Vector identities (curl of grad, div of curl, product rules)

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2018, 2023, 2025)
• Priority tier: T3
• Marks (count): 10 (1), 12 (1), 15 (1)
• Average solve time: ~11 min
• Difficulty mix: medium 3
• Section: B | Dominant type: proof

Why This Chapter Matters

Vector identity proofs appear in three consecutive years (2018, 2023, 2025) with marks ranging from 10 to 15, making this one of the more consistent T3 items in Paper 1. All three questions use the same technique: expand in Cartesian coordinates, regroup terms, identify the target expressions. The 2025 question applies the curl-curl identity to Maxwell’s equations to derive the electromagnetic wave equation — a 10-mark bonus that becomes trivial once you know the identity cold. Together these three questions cover 37 marks across three examinations.

Minimum Theory

Core identities (memorise these). For smooth scalar field $\phi$ and vector fields $\mathbf{F},\mathbf{G}$:

$\nabla\times(\nabla\phi) = \mathbf{0}$

$\nabla\cdot(\nabla\times\mathbf{F}) = 0$

$\nabla\times(\nabla\times\mathbf{F}) = \nabla(\nabla\cdot\mathbf{F}) - \nabla^2\mathbf{F}$

$\nabla\cdot(\phi\mathbf{F}) = \nabla\phi\cdot\mathbf{F} + \phi\,\nabla\cdot\mathbf{F}$

$\nabla\times(\phi\mathbf{F}) = \nabla\phi\times\mathbf{F} + \phi\,\nabla\times\mathbf{F}$

The curl-curl identity is the “BAC–CAB” rule applied formally: $\nabla\times(\nabla\times\mathbf{F}) = \nabla(\nabla\cdot\mathbf{F}) - (\nabla\cdot\nabla)\mathbf{F}$.

Proof technique. Every proof proceeds the same way: (1) write out the $\hat\imath$-component using the definitions of curl and divergence; (2) add and subtract a term (typically $\pm\partial^2 F_1/\partial x^2$) to complete the target expressions; (3) invoke cyclic symmetry $x\to y\to z$ to extend to all components.

Radial vector identities. In 3D: $\nabla\cdot\mathbf{r} = 3$; $\nabla r = \hat r = \mathbf{r}/r$; $\nabla(f(r)) = f'(r)\hat r = f'(r)\mathbf{r}/r$. The chain rule for $g(r) = f(r)/r$ gives $g'(r) = (rf'(r)-f(r))/r^2$.

Question Archetypes

vector-identity-proof (2 question(s); 2018, 2023)

Recognition Cues

• “Let $\mathbf{v} = v_1\hat\imath + v_2\hat\jmath + v_3\hat k$; show $\nabla\times(\nabla\times\mathbf{v}) = \nabla(\nabla\cdot\mathbf{v}) - \nabla^2\mathbf{v}$.”
• “Prove $\nabla\cdot(\phi\mathbf{F}) = \nabla\phi\cdot\mathbf{F} + \phi\nabla\cdot\mathbf{F}$; find $\nabla\cdot(f(r)/r\cdot\mathbf{r})$.”
• Both require a Cartesian expansion proof followed by an application to a specific radial field.

Solution Template

For curl-curl:

1. Write the $\hat\imath$-component of $\nabla\times(\nabla\times\mathbf{v})$ by expanding the outer curl applied to $\nabla\times\mathbf{v}$.
2. Add and subtract $\partial^2 v_1/\partial x^2$ to complete $\partial_x(\nabla\cdot\mathbf{v})$ and $-\nabla^2 v_1$.
3. State the result; invoke cyclic symmetry for $\hat\jmath$ and $\hat k$ components.

For divergence product rule:

1. Write $\nabla\cdot(\phi\mathbf{F}) = \sum_i \partial_i(\phi F_i)$; apply product rule to each term; regroup.
2. For the application $\phi = f(r)/r$, $\mathbf{F} = \mathbf{r}$: compute $\nabla(f(r)/r)$ via chain rule, $\nabla\cdot\mathbf{r} = 3$, and dot product.

Worked Example

2018 Paper 1, 2018-P1-Q8a (12 marks)

Let $\mathbf{v}=v_1\hat\imath+v_2\hat\jmath+v_3\hat k$. Show that $\nabla\times(\nabla\times\mathbf{v})=\nabla(\nabla\cdot\mathbf{v})-\nabla^2\mathbf{v}$.

Step 1 — Compute $\nabla\times\mathbf{v}$. The curl $\mathbf{w} = \nabla\times\mathbf{v}$ has components

$w_1 = \partial_y v_3 - \partial_z v_2,\quad w_2 = \partial_z v_1 - \partial_x v_3,\quad w_3 = \partial_x v_2 - \partial_y v_1.$

Step 2 — $\hat\imath$-component of $\nabla\times\mathbf{w}$. Using $[\nabla\times\mathbf{w}]_1 = \partial_y w_3 - \partial_z w_2$:

$[\nabla\times(\nabla\times\mathbf{v})]_1 = \partial_y(\partial_x v_2 - \partial_y v_1) - \partial_z(\partial_z v_1 - \partial_x v_3)$

$= \partial_x\partial_y v_2 - \partial_y^2 v_1 - \partial_z^2 v_1 + \partial_x\partial_z v_3.$

Step 3 — Add and subtract $\partial_x^2 v_1$.

