Problem 10 - Calculus Early Transcendentals

Question

Show that the integral is independent of the path, and use Theorem 15.3.1 to find its value. $$ \int_{(0,0)}^{(1, \pi / 2)} e^{x} \sin y d x+e^{x} \cos y d y $$

Step-by-Step Solution

Answer

The integral is path-independent and its value is \( e \).

Step 1: Understand path independence

We need to show that the integral \( \int_{(0,0)}^{(1, \pi / 2)} e^{x} \sin y \, dx + e^{x} \cos y \, dy \) is independent of the path. According to vector calculus, if a vector field \( \mathbf{F} = (M, N) = (e^x \sin y, e^x \cos y) \) is conservative, then its line integral is path-independent. A vector field is conservative if there exists a scalar potential function \( f \) such that \( abla f = \mathbf{F} \). This implies that \( M = \frac{\partial f}{\partial x} \) and \( N = \frac{\partial f}{\partial y} \) must satisfy the equality \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

Step 2: Verify Path Conditions

Calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \):\[ \frac{\partial}{\partial y}(e^x \sin y) = e^x \cos y \]\[ \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y \]Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = e^x \cos y \), the vector field is conservative, indicating that the integral is independent of the path.

Step 3: Identify a Potential Function

To find the potential function \( f \), integrate \( M = e^x \sin y \) with respect to \( x \):\[ f(x, y) = \int e^x \sin y \, dx = e^x \sin y + g(y) \]Differentiate \( f(x, y) \) with respect to \( y \) and set equal to \( N = e^x \cos y \):\[ \frac{\partial}{\partial y}(e^x \sin y + g(y)) = e^x \cos y + g'(y) = e^x \cos y \]This implies \( g'(y) = 0 \), so \( g(y) \) is a constant. Therefore, \( f(x, y) = e^x \sin y \).

Step 4: Evaluate the Potential at the ending points

Since the integral is path-independent, compute the value using the potential function found. Evaluate \( f \) at the upper limit \((1, \pi/2)\) and lower limit \((0, 0)\):\[ f(1, \pi/2) = e^1 \sin(\pi/2) = e \]\[ f(0, 0) = e^0 \sin(0) = 0 \]

Step 5: Calculate the Integral Value

Subtract the potential at the initial point from the potential at the terminal point:\[ f(1, \pi/2) - f(0, 0) = e - 0 = e \]Thus, the value of the integral is \( e \).

Key Concepts

Path Independence

Path independence is a key concept in vector calculus, especially when dealing with line integrals. It implies that the value of the integral from one point to another does not depend on the path taken between the two points.

For a line integral to be path independent, the underlying vector field must be conservative. Practically, this means that you can take any route and still end up with the same integral value, simplifying calculations significantly.

To check for path independence, we verify the equality of mixed partial derivatives. In mathematical terms, for a 2D vector field \( \mathbf{F} = (M, N) \), the condition for path independence is that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

When this condition holds, the vector field has a potential function, ensuring that the line integral is independent of the path selected.

Conservative Vector Field

A conservative vector field is one where, each direction and magnitude of the vector field is derived from some scalar potential function. This implies that the work done by the field is stored and can be completely recovered within a system.

In the context of integrals, a conservative vector field signifies that moving an object from one point to another within this field requires the same amount of work, regardless of the path taken. This is an inherent property of path independence.

To determine if a given vector field \( \mathbf{F} = (M, N) \) is conservative:

• Check that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
• If true, there exists a function \( f(x, y) \) such that \( abla f = \mathbf{F} \), confirming the field is conservative.
  

Potential functions are the next step toward simplifying the evaluation of line integrals over conservative vector fields.

Potential Function

The potential function is a scalar function whose gradient matches the given vector field. It simplifies the calculation of line integrals greatly, because once found, the integral over a conservative vector field becomes independent of the path taken.

To find a potential function from a vector field \( \mathbf{F} = (M, N) \):

• Integrate \( M \) concerning \( x \) to form part of the potential function.
• Integrate \( N \) concerning \( y \) and ensure consistency with the function found by integrating \( M \).
• Adjust for any constant functions or variables consistent between these integrations.

Once identified, a potential function can transform a complicated path-dependent integral into a simple subtraction problem. You compute it at two endpoints, reflecting the same work done for varying paths in the field.

Using this approach, the line integral \ over the vector field becomes markedly simpler, saving time and resources in analyzing complex paths.