Transform explicit function to polar form

In summary, the conversation revolves around the transformation of a cartesian equation into polar form. The original equation is y=2.sin(x)-1 and the attempted transformation is r.sin(\theta)=2.sin(r.cos(\theta))-1. However, it is pointed out that x and y may not be the traditional Cartesian coordinates and thus, the transformation may not make sense. The question is raised if there is an equation in polar form that can describe the original cartesian equation. Ultimately, it is concluded that for this particular example, it is not possible.
  • #1
nietschje
7
0
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([itex]\theta[/itex]))
- x=r.cos([itex]\theta[/itex])
- y=r.sin([itex]\theta[/itex])

So I get: r.sin([itex]\theta[/itex])=2.sin(r.cos([itex]\theta[/itex]))-1
Now you see the problem: how can I isolate r in this equation ?
 
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  • #2
You can't, or at least not with standard functions.
 
  • #3
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
 
  • #4
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 [itex]\theta[/itex] plane. Being [itex]\theta[/itex] tha angle between r and de x axle. In my case: r=3.cos([itex]\theta[/itex]).

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([itex]\theta[/itex])
y = r.sin([itex]\theta[/itex])

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
 
  • #5
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.
 
  • #6
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
 
  • #7
I guess for my example it's not possible, thank for your replies !
 

FAQ: Transform explicit function to polar form

1. What is the difference between Cartesian and polar coordinates?

Cartesian coordinates use the x and y axes to plot points on a two-dimensional plane, while polar coordinates use a distance from the origin and an angle to locate points on a circular grid.

2. How do you convert from Cartesian to polar coordinates?

To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the following formulas: r = √(x² + y²) and θ = tan⁻¹(y/x). Note that the angle θ may need to be adjusted based on the quadrant the point is located in.

3. How do you write an explicit function in polar form?

An explicit function in polar form will have r expressed in terms of θ. For example, an explicit function could be r = 2 + 3cos(θ). This means that for any given angle θ, the value of r will be 2 plus 3 times the cosine of θ.

4. Can you transform a polar function into Cartesian form?

Yes, you can transform a polar function into Cartesian form by using the following substitutions: x = rcos(θ) and y = rsin(θ). This will give you an equation in terms of x and y, which can be plotted on a Cartesian plane.

5. How can transforming a function to polar form be useful in scientific calculations?

Polar coordinates can be useful in situations where there is circular or rotational symmetry, such as calculating the motion of objects in circular orbits. It can also make certain calculations, such as finding the area of a sector or the length of a curve, easier to solve using polar coordinates.

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