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

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

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Answer 1 · 2018-04-08T09:27:56

Please see below.

Explanation:

As $r = 2 sin θ$ $r=2sintheta$ , at $θ = 0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}$ $theta=0,pi/4,pi/2,(3pi)/4$ and $π$ $pi$

$r$ $r$ takes the values $0, \sqrt{2}, 2, \sqrt{2}, 0$ $0,sqrt2,2,sqrt2,0$

Thus these represents points $(0, 0)$ $(0,0)$ , $(\sqrt{2}, \frac{π}{4})$ $(sqrt2,pi/4)$ , $(2, \frac{π}{2})$ $(2,pi/2)$ , $(\sqrt{2}, \frac{3 p u}{4})$ $(sqrt2,(3pu)/4)$ and $(0, π)$ $(0,pi)$ .

We can select more such points, say by having $θ = \frac{π}{6}, \frac{π}{3}, \frac{2 π}{3}, \frac{5 π}{6}$ $theta=pi/6,pi/3,(2pi)/3, (5pi)/6$ and corresponding value of $r$ $r$ would be $r = 1, \sqrt{3}, \sqrt{3}, 1$ $r=1,sqrt3,sqrt3,1$ and points are $(1, \frac{π}{6})$ $(1,pi/6)$ , $(\sqrt{3}, \frac{π}{3})$ $(sqrt3,pi/3)$ , $(\sqrt{3}, \frac{2 π}{3})$ $(sqrt3,(2pi)/3)$ and $(1, \frac{5 π}{6})$ $(1,(5pi)/6)$ .

The gaph will appear as follows:

created using utility at desmos

It is a circle with center at $(1, \frac{π}{2})$ $(1,pi/2)$ and radius $1$ $1$

and as $r = 2 sin θ$ $r=2sintheta$ means $r^{2} = 2 r sin θ$ $r^2=2rsintheta$

in rectangular coordinates, it is equivalent to $x^{2} + y^{2} = 2 y$ $x^2+y^2=2y$

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

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Explanation:

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