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Question Number 225768    Answers: 0   Comments: 0

WeinGarten Equation r_u ∙r_u =E , r_u ∙r_v =F , r_v ∙r_v =G N^ ∙N^ =1 ((∂(N^ ∙N^ ))/∂u)=N_u ^ ∙N^ +N^ ∙N_u ^ =0 ((∂(N^ ∙N^ ))/∂v)=N_v ^ ∙N^ +N^ ∙N_v ^ =0 N_μ ^ ∙N^ =0 ⇋ N_μ ^ ⊥N^ N_u ^ ∙r_u ^ =(Ar_u ^ +Br_v ^ )∙r_u ^ → =Ar_u ^ ∙r_u ^ +Br_u ^ ∙r_v ^ =AE+BF N_u ^ ∙r_v ^ =(Ar_u ^ +Br_v ^ )∙r_v ^ → =Ar_u ^ ∙r_v ^ +Br_v ^ ∙r_v ^ =AF+BG N_v ^ ∙r_u ^ =(Cr_u ^ +Dr_v ^ )∙r_u ^ →=Cr_u ^ ∙r_u ^ +Dr_v ^ ∙r_u ^ =CE+DF N_u ^ ∙r_v ^ =(Cr_u ^ +Dr_v ^ )∙r_v ^ → =Cr_u ^ ∙r_v ^ +Dr_v ^ ∙r_v ^ =CF+DG N_u ^ ∙r_u ^ =−L N_u ^ ∙r_v ^ =−M N_v ^ ∙r_u ^ =−M N_v ^ ∙r_v ^ =−N −L=AE+BF −M=AF+BG −M=CE+DF −N=CF+DG (((−L)),((−M)) )= ((E,F),(F,G) ) ((A),(B) ) (((−M)),((−N)) )= ((E,F),(F,G) ) ((C),(D) ) n_u ^ =((M′F′−G′L′)/(E′G′−F′^2 )) r_u ^ +((L′F′−E′M′)/(E′G′−F′^2 )) r_v ^ n_v ^ =((N′F′−G′M′)/(E′G′−F′^2 )) r_u ^ +((M′F′−E′N′)/(E′G′−F′^2 )) r_v ^

