Gradient: definition, geometric meaning, computation

At a Glance

• Frequency: 6 sub-parts across 4 of 13 years (2015, 2016, 2019, 2025)
• Priority tier: T3
• Marks (count): 10 (5), 12 (1)
• Average solve time: ~7 min
• Difficulty mix: easy 6
• Section: B | Dominant type: computation

Why This Chapter Matters

Gradient questions are guaranteed marks — every question in this atom is classified easy, and six of them have appeared across four years. The questions are short: compute the gradient, evaluate at a point, take a dot product with a unit vector, or check perpendicularity. The only pitfall is forgetting to normalise the direction vector. Mastering the four gradient archetypes here also underlies all of divergence, curl, Stokes, and divergence theorem, so the 30 minutes you spend here pays off across the whole vector calculus chapter.

Minimum Theory

Gradient. For a scalar field $\phi(x,y,z)$: $\nabla\phi = \frac{\partial\phi}{\partial x}\,\hat i + \frac{\partial\phi}{\partial y}\,\hat j + \frac{\partial\phi}{\partial z}\,\hat k.$ $\nabla\phi$ at a point $P$ is perpendicular to the level surface $\phi = \text{const}$ through $P$, and points in the direction of steepest increase of $\phi$.

Directional derivative. The rate of change of $\phi$ in the direction of a unit vector $\hat n$: $D_{\hat n}\phi = \nabla\phi \cdot \hat n.$ If the direction is given as a non-unit vector $\vec v$, always normalise first: $\hat n = \vec v / |\vec v|$.

Radial gradient. For a function $f(r)$ where $r = |\vec r| = \sqrt{x^2+y^2+z^2}$: $\nabla f(r) = f'(r)\,\hat r = f'(r)\,\frac{\vec r}{r}.$ To recover $f$ from $\nabla f = g(r)\,\hat r$: solve $f'(r) = r\,g(r)$ and integrate.

Question Archetypes

directional-derivative (2 questions; 2019, 2025)

Recognition Cues

• “Find the directional derivative of $\phi$ along the tangent to the curve $x=x(t), y=y(t), z=z(t)$ at the point …”
• The direction is given implicitly as the tangent to a parametric curve — you must compute and normalise the tangent vector.

Solution Template

1. Compute $\nabla\phi$. Differentiate $\phi$ with respect to $x$, $y$, $z$.
2. Evaluate $\nabla\phi$ at the given point $P$. Substitute the coordinates.
3. Find the tangent direction. Compute $\vec r\,'(t) = (x'(t), y'(t), z'(t))$. Identify the value of $t$ at which the curve passes through $P$.
4. Normalise: $\hat T = \vec r\,'(t_0) / |\vec r\,'(t_0)|$.
5. Dot product: $D_{\hat T}\phi = \nabla\phi|_P \cdot \hat T$.

Worked Example(s)

2019 Paper 1, 2019-P1-Q5e (10 marks)

Find the directional derivative of $\phi = xy^2+yz^2+zx^2$ along the tangent to $x=t, y=t^2, z=t^3$ at $(1,1,1)$.

Step 1 — Gradient: $\nabla\phi = (y^2+2zx,\; 2xy+z^2,\; 2yz+x^2).$ At $(1,1,1)$: $\nabla\phi = (1+2,\; 2+1,\; 2+1) = (3,3,3)$.

Step 2 — Tangent. $\vec r\,'(t) = (1, 2t, 3t^2)$. The point $(1,1,1)$ corresponds to $t=1$, so $\vec r\,'(1) = (1,2,3)$, $|\vec r\,'(1)| = \sqrt{14}$.

Step 3 — Directional derivative: $D_{\hat T}\phi = (3,3,3) \cdot \frac{(1,2,3)}{\sqrt{14}} = \frac{3+6+9}{\sqrt{14}} = \frac{18}{\sqrt{14}} = \frac{9\sqrt{14}}{7}.$

$\boxed{\frac{18}{\sqrt{14}} = \frac{9\sqrt{14}}{7} \approx 4.81.}$

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

Find $|D_{\hat n}\phi|$ for $\phi = x^2y^2z^2$ at $(1,1,-1)$, direction = tangent to $x=e^t, y=2\sin t+1, z=t-\cos t$ at $t=0$.

Gradient: $\nabla\phi = (2xy^2z^2, 2x^2yz^2, 2x^2y^2z)$. At $(1,1,-1)$: $(2,2,-2)$.

Tangent at $t=0$: $\vec r\,'(t) = (e^t, 2\cos t, 1+\sin t)$, so $\vec r\,'(0) = (1,2,1)$. Verify: $\vec r(0) = (1,1,-1)$ ✓. Unit tangent: $(1,2,1)/\sqrt{6}$.

