Problem 8 - Calculus Early Transcendentals

Question

(a) Show that the line integral \(\int_{C} y \sin x d x-\cos x d y\) is independent of the path. (b) Evaluate the integral in part (a) along the line segment from \((0,1)\) to \((\pi,-1) .\) (c) Evaluate the integral \(\int_{(0,1)}^{(\pi,-1)} y \sin x d x-\cos x d y\) using Theorem \(15.3 .1,\) and confirm that the value is the same as that obtained in part (b).

Step-by-Step Solution

Answer

The integral evaluates to 2 in both parts (b) and (c).

Step 1: Identify the vector field

The given line integral is \(\int_{C} y \sin x \, dx - \cos x \, dy\). We identify the vector field \( \mathbf{F} = (y \sin x, -\cos x) \).

Step 2: Check if the vector field is conservative

For a vector field \( \mathbf{F} = (M, N) \) to be conservative, \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \). Here, \( M = y \sin x \) and \( N = -\cos x \). Therefore, \( \frac{\partial N}{\partial x} = \sin x \) and \( \frac{\partial M}{\partial y} = \sin x \). Since both partial derivatives are equal, the vector field is conservative, meaning the line integral is path independent.

Step 3: Find a potential function for the field

A conservative vector field has a potential function \( f(x, y) \) where \( abla f = \mathbf{F} \). Integrate \( M = y \sin x \) with respect to \( x \): \[ f(x, y) = \int y \sin x \, dx = -y \cos x + g(y) \]. Differentiate this expression with respect to \( y \) and equate to \( N = -\cos x \): \( -\cos x + g'(y) = -\cos x \), hence \( g'(y) = 0 \), and so \( g(y) = C \). The potential function is \( f(x, y) = -y \cos x \).

Step 4: Evaluate the path-independent integral

Since the line integral is path independent, use the potential function \( f(x, y) \) to evaluate it. So, \( \int_{C} y \sin x dx - \cos x dy = f(\pi, -1) - f(0, 1) \). Compute \( f(\pi, -1) = -(-1)\cos(\pi) = 1 \) and \( f(0, 1) = -1\cos(0) = -1 \). Therefore, the integral evaluates to \( 1 - (-1) = 2 \).

Step 5: Evaluate the integral using Theorem 15.3.1

Theorem 15.3.1 states the line integral of a conservative field \( \mathbf{F} \) along a curve \( C \) from \( A \) to \( B \) is \( f(B) - f(A) \) if \( abla f = \mathbf{F} \). From (0,1) to (\pi,-1), this gives \( f(\pi, -1) - f(0, 1) = 2 \). This is the same result as in step 4, confirming the consistency of our finding.

Key Concepts

Conservative Vector Fields

A vector field \( \mathbf{F} = (M, N) \) is considered conservative if it satisfies a certain condition: the partial derivative of \( N \) with respect to \( x \) must equal the partial derivative of \( M \) with respect to \( y \). This means that the vector field is derived from a gradient of a scalar potential function.
Conservative vector fields have several nice properties. For example:

• Line integrals over these fields are path-independent, which means no matter which path you take between two points, the integral will have the same value.
• They have associated potential functions, which is a great tool for evaluating integrals.

In our problem, we identified the vector field as \( \mathbf{F} = (y \sin x, -\cos x) \). By checking the condition for conservativity, \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \), we find that both derivatives equal \( \sin x \). Thus, the vector field is indeed conservative.

Potential Functions

A potential function is a scalar function \( f(x, y) \) such that its gradient \( abla f \) is equal to the vector field \( \mathbf{F} \). For conservative vector fields, finding the potential function allows us to evaluate line integrals easily.
To find the potential function:

• Integrate \( M = y \sin x \) with respect to \( x \), yielding \( f(x, y) = -y \cos x + g(y) \), where \( g(y) \) is a function of \( y \) alone.
• Differentiate this expression with respect to \( y \) and equate it to \( N = -\cos x \). This step helps us to determine \( g(y) \).
• Solving, we find \( g'(y) = 0 \) which implies \( g(y) = C \), a constant.

Thus, the potential function is \( f(x, y) = -y \cos x \). With this potential function, we can easily compute line integrals by applying the difference \( f(B) - f(A) \) for points \( A \) to \( B \).

Path Independence

Path independence is a key feature of conservative vector fields. It implies that the line integral of a vector field between two points is the same, regardless of the path taken. This is because the line integral is equivalent to the change in the potential function along the path.
The steps for confirming path independence are:

• Check if the vector field is conservative. As discussed, if \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \), it is conservative.
• If a potential function exists, then the integral is independent of the path.

In the given exercise, path independence allowed us to compute the integral simply by evaluating the potential function at two points \((\pi, -1)\) and \((0, 1)\). This confirmed that the integral's value is the same through any path between these points, making the computation straightforward and efficient.