Question

If $\sqrt{1-x^6}+\sqrt{1-y^6}=a^3.(x^3-y^3)$, prove that $\frac{dy}{dx}=\frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}$

Answer 1

To prove, $\frac{dy}{x}=\frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}$

Putting $x^3=\sin A$ & $y^3=\sin B$, we get

$\sqrt{1-{\sin }^{2}A}+\sqrt{1-{\sin }^{2}B}=a(\sin A-\sin B)$

$\cos A+\cos B=a(\sin A-\sin B)$

$\Rightarrow 2\cos (A+B)\cos (A-B)=2a\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)$

$\Rightarrow \cot \left(\frac{A-B}{2}\right)=a\Rightarrow \frac{A-B}{2}={\cot }^{-1}a$

$\Rightarrow {\sin }^{-1}{x}^3-{\sin }^{-1}{y}^3=2{\cot }^{-1}a$

Differentiating both side, we get

$\frac{1}{\sqrt{1-x^6}}\frac{d}{dx}\left(x^3\right)-\frac{1}{\sqrt{1-y^6}}\frac{d}{dx}\left(y^3\right)=0$

$\Rightarrow \frac{3x^2}{\sqrt{1-x^6}}-\frac{3y^2}{\sqrt{1-y^6}}\times \frac{dy}{dx}=0$

$\frac{3x^2}{\sqrt{1-x^6}}=\frac{3y^2}{\sqrt{1-y^6}}\frac{dy}{dx}$

$\frac{x^2}{y^2}\times \sqrt{\frac{1-y^6}{1-x^6}}=\frac{dy}{dx}$.