written 7.8 years ago by
teamques10
★ 64k
|
•
modified 7.8 years ago
|
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)
$
Adding (v) and (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.