How do you graph #r^2= - cos theta#?

Question

Samantha Abate · Accepted Answer

Nolan Alexander · Answer

The pass equations are: 
#x = r*cos(theta), y = r*sin(theta)#
squaring them, #x^2=r^2*cos(theta)^2,y^2=r^2*sin(theta)#. Substituting now, #x^2=-cos(theta)*cos(theta)^2,y^2=-cos(theta)*sin(theta)^2#. Those equations make sense only for #cos(theta) le 0# so for #pi le theta le 2pi#. Having this in mind we can draw: #x = pm cos(theta)*sqrt(-cos(theta)),y = pm sin(theta)*sqrt(-cos(theta))# for #pi le theta le 2pi#

Delilah Aguilar · Answer

undefined To graph $r^2 = -\cos(\theta)$, we first need to recognize that $r$ is the distance from the origin to a point on the graph, and $\theta$ is the angle measured counterclockwise from the positive x-axis to the point.

1. **Consider the Range of $\cos(\theta)$:** The range of the cosine function is $[-1, 1]$. Since we have $-\cos(\theta)$, the range will be $[-1, 0]$.

2. **Square Root:** Notice that $r^2$ will always be non-negative, so $-\cos(\theta)$ must also be non-negative. This occurs when $\cos(\theta) = 0$, which happens at $\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots$.

3. **Plotting the Graph:** At these values of $\theta$, $r^2 = -\cos(\theta) = 0$, so $r = 0$. This means the graph consists of points where $r = 0$, which is just the origin.

4. **Summary:** The graph of $r^2 = -\cos(\theta)$ is a single point at the origin (0, 0).

So, the graph is just a single point at the origin (0, 0).