Problem 22 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 30, show that the value of the line integral is independent of the path and compute the value in any convenient manner. In each exercise, \(C\) is any sectionally smooth curve from the point \(A\) to the point \(B\). \(\int_{C}(\ln x+2 y) d x+\left(e^{y}+2 x\right) d y ; A\) is \((3,1)\) and \(B\) is \((1,3)\)

Step-by-Step Solution

Answer

The value of the line integral is \3 + e^3 - 3 \ln 3 - e\. The integral is path-independent.

Step 1: Identify the given vector field components

Given, the line integral involves the vector field \(\textbf{F}(x,y) = (\ln x + 2y, e^y + 2x)\). Identify the components of this vector field as \(M(x,y) = \ln x + 2y\) and \(N(x,y) = e^y + 2x\).

Step 2: Check for a conservative vector field

Calculate the partial derivatives to check if the vector field is conservative. If \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the field is conservative. Compute: \ \frac{\partial M}{\partial y} = 2 \ and \ \frac{\partial N}{\partial x} = 2 \.

Step 3: Proof that the vector field is conservative

Since \(\frac{\partial M}{\partial y} = 2\) and \(\frac{\partial N}{\partial x} = 2\), the conditions for a conservative field are satisfied. Therefore, the vector field is conservative and the line integral is path-independent.

Step 4: Find the potential function \(f(x,y)\)

To find the potential function, integrate \(M(x,y) = \ln x + 2y\) with respect to \(x\): \[ \frac{\partial f}{\partial x} = \ln x + 2y \Rightarrow f(x,y) = x \ln x - x + 2xy + g(y) \] Now, differentiate this with respect to \(y\) and compare with \(N(x,y)\): \[ \frac{\partial f}{\partial y} = 2x + g'(y) \] Equate this to \(N(x,y) = e^y + 2x\): \[ 2x + g'(y) = e^y + 2x \Rightarrow g'(y) = e^y \] Integrate \(g'(y)\): \[ g(y) = e^y + C \] Thus, the potential function \(f(x,y)\) is: \[ f(x,y) = x \ln x - x + 2xy + e^y + C \]

Step 5: Evaluate the potential function at points \(A = (3,1)\) and \(B = (1,3)\)

Compute \(f(3,1)\) and \(f(1,3)\): \[ f(3,1) = 3 \ln 3 - 3 + 2 \cdot 3 \cdot 1 + e^1 = 3 \ln 3 + 3 + e \] \[ f(1,3) = 1 \ln 1 - 1 + 2 \cdot 1 \cdot 3 + e^3 = 6 + e^3 \]

Step 6: Compute the value of the line integral

The value of the line integral is \[ f(B) - f(A) = (6 + e^3) - (3 \ln 3 + 3 + e ) = 3 + e^3 - 3 \ln 3 - e \]

Key Concepts

Conservative Vector Field

A vector field is termed as 'conservative' if the line integral of the field from one point to another is independent of the path taken. This is a key concept in vector calculus, as it simplifies the computation of line integrals.
To determine if a vector field \(\textbf{F} = (M, N)\) is conservative, we must check if it satisfies a certain condition: the partial derivatives of its components.
Specifically:

\(\frac{\frac{\boldsymbol{\textbf{M}}}{{\boldsymbol{\textbf{y}}}}}{{}} = \frac{\boldsymbol{\textbf{2}}}{{}} \) should be equal to \(\frac{\frac{\boldsymbol{\textbf{N}}}{{\boldsymbol{\textbf{x}}}}}{{}} = \) \(\frac{\boldsymbol{\textbf{2}}}{{}}\)

If the equality holds true, the field is conservative. This was verified for the given vector field \(\textbf{F}(x, y) = (\textbf{ln}\textbf{x}+2\textbf{y}, \textbf{e}^{\textbf{y}} + 2\textbf{x})\) in the original solution, making the field conservative.

Potential Function

If a vector field is conservative, it implies that there exists a potential function \(\textbf{f(x, y)}\) such that \(\textbf{F} = \textbf{grad}{\textbf{f}}\). In other words, the vector field can be expressed as the gradient of a scalar function.
In our exercise, to find \(\textbf{f(x, y)}\), we integrate the component \(\textbf{M(x, y)} = \textbf{ln{x}} + 2\textbf{y}\) with respect to \(\textbf{x}\):
\(\frac{{\bigtriangleup}}{{\bigtriangleup \textbf{x}}} \textbf{f} = \textbf{ln}\textbf{x} + 2\textbf{y}\)
Solving this, \(\textbf{f(x, y)}\) is: \(\textbf{xln}\textbf{x} - \textbf{x} + 2\textbf{xy} + \textbf{g(y)}\), where \(\textbf{g(y)}\) is a function of \(\textbf{y}\) alone.
We then differentiate \(\textbf{f(x, y)}\) with respect to \(\textbf{y}\) and equate it to \(\textbf{N(x, y)}\): \(\frac{\bigtriangleup \textbf{f}}{\bigtriangleup \textbf{y}} = 2\textbf{x} + \textbf{g(y)}^{\textbf{y}}\)
Solving this integration problem, we get \(\textbf{g(y)} = \textbf{e^y} \). Thus, our potential function is:
\(\textbf{f(x, y)} = \textbf{xln}\textbf{x} - \textbf{x} + 2\textbf{xy} + \textbf{e^y} + \textbf{C}\).
This potential function helps in evaluating the final line integral.

Partial Derivatives

Partial derivatives measure how a function changes as each variable changes, holding the other variables constant. In the context of vector fields, they help in verifying whether a field is conservative.
For our vector field components \(\textbf{M(x, y)} = \textbf{ln{x}} + 2\textbf{y}\) and \(\textbf{N(x, y)} = \textbf{e^{\textbf{y}}} + 2\textbf{x}\):
We compute:

\(\frac{\bigtriangleup \textbf{M}}{\bigtriangleup \textbf{y}} = 2 \)
\(\frac{\bigtriangleup \textbf{N}}{\bigtriangleup \textbf{x}} = 2 \)

Since they are equal, the field is conservative. This simplification allowed us to use a potential function to compute the integral.
Remember, partial derivatives are essential tools in multivariable calculus, simplifying many problems by breaking them into smaller, more manageable parts.