Question

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

Short answer

The length of the curve is 8 units.

Step-by-step solution

Recall the Formula for Arc Length in Polar Coordinates

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

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

Given \( r = 1 - \sin \theta \), we need to find \( \frac{dr}{d\theta} \). Using the derivative of sine, we get:\[ \frac{dr}{d\theta} = -\cos \theta \]

Substitute \( r \) and \( \frac{dr}{d\theta} \) into the Arc Length Formula

Substitute \( r = 1 - \sin \theta \) and \( \frac{dr}{d\theta} = -\cos \theta \) into the arc length formula:\[ L = \int_{0}^{2\pi} \sqrt{(-\cos \theta)^2 + (1-\sin \theta)^2} \, d\theta \]

Simplify the Integrand

Simplify the expression under the square root:\[ (-\cos \theta)^2 + (1-\sin \theta)^2 = \cos^2 \theta + (1 - 2\sin \theta + \sin^2 \theta) \]This simplifies to:\[ \cos^2 \theta + \sin^2 \theta - 2\sin \theta + 1 = 2 - 2\sin \theta \]

Evaluate the Integral for Arc Length

The integral becomes:\[ L = \int_{0}^{2\pi} \sqrt{2 - 2\sin \theta} \, d\theta \]Using the identity \( \sqrt{2 - 2\sin \theta} = 2\cos\left(\frac{\theta}{2}\right) \), the integral simplifies to:\[ 2\int_{0}^{2\pi} \cos\left(\frac{\theta}{2}\right) \, d\theta \]After integrating and evaluating the limits, the result is:\[ L = 8 \]

Conclusion

After solving the integral, we determine that the length of the curve \( r = 1 - \sin \theta \) within the interval \( 0 \leq \theta \leq 2\pi \) is \( 8 \) units.

Key concepts

Polar Coordinates

Polar coordinates offer a unique way of describing a point in a plane using a distance from a reference point and an angle from a reference direction. Instead of the usual Cartesian coordinates (x, y), polar coordinates use the pair (r, θ), where:

• \( r \): The radial distance from the origin (also known as the pole).
• \( \theta \): The angular coordinate, representing the angle measured from the positive x-axis (counterclockwise direction is typically considered positive).

Polar coordinates are particularly useful in situations with circular or rotational symmetry, where representing curves can often be simplified. For instance, in our given curve \( r = 1 - \sin \theta \) on the interval \( 0 \leq \theta \leq 2 \pi \), each value of \( \theta \) determines a specific point along the curve by altering the radius \( r \). Thus, studying curves within the polar coordinate system often helps in simplifying complex geometric problems.

Arc Length Formula

To find the length of a curve using polar coordinates, we need the arc length formula specifically designed for these coordinates. The formula for the arc length \( L \) from an angle \( \theta = a \) to \( \theta = b \) is:

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

This formula accounts for the continuous variation of both the angle \( \theta \) and the radius \( r \). It combines the changes in the radius with how sharply the curve is turning (by incorporating the derivative \( \frac{dr}{d\theta} \)). Without this formula, calculating the arc length of a curve in polar coordinates would be significantly more complex, as it effectively considers small segments of the curve and sums up their lengths to find the total arc length.

Integral Calculus

Integral calculus plays a crucial role when we need to accumulate quantities, such as the total length of a curve as in our problem. In the context of finding the arc length in polar coordinates, we set up an integral that considers infinitesimally small segments of the curve. The arc length integral:

• \[ L = \int_{0}^{2\pi} \sqrt{2 - 2\sin \theta} \, d\theta \]

Here, the integral evaluates the sum of the lengths of infinitely small line segments that make up the entire curve. Calculating the integral requires careful simplification and sometimes using trigonometric identities, such as the one used here to transform \( \sqrt{2 - 2\sin \theta} \) to \( 2\cos\left(\frac{\theta}{2}\right) \). Ultimately, integration allows us to find the arc length without manually measuring each point, providing a powerful mathematical tool for solving complex problems.

Derivatives

A derivative signifies how a function changes as its input changes. This concept is fundamental when working with the arc length formula in polar coordinates, specifically needing \( \frac{dr}{d\theta} \). For the curve \( r = 1 - \sin \theta \), we calculate the derivative:

• \[ \frac{dr}{d\theta} = -\cos \theta \]

Understanding how \( r \) changes with \( \theta \) is critical because every small change in the angle affects the radial distance. The derivative \( \frac{dr}{d\theta} \) helps us quantify these changes, allowing the arc length formula to incorporate these radial changes along with angular changes. The precise calculation of this derivative is crucial to accurately determining the overall curve length. Without derivatives, understanding the varying radius in polar curves would be incomplete, thereby complicating the arc length calculations.