Sample Midterm Problem 1
edit
Given:
The vectors
a
{\displaystyle \mathbf {a} \,}
b
{\displaystyle \mathbf {b} \,}
c
{\displaystyle \mathbf {c} \,}
(
e
^
1
,
e
^
2
,
e
^
3
)
{\displaystyle ({\widehat {\mathbf {e} }}_{1},{\widehat {\mathbf {e} }}_{2},{\widehat {\mathbf {e} }}_{3})}
a
=
5
e
^
1
−
3
e
^
2
+
10
e
^
3
;
b
=
4
e
^
1
+
6
e
^
2
−
2
e
^
3
;
c
=
10
e
^
1
+
6
e
^
2
{\displaystyle \mathbf {a} =5~{\widehat {\mathbf {e} }}_{1}-3~{\widehat {\mathbf {e} }}_{2}+10~{\widehat {\mathbf {e} }}{3}~;~~\mathbf {b} =4~{\widehat {\mathbf {e} }}_{1}+6~{\widehat {\mathbf {e} }}_{2}-2~{\widehat {\mathbf {e} }}_{3}~;~~\mathbf {c} =10~{\widehat {\mathbf {e} }}_{1}+6~{\widehat {\mathbf {e} }}_{2}}
Find:
(a) Evaluate
d
=
a
m
c
m
b
1
{\displaystyle d=a_{m}~c_{m}~b_{1}}
(b) Evaluate
D
=
a
⊗
c
{\displaystyle \mathbf {D} =\mathbf {a} \otimes \mathbf {c} }
D
{\displaystyle \mathbf {D} \,}
(c) Name and define
δ
i
j
{\displaystyle \delta _{ij}\,}
e
i
j
k
{\displaystyle e_{ijk}\,}
(d) Evaluate
g
=
D
i
j
δ
i
j
{\displaystyle g=D_{ij}\delta _{ij}\,}
(e) Show that
δ
i
k
e
i
k
m
=
0
{\displaystyle \delta _{ik}e_{ikm}=0\,}
(f) Rotate the basis
(
e
^
1
,
e
^
2
,
e
^
3
)
{\displaystyle ({\widehat {\mathbf {e} }}_{1},{\widehat {\mathbf {e} }}_{2},{\widehat {\mathbf {e} }}_{3})}
e
^
3
{\displaystyle {\widehat {\mathbf {e} }}_{3}}
(
e
1
′
,
e
2
′
,
e
3
′
)
{\displaystyle (\mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'})}
b
{\displaystyle \mathbf {b} \,}
(
e
1
′
,
e
2
′
,
e
3
′
)
{\displaystyle (\mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'})}
(g) Find the component
D
12
{\displaystyle D_{12}\,}
D
{\displaystyle \mathbf {D} \,}
(
e
1
′
,
e
2
′
,
e
3
′
)
{\displaystyle (\mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'})}
d
=
[
(
5
)
(
10
)
+
(
−
3
)
(
6
)
+
(
10
)
(
0
)
]
(
4
)
=
128
{\displaystyle d=[(5)(10)+(-3)(6)+(10)(0)](4)=128}
d
=
128
{\displaystyle {d=128}}
D
=
a
i
c
j
=
[
(
5
)
(
10
)
(
5
)
(
6
)
(
5
)
(
0
)
(
−
3
)
(
10
)
(
−
3
)
(
6
)
(
−
3
)
(
0
)
(
10
)
(
10
)
(
10
)
(
6
)
(
10
)
(
0
)
]
{\displaystyle \mathbf {D} =a_{i}~c_{j}={\begin{bmatrix}(5)(10)&(5)(6)&(5)(0)\\(-3)(10)&(-3)(6)&(-3)(0)\\(10)(10)&(10)(6)&(10)(0)\end{bmatrix}}}
D
=
[
50
30
0
−
30
−
18
0
100
60
0
]
{\displaystyle {\mathbf {D} ={\begin{bmatrix}50&30&0\\-30&-18&0\\100&60&0\end{bmatrix}}}}
D
is a second-order tensor
.
