How do you graph $r = 10 sin 4 θ$ $r=10sin4theta$ ?

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How do you graph $r = 10 sin 4 θ$ $r=10sin4theta$ ?

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Answer 1 · 2018-08-10T16:41:03

See explanation and graph.

Explanation:

$0 \leq r = 10 sin 4 θ \in [0, 10]$ $0 <= r = 10 sin 4theta in [ 0, 10 ]$ and the period = $\frac{2 π}{4} = \frac{π}{2}$ $(2pi)/4 = pi/2$ .

As r is non-negative, so is $sin 4 θ$ $sin 4theta$ , and so,

$4 θ \notin [π, 2 π] \Rightarrow θ \notin [\frac{π}{4}, \frac{π}{2}] \Rightarrow r \geq 0$ $4theta notin [ pi, 2pi ] rArr theta notin [ pi/4, pi/2 ] rArr r >= 0$ for

only half of every period.

Four loops are created for

$θ \in [0, \frac{π}{4}], [\frac{π}{2}, \frac{3}{4} π] . [π, \frac{5}{4} π] and [\frac{3}{2} π, \frac{7}{4} π]$ $theta in [ 0, pi/4 ], [ pi/2, 3/4pi ]. [ pi, 5/4pi ] and [ 3/2pi, 7/4pi ]$ .

See graph depicting these aspects now, for the converted

equation, using

$sin 4 θ = 4 ({cos}^{3} θ sin θ - cos θ {sin}^{3} θ)$ $sin 4theta = 4 (cos^3theta sin theta - cos theta sin^3theta )$

${(x^{2} + y^{2})}^{2.5} = 10 (4 (x^{3} y - x y^{3}))$ $( x^2 + y^2 )^2.5 = 10 ( 4(x^3y - xy^3 ))$
graph{( x^2 + y^2 )^2.5 - 40 (x^3y - xy^3 ) = 0 [ -20 20 -10 10]}

I think, after reading again and again that scalar $r \geq 0$ $r >= 0$ in my

answers, the centuries old practice of showing ( 4 r-positive + 4 r-

negative ) 8 loops, for this equation, is given a go.

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How do you graph $r = 10 sin 4 θ$ $r=10sin4theta$ ?

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Explanation:

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How do you graph r=10sin4θr=10sin4theta?

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How do you graph $r = 10 sin 4 θ$ $r=10sin4theta$ ?