Gradient as direction of steepest ascent: Worked examples

This page assumes you have read the parent note. We reuse three things from it, and nothing more:

$\nabla f = (f_{x}, f_{y})$ , the vector of partial derivatives (the slopes in the $x$ and $y$ directions).
$D_{u} f = \nabla f \cdot u$ for a unit direction $u$ — see Directional Derivative and Dot Product and Angle.
Steepest ascent along $u^{⋆} = \nabla f /∥\nabla f ∥$ , at rate $∥\nabla f ∥$ .

The scenario matrix

Every problem this topic throws at you falls into one of these cells. Each example below is tagged with the cell it kills.

#	Case class	What makes it tricky	Example
A	Both partials positive	the "clean" textbook case	Ex 1
B	Mixed signs in $\nabla f$	direction points into quadrant II/IV	Ex 2
C	Direction given downhill	dot product comes out negative	Ex 3
D	Zero gradient (flat point / peak)	steepest direction is undefined	Ex 4
E	Perpendicular to a level curve	dot product must be exactly $0$	Ex 5
F	Real-world word problem (units!)	translate temperature/height into $f$	Ex 6
G	Exam twist — "which way keeps $f$ constant?"	must find the $9 0^{\circ}$ direction	Ex 7
H	Limiting / degenerate ridge (partial = 0)	gradient points along an axis	Ex 8

Case A — both partials positive

Worked example Ex 1 (cell A) — steepest ascent, clean case

$f (x, y) = x^{2} + 3 y$ at the point $P = (2, 1)$ . Find the direction of steepest ascent and the maximum rate.

Forecast: guess before computing — will the arrow lean more towards $x$ or towards $y$ ? (At $x = 2$ the $x$ -slope $2 x = 4$ beats the constant $y$ -slope $3$ , so lean towards $x$ .)

$f_{x} = 2 x, f_{y} = 3$ . Why this step? the gradient is the vector of partial derivatives; differentiate $f$ in $x$ holding $y$ fixed, then in $y$ holding $x$ fixed.
$\nabla f (2, 1) = (4, 3)$ . Why? plug the point into each partial.
$∥\nabla f ∥ = 4^{2} + 3^{2} = 25 = 5$ . Why? the max rate of increase equals the gradient's length (parent's "three big facts").
$u^{⋆} = \frac{( 4 , 3 )}{5} = (0.8, 0.6)$ . Why? steepest ascent is the gradient normalised to unit length.

Verify: $∥ u^{⋆} ∥ = 0. 8^{2} + 0. 6^{2} = 0.64 + 0.36 = 1$ ✓, and $\nabla f \cdot u^{⋆} = 4 (0.8) + 3 (0.6) = 3.2 + 1.8 = 5 = ∥\nabla f ∥$ ✓ — dotting the gradient with its own unit direction returns its length, exactly as steepest ascent demands. The arrow does lean towards $x$ ( $0.8 > 0.6$ ), matching the forecast.

Figure — Gradient as direction of steepest ascent

The figure shows the amber gradient arrow $(4, 3)$ at $P$ , its cyan unit version, and the two white component slopes it is built from.

Case B — mixed signs in the gradient

Worked example Ex 2 (cell B) — a gradient pointing into quadrant IV

$f (x, y) = x^{2} - y^{2}$ at $P = (3, 2)$ . Direction of steepest ascent?

Forecast: $f$ grows as $x$ grows but grows as $y$ shrinks (because of $- y^{2}$ ). So the uphill arrow should point right and down.

$f_{x} = 2 x, f_{y} = - 2 y$ . Why? differentiate; the minus sign on $y^{2}$ carries through.
$\nabla f (3, 2) = (6, - 4)$ . Why? substitute $P$ . The negative $y$ -component means "go downward in $y$ to climb."
$∥\nabla f ∥ = 6^{2} + (- 4)^{2} = 36 + 16 = 52 = 213$ . Why? magnitude uses squares, so signs vanish — the length is still well-defined.
$u^{⋆} = \frac{( 6 , - 4 )}{2 13} = \frac{( 3 , - 2 )}{13}$ . Why? normalise; factor of 2 cancels.

Verify: $∥ u^{⋆} ∥^{2} = \frac{9 + 4}{13} = 1$ ✓. The arrow points into quadrant IV (right, down), matching the forecast — a common trap is to ignore the sign and point up-right.

The figure shows $P = (3, 2)$ with the amber gradient $(6, - 4)$ leaning down-right into quadrant IV, and its cyan unit version.

