Problem 43 - Thomas Calculus

Question

In Exercises \(43-46,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array}{l}{\text { a. Find } d s=|\mathbf{v}(t)| d t \text { for the path } \mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k} \text { . }} \\ {\text { b. Express the integrand } f(g(t), h(t), k(t))|\mathbf{v}(t)| \text { as a function of }} \\ {\text { the parameter } t .} \\ {\text { c. Evaluate } \int_{C} f d s \text { using Equation }(2) \text { in the text. }}\end{array} $$ $$ \begin{array}{l}{f(x, y, z)=\sqrt{1+30 x^{2}+10 y} ; \quad \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+3 t^{2} \mathbf{k}} \\ {0 \leq t \leq 2}\end{array} $$

Step-by-Step Solution

Answer

The line integral evaluates to \( \frac{326}{3} \).

Step 1: Calculate the velocity vector and its magnitude

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 3t^2 \mathbf{k} \). Compute the derivatives: \( \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + 3t^2 \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + 6t \mathbf{k} \). Then, compute the magnitude: \( |\mathbf{v}(t)| = \sqrt{1^2 + (2t)^2 + (6t)^2} = \sqrt{1 + 4t^2 + 36t^2} = \sqrt{1 + 40t^2} \).

Step 2: Express the integrand as a function of parameter \( t \)

The integrand is \( f(g(t), h(t), k(t)) |\mathbf{v}(t)| \). Given \( g(t) = t \), \( h(t) = t^2 \), and \( k(t) = 3t^2 \), we find \( f(g(t), h(t), k(t)) = \sqrt{1 + 30(t)^2 + 10(t^2)} = \sqrt{1 + 30t^2 + 10t^2} = \sqrt{1 + 40t^2} \). Thus, the integrand becomes \( \sqrt{1 + 40t^2} \cdot \sqrt{1 + 40t^2} = 1 + 40t^2 \).

Step 3: Evaluate the line integral

The line integral \( \int_{C} f \, d s = \int_{0}^{2} (1 + 40t^2) \, dt \). First, integrate the function: \( \int (1 + 40t^2) \, dt = \int 1 \, dt + \int 40t^2 \, dt = t + \frac{40}{3}t^3 \). Evaluate from 0 to 2: \( \left[ t + \frac{40}{3}t^3 \right]_0^2 = \left(2 + \frac{40}{3}(8)\right) - (0 + 0) = 2 + \frac{320}{3} = \frac{326}{3} \).

Key Concepts

Vector Calculus

Vector Calculus is a branch of mathematics that deals with vectors and how they can be applied to solve complex mathematical problems. One crucial concept in vector calculus is the line integral, which helps to calculate quantities over a curve or path.

A line integral allows us to combine functions and curves, evaluating them simultaneously over a given path. This is particularly useful in fields like physics, where we often need to compute work done by a force along a curve. In the given exercise, the line integral is used to evaluate the function over the curve defined by the vector path \( \mathbf{r}(t) \).

The approach starts by determining the velocity vector \( \mathbf{v}(t) \), which is the derivative of the position vector \( \mathbf{r}(t) \). Then, the magnitude \( |\mathbf{v}(t)| \) is calculated, providing the length element necessary for integration over the path. This step is essential for converting the curve into a computable form that interacts with the scalar function \( f(x, y, z) \), resulting in the line integral \( \int_{C} f \, d s \).

Parametric Equations

Parametric equations are a powerful tool in mathematics to describe a wide variety of curves by expressing coordinates as functions of a parameter \( t \). Instead of using traditional Cartesian equations, which rely on \( x \) and \( y \) solely, parametric equations give more flexibility and are crucial for describing paths in vector calculus.

In the exercise, the path is given by a set of parametric equations: \( \mathbf{r}(t) = g(t) \mathbf{i} + h(t) \mathbf{j} + k(t) \mathbf{k} \). Here, \( g(t) = t \), \( h(t) = t^2 \), and \( k(t) = 3t^2 \) effectively map out the space curve along which the integral is evaluated. With the range \( 0 \leq t \leq 2 \), parametric equations define the trajectory and ensure the vector calculus operations can be performed effectively.

Using parametric representations of paths allows us to integrate complex integrands easily, as it simplifies multiplying function evaluations along with the differentials, streamlining the process of solving line integrals.

Calculus Applications

The practical applications of calculus, especially involving line integrals, are vast and touch many scientific fields such as engineering, physics, and computer graphics. In our exercise, we see how calculus methods evaluate a line integral over a defined parametric path, producing tangible results.

One typical application is computing the work done by a given force field along a path, as in physics scenarios. By using the line integral, the work can be quantified, helping to solve mechanical problems or understand the behavior of physical systems. In this exercise, the integral \( \int_{C} f ds \) evaluates to \( \frac{326}{3} \), illustrating the force's effect or other quantities derived using similar setups.

This showcases how calculus, especially line integrals, plays crucial roles in determining and understanding multi-dimensional phenomena, making it a cornerstone of modern mathematical applications. Mastering these concepts enhances problem-solving skills and deepens comprehension of complex systems.