How do you graph $r = 4 sin (5 θ)$ $r=4sin(5theta)$ ?

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How do you graph $r = 4 sin (5 θ)$ $r=4sin(5theta)$ ?

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Answer 1 · 2017-10-21T04:32:17

Please see below.

Explanation:

$r = a sin n θ$ $r=asinntheta$ is a typical rose curve though it looks more like daisy. If $n$ $n$ is odd, number of petals are $n$ $n$ , but if $n$ $n$ is even, number of petals are $2 n$ $2n$ . For example $r = 3 sin 2 θ$ $r=3sin2theta$ looks like

graph{((x^2+y^2)^(3/2)-6xy)((x^2+y^2)^(3/2)+6xy)=0 [-10, 10, -5, 5]}

But $r = 4 sin 5 θ$ $r=4sin5theta$ is

graph{(x^2+y^2)^3=20y(x^2+y^2)^2-80y^3(x^2+y^2)+64 y^5 [-10, 10, -5, 5]}

Observe that when $θ = \frac{π}{10}$ $theta=pi/10$ we have $r = 4$ $r=4$ and when $θ = \frac{π}{5}$ $theta=pi/5$ , $r = 0$ $r=0$ . Again at $θ = \frac{3 π}{10}$ $theta=(3pi)/10$ $r = - 4$ $r=-4$ i.e. maximum on the opposite side and when $θ = \frac{4 π}{10}$ $theta=(4pi)/10$ , $r = 0$ $r=0$ and so on and thus forming $5$ $5$ petals.

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How do you graph $r = 4 sin (5 θ)$ $r=4sin(5theta)$ ?

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Explanation:

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How do you graph r=4sin(5θ)r=4sin(5theta)?

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Explanation:

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How do you graph $r = 4 sin (5 θ)$ $r=4sin(5theta)$ ?