Directional derivative: $D_{\hat T}\phi = (2,2,-2)\cdot\frac{(1,2,1)}{\sqrt{6}} = \frac{2+4-2}{\sqrt{6}} = \frac{4}{\sqrt{6}} = \frac{2\sqrt{6}}{3} \approx 1.63.$

$\boxed{\left|D_{\hat T}\phi\right| = \frac{2\sqrt{6}}{3}.}$

Common Traps

• The direction is specified as the tangent to a curve — it is not a unit vector. Dividing by $|\vec r\,'(t_0)|$ is mandatory; omitting normalisation gives a wrong answer.
• Evaluate both $\nabla\phi$ and $\vec r\,'$ at the same point: $\nabla\phi$ at the Cartesian coordinates, $\vec r\,'$ at the corresponding parameter value $t_0$.
• In 2025, the point $(1,1,-1)$ has $z = -1$, so $z^2 = 1$ — the gradient components are $(2,2,-2)$, not $(2,2,2)$.

angle-between-surfaces (1 question; 2015)

Recognition Cues

• “Find the angle between the surfaces $S_1: F(x,y,z)=0$ and $S_2: G(x,y,z)=0$ at the point $P$.”
• The normal to a level surface $F = c$ is $\nabla F$. The angle between surfaces at $P$ is the angle between their normals.

Solution Template

1. Verify $P$ lies on both surfaces (quick check; one mark).
2. Compute $\nabla F|_P$ and $\nabla G|_P$.
3. $\cos\theta = |\nabla F \cdot \nabla G| / (|\nabla F|\,|\nabla G|)$ (use absolute value for the acute angle).

Worked Example

2015 Paper 1, 2015-P1-Q5e (10 marks)

Angle between $x^2+y^2+z^2 = 9$ and $z = x^2+y^2-3$ at $(2,-1,2)$.

Verify: $4+1+4 = 9$ ✓; $2-4-1+3 = 0$ ✓.

Normals. $\nabla(x^2+y^2+z^2-9)|_{(2,-1,2)} = (4,-2,4)$; $\nabla(z-x^2-y^2+3)|_{(2,-1,2)} = (-4,2,1)$.

Angle: $\cos\theta = \frac{|(4,-2,4)\cdot(-4,2,1)|}{|(4,-2,4)|\,|(-4,2,1)|} = \frac{|-16-4+4|}{6\cdot\sqrt{21}} = \frac{16}{6\sqrt{21}} = \frac{8}{3\sqrt{21}}.$ $\boxed{\theta = \cos^{-1}\!\left(\frac{8}{3\sqrt{21}}\right) \approx 54.4^\circ.}$

Common Traps

• The angle between surfaces uses the absolute value of $\cos\theta$ to get the acute angle; the sign of $\nabla F$ depends on which side you define $F = 0$.
• Double-check that $|\vec n_1|$ and $|\vec n_2|$ are computed at the given point after substitution.

orthogonal-surfaces (1 question; 2015)

Recognition Cues

• “Find $\lambda$, $\mu$ (or similar parameters) so that two surfaces intersect orthogonally at a point.”
• Two unknowns → two equations: (i) point on Surface 1, (ii) $\nabla F_1 \cdot \nabla F_2 = 0$.

Solution Template

1. Equation (i): point on Surface 1. Substitute the coordinates; this gives one equation in the unknowns.
2. Equation (ii): $\nabla F_1|_P \cdot \nabla F_2|_P = 0$. Compute both gradients at $P$ using values from step 1; set the dot product to zero; solve for the remaining unknown.

Worked Example

2015 Paper 1, 2015-P1-Q6c (12 marks)

Find $\lambda$, $\mu$ so $\lambda x^2 - \mu yz = (\lambda+2)x$ and $4x^2y+z^3=4$ intersect orthogonally at $(1,-1,2)$.

Step 1 — Point on Surface 1: $\lambda(1) - \mu(-1)(2) - (\lambda+2)(1) = \lambda + 2\mu - \lambda - 2 = 2\mu - 2 = 0 \Rightarrow \mu = 1$.

Step 2 — Orthogonality. With $\mu = 1$: $\nabla F_1|_{(1,-1,2)} = (2\lambda - (\lambda+2),\; -2,\; 1) = (\lambda-2,\; -2,\; 1).$ $\nabla F_2|_{(1,-1,2)} = (8xy,\; 4x^2,\; 3z^2)|_{(1,-1,2)} = (-8,\; 4,\; 12).$ $(\lambda-2)(-8) + (-2)(4) + (1)(12) = 0 \;\Rightarrow\; -8\lambda + 16 - 8 + 12 = 0 \;\Rightarrow\; \lambda = \tfrac{5}{2}.$

$\boxed{\lambda = \tfrac{5}{2},\; \mu = 1.}$

Common Traps

• Surface 2 contains no free parameters — its gradient just needs to be evaluated at $P$ (which you can verify is on it).
• The gradient of $F_1 = \lambda x^2 - \mu yz - (\lambda+2)x$ involves the unknowns; first solve for $\mu$ from condition (i), then substitute before computing the dot product.

recover-scalar-field (1 question; 2016)

Recognition Cues

• “Find $f(r)$ such that $\nabla f = \vec r / r^n$” (or similar radial form).
• The boundary condition $f(r_0) = \text{const}$ pins the integration constant.