{\displaystyle {\mathbf {D} ~{\text{is a second-order tensor}}.}}
δ
i
j
=
Kronecker delta
{\displaystyle {\delta _{ij}={\text{Kronecker delta}}}}
e
i
j
k
=
Permutation symbol
{\displaystyle {e_{ijk}={\text{Permutation symbol}}}}
δ
i
j
=
{
1
i
f
i
=
j
0
o
t
h
e
r
w
i
s
e
{\displaystyle {\delta _{ij}={\begin{cases}1&{\rm {{if}~i=j}}\\0&{\rm {otherwise}}\end{cases}}}}
e
i
j
k
=
{
1
i
f
i
j
k
=
123
,
231
,
312
−
1
i
f
i
j
k
=
321
,
213
,
132
0
o
t
h
e
r
w
i
s
e
{\displaystyle {e_{ijk}={\begin{cases}1&{\rm {{if}~ijk=123,~231,~312}}\\-1&{\rm {{if}~ijk=321,~213,~132}}\\0&{\rm {otherwise}}\end{cases}}}}
g
=
D
k
k
=
D
11
+
D
22
+
D
33
=
50
−
18
+
0
=
32
{\displaystyle g=D_{kk}=D_{11}+D_{22}+D_{33}=50-18+0=32\,}
g
=
32
{\displaystyle {g=32}\,}
δ
i
k
e
i
k
m
=
e
j
j
m
=
0
{\displaystyle {\delta _{ik}e_{ikm}=e_{jjm}=0}}
Because
j
j
m
{\displaystyle jjm}
1
,
2
,
3
{\displaystyle 1,2,3}
The basis transformation rule for vectors is
v
i
′
=
l
i
j
v
j
{\displaystyle v_{i}^{'}=l_{ij}v_{j}}
where
l
i
j
=
e
^
i
′
∙
e
^
j
=
cos
(
e
^
i
′
,
e
^
j
)
{\displaystyle l_{ij}={\widehat {\mathbf {e} }}{i}^{'}\bullet {\widehat {\mathbf {e} }}{j}=\cos({\widehat {\mathbf {e} }}{i}^{'},{\widehat {\mathbf {e} }}{j})}
Therefore,
[
L
]
=
[
cos
(
30
o
)
cos
(
90
o
−
30
o
)
cos
(
90
o
)
cos
(
90
o
+
30
o
)
cos
(
30
o
)
cos
(
90
o
)
cos
(
90
o
)
cos
(
90
o
)
cos
(
0
o
)
]
=
[
cos
(
30
o
)
sin
(
30
o
)
cos
(
90
o
)
−
sin
(
30
o
)
cos
(
30
o
)
cos
(
90
o
)
cos
(
90
o
)
cos
(
90
o
)
cos
(
0
o
)
]
=
[
3
/
2
1
/
2
0
−
1
/
2
3
/
2
0
0
0
1
]
{\displaystyle {\begin{aligned}\left[L\right]&={\begin{bmatrix}\cos(30^{o})&\cos(90^{o}-30^{o})&\cos(90^{o})\\\cos(90^{o}+30^{o})&\cos(30^{o})&\cos(90^{o})\\\cos(90^{o})&\cos(90^{o})&\cos(0^{o})\end{bmatrix}}\\&={\begin{bmatrix}\cos(30^{o})&\sin(30^{o})&\cos(90^{o})\\-\sin(30^{o})&\cos(30^{o})&\cos(90^{o})\\\cos(90^{o})&\cos(90^{o})&\cos(0^{o})\end{bmatrix}}\\&={\begin{bmatrix}{\sqrt {3}}/2&1/2&0\\-1/2&{\sqrt {3}}/2&0\\0&0&1\end{bmatrix}}\end{aligned}}}
Hence,
b
1
′
=
l
11
b
