Transform explicit function to polar form

[nietschje]

Hello,

I'am new here and happy to find this great forum!
Here's my first question: there's an explicit function as follows:
y=2.sin(x)-1

The transformation to polar form (r=3cos(\theta))
- x=r.cos(\theta)
- y=r.sin(\theta)

So I get: r.sin(\theta)=2.sin(r.cos(\theta))-1
Now you see the problem: how can I isolate r in this equation ?

 

[HallsofIvy]

You can't, or at least not with standard functions.

 

[gsal]

I am confused...just because you are used to x and y being Cartesian coordinates, does not mean that they always are...

what I am trying to say is that x and y in your original equation may not be what you think they are. For example, x in sin(x) is not an x-coordinate from the cartesian plane...for example x is not 42 inches...after all, you cannot take the sin(42 inches)...if x is radians, that's another story, then sin(42radians) is possible...

so, you need to get your ducks in a row, here, and figure which way you are supposed to do your transformation...as it is, your attempt does not make sense at all...not to me, anyway.

my 2 cents

 

[nietschje]

Hello,

thank you all for quick replies !
The cartesian equation y = 2.sin(x)-3 has indeed x as argument; ofcourse in radians.
But what I know form transformation to polar form is that the y x plane transforms to a
r \theta plane. Being \theta tha angle between r and de x axle. In my case: r=3.cos(\theta).

So I suppose that every variable (also x in radians) in the equation has to be transformed by definition of the polar function. The value of y and x are described by
x = r.cos(\theta)
y = r.sin(\theta)

If I can transform f(x) to paramatric function, I also can it transform to polar function (this is essentially a parameteric function). Am I wrong somewhere.

I do realize that this equation isn't easy to write in polar form!

greets

 

[disregardthat]

You can't get an equation such as r = g(theta) from this, g elementary or not. g won't be well-defined, since there can be many points (x,y) on the original graph on a line with angle theta between it and the x-axis.

 

[nietschje]

Oké, but there exist an equation in polar form to describe the cartesian equation ? Because there al multiple x,y coordinates, and one of them wil point to the cartesian equation as you explained...

greetz

 

[nietschje]

I guess for my example it's not possible, thank for your replies !