Gradient vector ∇f — definition, properties: Exercises

Reminders you will lean on (all built in the parent):

$\nabla f = (f_{x}, f_{y})$ — the vector of partial derivatives.
$D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = ∣\nabla f ∣ cos θ$ — the directional derivative, where $θ$ is the angle between the step direction $\overset{u}{^}$ and $\nabla f$ .
$\overset{u}{^}$ must be a unit vector (length $1$ ), else the answer is scaled.
$\nabla f$ points steepest uphill; $- \nabla f$ steepest downhill; $\nabla f ⊥$ level curves.

The one picture to keep in your head for the whole page:

Figure — Gradient vector ∇f — definition, properties

The coral arrow is $\nabla f$ (steepest uphill). Turn it $9 0^{\circ}$ and you get the mint arrow, the tangent to the level curve where $f$ never changes. Flip the coral arrow and you get steepest downhill. Almost every exercise below is one of these three moves.

Level 1 — Recognition

L1.1

For $f (x, y) = 3 x - 5 y + 7$ , write $\nabla f$ .

Recall Solution

What. The gradient is the list of partial slopes $(f_{x}, f_{y})$ . Why. $f_{x}$ = slope when we wiggle $x$ and freeze $y$ ; treat $y$ as a constant. The constant $7$ and the $- 5 y$ term vanish under $\partial / \partial x$ .

$f_{x} = 3$ , $f_{y} = - 5$ . $\nabla f = (3, - 5) .$ Note it is the same everywhere — a linear function has a constant gradient (its picture is a flat tilted plane).

L1.2

Which of these is the gradient of $f (x, y) = x^{2} + y^{2}$ : (a) the scalar $2 x + 2 y$ , (b) the vector $(2 x, 2 y)$ , (c) the number $∣\nabla f ∣$ ?

Recall Solution

(b). The gradient is a vector — one number per input variable. Choice (a) illegally adds two different slopes into a scalar; (c) is the magnitude of the gradient, a length, not the gradient itself.

L1.3

True or false: $\nabla f$ is tangent to the level curve $f = c$ .

Recall Solution

False. $\nabla f$ is perpendicular (normal) to the level curve. Walking along a level curve keeps $f$ constant, so the directional derivative there is $0$ ; and $0 = \nabla f \cdot \overset{u}{^}$ forces $\nabla f ⊥ \overset{u}{^}$ for the tangent direction $\overset{u}{^}$ . (This is exactly the coral-vs-mint relationship in the picture above.)

Level 2 — Application

L2.1

$f (x, y) = x^{2} y$ . Compute $\nabla f$ at $(2, 3)$ .

Recall Solution

$f_{x} = 2 x y$ (freeze $y$ ; $\frac{d}{d x} x^{2} = 2 x$ times the constant $y$ ).
$f_{y} = x^{2}$ (freeze $x$ ; $\frac{d}{d y} y = 1$ times the constant $x^{2}$ ). At $(2, 3)$ : $f_{x} = 2 (2) (3) = 12$ , $f_{y} = 2^{2} = 4$ . $\nabla f (2, 3) = (12, 4) .$

L2.2

For the same $f$ at $(2, 3)$ , find the maximum rate of change and the direction achieving it.

Recall Solution

Max rate $= ∣\nabla f ∣ = 1 2^{2} + 4^{2} = 144 + 16 = 160 = 410 \approx 12.649$ . Direction = the gradient's own unit vector: $\hat{d} = \frac{( 12 , 4 )}{4 10} = (\frac{3}{10}, \frac{1}{10}) .$ Why $∣\nabla f ∣$ ? Because $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ is largest when $cos θ = 1$ , i.e. $θ = 0$ (walk straight along $\nabla f$ ); then the rate equals $∣\nabla f ∣$ . The figure below shows this "shadow" $\nabla f cos θ$ shrinking as $\overset{u}{^}$ swings away from the gradient.

L2.3

$f (x, y) = x^{2} y$ , point $(2, 3)$ , direction toward $(2 + 3, 3 + 4) = (5, 7)$ . Find $D_{\overset{u}{^}} f$ .

Recall Solution

Step 1 — build the direction. Displacement $= (5 - 2, 7 - 3) = (3, 4)$ , length $3^{2} + 4^{2} = 5$ . Step 2 — normalise (must be unit length): $\overset{u}{^} = (3/5, 4/5)$ . Step 3 — dot with the gradient (from L2.1, $\nabla f = (12, 4)$ ): $D_{\overset{u}{^}} f = (12, 4) \cdot (3/5, 4/5) = \frac{36}{5} + \frac{16}{5} = \frac{52}{5} = 10.4.$

L2.4

$f (x, y, z) = x y z$ . Find $\nabla f$ at $(1, 2, 3)$ .

Recall Solution

Same rule in three variables — the gradient just gets a third slot.

$f_{x} = y z$ , $f_{y} = x z$ , $f_{z} = x y$ . At $(1, 2, 3)$ : $f_{x} = 6$ , $f_{y} = 3$ , $f_{z} = 2$ . $\nabla f (1, 2, 3) = (6, 3, 2) .$

Level 3 — Analysis

L3.1

$f (x, y) = x^{2} + y^{2}$ . At $(3, 4)$ , show $\nabla f$ is perpendicular to the level circle, and give a unit tangent to the circle there.

