Chapter H.1, Problem 38E

To determine

To sketch: the curve with the given equation

Answer to Problem 38E

  

Explanation of Solution

Given:

: $r^2=\cos 4\theta$

Calculation:

  $r^2=\cos 4\theta$

First make a Cartesian graph with $\theta$ on the horizontal axis and $r$ on the vertical axis. Split into two functions

  $r=\sqrt{\cos 4\theta },0\le r\le 1$

  $r=-\sqrt{\cos 4\theta },-1\le r\le 0$

The negative one is the reflection of the positive one about the $x$ -axis.

Take the 4 in front of $\theta$ and divide $2\pi$ by it to get the period

period $=\frac{2\pi }{4}=\frac{\pi }{2}\approx 1.5708$

Values that make cos $4\theta <0$ are also not in the domain.

  $\begin{array}{l}\frac{\pi }{2}<4\theta <\frac{3\pi }{2}\hfill \\ \frac{\pi }{8}<\theta <\frac{3\pi }{8}\hfill \end{array}$

or in general

  $\frac{\pi }{8}+n\frac{\pi }{2}<\theta <\frac{3\pi }{8}+n\frac{\pi }{2}n$ is an integer

Start at the middle value $0,$ go up to the maximum value $2,$ go back up and cross the $x$ -axis at $\theta =\frac{\pi }{6},$ then go back up to 0 at $\theta =\frac{\pi }{3},$ repeat a few more times.

Each interval that has $r=0,1,0$ you can draw a petal that gets as far as 1 unit distance from the pole. Each interval that has $0,-----,0$ has nothing.

One full figure is completed for $0\le \theta \le 2\pi ,$ and it has 4 petals.

The $r=-\sqrt{\cos 4\theta }$ part of the original draws the same petals, but in a different order

  

Conclusion:

Thus, the curve with the given equation is drawn.