Case C — a direction given, and it's downhill

Worked example Ex 3 (cell C) — negative directional derivative

With $f (x, y) = x^{2} + 3 y$ at $P = (2, 1)$ (same as Ex 1, so $\nabla f = (4, 3)$ ), how fast does $f$ change if you walk towards $Q = (- 2, - 2)$ ?

Forecast: $Q$ is down-and-left of $P$ , roughly opposite the uphill arrow $(4, 3)$ . Expect a negative rate (you're descending).

Raw direction $v = Q - P = (- 2 - 2, - 2 - 1) = (- 4, - 3)$ . Why? tip minus tail gives the direction of travel.
$∥ v ∥ = 16 + 9 = 5$ , so $u = \frac{( - 4 , - 3 )}{5} = (- 0.8, - 0.6)$ . Why? you must normalise, or the rate is scaled by $∥ v ∥$ and no longer "per unit distance" (parent's headline mistake).
$D_{u} f = \nabla f \cdot u = 4 (- 0.8) + 3 (- 0.6) = - 3.2 - 1.8 = - 5$ . Why? apply $D_{u} f = \nabla f \cdot u$ .

Verify: $u = - u^{⋆}$ from Ex 1, i.e. exactly the steepest-descent direction, so we should get $- ∥\nabla f ∥ = - 5$ ✓. A negative directional derivative simply means " $f$ decreases as you step" — the sign is information, not an error.

The figure shows the amber uphill gradient at $P$ and the cyan walk-direction pointing the opposite way towards $Q$ — the $18 0^{\circ}$ between them is why the rate is $- 5$ .

Case D — the zero gradient (degenerate!)

Worked example Ex 4 (cell D) — every direction is equally steep

$f (x, y) = x^{2} + y^{2}$ at the origin $P = (0, 0)$ . Which way is steepest ascent?

Forecast: the origin is the bottom of a bowl. Every direction goes uphill equally — so is there a unique steepest way? (No.)

$f_{x} = 2 x, f_{y} = 2 y \Rightarrow \nabla f (0, 0) = (0, 0)$ . Why? both partials vanish at the centre of the bowl.
$∥\nabla f ∥ = 0$ . Why? max rate of increase is zero to first order — the surface is flat at the very bottom.
$u^{⋆} = \frac{( 0 , 0 )}{0}$ cannot be formed. Why? you cannot normalise the zero vector, so the formula gives no single preferred direction.

Verify: check any unit direction, say $u = (1, 0)$ : $D_{u} f = \nabla f \cdot u = 0 (1) + 0 (0) = 0$ . Repeat for $(0, 1)$ , $(\frac{1}{2}, \frac{1}{2})$ , anything — you always get $0$ ✓.

The nuance: "undefined" is a statement about the formula $\nabla f /∥\nabla f ∥$ (division by zero). Geometrically the honest reading is "every direction is equally steep" — the first-order rate is $0$ in all of them, so no direction wins and no direction loses. This is a critical point; report it as " $\nabla f = 0$ , all directions tie at rate $0$ ."

Case E — perpendicular to a level curve

Worked example Ex 5 (cell E) — the

$9 0^{\circ}$ fact, checked with a figure $f (x, y) = x y$ at $P = (2, 3)$ ; the level curve through it is $x y = 6$ . Show $\nabla f$ is perpendicular to that curve.

Forecast: the parent note claims the gradient is $⊥$ to level curves, so the dot product of $\nabla f$ with the tangent should be exactly $0$ .

$\nabla f = (y, x) = (3, 2)$ at $P$ . Why? $\partial_{x} (x y) = y$ , $\partial_{y} (x y) = x$ .
Tangent to $x y = 6$ : differentiate implicitly, $y + x y^{'} = 0 \Rightarrow y^{'} = - \frac{3}{2}$ . Why? along a level curve $f$ is constant, so its total change is zero; $y^{'}$ is the curve's slope. A tangent vector reads its components as $(run, rise) = (1, y^{'}) = (1, - \frac{3}{2})$ . To clear the fraction we scale by 2 (scaling a direction by a positive number does not change the direction), giving the tidy integer form $(2, - 3)$ .
$\nabla f \cdot (2, - 3) = 3 (2) + 2 (- 3) = 6 - 6 = 0$ . Why? if the dot product is $0$ , the angle between them is $9 0^{\circ}$ .

Verify: $cos θ = \frac{\nabla f \cdot tangent}{∥\nabla f ∥ ∥ tangent ∥} = \frac{0}{13 \cdot 13} = 0 \Rightarrow θ = 9 0^{\circ}$ ✓. Along the tangent the directional derivative is $0$ (you stay on $x y = 6$ ), so $f$ doesn't change — precisely why the uphill arrow must leave the curve at a right angle.

