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Question Number 1858 by 123456 last updated on 16/Oct/15

$$\left[\begin{array}{c}x(\rho ,\theta ,\xi )\\ y(\rho ,\theta ,\xi )\\ z(\rho ,\theta ,\xi )\end{array}\right]=\left[\begin{array}{c}\rho e^{\xi }\cos \theta \\ \rho e^{\xi }\sin \theta \\ \xi \end{array}\right]\{\begin{array}{l}\rho \in [0,+\mathrm{\infty })\\ \theta \in [0,2\pi )\\ \xi \in \mathbb{R}\end{array}$$$$\mathit{r}(\rho ,\theta ,\xi )=x(\rho ,\theta ,\xi )\mathit{i}+y(\rho ,\theta ,\xi )\mathit{j}+z(\rho ,\theta ,\xi )\mathit{k}$$$${\mathit{a}}_{\rho }=\frac{\partial \mathit{r}}{\partial \rho }$$$${\mathit{a}}_{\theta }=\frac{\partial \mathit{r}}{\partial \theta }$$$${\mathit{a}}_{\xi }=\frac{\partial \mathit{r}}{\partial \xi }$$$${\mathit{a}}_{\rho }\cdot {\mathit{a}}_{\rho }+{\mathit{a}}_{\theta }\cdot {\mathit{a}}_{\theta }+{\mathit{a}}_{\xi }\cdot {\mathit{a}}_{\xi }=???$$$${\mathit{a}}_{\rho }\cdot {\mathit{a}}_{\theta }+{\mathit{a}}_{\rho }\cdot {\mathit{a}}_{\xi }+{\mathit{a}}_{\theta }\cdot {\mathit{a}}_{\xi }=???$$$${\mathit{a}}_{\rho }\times {\mathit{a}}_{\theta }+{\mathit{a}}_{\rho }\times {\mathit{a}}_{\xi }+{\mathit{a}}_{\theta }\times {\mathit{a}}_{\xi }=???$$

Answered by 112358 last updated on 16/Oct/15

$$\mathit{r}(\rho ,\theta ,\xi )=\left(\rho e^{\xi }cos\theta \right)\mathit{i}+\left(\rho e^{\xi }sin\theta \right)\mathit{j}+\xi \mathit{k}$$$${\mathit{a}}_{\rho }=\frac{\partial \mathit{r}}{\partial \rho }=\frac{\partial }{\partial \rho }\left(\rho e^{\xi }cos\theta \right)\mathit{i}+\frac{\partial }{\partial \rho }\left(\rho e^{\xi }sin\theta \right)\mathit{j}+\frac{\partial }{\partial \rho }\left(\xi \right)\mathit{k}$$$${\mathit{a}}_{\rho }=\left(e^{\xi }cos\theta \right)\mathit{i}+\left(e^{\xi }sin\theta \right)\mathit{j}+0\mathit{k}$$$$Bythesimilarprocessinvolving$$$$partialdifferentiationweget$$$${\mathit{a}}_{\theta }=(-\rho e^{\xi }sin\theta )\mathit{i}+\left(\rho e^{\xi }cos\theta \right)\mathit{j}+0\mathit{k}$$$${\mathit{a}}_{\xi }=\left(\rho e^{\xi }cos\theta \right)\mathit{i}+\left(\rho e^{\xi }sin\theta \right)\mathit{j}+\mathit{k}$$$$Foranyvectorwehave\mathit{x}.\mathit{x}=\mid \mathit{x}{\mid }^2.$$$$\therefore S_1=\mathrm{\Sigma }\mathit{a}.\mathit{a}=\mid {\mathit{a}}_{\theta }{\mid }^2+\mid {\mathit{a}}_{\rho }{\mid }^2+\mid {\mathit{a}}_{\xi }{\mid }^2$$$$S_1=e^{2\xi }({sin}^2\theta +{cos}^2\theta )+2{\rho }^2{e}^{2\xi }({sin}^2\theta +{cos}^2\theta )+1$$$$S_1=e^{2\xi }(2{\rho }^2+1)+1$$$$Let{S}_2={\mathit{a}}_{\theta }.{\mathit{a}}_{\rho }+{\mathit{a}}_{\xi }.{\mathit{a}}_{\rho }+{\mathit{a}}_{\theta }.{\mathit{a}}_{\xi }$$$${\mathit{a}}_{\rho }.{\mathit{a}}_{\theta }=-\rho e^{2\xi }cos\theta sin\theta +\rho e^{2\xi }sin\theta cos\theta =0$$$${\mathit{a}}_{\xi }.{\mathit{a}}_{\rho }=\rho e^{2\xi }{cos}^2\theta +\rho e^{2\xi }{sin}^2\theta =\rho e^{2\xi }$$$${\mathit{a}}_{\theta }.{\mathit{a}}_{\xi }=-{\rho }^2{e}^{2\xi }sin\theta cos\theta +{\rho }^2{e}^{2\xi }sin\theta cos\theta =0$$$$\therefore S_2=0+\rho e^{2\xi }+0=\rho e^{2\xi }$$$${\mathit{a}}_{\rho }\times {\mathit{a}}_{\theta }=\mathit{i}(0-0)-\mathit{j}(0-0)+\mathit{k}(e^{\xi }cos\theta \times \rho e^{\xi }cos\theta -e^{\xi }sin\theta (-\rho e^{\xi }sin\theta ))$$$${\mathit{a}}_{\rho }\times {\mathit{a}}_{\theta }=0\mathit{i}+0\mathit{j}+\rho e^{2\xi }\mathit{k}$$$${\mathit{a}}_{\rho }\times {\mathit{a}}_{\xi }=\mathit{i}(e^{\xi }sin\theta \times 1-0)-\mathit{j}(1\times e^{\xi }cos\theta -0)+\mathit{k}(\rho e^{2\xi }cos\theta sin\theta -\rho e^{2\xi }cos\theta sin\theta )$$$${\mathit{a}}_{\rho }\times {\mathit{a}}_{\xi }=\left(e^{\xi }sin\theta \right)\mathit{i}-\left(e^{\xi }cos\theta \right)\mathit{j}+0\mathit{k}$$$${\mathit{a}}_{\theta }\times {\mathit{a}}_{\xi }=\mathit{i}(\rho e^{\xi }cos\theta -0)-\mathit{j}(-\rho e^{\xi }sin\theta -0)+\mathit{k}(-{\rho }^2{e}^{2\xi }{sin}^2\theta -{\rho }^2{e}^{2\xi }{cos}^2\theta )$$$${\mathit{a}}_{\theta }\times {\mathit{a}}_{\xi }=\left(\rho e^{\xi }cos\theta \right)\mathit{i}+\left(\rho e^{\xi }sin\theta \right)\mathit{j}-\left({\rho }^2{e}^{2\xi }\right)\mathit{k}$$$$\therefore {\mathit{a}}_{\rho }\times {\mathit{a}}_{\theta }+{\mathit{a}}_{\rho }\times {\mathit{a}}_{\xi }+{\mathit{a}}_{\theta }\times {\mathit{a}}_{\xi }$$$$=e^{\xi }(\rho cos\theta +sin\theta )\mathit{i}+e^{\xi }(\rho sin\theta -cos\theta )\mathit{j}+\rho e^{2\xi }(1-\rho )\mathit{k}$$

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