Lösungen Modulprüfung Analysis E+M 07

1E+M

a

Remove["Global`*"]

v[t_]:= {Cos[t],Sin[t],t};
w[t_]:= {Cos[t],Sin[t],t+(t/10)^2};

p1 = ParametricPlot3D[v[t],{t,0,4Pi},DisplayFunction->Identity];
p2 = ParametricPlot3D[w[t],{t,0,4Pi},DisplayFunction->Identity];
Show[p2,p1,DisplayFunction->$DisplayFunction];

b

D[v[t],t]/.t->0

D[w[t],t]/.t->0

ArcCos[D[v[t],t].D[w[t],t]/Norm[D[v[t],t]]/Norm[D[w[t],t]]]/.t->0

c

D[v[t],t]/.t->4Pi

D[w[t],t]/.t->4Pi

N[%]

ArcCos[D[v[t],t].D[w[t],t]/Norm[D[v[t],t]]/Norm[D[w[t],t]]]/.t->4Pi

N[%]

%/Degree

d

len1 = Integrate[Evaluate[Sqrt[D[v[t],t].D[v[t],t]]],{t,0,4Pi}]

N[%]

len2 = Integrate[Evaluate[Sqrt[D[w[t],t].D[w[t],t]]],{t,0,4Pi}]

N[%]

NIntegrate[Evaluate[Sqrt[D[w[t],t].D[w[t],t]]],{t,0,4Pi}]

e

Integrate[Evaluate[D[v[t],t].D[w[t],t] * Sqrt[D[v[t],t].D[v[t],t]]],{t,0,4Pi}]

N[%]

NIntegrate[Evaluate[D[v[t],t].D[w[t],t] * Sqrt[D[v[t],t].D[v[t],t]]],{t,0,4Pi}]

2E+M

Remove["Global`*"];

a

f[x_,y_]:= Sin[x+y] + Cos[x-y]

D[f[x,y],x] == 0

D[f[x,y],y] == 0

Solve[Evaluate[{D[f[x,y],x] == 0, D[f[x,y],y] == 0}],{x,y}]

N[%]

b

f[x,y] //TrigFactor

Sin[x+Pi/4] Sqrt[2]//TrigFactor //Simplify

(Sin[x+Pi/4] Sqrt[2])(Sin[y+Pi/4] Sqrt[2])//TrigFactor //Simplify

( (Sin[x+Pi/4] Sqrt[2])(Sin[y+Pi/4] Sqrt[2])//TrigFactor //Simplify ) == ( f[x,y] //TrigFactor )

c

ContourPlot[f[x,y],{x,-Pi,Pi},{y,-Pi,Pi}];

Plot3D[f[x,y],{x,-Pi,Pi},{y,-Pi,Pi}];

p1= Plot3D[Sin[x+Pi/4] 2 Sin[y+Pi/4],{x,-Pi,Pi},{y,-Pi,Pi}];

(* Maximum mit Differentialrechnung? *)

D[Sin[x+Pi/4] 2 Sin[y+Pi/4],x]

Solve[Evaluate[{D[Sin[x+Pi/4] 2 Sin[y+Pi/4],x]==0, D[Sin[x+Pi/4] 2 Sin[y+Pi/4],y]==0}],{x,y}]

d

h[x_,y_]:= Sin[x+y]-Cos[x-y];
Plot3D[h[x,y],{x,-Pi,Pi},{y,-Pi,Pi}];

Plot3D[h[x,y],{x,-Pi,Pi},{y,-Pi,Pi},PlotRange->{-10,0}];

grad2[v_]:={D[v,x],D[v,y]};
grad2[f[x,y]]

grad2[h[x,y]]

?Reduce

Reduce[Evaluate[grad2[f[x,y]] == λ grad2[h[x,y]] &&  grad2[h[x,y]]==0],{x,y}]

N[%]

Reduce[{grad2[f[x,y]] == λ grad2[h[x,y]],  grad2[h[x,y]]==0}]

Solve[{grad2[f[x,y]] == λ grad2[h[x,y]],  grad2[h[x,y]]==0},{x,y}]

N[%]

f[((3 π)/4), ((3 π)/4)]

Plot[f[x,x],{x,-2Pi,2Pi}];

Solve[h[x,y]==0,{y}]

N[%]

Expand[TrigExpand[Sin[x+y]] /Cos[y]/Sin[x]]==Expand[TrigExpand[Cos[x-y]]/Cos[y]/Sin[x]]

Solve[1+(Cot[x]-1) Tan[y] == Cot[x],{Tan[y]}]

(*  Alternativen *)

Cos[y]==0

Cos[x]==0

h1[x_,y_]:= If[h[x,y] <= 0,  h[x,y], 0];

p2= Plot3D[h1[x,y],{x,-Pi,Pi},{y,-Pi,Pi},PlotPoints->40];

Show[p1,p2];

