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

1 Answer
Shwetank Mauria
Oct 21, 2017

Please see below.

Explanation:

#r=asinntheta# is a typical rose curve though it looks more like daisy. If #n# is odd, number of petals are #n#, but if #n# is even, number of petals are #2n#. For example #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=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 #theta=pi/10# we have #r=4# and when #theta=pi/5#, #r=0#. Again at #theta=(3pi)/10# #r=-4# i.e. maximum on the opposite side and when #theta=(4pi)/10#, #r=0# and so on and thus forming #5# petals.

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