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# If Z=F(X.Y). X=R Cos θ, Y=R Sinθ. Prove that ( ∂ Z ∂ X ) 2 + ( ∂ Z ∂ Y ) 2 = ( ∂ Z ∂ R ) 2 + 1 R 2 ( ∂ Z ∂ θ ) 2 - Applied Mathematics 1

ConceptDifferentiation of Implicit Functions

#### 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

#### Solution

= 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

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#### APPEARS IN

Solution If Z=F(X.Y). X=R Cos θ, Y=R Sinθ. Prove that ( ∂ Z ∂ X ) 2 + ( ∂ Z ∂ Y ) 2 = ( ∂ Z ∂ R ) 2 + 1 R 2 ( ∂ Z ∂ θ ) 2 Concept: Differentiation of Implicit Functions.
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