Cartesian and Polar coordinate system increments

In summary, the conversation discusses the correspondence between Cartesian and Polar coordinate systems and how to calculate what corresponds to Δx * Δy in Polar system. It also delves into the concept of Jacobian determinant and its application in calculating surface area. The participants also mention the use of wedges in understanding integration and provide examples of surfaces defined by parametric equations. Finally, they mention the use of the term "correspond" and the origin of the username "Serenaphile."
  • #1
21
0
we know that for any (x, y) in Cartesian system, there is such (r, θ) in Polar system, so

x = r * cosθ and
y = r * sinθ

how you can calculate what corresponds to (Δx, Δy) in polar system?


how come Δx * Δy = r * Δr * Δθ?

Maybe this is very stupid question and has obvious answer, but I haven't been practicing higher math since the university (8 years since graduation), but now I need this proof to prove that for normal distribution the sum of all probabilities in (-∞; ∞) equals to 1.
 
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  • #2
I guess this is which area on polar coordinate system corresponds to the area of Δx * Δy...
But which and how you calculate that?
 
  • #3
I attach graphical presentation of the problem. The shadowed area needs to be presented in polar coordinates...
attachment.php?attachmentid=40448&stc=1&d=1319887583.jpg
 

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  • dxdy polar.jpg
    dxdy polar.jpg
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  • #4
Welcome to PF, XYZ313! :smile:

Your problem statement uses the word "correspond".
Actually, in polar coordinates you would not use the rectangle you have in your drawing, since it won't help you.
What you would do is use something that is approximately (in the first order) a rectangle that aligns with the polar coordinates.
This is shown in the leftmost picture of this drawing (just one I plucked somewhere to illustrate my point):
270px-Passaggio_in_coordinate_polari.svg.png


Can you tell what the lengths of sides of the figure are?
Furthermore, your equation Δx * Δy = r * Δr * Δθ is not correct.
These values are not equal.
 
  • #5
That is from calculus it is called Jacobian determinant.
It works linearizing the change of variable (x,y)->(u,v)
dudv=(uxdx+uydy)(vxdx+vydy)
=(uxvy-uyvx)dxdy
 
  • #6
Looks like:
[tex]dx.dy → r.dr.d\theta[/tex]

it's approximate except where the area is very small, like you use in integration ...

dx.dy is a very small area, so small it fits in all the weird corners of whatever your shape is. The idea of integration is that you divide your shape into lots of these and then count how many you have and the total is the area of the shape.

for some shapes it is easier on the math to turn them into lots of little wedge-segments like the one in the Serenaphile's post. If the area is very small then it is very nearly a wee rectangle with a base of r.dθ and a height dr which gives a total area of r.dr.dθ

the Jacobean, as mentioned, works for any change in coordinates but is pretty abstract. The wedge helps you understand better.

For your rectangle, it should be clear that for large ΔxΔy the polar representation will depend on where it is.
 
  • #7
Another, more advanced, way to look at it. If you have a surface defined by parametric equations, x= f(u,v), y= g(u,v), z= h(u,v), then you can write the "position vector" for any point on the surface as [itex]\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex]. The derivatives with respect to the two parameters,
[tex]\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}[/tex]
and
[tex]\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}[/tex]
lie in the tangent plane to the surface and describe how "measurement" changes in each parameter direction. Their cross product, [itex]\vec{r}_u\times\vec{r}_v[/itex], is perpendicular to the surface at each point and its length gives the differential of surface area of the surface: [itex]|\vec{r}_u\times\vec{r}_v|dudv[/itex].

In particular, we can write the xy-plane as [itex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}[/itex] or, in polar coordinates, [itex]\vec{r}(r, \theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}[/itex]. The derivatives with respect to the parameters are:
[tex]\vec{r}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/tex]
and\
[tex]\vec{r}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}[/tex]

The cross product of those two vectors is
[tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & sin(\theta) & 0 \\ -r sin(\theta) & r cos(\theta) & 0 \end{array}\right|= 0\vec{i}+ 0\vec{j}+ (rcos^2(\theta)+ rsin^2(\theta))\vec{k}= r\vec{k}[/tex]
So that the differential of area is [itex]r drd\theta[/itex].
 
  • #8
Thank you all!

I like Serena: yes you are right, I should have used the word "correspond";

Simon Bridge: yes, the changes are very small, just wrote with delta to make it more readable.

Simon Bridge: Serenaphile :smile:

I like Serena: which Serena is that? Gossip Girl or Gomez :)?

Thank you again I like Serena, lurflurf, Simon Bridge, HallsofIvy!
 
  • #9
XYZ313 said:
Simon Bridge: Serenaphile :smile:

I like Serena: which Serena is that? Gossip Girl or Gomez :)?

Thank you again I like Serena, lurflurf, Simon Bridge, HallsofIvy!

First time I've been called Serenaphile. I'll remember and use that! :biggrin:

Neither Serena, nor Serena "Black Booty" Williams.

It's this one:
http://masterofmagic.info/node/19 [Broken]

[PLAIN]http://masterofmagic.info/sites/default/files/serena.gif [Broken]

It's from a very old computer game. :redface:

--I like Serenaphile
 
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1. What are the differences between Cartesian and Polar coordinate systems?

The main difference between Cartesian and Polar coordinate systems is the way they represent points on a plane. In Cartesian coordinates, points are represented by their horizontal and vertical distance from the origin. In Polar coordinates, points are represented by their distance from the origin and the angle they make with the positive x-axis.

2. How do you convert a point from Cartesian coordinates to Polar coordinates?

To convert a point from Cartesian coordinates (x,y) to Polar coordinates (r,θ), you can use the formulas r = √(x² + y²) and θ = tan⁻¹(y/x). Simply plug in the values for x and y to find the corresponding values for r and θ.

3. How do you determine the direction in Polar coordinates?

In Polar coordinates, the direction of a point is given by the angle θ, also known as the polar angle. This angle is measured counterclockwise from the positive x-axis to the line connecting the origin and the point.

4. What is the purpose of Cartesian and Polar coordinate system increments?

The purpose of Cartesian and Polar coordinate system increments is to determine the distance between two points on a plane. In Cartesian coordinates, the increments are measured in units along the x and y axes. In Polar coordinates, the increments are measured in units of distance and angle.

5. How do you plot a point in Cartesian and Polar coordinates?

In Cartesian coordinates, a point is plotted by moving horizontally from the origin by the value of x, and then vertically from that point by the value of y. In Polar coordinates, a point is plotted by moving a certain distance from the origin in the direction of the angle θ. This can be done using a compass and protractor or by using the formulas r = √(x² + y²) and θ = tan⁻¹(y/x) to find the coordinates of the point.

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