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 polar curve can be sketched by evaluating the behavior of \(r\) as \(\theta\) varies within the given range. Upon conversion, the rectangular form of the equation is obtained. The area enclosed by the loop can be calculated using a definite integral with limits derived from the points where \(r=0\).

Step-by-step solution

Sketch the Polar Curve

To sketch a polar curve, it is advisable to first consider the behaviour of \(r\) as \(\theta\) varies within the given range. Since \(\cos\theta \leq 1\), then, \(\sec\theta-2\cos\theta \geq 0\), for all values of \(\theta\) within \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\). This tells us that \(r\) is never negative within the given range, and that the graph lies entirely to the right of the y-axis. By assessing particular cardinal points such as \(\theta = 0\), \(\theta = \frac{\pi}{4}\), and \(\theta = \frac{\pi}{2}\), we can plot a reasonably accurate sketch of the polar curve.

Convert the Equation to Rectangular Coordinates

Polar and rectangular coordinates are related by the formulas \(x = r\cos \theta\) and \(y = r\sin \theta\). Substitute \(r = \sec\theta - 2\cos\theta\) into these formulas to get the rectangular coordinates equations.

Find the Area Enclosed by the Loop

The area \(A\) enclosed by a polar curve \(r=f(\theta)\) between \(\theta=a\) and \(\theta=b\) is given by \[A = \frac{1}{2}\int_{a}^{b}(f(\theta))^2 d\theta\]. By locating where the loop starts and ends (those are the points where \(r=0\)), we can set the limits of integration a and b accordingly and calculate the area by integrating \((\sec\theta - 2\cos\theta)^2\).