Applications Of Derivatives Question 203
Question: If $ \sqrt{(1-x^{6})}+\sqrt{(1-y^{6})}=a^{3}(x^{3}-y^{3}) $ , then $ \frac{dy}{dx}= $
[Roorkee 1994]
Options:
A) $ \frac{x^{2}}{y^{2}}\sqrt{\frac{1-x^{6}}{1-y^{6}}} $
B) $ \frac{y^{2}}{x^{2}}\sqrt{\frac{1-y^{6}}{1-x^{6}}} $
C) $ \frac{x^{2}}{y^{2}}\sqrt{\frac{1-y^{6}}{1-x^{6}}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ x^{3}=\sin \theta ,y^{3}=\sin \varphi $
$ \therefore \sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}(x^{3}-y^{3}) $
therefore $ \cos \theta +\cos \varphi =a^{3}(\sin \theta -\sin \varphi ) $
or $ 2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=2a^{3}\sin \frac{\theta -\varphi }{2}\cos \frac{\theta +\varphi }{2} $
or $ \cos \frac{\theta +\varphi }{2}[ \cos \frac{\theta -\varphi }{2}-a^{3}\sin \frac{\theta -\varphi }{2} ]=0 $
If $ \cos \frac{\theta +\varphi }{2}=0, $ then $ \frac{\theta +\varphi }{2}=\frac{\pi }{2} $
$ \therefore \theta =\pi -\varphi $ or $ \sin \theta =\sin \varphi $ or $ x=y $
But if we put $ x=y $ in the given equation it is not satisfied and
Hence we must have
$ \cos \frac{\theta -\varphi }{2}-a^{3}\sin \frac{\theta -\varphi }{2}=0 $ or $ \cot \frac{\theta -\varphi }{2}=a^{3} $
$ \therefore \theta -\varphi =2{{\cot }^{-1}}a^{3} $ or $ {{\sin }^{-1}}x^{3}-{{\sin }^{-1}}y^{3}=2{{\cot }^{-1}}a^{3} $
Differentiating w.r.t. x, we get $ \frac{3x^{2}}{\sqrt{1-x^{6}}}-\frac{3y^{2}}{\sqrt{1-y^{6}}}\frac{dy}{dx}=0 $
$ \Rightarrow \frac{dy}{dx}=\frac{x^{2}}{y^{2}}\sqrt{\frac{1-y^{6}}{1-x^{6}}} $ .
BETA
