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UPSC 2026 Maths Optional Paper 1 Q8c-i — Step-by-Step Solution 10 marks · Section B
Vector identities (curl of grad, div of curl, product rules) · Vector Analysis · asked 4× in 14 yrs · Read the full method →
Question
For a scalar point function ϕ \phi ϕ F ⃗ \vec{F} F
∇ × ( ϕ F ⃗ ) = ( ∇ ϕ ) × F ⃗ + ϕ ( ∇ × F ⃗ ) . \nabla \times (\phi\vec{F}) = (\nabla\phi) \times \vec{F} + \phi(\nabla \times \vec{F}). ∇ × ( ϕ F ) = ( ∇ ϕ ) × F + ϕ ( ∇ × F ) . If r ⃗ = x i ^ + y j ^ + z k ^ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} r = x i ^ + y j ^ + z k ^ r = ∣ r ⃗ ∣ r = |\vec{r}| r = ∣ r ∣ ϕ = 1 r \phi = \dfrac{1}{r} ϕ = r 1 F ⃗ = r ⃗ \vec{F} = \vec{r} F = r
Technique
Prove the identity by expanding one Cartesian component in full and splitting each product derivative by the Leibniz rule; the two halves reassemble as the two terms on the right. (The suffix/Levi-Civita form gives the same proof in one line and is worth adding.) For the verification, compute both sides independently for ϕ F ⃗ = r ⃗ / r \phi\vec F = \vec r/r ϕ F = r / r
Solution
Step 1 — Statement in components
Write F ⃗ = F 1 i ^ + F 2 j ^ + F 3 k ^ \vec F = F_1\hat i + F_2\hat j + F_3\hat k F = F 1 i ^ + F 2 j ^ + F 3 k ^ ϕ , F 1 , F 2 , F 3 \phi, F_1,F_2,F_3 ϕ , F 1 , F 2 , F 3
∇ × G ⃗ = ∣ i ^ j ^ k ^ ∂ x ∂ y ∂ z G 1 G 2 G 3 ∣ . \nabla\times\vec G = \begin{vmatrix}\hat i & \hat j & \hat k\\[2pt] \partial_x & \partial_y & \partial_z\\[2pt] G_1 & G_2 & G_3\end{vmatrix}. ∇ × G = i ^ ∂ x G 1 j ^ ∂ y G 2 k ^ ∂ z G 3 . Step 2 — The i ^ \hat i i ^
[ ∇ × ( ϕ F ⃗ ) ] 1 = ∂ ∂ y ( ϕ F 3 ) − ∂ ∂ z ( ϕ F 2 ) . \bigl[\nabla\times(\phi\vec F)\bigr]_1 = \frac{\partial}{\partial y}\bigl(\phi F_3\bigr) - \frac{\partial}{\partial z}\bigl(\phi F_2\bigr). [ ∇ × ( ϕ F ) ] 1 = ∂ y ∂ ( ϕ F 3 ) − ∂ z ∂ ( ϕ F 2 ) . Apply the product rule to each term:
= ( ∂ ϕ ∂ y F 3 + ϕ ∂ F 3 ∂ y ) − ( ∂ ϕ ∂ z F 2 + ϕ ∂ F 2 ∂ z ) . = \left(\frac{\partial\phi}{\partial y}F_3 + \phi\frac{\partial F_3}{\partial y}\right) - \left(\frac{\partial\phi}{\partial z}F_2 + \phi\frac{\partial F_2}{\partial z}\right). = ( ∂ y ∂ ϕ F 3 + ϕ ∂ y ∂ F 3 ) − ( ∂ z ∂ ϕ F 2 + ϕ ∂ z ∂ F 2 ) . Regroup into “derivative on ϕ \phi ϕ F ⃗ \vec F F
