Problem 15 - 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} y e^{x y} d x+\left(x e^{x y}-2 y\right) d y, \text { where \(C\) runs from \((1,0)\) to \((0,4)\) }$$

Step-by-Step Solution

Answer

-17

Step 1: Check If Vector Field is Conservative

A two-dimensional vector field \(F(x, y) = P(x, y)i + Q(x, y)j\) is conservative if and only if \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\). Here, \(P(x, y) = ye^{xy}\) and \(Q(x, y) = xe^{xy} - 2y\). So, check whether \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\)

Step 2: Calculate Partial Derivatives

Calculate \(\frac{\partial P}{\partial y} = xe^{xy} + e^{xy}\) and \(\frac{\partial Q}{\partial x} = ye^{xy} + e^{xy} - 0\). This implies that \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\).

Step 3: Potential Function

Since \(F(x, y)\) is conservative, there exists a potential function \(f(x, y)\) such that \(P(x, y) = \frac{\partial f}{\partial x}\) and \(Q(x, y) = \frac{\partial f}{\partial y}\). Integrate \(P(x, y)\) with respect to \(x\) and \(Q(x, y)\) with respect to \(y\) and compare the two integrals. The shared term is the potential function, in this case is \(f(x, y) = e^{xy} - y^2\).

Step 4: Evaluate Integral using Potential Function

The line integral over any path from point A to point B of a conservative vector field F is simply the difference in the potential function \(f(x, y)\) evaluated at A and B. So, evaluate \(f\) at the endpoints \(A(1, 0)\) and \(B(0, 4)\). Thus, \(\int_C F \cdot dr = f(B) - f(A) = e^{(0)(4)} - {(4)^2} - (e^{(1)(0)} - {(0)^2}) = -16 - (1 - 0) = -17.\)

Key Concepts

Conservative Vector Field

In vector calculus, a vector field is considered conservative if its line integral does not depend on the path taken. This is an important property because it simplifies the calculation of line integrals significantly.
Moreso, a vector field is conservative if it can be expressed as the gradient of some scalar potential function. A conventional condition to determine if a vector field \( F(x, y) = P(x, y)i + Q(x, y)j \) is conservative involves checking if the partial derivatives satisfy \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \).

In our exercise, we have \( P(x, y) = ye^{xy} \) and \( Q(x, y) = xe^{xy} - 2y \). By calculating the partial derivatives, \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \), and confirming they are equal, we establish that the vector field is indeed conservative.

Potential Function

Once we know a vector field is conservative, our next step is to find the potential function. The potential function \( f(x, y) \) is a scalar field such that its gradient yields the original vector field.
This means:

• \( P(x, y) = \frac{\partial f}{\partial x} \)
• \( Q(x, y) = \frac{\partial f}{\partial y} \)

To find \( f(x, y) \), we integrate \( P(x, y) \) with respect to \( x \) and \( Q(x, y) \) with respect to \( y \), ensuring consistency in the integration constants or terms that result.

In the exercise, this process gives us the potential function \( f(x, y) = e^{xy} - y^2 \). This function helps evaluate the line integral easily because it connects the vector field directly with a scalar description.

Path Independence

One of the key features of a conservative vector field is path independence. This means that the line integral through such a field depends only on the start and end points, not on the path taken.
This characteristic simplifies the computation of line integrals dramatically, especially when using potential functions. For example, when you have a potential function \( f(x, y) \), the line integral along any curve \( C \) from point \( A \) to point \( B \) is simply \( f(B) - f(A) \).

In our specific problem, by evaluating the potential function \( f(x, y) \) at the endpoints, \( (1, 0) \) and \( (0, 4) \), and subtracting, we calculated the line integral as \( -17 \). This process confirms the path independence property of the conservative vector field.