In a nutshell you have to restrict the domain of theta to the values that give you real numbers.
Also if you were going to look at specific values of theta you would want to look at the ones that make 4 theta equal to, say, pi/6, pi/3, pi/2, etc.. Otherwise your values of 4 theta will skip over large parts of the plane and fail to give you sufficient information.
See the analysis I've included below.
Sir,
There is something strange going on with the following question.
Question:
Identify and graph the polar equation… r^2-7sin(4*theta)..
Answer:
r^2=x^2+y^2 , So, r=sqrt (x^2+y^2 )
Therefore,
r = sqrt(7*sin(4*theta))
I tried setting up a table of theta vs. sqrt(7*sin(4*theta)) but encountered complex numbers, ie pi/3,pi/2 ,5pi/6 &pi. having done so I thought of graphing r = sqrt(7*sin(4*theta)) and got a verity of lines scattered around the pole.
This function is defined only where sin(4 theta) is non-negative. This occurs when 4 theta lies between 0 and pi, between 2 pi and 3 pi, and in general between 2 n pi and (2 n + 1) pi.
If 2 n pi < 4 theta < (2 n + 1) pi then n/2 pi < theta < (n/2 + 1/4) pi. So (using n = 0, 1, 2, ...) we see that the graph is defined from 0 to pi/4, then from pi/2 to 3 pi/4, then from pi to 5 pi/4, then from 2 pi/3 to 7 pi/4, then from 2 pi to 9 pi/4 which is the same region of the graph as the region from 0 to pi/4. From this interval on, the regions start repeating.
Now in any given region, 4 theta will go from 2 n pi to (2 n + 1) pi. For example for n = 0 the values of 4 theta will be from 0 to pi. The values of sin(4 theta) therefore go from 0 to 1 and back to 0. For n = 0 this occurs between theta = 0 and theta = pi/4, so you will get a single 'petal' of a 'rose'; however because of the square root the 'petal' will be a bit 'fatter' than for a standard 'rose' (e.g., for theta = pi/16 we have 4 theta = pi/4 so the sine is sqrt(2)/2, about .7, but the square root of the sine is more like .83, which puts the pi/16 point of the graph a bit further from the origin and makes the 'petal' a bit fatter).
The same sort of thing will occur for any value of n, so we have a 'fat rose' with 'petals' between theta= 0 and pi/4, between theta = pi/2 and 3 pi/4, etc.. The 'petals' that go in between (e.g., between pi/4 and pi/2, between 3 pi/4 and pi, etc.) are 'missing' because you can't take the square root of the negative sine values that occur for these values of theta.
When I look at the Equation and try to convert to rectangular coordinates I get :
x^2 +y^2 =(7*sin(4*theta))
y=sqrt(-x+(7*sin(4*theta))
I see the negative x which confirms the imaginary number..
You would also want to change sin(4 theta) to x and y form. However the polar form of the graph is easier to analyze than the rectangular.
Can you tell me what is going on here or have I just messed this up.