How do you graph #r=4sintheta-2#?

Question

Caroline Adams · Accepted Answer

See explanation.

#r=4 sin theta - 2 >=0 to sin theta >=1/2 to theta in (pi/6, 5/6pi)# A Table for making one half of the graph: #(r, theta): (0, pi/5) (2(sqrt2-1), pi/4) ((2(sqrt3-1), pi/3) (2, pi/2)# The graph is symmetrical about the vertical #theta=pi/2#. I have inserted graph for the cartesian double #4y -(x^2+y^2)=+-2sqrt(x^2+y^2)# that is supposed to be the equivalent. As #r=sqrt(x^2+y^2)>=0#, my graph chooses + sign only. For the given polar equation, the graph would be the inner loop only. Please feel such nuances while making graphs, when you make conversions.. graph{x^4+y^4-8y^3+2x^2y^2-8x^2y+12y^2-4x^2=0 [-10, 10, -5, 5]}

Daniel Acosta · Answer

undefined To graph $ r = 4\sin(\theta) - 2 $, follow these steps:

1. Plot the initial value of $ r $ when $ \theta = 0 $. Substitute $ \theta = 0 $ into the equation to find $ r $. In this case, $ r = 4\sin(0) - 2 = 0 - 2 = -2 $.
2. Plot the point (0, -2) on the polar coordinate plane.
3. Determine the behavior of the graph by analyzing the equation. Since $ \sin(\theta) $ ranges from -1 to 1, $ r $ will vary between $ 4(-1) - 2 = -6 $ and $ 4(1) - 2 = 2 $.
4. Plot additional points by choosing different values of $ \theta $, such as $ \theta = \frac{\pi}{6} $, $ \theta = \frac{\pi}{4} $, $ \theta = \frac{\pi}{3} $, and so on, and calculating the corresponding $ r $ values using the equation.
5. Connect the plotted points to form the graph of $ r = 4\sin(\theta) - 2 $.