Gradient vector ∇f — definition, properties: Worked examples

Before we start, three tiny reminders so every symbol is earned:

The scenario matrix

Every gradient problem is one of these cells. Each example below is tagged with the cell(s) it covers.

Cell	Scenario class	What is tricky about it	Example
A	All positive components, plain compute	nothing — the warm-up	Ex 1
B	Mixed signs $(+, -)$ , direction points downhill	negative $D_{\overset{u}{^}} f$ ; sign of $cos θ$	Ex 2
C	Degenerate: $\nabla f = 0$ (a flat/critical point)	every direction gives rate $0$ ; steepest direction undefined	Ex 3
D	Perpendicularity to a non-circular level curve	tangent isn't obvious; must use $\nabla f \cdot (tangent) = 0$	Ex 4
E	Non-unit vector given (the trap)	must normalise first	Ex 5
F	Real-world word problem (temperature/hill)	translate words → $\nabla f$ , units	Ex 6
G	Limiting/extreme direction: fastest descent & the zero-change direction	signs of all three key angles $0, π, π /2$	Ex 7
H	Exam twist: work backwards — given a directional derivative, find unknown	solve for a component	Ex 8

See Partial derivatives for computing each $f_{x}, f_{y}$ , and Directional derivative for the dot-product rule.

Worked examples

Worked example Ex 1 — Cell A: all-positive warm-up

$f (x, y) = 3 x^{2} + y^{3}$ . Find $\nabla f$ at $(1, 2)$ and the steepest-ascent rate.

Forecast: guess — will the steepest rate be bigger or smaller than $10$ ?

$f_{x} = 6 x$ , $f_{y} = 3 y^{2}$ . Why this step? A partial derivative freezes the other variable, so $3 x^{2}$ differentiates like a one-variable function and $y^{3}$ is treated as a constant (its $x$ -derivative is $0$ ).
At $(1, 2)$ : $\nabla f = (6 \cdot 1, 3 \cdot 2^{2}) = (6, 12)$ . Why? Just plug the point into each slope.
Steepest rate $= ∣\nabla f ∣ = 6^{2} + 1 2^{2} = 180 = 65 \approx 13.42$ . Why? Property 2: max rate equals the length of the gradient, and length $= sum of squares$ .

Verify: both components positive, so the arrow points up-and-right into the first quadrant, and $180 > 10$ as expected. Direction of climb $= (6, 12) / 180$ , a unit vector: $(6^{2} + 1 2^{2}) /180 = 1$ . ✓

Worked example Ex 2 — Cell B: mixed signs, stepping downhill

$f (x, y) = x^{2} - y^{2}$ at $(2, 1)$ . Step in direction $\overset{u}{^} = (0, 1)$ (straight up in $y$ ). Find $D_{\overset{u}{^}} f$ and say if you go up or downhill.

Forecast: the $y^{2}$ has a minus sign. Do you gain or lose height moving in $+ y$ ?

$\nabla f = (2 x, - 2 y)$ , so at $(2, 1)$ : $\nabla f = (4, - 2)$ . Why the minus? $- y^{2}$ differentiates to $- 2 y$ ; the sign carries through.
$\overset{u}{^} = (0, 1)$ has length $0^{2} + 1^{2} = 1$ — already unit. Why check? $D_{\overset{u}{^}} f$ is only a true "per-step" rate for unit vectors.
$D_{\overset{u}{^}} f = (4, - 2) \cdot (0, 1) = 4 \cdot 0 + (- 2) \cdot 1 = - 2$ . Why? Dot product multiplies matching pieces and adds; here only the $y$ -piece survives.

Verify: the result is negative, meaning moving in $+ y$ takes you downhill — makes sense since $- y^{2}$ decreases as $∣ y ∣$ grows. Also $∣ D_{\overset{u}{^}} f ∣ = 2 \leq ∣\nabla f ∣ = 20 \approx 4.47$ , obeying "no direction beats the gradient's length". ✓

Worked example Ex 3 — Cell C: degenerate, gradient is zero

$f (x, y) = x^{2} + y^{2}$ at the origin $(0, 0)$ . Find $\nabla f$ and the steepest-ascent direction.

Forecast: the bowl's bottom is at the origin. Which way is "uphill" from the very bottom?

