(1- X2)1/2 + (1-Y2)1/2 =A (X - Y) FIND DY / DX = (1-Y2 /1-X2) 1/2 - Maths -

Question

(1- X2)1/2 +
(1-Y2)1/2 =A (X - Y) FIND DY / DX =
(1-Y2 /1-X2) 1/2

Ajanta Trivedi · Accepted Answer

the given equation is $\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = A (x - y) f o r r e a l v a l u e s - 1 \leq x \leq 1 a n d - 1 \leq y \leq 1$

where A is a constant.

substitute $x = \sin B a n d y = \sin C$ therefore the equation becomes:

$\sqrt{1 - \sin^{2} B} + \sqrt{1 - \sin^{2} C} = A (\sin B - \sin C) \cos B + \cos C = A (\sin B - \sin C) A = \frac{\cos B + \cos C}{\sin B - \sin C} = \frac{2 \cos \frac{B + C}{2} \cdot \cos \frac{B - C}{2}}{2 \cos \frac{B + C}{2} \cdot \sin \frac{B - C}{2}} = \cot \frac{B - C}{2}$

thus $A = \cot (\frac{\sin^{- 1} x - \sin^{- 1} y}{2})$

now differentiate wrt x,

$0 = - \cos e c^{2} (\frac{\sin^{- 1} x - \sin^{- 1} y}{2}) \cdot \frac{1}{2} (\frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} \frac{d y}{d x})$

since the value of $\cos e c θ \geq 1 a n d \cos e c θ \leq - 1$ it cannot be zero.

therefore

$\frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} \frac{d y}{d x} = 0 \frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$

hope this helps you.

cheers!!!

(1- X2)1/2 + (1-Y2)1/2 =A (X - Y) .FIND DY / DX = (1-Y2 /1-X2) 1/2

(1- X²)^1/2 + (1-Y²)^1/2 =A (X - Y) .FIND DY / DX = (1-Y²/1-X²) ^1/2