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Ingenieurmathematik Lösungen zur Prüfung 1
11.Mai2010
Zeit 90 Minuten, Reihenfolge beliebig, 8 Punkte pro Hauptaufgabe,
40 Pt. = N.6.
plot(x,y,'b*-')
U=[4 0 0]' ; V=[0 6 0]' ; W=[0 0 3]' ;
lin = [U V W U]
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(1,5)= 1; Pl(2,4) = 1; Pl(4,3) = 1; Pl(5,1)=1; Pr(5,1)= 1; Pr(4,2) = 1; Pr(2,5) = 1; Rs = Pl * A * Pr
ntur = 3; h = 2; r = 1.5; w = (0:0.002:3)* 2 * pi; z = w*h/(2*pi); xl = r*cos(w) -1.5; yl = -r*sin(w); xr = -r*cos(w) + 1.5; yr = -r*sin(w); 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 2 4 6],'ko') axis equal view(-28,40) hold off
A=[0 0 0]' ; B = [2 0 0]' ; D = [0 3 0]'; E =[0 0 2]' 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) % [ 1.5 1 1.5 eng = N / norm(N) % [0.6396 0.4268 0.6396]' dkritg= eng'*MAB % 0.6396 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 = 16 * exp(i*pi) z1 = 16^(1/4) * exp(i*pi/4) z2 = 16^(1/4) * exp(i*(pi/4 +2*pi/4)) z3 = 16^(1/4) * exp(i*(pi/4 +2*2*pi/4)) z4 = 16^(1/4) * exp(i*(pi/4 +3*2*pi/4)) p1 = 8*exp(i*3*pi/4) p2 = 8 * exp(i*3*(pi/4 +2*pi/4)) p3 = 8 * exp(i*3*(pi/4 +2*2*pi/4)) p4 = 8 * exp(i*3*(pi/4 +3*2*pi/4))