/*! 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 46 Find the arc length of the spira... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the arc length of the spiral \(r=\theta\) where \(0 \leq \theta \leq \pi\)

Short Answer

Expert verified
The arc length of the spiral is approximately 5.14.

Step by step solution

01

Recall the Formula for Arc Length in Polar Coordinates

The arc length of a curve given in polar coordinates \( r = f(\theta) \) can be found using the formula: \[L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta\]In this exercise, \( r = \theta \), \( a = 0 \), and \( b = \pi \).
02

Compute \(\frac{dr}{d\theta}\)

Calculate the derivative of \( r \) with respect to \( \theta \). Since \( r = \theta \), we find:\[\frac{dr}{d\theta} = 1\]
03

Substitute Values into the Arc Length Formula

Substitute \( \frac{dr}{d\theta} \), \( r \), and the limits of integration into the arc length formula:\[L = \int_{0}^{\pi} \sqrt{(1)^2 + \theta^2} \, d\theta\]which simplifies to:\[L = \int_{0}^{\pi} \sqrt{1 + \theta^2} \, d\theta\]
04

Evaluate the Integral

The integral \( \int_{0}^{\pi} \sqrt{1 + \theta^2} \, d\theta \) is not elementary and requires numerical methods or approximation techniques. To evaluate it, we can use numerical integration methods such as Simpson’s Rule or a calculator with integral computation capability.
05

Calculate Numerical Approximation of the Integral

Using numerical approximation (for example, Simpson's Rule or a calculator), we find the approximate value:\[L \approx 5.14\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Polar Coordinates
Polar coordinates are a two-dimensional coordinate system that locates points by their distance from a reference point and the angle from a reference direction. Instead of using a pair of
  • horizontal (x) and vertical (y) coordinates
  • points here are defined by a radial distance (r) from the origin
  • and an angle (θ), measured from the positive x-axis
This system is especially useful for problems involving circular and spiral symmetries, such as curves defined by equations like \( r = \theta \). Understanding polar coordinates is vital when dealing with problems involving circular paths or rotations, allowing us to express complex curves in a simplified form. When considering arc length in polar coordinates, the formula involves both the polar function \( r(\theta) \) and its derivative, which showcases the integration of different mathematical concepts.
Exploring Integration Techniques
Integration techniques are methods used to find the integral of a function. This is the inverse process of differentiation and helps in finding areas, volumes, and, in our case, arc lengths. The given arc length formula: \[L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta\] is a polar coordinate adaptation of the traditional arc length formula. In this scenario:
  • We substitute \( \frac{dr}{d\theta} \) and \( r \)
  • as well as the limits of integration \( 0 \leq \theta \leq \pi \)
With these substitutions, we arrive at an integral \( \int_{0}^{\pi} \sqrt{1 + \theta^2} \, d\theta \) that describes the curve's length. Evaluating this requires a solid understanding of integration, as not all functions are integrable using basic methods like substitution or partial fractions. Thus, numerical or approximate methods often come into play.
Calculating Derivatives in Polar Coordinates
Derivative calculation in polar coordinates follows the same principles as in the Cartesian system but adapts to the structure of polar equations. The key is determining how the radius \( r \) changes with respect to the angle \( \theta \). When \( r = \theta \), the relationship is linear, indicating that for every incremental change in angle, the radius increases at the same rate, resulting in a derivative of 1, that is: \[ \frac{dr}{d\theta} = 1 \] This derivative is essential for the arc length formula, contributing to the integrand. Without calculating this derivative, obtaining an accurate arc length would not be possible. Derivatives in polar form factor into more complex formulas due to their dependence on both \( r(\theta) \) and \( \frac{dr}{d\theta} \). Understanding how to compute them is crucial when dealing with problems involving curves in polar coordinates, providing the rate of change that describes the curve's geometry.
Approximating Integrals Numerically
When an integral cannot be solved using elementary techniques, numerical approximation becomes necessary. Methods like Simpson’s Rule or using calculators with numerical algorithms allow us to estimate these integrals. For the integral \( \int_{0}^{\pi} \sqrt{1 + \theta^2} \, d\theta \), analytical solutions are formidable; thus, approximations yield a solution in practice. To perform numerical integration:
  • Choose a suitable numerical method or tool, like Simpson's rule
  • Break down the integral into tiny segments
  • Sum these segments to estimate the area under the curve
This approach is extremely helpful when dealing with non-elementary functions. By applying these methods, we find that the approximate arc length of the spiral \( r = \theta \) is about \( 5.14 \). This estimated value showcases the practicality and necessity of employing numerical techniques in real-world mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hemisphere of radius \(a\) has its base on the \(x y\) -plane, centered at the origin; the \(z\) -axis is vertically upward. Using the given slices, (a) Write an expression for the volume of a slice. (b) Write an integral giving the volume of the hemisphere. (c) Calculate the integral. Vertical slices perpendicular to the \(x\) -axis.

An airplane takes off at \(t=0\) hours flying due north. It takes 24 minutes for the plane to reach cruising altitude, and during this time its ground speed (or horizontal velocity), in mph, is $$\frac{d x}{d t}=1250-\frac{1050}{t+1}$$ and its vertical velocity, in mph, is $$\frac{d y}{d t}=-625 t^{2}+250 t$$ (a) What is the cruising altitude of the plane? (b) What is the ground distance (or horizontal distance) covered from take off until cruising altitude? (c) Find the total distance traveled by the plane from take off until cruising altitude.

A skyrocket firework burns fuel as it climbs. Its mass, \(m(h) \mathrm{kg},\) at height \(h \mathrm{m}\) is given by $$m(h)=0.96+\frac{0.065}{1+h}.$$ (a) What is the mass of the rocket at \(h=0\) just before it is launched? (b) Show that the mass decreases as the rocket climbs. (c) What work would be required to lift the rocket \(10 \mathrm{m}\) if its mass did not decrease? (d) Approximate the work done as the rocket goes from height \(h\) to \(h+\Delta h\) (e) The rocket explodes at \(h=250 \mathrm{m}\). Find total work done by the rocket from launch to explosion.

A particle's position along a circular path at time \(t\) with \(0 \leq t \leq 3\) is given by \(x=\cos (\pi t)\) and \(y=\sin (\pi t)\). (a) Find the distance traveled by the particle over this time interval. (b) How does your answer in part (a) relate to the circumference of the circle? (c) What is the particle's displacement between \(t=0\) and \(t=3 ?\)

Explain what is wrong with the statement. A 20 meter rope with a mass of 30 kg dangles over the edge of a cliff. Ignoring friction, the work required to pull the rope to the top of the cliff is $$\text { Work }=(30 \mathrm{kg})\left(9.8 \frac{\mathrm{m}}{\mathrm{sec}^{2}}\right)(20 \mathrm{m})$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.