How do you sketch one cycle of #y=-3sin4theta#?

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Answer 1 · 2018-07-22T15:20:18

One petal graph only, for one cyle #theta in( 0, pi/2 )#.

#r = - 3 sin 4theta >= 0 rArr sin 4theta < 0 #

#rArr 3theta in Q_1#or #Q_2 rArr theta in Q_1#

If cycle means period, here, the period = (2pi)/4 = pi/2#.

Considering one cycle #theta in [ 0, pi/2 ]#,

#r >= 0 in ( 0, pi/4 ) and r < 0 in ( pi/4, pi/2 )#.

For creating a graph, convert to Cartesian form, usimg

#r = sqrt ( x^2 + y^2 )>= 0, r ( cos theta, sin theta)# and

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

#( x^2 + y^2 )^2.5 + 12 xy( ( x^2 - y^2 ) = 0#.

Now, the Socratic graph is ready.

graph{( x^2 + y^2 )^2.5 + 12 xy ( x^2 - y^2 ) = 0[-0.1 8 -0.1 4] }

Slide the graph #rarr uarr#, to view the other three petals.