3E+M

Remove["Global`*"];

a

f[x_,y_]:= x^2+y^2

p1 = Plot3D[f[x,y],{x,-5,5},{y,-5,5}];

p2 = Plot3D[50,{x,-5,5},{y,-5,5}];

Show[p2,p1,ViewPoint->{-2.418, -1.626, 0.262}];

Show[p2,p1,ViewPoint->{-2.245, -1.510, -1.114}];

p3=ParametricPlot3D[{2 Cos[t],2 Sin[t],h},{t,0,2Pi},{h,0,50}];

Show[p2,p1,p3,ViewPoint->{-2.418, -1.626, 0.262}];

b

Integrate[50-f[x,y],{x,-5,5},{y,-5,5}]

N[%]

c

f[x,5]

len = Integrate[Evaluate[Sqrt[1+D[f[x,5],x]^2]],{x,-5,5}]

N[%]

d

Integrate[Sqrt[1+D[Sqrt[x],x]^2] 2 Sqrt[x] Pi,{x,0,t},GenerateConditions->False]

Integrate[Sqrt[1+D[Sqrt[x],x]^2] 2 Sqrt[x] Pi,{x,0,2}]

N[%]

Integrate[Sqrt[1+D[Sqrt[x],x]^2] 2 Sqrt[x] Pi,{x,0,4}]

N[%]

e

v[x_,y_]:={x,y,f[x,y]};
Norm[Cross[D[v[x,y],x],D[v[x,y],y]]]

Sqrt[Cross[D[v[x,y],x],D[v[x,y],y]].Cross[D[v[x,y],x],D[v[x,y],y]]]

Integrate[Evaluate[Sqrt[Cross[D[v[x,y],x],D[v[x,y],y]].Cross[D[v[x,y],x],D[v[x,y],y]]]], {x,-5,5}]

Integrate[Integrate[Evaluate[Sqrt[Cross[D[v[x,y],x],D[v[x,y],y]].Cross[D[v[x,y],x],D[v[x,y],y]]]], {x,-5,5}], {y,-5,5}]

N[%]

NIntegrate[Evaluate[Sqrt[Cross[D[v[x,y],x],D[v[x,y],y]].Cross[D[v[x,y],x],D[v[x,y],y]]]], {x,-5,5}, {y,-5,5}]

Solve[h == r^2,{r}]

Solve[h == 2^2,{h}]

4E+M

a

Series[Sin[x],{x,2Pi,10}]

N[%]

b

u[t_]:=(Series[Sin[x],{x,2Pi,10}]//Normal)/.x->1/t

u[t]

N[%]

c

Plot[u[t],{t,0.1,4Pi-0.1}];

Plot[u[t],{t,0.09,1}];

Plot[{u[t],Sin[1/t]},{t,0.1,0.4}];

d

u[0.4]

Sin[1/0.4]

Sin[1/0.4]-u[0.4]

100 (Sin[1/0.4]-u[0.4])/Sin[1/0.4]

e

su[x_]:=Integrate[u[t],{t,0.1,x},GenerateConditions->False]// Expand;
su[x]

N[%]

(* Terme mit Koeffizienten absolut grösser 0.1 *)

tab1=Union[Table[If[Abs[su[x][[n]][[1]]]>0.1,su[x][[n]],0],{n,2,11}], {If[Abs[su[x][[1]]]>0.1,su[x][[1]],0]}]

Apply[Plus,tab1]

5E+M

a

Remove["Global`*"];
<<Graphics`PlotField`

g1=PlotVectorField[{1, 16/12 x/y},{x,-2,2,0.2},{y,0.2,2,0.2},Epilog->{Hue[1],Line[{{-2,0},{2,0}}],Line[{{0,-2},{0,2}}]},AspectRatio->Automatic, DisplayFunction->Identity];
g2=PlotVectorField[{1, 16/12 x/y},{x,-2,2,0.2},{y,-2,-0.2,0.2},Epilog->{Hue[1],Line[{{-2,0},{2,0}}],Line[{{0,-2},{0,2}}]},AspectRatio->Automatic, DisplayFunction->Identity];
Show[g1,g2, DisplayFunction->$DisplayFunction];

b

Remove[x,y];
solv = Flatten[DSolve[{12 y'[x] - 16 x/y[x] == 0, y[0]==1},y,x]];
y = y/.solv;
Print["y(x) = ",Simplify[y[x]]];
g3=Plot[y[x],{x,-2,2}];

y[x]

Show[g1,g2,g3, DisplayFunction->$DisplayFunction];

c

Limit[y[x]/x,x->Infinity]

(* ===> Linear *)

6E+M

Remove["Global`*"];

a

solv1 = Flatten[DSolve[{y''[t] + 2 y'[t]+ y[t] == 0},y,t]]

b

y[t_]:= a t^2 E^(-t);
(y''[t] +2 y'[t]+ y[t] //Simplify) == E^(-t)

