91Ó°ÊÓ

Understanding polar equations is crucial when working with polar coordinates. In a polar coordinate system, we express curves in terms of the radius \( r \) and the angle \( \theta \). This is different from the Cartesian coordinate system where curves are defined by \( x \) and \( y \).
In polar equations, a point \( (r, \theta) \) is described by the distance from the origin, known as the radius \( r \), and the angle \( \theta \), measured from a reference direction, usually the positive x-axis.
These equations are very versatile and can describe complex curves like spirals, circles, and, in our case, a parabola. The polar equation given in the exercise is \( r = \frac{2}{1-\cos \theta} \). This is an example of a conic section, which describes a parabolic curve in the polar coordinate system.

To find arc lengths or work with motion in polar coordinates, calculating derivatives with respect to \( \theta \) is an essential skill.
For the given polar equation \( r = \frac{2}{1-\cos \theta} \), we need to derive with respect to \( \theta \) using the quotient rule. The quotient rule is helpful when deriving a function expressed as the ratio of two other functions: \( \frac{u}{v} \). The derivative is given by \( \frac{d}{d\theta}\left(\frac{u}{v}\right) = \frac{v \frac{du}{d\theta} - u \frac{dv}{d\theta}}{v^2} \).
Applying the quotient rule to our \( r \), the derivative \( \frac{dr}{d\theta} = \frac{2 \sin \theta}{(1 - \cos \theta)^2} \) is achieved. This derivative is vital for plugging into the arc length formula to compute the total length of the curve.

The arc length formula is central to our problem. It allows us to integrate over a defined interval to determine the length of a curve.
For a polar curve described by \( r(\theta) \), the arc length \( L \) from \( \theta = a \) to \( \theta = b \) is calculated by the formula: \[ L = \int_a^b \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta. \]
This formula is derived from the Pythagorean theorem and accounts for both radial and angular changes. We used it with our derived equation \( \frac{dr}{d\theta} \) and original \( r \) to set up the integral necessary for finding the curve's length. Understanding each component: radial change \( r \) squared and the derivative squared, helps connect geometric intuition to calculus in polar coordinates.

To get the arc length, curve integration is performed based on the arc length formula. The process involves setting up and solving an integral over the specified interval \( \theta = \frac{\pi}{2} \) to \( \pi \).
In our exercise, the integrand is simplified to \( 2 \sqrt{2}(1 - \cos \theta)^{-3/2} \).

• The substitution \( u = 1 - \cos \theta \) was used, transforming the limits of integration to match the new variable \( u \).
• Finally, an elementary integral remains, which we evaluate and adjust the limits accordingly.

Ultimately, solving the integral gives the arc length of the curve as \( 4\sqrt{2} \). Curve integration links several calculus concepts, making it powerful yet requiring detailed step-by-step application.