How do you graph $r = 7 - 2 sin θ$ $r=7-2sintheta$ ?

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Answer 1 · 2018-07-18T07:38:15

See graph and explanation.

$r = 7 - 2 sin θ \Rightarrow r^{2} = 7 r - 2 r sin θ$ $r = 7 - 2 sin theta rArr r^2 = 7 r - 2 r sin theta$

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

$θ = \frac{π}{2}$ $theta = pi/2$ .

$r \in [5, 7]$ $r in [ 5, 7 ]$ ..

Using

$(x, y) = r (cos θ, sin θ) and r = \sqrt{x^{2} + y^{2}} \geq 0$ $( x, y ) = r ( cos theta, sin theta ) and r = sqrt( x^2 + y^2 ) >= 0$ .

the Cartesian form of the given equation is

$x^{2} + y^{2} = 7 \sqrt{x^{2} + y^{2}} - 2 y$ $x^2 + y^2 = 7 sqrt( x^2 + y^2 ) -2 y$ .

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

$(14_{+}$ $(14_+$ and 14 ) are not equal.

How do you graph r=7-2sinthetar=7−2sinθr=7-2sintheta?