SOLUTION: The equation r^2 = 4cos2θ is of the form r^2 = a^2 cos2θ, so the graph is a lemniscate. What is the length of each loop? At what values of θ do the endpoints of each

Algebra ->  Trigonometry-basics -> SOLUTION: The equation r^2 = 4cos2θ is of the form r^2 = a^2 cos2θ, so the graph is a lemniscate. What is the length of each loop? At what values of θ do the endpoints of each      Log On


   



Question 1102813: The equation r^2 = 4cos2θ is of the form r^2 = a^2 cos2θ, so the graph is a lemniscate. What is the length of each loop? At what values of θ do the endpoints of each loop occur? Sketch the graph of r^2 = 4cos2θ.
https://www.dropbox.com/s/0q9rpf0ywzip2tx/Screen%20Shot%202017-11-21%20at%209.17.53%20AM.png?dl=0

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
it must be cos%282theta%29%3E=0 , so -pi%2F2%3C=2theta%3C=pi%2F2 <--> -pi%2F4%3C=theta%3C=pi%2F4 .
That means that the r%3E0 part of the curve stays within
a pi%2F4 angle from the horizontal x-axis.
As cos%28anything%29%3C=1 , r%5E2=4cos%282theta%29%3C=4 ,
and only for theta=0 is
r%5E2=4cos%280%29=4%2A1=4 and abs%28r%29=2 .
The endpoints of each loop occur for theta=0 , with r=+%22+%22+%2B-+2 .
For all other possible values of theta ,
0%3C=cos%282theta%29%3C1 , so 0%3C=r%5E2%3C4 , and 0%3C=abs%28r%29%3C2 .
As theta approaches %22+%22+%2B-+pi%2F4 ,
2theta approaches %22+%22+%2B-+pi%2F2 ,
and r%5E2=4cos%282theta%29 approaches 0 .
The curve looks like an infinity symbol centered at the origin,
tangent to the slanted lines y=%22+%22+%2B-+x .
For a better sketch, tabulate some theta , r , x , and y values, like
.