# Edit Edit an Image Page: Change of Coordinate Systems

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 Image Title*: Upload a Math Image The same object, here a disk, can look completely different depending on which coordinate system is used. It is a common practice in mathematics to use different coordinate systems to solve different problems. An example of a switch between coordinate systems follows: suppose we take a set of points in regular x-y '''Cartesian Coordinates''', represented by ordered pairs such as (1,2), then multiply their x-components by two, meaning (1,2) in the old coordinates is matched with (2,2) in the new coordinates.

This transformation is shown below in the following to images. In the image on the left we have a square with four points marked: (0.2,0.7), (0.3,0.9), (0.4,0.3), and (0.9,0.3). The image on the right has undergone the transformation: instead of having a square, we now have a rectangle with the points (0.4,0.7), (0.6,0.9), (0.8,0.3), and (1.8,0.3). We can see that in call cases, y dimensions and coordinates remain the same, while all x coordinates and dimensions are doubled. [[Image:Unstretched.png|left|220px]] [[Image:Stretched.png|400px]]

Under this transformation, a set of points would be stretched out in the horizontal x-direction since each point becomes further from the vertical y-axis (except for points originally on the y-axis, which remain on the axis). We can also see that a set of points that was originally contained in a circle in the old coordinates would be contained by a stretched-out ellipse in the new coordinate system, as shown in the top two figures of this page's main image. Many other such transformations exist and are useful throughout mathematics, such as mapping the points in a disk to a rectangle. Some of these mappings can be neatly represented by [[Vector|vectors]] and [[Matrix|matrices]], in the form $A\vec{x}=\vec{x'}$ Where $\vec{x}$ is the coordinate vector of our point in the original coordinate system and $\vec{x'}$ is the coordinate vector of our point in the new coordinate system. For example the transformation in the basic description, doubling the value of the x-coordinate, is represented in this notation by : $\begin{bmatrix} 2 & 0 \\ 0 & 1 \\ \end{bmatrix}\vec{x} = \vec{x'}$ As can be easily verified. In the main image of the page, the ellipse that is tilted relative to the coordinate axes is created by a combination of rotation and stretching, represented by the matrix : $\begin{bmatrix} 2cos(\theta) & -sin(\theta)\\ 2sin(\theta) & cos(\theta) \\ \end{bmatrix}\vec{x} = \vec{x'}$ Some very useful mappings cannot be represented in matrix form, such as mapping points from '''Cartesian Coordinates''' to '''[[Polar Coordinates]]'''. Such a mapping, as shown in this page's main image, can map a disk to a rectangle. We can think of the disk as a series of rings wrapped around the origin, and the rectangle as a series of lines. Each of these rings is a different distance from the origin, and gets mapped to a different line within the rectangle. Each origin-centered ring in the disk consists of points at constant distance from the origin and angles ranging from 0 to $2\pi$. These points create a vertical line in Polar Coordinates. Each ring at a different distance from the origin creates its own line in the polar system, and the collection of these lines creates a rectangle. The transformation from Cartesian coordinates to Polar Coordinate can be represented algebraically by $\begin{bmatrix} r\\ \theta\\ \end{bmatrix} = \begin{bmatrix} \sqrt{x^2 + y^2}\\ \arctan{y/x}\\ \end{bmatrix}$ ===Three-Dimensional Coordinates=== In 3 dimensions, similar coordinate systems and transformations between them exist. Three common systems are rectangular, cylindrical and spherical coordinates: :*Rectangular Coordinates use standard $x,y,z$ coordinates, where the three coordinates represent left-right position, up-down position, and forward-backward position, respectively. These three directions are mutually perpendicular .
[[Image:Cylindrical.png|200px|right]] :*Cylindrical Coordinates use $r,\theta,z$, where $r, \theta$ are the same as two-dimensional polar coordinates and ''z'' is distance from the x-y plane as shown on the right.

[[Image:Spherical.png|200px|left]]

:*Spherical Coordinates use $\rho, \theta, \phi$, where $\rho$ is the distance from the origin, $\theta$ is rotation from the positive x-axis as in polar coordinates, $\phi$ and is rotation from the positive z-axis. Note that this standard varies from discipline to discipline. For example, the standard in physics is to switch the $\phi$ and $\theta$ labeling. Always be aware of what standard you should be using given a particular textbook or course. The mathematics standard noted above and shown in the image on the left is used for this page.

