Question

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Answer 1

Question:-

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

Proof:-

➡ This is the required proof. I have also attached images. you can also check it from there.

Given that,

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

Let us assume that,

$\sf {x}^{3} = \cos(p)$

and,

$\sf {y}^{3} = \cos(q)$

So,

$\sf \implies \sqrt{1 - { \cos }^{2}(p)} + \sqrt{1 - { \cos}^{2}(q) } = a( \cos(p) - \cos(q) )$

$\sf \implies \sqrt{ { \sin}^{2}(p)} + \sqrt{ { \sin}^{2}(q)} = a( \cos(p) - \cos(q) )$

$\sf \implies\sin(p) + \sin(q) = a( \cos(p) - \cos(q) )$

We know that,

$\boxed{ \small\sf \sin(x) + \sin(y) = 2 \sin \bigg( \frac{x + y}{2} \bigg) \cos \bigg( \frac{x - y}{2} \bigg) }$

So,

$\sf \implies 2 \sin( \frac{p + q}{2} ) \cos( \frac{p - q}{2} ) = a( \cos(p) - \cos(q) )$

Again, we know that,

$\boxed{ \small\sf \cos(x) - \cos(y) = - 2 \sin \bigg( \frac{x + y}{2} \bigg) \sin \bigg( \frac{x - y}{2} \bigg) }$

So,

$\sf \implies \cancel{2\sin( \frac{p + q}{2} )} \cos( \frac{p - q}{2} ) = \cancel{- 2}a \cancel{\sin( \frac{p + q}{2} ) }\sin( \frac{p - q}{2} )$

$\sf \implies \cos( \frac{p - q}{2} ) = -a \sin( \frac{p - q}{2} )$

$\sf \implies \tan( \frac{p - q}{2} ) = \frac{ - 1}{a}$

$\sf \implies \frac{p - q}{2} = \tan^{ - 1} \big( \frac{ - 1}{a} \big)$

$\sf \implies p - q= 2\tan^{ - 1} \big( \frac{ - 1}{a} \big)$

Now,

$\sf {x}^{3} = \cos(p) \implies p = { \cos }^{ - 1} ( {x}^{3} )$

Also,

$\sf {y}^{3} = \cos(q) \implies q = { \cos }^{ - 1} ( {y}^{3} )$

$\sf \therefore { \cos}^{ - 1} ( {x}^{3} ) - { \cos}^{ - 1} ( {y}^{3} ) = 2 \ { \tan}^{ - 1} \big( \frac{ - 1}{a} \big)$

Differentiate...

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

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

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

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

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

Hence, Proved.

Answer 2

Answer:

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