How do you graph #r=7-2sintheta#?

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Answer 1 · 2018-07-18T07:38:15

See graph and explanation.

#r = 7 - 2 sin theta rArr r^2 = 7 r - 2 r sin theta #

As #r = f (sin theta )#, the limacon is symmetrical about ( y-axis )

#theta = pi/2 #.

#r in [ 5, 7 ]#..

Using

#( x, y ) = r ( cos theta, sin theta ) and r = sqrt( x^2 + y^2 ) >= 0#.

the Cartesian form of the given equation is

#x^2 + y^2 = 7 sqrt( x^2 + y^2 ) -2 y#.

The Socratic graph is immediate.
graph{(x^2 + y^2 - 7 sqrt( x^2 + y^2 ) + 2 y)(y+2) = 0[-16 16 -10 6]}

The graph is not a circle.

There is no ( tangent-crossing-curve ) dimple, at the lowest point,

The perpendicular horizontal and vertical diameters

#(14_+# and 14 ) are not equal.