Question

How do you graph `r=4cos6theta`?

Answer 1

Th Socratic graph of this 6-petal rose is inserted.

Explanation:

`r=4 cos 6theta>=0`. So, `cos 6theta >=0`.

It follows that `6theta in Q_1 or Q_4`.

The period for the graph is `(2pi)/6 =pi/3`.

So, there are 6 cycles for `theta in [-pi, pi]`.

For one cycle `theta in [-pi/6, pi/6]`

Yet, `r >=0`, only in the middle half `[-pi/12, pi/12]`, wherein one

petal is completed..

In the remaining half`r <0` and the graphics software designers

have kept off negative r.

For Socratic graphing, I have used the cartesian form of the

equation obtained from

`cos 6theta`

`=cos^6theta-6C_2cos^4thetasin^2theta`

`+6C_4cos^2thetasin^4theta-6C_6sin^6theta`

graph{(x^2+y^2)^3.5-4(x^6-y^6-15x^2y^2(x^2-y^2))=0 [-10, 10, -5, 5]}