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x-y-z-e-cos-e-sin-0-0-2pi-R-r-x-i-y-j-z-k-a-r-a-r-a-r-a-




Question Number 1858 by 123456 last updated on 16/Oct/15
 [((x(ρ,θ,ξ))),((y(ρ,θ,ξ))),((z(ρ,θ,ξ))) ]= [((ρe^ξ cos θ)),((ρe^ξ sin θ)),(ξ) ] { ((ρ∈[0,+∞))),((θ∈[0,2π))),((ξ∈R)) :}  r(ρ,θ,ξ)=x(ρ,θ,ξ)i+y(ρ,θ,ξ)j+z(ρ,θ,ξ)k  a_ρ =(∂r/∂ρ)  a_θ =(∂r/∂θ)  a_ξ =(∂r/∂ξ)  a_ρ ∙a_ρ +a_θ ∙a_θ +a_ξ ∙a_ξ =???  a_ρ ∙a_θ +a_ρ ∙a_ξ +a_θ ∙a_ξ =???  a_ρ ×a_θ +a_ρ ×a_ξ +a_θ ×a_ξ =???
[x(ρ,θ,ξ)y(ρ,θ,ξ)z(ρ,θ,ξ)]=[ρeξcosθρeξsinθξ]{ρ[0,+)θ[0,2π)ξRr(ρ,θ,ξ)=x(ρ,θ,ξ)i+y(ρ,θ,ξ)j+z(ρ,θ,ξ)kaρ=rρaθ=rθaξ=rξaρaρ+aθaθ+aξaξ=???aρaθ+aρaξ+aθaξ=???aρ×aθ+aρ×aξ+aθ×aξ=???
Answered by 112358 last updated on 16/Oct/15
r(ρ,θ,ξ)=(ρe^ξ cosθ)i+(ρe^ξ sinθ)j+ξk  a_ρ =(∂r/∂ρ)=(∂/∂ρ)(ρe^ξ cosθ)i+(∂/∂ρ)(ρe^ξ sinθ)j+(∂/∂ρ)(ξ)k  a_ρ =(e^ξ cosθ)i+(e^ξ sinθ)j+0k  By the similar process involving  partial differentiation we get  a_θ =(−ρe^ξ sinθ)i+(ρe^ξ cosθ)j+0k  a_ξ =(ρe^ξ cosθ)i+(ρe^ξ sinθ)j+k  For any vector we have x.x=∣x∣^2 .  ∴S_1 = Σa.a=∣a_θ ∣^2 +∣a_ρ ∣^2 +∣a_ξ ∣^2   S_1 =e^(2ξ) (sin^2 θ+cos^2 θ)+2ρ^2 e^(2ξ) (sin^2 θ+cos^2 θ)+1  S_1 =e^(2ξ) (2ρ^2 +1)+1  Let S_2 =a_θ .a_ρ +a_ξ .a_ρ +a_θ .a_ξ   a_ρ .a_θ =−ρe^(2ξ) cosθsinθ+ρe^(2ξ) sinθcosθ=0  a_ξ .a_ρ =ρe^(2ξ) cos^2 θ+ρe^(2ξ) sin^2 θ=ρe^(2ξ)   a_θ .a_ξ =−ρ^2 e^(2ξ) sinθcosθ+ρ^2 e^(2ξ) sinθcosθ=0  ∴ S_2 =0+ρe^(2ξ) +0=ρe^(2ξ)   a_ρ ×a_θ =i(0−0)−j(0−0)+k(e^ξ cosθ×ρe^ξ cosθ−e^ξ sinθ(−ρe^ξ sinθ))  a_ρ ×a_θ =0i+0j+ρe^(2ξ) k  a_ρ ×a_ξ =i(e^ξ sinθ×1−0)−j(1×e^ξ cosθ−0)+k(ρe^(2ξ) cosθsinθ−ρe^(2ξ) cosθsinθ)  a_ρ ×a_ξ =(e^ξ sinθ)i−(e^ξ cosθ)j+0k  a_θ ×a_ξ =i(ρe^ξ cosθ−0)−j(−ρe^ξ sinθ−0)+k(−ρ^2 e^(2ξ) sin^2 θ−ρ^2 e^(2ξ) cos^2 θ)  a_θ ×a_ξ =(ρe^ξ cosθ)i+(ρe^ξ sinθ)j−(ρ^2 e^(2ξ) )k  ∴a_ρ ×a_θ +a_ρ ×a_ξ +a_θ ×a_ξ   =e^ξ (ρcosθ+sinθ)i+e^ξ (ρsinθ−cosθ)j+ρe^(2ξ) (1−ρ)k
r(ρ,θ,ξ)=(ρeξcosθ)i+(ρeξsinθ)j+ξkaρ=rρ=ρ(ρeξcosθ)i+ρ(ρeξsinθ)j+ρ(ξ)kaρ=(eξcosθ)i+(eξsinθ)j+0kBythesimilarprocessinvolvingpartialdifferentiationwegetaθ=(ρeξsinθ)i+(ρeξcosθ)j+0kaξ=(ρeξcosθ)i+(ρeξsinθ)j+kForanyvectorwehavex.x=∣x2.S1=Σa.a=∣aθ2+aρ2+aξ2S1=e2ξ(sin2θ+cos2θ)+2ρ2e2ξ(sin2θ+cos2θ)+1S1=e2ξ(2ρ2+1)+1LetS2=aθ.aρ+aξ.aρ+aθ.aξaρ.aθ=ρe2ξcosθsinθ+ρe2ξsinθcosθ=0aξ.aρ=ρe2ξcos2θ+ρe2ξsin2θ=ρe2ξaθ.aξ=ρ2e2ξsinθcosθ+ρ2e2ξsinθcosθ=0S2=0+ρe2ξ+0=ρe2ξaρ×aθ=i(00)j(00)+k(eξcosθ×ρeξcosθeξsinθ(ρeξsinθ))aρ×aθ=0i+0j+ρe2ξkaρ×aξ=i(eξsinθ×10)j(1×eξcosθ0)+k(ρe2ξcosθsinθρe2ξcosθsinθ)aρ×aξ=(eξsinθ)i(eξcosθ)j+0kaθ×aξ=i(ρeξcosθ0)j(ρeξsinθ0)+k(ρ2e2ξsin2θρ2e2ξcos2θ)aθ×aξ=(ρeξcosθ)i+(ρeξsinθ)j(ρ2e2ξ)kaρ×aθ+aρ×aξ+aθ×aξ=eξ(ρcosθ+sinθ)i+eξ(ρsinθcosθ)j+ρe2ξ(1ρ)k

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