= ( ∂ ϕ ∂ y F 3 − ∂ ϕ ∂ z F 2 ) ⏟ = [ ( ∇ ϕ ) × F ⃗ ] 1 + ϕ ( ∂ F 3 ∂ y − ∂ F 2 ∂ z ) ⏟ = [ ∇ × F ⃗ ] 1 . = \underbrace{\left(\frac{\partial\phi}{\partial y}F_3 - \frac{\partial\phi}{\partial z}F_2\right)}_{=\;[(\nabla\phi)\times\vec F]_1} \;+\; \phi\underbrace{\left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}\right)}_{=\;[\nabla\times\vec F]_1}. = = [( ∇ ϕ ) × F ] 1 ( ∂ y ∂ ϕ F 3 − ∂ z ∂ ϕ F 2 ) + ϕ = [ ∇ × F ] 1 ( ∂ y ∂ F 3 − ∂ z ∂ F 2 ) . The first bracket is precisely the i ^ \hat i i ^ ( ∇ ϕ ) × F ⃗ (\nabla\phi)\times\vec F ( ∇ ϕ ) × F
( ∇ ϕ ) × F ⃗ = ∣ i ^ j ^ k ^ ϕ x ϕ y ϕ z F 1 F 2 F 3 ∣ , (\nabla\phi)\times\vec F = \begin{vmatrix}\hat i & \hat j & \hat k\\[2pt] \phi_x & \phi_y & \phi_z\\[2pt] F_1 & F_2 & F_3\end{vmatrix}, ( ∇ ϕ ) × F = i ^ ϕ x F 1 j ^ ϕ y F 2 k ^ ϕ z F 3 , whose i ^ \hat i i ^ ϕ y F 3 − ϕ z F 2 \phi_y F_3 - \phi_z F_2 ϕ y F 3 − ϕ z F 2
[ ∇ × ( ϕ F ⃗ ) ] 1 = [ ( ∇ ϕ ) × F ⃗ ] 1 + ϕ [ ∇ × F ⃗ ] 1 . \bigl[\nabla\times(\phi\vec F)\bigr]_1 = \bigl[(\nabla\phi)\times\vec F\bigr]_1 + \phi\bigl[\nabla\times\vec F\bigr]_1 . [ ∇ × ( ϕ F ) ] 1 = [ ( ∇ ϕ ) × F ] 1 + ϕ [ ∇ × F ] 1 . Step 3 — The other two components, and the conclusion
The j ^ \hat j j ^ k ^ \hat k k ^ x → y → z → x x\to y\to z\to x x → y → z → x F 1 → F 2 → F 3 → F 1 F_1\to F_2\to F_3\to F_1 F 1 → F 2 → F 3 → F 1
[ ∇ × ( ϕ F ⃗ ) ] 2 = ( ϕ z F 1 − ϕ x F 3 ) + ϕ ( ∂ z F 1 − ∂ x F 3 ) , \bigl[\nabla\times(\phi\vec F)\bigr]_2 = \bigl(\phi_z F_1 - \phi_x F_3\bigr) + \phi\bigl(\partial_z F_1 - \partial_x F_3\bigr), [ ∇ × ( ϕ F ) ] 2 = ( ϕ z F 1 − ϕ x F 3 ) + ϕ ( ∂ z F 1 − ∂ x F 3 ) , [ ∇ × ( ϕ F ⃗ ) ] 3 = ( ϕ x F 2 − ϕ y F 1 ) + ϕ ( ∂ x F 2 − ∂ y F 1 ) . \bigl[\nabla\times(\phi\vec F)\bigr]_3 = \bigl(\phi_x F_2 - \phi_y F_1\bigr) + \phi\bigl(\partial_x F_2 - \partial_y F_1\bigr). [ ∇ × ( ϕ F ) ] 3 = ( ϕ x F 2 − ϕ y F 1 ) + ϕ ( ∂ x F 2 − ∂ y F 1 ) . All three components agree, therefore
