How do you graph #r=10sin4theta#?

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Answer 1 · 2018-08-10T16:41:03

See explanation and graph.

#0 <= r = 10 sin 4theta in [ 0, 10 ]# and the period = #(2pi)/4 = pi/2#.

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

#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

#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 4theta = 4 (cos^3theta sin theta - cos theta sin^3theta )#

#( 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 >= 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.