Solution Template

1. Use the radial gradient formula. $\nabla f(r) = f'(r)\hat r = f'(r)\,\vec r/r$. Equate to $\vec r / r^n$ to get $f'(r) = r^{1-n}/1 = r^{1-n}$.

   More precisely: $\nabla f = f'(r)\,\vec r/r$ and $\vec r/r^n = (1/r^{n-1})\,\vec r/r$, so $f'(r) = r^{1-n}$.

2. Integrate: $f(r) = \int r^{1-n}\,dr$; simplify.

3. Apply the boundary condition to find the integration constant.

Worked Example

2016 Paper 1, 2016-P1-Q8a (10 marks)

Find $f(r)$ with $\nabla f = \vec r/r^5$ and $f(1)=0$.

$\nabla f = f'(r)\hat r$ and $\vec r/r^5 = r^{-4}\hat r$, so $f'(r) = r^{-4}$.

$f(r) = \int r^{-4}\,dr = -\frac{1}{3r^3} + C.$

$f(1) = 0 \Rightarrow C = 1/3$.

$\boxed{f(r) = \frac{1}{3}\!\left(1 - \frac{1}{r^3}\right).}$

Common Traps

• $\int r^{-4}\,dr = -\tfrac{1}{3}r^{-3}$, not $-r^{-3}$ (forget the factor of $1/3$).
• The formula $\nabla f = f'(r)\,\vec r/r$ requires writing $\vec r/r^5 = r^{-4}\,(\vec r/r)$ to extract $f'(r) = r^{-4}$ cleanly.

gradient-coplanarity (1 question; 2025)

Recognition Cues

• “Show $\nabla u$, $\nabla v$, $\nabla w$ are coplanar.”
• Three vectors are coplanar $\Leftrightarrow$ their scalar triple product is zero $\Leftrightarrow$ the $3\times3$ determinant whose rows are the three gradient vectors vanishes.

Solution Template

1. Compute the three gradient vectors explicitly.
2. Form the $3\times3$ matrix $M$ with rows $\nabla u$, $\nabla v$, $\nabla w$.
3. Show $\det M = 0$ (use row operations to reveal a linear dependence).

Worked Example

2025 Paper 1, 2025-P1-Q5e (10 marks)

If $u=x+y+z$, $v=x^2+y^2+z^2$, $w=xy+yz+zx$, show $\nabla u$, $\nabla v$, $\nabla w$ are coplanar.

Gradients: $\nabla u = (1,1,1),\quad \nabla v = (2x,2y,2z),\quad \nabla w = (y+z, z+x, x+y).$

Scalar triple product (= det $M$): $\begin{vmatrix}1 & 1 & 1 \\ 2x & 2y & 2z \\ y+z & z+x & x+y\end{vmatrix} = 2\begin{vmatrix}1 & 1 & 1 \\ x & y & z \\ y+z & z+x & x+y\end{vmatrix}.$ Apply $R_3 \to R_3 + R_2$: the new third row is $(x+y+z, x+y+z, x+y+z) = (x+y+z)(1,1,1)$.

Factor out $(x+y+z)$: $= 2(x+y+z)\begin{vmatrix}1 & 1 & 1 \\ x & y & z \\ 1 & 1 & 1\end{vmatrix} = 0,$ since rows 1 and 3 are identical.

Therefore the scalar triple product is zero, so $\nabla u$, $\nabla v$, $\nabla w$ are coplanar. (Explicit linear relation: $v = u^2 - 2w$, so $\nabla v = 2u\,\nabla u - 2\,\nabla w$.)

Common Traps

• Three vectors coplanar = scalar triple product = determinant = 0. Do not confuse with parallel (which would be a stronger condition).
• The row operation $R_3 \to R_2 + R_3$ is the key step; without it the determinant looks messy.
• The geometric insight ($v = u^2-2w$) is worth one sentence — it explains the coplanarity without computation.

Marks-Aware Writing

10-mark directional derivative: Five steps, each one line: gradient → evaluate at P → tangent vector → normalise → dot product. Write the intermediate vectors explicitly; the examiner checks each step. Final answer in rationalised surd form.

12-mark orthogonal surfaces: Two clearly labelled conditions (i) and (ii), each yielding one equation. Show the algebra to find $\mu$ from (i) before substituting into (ii).

Practice Set

• 2015-P1-Q7c (12 m) — related gradient computation;
• 2022-P1-Q5e (10 m) — directional derivative variant;
• 2020-P1-Q5c (10 m) — gradient and normal;