∇ × ( ϕ F ⃗ ) = ( ∇ ϕ ) × F ⃗ + ϕ ( ∇ × F ⃗ ) . ■ \nabla\times(\phi\vec F) = (\nabla\phi)\times\vec F + \phi\,(\nabla\times\vec F). \qquad\blacksquare ∇ × ( ϕ F ) = ( ∇ ϕ ) × F + ϕ ( ∇ × F ) . ■ Suffix form (same proof, one line). With the summation convention and the alternating symbol ε i j k \varepsilon_{ijk} ε ij k
[ ∇ × ( ϕ F ⃗ ) ] i = ε i j k ∂ j ( ϕ F k ) = ε i j k ( ∂ j ϕ ) F k + ϕ ε i j k ∂ j F k = [ ( ∇ ϕ ) × F ⃗ ] i + ϕ [ ∇ × F ⃗ ] i . \bigl[\nabla\times(\phi\vec F)\bigr]_i = \varepsilon_{ijk}\,\partial_j(\phi F_k) = \varepsilon_{ijk}(\partial_j\phi)F_k + \phi\,\varepsilon_{ijk}\partial_j F_k = \bigl[(\nabla\phi)\times\vec F\bigr]_i + \phi\bigl[\nabla\times\vec F\bigr]_i. [ ∇ × ( ϕ F ) ] i = ε ij k ∂ j ( ϕ F k ) = ε ij k ( ∂ j ϕ ) F k + ϕ ε ij k ∂ j F k = [ ( ∇ ϕ ) × F ] i + ϕ [ ∇ × F ] i . The whole identity is nothing but the Leibniz rule, with ε i j k \varepsilon_{ijk} ε ij k
Step 4 — Verification for ϕ = 1 / r \phi = 1/r ϕ = 1/ r F ⃗ = r ⃗ \vec F = \vec r F = r r ≠ 0 r\neq 0 r = 0
First the two ingredients.
(a) ∇ ( 1 r ) \nabla\left(\dfrac1r\right) ∇ ( r 1 ) Since r = ( x 2 + y 2 + z 2 ) 1 / 2 r = (x^2+y^2+z^2)^{1/2} r = ( x 2 + y 2 + z 2 ) 1/2
∂ r ∂ x = x r , ∂ ∂ x ( 1 r ) = − 1 r 2 ⋅ x r = − x r 3 , \frac{\partial r}{\partial x} = \frac{x}{r},\qquad \frac{\partial}{\partial x}\left(\frac1r\right) = -\frac{1}{r^2}\cdot\frac{x}{r} = -\frac{x}{r^3}, ∂ x ∂ r = r x , ∂ x ∂ ( r 1 ) = − r 2 1 ⋅ r x = − r 3 x , and similarly for y , z y,z y , z
∇ ( 1 r ) = − x i ^ + y j ^ + z k ^ r 3 = − r ⃗ r 3 . \nabla\left(\frac1r\right) = -\frac{x\hat i + y\hat j + z\hat k}{r^3} = -\frac{\vec r}{r^3}. ∇ ( r 1 ) = − r 3 x i ^ + y j ^ + z k ^ = − r 3 r . (b) ∇ × r ⃗ \nabla\times\vec r ∇ × r
∇ × r ⃗ = ∣ i ^ j ^ k ^ ∂ x ∂ y ∂ z x y z ∣ = i ^ ( ∂ y z − ∂ z y ) − j ^ ( ∂ x z − ∂ z x ) + k ^ ( ∂ x y − ∂ y x ) = 0 ⃗ . \nabla\times\vec r = \begin{vmatrix}\hat i & \hat j & \hat k\\ \partial_x & \partial_y & \partial_z\\ x & y & z\end{vmatrix}
= \hat i(\partial_y z - \partial_z y) - \hat j(\partial_x z - \partial_z x) + \hat k(\partial_x y - \partial_y x) = \vec 0 . ∇ × r = i ^ ∂ x x j ^ ∂ y y k ^ ∂ z z = i ^ ( ∂ y z − ∂ z y ) − j ^ ( ∂ x z − ∂ z x ) + k ^ ( ∂ x y − ∂ y x ) = 0 . Right-hand side.
