Problem 13 - The Calculus with Analytic Geometry

Question

\(\int e^{x} \sin y d x+e^{x} \cos y d y ; A\) is \((0,0)\) and \(B\) is \(\left(2, \frac{1}{2} \pi\right) ;\) Exercise 3

Step-by-Step Solution

Answer

The value of the integral is \( e^{2} \).

Step 1: Identify the Vector Field

Recognize that the given integral \(\int e^{x} \sin y dx + e^{x} \cos y dy \) can be written in the form of a line integral of a vector field \(F(x,y) = (e^{x} \sin y, e^{x} \cos y) \).

Step 2: Check for Conservativeness

Check if the vector field is conservative by verifying that \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(e^{x} \sin y) = e^{x} \cos y \] and \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(e^{x} \cos y) = e^{x} \cos y \]. Since \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\), the vector field is conservative.

Step 3: Find the Potential Function

Find a potential function \(\phi(x,y)\) such that \(\frac{\partial \phi}{\partial x} = P\) and \(\frac{\partial \phi}{\partial y} = Q\). Integrate \(P = e^{x} \sin y\) with respect to \(x\) to get \[ \phi(x,y) = \int e^{x} \sin y dx = e^{x} \sin y + h(y) \]. Next, differentiate \(\phi\) with respect to \(y\) to find \[ \frac{\partial \phi}{\partial y} = e^{x} \cos y + h'(y) \]. Set this equal to \(Q = e^{x} \cos y\), thus \[ h'(y) = 0 \Rightarrow h(y) = C \]. Therefore, the potential function is \(\phi(x,y) = e^{x} \sin y + C\).

Step 4: Evaluate the Potential Function at Points A and B

Evaluate \(\phi(x,y)\) at points \(A = (0,0)\) and \(B = (2, \frac{1}{2} \pi)\). \(\phi(0,0) = e^{0} \sin 0 = 0\) and \(\phi(2, \frac{1}{2} \pi) = e^{2} \sin (\frac{1}{2} \pi) = e^{2}\).

Step 5: Compute the Line Integral

Use the Fundamental Theorem for Line Integrals to find \(\int_{A}^{B} e^{x} \sin y dx + e^{x} \cos y dy = \phi(B) - \phi(A) = e^{2} - 0 = e^{2}\).

Key Concepts

vector field

A vector field is a mathematical construct where each point in a space is associated with a vector. In this exercise, the vector field is represented by the function \(F(x,y) = (e^{x} \sin y, e^{x} \cos y)\). This means at every point \( (x, y) \) in the plane, a vector is defined with components \( F_1 = e^{x} \sin y \) and \( F_2 = e^{x} \cos y \).
It is important to understand that the components of the vector field can change for different points in the space, illustrating how the vectors can vary in direction and magnitude throughout the region.

conservative field

A vector field is called conservative if it can be expressed as the gradient of a potential function. One of the main properties of a conservative vector field is that its line integral only depends on the endpoints of the curve, not the path taken. To check if a vector field is conservative, we verify that the mixed partial derivatives are equal: \( \frac{\text{\partial } P}{\text {\partial } y} = \frac{\text {\partial } Q}{\text {\partial } x} \).
In this exercise, we checked the vector field \( F(x,y) = (e^{x} \sin y, e^{x} \cos y) \) by computing these partial derivatives and confirming they are equal, thus ensuring that the field is conservative.

potential function

A potential function \( \phi(x,y)\) for a conservative vector field gives us a way to simplify the computation of line integrals. The potential function is determined such that the components of the vector field correspond to its partial derivatives: \( \frac{\partial \phi}{\partial x} = P \;\text {and}\frac{\partial \phi}{\partial y} = Q \. \).
In this exercise, we found the potential function by integrating \( P = e^{x} \sin y \) with respect to \( x \), leading to \( \phi(x,y) = e^{x} \sin y + h(y) \. \). By equating the derivative of \( \phi(x,y) \) with respect to \( y \) to \( Q = e^{x} \cos y \), we determined that \( h(y) \) is a constant function, resulting in \( \phi(x,y) = e^{x} \sin y + c \. \).

fundamental theorem for line integrals

The Fundamental Theorem for Line Integrals states that if a vector field is conservative, the line integral from point \( A \) to point \( B \) of the vector field can be calculated using the potential function. This can be expressed as \( \int_A^B \abla \phi \cdot d\mathbf{r} = \phi(B) - \phi(A) \. \).
This theorem significantly simplifies the computation of line integrals for conservative fields. In this exercise, we used the potential function \( \phi(x,y) = e^{x} \sin y \) and evaluated it at the endpoints \( A = (0,0) \) and \( B = (2, \frac{1}{2} \pi) \). By calculating \( \phi(B) - \phi(A) \), we found the value of the line integral to be \( e^{2} \).