Question

Use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.\(r=3 \sin \theta\) on the interval \(0 \leq \theta \leq \pi\)

Short answer

The length of the curve is \(3\pi\).

Step-by-step solution

Understanding the Problem

We need to find the length of the curve defined by the polar equation \( r = 3 \sin \theta \) over the range \( 0 \leq \theta \leq \pi \). We will then confirm this length using a definite integral.

Formula for Curve Length in Polar Coordinates

The length \( L \) of a curve described in polar coordinates \( r(\theta) \) from \( \theta = a \) to \( \theta = b \) is given by the integral:\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

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

Given \( r = 3 \sin \theta \), its derivative with respect to \( \theta \) is \( \frac{dr}{d\theta} = 3 \cos \theta \).

Substitute Back into the Length Formula

Substitute \( r = 3 \sin \theta \) and \( \frac{dr}{d\theta} = 3 \cos \theta \) into the length formula:\[ L = \int_{0}^{\pi} \sqrt{ (3 \cos \theta)^2 + (3 \sin \theta)^2 } \, d\theta \]

Simplify the Integral

Further simplify the integral expression:\[ L = \int_{0}^{\pi} \sqrt{ 9 \cos^2 \theta + 9 \sin^2 \theta } \, d\theta \]Notice that \( \cos^2 \theta + \sin^2 \theta = 1 \). Thus:\[ L = \int_{0}^{\pi} \sqrt{ 9 } \, d\theta = \int_{0}^{\pi} 3 \, d\theta \]

Evaluate the Definite Integral

Integrate and evaluate from \( 0 \) to \( \pi \):\[ L = \left[ 3 \theta \right]_{0}^{\pi} = 3(\pi) - 3(0) = 3\pi \].

Key concepts

Curve Length in Polar Coordinates

Finding the length of a curve in polar coordinates can be a fascinating concept to explore. In this context, polar coordinates describe the position of a point based on its distance from a fixed point (the origin) and the angle with respect to a fixed direction. For any curve expressed in polar coordinates as\( r(\theta) \), the curve length over an interval \( a \leq \theta \leq b \) is determined by the formula:

• \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

This formula accounts for all angular changes and distances, essentially summing up tiny segments along the curve to find the total length.
For example, if you're dealing with \( r = 3 \sin \theta \), this function beautifully represents a loop, where you can witness the curve's formation by calculating its length over the specified interval. Plugging the function and its derivative into the length formula allows you to determine how far the curve actually runs between those angular bounds.

Understanding Definite Integrals

A definite integral is a fundamental concept in calculus used to find the exact accumulation of quantities, such as areas under curves or lengths of curves, over certain intervals. When you encounter the curve length integral \( \int_{a}^{b} \sqrt{9} \, d\theta \) , it might seem simple, yet this process is deeply rooted in the principles of definite integrals.
The evaluation of such an integral involves integrating a constant over an interval. Once simplified, our integral for the curve length shows that it reduces to \( 3 \int_{0}^{\pi} 1 \, d\theta \). This represents the summation of constant values over an interval, yielding the total length. Definite integrals provide a valuable tool for transitioning from a continuous curve to a comprehensive numerical value, as seen when converting the integral into the expression \( 3\pi \).
This efficient process showcases how integral calculus can deliver useful quantities in physics and engineering, such as curve lengths, through meticulous calculations.

Derivative Calculation and its Role

The derivative plays a crucial role in many calculus problems, including determining the curve length in polar coordinates. For polar coordinate problems, the expression for \( \frac{dr}{d\theta} \) is essential in capturing how \( r \), the radial distance, changes as \( \theta \) varies.
Using the example function \( r = 3 \sin \theta \), the derivative is calculated as \( \frac{dr}{d\theta} = 3 \cos \theta \). This derivative reflects how the radius grows or shrinks with each infinitesimal rotation around the origin.
In curve length calculation, substituting the derivative \( \frac{dr}{d\theta} \) into the length formula helps illustrate how distance components vary along the entire curve. The components \( (3 \cos \theta)^2 \) and \( (3 \sin \theta)^2 \) essentially map the changing directions and magnitudes into a singular measurement, allowing the integral to ultimately simplify to a linear quantity.
By understanding derivative calculation in this context, students can appreciate its pivotal role in bridging mathematical theory with practical measurements.