( ∇ ϕ ) × F ⃗ + ϕ ( ∇ × F ⃗ ) = ( − r ⃗ r 3 ) × r ⃗ + 1 r 0 ⃗ . (\nabla\phi)\times\vec F + \phi(\nabla\times\vec F) = \left(-\frac{\vec r}{r^3}\right)\times\vec r + \frac1r\,\vec 0 . ( ∇ ϕ ) × F + ϕ ( ∇ × F ) = ( − r 3 r ) × r + r 1 0 . The first term vanishes because − r ⃗ r 3 -\dfrac{\vec r}{r^3} − r 3 r r ⃗ \vec r r r ⃗ × r ⃗ = 0 ⃗ \vec r\times\vec r = \vec 0 r × r = 0
RHS = 0 ⃗ . \text{RHS} = \vec 0 . RHS = 0 . Left-hand side (computed independently, not inferred). Here ϕ F ⃗ = r ⃗ r = x r i ^ + y r j ^ + z r k ^ \phi\vec F = \dfrac{\vec r}{r} = \dfrac{x}{r}\hat i + \dfrac{y}{r}\hat j + \dfrac{z}{r}\hat k ϕ F = r r = r x i ^ + r y j ^ + r z k ^ i ^ \hat i i ^
∂ ∂ y ( z r ) − ∂ ∂ z ( y r ) = z ( − y r 3 ) − y ( − z r 3 ) = − y z r 3 + y z r 3 = 0. \frac{\partial}{\partial y}\!\left(\frac{z}{r}\right) - \frac{\partial}{\partial z}\!\left(\frac{y}{r}\right)
= z\left(-\frac{y}{r^3}\right) - y\left(-\frac{z}{r^3}\right) = -\frac{yz}{r^3} + \frac{yz}{r^3} = 0 . ∂ y ∂ ( r z ) − ∂ z ∂ ( r y ) = z ( − r 3 y ) − y ( − r 3 z ) = − r 3 y z + r 3 y z = 0. Cyclically, the j ^ \hat j j ^ ∂ z ( x r ) − ∂ x ( z r ) = − z x r 3 + z x r 3 = 0 \partial_z\!\left(\frac{x}{r}\right) - \partial_x\!\left(\frac{z}{r}\right) = -\dfrac{zx}{r^3} + \dfrac{zx}{r^3} = 0 ∂ z ( r x ) − ∂ x ( r z ) = − r 3 z x + r 3 z x = 0 k ^ \hat k k ^ ∂ x ( y r ) − ∂ y ( x r ) = − x y r 3 + x y r 3 = 0 \partial_x\!\left(\frac{y}{r}\right) - \partial_y\!\left(\frac{x}{r}\right) = -\dfrac{xy}{r^3} + \dfrac{xy}{r^3} = 0 ∂ x ( r y ) − ∂ y ( r x ) = − r 3 x y + r 3 x y = 0
LHS = ∇ × ( r ⃗ r ) = 0 ⃗ . \text{LHS} = \nabla\times\left(\frac{\vec r}{r}\right) = \vec 0 . LHS = ∇ × ( r r ) = 0 . LHS = = = = 0 ⃗ = \vec 0 = 0 ϕ = 1 / r \phi = 1/r ϕ = 1/ r F ⃗ = r ⃗ \vec F = \vec r F = r r ≠ 0 r\neq 0 r = 0
(Remark worth one line: the vanishing is not an accident — r ⃗ r = ∇ r \dfrac{\vec r}{r} = \nabla r r r = ∇ r
Answer
∇ × ( ϕ F ⃗ ) = ( ∇ ϕ ) × F ⃗ + ϕ ( ∇ × F ⃗ ) ; for ϕ = 1 r , F ⃗ = r ⃗ : ∇ × ( r ⃗ r ) = 0 ⃗ = ( − r ⃗ r 3 ) × r ⃗ + 1 r 0 ⃗ ( r ≠ 0 ) . \boxed{\;\nabla\times(\phi\vec F) = (\nabla\phi)\times\vec F + \phi(\nabla\times\vec F);\qquad
\text{for }\phi=\tfrac1r,\ \vec F=\vec r:\ \ \nabla\times\!\left(\tfrac{\vec r}{r}\right) = \vec 0 = \left(-\tfrac{\vec r}{r^{3}}\right)\times\vec r + \tfrac1r\,\vec 0 \quad (r\neq 0).\;} ∇ × ( ϕ F ) = ( ∇ ϕ ) × F + ϕ ( ∇ × F ) ; for ϕ = r 1 , F = r : ∇ × ( r r ) = 0 = ( − r 3 r ) × r + r 1 0 ( r = 0 ) .