simplify the computation of line integrals in conservative fields because they allow the evaluation of the integral based only on the start and end points of the path, rather than the specific path itself. This property significantly reduces computational complexity when dealing with such integrals.

Line Integrals

Line integrals are a tool in vector calculus used to integrate functions along a curve or path in a vector field. For a vector field \( \mathbf{F} \) and a curve \( C \), the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) represents the work done by \( \mathbf{F} \) as an object moves along the curve, or the total "flow" of \( \mathbf{F} \) through \( C \).

In a conservative vector field, calculating a line integral can be considerably easier, as the field has a potential function. The fundamental theorem of line integrals states that for a conservative vector field, the line integral of \( \mathbf{F} \) over a curve \( C \) that goes from point \( A \) to point \( B \) is simply \( f(B) - f(A) \), where \( f \) is a potential function of \( \mathbf{F} \). This allows us to ignore the actual path taken, focusing only on the endpoints of the curve.

• If the path forms a closed loop, the line integral of a conservative vector field over it is zero.
• This property makes evaluating line integrals over conservative vector fields straightforward and efficient.

When solving problems involving line integrals, always check if the vector field is conservative first, as this can simplify the work substantially.

Vector Calculus

Vector calculus is a branch of mathematics dealing with vector fields and differential operators. It extends concepts from standard calculus to multi-dimensional spaces, allowing for a deeper understanding of various physical phenomena, such as fluid flow, electromagnetic fields, and gravitational forces.

Key operations in vector calculus include differentiation and integration of vector fields. The differentiation involves operations like gradient, divergence, and curl. In a conservative field, the curl \( abla \times \mathbf{F} \) is zero, confirming that the field can be associated with a potential function. Conversely, non-zero curl indicates rotational characteristics in the field.

Vector calculus provides the tools to explore and analyze fields in three dimensions, often employing these operators:

• Gradient (\( abla f \)): Yields a vector indicating the direction of the greatest rate of increase of the scalar function \( f \).
• Divergence (\( abla \cdot \mathbf{F} \)): Quantifies a field's rate of expansion at any point.
• Curl (\( abla \times \mathbf{F} \)): Measures the rotation or "twisting" tendency of a field around a point.

These concepts underpin much of physics and engineering, helping to model and solve problems involving complex field dynamics and interactions.