How do you graph #r^2=3sin2θ#?

Question

Carter Anderson · Accepted Answer

Please see below.

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is #x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#. We can use this to convert equation in polar coordinates to an equation with Cartesian coordinates. #r^2=3sin2theta# #hArrr^2=3xx2sinthetacostheta# or #r^2xxr^2=6xxrsinthetaxxrcostheta# or #(x^2+y^2)^2=6xy# Note that (a) As #6xy# is a complete square, it is positive and hence curve can lie only in first and third quadrant. (b) Further as maximum value of #r^2=3sin2theta#, maximum possible value for #r^2# is #3# and so #r# cannot be more than #sqrt3=1.732....# (c) As replacing #x# and #y# with each other does not change the equation, it is symmetric along #x=y#. Now we can put different values of #x# to get #y# (both less than #sqrt3#) and draw the graph. The function appears as follows. graph{((x^2+y^2)^2-6xy)(x-y)=0 [-5, 5, -2.5, 2.5]}

Nolan Alexander · Answer

undefined To graph the polar equation $ r^2 = 3 \sin(2\theta) $, follow these steps:

1. Rewrite the equation in terms of $ r $ and $ \theta $ separately.
2. Plot points for various values of $ \theta $, calculate the corresponding values of $ r $, and plot them on the polar coordinate system.
3. Connect the plotted points to form the graph.

Since $ r^2 = 3 \sin(2\theta) $, we can rewrite it as $ r = \sqrt{3 \sin(2\theta)} $.

Now, for various values of $ \theta $, calculate $ r $ using the equation $ r = \sqrt{3 \sin(2\theta)} $. Plot these points on the polar coordinate system.
Ensure that $ \theta $ varies from $ 0 $ to $ 2\pi $ to cover one complete cycle of the graph.

After plotting a sufficient number of points, connect them to form the graph of the polar equation $ r = \sqrt{3 \sin(2\theta)} $. This graph represents the polar curve corresponding to the given equation.