Problem 91 - Algebra and Trigonometry

Question

Determine whether each statement makes sense or does not make sense, and explain your reasoning. There are no points on my graph of \(r^{2}=9 \cos 2 \theta\) for which \(\frac{\pi}{4}<\theta<\frac{3 \pi}{4}\)

Step-by-Step Solution

Answer

The statement makes sense because there are indeed no points on the graph of \(r^{2}=9 \cos 2 \theta\) for which \(\frac{\pi}{4}<\theta<\frac{3 \pi}{4}\). This is due to the cosine function in the given equation providing negative values in this range of \(\theta\), yet the term \(r^{2}\) cannot be negative.

Step 1: Understand the Graph

The graph of \(r^{2}=9 \cos 2 \theta\) is a limaçon, which is a type of polar plot. The angular coefficient 2 indicates a faster variation of the radian than normal, meaning this limaçon will have 'corners' or loops.

Step 2: Understand the Range

For polar coordinates, \(\theta\) is the angle from the positive x-axis, and ranges from \(-\pi\) to \(\pi\) or from 0 to \(2\pi\). The given range of \(\frac{\pi}{4}<\theta<\frac{3 \pi}{4}\) corresponds to the second quadrant of the coordinate plane.

Step 3: Determine the absence of points

In the equation \(r^{2}=9 \cos 2 \theta\), we can see that 'r' depends on the cosine function, which will be negative in the second quadrant, since \(\cos(2\theta)\) in the given range will fall into the third and fourth quadrant in unit circle, where cosine is either negative or zero. But, \(r^{2}\) can't be negative implying that there are no points in the given range of \(\theta\).