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

Given,

$\sqrt{1-x^6}+\sqrt{1-y^6}=a^3(x^3-y^3)$

substitute $x^3=\sin a,y^3=\sin b$

$\sqrt{1-{\sin }^{3}a}+\sqrt{1-{\sin }^{3}b}=a^3(\sin a-\sin b)$

$\cos a+\cos b=a^3(\sin a-\sin b)$

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

$\cot \left(\frac{a-b}{2}\right)=a$

$\frac{a-b}{2}={\cot }^{-1}a$

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

differentiating on both sides, we get,

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

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

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

Hence proved.