prova de física matemática resolvida
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Prova de Física Matemática resolvidaTRANSCRIPT
![Page 1: Prova de física matemática resolvida](https://reader036.vdocuments.net/reader036/viewer/2022081821/5535ae074a795919158b46a6/html5/thumbnails/1.jpg)
P1
1)
2
22 3
2 2 3 3 3 4
1
2 2 1 2 2 1 2 21 1 2
2! 3!
1 1 2 3 4 2 3 4
x x
xx x x
x x x x x x x x x x
2)
a)
. . 0
. 0.......... 3.53
s v
V ndS V dv
V
b)
. . 0
0
s c
r ndS r dr
r
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3)
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2 2
22 2
cos
12
1 cos2
12
12
1 1 1 11 cos cos 12 2 4 2
11 12
x
y sen
z
x
y sen
z
nh sen sen
h
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2 2
22 2
1 cos2
12
12
1 1 1 12 cos cos 12 2 4 2
12 12
x
y sen
z
h sen sen
h
2 2
cos
0
3 cos
x sen
y
z
h sen
4)
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2 22
2 2
2 22
2 2
2 22
2 2
2 22
2 2
2
, ,
0
0......... ( .. , ,
0.........( ... )
u u ux y t
u X Y T X x Y y T t
XYT XYT XYTx y t
YT X XT Y XY Tdividindo por u X Y T
x y t
X Y Tarrumando X
X x Y y T t
2 22
2 2
2 2
2 2 2
22
2 2
22
2
22
2 2
2 22
22
2
2 22
2 2 2
0
0
0
0
X Y TX x Y y T t
X Y TX
x Y y T t
Y TY y T t
XX
x
Y TY
y T t
TT t
YY
y
TT t
TT
t
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5)
5)
2
2
1 1'' '
1( )
1( )
1( )
y y y xx x
y x x
P xx
Q xx
Na equação (5.22)
ln
2 2 22( ) ln
2( ) ln
2( ) ln
dxxxy x e e xdx dx dx x
x x x x
y x xx
y x x x
Agora usando (5.13)
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2 2 2 2
2
( ) 1( ) '2( ) '1( ) 2( ) 1 ln ln
ln .1( ) ln
ln lnln2 2 2 4
.2( )
2
W x y x y x y x y x x x x x x
x x x dxv x x x
x
x x x dx x x xx xx
x x dx xv xx
Esssas foram a equação (5.35)
Obs:Tivemos uma integral por partes de xlnx(tem que considerar a troca de sinal de v1)
Agora usamos o resultado da equação (5.24)
2 2 2 3ln( ) ln2 4 2 4
x x x x xu x x x x