Question

If x²=au + bu, y² =au - bv, prove that (du/dx)y.(dx/du)v=1/2=(dv/dy)x.(dy/dv)u​

Answer 1

Answer:

Please refer to the image attached

Answer 2

Step-by-step explanation:

Given:$x^{2} =au + bu, y^{2} =au - bv$

To prove:$(\frac{2v}{2y})x . (\frac{2y}{2v})u = \frac{-b}{2y} . \frac{-y}{b} = \frac{1}{2}$

For proving the above equations,

Let's assume that $x^{2} = au+bv$ (I) and $y^{2}=au-bv$ (II)

Differentiating equation I w.r.t. 'u' and keeping 'v' constant,

⇒ $\frac{2x^{2} }{2u} = \frac{2(au)}{2u} + 0$

⇒ $\frac{2x^{2} }{2u} \frac{2x}{2u} = a$ ; or $\frac{2x2x}{2u} = a$

∴ $(\frac{2x}{2u})v = \frac{a}{2u}$

⇒ $bv=x^{2} - au$ ; or $bv=au-y^{2}$

∴ $x^{2} -au=au-y^{2}$

Thus, we have  $2au=x^{2} + y^{2}$ (III)

Differentiating equation III w.r.t. 'x' and keeping 'y' constant,

⇒ $\frac{2(2au)}{2u} \frac{2u}{2x} = 2x+0$ ;

⇒ $2a \frac{2u}{2x} = 2v$

⇒ $(\frac{2u}{2x}) y = \frac{x}{a}$;

∴ $(\frac{2u}{2x}) y . \frac{2x}{2u} v= \frac{1}{2}$

Differentiating equation III w.r.t. 'v' and keeping 'u' constant,

⇒ $\frac{2y^{2} }{2v} = \frac{2(au)}{2v} - \frac{2(bv)}{2v}$ ;

⇒ $\frac{2y^{2} }{2y} . \frac{2y}{2v} = 0-b$ ; or $2y\frac{2y}{2v} = -b$;

⇒ $(\frac{2y}{2v})u=\frac{-b}{2y}$

From equations II and I, we have $x^{2} - bv= y^{2} + bv$ ; or $x^{2} - y^{2} =2bv$ (IV)

Differentiating equation IV w.r.t. 'y' and keeping 'x' constant,

⇒ $-2y= \frac{2(2bv)}{2y}$ ; or $-2y=\frac{2(2bv)}{2v} = \frac{2v}{2y}$

⇒ $-2y =2b \frac{2v}{2y}$ or $(\frac{2v}{2y})x = \frac{-y}{v}$

∴ $(\frac{2v}{2y})x . (\frac{2y}{2v})u = \frac{-b}{2y} . \frac{-y}{b} = \frac{1}{2}$

Hence, proved $(\frac{2v}{2y})x . (\frac{2y}{2v})u = \frac{-b}{2y} . \frac{-y}{b} = \frac{1}{2}$.