UPSC 2024 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

Jacobian · Calculus · asked 4× in 13 yrs · Read the full method →

Question

If $u=(x+y)/(1-xy)$ and $v=\tan^{-1}x+\tan^{-1}y$ , find $\partial(u,v)/\partial(x,y)$ . Are $u$ and $v$ functionally related? If yes, find the relationship.

Technique

Compute the Jacobian directly; it vanishes iff there is functional dependence; identify the dependence using the $\tan^{-1}$ addition formula.

Solution

Step 1 — Partial derivatives.

u_x=\frac{(1-xy)\cdot 1-(x+y)(-y)}{(1-xy)^2}=\frac{1+y^2}{(1-xy)^2}.

By symmetry of $u$ in $x,y$ : $u_y=\dfrac{1+x^2}{(1-xy)^2}$ .

v_x=\frac{1}{1+x^2},\qquad v_y=\frac{1}{1+y^2}.

Step 2 — Jacobian.

\frac{\partial(u,v)}{\partial(x,y)}=u_x v_y-u_y v_x=\frac{1+y^2}{(1-xy)^2}\cdot\frac{1}{1+y^2}-\frac{1+x^2}{(1-xy)^2}\cdot\frac{1}{1+x^2}=\frac{1}{(1-xy)^2}-\frac{1}{(1-xy)^2}=0.

Step 3 — Functional relationship.

Since the Jacobian is identically zero on $xy<1$ , $u$ and $v$ are functionally related there. The standard identity (valid when $xy<1$ ) gives

\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\Bigl(\frac{x+y}{1-xy}\Bigr),

so $v=\tan^{-1}u$ , equivalently $u=\tan v$ .

Answer

\boxed{\;\frac{\partial(u,v)}{\partial(x,y)}=0;\quad u\text{ and }v\text{ are related by }v=\tan^{-1}u\;\;(\text{i.e., }u=\tan v).\;}