$\nabla f = (2 x, 2 y)$ , so at $(0, 0)$ : $\nabla f = (0, 0) = 0$ . Why? Both slopes vanish — the surface is momentarily flat at the low point.
Steepest rate $= ∣ 0 ∣ = 0$ . Why? The length of the zero vector is $0$ .
Steepest direction is undefined. Why this step matters? Every direction gives $D_{\overset{u}{^}} f = 0 \cdot \overset{u}{^} = 0$ ; there is no "most uphill" way because it's equally flat everywhere here. This is exactly a critical point — the flag Gradient descent and Lagrange multipliers both hunt for.

Verify: try any $\overset{u}{^} = (u_{1}, u_{2})$ : $D_{\overset{u}{^}} f = 0 \cdot u_{1} + 0 \cdot u_{2} = 0$ for all of them. Consistent — no preferred uphill. ✓

Worked example Ex 4 — Cell D: perpendicular to a non-circular level curve

$f (x, y) = x^{2} + 4 y^{2}$ . At $(2, 1)$ (which lies on $f = 8$ , an ellipse), show $\nabla f$ is perpendicular to the level curve.

Forecast: the level curve is a squashed circle. Will $\nabla f$ still point exactly across it?

$\nabla f = (2 x, 8 y)$ , so at $(2, 1)$ : $\nabla f = (4, 8)$ . Why the $8$ ? $4 y^{2}$ differentiates to $8 y$ .
Find a tangent to the ellipse at $(2, 1)$ . Differentiate $x^{2} + 4 y^{2} = 8$ implicitly: $2 x + 8 y y^{'} = 0 \Rightarrow y^{'} = - \frac{2 x}{8 y} = - \frac{2 \cdot 2}{8 \cdot 1} = - \frac{1}{2}$ . So a tangent direction is $(1, - \frac{1}{2})$ , or cleared of fractions $t = (2, - 1)$ . Why implicit differentiation? The curve isn't a function $y (x)$ we can solve, so we differentiate the equation as-is to get the slope.
Dot the gradient with the tangent: $\nabla f \cdot t = (4, 8) \cdot (2, - 1) = 8 - 8 = 0$ . Why? Property 3 says $\nabla f ⊥$ level set; a zero dot product is exactly perpendicularity.

Verify: the dot is $0$ , so the red arrow (gradient) meets the ellipse at a right angle even though the curve is stretched. See Level curves and surfaces. ✓

Worked example Ex 5 — Cell E: the non-unit trap

$f (x, y) = x y$ at $(3, 2)$ . Find the rate of change toward the point $(6, 6)$ .

Forecast: the step vector to $(6, 6)$ is $(3, 4)$ , length $5$ . If you skip normalising, your answer is scaled by $5$ — spot the error.

$\nabla f = (y, x)$ , so at $(3, 2)$ : $\nabla f = (2, 3)$ . Why $(y, x)$ ? $\partial (x y) / \partial x = y$ (treat $y$ constant), $\partial (x y) / \partial y = x$ .
Step vector $v = (6, 6) - (3, 2) = (3, 4)$ , with $∣ v ∣ = 3^{2} + 4^{2} = 5$ . Why? Direction = destination minus start.
Normalise: $\overset{u}{^} = (3, 4) /5 = (0.6, 0.8)$ . Why this step? $D_{\overset{u}{^}} f$ is only a genuine per-unit-length rate for a length- $1$ arrow (mistake (a) in the parent note).
$D_{\overset{u}{^}} f = (2, 3) \cdot (0.6, 0.8) = 1.2 + 2.4 = 3.6$ . Why? Multiply matching pieces, add.

Verify: the wrong (unnormalised) answer would be $(2, 3) \cdot (3, 4) = 6 + 12 = 18 = 5 \times 3.6$ — exactly $5 \times$ too big, confirming the scale factor is $∣ v ∣ = 5$ . ✓

Worked example Ex 6 — Cell F: real-world word problem

A metal plate has temperature $T (x, y) = 100 - x^{2} - 3 y^{2}$ degrees, with $x, y$ in metres. An ant sits at $(1, 2)$ . (a) Which way should it walk to warm up fastest, and how fast (°C per metre)? (b) Which way to keep the same temperature?

Forecast: the plate is hottest at the centre $(0, 0)$ ; the ant is off-centre. Will "warmest way" point toward or away from the centre?

$\nabla T = (- 2 x, - 6 y)$ , at $(1, 2)$ : $\nabla T = (- 2, - 12)$ . Why negatives? Temperature drops as you move out, so the slopes are negative.
Fastest warming = direction of $\nabla T = (- 2, - 12)$ ; rate $= ∣\nabla T ∣ = (- 2)^{2} + (- 12)^{2} = 148 = 237 \approx 12.17$ °C/m. Why? Property 2 again — steepest ascent is along the gradient, rate is its length. Units: (°C)/(m) since $T$ is °C and steps are metres.
Constant-temperature direction: perpendicular to $\nabla T$ . Swap and negate one component: $(- 12, 2)$ (or $(12, - 2)$ ). Why? Property with $θ = π /2$ gives $D_{\overset{u}{^}} T = 0$ ; a vector perpendicular to $(a, b)$ is $(- b, a)$ .