$$\mathrm{WeinGarten}\:\mathrm{Equation} \\ $$$$\:\boldsymbol{\mathrm{r}}_{{u}} \centerdot\boldsymbol{\mathrm{r}}_{{u}} ={E}\:,\:\boldsymbol{\mathrm{r}}_{{u}} \centerdot\boldsymbol{\mathrm{r}}_{{v}} ={F}\:,\:\boldsymbol{\mathrm{r}}_{{v}} \centerdot\boldsymbol{\mathrm{r}}_{{v}} ={G} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat {\boldsymbol{\mathrm{N}}}=\mathrm{1} \\ $$$$\frac{\partial\left(\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat {\boldsymbol{\mathrm{N}}}\right)}{\partial{u}}=\hat {\boldsymbol{\mathrm{N}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{N}}}+\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat {\boldsymbol{\mathrm{N}}}_{{u}} =\mathrm{0} \\ $$$$\frac{\partial\left(\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat {\boldsymbol{\mathrm{N}}}\right)}{\partial{v}}=\hat {\boldsymbol{\mathrm{N}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{N}}}+\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat {\boldsymbol{\mathrm{N}}}_{{v}} =\mathrm{0} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{\mu} \centerdot\hat {\boldsymbol{\mathrm{N}}}=\mathrm{0}\:\leftrightharpoons\:\hat {\boldsymbol{\mathrm{N}}}_{\mu} \bot\hat {\boldsymbol{\mathrm{N}}} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} =\left({A}\hat {\boldsymbol{\mathrm{r}}}_{{u}} +{B}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \right)\centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} \:\rightarrow\:={A}\hat {\boldsymbol{\mathrm{r}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} +{B}\hat {\boldsymbol{\mathrm{r}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} ={AE}+{BF} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} =\left({A}\hat {\boldsymbol{\mathrm{r}}}_{{u}} +{B}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \right)\centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} \:\rightarrow\:={A}\hat {\boldsymbol{\mathrm{r}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} +{B}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} ={AF}+{BG} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} =\left({C}\hat {\boldsymbol{\mathrm{r}}}_{{u}} +{D}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \right)\centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} \:\rightarrow={C}\hat {\boldsymbol{\mathrm{r}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} +{D}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} ={CE}+{DF} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} =\left({C}\hat {\boldsymbol{\mathrm{r}}}_{{u}} +{D}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \right)\centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} \rightarrow\:={C}\hat {\boldsymbol{\mathrm{r}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} +{D}\hat {\boldsymbol{\mathrm{r}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} ={CF}+{DG} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} =−{L} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{u}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} =−{M} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{u}} =−{M} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}_{{v}} \centerdot\hat {\boldsymbol{\mathrm{r}}}_{{v}} =−{N} \\ $$$$−{L}={AE}+{BF}\:\:\:\:\:\:\:\: \\ $$$$−{M}={AF}+{BG} \\ $$$$−{M}={CE}+{DF} \\ $$$$−{N}={CF}+{DG} \\ $$$$\begin{pmatrix}{−{L}}\\{−{M}}\end{pmatrix}=\begin{pmatrix}{{E}}&{{F}}\\{{F}}&{{G}}\end{pmatrix}\begin{pmatrix}{{A}}\\{{B}}\end{pmatrix} \\ $$$$\begin{pmatrix}{−{M}}\\{−{N}}\end{pmatrix}=\begin{pmatrix}{{E}}&{{F}}\\{{F}}&{{G}}\end{pmatrix}\begin{pmatrix}{{C}}\\{{D}}\end{pmatrix} \\ $$$$\hat {\boldsymbol{\mathrm{n}}}_{{u}} =\frac{{M}'{F}'−{G}'{L}'}{{E}'{G}'−{F}'^{\mathrm{2}} }\:\hat {\boldsymbol{\mathrm{r}}}_{{u}} +\frac{{L}'{F}'−{E}'{M}'}{{E}'{G}'−{F}'^{\mathrm{2}} }\:\hat {\boldsymbol{\mathrm{r}}}_{{v}} \\ $$$$\hat {\boldsymbol{\mathrm{n}}}_{{v}} =\frac{{N}'{F}'−{G}'{M}'}{{E}'{G}'−{F}'^{\mathrm{2}} }\:\hat {\boldsymbol{\mathrm{r}}}_{{u}} +\frac{{M}'{F}'−{E}'{N}'}{{E}'{G}'−{F}'^{\mathrm{2}} }\:\hat {\boldsymbol{\mathrm{r}}}_{{v}} \\ $$

Question Number 225767    Answers: 0   Comments: 4

Question Number 225758    Answers: 4   Comments: 1

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Question Number 225740    Answers: 1   Comments: 0

Question Number 225730    Answers: 0   Comments: 0

Question Number 225726    Answers: 1   Comments: 1

Question Number 225716    Answers: 1   Comments: 4

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Question Number 225700    Answers: 0   Comments: 3

Question Number 225703    Answers: 3   Comments: 0

Question Number 225691    Answers: 1   Comments: 0

((−1))^(1/i)

$$\sqrt[{{i}}]{−\mathrm{1}} \\ $$

Question Number 225676    Answers: 0   Comments: 0

Question Number 225666    Answers: 1   Comments: 2

Question Number 225664    Answers: 0   Comments: 0

If a(x)=1 and W_(n=−1) ^(Qw_(fr.) ((1/(x−1))×x)) a′′(ust^n x)=0; What value of Tk(x^2 )? (This is nonstandartmath exercise)

$${If}\:{a}\left({x}\right)=\mathrm{1}\:{and}\:\underset{{n}=−\mathrm{1}} {\overset{{Qw}_{{fr}.} \left(\frac{\mathrm{1}}{{x}−\mathrm{1}}×{x}\right)} {{W}}a}''\left({ust}^{{n}} {x}\right)=\mathrm{0}; \\ $$$${What}\:{value}\:{of}\:{Tk}\left({x}^{\mathrm{2}} \right)? \\ $$$$\left({This}\:{is}\:{nonstandartmath}\:{exercise}\right) \\ $$

