Question

If Z=f(x.y). x=r cos θ, y=r sinθ. prove that `((delz)/(delx))^2+((delz)/(dely))^2=((delz)/(delr))^2+1/r^2((delz)/(delθ))^2`

Answer 1

= r cos θ and y = r sin θ                                  …(1)
Differentiating partially w.r.t. ‘θ’.`(delx)/(delθ)=-rsinθ; (dely)/(delθ)=rcosθ`                    …(2) 

Differentiating partially w.r.t.’ r’,`(delx)/(delr)=cosθ; (dely)/(delr)=sinθ`                  …(3) 

Now, z → x, y →r , θ 

By Chain Rule, `(delz)/(delr)=(delz)/(delx)xx(delx)/(delr)+(delz)/(dely)xx(dely)/(delr)`

∴`(delz)/(delr)=cosθ (delz)/(delx)+sinθ(delz)/(dely)`  (From 3) …(4)

Similarly, By Chain Rule,`(delz)/(delθ)=(delz)/(delx)xx(delx)/(delθ)+(delz)/(dely)xx(dely)/(delθ)`

∴`(delz)/(delθ)=rsinθ (delz)/(delx)+rcosθ (delz)/(dely)`  (From 2) …(5)

RHS = `((delZ)/(delr))^2+1/r^2((delz)/(delθ))^2 `

= `(cosθ (delz)/(delx)+sinθ (delz)/(dely))^2 1/r^2 (-rsinθ(delz)/(delx)+rcosθ(delz)/(dely))^2` (From 4 & 5) 

=`cos^2θ ((delz)/(delx))^2+2sinθ (delz)/(delx).cosθ (delz)/(dely) +sin^2θ ((delz)/(dely))^2+sin^2θ ((delz)/(delx))^2-2sinθ(delz)/(delx) cosθ (delz)/(dely)+cos^2θ ((delz)/(dely))^2` 

=`((delz)/(delx))^2(cos^2θ +sin^2θ )+((delz)/(dely))^2 (cos^2θ +sin^2θ )`

= `((delz)/(delx))^2+((delz)/(dely))^2`

= LHS 

Hence, `((delz)/(delx))^2+((delz)/(dely))^2=((delz)/(delr))^2+1/r^2 (delz)/(delθ)^2`