Problem 14 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. \(\int_{c}(x y-z) d x+e^{x} d y+y d z ;\) C: the line segment from \((1,0,0)\) to \((3,4,8)\)

Step-by-Step Solution

Answer

\frac{8}{3} + 2e^3 - 2e + 10

Step 1: Parameterize the Line Segment

Let's parameterize the line segment from \( (1,0,0) \) to \( (3,4,8) \). A linear parameterization can be given by \( x(t) = 1 + 2t \), \( y(t) = 4t \), and \( z(t) = 8t \), where \( 0 \leq t \leq 1 \).

Step 2: Compute Derivatives of the Parameterization

Next, calculate the derivatives with respect to \( t \): \[ \frac{dx}{dt} = 2 \] \[ \frac{dy}{dt} = 4 \] \[ \frac{dz}{dt} = 8 \]

Step 3: Substitute Parameterized Expressions into the Integral

Substitute \ x(t) \, \ y(t) \, \ z(t) \ and the derivatives into the integral: \[ \int_{0}^{1} \left[ (x(t) y(t) - z(t)) \frac{dx}{dt} + e^{x(t)} \frac{dy}{dt} + y(t) \frac{dz}{dt} \right] dt \] becomes \[ \int_{0}^{1} \left[ (1 + 2t)(4t) - 8t) \cdot 2 + e^{1 + 2t} \cdot 4 + 4t \cdot 8 \right] dt \]

Step 4: Simplify the Integrand

Simplify the integrand: \[ \int_{0}^{1} \left[ (4t + 8t^2 - 16t) + 4e^{1+2t} + 32t \right] dt \] \[ \int_{0}^{1} \left[ 8t^2 + 4e^{1+2t} + 20t \right] dt \]

Step 5: Integrate Each Term

Integrate each term separately: \[ 8 \int_{0}^{1} t^2 dt + 4 \int_{0}^{1} e^{1+2t} dt + 20 \int_{0}^{1} t dt \] For the first term: \[ 8 \left[ \frac{t^3}{3} \right]_0^1 = 8 \frac{1}{3} = \frac{8}{3} \] For the second term: \[ 4 \[ \frac{e^{1+2t}}{2} \]_0^1 = 2(e^3 - e) \] For the third term: \[ 20 \[ \frac{t^2}{2} \]_0^1 = 10 \] The result is: \[ \frac{8}{3} + 2e^3 - 2e + 10 \]

Key Concepts

parameterization

Parameterization is a method in mathematics where you express a curve or surface in terms of one or more parameters. This allows you to simplify complex shapes and functions by representing them with simple variables. For example, in our exercise, we parameterize the line segment connecting point \( (1,0,0) \) and point \( (3,4,8) \) using functions of \( t \). This can be done with the equations: \(x(t) = 1 + 2t, y(t) = 4t, z(t) = 8t\). The parameter \( t \) usually varies from 0 to 1.

• This converts the 3D curve into a simpler 1D representation.
• Makes it easy to compute integrals and derivatives.
• Helps in visualizing complex shapes.

When you parameterize a curve, you also need to calculate the derivatives of the parameterized functions with respect to the parameter. This step is crucial for setting up integrals in the next stages of calculus problems.

integral calculus

Integral calculus is the study of integrals and their properties. It is a core part of calculus and helps in finding areas, volumes, central points, among others. In our exercise, we deal with a line integral which is an integral where the function to be integrated is evaluated along a curve created by parametric equations.

• This allows you to calculate 'weighted' sums across curves.
• Useful for many physical applications, like finding work done by a force field.

Integrals in this scenario often involve setting up the integrand based on the parameterized equations. Next comes substitution and simplification, turning a multi-variable integral into a single-variable one, as seen with the transformation to \( \int_{0}^{1} \left[8t^2 + 4e^{1+2t} + 20t \right] dt \).

vector calculus

Vector calculus extends calculus concepts to higher dimensions. This includes studying vector fields and using vectors to solve problems involving differential and integral calculus. In the context of our exercise, vector calculus allows us to integrate scalar and vector fields over curves like the line segments we parameterized earlier.

• Useful for analyzing physical phenomena like fluid flow and electromagnetic fields.
• Helps in evaluating line integrals and surface integrals.

The key here is to convert the physical problem into a parameterized form which can then be integrated. Understanding vector calculus helps in evaluating expressions such as \( \left[ (x(t) y(t) - z(t)) \frac{dx}{dt} + e^{x(t)} \frac{dy}{dt} + y(t) \frac{dz}{dt} \right] \).

calculus problem solving

Solving calculus problems, especially involving integrals, requires a systematic approach. Whether you're dealing with line integrals, surface integrals, or basic integrals in one or two dimensions:

• Start by understanding the problem.
• Identify what needs to be parameterized.
• Write down parametric equations and compute their derivatives.
• Substitute these into the integral.
• Simplify the integrand before solving the integral.

For example, in our problem, after parameterizing the line segment and substituting the expressions, we simplified the integrand to get \( \int_{0}^{1} \left[8t^2 + 4e^{1+2t} + 20t \right] dt \). This is broken down into simpler integrals which can be easily computed individually before adding them back together. Following these steps ensures you address each part of the problem methodically.