1
+
l
12
b
2
+
l
13
b
3
=
(
3
/
2
)
(
4
)
+
(
1
/
2
)
(
6
)
+
(
0
)
(
−
2
)
=
2
3
+
3
=
6.46
b
2
′
=
l
21
b
1
+
l
22
b
2
+
l
23
b
3
=
(
−
1
/
2
)
(
4
)
+
(
3
/
2
)
(
6
)
+
(
0
)
(
−
2
)
=
−
2
+
3
3
=
3.2
b
3
′
=
l
31
b
1
+
l
32
b
2
+
l
33
b
3
=
(
0
)
(
4
)
+
(
0
)
(
6
)
+
(
1
)
(
−
2
)
=
−
2
{\displaystyle {\begin{aligned}b_{1}^{'}&=l_{11}b_{1}+l_{12}b_{2}+l_{13}b_{3}=({\sqrt {3}}/2)(4)+(1/2)(6)+(0)(-2)=2{\sqrt {3}}+3=6.46\\b_{2}^{'}&=l_{21}b_{1}+l_{22}b_{2}+l_{23}b_{3}=(-1/2)(4)+({\sqrt {3}}/2)(6)+(0)(-2)=-2+3{\sqrt {3}}=3.2\\b_{3}^{'}&=l_{31}b_{1}+l_{32}b_{2}+l_{33}b_{3}=(0)(4)+(0)(6)+(1)(-2)=-2\end{aligned}}}
Thus,
b
′
=
6.46
e
1
′
+
3.2
e
2
′
−
2
e
3
′
{\displaystyle {\mathbf {b} ^{'}=6.46~\mathbf {e} _{1}^{'}~+~3.2~\mathbf {e} _{2}^{'}~-~2\mathbf {e} _{3}^{'}}}
The basis transformation rule for second-order tensors is
D
i
j
′
=
l
i
p
l
j
q
D
p
q
{\displaystyle D_{ij}^{'}=l_{ip}l_{jq}D_{pq}\,}
Therefore,
D
12
′
=
l
11
l
21
D
11
+
l
12
l
21
D
21
+
l
13
l
21
D
31
+
l
11
l
22
D
12
+
l
12
l
22
D
22
+
l
13
l
22
D
32
+
l
11
l
23
D
13
+
l
12
l
23
D
23
+
l
13
l
23
D
33
=
l
11
(
l
21
D
11
+
l
22
D
12
+
l
23
D
13
)
+
l
12
(
l
21
D
21
+
l
22
D
22
+
l
23
D
23
)
+
l
13
(
l
21
D
31
+
l
22
D
32
+
l
23
D
33
)
=
(
3
2
)
[
(
−
1
2
)
(
50
)
+
(
3
2
)
(
30
)
+
(
0
)
(
0
)
]
+
(
1
2
)
[
(
−
1
2
)
(
−
30
)
+
(
3
2
)
(
−
18
)
+
(
0
)
(
0
)
]
+
(
0
)
[
(
−
1
2
)
(
100
)
+
(
3
2
)
(
60
)
+
(
0
)
(
0
)
]
=
(
3
2
)
[
−
25
+
15
3
]
+
(
1
2
)
[
15
−
9
3
]
=
−
25
3
2
+
45
2
+
15
2
−
9
3
2
=
−
17
3
+
30
{\displaystyle {\begin{aligned}D_{12}^{'}=&l_{11}l_{21}D_{11}+l_{12}l_{21}D_{21}+l_{13}l_{21}D_{31}+l_{11}l_{22}D_{12}+l_{12}l_{22}D_{22}+l_{13}l_{22}D_{32}+\\&l_{11}l_{23}D_{13}+l_{12}l_{23}D_{23}+l_{13}l_{23}D_{33}\\=&l_{11}(l_{21}D_{11}+l_{22}D_{12}+l_{23}D_{13})+l_{12}(l_{21}D_{21}+l_{22}D_{22}+l_{23}D_{23})+\\&l_{13}(l_{21}D_{31}+l_{22}D_{32}+l_{23}D_{33})\\=&({\frac {\sqrt {3}}{2}})\left[(-{\frac {1}{2}})(50)+({\frac {\sqrt {3}}{2}})(30)+(0)(0)\right]+({\frac {1}{2}})\left[(-{\frac {1}{2}})(-30)+({\frac {\sqrt {3}}{2}})(-18)+(0)(0)\right]+\\&(0)\left[(-{\frac {1}{2}})(100)+({\frac {\sqrt {3}}{2}})(60)+(0)(0)\right]\\=&({\frac {\sqrt {3}}{2}})\left[-25+15{\sqrt {3}}\right]+({\frac {1}{2}})\left[15-9{\sqrt {3}}\right]\\=&-25{\frac {\sqrt {3}}{2}}+{\frac {45}{2}}+{\frac {15}{2}}-9{\frac {\sqrt {3}}{2}}\\=&-17{\sqrt {3}}+30\end{aligned}}}
D
12
′
=
−
17
3
+
30
=
0.55
{\displaystyle {D_{12}^{'}=-17{\sqrt {3}}+30=0.55}}
=128}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e92f1c5e73ae5ea6db427b855f94dc188b9b495)
![{\displaystyle {\begin{aligned}\left[L\right]&={\begin{bmatrix}\cos(30^{o})&\cos(90^{o}-30^{o})&\cos(90^{o})\\\cos(90^{o}+30^{o})&\cos(30^{o})&\cos(90^{o})\\\cos(90^{o})&\cos(90^{o})&\cos(0^{o})\end{bmatrix}}\\&={\begin{bmatrix}\cos(30^{o})&\sin(30^{o})&\cos(90^{o})\\-\sin(30^{o})&\cos(30^{o})&\cos(90^{o})\\\cos(90^{o})&\cos(90^{o})&\cos(0^{o})\end{bmatrix}}\\&={\begin{bmatrix}{\sqrt {3}}/2&1/2&0\\-1/2&{\sqrt {3}}/2&0\\0&0&1\end{bmatrix}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba5d7072279f064f7730757bda8b1c72ebfeedb4)
![{\displaystyle {\begin{aligned}D_{12}^{'}=&l_{11}l_{21}D_{11}+l_{12}l_{21}D_{21}+l_{13}l_{21}D_{31}+l_{11}l_{22}D_{12}+l_{12}l_{22}D_{22}+l_{13}l_{22}D_{32}+\\&l_{11}l_{23}D_{13}+l_{12}l_{23}D_{23}+l_{13}l_{23}D_{33}\\=&l_{11}(l_{21}D_{11}+l_{22}D_{12}+l_{23}D_{13})+l_{12}(l_{21}D_{21}+l_{22}D_{22}+l_{23}D_{23})+\\&l_{13}(l_{21}D_{31}+l_{22}D_{32}+l_{23}D_{33})\\=&({\frac {\sqrt {3}}{2}})\left[(-{\frac {1}{2}})(50)+({\frac {\sqrt {3}}{2}})(30)+(0)(0)\right]+({\frac {1}{2}})\left[(-{\frac {1}{2}})(-30)+({\frac {\sqrt {3}}{2}})(-18)+(0)(0)\right]+\\&(0)\left[(-{\frac {1}{2}})(100)+({\frac {\sqrt {3}}{2}})(60)+(0)(0)\right]\\=&({\frac {\sqrt {3}}{2}})\left[-25+15{\sqrt {3}}\right]+({\frac {1}{2}})\left[15-9{\sqrt {3}}\right]\\=&-25{\frac {\sqrt {3}}{2}}+{\frac {45}{2}}+{\frac {15}{2}}-9{\frac {\sqrt {3}}{2}}\\=&-17{\sqrt {3}}+30\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc484e1654f1f4850d9d4e6f0ca078ff9c259d7e)
