Question

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

• A. $\begin{array}{rcl}& & \frac{x^2\sqrt{1-x^6}}{y^2\sqrt{1-y^6}}\end{array}$
• B. $\begin{array}{rcl}\frac{dy}{dx}& =& \frac{y^2\sqrt{1-y^6}}{x^2\sqrt{1-x^6}}\end{array}$
• C. $\begin{array}{rcl}\frac{dy}{dx}& =& \frac{x^2\sqrt{1-y^6}}{y^2\sqrt{1-x^6}}\end{array}$
• D. None of these

Answer 1

The correct option is C

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{x^2\sqrt{1-y^6}}{y^2\sqrt{1-x^6}}\end{array}$

Explanation for the correct option:

Step 1: Change the given terms into other parameters.

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

Put$\begin{array}{rcl}{x}^3& =& \sin \left(\theta \right)\&y^3=\sin \left(\phi \right)\end{array}$

$\begin{array}{rcl}\therefore \sqrt{1-{\sin }^2\left(\theta \right)}+\sqrt{1-{\sin }^2\left(\phi \right)}& =& a^3(\sin \left(\theta \right)-\sin \left(\phi \right))\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\cos \left(\theta \right)+\cos \left(\varphi \right)& =& a^3\left[2\cos \left(\frac{\theta +\phi }{2}\right)\sin \left(\frac{\theta -\phi }{2}\right)\right]\end{array}$

$\Rightarrow$ $\begin{array}{rcl}2\cos \left(\frac{\theta +\phi }{2}\right)\cos \left(\frac{\theta -\phi }{2}\right)& =& a^3\left[2\cos \left(\frac{\theta +\phi }{2}\right)\sin \left(\frac{\theta -\phi }{2}\right)\right]\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\cot \left(\frac{\theta -\phi }{2}\right)& =& a^3\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{\theta -\phi }{2}& =& {\cot }^{-1}\left(a^3\right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\sin }^{-1}\left(x^3\right)-{\sin }^{-1}\left(y^3\right)& =& 2{\cot }^{-1}\left(a^3\right).......\left(1\right)\end{array}$

Step 2: Differentiating both sides of the equation$\left(1\right)$ with respect to$\text{'}x\text{'}$.

$\begin{array}{rcl}\frac{d}{dx}({\sin }^{-1}\left(x^3\right)-{\sin }^{-1}\left(y^3\right))& =& \frac{d}{dx}\left(2{\cot }^{-1}\left(a^3\right)\right)\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{1}{\sqrt{1-x^6}}.3x^2-\frac{1}{\sqrt{1-y^6}}.3y^2.\frac{dy}{dx}& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{x^2\sqrt{1-y^6}}{y^2\sqrt{1-x^6}}& =& \frac{dy}{dx}\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}& \mathbf{=}& \frac{{\mathbf{x}}^{\mathbf{2}}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{6}}}}{{\mathbf{y}}^{\mathbf{2}}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{6}}}}\end{array}$

Hence, the correct option is$\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{.}$.