How do you graph $r = 1 + 4 sin θ$ $r=1+4sintheta$ ?

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How do you graph $r = 1 + 4 sin θ$ $r=1+4sintheta$ ?

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Answer 1 · 2016-10-21T06:51:25

I strictly adhere to length $r = \sqrt{x^{2} + y^{2}} \geq 0$ $r = sqrt(x^2+y^2)>=0$

So, despite that r is periodic with period $2 π$ $2pi$ , I have to exclude

$θ \in (π - {sin}^{- 1} (- \frac{1}{4}), 2 π + {sin}^{- 1} (- \frac{1}{4}))$ $theta in (pi-sin^(-1)(-1/4), 2pi+sin ^(-1) (-1/4))$ for which $r < 0$ $r <0$
.
Use this Table for making the graph in one period that excludes the

above subinterval.

$(r, θ)$ $(r, theta)$ :

$(1, 0) (3, \frac{π}{3}) (5, \frac{π}{2}) (3, \frac{2}{3} π) (1, π) (0, π - {sin}^{- 1} (- \frac{1}{4})$ $(1, 0) (3, pi/3) (5, pi/2) (3, 2/3pi) (1, pi) (0, pi-sin^(-1)(-1/4)$

After discontinuity with respect to $θ$ $theta$ ,

$(0, 2 π + {sin}^{- 1} (- \frac{1}{4})) (1, 2 π)$ $(0, 2pi+sin^(-1)(-1/4)) (1, 2pi)$

The graph looks like the cardioid $r = 1 + sin θ$ $r=1+sin theta$ that is elongated

in the $θ = \frac{π}{2}$ $theta= pi/2$ direction.

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How do you graph $r = 1 + 4 sin θ$ $r=1+4sintheta$ ?

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How do you graph r=1+4sinθr=1+4sintheta?

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How do you graph $r = 1 + 4 sin θ$ $r=1+4sintheta$ ?