Moving center of coordinates in the polar graph

In summary, a function in polar coordinates is given by t(rho, phi) = H^2 / (H^2 + rho^2). After moving the center to the right and using cartesian coordinates to simplify the transformation, the correct formula is t(rho, phi) = H^2 / (H^2 + rho^2 - 2 * rho * L * cos(phi) )**2. The mistake was omitting the L^2 term in the denominator. The first picture corresponds to the original function (1), while the second picture corresponds to the corrected function (3). The mistake was corrected and the change made a significant difference. The use of LaTeX was also recommended for better readability.
  • #1
ektov_konstantin
5
0
TL;DR Summary
Please, check if made a mistake. I get bad result but can`t understand where is my mistake
I have a function in polar coordinates:

t (rho, phi) = H^2 / (H^2 + rho^2) (1)

I have moved the center to the right and want to get the new formulae.

I use cartesian coordinates to simplify the transformation (L = 232.5).

rho^2 = (x')^2+(y')^2
x' = x-L (2)
y' = y

Then I substitute expression (2) into (1) and go back to the polar coordinates (using x=rho*cos(phi) and y=rho*sin(phi) ). The result is:

t (rho, phi) = H^2 / (H^2 + rho^2 - 2 * rho * L * cos(phi) )**2 (3)

The first picture is for (1) function.

The second picture is for (3).
 

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  • #2
ektov_konstantin said:
Summary:: Please, check if made a mistake. I get bad result but can`t understand where is my mistake

I have a function in polar coordinates:

t (rho, phi) = H^2 / (H^2 + rho^2) (1)
Or as a bit more readable,
$$t(\rho, \phi) = \frac{H^2}{H^2 + \rho^2}$$
If you click what I wrote above, you can see my LaTeX script. There's more information about LaTeX in the LaTeX Guide line at the lower left of the input pane.
ektov_konstantin said:
I have moved the center to the right and want to get the new formulae.

I use cartesian coordinates to simplify the transformation (L = 232.5).

rho^2 = (x')^2+(y')^2
x' = x-L (2)
y' = y

Then I substitute expression (2) into (1) and go back to the polar coordinates (using x=rho*cos(phi) and y=rho*sin(phi) ). The result is:

t (rho, phi) = H^2 / (H^2 + rho^2 - 2 * rho * L * cos(phi) )**2 (3)
It looks to me like you are omitting the ##L^2## term in the denominator. After the substitution -- in equation (2) -- I get this:
$$t(\rho, \phi) = \frac{H^2}{H^2 + (x - L)^2 + y^2} = \frac{H^2}{H^2 + x^2 - 2Lx + L^2 + y^2} $$
$$ = \frac{H^2}{H^2 +\rho^2 - 2L\rho\cos(\phi) + L^2}$$
ektov_konstantin said:
The first picture is for (1) function.

The second picture is for (3).
Does my change make a difference?
 
  • #3
Mark44 said:
Or as a bit more readable,
$$t(\rho, \phi) = \frac{H^2}{H^2 + \rho^2}$$
If you click what I wrote above, you can see my LaTeX script. There's more information about LaTeX in the LaTeX Guide line at the lower left of the input pane.
It looks to me like you are omitting the ##L^2## term in the denominator. After the substitution -- in equation (2) -- I get this:
$$t(\rho, \phi) = \frac{H^2}{H^2 + (x - L)^2 + y^2} = \frac{H^2}{H^2 + x^2 - 2Lx + L^2 + y^2} $$
$$ = \frac{H^2}{H^2 +\rho^2 - 2L\rho\cos(\phi) + L^2}$$

Does my change make a difference?
Yes. I made such a stupid mistake. Thanks for advise about LaTeX
 

1. What is the purpose of moving the center of coordinates in a polar graph?

Moving the center of coordinates in a polar graph allows for a more accurate representation of data and makes it easier to analyze the data. It also allows for the graph to be rotated or shifted without affecting the data.

2. How do you move the center of coordinates in a polar graph?

To move the center of coordinates in a polar graph, you can either physically move the graph or use mathematical equations to shift the center. This is usually done by adding or subtracting a constant value to the angle or radius measurements.

3. Can the center of coordinates be moved to any point in a polar graph?

Yes, the center of coordinates can be moved to any point in a polar graph. This allows for greater flexibility in graphing and analyzing data.

4. What effect does moving the center of coordinates have on the shape of a polar graph?

Moving the center of coordinates does not change the shape of the polar graph. It only changes the position of the graph in relation to the new center.

5. Are there any limitations to moving the center of coordinates in a polar graph?

One limitation of moving the center of coordinates in a polar graph is that it may affect the readability of the graph if the new center is too far from the original center. It is important to choose a new center that still allows for clear visualization of the data.

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