Question

If x=rcos⁡θ, y=rsin⁡θ then the value of ∂(x,y)/∂(r,θ) is

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x = r \: cos\theta$

$\rm :\longmapsto\:y = rsin\theta$

Consider,

$\rm :\longmapsto\:\dfrac{\partial x}{\partial r} = cos\theta$

$\rm :\longmapsto\:\dfrac{\partial y}{\partial r} = sin\theta$

$\rm :\longmapsto\:\dfrac{\partial x}{\partial \theta } = - r \: sin\theta$

$\rm :\longmapsto\:\dfrac{\partial y}{\partial \theta } = r \: cos\theta$

Now, we know that

$\rm :\longmapsto\:\dfrac{\partial (x,y)}{\partial (r,\theta )}$

$\rm \: = \: \: \begin{array}{|cc|}\sf \dfrac{\partial x}{\partial r} &\sf \dfrac{\partial x}{\partial \theta } \\ \\ \sf \dfrac{\partial y}{\partial r } &\sf \dfrac{\partial y}{\partial \theta } \\\end{array}$

$\rm \: = \: \: \begin{array}{|cc|}\sf cos\theta &\sf - r \: sin\theta \\ \sf sin\theta &\sf r \: cos\theta \\\end{array}$

$\rm \: = \: \: {rcos}^{2}\theta + {rsin}^{2} \theta$

$\rm \: = \: \: r( {cos}^{2}\theta + {sin}^{2} \theta )$

$\rm \: = \: \: r \times 1$

$\rm \: = \: \: r$

Hence,

$\red{\bf :\longmapsto\:\dfrac{\partial (x,y)}{\partial (r,\theta )} \: = \: r}$

Additional Information :-

Properties of Jacobian

$\red{\bf :\longmapsto\:\dfrac{\partial (x,y)}{\partial (r,\theta )} \: = \: J}$

$\red{\bf :\longmapsto\:\dfrac{\partial (r,\theta )}{\partial (x,y )} \: = \: J'}$

$\red{\bf :\longmapsto\: \: J J' \: = \: 1}$

$\red{\bf :\longmapsto\: \: \dfrac{\partial (x,y)}{\partial (r,\theta )} \times \dfrac{\partial (r,\theta )}{\partial (x,y)} = \: 1}$