Problem 42580. Conic equation

Created by Tim

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{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>A conic of revolution (around the</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>z</w:t></w:r><w:r><w:t> axis) can be defined by the equation</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ s^2 – 2*R*z + (k+1)*z^2 = 0]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>where</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>s^2=x^2+y^2</w:t></w:r><w:r><w:t>,</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R</w:t></w:r><w:r><w:t> is the vertex radius of curvature, and</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k</w:t></w:r><w:r><w:t> is the conic constant:</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k&lt;-1</w:t></w:r><w:r><w:t> for a hyperbola,</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k=-1</w:t></w:r><w:r><w:t> for a parabola,</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>-1&lt;k&lt;0</w:t></w:r><w:r><w:t> for a tall ellipse,</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k=0</w:t></w:r><w:r><w:t> for a sphere, and</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k&gt;0</w:t></w:r><w:r><w:t> for a short ellipse.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Write a function</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>z=conic(s,R,k)</w:t></w:r><w:r><w:t> to calculate height</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>z</w:t></w:r><w:r><w:t> as a function of radius</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>s</w:t></w:r><w:r><w:t> for given</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R</w:t></w:r><w:r><w:t> and</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k</w:t></w:r><w:r><w:t>. Choose the branch of the solution that gives</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>z=s^2/(2*R)+...</w:t></w:r><w:r><w:t> for small values of</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>s</w:t></w:r><w:r><w:t>. This defines a concave surface for</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R&gt;0</w:t></w:r><w:r><w:t> and a convex surface for</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R&lt;0</w:t></w:r><w:r><w:t>.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>The trick is to get full machine precision for all values of</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>s</w:t></w:r><w:r><w:t> and</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R</w:t></w:r><w:r><w:t>. The test suite will require a relative error less than</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>4*eps</w:t></w:r><w:r><w:t>, where</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>eps</w:t></w:r><w:r><w:t> is the machine precision.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Hint (added 2015/09/03): the straightforward solution is</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ z = (R-sqrt(R^2-(k+1)*s^2))/(k+1),]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>but this does not work if</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k=-1</w:t></w:r><w:r><w:t>, gives the wrong branch of the solution if</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R&lt;0</w:t></w:r><w:r><w:t>, and is subject to severe roundoff error if</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>s^2</w:t></w:r><w:r><w:t> is small compared to</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>R^2</w:t></w:r><w:r><w:t>. It is possible, however, to find a mathematically equivalent form of the solution that solves all three problems at once.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

A conic of revolution (around the <tt>z</tt> axis) can be defined by the equation<pre> s^2 – 2*R*z + (k+1)*z^2 = 0</pre>where <tt>s^2=x^2+y^2</tt>, <tt>R</tt> is the vertex radius of curvature, and <tt>k</tt> is the conic constant: <tt>k&lt;-1</tt> for a hyperbola, <tt>k=-1</tt> for a parabola, <tt>-1&lt;k&lt;0</tt> for a tall ellipse, <tt>k=0</tt> for a sphere, and <tt>k&gt;0</tt> for a short ellipse.Write a function <tt>z=conic(s,R,k)</tt> to calculate height <tt>z</tt> as a function of radius <tt>s</tt> for given <tt>R</tt> and <tt>k</tt>. Choose the branch of the solution that gives <tt>z=s^2/(2*R)+...</tt> for small values of <tt>s</tt>. This defines a concave surface for <tt>R&gt;0</tt> and a convex surface for <tt>R&lt;0</tt>.The trick is to get full machine precision for all values of <tt>s</tt> and <tt>R</tt>. The test suite will require a relative error less than <tt>4*eps</tt>, where <tt>eps</tt> is the machine precision.Hint (added 2015/09/03): the straightforward solution is<pre> z = (R-sqrt(R^2-(k+1)*s^2))/(k+1), </pre>but this does not work if <tt>k=-1</tt>, gives the wrong branch of the solution if <tt>R&lt;0</tt>, and is subject to severe roundoff error if <tt>s^2</tt> is small compared to <tt>R^2</tt>. It is possible, however, to find a mathematically equivalent form of the solution that solves all three problems at once.

A conic of revolution (around the |z| axis) can be defined by the equation

s^2 – 2*R*z + (k+1)*z^2 = 0

where |s^2=x^2+y^2|, |R| is the vertex radius of curvature, and |k| is the conic constant: |k<-1| for a hyperbola, |k=-1| for a parabola, |-1<k<0| for a tall ellipse, |k=0| for a sphere, and |k>0| for a short ellipse.

Write a function |z=conic(s,R,k)| to calculate height |z| as a function of radius |s| for given |R| and |k|. Choose the branch of the solution that gives |z=s^2/(2*R)+...| for small values of |s|. This defines a concave surface for |R>0| and a convex surface for |R<0|.

The trick is to get full machine precision for all values of |s| and |R|.  The test suite will require a relative error less than |4*eps|, where |eps| is the machine precision.

Hint (added 2015/09/03): the straightforward solution is

z = (R-sqrt(R^2-(k+1)*s^2))/(k+1),

but this does not work if |k=-1|, gives the wrong branch of the solution if |R<0|, and is subject to severe roundoff error if |s^2| is small compared to |R^2|. It is possible, however, to find a mathematically equivalent form of the solution that solves all three problems at once.

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Last Solution submitted on May 05, 2021

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Rafael S.T. Vieira on 30 Jun 2020

good one.

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conic quadratic

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