How do you convert #y=-y^2+3x^2-xy # into a polar equation?

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Answer 1 · 2017-08-13T16:14:34

#r = (2sintheta)/(6cos^2theta-2sin^2theta -sin2theta)#

In order to convert between Cartesian and polar equations, we must make use of the following three identities:

#x -=rcostheta#

#y -= r sintheta#

#x^2+y^2 -= r^2#

#y= -y^2+3x^2-xy#

#rsintheta = -r^2sin^2theta +3r^2cos^2theta-r^2sinthetacostheta#

#sintheta = r(-sin^2theta+3cos^2theta -sinthetacostheta)#

#r = (2sintheta)/(6cos^2theta-2sin^2theta -sin2theta)#

Answer 2 · 2017-08-13T16:31:06

Use the formulas #x = rcos(theta) and y = rsin(theta)#
Use algebraic methods to write the radius #r# as a function of #theta#

Before we begin, let's take a look at the graph of the Cartesian equation:

www.desmos.com/calculator

Substitute #rcos(theta)# for every x and #rsin(theta)# for every y:

#rsin(theta) = -(rsin(theta))^2+3(rcos(theta))^2 - (rcos(theta))(rsin(theta)#

Remove the common factor, #r^2#, from the right:

#rsin(theta) = r^2(-sin^2(theta)+3cos^2(theta) - cos(theta)sin(theta))#

We can divide both sides by r, because that only discards the degenerative root #r=0# and the graph will still contain that point when #theta = 0#:

#sin(theta) = r(-sin^2(theta)+3cos^2(theta) - cos(theta)sin(theta))#

Divide both sides by the coefficient of r:

#r= sin(theta)/(-sin^2(theta)+3cos^2(theta) - cos(theta)sin(theta))#

The following is a graph of the polar equation:

www.desmos.com/calculator

Please observe that the graphs are identical.