Gradients and directional derivatives: Worked examples

Before we start, three symbols we will lean on — defined in plain words:

The scenario matrix

Every worked example below is tagged with the cell of this matrix it covers. Together they fill the whole grid — you will never meet an unshown case.

Cell	Scenario class	What is special / what can trip you
A	$\nabla f$ in Quadrant I (both partials $+$ )	plain baseline, positive dot product
B	$\nabla f$ with a negative component (Q II / IV)	signs must survive the dot product
C	Both components negative (Q III)	steepest ascent points "down-left"
D	Direction $u$ not yet unit	must normalize before dotting
E	Walking along a contour ( $θ = 90°$ )	answer is exactly $0$
F	Walking opposite the gradient ( $θ = 180°$ )	most negative slope = $- ∥\nabla f ∥$
G	Degenerate: $\nabla f = 0$	a flat / critical point, every direction gives $0$
H	Limiting behaviour of the slope as $u$ rotates	slope traces a cosine wave
I	Word problem (temperature field)	attach units, interpret sign
J	Exam twist: find the direction with a given slope	invert the cosine relation

Cell A — Quadrant I gradient (baseline)

Worked example A · both partials positive

$f (x, y) = x^{2} + x y + y^{2}$ at $a = (1, 1)$ , step direction $u = (0, 1)$ (due "north").

Forecast: north-slope should be positive (surface rises going north). Guess a number before reading.

Step 1 — partials. $\partial f / \partial x = 2 x + y$ , $\partial f / \partial y = x + 2 y$ . Why this step? The gradient is built from Partial derivatives; freeze the other variable and differentiate.

Step 2 — evaluate. $\nabla f (1, 1) = (2 + 1, 1 + 2) = (3, 3)$ . Both components positive ⇒ arrow in Quadrant I, pointing up-right.

Step 3 — is $u$ unit? $∥ (0, 1) ∥ = 0^{2} + 1^{2} = 1$ . ✓ No normalization needed. Why check? $D_{u} f$ only means "slope per unit distance" when $∥ u ∥ = 1$ .

Step 4 — dot product. $D_{u} f = (3, 3) \cdot (0, 1) = 3 \cdot 0 + 3 \cdot 1 = 3$ . Why a dot product? From the chain-rule derivation in the parent, $g^{'} (0) = \sum_{i} \partial_{i} f u_{i}$ , which is the dot product $\nabla f \cdot u$ .

Verify: north step changes only $y$ ; $\partial f / \partial y (1, 1) = 3$ — matches exactly. Positive, as forecast. ✓

Figure — Gradients and directional derivatives

Look at the figure: the blue arrow is $\nabla f = (3, 3)$ ; the yellow arrow is your northward step; the number $3$ is the shadow (projection) of your step onto the gradient scaled by $∥\nabla f ∥$ — that is what the dot product measures.

Cell B & J — a negative component, then the exam twist

Worked example J · exam twist — find a direction with a

prescribed slope Same $f$ , same point ( $\nabla f = (4, - 2)$ , so $∥\nabla f ∥ = 20 \approx 4.472$ ). Find a unit direction $u$ along which $D_{u} f = 3$ .

Forecast: since $3 < ∥\nabla f ∥ = 4.472$ , such a direction must exist (you can always get any slope between $- ∥\nabla f ∥$ and $+ ∥\nabla f ∥$ ). Guess whether one or two directions work.

Step 1 — geometric relation. $D_{u} f = ∥\nabla f ∥ cos θ$ where $θ$ is the angle from $\nabla f$ to $u$ . Why cosine? The dot product's geometric form; it earns $θ$ as "how aligned are the two arrows".

Step 2 — solve for $θ$ . $cos θ = 3/ 20 = 0.6708$ , so $θ = \pm 47.87°$ . Why $\pm$ ? Cosine is even: rotating $u$ that angle either clockwise or counter-clockwise gives the same slope. Two directions work.

Verify: $∥\nabla f ∥ cos (47.87°) = 4.472 \times 0.6708 = 3.00$ . ✓ Both the $+$ and $-$ rotations satisfy it.

Cell C — both components negative (Quadrant III)

Cell D — direction not yet a unit vector

Cell E & F — along the contour, and dead against the gradient

Cell G — the degenerate case: $\nabla f = 0$

Worked example G · a flat / critical point

$f (x, y) = x^{2} + y^{2}$ at $a = (0, 0)$ (the bowl's bottom).

Forecast: at the very bottom, is there any uphill direction? Guess before computing.

Step 1 — gradient. $\nabla f = (2 x, 2 y) = (0, 0)$ at the origin. The arrow has length zero.

Step 2 — directional derivative in any $u$ . $D_{u} f = (0, 0) \cdot u = 0$ for every unit $u$ . Why every direction gives 0? A zero vector dotted with anything is $0$ — there is no preferred uphill.

Step 3 — what "steepest ascent direction" means here. It is undefined: $\nabla f /∥\nabla f ∥ = 0/0$ . This is exactly the signal ML uses — $\nabla f = 0$ means a stationary point (min, max, or saddle) where gradient descent stops moving.