Question Number 225661    Answers: 1   Comments: 5

Question Number 225658    Answers: 1   Comments: 0

Question Number 225657    Answers: 0   Comments: 0

Question Number 225656    Answers: 0   Comments: 0

Question Number 225652    Answers: 2   Comments: 0

∫_0 ^(π/2) e^(iπx) dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{i}\pi{x}} \:{dx} \\ $$

Question Number 225651    Answers: 0   Comments: 0

Question Number 225629    Answers: 2   Comments: 0

Question Number 225625    Answers: 0   Comments: 4

quick short Q 2 things are mixed 1)both same volume 2)both same mass density⇒d_v and d_m d_v ? d_m [=,< or>]

$${quick}\:{short}\:{Q} \\ $$$$\mathrm{2}\:{things}\:{are}\:{mixed}\: \\ $$$$\left.\mathrm{1}\right){both}\:{same}\:{volume} \\ $$$$\left.\mathrm{2}\right){both}\:{same}\:{mass} \\ $$$${density}\Rightarrow{d}_{{v}} \:{and}\:{d}_{{m}} \\ $$$${d}_{{v}} \:?\:{d}_{{m}} \left[=,<\:{or}>\right] \\ $$

Question Number 225613    Answers: 7   Comments: 0

Question Number 225610    Answers: 0   Comments: 0

prove Gauss curvature K is intrinsic by showing K=(( determinant (((−(1/2)E_(vv) +F_(uv) −G_(uu) ),((1/2)E_u ),(F_u −(1/2)E_v )),(( F_v −(1/2)G_u ),( E),( F)),(( (1/2)G_v ),( F),( G)))− determinant ((( 0),((1/2)E_v ),((1/2)G_u )),(((1/2)E_v ),( E),( F)),(((1/2)G_v ),( F),( G))))/((EG−F^( 2) )^2 )) E,F,G is First Fundametal form of metric tensor.

$$\mathrm{prove} \\ $$$$\mathrm{Gauss}\:\mathrm{curvature}\:{K}\:\mathrm{is}\:\mathrm{intrinsic}\:\mathrm{by}\:\mathrm{showing} \\ $$$${K}=\frac{\begin{vmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}{E}_{{vv}} +{F}_{{uv}} −{G}_{{uu}} }&{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{u}} }&{{F}_{{u}} −\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }\\{\:\:\:\:\:\:\:\:\:\:\:\:{F}_{{v}} −\frac{\mathrm{1}}{\mathrm{2}}{G}_{{u}} }&{\:\:\:{E}}&{\:\:\:\:\:\:{F}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{G}_{{v}} }&{\:\:\:{F}}&{\:\:\:\:\:{G}}\end{vmatrix}−\begin{vmatrix}{\:\:\:\:\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }&{\frac{\mathrm{1}}{\mathrm{2}}{G}_{{u}} }\\{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }&{\:\:\:\:{E}}&{\:\:\:\:{F}}\\{\frac{\mathrm{1}}{\mathrm{2}}{G}_{{v}} }&{\:\:\:\:\:{F}}&{\:\:\:{G}}\end{vmatrix}}{\left({EG}−{F}^{\:\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${E},{F},{G}\:\mathrm{is}\:\mathrm{First}\:\mathrm{Fundametal}\:\mathrm{form}\:\mathrm{of}\:\mathrm{metric}\:\mathrm{tensor}. \\ $$

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