Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 33-36, set up a definite integral for the arc length of the given curve. Use the Parabolic Rule with \(n=10\) or a CAS to approximate the integral. $$ x=\sin t, y=\cos t, z=\sin t ; 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

The arc length integral is \(\int_{0}^{2\pi} \sqrt{1 + \cos^2 t} \; dt\). Approximate using numerical methods.

Step 1: Define the Problem

We need to find the arc length of the curve described by the parametric equations \(x = \sin t\), \(y = \cos t\), and \(z = \sin t\) over the interval \(0 \leq t \leq 2\pi\).

Step 2: Understand the Arc Length Formula

The arc length \(L\) for a 3D parametric curve \((x(t), y(t), z(t))\) over \(a \leq t \leq b\) is given by the integral: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \; dt \]

Step 3: Compute Derivatives

Find the derivatives: \(\frac{dx}{dt} = \cos t\), \(\frac{dy}{dt} = -\sin t\), \(\frac{dz}{dt} = \cos t\).

Step 4: Set Up the Integral

Substitute the derivatives into the arc length formula: \[ L = \int_{0}^{2\pi} \sqrt{(\cos t)^2 + (-\sin t)^2 + (\cos t)^2} \; dt \]

Step 5: Simplify the Expression

Simplify the expression inside the square root: \((\cos t)^2 + (-\sin t)^2 + (\cos t)^2 = 2\cos^2 t + \sin^2 t\). Using the Pythagorean identity, \(\cos^2 t + \sin^2 t = 1\), so this becomes \(1 + \cos^2 t\).

Step 6: Evaluate the Integral Using the Parabolic Rule

Since we have \(1 + \cos^2 t\) inside the integral, approximate \(\int_{0}^{2\pi} \sqrt{1 + \cos^2 t} \; dt\) using a numerical method like the Parabolic Rule (SIMD integration method) with \(n=10\). This procedure will involve computing contribution of each segment and summing them.

Key Concepts

Definite Integral

A definite integral is a mathematical tool used to calculate the area under a curve on a graph, and it can also measure other quantities, such as work done by a force or the cumulative sum of a function's values. In calculus, when we talk about a definite integral, we are referring to the integral of a function over a specific interval. For instance, if we want to find the definite integral of a function \(f(t)\) over the interval \[ a \le t \le b \], it is expressed as \( \int_{a}^{b} f(t) \, dt \).

• The limits \(a\) and \(b\) are called the bounds of the integral.
• The process of integration gives us a number that represents the cumulative effect between these bounds.

In our exercise, we set up a definite integral to compute the arc length of a curve defined by given parametric equations over the interval \( 0 \leq t \leq 2 \pi \). This use case is an excellent demonstration of how definite integrals can be applied to measure distances, specifically the length of a curve in this context.

Arc Length

Arc length is the measure of the distance along a curved path. In the context of parametric equations in three dimensions, it is the length of a path described by these equations over a certain interval.To calculate the arc length \(L\) for a curve defined by parametric equations \((x(t), y(t), z(t))\), you can use the formula:\[L = \int_{a}^{b} \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 } \, dt\]This formula calculates the infinitesimal distances along the curve and sums them up over the interval from \(a\) to \(b\).

• Each derivative represents how fast each coordinate changes with respect to \(t\).
• The expression inside the square root, \(\sqrt{\cdots}\), represents the differential arc length.

In our example, we used this formula to find the arc length of the given parametric curve, helping us understand how arc length can be computed in three dimensions.

Parametric Equations

Parametric equations describe a group of quantities as functions of a number of independent variables, called parameters. In our example, we dealt with parametric equations in three dimensions:

• \(x = \sin t\)
• \(y = \cos t\)
• \(z = \sin t\)

This particular curve takes the parameter \(t\) and maps it to a point in 3D space as \(t\) varies over a given interval.

• These equations are especially useful for modeling curves that do not easily fit into a single function of \(x\) and \(y\).
• Each equation describes one dimension, making it straightforward to understand how the position changes as \(t\) changes.

We used these parametric equations to explore how the curve creates a path in space, and then calculated the arc length of this path to get a clearer understanding of its geometry.

Numerical Integration

Numerical integration is a technique used to approximate the value of integrals when they cannot be evaluated analytically. This is particularly useful for complex functions or when dealing with functions for which no antiderivative can be found.In our exercise, we used a specific numerical integration method known as the Parabolic Rule or Simpson's Rule.

• This involves dividing the integral into a number of equally spaced intervals.
• Each segment contributes to the overall integral with its own calculated approximation.

Given \(n=10\) subintervals for our curve’s arc length approximation, the method applies calculations at specific points to sum up small "areas" under the curve, yielding an overall arc length approximation. Numerical integration methods like this allow us to solve practical real-world problems where exact calculations are challenging.