Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each line integral. \(\int_{C}(\sin x+\cos y) d s ; C\) is the line segment from \((0,0)\) to \((\pi, 2 \pi)\).

Step-by-Step Solution

Answer

The value of the line integral is \(2\sqrt{5}\).

Step 1: Parameterize the Curve

The line segment from \((0,0)\) to \((\pi, 2\pi)\) can be parameterized by \(\mathbf{r}(t) = (t, 2t)\), where \(t\) varies from 0 to \(\pi\). This parameterization covers the entire line segment as \(t\) passes through the necessary points.

Step 2: Express the Integral in Terms of t

The line integral is \(\int_{C}(\sin x+\cos y)\, ds\). Substituting \(x(t) = t\) and \(y(t) = 2t\) into the integrand, we have \(\sin(t) + \cos(2t)\).

Step 3: Find the Differential Element ds

The differential arc length element \(ds\) along the curve is given by \(|\mathbf{r}'(t)| \cdot dt\). The derivative \(\mathbf{r}'(t)\) is \((1, 2)\), and its magnitude is \(\sqrt{1^2 + 2^2} = \sqrt{5}\). Therefore, \(ds = \sqrt{5}\, dt\).

Step 4: Set Up the Integral in Terms of t

The integral transforms to \(\int_{0}^{\pi}(\sin(t) + \cos(2t))\sqrt{5}\, dt\), where \(\sqrt{5}\) is a constant factor.

Step 5: Evaluate the Integral

Evaluate the integral \(\sqrt{5} \int_{0}^{\pi} (\sin(t) + \cos(2t))\, dt = \sqrt{5}\left(\int_{0}^{\pi} \sin(t)\, dt + \int_{0}^{\pi} \cos(2t)\, dt\right)\). The first integral \(\int_{0}^{\pi} \sin(t) \ dt = [-\cos(t)]_{0}^{\pi} = 2\), and the second integral \(\int_{0}^{\pi} \cos(2t) \ dt = \frac{1}{2} \sin(2t)\mid_{0}^{\pi} = 0\). Adding these results gives 2. Thus the integral evaluates to \(2\sqrt{5}\).

Key Concepts

Parametrization

In order to tackle problems involving line integrals, parametrization is an essential step. Parametrization involves expressing a curve in terms of a parameter, usually denoted as \(t\). For our exercise, the curve is a straight line segment connecting the points \((0,0)\) to \((\pi, 2\pi)\). To parameterize this line, we define a vector function \(\mathbf{r}(t) = (t, 2t)\), where \(t\) ranges from 0 to \(\pi\). This essentially maps every value of \(t\) to a point on the line, ensuring we cover the entire curve from start to finish.

• Parametrization simplifies the process of dealing with curves, making integration possible.
• For every line segment, there's typically a straightforward parametrization reflecting its linear nature.

By representing the coordinates \(x\) and \(y\) in terms of \(t\), we gain a clearer understanding of the curve's orientation and behavior, easing the transition to evaluating a line integral.

Arc Length

Understanding arc length is crucial when working with line integrals, as it determines how we integrate along a curve. Specifically, for a curve parameterized by \(\mathbf{r}(t)\), the differential arc length element \(ds\) is computed as \(|\mathbf{r}'(t)| \cdot dt\). This describes a tiny piece of the curve being traversed. In our specific problem:

• Determine the derivative of \(\mathbf{r}(t)\), which is \(\mathbf{r}'(t) = (1, 2)\).
• Calculate the magnitude of this derivative: \(\sqrt{1^2 + 2^2} = \sqrt{5}\).
• Thus, \(ds = \sqrt{5}\, dt\), acting as the uniform length of the infinitesimal segments you're integrating over.

Arc length ensures your integration respects the true geometry of the curve regardless of its shape or direction.

Trigonometric Integrals

Trigonometric integrals often appear in line integrals, especially when dealing with sinusoidal functions within the integrands. In our exercise, the function \((\sin x + \cos y) \) becomes a trigonometric integral after substituting the parameterized equations \(x = t\) and \(y = 2t\). This transforms the integrand into \(\sin(t) + \cos(2t)\).

• The integral of \(\sin(t)\) over \([0, \pi]\) evaluates to \([-\cos(t)]_{0}^{\pi} = 2\).
• The integral of \(\cos(2t)\) results in \(\frac{1}{2} \sin(2t)\mid_{0}^{\pi} = 0\).

When combined, these trigonometric behaviors deliver the integral's overall value. Understanding trigonometric integrals often involves familiar rules like integrating sine and cosine, which are foundational in calculus.

Curve Analysis

Curve analysis is the comprehensive examination of a curve's attributes, such as its path, how it evolves, and its geometric properties. In the context of line integrals, analyzing a curve helps in understanding how the integral function behaves along it.

• Begin by analyzing the parameterization \(\mathbf{r}(t) = (t, 2t)\); this defines the linear path.
• Check the derivative \(\mathbf{r}'(t)\) for straightforward interpretation of the curve's direction and rate of change, which in this case, remains constant.
• Understand that a line segment offers relatively simple behavior in curve analysis, as opposed to more complex curves like circles or ellipses.

Analyzing these properties streamlines problem-solving, as you can predict how the integral's value manifests along different points of the curve. It's a pivotal step towards mastering line integrals and their applications across different fields.