Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{C} x z d x+(y+z) d y+x d z ; C\) is the curve \(x=e^{t}\), \(y=e^{-t}, z=e^{2 t}, 0 \leq t \leq 1 .\)

Step-by-Step Solution

Answer

The value of the integral is \(\frac{1}{4}e^4 + \frac{2}{3}e^3 - e - \frac{3}{4}\).

Step 1: Parameterize the Curve

The curve is given by the parameterization:\[x = e^t, \quad y = e^{-t}, \quad z = e^{2t}\]with the parameter \(t\) ranging from 0 to 1.

Step 2: Determine the Derivatives

Calculate the derivatives of the parameterized functions with respect to \(t\):\[\frac{dx}{dt} = e^t, \quad \frac{dy}{dt} = -e^{-t}, \quad \frac{dz}{dt} = 2e^{2t}\]

Step 3: Substitute and Set Up the Integral

Substitute the parameterized expressions and their derivatives into the integral:\[\int_{0}^{1} \left( e^t e^{2t} \cdot e^t + (e^{-t} + e^{2t})(-e^{-t}) + e^t \cdot 2e^{2t} \right) dt\]Simplify the expression inside the integral:\[\int_{0}^{1} \left( e^{4t} - 1 - e^t + 2e^{3t} \right) dt\]

Step 4: Integrate Each Term

Integrate each term separately:\[\int_{0}^{1} e^{4t} \, dt = \left[ \frac{1}{4} e^{4t} \right]_{0}^{1} = \frac{1}{4}(e^4 - 1)\]\[\int_{0}^{1} (-1) \, dt = \left[-t\right]_{0}^{1} = -1\]\[\int_{0}^{1} (-e^t) \, dt = \left[-e^t\right]_{0}^{1} = -(e - 1)\]\[\int_{0}^{1} 2e^{3t} \, dt = \left[\frac{2}{3}e^{3t}\right]_{0}^{1} = \frac{2}{3}(e^3 - 1)\]

Step 5: Sum the Results

Combine the results of the integrals:\[\frac{1}{4}(e^4 - 1) - 1 - (e - 1) + \frac{2}{3}(e^3 - 1)\]Simplify:\[\frac{1}{4}e^4 + \frac{2}{3}e^3 - e - \frac{3}{4}\]

Key Concepts

Line Integrals

Line integrals are an important concept in vector calculus. They allow us to integrate functions along a curve or path. Specifically, they help in calculating the accumulative effect of a vector field on a moving particle along a defined path. Instead of just considering a function in a space, line integrals consider how the function behaves along a specific path.

• In mathematics, a line integral involves adding up values throughout a curve.
• This is useful in fields like physics where you might want to measure things like work done by a force along a path.

In this exercise, the curve is parameterized, which simplifies setting up and evaluating the integral over the path. By substituting the parametric equations into the integral, you can evaluate complex paths, including those that curve through three-dimensional space.

Parametric Equations

Parametric equations allow us to describe a path or curve in space using parameters. Each spatial coordinate is expressed as a function of one or more parameters. This makes it easier to describe complex curves or shapes that do not follow a simple algebraic equation.

• In our problem, the curve is described with respect to the parameter \( t \):
• \(x = e^t\), \(y = e^{-t}\), \(z = e^{2t}\), for \(0 \leq t \leq 1\).

This approach simplifies calculations involving the curve, such as finding tangent vectors or when performing line integrals. Each coordinate depends on \( t \), making it easier to track changes as \( t \) progresses.

Differentiation

Differentiation is a fundamental mathematical process in calculus. It involves finding the derivative of a function, which is essentially finding the rate at which the function changes at any point. Differentiation is crucial for understanding change and motion in mathematics and physics.

• In this exercise, differentiation helps find derivatives of the parametric equations.
• The derivatives are: \( \frac{dx}{dt} = e^t \), \( \frac{dy}{dt} = -e^{-t} \), and \( \frac{dz}{dt} = 2e^{2t} \).

These derivatives tell us how each coordinate changes as the independent parameter \( t \) changes. They are used to substitute into the integral, providing the path's trajectory necessary for the integration process.

Exponential Functions

Exponential functions are a class of mathematical functions that model exponential growth or decay. These functions have the general form \( f(x) = a \, e^{bx} \) where \( e \) is the base of the natural logarithm. They are prominent in various scientific fields like biology, physics, and finance.

• The given parameterization uses exponential functions: \( x = e^t \), \( y = e^{-t} \), and \( z = e^{2t} \).
• Each represents constant growth or decay along the curve as \( t \) changes.

These functions are easy to differentiate and integrate, which makes them handy for solving problems involving growth rates, decay processes, and continuous compound interest.