Problem 13 - Calculus: Early Transcendental Functions

Question

Show that the line integral is independent of path and use a potential function to evaluate the integral. $$\int_{C} 2 x y d x+\left(x^{2}-1\right) d y, \text { where \(C\) runs from \((1,0)\) to \((3,1)\) } $$

Step-by-Step Solution

Answer

The line integral is path independent and its value is 8.

Step 1: Check for Conservative Vector Field

The given vector field is \(F(x,y) = 2xy\hat{i} + (x^2-1)\hat{j}\). A necessary (but not sufficient) condition for \(F\) to be conservative in a simply connected domain is that the mixed partial derivatives of its components are equal, i.e., \(\frac{\partial}{\partial y}(2xy) = \frac{\partial}{\partial x}(x^2-1)\). After calculating both sides, it is found that both equal \(2x\), verifying that the given vector field is indeed conservative.

Step 2: Find the Potential Function

Since the given vector field is conservative, there exists a scalar potential function φ(x, y) such that \(F = \nabla φ\), meaning φ satisfies the equations \(\frac{\partial φ}{\partial x} = 2xy\) and \(\frac{\partial φ}{\partial y} = x^2 - 1\). Integrating of \(\frac{\partial φ}{\partial x}\) with respect to \(x\) gives us \(φ = x^2y + g(y)\) where \(g(y)\) is the constant of integration which may be a function of \(y\). To find \(g(y)\), substitute \(φ\) into \(\frac{\partial φ}{\partial y} = x^2 - 1\) and solve for \(g'(y)\). This yields \(φ(x,y) = x^2y - y\).

Step 3: Evaluate the Line Integral

Since the vector field is conservative, the line integral between two points is simply the potential function evaluated at these points. Therefore, we have \(\int_{C} F \cdot dr = φ(3,1) - φ(1,0) = 3^2*1-1 - 1^2*0-0 = 8\).

Key Concepts

Conservative Vector Fields

A vector field is considered conservative when it can be expressed as the gradient of a potential function. Understanding this concept can simplify many calculations, especially in evaluating line integrals, by verifying that a vector field is conservative.

Some properties of conservative vector fields include:

• The line integral of a conservative vector field is path-independent. This means it depends only on the endpoints, not on the path taken between them.
• In a conservative field, any closed loop integral is zero. This is because the starting and ending points are the same, so the potential function's value doesn't change.
• A mathematical check for conservativeness involves equating the mixed partial derivatives of the field's components. For instance, with our example vector field, verifying that \(rac{\partial}{\partial y}(2xy) = \frac{\partial}{\partial x}(x^2-1)\) allowed us to confirm conservativeness.

Recognizing conservative vector fields paves the way to using potential functions effectively in calculations.

Potential Functions

A potential function is directly related to a conservative vector field. If a vector field is conservative, there exists a potential function \( \phi(x, y) \) such that the vector field \( F \) is the gradient of \( \phi \). This relationship enables us to evaluate line integrals more simply.

Here's how you find the potential function:

• Integrate the components of the vector field. For example, with \( \frac{\partial \phi}{\partial x} = 2xy \), integrating with respect to \( x \) gives \( \phi = x^2y + g(y) \), where \( g(y) \) is an unknown function of \( y \).
• Use the second component, \( \frac{\partial \phi}{\partial y} = x^2 - 1 \), to find \( g(y) \). Here, solving gives \( g(y) = -y \), leading to the potential function \( \phi(x, y) = x^2y - y \).

Once you have the potential function, use it to evaluate line integrals, by subtracting its values at the terminal and initial points.

Partial Derivatives

Partial derivatives play a critical role in vector calculus, particularly when dealing with concepts like conservative vector fields and potential functions. They measure how a function changes as one variable changes, while keeping other variables constant.

• In our context, evaluate the mixed partial derivatives to determine if a vector field is conservative. For our problem, the check \( \frac{\partial}{\partial y}(2xy) = \frac{\partial}{\partial x}(x^2-1) \) was fulfilled, confirming the field's conservativeness.
• Partial derivatives are also used to find potential functions. By integrating one variable's partial derivative while treating other variables as constants, we construct potential functions. The integration process might introduce functions of integration for other variables, which are determined using the second set of partial derivatives.

Understanding how to handle partial derivatives is essential for tackling more advanced problems in calculus and related fields.