Problem 222 - Calculus Volume 3

Question

Find the length of the curve over the given interval. \(r=1-\sin \theta\) on the interval \(0 \leq \theta \leq 2 \pi\)

Step-by-Step Solution

Answer

The length of the curve is 8.

Step 1: Understanding the Problem

We need to find the length of the curve defined by the polar equation \(r = 1 - \sin \theta\) over the interval \(0 \leq \theta \leq 2 \pi\). The formula for the length of a polar curve \(r = f(\theta)\) is given by \(L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta\).

Step 2: Differentiating the Function

To use the length formula, we first need to differentiate \(r = 1 - \sin \theta\) with respect to \(\theta\). The derivative is \(\frac{dr}{d\theta} = -\cos \theta\).

Step 3: Setting Up the Integral

Substitute \(r = 1 - \sin \theta\) and \(\frac{dr}{d\theta} = -\cos \theta\) into the length formula. The integral becomes \(L = \int_{0}^{2\pi} \sqrt{(-\cos \theta)^2 + (1 - \sin \theta)^2} \, d\theta\).

Step 4: Simplifying the Integrand

Simplify the expression inside the integral: \((-\cos \theta)^2 + (1-\sin \theta)^2 = \cos^2 \theta + 1 - 2\sin \theta + \sin^2 \theta\). Using the Pythagorean identity, \(\cos^2 \theta + \sin^2 \theta = 1\), the expression simplifies to \(2 - 2\sin \theta\).

Step 5: Further Simplifying and Evaluating

Notice that \(\sqrt{2 - 2\sin \theta} = \sqrt{2(1 - \sin \theta)}\) which simplifies to \(\sqrt{2}\sqrt{1 - \sin \theta}\). Factor the \(\sqrt{2}\) out of the integral: \(L = \sqrt{2}\int_{0}^{2\pi} \sqrt{1 - \sin \theta} \, d\theta\). This integral can be solved using substitution or known integral formulas, resulting in \(8\).

Step 6: Finding the Length of the Curve

After evaluating the integral, we find that the length of the curve over the interval \(0 \leq \theta \leq 2\pi\) is \(8\).

Key Concepts

Polar Coordinates

Polar coordinates are a way to represent points in a plane using a distance from a reference point and an angle from a reference direction. This is different from Cartesian coordinates, where points are identified by x and y distances from the origin. In polar coordinates, each point is given by a pair \(r, \theta\), where \(r\) is the radial distance from the origin (or pole), and \(\theta\) is the angle measured counterclockwise from the positive x-axis.

• The radial distance \(r\) tells us how far away the point is from the origin.
• The angle \(\theta\) indicates the direction of the point from the origin.

These coordinates are particularly useful in situations where the relation between variables is more naturally circular or spiral in nature. For example, many curves, like spirals, are easier to describe using polar coordinates.

Polar Equation

A polar equation expresses a relationship between \(r\) and \(\theta\). In the given exercise, the curve is defined by the polar equation \(r = 1 - \sin \theta\). This means that for each angle \(\theta\), the distance \(r\) from the origin is determined by subtracting the sine of \(\theta\) from 1.

• This specific equation describes a limacon, a type of polar curve that may have a loop.
• Understanding polar equations helps in analyzing how curves behave as the angle changes.

The exercise involves determining the length of this type of curve over a full rotation from \(0\) to \(2\pi\). Polar equations can create interesting patterns, making them valuable in fields like astronomy, navigation, and even art.

Integration Techniques

Finding the length of a curve using polar coordinates often requires integration techniques. The formula for calculating the length of a polar curve is \(L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta\). This formula stems from the arc length in calculus.

To apply this, the steps are:

• Differentiate the polar equation \(r = f(\theta)\) to find \(\frac{dr}{d\theta}\).
• Substitute \(r\) and \(\frac{dr}{d\theta}\) into the formula.
• Simplify the expression under the square root.
• Integrate over the given interval.

Solving this often requires substitution methods or recognizing special integral forms. Mastering these techniques enhances problem-solving skills in calculus, allowing you to tackle complex integrals.

Calculus Problem-Solving

Calculus problem-solving involves breaking down complex problems into manageable steps, as seen in this exercise. The process uses mathematical concepts like differentiation and integration to analyze curves. Let's reexamine how these concepts were applied here:

• The function \(r = 1 - \sin \theta\) was differentiated to find \(\frac{dr}{d\theta}\).
• The expression inside the integral was simplified using identities and algebraic manipulation.
• Finally, the integral is evaluated, revealing the curve's length.

Advanced problem-solving in calculus often involves interpreting results and ensuring calculations account for the full context. Gaining experience in these methods enhances mathematical reasoning and the ability to apply calculus to real-world challenges.