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Topic: x and y plot finding maximum
Replies: 4   Last Post: Nov 19, 2011 6:18 AM

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Dr. Peter Klamser

Posts: 23
Registered: 6/11/11
Integrating a rotated function inbetween two circles
Posted: Nov 19, 2011 6:18 AM
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Hi,

when you execute the following code you get a rotated parabolic function
a x^2 rotated around the centre of coordinate system:

becher =.
becher[a_, x_] = a x^2;
rot = RotationTransform[(\[Theta]), {0, 0}] // FullSimplify
bechergedreht =.
bechergedreht[\[Theta]_, a_, r_, x_] = rot[{(x), becher[a, x - r]}]
x$fuer$r[a_, b_,
r_] = (x /. Solve[becher[a, x - r]^2 + x^2 == (r + b)^2, x])[[2]] //
Simplify;
ParametricPlot[
Table[bechergedreht[2 \[Pi] i/10, 1, 3, x], {i, 0, 9}], {x, 3,
x$fuer$r[1, 2, 3]}, PlotRange -> 1.01 {{-5, 5}, {-5, 5}},
Epilog -> {Circle[{0, 0}, 3], Circle[{0, 0}, 5]}]
{xstrich, ystrich} = bechergedreht[\[Theta], a, r, x];
loesungx = x /. Solve[xstrich == xs, x] // TrigFactor;
bechergedreht$xstrich[\[Theta]_, a_, r_, xs_] =
ystrich /. x -> loesungx // FullSimplify;
Print["Mit n wird für Table die Anzahl der Schritte festgelegt."]
Print["Mit n = 4 bekomme ich das erwartete Ergebnis."]
n = 4;
Plot[Table[
bechergedreht$xstrich[2 \[Pi] i/10, 1, 3, x], {i, 1, n}], {x, -4,
4}, AspectRatio -> 1, Epilog -> {Circle[{0, 0}, 3]},
PlotRange -> 1.01 {{-5, 5}, {-5, 5}}]
Print["Mit n = 5 bekomme ich keinen Plot mehr."]
n = 5;
Plot[Table[
bechergedreht$xstrich[2 \[Pi] i/10, 1, 3, x], {i, 1, n}], {x, -4,
4}, AspectRatio -> 1, Epilog -> {Circle[{0, 0}, 3]},
PlotRange -> 1.01 {{-5, 5}, {-5, 5}}]

If you fill water into the cups on the right side you get a water wheel.

How do I manage the rotated function most efficiently, that is the well
known pair of functions regrading to the rotation vector?

As you see I developed the function bechergedreht$xstrich, solving the
x´(x) part of the rotating function to x(x´) and inserting this into the
y´(x) part of the rotation vector. But this does not handle to well.

How can I calculate the surface oft the cups on the right side?

Kind regards

Peter




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