Problem 18 - Calculus Early Transcendentals

Question

Evaluate the line integral with respect to \(s\) along the curve \(C .\) $$ \begin{array}{l}{\int_{C} \frac{e^{-z}}{x^{2}+y^{2}} d s} \\ {C: \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k} \quad(0 \leq t \leq 2 \pi)}\end{array} $$

Step-by-Step Solution

Answer

The value of the line integral is \( \frac{\sqrt{5}}{4} (1 - e^{-2\pi}) \).

Step 1: Understand the Curve

The curve \( C \) is given by the vector function \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + 2 \sin t \mathbf{j} + t \mathbf{k} \), where \( 0 \leq t \leq 2\pi \). This describes a helix with a circular base of radius 2 on the \( xy \)-plane, and it extends upwards with \( z = t \).

Step 2: Compute the Derivative of the Curve

Find the derivative of \( \mathbf{r}(t) \) with respect to \( t \), which will be the tangent vector. \[ \mathbf{r}'(t) = \frac{d}{dt} \left( 2 \cos t \mathbf{i} + 2 \sin t \mathbf{j} + t \mathbf{k} \right) = -2\sin t \mathbf{i} + 2\cos t \mathbf{j} + \mathbf{k} \]

Step 3: Calculate the Magnitude of the Tangent Vector

Find the magnitude of \( \mathbf{r}'(t) \) to get \( ds \), the differential arc length. \[ \left| \mathbf{r}'(t) \right| = \sqrt{(-2\sin t)^2 + (2\cos t)^2 + 1^2} = \sqrt{4\sin^2 t + 4\cos^2 t + 1} = \sqrt{4 + 1} = \sqrt{5} \]

Step 4: Express the Integrand Using Parameterization

Substitute the parameterization expressions for \( x, y, z \) into the integrand. We have \( x = 2\cos t, y = 2\sin t, z = t \), so the integrand becomes: \[ \frac{e^{-z}}{x^2 + y^2} = \frac{e^{-t}}{(2\cos t)^2 + (2\sin t)^2} = \frac{e^{-t}}{4} \]

Step 5: Formulate the Integral

Substitute the integrand and \( ds \) into the line integral and use the parameter \( t \). The integral becomes: \[ \int_C \frac{e^{-z}}{x^2 + y^2} \, ds = \int_0^{2\pi} \frac{e^{-t}}{4} \sqrt{5} \, dt \] Simplify it to: \[ \frac{\sqrt{5}}{4} \int_0^{2\pi} e^{-t} \, dt \]

Step 6: Evaluate the Integral

Compute the integral \( \int_0^{2\pi} e^{-t} \, dt \). This is a simple exponential integral with a straightforward antiderivative. \[ \int e^{-t} \, dt = -e^{-t} + C \] Evaluate from 0 to \( 2\pi \): \[ \left[ -e^{-t} \right]_0^{2\pi} = -(e^{-2\pi} - e^{0}) = -e^{-2\pi} + 1 \]

Step 7: Combine Results

Multiply the result of the definite integral by the constant factor from \( ds \): \[ \frac{\sqrt{5}}{4} \left( 1 - e^{-2\pi} \right) \] This is the value of the line integral along curve \( C \).

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that focuses on vector fields and differentiation and integration with respect to vector functions. It's crucial for understanding how to compute line integrals, like in the given task. A vector function, such as \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + 2 \sin t \mathbf{j} + t \mathbf{k} \), defines a path or curve in space.

• Vectors: Vectors have both magnitude and direction. In calculus, they're often used to represent physical quantities like velocity and force, along with lines and curves in space.
• Vector Derivatives: To analyze changes in vector functions, we differentiate them. The derivative of our curve \( \mathbf{r}(t) \) produces a tangent vector, \( \mathbf{r}'(t) \). This tells us the direction and rate of change of the curve at any point.
• Line Integrals: A line integral is a type of integral where we integrate across a curve or line in the vector field. It extends the concept of an integral to multi-dimensional paths, allowing calculations across curves and surfaces in spaces.

The result of our line integral provides information about the curve, often related to physical properties like work done by a force along a path. Understanding vector calculus is foundational for fields like physics and engineering where multi-dimensional calculations are necessary.

Parameterization

Parameterization is a technique where you express a mathematical object as a function of parameters. This helps describe complex curves or surfaces with simpler functions. For curves like the helix in this problem, parameterization is essential.

• Defining the Curve: We have \( \mathbf{r}(t) = 2\cos t \mathbf{i} + 2\sin t \mathbf{j} + t \mathbf{k} \), representing a helix. The \( t \) parameter smoothly traces the path of the curve from 0 to \( 2\pi \), as it links \( x, y, \) and \( z \) coordinates.
• Simplifying Calculations: By substituting the parameterized expressions, \( x = 2\cos t, y = 2\sin t, z = t \), into equations, we simplify the integrand when performing integrations.
• Applications: Parameterization is not exclusive to curves but used for surfaces and other complex shapes, aiding in visualization and calculation processes across various problems.

In this problem, the parameterization of \( x, y, \) and \( z \) into terms of \( t \) allows us to compute line integrals directly on the simpler expressions. It's a powerful tool, making complex geometries more approachable in calculus.

Arc Length

Arc length measures the distance along a curve from one point to another. When we talk about line integrals with respect to \( s \), \( ds \) represents the infinitesimal length of the curve.

• Computing Arc Length: For a curve defined by \( \mathbf{r}(t) \), the arc length from \( t = a \) to \( t = b \) is determined by integrating the magnitude of the tangent vector \( \mathbf{r}'(t) \). In this exercise, the arc length element \( ds \) is the integral's differential segment.
• Magnitude of the Tangent Vector: We computed \( \left| \mathbf{r}'(t) \right| = \sqrt{5} \). This magnitude gives the scale of change in the curve's length as we move over \( dt \), providing \( ds = \sqrt{5} \).
• Importance: Arc length is crucial in various applications, from calculating the work done along a path in physics to determining the geometry of curved structures in engineering.

Thus, understanding arc length in line integrals helps us measure the effects over a curve, enabling precise computation of quantities related to distance in vector fields.