How do you convert #5y= -3x^2-2x # into a polar equation?

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Answer 1 · 2017-03-06T10:00:26

#r=-1/3sectheta(5tantheta-2)#

for Cartesian to /from Polar form we use teh eqns.

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

#x=rcostheta#

#y=rsintheta#

#5y=-3x^2-2x#

becomes;
#5rsintheta=-3(rcostheta)^2-2rcostheta#

#5cancel(r)sintheta=-3cancel(r^2)^rcos^2theta-2cancel(r)costheta#

#5sintheta=-3rcos^2theta-2costheta#

#3rcos^2theta=-5sintheta-2costheta#

#r=(-5sintheta-2costheta)/(3cos^2theta)#

#r=(-5sintheta)/(3cos^2theta) -(2costheta)/(3cos^2theta)#

#r=-5/3tanthetasectheta-2/3sectheta#

#r=-1/3sectheta(5tantheta-2)#