Jacobian

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2014, 2019, 2021, 2024)
• Priority tier: T3
• Marks (count): 15 (2), 8 (1), 7 (1)
• Average solve time: ~8 min
• Difficulty mix: medium 3, easy 1
• Section: A | Dominant type: computation

Why This Chapter Matters

The Jacobian appears in three distinct roles in UPSC exams, each with its own algorithm. The simplest (compute a Jacobian determinant, 7 marks) is purely mechanical. The hardest (use a Jacobian substitution to evaluate a double integral, 15 marks) requires recognising which substitution transforms the integration region into a rectangle and separates the integrand into Beta functions. The middle archetype (functional dependence via vanishing Jacobian) is conceptual but short. Know all three roles.

Minimum Theory

Definition. For functions $u = u(x,y)$ and $v = v(x,y)$, the Jacobian is $\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix}\partial u/\partial x & \partial u/\partial y\\ \partial v/\partial x & \partial v/\partial y\end{vmatrix} = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}.$

Change of variables. If $(u,v) = T(x,y)$ and $T$ is one-to-one, then $\iint f(x,y)\,dx\,dy = \iint f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv.$

Functional dependence. If $\partial(u,v)/\partial(x,y) \equiv 0$ on an open region, then $u$ and $v$ are functionally dependent — there exists a function $\Phi$ such that $u = \Phi(v)$ (or $v = \Phi(u)$). To identify $\Phi$, evaluate at a convenient special point.

Question Archetypes

compute-jacobian (1 question(s); 2021)

Compute a Jacobian determinant directly

Recognition Cues

• Explicit formulae for $u$ and $v$ in terms of $(r,\theta)$ or other coordinates.
• May simplify by expressing $u,v$ in the new variables before computing partial derivatives.

Solution Template

1. Express $u$ and $v$ in terms of the independent variables (simplify using trig identities if possible).
2. Compute all four partial derivatives.
3. Evaluate the $2\times2$ determinant.

Worked Example

2021 Paper 1, 2021-P1-Q3a-i (7 marks)

$u=x^2+y^2$, $v=x^2-y^2$, $x=r\cos\theta$, $y=r\sin\theta$. Find $\partial(u,v)/\partial(r,\theta)$.

Step 1. $u = r^2$ (depends only on $r$); $v = r^2\cos2\theta$ (using $\cos^2\theta-\sin^2\theta=\cos2\theta$).

Step 2. $\partial u/\partial r=2r$, $\partial u/\partial\theta=0$; $\partial v/\partial r=2r\cos2\theta$, $\partial v/\partial\theta=-2r^2\sin2\theta$.

Step 3. Determinant: $(2r)(-2r^2\sin2\theta)-(0)(2r\cos2\theta) = \boxed{-4r^3\sin2\theta.}$

Common Traps

• Simplify first. Express $u$ and $v$ in $(r,\theta)$ before differentiating; computing $\partial/\partial\theta(x^2+y^2)$ directly is fine but more work.
• One row may be sparse. When $u=r^2$, we have $\partial u/\partial\theta=0$, making the determinant a $2\times1$ structure — use this to skip algebra.

jacobian-change-of-variable (1 question(s); 2014)

Use a Jacobian change of variables to transform a double integral

Recognition Cues

• A specific substitution is given (e.g. $u=x+y$, $v=y/(x+y)$).
• The integration region is a triangle or other non-rectangular shape.
• After substitution, the integrand must factorise and/or the limits become $[0,1]^2$.

Solution Template

1. Invert the substitution to express $(x,y)$ in terms of $(u,v)$.
2. Compute the Jacobian $\partial(x,y)/\partial(u,v)$.
3. Transform the limits: find the image of the original region in $(u,v)$ coordinates.
4. Rewrite the integrand in $(u,v)$; check for factorisation.
5. Evaluate using Beta functions if the integrand has the form $u^a(1-u)^b$.

Worked Example

2014 Paper 1, 2014-P1-Q2c (15 marks)

Using $x+y=u$, $y=uv$, evaluate $\displaystyle\iint\sqrt{xy(1-x-y)}\,dx\,dy$ over the triangle $x\ge0$, $y\ge0$, $x+y\le1$.

Step 1. Inversion: $x=u(1-v)$, $y=uv$.

Step 2. Jacobian: $\partial(x,y)/\partial(u,v) = \begin{vmatrix}1-v&-u\\v&u\end{vmatrix} = u$.

Step 3. Limits: the triangle maps to the unit square $0\le u\le1$, $0\le v\le1$.

