R
Ingenieurmathematik Lösungen zur Prüfung 1
11.Mai2010
Zeit 90 Minuten, Reihenfolge beliebig, 8 Punkte pro Hauptaufgabe,
40 Pt. = N.6.
plot(x,y,'k*-')
A=[4 0 0]' ; B=[0 6 0]' ; C=[0 0 3]' ;
lin = [A B C A]
plot3(lin(1,:) ,lin(2,:), lin(3,:) ); axis equal
A = [11:15 ; 21:25; 31:35; 41:45 ; 51:55 ] Pl = zeros(5); Pr = zeros(5); Pl(1,2)= 1; Pl(2,1) = 1; Pl(4,5) = 1; Pl(5,3)=1; Pr(2,3)= 1; Pr(3,4) = 1; Pr(5,5) = 1; Rs = Pl * A * Pr
ntur = 3; h = 1; r = 2; w = (0:0.002:3)* 2 * pi; z = w*h/(2*pi); xl = r*sin(w) ; yl = r*cos(w) -2; xr = r*sin(w) ; yr = -r*cos(w) +2; figure(1) clf hold on plot3(xl,yl,z,'g') plot3(xr,yr,z,'r') plot3([0 0 0 0], [0 0 0 0], [0 1 2 3 ],'ko') axis equal view(69,45) hold off
A=[0 0 0]' ; B = [6 0 0]' ; D = [0 4 0]'; E =[0 0 4]' drb = [A B A D A E] figure(1) clf plot3(drb(1,:), drb(2,:), drb(3,:), 'ko-' ) hold on axis equal MAB = (A+B)/2 MAD = (A+D)/2 MAE = (A+E)/2 eb = [MAB MAD MAE MAB]; plot3(eb(1,:), eb(2,:), eb(3,:), 'r' ) axis equal; hold off v = MAD - MAB w = MAE - MAB N = cross(v,w) % [4 6 6 ]' eng = N / norm(N) % [ 0.4268 0.6396 0.6396]' dkritg= eng'*MAB % 1.2782 dtsD = eng'*MAD -dkritg dtsE = eng'*MAE -dkritg enh = eng dkrith = 0 % Ebene durch (0/0/0)
L*y = b
, ohne
von Schleifenkonstruktionen Gebrauch zu machen.
function y = fix3forwsub(L,b) y = zeros(3,1); % y1 kann b1 direkt uebernehmen y(1) = b(1); % div durch 1 , weil L(1,1) = 1 % y1 wird gebraucht weil Element links der Diag nicht 0 y(2) = (b(2) - L(2,1)* y(1) ) % y1 und y2 werden gebraucht weil Elemente links der Diag nicht 0 y(3) = (b(3) - L(3,1)* y(1) - L(3,2)*y(2) )
Z0 = 81 * exp(i*pi) z1 = 81^(1/4) * exp(i*pi/4) z2 = 81^(1/4) * exp(i*(pi/4 +2*pi/4)) z3 = 81^(1/4) * exp(i*(pi/4 +2*2*pi/4)) z4 = 81^(1/4) * exp(i*(pi/4 +3*2*pi/4)) p1 = 243*exp(i*5*pi/4) p2 = 243 * exp(i*5*(pi/4 +2*pi/4)) p3 = 243 * exp(i*5*(pi/4 +2*2*pi/4)) p4 = 243 * exp(i*5*(pi/4 +3*2*pi/4))