How do you graph $r^2=3sin2θ$ ?

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Answer 1 · 2016-12-12T11:17:02

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]}

How do you graph r^2=3sin2θr^2=3sin2θ?