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If $u=x^2-y^2 \; , \; v=2xy \; and \; z=f(u,v)$ prove the following

$(\dfrac{\partial z}{\partial x})^2 + (\dfrac{\partial z}{\partial y})^2 \;=\; 4\sqrt{u^2+v^2} \bigg[ (\dfrac{\partial z}{\partial u})^2 + (\dfrac{\partial z}{\partial v})^2 \bigg] \ \; \ \; \ \; \$

partial differentiation • 3.2k  views
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u$=x^2-y^2 \; \; \; ∴ \dfrac{∂u}{∂x}=2x \; \; and \; \; \dfrac{∂u}{∂y} =-2y \; \; \; …(i) \\ \; \\ \; \\ v=2xy \; \; \; ∴ \dfrac{∂v}{∂x} =2y \; \; and \; \; \dfrac{∂v}{∂y}=2x \; \; \; …(ii) \\ \; \\ \; \\ \; \\ \dfrac{\partial z}{\partial x} \;=\; \dfrac{\partial z}{\partial u} . \dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v} . \dfrac{\partial v}{\partial x} \;=\; (2x)\dfrac{\partial z}{\partial u} + (2y) \dfrac{\partial z}{\partial v} \; \; \; \ldots (iii) \; \; \; from \; (i) \; and \; (ii) \\ \; \\ \; \\ \; \\ \dfrac{\partial z}{\partial y} \;=\; \dfrac{\partial z}{\partial u} . \dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v} . \dfrac{\partial v}{\partial y} \;=\; (-2y)\dfrac{\partial z}{\partial u} + (2x) \dfrac{\partial z}{\partial v} \; \; \; \ldots (iv) \; \; \; from \; (i) \; and \; (ii)$

Squaring equation (iii) and (iv),

$\therefore (\dfrac{\partial z}{\partial x})^2 \;=\; 4x^2 (\dfrac{\partial z}{\partial u})^2 + 8xy (\dfrac{\partial z}{\partial u}) (\dfrac{\partial z}{\partial v}) + 4y^2 (\dfrac{\partial z}{\partial v})^2 \; \; \; \; \ldots (v) \; \; and \\ \; \\ \; \\ (\dfrac{\partial z}{\partial y})^2 \;=\; 4y^2 (\dfrac{\partial z}{\partial u})^2 - 8xy (\dfrac{\partial z}{\partial u}) (\dfrac{\partial z}{\partial v}) + 4x^2 (\dfrac{\partial z}{\partial v})^2 \; \; \; \; \ldots (vi)$

$\therefore (\dfrac{\partial z}{\partial x})^2 + (\dfrac{\partial z}{\partial y})^2 \;=\; 4(x^2+y^2) (\dfrac{\partial z}{\partial u})^2 + 4(x^2+y^2) (\dfrac{\partial z}{\partial v})^2 \;=\; 4(x^2+y^2) \bigg[ (\dfrac{\partial z}{\partial u})^2 + (\dfrac{\partial z}{\partial v})^2 \bigg] \; \; \; \; \ldots (vii)$

Now, squaring and adding equations (i) and (ii),

$u^2+v^2=x^4-2x^2 y^2+y^4+4x^2 y^2=x^4+2x^2 y^2+y^4=(x^2+y^2 )^21 \\ \; \\ \; \\ ∴x^2+y^2= \sqrt{u^2+v^2 }$

Substituting this value in equation (vii),

$(\dfrac{\partial z}{\partial x})^2 + (\dfrac{\partial z}{\partial y})^2 \;=\; 4\sqrt{u^2+v^2} \bigg[ (\dfrac{\partial z}{\partial u})^2 + (\dfrac{\partial z}{\partial v})^2 \bigg]$

Hence Proved.

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