/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 (a) Show that the are length of ... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Show that the are length of one petal of the rose \(r=\cos n \theta\) is given by $$ 2 \int_{0}^{\pi /(2 n)} \sqrt{1+\left(n^{2}-1\right) \sin ^{2} n \theta} d \theta $$ (b) Use the numerical integration capability of a calculating utility to approximate the are length of one petal of the four-petal rose \(r=\cos 2 \theta\) (c) Use the numerical integration capability of a calculating utility to approximate the are length of one petal of the \(n\) -petal rose \(r=\cos n \theta\) for \(n=2,3,4, \ldots, 20\) then make a conjecture about the limit of these are Iengths as \(n \rightarrow+\infty\)

Short Answer

Expert verified
The arc length is given by the integral. Numerically evaluating it for specific \(n\), it approaches a limit.

Step by step solution

01

Calculate the Derivative of the Parametric Equations

The parametric equations for the polar curve \(r = \cos n\theta\) can be expressed as \(x = r \cos \theta\) and \(y = r \sin \theta\). Thus, the expressions for \(x\) and \(y\) become: \[ x = \cos n\theta \cos \theta \quad \text{and} \quad y = \cos n\theta \sin \theta \]We need to find \(x' = \frac{dx}{d\theta}\) and \(y' = \frac{dy}{d\theta}\). These derivatives are:\[ x' = -n\sin n\theta \cos \theta + \cos n\theta(-\sin \theta) \]\[ y' = -n\sin n\theta \sin \theta + \cos n\theta \cos \theta \]
02

Calculating the Integrand for the Arc Length

The arc length in polar coordinates expressed in terms of \(\theta\) is given by:\[ \int \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta \]Substitute \(x'\) and \(y'\) into this equation. Simplifying, we obtain:\[ \sqrt{n^2\sin^2 n\theta + \cos^2 n\theta \sin^2 \theta + \cos^2 n\theta \cos^2 \theta} = \sqrt{1 + (n^2 - 1) \sin^2 n\theta} \]
03

Evaluate the Given Integral for Arc Length

The integral for the arc length of one petal is:\[ 2 \int_0^{\pi/(2n)} \sqrt{1 + (n^2 - 1) \sin^2 n\theta} \, d\theta \]This confirms the given expression for the arc length of one petal of the rose curve.
04

Numerical Integration for Specific Values

For part (b), use a numerical integration tool to approximate the arc length of one petal of the rose \(r = \cos 2\theta\). Input the function into the utility and evaluate:\[ 2 \int_0^{\pi/4} \sqrt{1 + 3 \sin^2 2\theta} \, d\theta \]
05

Numerical Integration for Different Petals

For part (c), repeat the numerical integration for \(n\) ranging from 2 to 20:\[ 2 \int_0^{\pi/(2n)} \sqrt{1 + (n^2 - 1) \sin^2 n\theta} \, d\theta \]Compute these values with a calculator to analyze the pattern and make a conjecture about the limit as \(n \rightarrow +\infty\).
06

Conjecture on Limit Behavior

After running the calculations, you will observe the data stabilization as \(n\) increases. If an idea emerges that the limit approaches a constant value, make a conjecture based on this observation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Integration
Numerical integration is a powerful method used to estimate the value of integrals, especially when an integral cannot be solved analytically. In our case, the arc length of the rose curve expressed as an integral can be numerically approximated. Numerical methods like the Trapezoidal Rule or Simpson’s Rule are commonly used because they split the integral into smaller, more manageable parts.
Using a calculating utility, we can input our function and use these methods to find approximate solutions. For example, when we approximate the arc length of one petal for the four-petal rose \( r = \cos 2\theta \), we evaluate the integral:
\[ 2 \int_0^{\pi/4} \sqrt{1 + 3 \sin^2 2\theta} \, d\theta \] By applying a numerical method, we obtain an estimated value that helps us understand the behavior of these mathematical curves without relying on paper calculations alone.
Derivatives of Parametric Equations
The derivatives of parametric equations are crucial when it comes to calculating arc lengths for polar and parametric curves. When dealing with parametric curves such as our rose curve, the equations are given in terms of \( \theta \), involving both \( x \) and \( y \).
In this scenario, we are working with these parametric equations:
  • \( x = \cos n\theta \cos \theta \)
  • \( y = \cos n\theta \sin \theta \)

Finding the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) is necessary to form the integrand used in calculating the arc length. After obtaining these derivatives, they are substituted into the formula for arc length, using:
\[ \int \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta \] This critical step ensures we are able to accurately measure the change along the curve, providing the groundwork needed for subsequent numerical calculations and integrations.
Limit Behavior of Sequences
When exploring the arc length of one petal of an \( n \)-petal rose curve, a notable observation emerges regarding the limit behavior as \( n \rightarrow +\infty \). As we calculate the arc lengths for increasing values of \( n \) using numerical integration, a pattern materializes.
We collect data points by evaluating:
\[ 2 \int_0^{\pi/(2n)} \sqrt{1 + (n^2 - 1) \sin^2 n\theta} \, d\theta \] for each \( n \) from 2 to 20. As \( n \) grows, the results seem to stabilize and approach a limiting value. This is indicative of a sequence's behavior approaching a finite limit; often described as convergence. Once stable values are observed consistently, we can make an educated conjecture that the arc length of one petal approaches a constant value, reflecting typical behavior as curves become more densely petaled.

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