Verify: the warming direction $(- 2, - 12)$ points toward the hot centre — sensible. Perpendicular check: $(- 2, - 12) \cdot (- 12, 2) = 24 - 24 = 0$ . ✓

Worked example Ex 7 — Cell G: all three limiting angles at once

$f (x, y) = 2 x + y$ (a tilted plane) at any point. Give the fastest-ascent, fastest-descent, and zero-change directions, with rates.

Forecast: for a plane the gradient is the same everywhere. What are the three special rates?

$\nabla f = (2, 1)$ , constant. Why constant? Partial derivatives of $2 x + y$ are the constants $2$ and $1$ .
$∣\nabla f ∣ = 2^{2} + 1^{2} = 5 \approx 2.236$ . Fastest ascent: direction $(2, 1) / 5$ , rate $+ 5$ ( $θ = 0$ , $cos θ = 1$ ). Why? $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ is maxed when the step aligns with $\nabla f$ .
Fastest descent: direction $- (2, 1) / 5$ , rate $- 5$ ( $θ = π$ , $cos θ = - 1$ ). Why? Opposite direction flips the sign; this is the arrow Gradient descent follows.
Zero change: any $\overset{u}{^} ⊥ (2, 1)$ , e.g. $(1, - 2) / 5$ , rate $0$ ( $θ = π /2$ , $cos θ = 0$ ). Why? Perpendicular to the gradient keeps you on a level line.

Verify: check the zero direction: $(2, 1) \cdot (1, - 2) / 5 = (2 - 2) / 5 = 0$ . And ascent rate: $(2, 1) \cdot (2, 1) / 5 = 5/ 5 = 5$ . ✓

Worked example Ex 8 — Cell H: exam twist, work backwards

At a point $P$ , $f_{x} = 3$ and $f_{y} = a$ is unknown. You are told the directional derivative in the direction $\overset{u}{^} = (\frac{3}{5}, \frac{4}{5})$ equals $\frac{29}{5}$ . Find $a$ , then the steepest-ascent rate at $P$ .

Forecast: you know the output of the dot product and must recover a missing input. One linear equation, one unknown — solvable.

Write the rule: $D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = 3 \cdot \frac{3}{5} + a \cdot \frac{4}{5} = \frac{9}{5} + \frac{4 a}{5}$ . Why? Same dot-product formula, just with $a$ symbolic.
Set equal to the given value: $\frac{9}{5} + \frac{4 a}{5} = \frac{29}{5} \Rightarrow \frac{4 a}{5} = \frac{20}{5} = 4 \Rightarrow a = 5$ . Why? Solve the single linear equation for $a$ .
Now $\nabla f = (3, 5)$ , so steepest rate $= ∣\nabla f ∣ = 3^{2} + 5^{2} = 34 \approx 5.83$ . Why? Once the gradient is known, its length is the max rate.

Verify: put $a = 5$ back: $(3, 5) \cdot (\frac{3}{5}, \frac{4}{5}) = \frac{9}{5} + \frac{20}{5} = \frac{9}{5} + \frac{20}{5} = \frac{29}{5}$ . ✓ Matches the given data.

Recall Quick self-test (reveal after guessing)

Which cell would "find the coldest-approach direction on a temperature map" be? ::: Cell F/G — fastest descent, direction $- \nabla T$ . If $\nabla f = (0, 0)$ at $P$ , what is $D_{\overset{u}{^}} f$ for every $\overset{u}{^}$ ? ::: Always $0$ ; the point is critical. Given a direction to a target point, what is the FIRST thing you must do? ::: Normalise the step vector to length $1$ .

Connections

4.4.09 Gradient vector ∇f — definition, properties (Hinglish) — the parent topic.
Partial derivatives — every example starts by computing these.
Directional derivative — the dot-product rule used throughout (Ex 2, 5, 8).
Chain rule (multivariable) — origin of that dot-product rule.
Level curves and surfaces — Ex 4's ellipse and Ex 7's level lines.
Tangent plane and linearisation — the plane in Ex 7.
Gradient descent — Ex 3's critical point and Ex 7's descent direction.
Lagrange multipliers — uses the same perpendicularity as Ex 4.