Question

Consider the family of limaçons \(r=1+b \cos \theta .\) Describe how the curves change as \(b \rightarrow \infty\)

Short answer

Answer: As \(b\) approaches infinity, the limaçon will tend to flatten along the horizontal axis, forming infinite arcs along the positive and negative \(x\)-axes.

Step-by-step solution

Understand the limaçon equation

A limaçon is a curve described by the polar equation \(r = 1 + b\cos\theta\), where \(r\) is the radial distance, \(\theta\) is the angle, and \(b\) is a parameter that affects the shape of the curve.

Identify the characteristics of a limaçon

Limaçons have some interesting properties depending on the value of the parameter \(b\): 1. When \(b=0\), the curve becomes a circle with radius 1 centered at the pole \((0,0)\). 2. When \(0 < b < 1\), the limaçon has no loops, and it forms a "dimple". 3. When \(b = 1\), the limaçon forms a "cardioid" shape, with a cusp at the pole \((0,0)\). 4. When \(b > 1\), the limaçon forms an inner loop.

Analyze the behavior of the limaçon as \(b \rightarrow \infty\)

To understand how the curve changes as \(b \rightarrow\infty\), consider the behavior of \(r = 1 + b\cos\theta\) for increasing values of \(b\) for different values of \(\theta\): 1. For \(\theta = 0\), \(r = 1 + b\cos(0) = 1 + b\), which approaches infinity as \(b\) increases. 2. For \(\theta = \pi\), \(r = 1 + b\cos(\pi) = 1 - b\), which approaches negative infinity as \(b\) increases. 3. For \(\theta = \frac{\pi}{2}\) or \(\theta = \frac{3\pi}{2}\), \(r = 1 + b\cos(\frac{\pi}{2}) \approx 1\) and \(r = 1 + b\cos(\frac{3\pi}{2}) \approx 1\). As \(b \rightarrow \infty\), the curves will tend to "[[http://mathworld.wolfram.com/Limacon.html|flatten]]" along the horizontal axis. The portion between \(0\) and \(\pi\) will tend to an infinite arc along the positive \(x\)-axis, while the portion between \(\pi\) and \(2\pi\) will tend to an infinite arc along the negative \(x\)-axis.