Verify: to first order $f$ is flat at a critical point in every direction, so all directional derivatives must vanish. ✓ (Curvature — who's a min vs saddle — is decided by the Hessian, not the gradient.)

Cell H — limiting behaviour as the direction rotates

Worked example H · slope traces a cosine as

$u$ spins $f (x, y) = x^{2} + 3 x y + y^{2}$ at $(1, 2)$ so $\nabla f = (8, 7)$ , $∥\nabla f ∥ = 113 \approx 10.630$ . Let $u = (cos ϕ, sin ϕ)$ sweep all angles $ϕ$ .

Forecast: sketch how $D_{u} f$ behaves as $ϕ$ goes $0 \to 360°$ . Bounded? Where max/min?

Step 1 — write it as one wave. $D_{u} f = ∥\nabla f ∥ cos θ$ , where $θ = ϕ - ϕ_{g}$ and $ϕ_{g} = arctan (7/8) \approx 41.19°$ is the gradient's own angle. Why $θ = ϕ - ϕ_{g}$ ? $θ$ is the gap between your heading $ϕ$ and the gradient's heading $ϕ_{g}$ .

Step 2 — read off the limits.

Max $= + 113 \approx 10.630$ at $ϕ = ϕ_{g}$ ( $θ = 0$ ).
Min $= - 113$ at $ϕ = ϕ_{g} + 180°$ ( $θ = 180°$ ).
Two zeros at $ϕ = ϕ_{g} \pm 90°$ (the contour directions).

Step 3 — check one intermediate value. At $ϕ = 90°$ (due north), $θ = 90° - 41.19° = 48.81°$ , $cos θ = 0.6585$ , slope $= 10.630 \times 0.6585 = 7.0$ . And indeed $\nabla f \cdot (0, 1) = 7$ . ✓

Verify: the slope is a pure cosine of amplitude $∥\nabla f ∥$ — it can never exceed $∥\nabla f ∥$ nor drop below $- ∥\nabla f ∥$ . That bound is the whole reason $\nabla f$ deserves the name "steepest". ✓

Cell I — word problem with real units

Worked example I · a temperature field

A metal plate has temperature $T (x, y) = 100 - x^{2} - 2 y^{2}$ degrees Celsius, with $x, y$ in metres. An ant sits at $(3, 1)$ and walks toward $(3, 4)$ at $1 m/s$ . How fast (in $° C/s$ ) does it feel the temperature change?

Forecast: more $y$ ⇒ the $- 2 y^{2}$ term drops fast ⇒ expect the ant to cool. Sign should be negative.

Step 1 — gradient (units: $° C/m$ ). $\nabla T = (- 2 x, - 4 y)$ ; at $(3, 1)$ : $\nabla T = (- 6, - 4)$ .

Step 2 — unit direction. Toward $(3, 4)$ : raw $(0, 3)$ , length $3$ , so $u = (0, 1)$ . Why unit? So $D_{u} T$ comes out per metre, ready to multiply by speed.

Step 3 — directional derivative. $D_{u} T = (- 6, - 4) \cdot (0, 1) = - 4 ° C/m$ .

Step 4 — attach the speed. Rate in time $= D_{u} T \times speed = (- 4 ° C/m) (1 m/s) = - 4 ° C/s$ . Why multiply? $D_{u} T$ is change per metre travelled; the chain rule turns metres into seconds via speed.

Verify: negative, matching the forecast — the ant cools at $4° C$ per second. Units check: $\frac{° C}{m} \cdot \frac{m}{s} = \frac{° C}{s}$ . ✓

Recall checkpoint

Recall Did every cell get covered?

Cell A (Q I) ::: Example A — $\nabla f = (3, 3)$ , slope $3$ north. Cell B (one negative component) ::: Example B — $\nabla f = (4, - 2)$ , slope $4$ . Cell C (both negative, Q III) ::: Example C — $\nabla f = (- 3, - 1)$ , max slope $10$ . Cell D (non-unit direction) ::: Example D — normalize $(3, 4)$ first, slope $10.4$ not $52$ . Cell E (along contour) ::: Example E — perpendicular step, slope $0$ . Cell F (opposite gradient) ::: Example F — slope $- 10 = - ∥\nabla f ∥$ . Cell G (zero gradient) ::: Example G — origin, every direction gives $0$ . Cell H (limiting cosine) ::: Example H — slope bounded in $[- 113, + 113]$ . Cell I (word problem, units) ::: Example I — ant cools at $- 4 ° C/s$ . Cell J (exam twist, invert cosine) ::: Example J — two directions at $\pm 47.87°$ give slope $3$ .

Connections

Partial derivatives — every gradient component here is one.
Chain rule — turns metres into seconds in Cell I.
Dot product and projections — the engine of $\nabla f \cdot u$ .
Level sets and contours — Cell E walks along one.
Gradient descent — Cell F is a single descent step's direction.
Jacobian and Hessian — decides min vs saddle when Cell G's gradient vanishes.
Hinglish version.