The figure shows the hyperbola $x y = 6$ , its cyan tangent at $P$ , and the amber gradient stabbing off it at exactly $9 0^{\circ}$ .

Case F — real-world word problem (mind the units!)

Worked example Ex 6 (cell F) — a hiker on a temperature field

On a metal plate, temperature is $T (x, y) = 20 + 4 x - 3 y$ degrees Celsius, with $x, y$ in metres. A bug sits at $(5, 5)$ . (a) Which way should it crawl to warm up fastest? (b) At what rate (°C per metre)? (c) It instead crawls east, direction $(1, 0)$ — how fast does its temperature change?

Forecast: $T$ rises with $x$ and falls with $y$ , so "warmest way" points east-and-south. Rate per metre should be around $4^{2} + 3^{2} = 5$ .

$T_{x} = 4, T_{y} = - 3$ , so $\nabla T = (4, - 3)$ everywhere (linear field). Why? differentiate; the field is planar, so the gradient is constant. Units: $^{\circ} C / m$ .
Warmest direction $u^{⋆} = \frac{( 4 , - 3 )}{16 + 9} = \frac{( 4 , - 3 )}{5} = (0.8, - 0.6)$ . Why? normalise the gradient — steepest ascent of $T$ .
Max rate $= ∥\nabla T ∥ = 5^{\circ} C/m$ . Why? magnitude of the gradient is the fastest possible rate of increase.
Crawling east: $D_{(1, 0)} T = \nabla T \cdot (1, 0) = 4 (1) + (- 3) (0) = 4^{\circ} C/m$ . Why? dot the gradient with the given (already unit) direction.

Verify: $u^{⋆}$ is a unit vector ( $0. 8^{2} + 0. 6^{2} = 1$ ) ✓. East gives $4 < 5$ ✓ — nothing beats the gradient direction, and the units stay $^{\circ} C per metre$ throughout, confirming we normalised correctly (a non-unit direction would have given nonsense units).

The figure shows the bug at $(5, 5)$ , the amber warmest-way arrow $(4, - 3)$ pointing east-south, and the cyan "east" arrow whose slower rate $4 < 5$ you can read from its shorter projection.

Case G — exam twist: "which way stays level?"

Worked example Ex 7 (cell G) — the direction of zero change

$f (x, y) = 3 x + 4 y$ at $P = (1, 1)$ . Find a unit direction along which $f$ does not change, and confirm it is a valid answer.

Forecast: "no change" means perpendicular to $\nabla f$ . There should be two such directions (opposite each other along the level line).

$\nabla f = (3, 4)$ . Why? partials of a linear field are its coefficients.
A vector perpendicular to $(3, 4)$ is $(4, - 3)$ (swap and negate one). Why? $(3) (4) + (4) (- 3) = 0$ , so it makes a $9 0^{\circ}$ angle — the flat direction.
Normalise: $u = \frac{( 4 , - 3 )}{16 + 9} = \frac{( 4 , - 3 )}{5} = (0.8, - 0.6)$ . Its opposite $(- 0.8, 0.6)$ works equally. Why? the direction must be unit length; both signs lie along the level curve.

Verify: $D_{u} f = \nabla f \cdot u = 3 (0.8) + 4 (- 0.6) = 2.4 - 2.4 = 0$ ✓. Zero rate means you walk along a level line of the plane, so $f$ is genuinely constant — the twist is just the $9 0^{\circ}$ fact read backwards.

The figure shows the amber gradient $(3, 4)$ and the two cyan level directions $\pm (0.8, - 0.6)$ crossing it at $9 0^{\circ}$ — walk either way and $f$ stays constant.

Case H — degenerate ridge (a partial is zero)

The figure shows $P = (1, 0)$ with the amber gradient $(2, 0)$ pointing purely along $+ x$ and the cyan ridge direction $(0, 1)$ along which $f$ stays flat.

Recall Which cell was which?

A positive-clean ::: Ex 1 B mixed signs ::: Ex 2 C downhill (negative $D_{u} f$ ) ::: Ex 3 D zero gradient (all directions tie) ::: Ex 4 E perpendicular to level curve ::: Ex 5 F word problem with units ::: Ex 6 G "stay level" twist ::: Ex 7 H degenerate ridge (a partial $= 0$ ) ::: Ex 8

Connections

Gradient as direction of steepest ascent (parent)
Directional Derivative
Partial Derivatives
Dot Product and Angle
Level Curves and Contour Maps
Gradient Descent (Machine Learning)
Tangent Plane and Linearization