If sin \(u = \frac{x^3 + y^3}{\sqrt{x} +\sqrt{y}}\), then xux + yuy will be equal to:

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  1. \(\frac{5}{2} tan \ u\)
  2. \(\frac{3}{2} tan \ u\)
  3. \(\frac{1}{2} cot \ u\)
  4. \(\frac{3}{2} cot \ u\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{5}{2} tan \ u\)
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Detailed Solution

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Explanation:

Given:

\(\sin u = \frac{{{x^3} + {y^3}}}{{\sqrt x + \sqrt y }}\)

\(x{u_x} + y{u_y} = ?\) 

Now,

u = f(x, y) but it is not homogenous but f(u) is homogenous of degree n

∴ \(x\frac{{dy}}{{dx}} + y\frac{{du}}{{dy}} = n\frac{{f\left( u \right)}}{{f'\left( u \right)}}\)

f(nu) = \(\frac{{{(nx)^3} + {(ny)^3}}}{{\sqrt nx + \sqrt ny }} = n^{5/2}.\frac{{{x^3} + {y^3}}}{{\sqrt x + \sqrt y }}\)

⇒  f(nu) = n5/2 f(u);

Therefore, homogeneous of degree 5/2;

∴ \(x\frac{{du}}{{dx}} + y\frac{{du}}{{dy}} = \frac{5}{2} \times \frac{{\sin u}}{{\cos u}}\)

\(x{u_x} + y{u_y} = \frac{5}{2}\tan u\)
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