$= \partial_x(\partial_x v_1 + \partial_y v_2 + \partial_z v_3) - (\partial_x^2 v_1 + \partial_y^2 v_1 + \partial_z^2 v_1)$

$= [\nabla(\nabla\cdot\mathbf{v})]_1 - [\nabla^2\mathbf{v}]_1.$

Step 4 — Cyclic symmetry. The $\hat\jmath$ and $\hat k$ components follow by the cyclic permutation $x\to y\to z$, $v_1\to v_2\to v_3$. Hence

$\boxed{\nabla\times(\nabla\times\mathbf{v}) = \nabla(\nabla\cdot\mathbf{v}) - \nabla^2\mathbf{v}.}$

(Here $\nabla^2\mathbf{v} = (\nabla^2 v_1,\nabla^2 v_2,\nabla^2 v_3)$ is the vector Laplacian.)

Common Traps

• Mixed partials commute only when $\mathbf{v}$ is $C^2$ — state this assumption.
• The insertion $\pm\partial_x^2 v_1$ is the key step; many students stall here without it.
• $\nabla^2\mathbf{v}$ is the vector Laplacian, not a scalar; each component is the scalar Laplacian of the corresponding component of $\mathbf{v}$.
• Do not try to apply the BAC–CAB rule directly without a Cartesian expansion; the examiner expects the proof.

2023 Paper 1, 2023-P1-Q8c (15 marks)

Prove $\nabla\cdot(\phi\mathbf{F})=\nabla\phi\cdot\mathbf{F}+\phi(\nabla\cdot\mathbf{F})$. Also find $\nabla\cdot\!\left(\dfrac{f(r)}{r}\mathbf{r}\right)$ and verify the identity.

Part 1 — Proof. In Cartesian coordinates:

$\nabla\cdot(\phi\mathbf{F}) = \sum_{i=1}^3 \frac{\partial(\phi F_i)}{\partial x_i} = \sum_i F_i\frac{\partial\phi}{\partial x_i} + \phi\sum_i\frac{\partial F_i}{\partial x_i} = \nabla\phi\cdot\mathbf{F} + \phi(\nabla\cdot\mathbf{F}).$

$\boxed{\nabla\cdot(\phi\mathbf{F}) = \nabla\phi\cdot\mathbf{F} + \phi\,\nabla\cdot\mathbf{F}.}$

Part 2 — Radial application. Apply with $\phi = f(r)/r$ and $\mathbf{F} = \mathbf{r}$:

$\nabla\!\left(\frac{f(r)}{r}\right) = g'(r)\hat r = \frac{rf'(r)-f(r)}{r^2}\cdot\frac{\mathbf{r}}{r} = \frac{(rf'(r)-f(r))\mathbf{r}}{r^3}.$

$\nabla\!\left(\frac{f(r)}{r}\right)\cdot\mathbf{r} = \frac{(rf'(r)-f(r))|\mathbf{r}|^2}{r^3} = \frac{rf'(r)-f(r)}{r}.$

$\frac{f(r)}{r}\nabla\cdot\mathbf{r} = \frac{3f(r)}{r}.$

$\boxed{\nabla\cdot\!\left(\frac{f(r)}{r}\mathbf{r}\right) = f'(r) + \frac{2f(r)}{r}.}$

Part 3 — Direct verification. Compute $\partial_x(xf(r)/r)$ directly using $\partial_x r = x/r$:

$\frac{\partial}{\partial x}\!\left(\frac{xf(r)}{r}\right) = \frac{f(r)}{r} + \frac{x^2(rf'(r)-f(r))}{r^3}.$

Sum over $x,y,z$: $3f(r)/r + r^2(rf'(r)-f(r))/r^3 = f'(r) + 2f(r)/r$ ✓.

Common Traps

• $\nabla\cdot\mathbf{r} = 3$ in 3D — a constant that is frequently forgotten.
• The chain rule $\nabla(g(r)) = g'(r)\hat r$ applies to any radial function; expand $g'(r)$ explicitly before substituting.
• Both the identity method and the direct method must give the same answer — state this and use it as a check.

wave-equation-derivation (1 question(s); 2025)

Recognition Cues

• Given: $\nabla\cdot\mathbf{E}=0$, $\nabla\cdot\mathbf{H}=0$, $\nabla\times\mathbf{E}=-\partial\mathbf{H}/\partial t$, $\nabla\times\mathbf{H}=\partial\mathbf{E}/\partial t$.
• “Show $\nabla^2\mathbf{H} = \partial^2\mathbf{H}/\partial t^2$” — apply curl-curl identity; use $\nabla\cdot\mathbf{H}=0$ to kill the grad-div term.