Remove["Global`*"];
solv2 = Flatten[DSolve[{y''[t] +2 y'[t]+ y[t] == E^(-t)},y,t]];
y = y/.solv2;
y[t]

c

y[0]

y'[t]/.t->0

(* Unendlich viele Möglichkeiten mit 2 Variablen C1, C2 *)

Remove["Global`*"];
y[t_,c1_,c2_]:= E^(-t)(1/2 t^2+c1+c2 t);
Plot[Evaluate[Table[y[t,c1,c1],{c1,-5,5}]],{t,-2,2}];

(* Kurven durch Ursprung *)

Solve[Evaluate[{y[0,c1,c2]==0, (D[y[t,c1,c2],t]/.t->0)==0}],{c1,c2}]

(* Eine Lösung, Eindeutikeit *)

d

Remove["Global`*"];
solv2 = Flatten[DSolve[{y''[t] +2 y'[t]+ y[t] == E^(-t)},y,t]];
y = y/.solv2;
Solve[Evaluate[D[y[t],t]==0],{t}]

(*  Unendlich viele Möglichkeiten  *)

% /. C[1]->0

(*  Nuch unendlich viele Möglichkeiten  *)

% /. C[2]->0

(*  Noch 2 Möglichkeiten  *)

e

y[-1] == E/2

y[0] == 1

Solve[{y[-1] == E/2,y[0] == 1},{C[1],C[2]}]

z[t_]:=(y[t]/.{C[1]->1,C[2]->1})//Evaluate;
z[t]

Plot[z[t],{t,-2,2}];

len= NIntegrate[Evaluate[Sqrt[1+D[z[t],t]^2]],{t,-1,1}]

7E+M

Remove["Global`*"];

a

solv = Flatten[DSolve[{y''[x] +3 y'[x] - 4 y[x] == 0, y[0]==1, y'[0]==0},y,x]]

Solve[x^2+3x-4==0,{x}]

(1/5) (E^(-4 x)) (1 + 
            4 (E^(5 x))) // Simplify

b

p1 = 
        Plot[(1/5) (E^(-4 x)) (1 + 
                4 (E^(5 x))), {x, -1.5, 
            4}];

c

solv = Flatten[DSolve[{y''[x] +3 y'[x] - 4 y[x] == 1, y[0]==1, y'[0]==0},y,x]]

(1/4) (E^(-4 x)) (1 - (E^(4 x)) + 4 (E^(5 x))) // Simplify

d

p2 = 
        Plot[(1/4) (E^(-4 x)) (1 - (E^(4 x)) + 4 (E^(5 x))), {x, -1.5, 
            4}];

e

Show[p1,p2];

d[x_] := 
        (1/4) (E^(-4 x)) (1 - (E^(4 x)) + 4 (E^(5 x))) - 
            (1/5) (E^(-4 x)) (1 + 
                  4 (E^(5 x))) // 
          Simplify;
d[x]

p3 = Plot[d[x], {x, -1.5, 4}];

(* Anteile von E^x und E^(-x) ähnlich. *)

8E+M

a

Sum[1/k! x^k,{k,0,Infinity}] -1-x

Integrate[Evaluate[Sum[1/k! x^k,{k,0,Infinity}] -1-x],{x,0,t}]

1/k! x^k

Integrate[1/k! x^k,x]//Simplify

(* Majorante ist Potenzreihe der e-Funktion - e^x-1-x-x^2/2 -C *)

b Diff., Indentität

(Sin[x/2]+Cos[x/2])^2 // Expand // TrigReduce

Cos[x] == Integrate[1-(Sin[x/2]+Cos[x/2])^2,x]

(* Die Identität ist aber falsch, da links eine bestimmte Funktion steht, rechts aber ein unbestimmtes Integral. Richtig nur für bestimmtes Integral mit spezieller Wahl der Grenzen. *)

c Partielle Integration

Integrate[(x-n+1)E^x,x] == E^x (x-n)

(* Die Identität ist aber falsch, da links eine bestimmte Funktion steht, rechts aber ein unbestimmtes Integral. Richtig nur für bestimmtes Integral mit spezieller Wahl der Grenzen. *)

d Diff, Indentität

D[Log[(x Sqrt[x])]-Log[(x Sqrt[x-n])],{x,1}]//Simplify

Limit[1/Evaluate[D[Log[(x Sqrt[x])]-Log[(x Sqrt[x-n])],{x,1}]//Simplify], {x ->n}]

Limit[Evaluate[1/D[Log[(x Sqrt[x])]-Log[(x Sqrt[x-n])],{x,1}]//Simplify], {x ->n^2}]

e

q = Sqrt[2];
sinE[x_]:=E^(x/q) Sin[x/q]; 
cosE[x_]:=E^(x/q) Cos[x/q];

Integrate[Integrate[sinE[x],x],x]

Integrate[Integrate[cosE[x],x],x]

Integrate[Integrate[sinE[x],x],x] == -cosE[x]

Integrate[Integrate[cosE[x],x],x] == sinE[x]