Problem 16 - 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}\left(2 x e^{x^{2}}-2 y\right) d x+(2 y-2 x) d y, \text { where \(C\) runs from \((1,2)\) to \((-1,1)\) }$$

Step-by-Step Solution

Answer

The line integral is -5.

Step 1: Verify if the Vector Field is Conservative

A vector field \((M, N) = (2x e^{x^{2}}-2y, 2y-2x)\) is conservative or path independent if the partial derivatives with respect to the other variable are equal i.e \(\frac{∂M}{∂y} = \frac{∂N}{∂x}\). In this case, \(\frac{∂M}{∂y} = -2\) and \( \frac{∂N}{∂x} = -2\). Since both partial derivatives are equal, the vector field is conservative.

Step 2: Find the Potential Function

For a conservative vector field \((M, N)\), we can find the potential function \(f\) such that \(M = \frac{∂f}{∂x}\) and \(N = \frac{∂f}{∂y}\). Solving these integrals respectively gives \(f(x,y) = x^{2}e^{x^{2}} - y^{2} + g(y)\). Now we differentiate this expression with respect to \(y\) and equating it to \(N\), we find that \(g'(y) = 0\). Therefore, our potential function simplifies to \(f(x,y) = x^{2}e^{x^{2}} - y^{2}\).

Step 3: Evaluate the Integral

With our potential function \(f(x,y) = x^{2}e^{x^{2}} - y^{2}\), we apply the fundamental theorem for line integrals, meaning we will only evaluate the function at the points (1,2) and (-1,1). Calculation gives \(f(-1,1) - f(1,2) = (-1 - 4) = -5\).

Key Concepts

Conservative Vector Field

Imagine a gentle breeze that flows in a specific direction across a field. In the world of mathematics, this is similar to the concept of a vector field. Specifically, a conservative vector field is a type of vector field with special characteristics. If a vector field is conservative, it means the work done by moving along a path only depends on the starting and ending points, not on the path taken. This is like walking between two points on a hill where the altitude difference is the same no matter which path you take—all you care about is the height difference, not the specific trail.

To determine if a vector field is conservative, we check a simple condition involving derivatives. Given a field described by functions \(M(x, y)\) and \(N(x, y)\), you take the derivative of \(M\) with respect to \(y\) and \(N\) with respect to \(x\). If these derivatives are equal, that means the vector field is conservative.

In the exercise example, the vector field \((M, N) = (2x e^{x^2} - 2y, 2y - 2x)\) is checked by comparing \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) which both equal \(-2\). Hence, the field is conservative.

Potential Function

Once we confirm that a vector field is conservative, the next step is to find its potential function. Think of a potential function as a magical map providing the landscape of our vector field.

For a vector field \((M, N)\), the potential function \(f(x, y)\) is found such that \(M = \frac{\partial f}{\partial x}\) and \(N = \frac{\partial f}{\partial y}\). The potential function stores all the energy or work details of the field. It's a bit like the altitude map of a hill.

In our example, the process involves integrating the function \(M\) with respect to \(x\) and integrating \(N\) with respect to \(y\). Solving these gives us a clue to what the potential function looks like. We adjust constants of integration through further comparison, ultimately leading us to the potential function.

For the given vector field, the potential function was found to be \(f(x, y) = x^2 e^{x^2} - y^2\), which beautifully describes its characteristics.

Fundamental Theorem for Line Integrals

In the grand scheme of calculus, the fundamental theorem for line integrals is like finding a shortcut. This theorem simplifies the process of calculating the work done by a force moving along a curve.

When we have a potential function, this theorem tells us that we only need to evaluate it at the two endpoints of the path, not along each individual step. How wonderful is that? It's like being able to calculate your uphill effort simply by knowing where you started and where you ended.

In our specific exercise, the fundamental theorem helps us quickly determine the value of the line integral. We simply need to compute the potential function values at point \((1, 2)\) and \((-1, 1)\) and then subtract. The computation gives us \(f(-1,1) - f(1,2) = -5\). This tells us that the net energy or work involved in moving from point \((1, 2)\) to \((-1, 1)\) is \(-5\), capturing the true essence of this powerful theorem.

• Saves calculation time
• Path-independent
• Focuses only on start and end points