Step 4. Transform: $xy = u^2v(1-v),\quad 1-x-y=1-u,\quad \sqrt{xy(1-x-y)}=u\sqrt{v(1-v)(1-u)}.$ Integral: $I=\int_0^1\int_0^1 u\sqrt{v(1-v)(1-u)}\cdot u\,du\,dv=\int_0^1 u^2\sqrt{1-u}\,du\cdot\int_0^1\sqrt{v(1-v)}\,dv.$

Step 5. Beta functions: $B(p,q)=\int_0^1t^{p-1}(1-t)^{q-1}dt = \Gamma(p)\Gamma(q)/\Gamma(p+q)$. $\int_0^1 u^2(1-u)^{1/2}\,du = B(3,3/2) = \frac{\Gamma(3)\Gamma(3/2)}{\Gamma(9/2)} = \frac{2\cdot(\sqrt\pi/2)}{105\sqrt\pi/16} = \frac{16}{105}.$ $\int_0^1 v^{1/2}(1-v)^{1/2}\,dv = B(3/2,3/2) = \frac{(\Gamma(3/2))^2}{\Gamma(3)} = \frac{\pi/4}{2} = \frac{\pi}{8}.$ $\boxed{I = \frac{16}{105}\cdot\frac{\pi}{8} = \frac{2\pi}{105}.}$

Common Traps

• The Jacobian is $u$, not $1$. Don’t forget the $u\,du\,dv$ factor — without it the integral has the wrong value.
• Half-integer Gamma values: $\Gamma(1/2)=\sqrt\pi$, $\Gamma(3/2)=\sqrt\pi/2$, $\Gamma(5/2)=3\sqrt\pi/4$, and so on by the recurrence $\Gamma(n+1)=n\Gamma(n)$.
• Integration region. The triangle $x\ge0$, $y\ge0$, $x+y\le1$ maps to the unit square — verify this by tracing each edge.

functional-dependence (2 question(s); 2019, 2024)

Use a vanishing Jacobian to detect and find a functional relation between $u$ and $v$

Recognition Cues

• Two functions $u(x,y)$ and $v(x,y)$ are given; the question asks whether they are “functionally related.”
• The Jacobian will be computed and found to be identically zero.
• The actual relationship is then identified (usually a standard identity).

Solution Template

1. Compute $\partial(u,v)/\partial(x,y)$ directly.
2. Show the Jacobian is identically zero — this proves functional dependence.
3. Identify $\Phi$: set one variable to a convenient value (e.g. $y=0$) to pin down $u=\Phi(v)$.

Worked Example 1

2024 Paper 1, 2024-P1-Q2b (15 marks)

$u=(x+y)/(1-xy)$, $v=\arctan x+\arctan y$. Find $\partial(u,v)/\partial(x,y)$. Are $u,v$ functionally related?

Step 1. $u_x = (1+y^2)/(1-xy)^2$, $u_y = (1+x^2)/(1-xy)^2$; $v_x = 1/(1+x^2)$, $v_y = 1/(1+y^2)$.

Step 2. Jacobian: $u_xv_y - u_yv_x = \dfrac{1+y^2}{(1-xy)^2}\cdot\dfrac{1}{1+y^2} - \dfrac{1+x^2}{(1-xy)^2}\cdot\dfrac{1}{1+x^2} = \dfrac{1}{(1-xy)^2}-\dfrac{1}{(1-xy)^2} = \boxed{0.}$

Step 3. Functionally dependent. By the $\arctan$ addition formula (valid for $xy<1$): $v = \arctan u$, i.e. $\boxed{u = \tan v.}$

Worked Example 2

2019 Paper 1, 2019-P1-Q4c-ii (8 marks)

If $f'(x)=1/(1+x^2)$ and $f(0)=0$, show $f(x)+f(y)=f\!\left(\dfrac{x+y}{1-xy}\right)$ using the Jacobian.

Step 1. Let $u=f(x)+f(y)$ and $w=(x+y)/(1-xy)$. Then $u_x=1/(1+x^2)$, $w_x=(1+y^2)/(1-xy)^2$.

Step 2. Jacobian $\partial(u,w)/\partial(x,y) = \frac{1}{1+x^2}\cdot\frac{1+x^2}{(1-xy)^2} - \frac{1}{1+y^2}\cdot\frac{1+y^2}{(1-xy)^2} = 0$.

Step 3. Functional dependence: $u=\Phi(w)$. Set $y=0$: $w=x$ and $u=f(x)+f(0)=f(x)$. So $\Phi=f$, giving $f(x)+f(y)=f\!\left(\dfrac{x+y}{1-xy}\right)$. $\blacksquare$

Common Traps

• The Jacobian cancels by design. In functional-dependence problems, the numerators of $u_x$ and $u_y$ are crafted to cancel exactly. If the terms don’t cancel, recheck the partial derivatives.
• Functional dependence gives only the form. The Jacobian proves $u=\Phi(w)$ exists, but doesn’t tell you what $\Phi$ is. You must find $\Phi$ separately by specialising variables — setting $y=0$ is almost always the right move.
• Domain restriction. The $\arctan$ addition formula (and the functional relation) hold only for $xy<1$; state this restriction in the conclusion.

Marks-Aware Writing

7-mark Jacobian computation: show each partial derivative computed, display the $2\times2$ determinant, then the evaluation. Three lines of algebra suffice.

15-mark change-of-variables: the examiner awards marks for (a) the Jacobian correctly computed, (b) the transformed limits (unit square), (c) the separation of the double integral, (d) the Beta function identification, and (e) the numerical answer. Missing the Jacobian factor (forgetting the $u$) or mis-identifying the Beta parameters each costs 3 marks.

8-mark functional dependence: the vanishing of the Jacobian (shown by explicit calculation) and the identification of $\Phi$ via $y=0$ are both required. A proof with only the Jacobian step and no identification of $\Phi$ earns half marks.

Practice Set

• 2021-P1-Q3a-i (7 m) already worked in full above.