New answers tagged coordinate-transformation
2
You can simply replace the function q using ReplaceAll (/.), e.g.
Simplify[L /. q -> Function[{t}, Q[t]/Sin[A t]]]
(* 1/2 m (A^2 Q[t]^2 + Q'[t]^2) *)
0
Suppose you have a vector p expressed as
p = p1 e1 + p2 e2 + p3 e3
where e1, e2, e3 are a dextral triad of unit vectors and p1, p2 and p3 are scalars.
Let n1, n2, n3 be another dextral triad of unit vectors.
Then p = p1 (e1.n1 n1 + e1.n2 n2 + e1.n3 n3) + p2 (e2.n1 n1 + e2.n2 n2 + e2.n3 n3) + p3 (e3.n1 n1 + e3.n2 n2 + e3.n3 n3)
You can collect terms to put ...
2
ClearAll[x];
n = 2;
Tuples@Array[x, {n, 2}]
{{x[1, 1], x[2, 1]}, {x[1, 1], x[2, 2]}, {x[1, 2], x[2, 1]}, {x[1, 2], x[2, 2]}}
3D:
ClearAll[x];
n = 3;
Tuples@Array[x, {n, 2}]
{{x[1, 1], x[2, 1], x[3, 1]}, {x[1, 1], x[2, 1], x[3, 2]}, {x[1, 1], x[2, 2], x[3, 1]}, {x[1, 1], x[2, 2], x[3, 2]}, {x[1, 2], x[2, 1], x[3, 1]}, {x[1, 2], x[2, 1], x[3, 2]}, {x[1,...
4
An alternative, and to get them in clock wise order, is to use Outer product, something like
canonicalizePolygon[{x1_, y1_}, {x2_, y2_}] := Module[{coord},
coord = Outer[List, {x1, x2}, {y1, y2}];
coord[[2]] = Reverse[coord[[2]]];
Flatten[coord, 1]
];
Call it as
canonicalizePolygon[{x1, y1}, {x2, y2}]
For a cubiod, same thing, find ...
3
Perhaps I didn't get the purpose of your question/application…
The function CanonicalizePolygon helps to find the 4 points of Rectangle[{x1,y1},{x2,y2}]
CanonicalizePolygon[Rectangle[ {x1, y1}, {x2, y2} ]][[ 1]]
(*{{x1, y1}, {x1, y2}, {x2, y1}, {x2, y2}}*)
or with assumptions
CanonicalizePolygon[Rectangle[ {x1, y1}, {x2, y2} ],Assumptions -> {x2 &...
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