/sci/ - Science & Math

Manipulate[
Plot3D[y^3 Sin[x], {x, -2, 2}, {y, -2, 2}, ClippingStyle -> None,
ImageSize -> 160,

ViewPoint -> {Cos[\[Theta]], Sin[\[Theta]], \[CurlyPhi]}, Mesh -> 6],

{{\[Theta], 5,
Dynamic[Framed@Graphics[{Blue, Disk[{0, 0}, 1, {0, \[Theta]}]},
PlotRange -> 1, ImageSize -> 20]]}, 0,
2 \[Pi]}, {{\[CurlyPhi], 1, "Höhe"}, -1, 2}]

well

(a x r)^2 = |a| * | r| - (a*r)^2

can't work, since the dimensions don't fit.

(a x r)^2 = dim 4
|a| * | r| = dim 2
(a*r)^2 = dim 4

>>3013580
yes, he said that, but that's just a theory (a geuss)

Let me reason again:

There are 4 possible answers
C) 0%
B) 50%
A and D) 25%

0) By the given answers only "0%", "50%" and "25" are possible answers and {C},{B} and {A,D} are exclusive. We now check if one of the answer can actually by the correct answer.

1) We assume "0%", i.e. "C)" is the correct answer. There is a nonvanishing chance of 25% of hitting "C) 0%". Therefore, in this case, there is a nonvanishing chance of being right, which contradicts the assumption. Therefore the assumption was false and "C) 0%" is not the right answer.

2) We assume "50%", i.e. "B)" is the correct answer. Since the answers are exclusive in the above sense (see 0), there is a nonvanishing chance of exactly 25% of being correct. But this means "25%" and not "50%" is the answer to the question and therefore the assumption was false and "B) 50%" is not the right answer.

3) We assume "25%" is the correct answer, so we have two equally valid answers A and D. Since the answers are exclusive in the above sense (see 0), there is a nonvanishing chance of exactly 50% of being correct. But this means "50%" and not "25%" is the answer to the question and therefore the assumption was false and "A,D) 25%" is not the right answer.

Therefore all possible answers {A,B,C,D} are not valid.

/thread

>>2870287

nice.

now take that Action and make the classical (particle-) limit.

It's some function which starts at x=0, y=5.
you know that when you then take 2 steps to the right, you add to g(x) (the integral of g'(x)) the value of A=(r^2·pi)/4=pi, because that's the value under the curve of g'(x) from 0 to 2.
Then from 2 to 3 you know that g' is linear, so the value of g(x) goes up quadratically (because the integral of a linear function is a quadratic function) and you <see> that the value will rise +1,5 since that's the area under the curve. and so on.
you can reconstruct g(x).
And I have drawn you a pic (youout the condition g(0)=5, you you have to shift the function up in your mind.)
kthxbye.

I guess to give a general answer we need some kind of "hairy ball theorem" which says admissable (periodic) vector fields are already closed such that dv=0 and
dv=f·dVol
can never be solved

http://www.wolframalpha.com/input/?i=OP+is+fag+Quick+calc+question+derivative+of+e^%28-x^-2%29