What the is the polar form of #y = y/x^2+(x-3)(y+1) #?

Question

1237 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-11T05:32:18

#r^3 sin theta cos^3theta#
# - r^2 cos^2theta ( 3 sin theta - cos theta ) - 3 r cos^2 theta + sin theta = 0#

#y =y/x^2 + ( x - 3 ) ( y + 1 )#. Reorganizing,

#y(x^3 - 3 x^2 + 1 ) = - x^2 ( x - 3 )#

Besides # y = - 1#, the asymptotes are given by{ x = root

of #x^3 - 3x^2 + 1 = 0# }.

See graph for the roots.
graph{y -(x^3-3x^2+1)=0}

Using #( x, y ) = r ( cos theta, sin theta )#, this converts to

#cos theta ( r^3 sin^3theta - 3 r^2 sin^2theta + 1 )#

r sin theta - 3 )# Reorganizing,

#r^3 sin theta cos^3theta#

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

# + sin theta = 0#

The period of #r = f ( sin theta# or #cos theta# ) is #2pi#.

Easily, on #theta = 0, r = 0, 3.#

See graph, for the Cartesian frame.
graph{y(x^3 - 3 x^2 + 1 ) + x^2 ( x - 3 )=0}

Compare with the first graph roots, giving asymptotes x = root.

,