====Converting between these coordinates==== The conversion from rectangular (Cartesian) coordinates to cylindrical coordinates is almost identical to the conversion between Crtesian coordinates and polar coordinates. : $r= \sqrt{x^2+y^2}$ : $\theta=\tan^{-1}\left(y/x\right)$ : $z=z$ {{Switch|link1=Click to show example|link2=Click to hide example |1= |2= We can calculate the cylindrical coordinates for the point given by (1,2,3) in Cartesian coordinates. : $r=\sqrt{x^2+y^2}=\sqrt{1^2+2^2}=\sqrt{5}$ : $\theta =\tan^{-1}\left(y/x\right)=\tan^{-1}\left(2/1\right)\approx 1.1$ radians : $z=3$ So we have the point $(\sqrt{5}, 1.1,3 )$ in cylindrical coordinates.}} The conversion from cylindrical coordinates to Cartesian coordinates is given by : $x=r \cos \theta$ : $y=r \sin \theta$ : $z=z$. {{Switch|link1=Click to show example|link2=Click to hide example |1= |2= Now we can convert the point $(\sqrt{5}, 1.1,3 )$ in cylindrical coordinates back to Cartesian coordinates. : $x=\sqrt{5} \cos (1.1)=\sqrt{5}(0.454)\approx 1$ : $y= \sqrt{5} \sin (1.1)=\sqrt{5}(0.891)\approx 2$ : $z=3$. We see that we do indeed get back the point (1,2,3). The approximately equal to signs are due to rounding involved in dealing with the square root of five and sine and cosine. }} In order to go from Cartesian to spherical coordinates, we have : $\rho=\sqrt{x^2+y^2+z^2}$ : $\tan\phi=\frac{\sqrt{x^2+y^2}}{z}$ : $\tan \theta = y/x$ {{Switch|link1=Click to show example|link2=Click to hide example |1= |2= Starting out with the point (2,1,2) in Cartesian coordinates, we find that : $\rho=\sqrt{x^2+y^2+z^2}=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3$ : $\tan\phi=\frac{\sqrt{x^2+y^2}}{z}=\frac{\sqrt{2^1+1^2}}{2}=\frac{\sqrt{5}}{2}$ ::$\rightarrow \phi=\tan^{-1}\left(\sqrt{5}/2\right)=0.841$ radians : $\tan \theta = y/x=1/2$ ::$\rightarrow \theta=\tan^{-1}=0.464$ radians So we have the point (3,0.841,0.464) in spherical coordinates}} The transformation from spherical coordinates to Cartesian coordinates is given by :$x=\rho \sin \phi \cos \theta$ :$y=\rho \sin \phi \sin \theta$ :$z= \rho \cos \phi$. {{Switch|link1=Click to show example|link2=Click to hide example |1= |2= We can convert our previous result back to Cartesian coordinates: :$x=\rho \sin \phi \cos \theta=3\sin(0.841)\cos(0.464)\approx 2$ :$y=\rho \sin \phi \sin \theta =3\sin(0.841)\sin(0.464)\approx 1$ :$z= \rho \cos \phi =3 cos(0.841)\approx 2$ Again, we retrieve our original ordered triple (2,1,2) by rounding.}} We can also write the transformation from cylindrical coordinates to spherical coordinates: :$\rho=\sqrt{r^2+z^2}$ :$\tan \phi = z/r$ :$\theta = \theta$. {{Switch|link1=Click to show example|link2=Click to hide example |1= |2= Beginning with the point (3,\pi/3,4) in cylindrical coordinates, we see that :$\rho=\sqrt{r^2+z^2}=\sqrt{3^2+4^2}=\sqrt{25}=5$ :$\tan \phi = z/r=4/3$ ::$\rightarrow \phi=\tan^{-1}(.75)=0.644$ :$\theta = \pi/3$. So the point in spherical coordinates is (5,\pi/3,0.644). }} Finally, the transformation from spherical to cylindrical coordinates is given by :$r=\rho \sin \phi$ :$\theta =\theta$ :$z=\rho \cos \phi$. {{Switch|link1=Click to show example|link2=Click to hide example |1= |2= For one last example, we will take our previous result and transform it back to cylindrical coordinates :$r=\rho \sin \theta=5 \sin (0.644)\approx 3$ :$\theta =\pi/3$ :$z=\rho \cos \theta5 \cos (0.644)\approx 4$ Therefore the point in cylindrical coordinates is (3, \pi/3, 4) as expected. }} ===Interactive Demonstration=== ChangeOfCoordsApp ====Future Ideas for this Page==== *add examples of transformations between three dimensional coordinate systems. Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Topology Other Yes, it is.