Question

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

Answer 1

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

Explanation:

`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.