If sqrt{1 - x^6} + sqrt{1 - y^6} = a^3(x^3 - | vedclass

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(C) Let $x^3 = \sin 	heta$ and $y^3 = \sin \phi$.Then the equation becomes $\sqrt{1 - \sin^2 	heta} + \sqrt{1 - \sin^2 \phi} = a^3(\sin 	heta - \si

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $x^3 = \sin 	heta$ and $y^3 = \sin \phi$.Then the equation becomes $\sqrt{1 - \sin^2 	heta} + \sqrt{1 - \sin^2 \phi} = a^3(\sin 	heta - \sin \phi)$.This simplifies to $\cos 	heta + \cos \phi = a^3(\sin 	heta - \sin \phi)$.Using trigonometric identities,$2 \cos \frac{	heta + \phi}{2} \cos \frac{	heta - \phi}{2} = 2 a^3 \sin \frac{	heta - \phi}{2} \cos \frac{	heta + \phi}{2}$.This implies $\cos \frac{	heta + \phi}{2} [\cos \frac{	heta - \phi}{2} - a^3 \sin \frac{	heta - \phi}{2}] = 0$.If $\cos \frac{	heta + \phi}{2} = 0$,then $	heta + \phi = \pi$,which leads to $x^3 = \sin 	heta = \sin(\pi - \phi) = \sin \phi = y^3$,so $x = y$. Substituting $x = y$ into the original equation gives $2\sqrt{1-x^6} = 0$,which is not true for all $x$.Thus,we must have $\cos \frac{	heta - \phi}{2} - a^3 \sin \frac{	heta - \phi}{2} = 0$,or $\cot \frac{	heta - \phi}{2} = a^3$.This means $	heta - \phi = 2 \cot^{-1}(a^3)$,or $\sin^{-1}(x^3) - \sin^{-1}(y^3) = 2 \cot^{-1}(a^3)$.Differentiating both sides with respect to $x$:$\frac{3x^2}{\sqrt{1 - x^6}} - \frac{3y^2}{\sqrt{1 - y^6}} \frac{dy}{dx} = 0$.Solving for $\frac{dy}{dx}$,we get $\frac{dy}{dx} = \frac{x^2}{y^2} \sqrt{\frac{1 - y^6}{1 - x^6}}$.

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

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