Kontrolle

Flächen

t

Füllmenge V(t)

Zuflussrate z(t)

t_0

Beispiel 1

Beispiel 2

Kontrolle

Deutung

z1(t) = If[0 ≤ t < 4, 40, If[4 ≤ t < 6, 20, If[6 ≤ t < 10, 60, If[10 ≤ t ≤ 10, 60]]]]
b = 2
f0(t) = If[0 ≤ t ≤ 4, If[b ≟ 1, 40, 40]]
f1(t) = If[4 ≤ t ≤ 6, If[b ≟ 1, 20, 0]]
f2(t) = If[6 ≤ t ≤ 10, If[b ≟ 1, 60, -20]]
z2(t) = If[0 ≤ t < 4, 40, If[4 ≤ t < 6, 0, If[6 ≤ t < 10, -20, If[10 ≤ t ≤ 10, -20]]]]
V1(t) = If[0 ≤ t < 4, (40 * t), If[4 ≤ t < 6, (60 / 3 * (t - 4)) + 160, If[6 ≤ t ≤ 10, (60 * (t - 6)) + 200]]]
d = true
V2(t) = If[0 ≤ t < 4, (40 * t), If[4 ≤ t < 6, 160, If[6 ≤ t ≤ 10, (-20 * (t - 6)) + 160]]]
V3(t) = If[0 ≤ t < 3, (10 * t^(2)), If[3 ≤ t < 8, (-10 * (t - 6)^(2)) + 180, If[8 ≤ t ≤ 10, (-40 * (t - 8)) + 140]]]
x0: 0
s = Segment((x0 | 0), (x0 + 10 | 0))
T = Point(s)
gL: -1
gR: 12
gUo: -50
gOo: 550
gUu: -70
gOu: 70
t0: x0
T0: (t0 | 0)
D = (3, 360)
Fuellmenge(t) = If[b ≟ 1, V1(t), V2(t)]
j1 = If(4 ≤ x(T) ≤ 6, Integral[f1, 4, x(T)], Integral[f1, 4, 6])
zuflussrate(t) = If[b ≟ 1, z1(t), z2(t)]
DPunkte = Sequence((i | Fuellmenge(i)), i, 0, 10)
F = (5, 390)
A = (0, 0)
B = (1, 350)
C = (2, 380)
E = (4, 370)
G = (6, 390)
H = (7, 360)
I = (8, 370)
J = (9, 380)
K = (10, 320)
a = true
c = true
j2 = If(6 ≤ x(T) ≤ 10, Integral[f2, 6, x(T)], Integral[f2, 6, 10])
j0 = If(0 ≤ x(T) ≤ 4, Integral[f0, 0, x(T)], Integral[f0, 0, 4])
DPunkt = Element(DPunkte, x(T) + 1)