Problem 23 - 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 \(B\). \(\int_{C} \tan y d x+x \sec ^{2} y d y ; A\) is \((-2,0)\) and \(B\) is \(\left(4, \frac{1}{4} \pi\right)\)

Step-by-Step Solution

Answer

4

Step 1 - Identify if the vector field is conservative

To show that the line integral is independent of the path, we first need to check if the given vector field is conservative. A vector field \(\textbf{F} = P(x,y) \mathbf{i} + Q(x,y) \mathbf{j}\) is conservative if there exists a scalar potential function \(f(x,y)\) such that \(abla f = \textbf{F} \). This means \(P = \frac{\text{∂}f}{\text{∂}x}\) and \(Q = \frac{\text{∂}f}{\text{∂}y}\). In this case, \(\textbf{F} = \tan y \mathbf{i} + x \sec^2 y \mathbf{j}\).

Step 2 - Apply the condition for a conservative vector field

The vector field \(\mathbf{F}\) is conservative if \(\frac{\text{∂}P}{\text{∂}y} = \frac{\text{∂}Q}{\text{∂}x}\). Find these partial derivatives: \(P = \tan y\) and \(Q = x \sec^2 y\). Calculate \(\frac{\text{∂}P}{\text{∂}y} = \frac{d}{dy} (\tan y) = \sec^2 y\) and \(\frac{\text{∂}Q}{\text{∂}x} = \frac{d}{dx} (x \sec^2 y) = \sec^2 y\). Since \(\frac{\text{∂}P}{\text{∂}y} = \frac{\text{∂}Q}{\text{∂}x}\), the given vector field is conservative.

Step 3 - Find the potential function

Since the vector field is conservative, find the potential function \(f(x, y)\). Integrate \(P = \tan y\) with respect to \(x\): \(\frac{\text{∂}f}{\text{∂}x} = \tan y \to f(x, y) = x\tan y + g(y)\). Differentiate this new potential function with respect to \(y\) and set it equal to \Q\ to find \(g(y)\): \( \frac{\text{∂}}{\text{∂}y}(x \tan y + g(y)) = x \sec^2 y + g'(y)\). Since \(x \sec^2 y\) must equal \Q\, \g'(y)\ must be 0, leading to \(g(y) = C\) where \(C\) is a constant. Thus, the potential function is \( f(x, y) = x\tan y + C \).

Step 4 - Compute the value of the integral

To evaluate the line integral, use the fact that the integral of a conservative field \- given by \(abla f\) \- from point \(A\) to \(B\) depends only on the values of the potential function at these points: \(\begin{aligned} \int_{C} \textbf{F} \cdot d\textbf{r} = f(4, \frac{1}{4} \pi) - f(-2, 0) \end{aligned}\). Calculate the values of \(f\) at these points: \(\begin{aligned} f(4, \frac{1}{4} \pi) &= 4\tan(\frac{1}{4} \pi) + C = 4 \times 1 + C = 4 + C \end{aligned}\); \(\begin{aligned} f(-2, 0) &= -2\tan(0) + C = -2 \times 0 + C = C \end{aligned}\). Subtract these values: \(\begin{aligned} f(4, \frac{1}{4} \pi) - f(-2, 0) &= (4+C) - C = 4 \end{aligned}\).

Key Concepts

conservative vector fields

In vector calculus, a conservative vector field is a vector field that is the gradient of some scalar potential function. This means there exists a function \( f(x,y) \) such that its gradient \( abla f = \textbf{F} \). The main property of conservative fields is that the line integral of the vector field between two points does not depend on the path taken but only on the endpoints. This simplifies calculations significantly when solving line integrals.
For example, if we confirm a vector field is conservative, we can find a potential function instead of manually computing complicated path integrals.

• Conservative fields ensure path independence in integrals.
• A scalar potential function exists such that \( abla f = \textbf{F} \).
• This property is useful for simplifying the evaluation of line integrals.

potential function

A potential function in a conservative vector field serves as a scalar field from which the vector field can be derived. If a vector field \( \textbf{F} = P(x,y) \,\mathbf{i} + Q(x,y) \,\mathbf{j} \) is conservative, then it has an associated potential function \( f(x,y) \) such that \( abla f = \textbf{F} \).
To find the potential function, we need to perform integration:

• Integrate \( P(x,y) \) with respect to \( x \) to find \( \frac{ \partial f}{ \partial x} = P(x,y) \).
• Integrate \( Q(x,y) \) with respect to \( y \) to find \( \frac{\partial f}{\partial y} = Q(x,y) \).

By comparing these partial derivatives, you can determine any additional functions of \( y \) (commonly denoted \( g(y) \)) that arise during the integration.
For instance, if you integrate \( \tan y \) with respect to \( x \), you obtain \( f(x,y) = x \, \tan y + g(y) \). Further differentiation and comparison with \( Q(x,y) \) will yield the complete potential function.

partial derivatives

Partial derivatives are fundamental in multivariable calculus, representing the rate of change of a function with respect to one variable, while keeping the other variables constant.
They are crucial for understanding vector fields and identifying conservative fields. For potential functions, partial derivatives help verify if a given vector field is conservative.
In the given vector field example, \( \textbf{F} = \tan y \, \mathbf{i} + x \, \sec^2 y \, \mathbf{j} \), the verification involves:

• Calculating \( \frac{ \partial P}{ \partial y} = \frac{d}{dy} (\tan y) = \sec^2 y \).
• Calculating \( \frac{ \partial Q}{ \partial x} = \frac{d}{dx} (x \, \sec^2 y) = \sec^2 y \).

Since these partial derivatives are equal, the field is confirmed to be conservative. This highlights how partial derivatives are used to establish key properties in vector calculus.