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Question Number 4922 by 123456 last updated on 22/Mar/16

 { ((x(ρ,θ,ψ)=ρ cos θ+ψ sin θ)),((y(ρ,θ,ψ)=ρ sin θ+ψ cos θ)),((z(ρ,θ,ψ)=ψ sin θ)) :}  r(ρ,θ,ψ)=x(ρ,θ,ψ) e_x +y(ρ,θ,ψ) e_y +z(ρ,θ,ψ) e_z   (∂r/∂ρ)=?  (∂r/∂θ)=?  (∂r/∂ψ)=?

$$\begin{cases}{{x}\left(\rho,\theta,\psi\right)=\rho\:\mathrm{cos}\:\theta+\psi\:\mathrm{sin}\:\theta}\\{{y}\left(\rho,\theta,\psi\right)=\rho\:\mathrm{sin}\:\theta+\psi\:\mathrm{cos}\:\theta}\\{{z}\left(\rho,\theta,\psi\right)=\psi\:\mathrm{sin}\:\theta}\end{cases} \\ $$$$\boldsymbol{{r}}\left(\rho,\theta,\psi\right)={x}\left(\rho,\theta,\psi\right)\:\boldsymbol{{e}}_{{x}} +{y}\left(\rho,\theta,\psi\right)\:\boldsymbol{{e}}_{{y}} +{z}\left(\rho,\theta,\psi\right)\:\boldsymbol{{e}}_{{z}} \\ $$$$\frac{\partial\boldsymbol{{r}}}{\partial\rho}=? \\ $$$$\frac{\partial\boldsymbol{{r}}}{\partial\theta}=? \\ $$$$\frac{\partial\boldsymbol{{r}}}{\partial\psi}=? \\ $$

Commented by prakash jain last updated on 22/Mar/16

(∂x/∂ρ)=cos θ,(∂y/∂ρ)=sin θ,(∂z/∂ρ)=0  (∂x/∂θ)=−ρsin θ+ψcos θ,(∂y/∂θ)=ρcos θ−ψsin θ,(∂z/∂θ)=ψcos θ  (∂x/∂ψ)=sin θ,(∂y/∂θ)=ψcos θ,(∂z/∂θ)=sin θ

$$\frac{\partial{x}}{\partial\rho}=\mathrm{cos}\:\theta,\frac{\partial{y}}{\partial\rho}=\mathrm{sin}\:\theta,\frac{\partial{z}}{\partial\rho}=\mathrm{0} \\ $$$$\frac{\partial{x}}{\partial\theta}=−\rho\mathrm{sin}\:\theta+\psi\mathrm{cos}\:\theta,\frac{\partial{y}}{\partial\theta}=\rho\mathrm{cos}\:\theta−\psi\mathrm{sin}\:\theta,\frac{\partial{z}}{\partial\theta}=\psi\mathrm{cos}\:\theta \\ $$$$\frac{\partial{x}}{\partial\psi}=\mathrm{sin}\:\theta,\frac{\partial{y}}{\partial\theta}=\psi\mathrm{cos}\:\theta,\frac{\partial{z}}{\partial\theta}=\mathrm{sin}\:\theta \\ $$

Commented by prakash jain last updated on 24/Mar/16

I thin e_x ,e_y  are unit vectors.

$$\mathrm{I}\:\mathrm{thin}\:\boldsymbol{{e}}_{\boldsymbol{{x}}} ,\boldsymbol{{e}}_{\boldsymbol{{y}}} \:\mathrm{are}\:\mathrm{unit}\:\mathrm{vectors}. \\ $$

Commented by 123456 last updated on 24/Mar/16

unit vetor in direction of x,y and z

$$\mathrm{unit}\:\mathrm{vetor}\:\mathrm{in}\:\mathrm{direction}\:\mathrm{of}\:{x},{y}\:\mathrm{and}\:{z} \\ $$

Commented by Yozzii last updated on 23/Mar/16

What exactly are e_x ,e_y  and e_z ?  Are they vectors such that r= (((x(ρ,φ,ψ))),((y(ρ,φ,ψ))),((z(ρ,φ,ψ))) ) ?

$${What}\:{exactly}\:{are}\:\boldsymbol{{e}}_{{x}} ,\boldsymbol{{e}}_{{y}} \:{and}\:\boldsymbol{{e}}_{{z}} ? \\ $$$${Are}\:{they}\:{vectors}\:{such}\:{that}\:\boldsymbol{{r}}=\begin{pmatrix}{{x}\left(\rho,\phi,\psi\right)}\\{{y}\left(\rho,\phi,\psi\right)}\\{{z}\left(\rho,\phi,\psi\right)}\end{pmatrix}\:? \\ $$

Commented by Yozzii last updated on 24/Mar/16

Then, you can combine Prakash′s   result and the vector form of r= ((a),(b),(c) )  to get (∂r/∂x_i )= (((∂a/∂x_i )),((∂b/∂x_i )),((∂c/∂x_i )) )     where x_i   is the i−th variable for a,b and c  being functions of i∈N variables.

$${Then},\:{you}\:{can}\:{combine}\:{Prakash}'{s}\: \\ $$$${result}\:{and}\:{the}\:{vector}\:{form}\:{of}\:\boldsymbol{{r}}=\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix} \\ $$$${to}\:{get}\:\frac{\partial\boldsymbol{{r}}}{\partial{x}_{{i}} }=\begin{pmatrix}{\frac{\partial{a}}{\partial{x}_{{i}} }}\\{\frac{\partial{b}}{\partial{x}_{{i}} }}\\{\frac{\partial{c}}{\partial{x}_{{i}} }}\end{pmatrix}\:\:\:\:\:{where}\:{x}_{{i}} \\ $$$${is}\:{the}\:{i}−{th}\:{variable}\:{for}\:{a},{b}\:{and}\:{c} \\ $$$${being}\:{functions}\:{of}\:{i}\in\mathbb{N}\:{variables}. \\ $$

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