Analysis

A33

RotationsOberfläche[f_, {x_, a_, b_}] := 
      2  π (∫_a^b) Simplify[
              f*(Sqrt[1 + D[f, x]^2])] \[DifferentialD]x

RotationsOberfläche[(Sqrt[1 - (x^2)]), {x, -1, 
        1}]

RotationsOberfläche[(Sqrt[(r^2) - (x^2)]), {x, -r, 
        r}]

A 34

A = 
      4 (∫_0^1) 
            b (Sqrt[ 1 - ((x^2)/(a^2))]) \[DifferentialD]x

A = FullSimplify[A, a > 0]

Plot[Evaluate[A /. b -> 1], {a, 0, 5}]

A 35

f = ((1 + y)/(1 - 3  x))

Plot3D[f, {x, -5, 0}, {y, -5, 5}]

(∫_0^10)(∫_((1/3) (1 - (E^y)))^0) f \[DifferentialD]x \[DifferentialD]y

(∫_0^b)(∫_((1/3) (1 - (E^y)))^0) f \[DifferentialD]x \[DifferentialD]y

Stammfunktion inneres Integral

∫f \[DifferentialD]x

Needs["Graphics`FilledPlot`"]

sol = 
      Solve[(1/3) (1 - (E^y)) == x, y]

FilledPlot[{y /. sol, 10}, {x, (1/3) (1 - (E^10)), 
        0}, {y, 0, 1}]

FilledPlot[{y /. sol, 2}, {x, (1/3) (1 - (E^2)), 
        0}, {y, 0, 1}]

Lineare Algebra

L30

Unprotect[C, D, E]; Clear[A, B, C, D, E];

M = {{A, (B/2), (D/2)}, {(B/2), C, (E/2)}, {(D/2), (E/2), F}}

v = {{x}, {y}, {1}}

Transpose[v] . M . v // Simplify

L31

Hilfsfunktionen

drehung[α_] := {{Cos[α], 
          Sin[α]}, {-Sin[α], Cos[α]}}

KoeffizientenZuweisung := 
      Module[{}, 
A = 
          Coefficient[G, x, 2]; 
B = 
          Coefficient[Coefficient[G, x], y]; 
C = 
          Coefficient[G, y, 2]; 
D = 
          Coefficient[Coefficient[G, x], y, 0]; 
E = 
          Coefficient[Coefficient[G, y], x, 0]; 
F = 
          Coefficient[Coefficient[G, x, 0], y, 
            0];
]

Typus := 
      Module[{}, 
KoeffizientenZuweisung[
          G, {x, y}]; 
If[
          (B^2) - 4  A C < 0, "Ellipse", 
          If[(B^2) - 4  A C > 
              0, "Hyperbel", "Parabel"]]
]

Drehwinkel := 
      If[A == C, (π/4), (1/2) ArcTan[(B/(A - C))]]

Needs["Graphics`ImplicitPlot`"]

(a)

G = 
      7  (x^2) - 2 (Sqrt[ 3]) x y + 5  (y^2) + 4  x + 4 (Sqrt[ 3]) y + 
        12

plot1 = 
      ImplicitPlot[
        G == 0, {x, -10, 10}, {y, -10, 10}]

KoeffizientenZuweisung

Typus

Drehwinkel

v = drehung[-Drehwinkel] . {{x}, {y}}

H = 
      G /. {x -> v[[1, 1]], y ->  v[[2, 1]]} // 
        Expand

Der Darstellung sieht man an, dass die quadratische Form positiv definit ist und daher die reelle Punktmenge leer ist.

(b)

G = -5 (x^2) + 6 (Sqrt[ 3]) x y + (y^2) + 4  x + 4 (Sqrt[ 3]) y + 
        12

plot1 = 
      ImplicitPlot[
        G == 0, {x, -10, 10}, {y, -10, 10}]

KoeffizientenZuweisung

Typus

Drehwinkel

v = drehung[-Drehwinkel] . {{x}, {y}}

H = 
      G /. {x -> v[[1, 1]], y ->  v[[2, 1]]} // 
        Expand

plot2 = 
      ImplicitPlot[H == 0, {x, -10, 10}, {y, -10, 10}, 
        PlotStyle -> RGBColor[1, 0, 0]]

Show[plot1, plot2]