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student:mathematics:coordinate-systems-2d [2018/08/10 15:21] – created bernstdh | student:mathematics:coordinate-systems-2d [2025/08/21 16:25] (current) – bernstdh | ||
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- | + | ===== Coordinate Systems for the Plane ===== | |
- | ===== Mathematical Foundations: | + | |
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In traditional Cartesian coordinates the origin is at located at the " | In traditional Cartesian coordinates the origin is at located at the " | ||
- | {{cartesian-coordinates.gif}} | + | {{cartesian-coordinates.gif|Cartesian coordinates}} |
The horizontal coordinate is traditionally denoted by x and the vertical coordinate is traditionally denoted by y. | The horizontal coordinate is traditionally denoted by x and the vertical coordinate is traditionally denoted by y. | ||
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Because of the way old hardware worked, most computer | Because of the way old hardware worked, most computer | ||
- | {{coordinates.gif}} | + | {{coordinates.gif|Screen coordinates}} |
==== Polar Coordinates ==== | ==== Polar Coordinates ==== | ||
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- | {{polar-coordinates.gif}} | + | {{polar-coordinates.gif|Polar coordinates}} |
While the angular quantity can be measured in radians or degrees, it is most commonly measured in radians. | While the angular quantity can be measured in radians or degrees, it is most commonly measured in radians. | ||
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In the following figure, the point p can be thought of as either having the Cartesian coordinates (p1,p2) or the polar coordinates (θ,d). When converting from polar coordinates to Cartesian coordinates we have (θ,d) and must calculate (p1,p2). On the other hand, when converting | In the following figure, the point p can be thought of as either having the Cartesian coordinates (p1,p2) or the polar coordinates (θ,d). When converting from polar coordinates to Cartesian coordinates we have (θ,d) and must calculate (p1,p2). On the other hand, when converting | ||
- | {{cartesian-to-polar.gif}} | + | {{cartesian-to-polar.gif|Converting from Cartesian coordinates to polar coordinates}} |
Given (θ,d) it is clear from the figure that cos(θ)=p1/d and, hence, that p1=dcos(θ). Similarly, it is clear from the figure that sin(θ)=p2/d and, hence, that p2=dsin(θ). | Given (θ,d) it is clear from the figure that cos(θ)=p1/d and, hence, that p1=dcos(θ). Similarly, it is clear from the figure that sin(θ)=p2/d and, hence, that p2=dsin(θ). |