Solution Template

1. Take curl of one Maxwell equation (e.g., $\nabla\times\mathbf{E} = -\partial\mathbf{H}/\partial t$).
2. Swap curl and time derivative on the right; substitute the other Maxwell equation.
3. Apply the curl-curl identity $\nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E}$ on the left.
4. Use $\nabla\cdot\mathbf{E} = 0$ to drop the first term; conclude $\nabla^2\mathbf{E} = \partial^2\mathbf{E}/\partial t^2$.
5. Repeat for $\mathbf{H}$.

Worked Example

2025 Paper 1, 2025-P1-Q6c-ii (10 marks)

If $\nabla\cdot\mathbf{E}=0$, $\nabla\cdot\mathbf{H}=0$, $\nabla\times\mathbf{E}=-\partial\mathbf{H}/\partial t$, $\nabla\times\mathbf{H}=\partial\mathbf{E}/\partial t$, show $\nabla^2\mathbf{E}=\partial^2\mathbf{E}/\partial t^2$ and $\nabla^2\mathbf{H}=\partial^2\mathbf{H}/\partial t^2$.

Wave equation for $\mathbf{E}$. Take $\nabla\times$ of $\nabla\times\mathbf{E} = -\partial\mathbf{H}/\partial t$:

$\nabla\times(\nabla\times\mathbf{E}) = -\frac{\partial}{\partial t}(\nabla\times\mathbf{H}) = -\frac{\partial}{\partial t}\!\left(\frac{\partial\mathbf{E}}{\partial t}\right) = -\frac{\partial^2\mathbf{E}}{\partial t^2}.$

Apply the curl-curl identity with $\nabla\cdot\mathbf{E} = 0$:

$\nabla(\underbrace{\nabla\cdot\mathbf{E}}_{0}) - \nabla^2\mathbf{E} = -\frac{\partial^2\mathbf{E}}{\partial t^2}.$

$\boxed{\nabla^2\mathbf{E} = \frac{\partial^2\mathbf{E}}{\partial t^2}.}$

Wave equation for $\mathbf{H}$. Take $\nabla\times$ of $\nabla\times\mathbf{H} = \partial\mathbf{E}/\partial t$:

$\nabla\times(\nabla\times\mathbf{H}) = \frac{\partial}{\partial t}(\nabla\times\mathbf{E}) = \frac{\partial}{\partial t}\!\left(-\frac{\partial\mathbf{H}}{\partial t}\right) = -\frac{\partial^2\mathbf{H}}{\partial t^2}.$

Apply curl-curl with $\nabla\cdot\mathbf{H}=0$:

$-\nabla^2\mathbf{H} = -\frac{\partial^2\mathbf{H}}{\partial t^2} \qquad\Longrightarrow\qquad \boxed{\nabla^2\mathbf{H} = \frac{\partial^2\mathbf{H}}{\partial t^2}.}$

Common Traps

• The step $\nabla\times(\partial\mathbf{H}/\partial t) = \partial(\nabla\times\mathbf{H})/\partial t$ relies on swapping spatial and temporal derivatives — valid for smooth fields; state this.
• After applying the curl-curl identity, the $\nabla(\nabla\cdot\mathbf{E})$ term vanishes because $\nabla\cdot\mathbf{E}=0$ — missing this step is the most common error.
• $\nabla^2\mathbf{E}$ here is the vector Laplacian, valid as written in Cartesian coordinates.
• Write the intermediate step $-\nabla^2\mathbf{E} = -\partial^2\mathbf{E}/\partial t^2$ before cancelling the minus signs — examiners expect to see it.

Marks-Aware Writing

A 12-mark curl-curl proof must: compute the curl of the curl by components; insert the $\pm\partial_x^2 v_1$ term with justification; complete both target expressions; invoke cyclic symmetry. Skipping the $\pm$ insertion and stating the result by pattern-matching scores at most 5 marks.

A 15-mark divergence product rule answer must: give the Cartesian proof (4–5 marks); compute $\nabla\cdot(f(r)\mathbf{r}/r)$ via the identity (5–6 marks); verify directly (3–4 marks).

A 10-mark wave-equation derivation must: take curl of one Maxwell equation; swap $\nabla\times$ and $\partial_t$; substitute the other equation; apply curl-curl; drop zero divergence. Each step is 2 marks.

Practice Set

• 2024-P1-Q5c (10 m) — Vector identity computation. ()
• 2023-P1-Q5e (10 m) — Radial vector identity application. ()
• 2015-P1-Q8c (12 m) — Curl and divergence identity proof. ()
• 2022-P1-Q6c (15 m) — Vector identity proof with application. ()
• 2021-P1-Q6c (15 m) — Vector identity proof. ()
• 2020-P1-Q6b (15 m) — Divergence identity and radial application. ()