Sample Midterm Problem 1
edit
Given:
The vectors a {\displaystyle \mathbf {a} \,} , b {\displaystyle \mathbf {b} \,} , and c {\displaystyle \mathbf {c} \,} are given, with respect to an orthonormal
basis ( e ^ 1 , e ^ 2 , e ^ 3 ) {\displaystyle ({\widehat {\mathbf {e} }}_{1},{\widehat {\mathbf {e} }}_{2},{\widehat {\mathbf {e} }}_{3})} , by
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} } . Is D {\displaystyle \mathbf {D} \,} a tensor? If not, why not? If yes, what is the order of the tensor?
(c) Name and define δ i j {\displaystyle \delta _{ij}\,} and 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})} by 30 degrees in the counterclockwise direction around e ^ 3 {\displaystyle {\widehat {\mathbf {e} }}_{3}} to obtain a new basis ( e 1 ′ , e 2 ′ , e 3 ′ ) {\displaystyle (\mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'})} . Find the components of the vector b {\displaystyle \mathbf {b} \,} in the new basis ( 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}\,} of D {\displaystyle \mathbf {D} \,} in the new basis ( e 1 ′ , e 2 ′ , e 3 ′ ) {\displaystyle (\mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'})} . Solution
edit
Part (a)
edit
d = [ ( 5 ) ( 10 ) + ( − 3 ) ( 6 ) + ( 10 ) ( 0 ) ] ( 4 ) = 128 {\displaystyle d=[(5)(10)+(-3)(6)+(10)(0)](4)=128}
d = 128 {\displaystyle {d=128}} Part (b)
edit
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}}.}} Part (c)
edit
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}}}} Part (d)
edit
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}\,} Part (e)
edit
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} cannot be an even or odd permutation of 1 , 2 , 3 {\displaystyle 1,2,3} .
Part (f)
edit
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}^{'}}} Part (g)
edit
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}}