but how to deduce it ? Thank you .
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$\begingroup$ For the record, this is the forum for the software Mathematica, you were probably looking for math.stackexchange.com $\endgroup$– FeyreCommented Aug 4, 2016 at 9:49
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$\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$– Michael E2Commented Aug 4, 2016 at 10:51
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$\begingroup$ See en.wikipedia.org/wiki/Envelope_(mathematics) $\endgroup$– Michael E2Commented Aug 4, 2016 at 11:03
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$\begingroup$ I don't think your lines on above illustrations are actually equal-length, since they seem to be from $(0, a)$ to $(1-a, 0)$. $\endgroup$– kirmaCommented Aug 10, 2016 at 7:43
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3 Answers
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It's an astroid and not a circle. In any event:
With[{r = 1, n = 30},
Graphics[{AbsoluteThickness[1],
Table[Line[r {{Cos[θ], 0}, {0, Sin[θ]}}], {θ, 0, π/2, π/(2 n)}]}]]
A short proof that the astroid is the correct envelope:
{x, InterpolatingPolynomial[{{Cos[θ], 0}, {0, Sin[θ]}}, x]} /.
First[Solve[D[InterpolatingPolynomial[{{Cos[θ], 0}, {0, Sin[θ]}}, x], θ] == 0, x]]
// FullSimplify
answered Aug 4, 2016 at 9:40
J. M.'s missing motivation♦J. M.'s missing motivation
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$\begingroup$ Fair enough, I'm too trusting with what people claim to know. $\endgroup$– FeyreCommented Aug 4, 2016 at 9:48
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Based on your image, but not your statement on lengths of lines:
With[{eqn = a - a x / (1 - a)},
Show[
Quiet@Plot[Table[eqn, {a, 0, 1, 1 / 20}], {x, 0, 1},
Evaluated -> True, AspectRatio -> Automatic, PlotRange -> {0, 1},
PlotStyle -> Gray],
Plot[MaxValue[{eqn, 0 <= a <= 1}, a], {x, 0, 1},
Evaluated -> True, PlotStyle -> Directive[Red, Thick]]]]
For the intended plot your can replace eqn with the following:
eqn = Sin[Pi a / 2] - Tan[Pi a / 2] x
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Solve[{#, D[#, a]} &[x/a + y/(1 - a) == 1], {x, y}, {a}]also yields the red parabola (1 + (x - y)^2 - 2 (x + y)). $\endgroup$ Commented Aug 10, 2016 at 19:00
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$$\sqrt{1-x^{2/3}}-\sqrt{1-x^{2/3}}x^{2/3}$$
Plot[Sqrt[1 - x^(2/3)] - Sqrt[1 - x^(2/3)] x^(2/3), {x, 0, 1},
AspectRatio -> 1, PlotStyle -> {Thick, Red, Dashed}
