Question

Sketch the strophoid \(r=\sec \theta-2 \cos \theta\) \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\). Convert this equation to rectangular coordinates. Find the area enclosed by the loop.

Short answer

The graph depicts a strophoid. The rectangular form of the equation can be obtained via the substitution of \(r = \sec \theta - 2\cos\theta\) into the conversion equations. The area within the loop can be determined by integrating \(\frac{1}{2}\int_{a}^{b} r^2 d\theta\) over the appropriate interval for \(\theta\).

Step-by-step solution

Sketch the polar equation

The polar equation \(r=\sec \theta-2 \cos \theta\), for \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\), describes a strophoid or looped shape. Use a graphing calculator or software to visualize it.

Convert to Rectangular Coordinates

To transition from polar coordinates to rectangular coordinates, use the conversion equations \(r^2 = x^2 + y^2\) and \(\tan \theta = \frac{y}{x}\). Substitute \(r = \sec \theta - 2 \cos \theta\) and express \(x\) and \(y\) in terms of \(\theta\). Hence \(x = r \cos \theta = (\sec \theta - 2\cos\theta)\cos\theta\) and \(y = r \sin \theta = (\sec \theta - 2\cos\theta)\sin\theta\).

Find the Area Enclosed by the Loop

In polar coordinates, the area \(A\) enclosed by the curve is given by the integral \(\frac{1}{2}\int_{a}^{b} r^2 d\theta\). Boundaries for this integral can be determined from the symmetries of the strophoid diagram. Once the area can be expressed as a definite integral, solve this integral to find the area enclosed by the strophoid.