Задачи по векторному анализу. Михайлов В.К - 144 стр.

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144
{} {} {}
ρρρ
=−
=
=−
eee
123
1
3
12 2
1
3
212
1
3
221,, , ,, , ,, .
5.31. 15.
5.32.
() .∇= ==
∑∑
ρρρ
a
a
x
adiva
ii
i
i
i
i
5.33. — .
5.34. Ñâåðòêîé áóäåò âåêòîð
[]c
i
, êîìïîíåíòû êîòîðîãî
cab
iijj
j
=
. Ïîäñòàâëÿÿ ÷èñëà, ïîëó÷àåì:
[] .c
i
=
8
17
26
5.35.
∇=
=
ρ
a
a
x
i
j
100
010
000
.
5.36.
,
=
j
j
i
j
i
x
a
l
r
a
ρ
ãäå
r
r
l
ρ
ρ
=
— åäèíè÷íûé ðàäèóñ-âåêòîð. Ïîäñòàâëÿÿ
êîìïîíåíòû âåêòîðîâ
ρ
a
è
l
ρ
, ïîëó÷àåì:
{}
ρ
a
r
xx
xx
=
+
1
1
2
2
2
21
,.
5.37.
.
–onstl
l
r
==
ρ
ρ
        ρ 1             ρ 1             ρ 1
        e1′ = {1,2,−2}, e 2′ = {2,1,2}, e3′ = {− 2,2,1}.
             3                3              3
5.31. 15.
               ρ              ∂a             ρ       ρ
5.32.   ∑ (∇a ) ii = ∑ ∂xi             = ∇ ⋅ a = div a .
        i                 i        i
5.33. — .

5.34. Ñâåðòêîé áóäåò âåêòîð [ ci ] , êîìïîíåíòû êîòîðîãî

                                                              8
        ci = ∑ a ij b j . Ïîäñòàâëÿÿ ÷èñëà, ïîëó÷àåì: [ c ] = 17 .
             j
                                                         i     
                                                               26

                         − 1 0 0
       ρ  ∂a i                
5.35. ∇a =          =  0 1 0 .
            ∂x j   0 0 0
                                
         ρ
       ∂a        ∂ai
5.36. ∂r  = ∑ l j ∂x ,
      
       i       j   j

            ρ rρ
     ãäå l = — åäèíè÷íûé ðàäèóñ-âåêòîð. Ïîäñòàâëÿÿ
               r
                             ρ      ρ
     êîìïîíåíòû âåêòîðîâ a è l , ïîëó÷àåì:
         ρ
        ∂a            1
           =                   {− x2 , x1 }.
        ∂r         x12 + x22
       ρ
      ∂r ρ
5.37.    = l = –onst.
      ∂l




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