Lituus: Difference between revisions
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A WASSERMANN (talk | contribs) (New page: a lituus is a spiral in which the angle is inversely proportional to the square of the radius (as expressed in polar coordinates). :<math>r^2\theta = k \,</math> <jsxgraph width="500" he...) |
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A '''lituus''' is a spiral in which the angle is inversely proportional to the square of the radius (as expressed in polar coordinates). | |||
:<math>r^2\theta = k \,</math> | :<math>r^2\theta = k \,</math> | ||
<jsxgraph width="500" height="500" box="box1"> | <jsxgraph width="500" height="500" box="box1"> | ||
b1 = JXG.JSXGraph.initBoard('box1', {axis:true, | var b1 = JXG.JSXGraph.initBoard('box1', {axis:true,boundingbox: [-10, 10, 10, -10]}); | ||
var k = b1. | var k = b1.create('slider', [[1,8],[5,8],[0,1,4]]); | ||
var c = b1. | var c = b1.create('curve', [function(phi){return Math.sqrt(k.Value()/phi); }, [0, 0],0, 8*Math.PI], | ||
{curveType:'polar', strokewidth: | {curveType:'polar', strokewidth:1}); | ||
</jsxgraph> | </jsxgraph> | ||
===The JavaScript code to produce this picture=== | |||
<source lang="javascript"> | |||
var b1 = JXG.JSXGraph.initBoard('box1', {axis:true,boundingbox: [-10, 10, 10, -10]}); | |||
var k = b1.create('slider', [[1,8],[5,8],[0,1,4]]); | |||
var c = b1.create('curve', [function(phi){return Math.sqrt(k.Value()/phi); }, [0, 0],0, 8*Math.PI], | |||
{curveType:'polar', strokewidth:1}); | |||
</source> | |||
===External links=== | |||
* [http://en.wikipedia.org/wiki/Lituus_(mathematics) http://en.wikipedia.org/wiki/Lituus_(mathematics)] | |||
[[Category:Examples]] | [[Category:Examples]] | ||
[[Category:Curves]] | [[Category:Curves]] |
Latest revision as of 08:35, 8 June 2011
A lituus is a spiral in which the angle is inversely proportional to the square of the radius (as expressed in polar coordinates).
- [math]\displaystyle{ r^2\theta = k \, }[/math]
The JavaScript code to produce this picture
var b1 = JXG.JSXGraph.initBoard('box1', {axis:true,boundingbox: [-10, 10, 10, -10]});
var k = b1.create('slider', [[1,8],[5,8],[0,1,4]]);
var c = b1.create('curve', [function(phi){return Math.sqrt(k.Value()/phi); }, [0, 0],0, 8*Math.PI],
{curveType:'polar', strokewidth:1});