Theoretishe Physik A WS 2000/01
Prof. Dr. J.Kuhn / Dr. W.Kilian Blatt 2 27.10.2000
Losungsvorshlage
1. Integrationsmethoden (4Punkte)
(a)
d
dx
1
2
(x+sinhxoshx)
= 1
2
(1+osh 2
x+sinh 2
x)=osh 2
x
(b)
Z
dx osh 2
x= Z
dx e
2x
+2+e 2x
4
= 1
2 x+
e 2x
e 2x
8
= 1
2 x+
(e x
+e x
)(e x
e x
)
8
= 1
2 x+
oshxsinhx
2
()
Z
dx osh 2
x=sinhxoshx Z
dx sinh 2
x=sinhxoshx Z
dx osh 2
x 1
=sinhxoshx+x Z
dx osh 2
x (Auosen nah Z
dx osh 2
x)
(a)
v(t)=
r!os!t
(b) v(t)= p
2
+r 2
! 2
os 2
!t
()
a(t)=
0
r!
2
sin!t
(d) a(t) =r!
2
jsin!tj
3. BahnkurveIII (7 Punkte)
(a)
v(t)=
t!os!t+sin!t
t!sin!t+os!t
(b) v(t)= p
1+! 2
t 2
()
a(t)=
t!
2
sin!t+2!os!t
t!
2
os!t 2!sin!t
(d) a(t) =! p
4+! 2
t 2
y y
AusKoeÆzientenvergleih in
ax 2
+2bx+=a(y 2
+y 2
0 )
folgtmity=x x
0 : x
0
= b=a und y
0
= p
a b 2
=a. Damitlautet das Integral:
Z
dx p
ax 2
+2bx+= p
a Z
dy q
y 2
+y 2
0
Substitution:
y=y
0
sinh; dy=y
0
oshd;
also
Z
dy q
y 2
+y 2
0
= Z
y 2
0 osh
2
d= y
2
0
2
(+sinhosh)
= y
2
0
2
Arsinh y
y
0 +
y
y
0 s
1+ y
2
y 2
0
!
= y
2
0
2
Arsinh y
y
0 +
y
2 q
y 2
+y 2
0
Einsetzen ergibt
Z
dx p
ax 2
+2bx+=
a b 2
2a 3=2
Arsinh
ax+b
p
2 +
ax+b
2a p
ax 2
+2bx+:
