Question

If \(\begin{array}{l} H = {\tan ^{ - 1}}\frac{x}{y}\\ \end{array}\), x = u + v, y = u - v then \(\begin{array}{l} \frac{{\partial H}}{{\partial V}}\\ \end{array}\)is 
1. \(\frac{u}{{\mathop u\nolimits^2 + \mathop v\nolimits^2 }}\)
2. \(\frac{{ - v}}{{\mathop u\nolimits^2 + \mathop v\nolimits^2 }}\)
3. \(\frac{u}{{\mathop x\nolimits^2 + \mathop y\nolimits^2 }}\)
4. \(\frac{{ - 2v}}{{\mathop x\nolimits^2 + \mathop y\nolimits^2 }}\)

Answer 1

Correct Answer - Option 1 : \(\frac{u}{{\mathop u\nolimits^2 + \mathop v\nolimits^2 }}\)

Concept:

\(\frac{{\partial H}}{{\partial v}} = \frac{{\partial H}}{{\partial t}} \times \frac{{\partial t}}{{\partial v}}\)

Calculation:

Given: 

\(\begin{array}{l} H = {\tan ^{ - 1}}\frac{x}{y}\\ \end{array}\)

x = u + v

y = u - v

substituting values of x and y in H, we get
\(H = {\tan ^{ - 1}}\left[ {\frac{{u + v}}{{u - v}}} \right]\)

let \(t = \frac{{u + v}}{{u - v}}\)------------(1)

\(H = {\tan ^{ - 1}}t\)

differentiating w.r.t 't'

\(\begin{array}{l} \frac{{\partial H}}{{\partial t}} = \frac{1}{{1 + \mathop t\nolimits^2 }}\\ \end{array}\)------------(2)

In equation (1), differentiate 't' w.r.t v

\(\frac{{\partial t}}{{\partial v}} = \frac{{(u - v)(1) - (u + v)(-1)}}{{\mathop {\left( {u - v} \right)}\nolimits^2 }}={\frac{{2u}}{{\mathop {\left( {u - v} \right)}\nolimits^2 }}} \)---------(3)

Multiplying (2) and (3), we get

\( \frac{{\partial H}}{{\partial v}} = \frac{1}{{1 + \mathop t\nolimits^2 }}\left[ {\frac{{2u}}{{\mathop {\left( {u - v} \right)}\nolimits^2 }}} \right]\\ \\\)

By substituting '\(t = \frac{{u + v}}{{u - v}}\)' we get

\(\frac{{\partial H}}{{\partial v}} = \frac{u}{{\mathop u\nolimits^2 + \mathop v\nolimits^2 }}\\ \\ \)