Recall Solution

$\nabla f = (2 x, 2 y) = (6, 8)$ at $(3, 4)$ — it points radially outward from the origin. This is the coral arrow in the page's opening figure: a radius is perpendicular to its circle, so the gradient is the normal. A tangent must satisfy $\nabla f \cdot \overset{u}{^} = 0$ . Rotate $(6, 8)$ by $9 0^{\circ}$ : $(- 8, 6)$ . Check: $(6) (- 8) + (8) (6) = - 48 + 48 = 0$ ✓ — this is the mint arrow. Unit tangent: $\frac{( - 8 , 6 )}{64 + 36} = \frac{( - 8 , 6 )}{10} = (- 4/5, 3/5)$ .

L3.2

At $(3, 4)$ for $f = x^{2} + y^{2}$ , in which direction does $f$ decrease fastest, and at what rate?

Recall Solution

Steepest descent is $- \nabla f = (- 6, - 8)$ — points straight toward the origin (downhill on the bowl), the flip of the coral arrow. Rate $= - ∣\nabla f ∣ = - 36 + 64 = - 100 = - 10$ . Why negative? $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ$ is smallest at $θ = π$ ( $cos π = - 1$ ), giving $- ∣\nabla f ∣$ .

L3.3

$f (x, y) = x^{2} y$ , point $(2, 3)$ , $\nabla f = (12, 4)$ . Find a unit direction along which $f$ is momentarily unchanged ( $D_{\overset{u}{^}} f = 0$ ).

Recall Solution

We need $\overset{u}{^} ⊥ \nabla f$ , i.e. $12 u_{1} + 4 u_{2} = 0 \Rightarrow u_{2} = - 3 u_{1}$ . Take $(1, - 3)$ , length $1 + 9 = 10$ , so $\overset{u}{^} = (\frac{1}{10}, \frac{- 3}{10}) (or its negative) .$ What it looks like: this $\overset{u}{^}$ is tangent to the level curve of $f$ through $(2, 3)$ — walk that way and your height doesn't change (the mint-arrow role again).

L3.4

For $f = x^{2} + y^{2}$ at $(3, 4)$ , compute $D_{\overset{u}{^}} f$ for $\overset{u}{^}$ at angle $6 0^{\circ}$ to $\nabla f$ , and check it matches the dot-product form.

Recall Solution

Cosine form: $D_{\overset{u}{^}} f = ∣\nabla f ∣ cos 6 0^{\circ} = 10 \cdot \frac{1}{2} = 5.$ This is the fast route: we only need the magnitude and the angle, not the exact $\overset{u}{^}$ . Any $\overset{u}{^}$ making $6 0^{\circ}$ with $(6, 8)$ gives the same value — the geometry (a shadow of $\nabla f$ onto $\overset{u}{^}$ , exactly the s02 figure) is all that matters.

Level 4 — Synthesis

L4.1

Find the tangent plane to $z = f (x, y) = x^{2} y$ at the point $p = (2, 3)$ .

Recall Solution

Notation first. Write $p = (2, 3)$ for the base point and $x = (x, y)$ for a nearby point, so $x - p = (x - 2, y - 3)$ is the small step away from $p$ . Why this formula. Near $p$ a smooth surface looks flat (a plane). A plane's height changes linearly with the step: move $Δ x$ in $x$ and the height rises by (slope in $x$ ) $\times Δ x$ , i.e. $f_{x} Δ x$ ; likewise $f_{y} Δ y$ . Adding both contributions is precisely the dot product $\nabla f (p) \cdot (x - p)$ . Start from the known height $f (p)$ and add this linear rise (see Tangent plane and linearisation): $z = f (p) + \nabla f (p) \cdot (x - p) .$ Plug in. $f (2, 3) = 2^{2} \cdot 3 = 12$ and $\nabla f (2, 3) = (12, 4)$ (from L2.1). $z = 12 + 12 (x - 2) + 4 (y - 3) = 12 + 12 x - 24 + 4 y - 12,$ $z = 12 x + 4 y - 24 .$ Sanity check at $p = (2, 3)$ : $12 (2) + 4 (3) - 24 = 24 + 12 - 24 = 12 = f (2, 3)$ ✓ — the plane touches the surface exactly at the base point.

L4.2

A hiker stands at $(1, 1)$ on the hill $h (x, y) = 4 - x^{2} - y^{2}$ . Which compass direction (unit vector) is steepest uphill, and how steep? In which direction would the hiker descend?

Recall Solution

$\nabla h = (- 2 x, - 2 y) = (- 2, - 2)$ at $(1, 1)$ . Direction of ascent = $\nabla h$ itself: unit form $\frac{( - 2 , - 2 )}{8} = (- \frac{1}{2}, - \frac{1}{2})$ — toward the origin (the summit at $(0, 0)$ ). Steepness $= ∣\nabla h ∣ = 4 + 4 = 8 = 22 \approx 2.828$ . Descent = $- \nabla h = (2, 2)$ , unit $(\frac{1}{2}, \frac{1}{2})$ — away from the summit. This $- \nabla h$ step is exactly what Gradient descent follows to go downhill.

L4.3

$f (x, y) = e^{x} sin y$ . Find $\nabla f$ at $(0, π /2)$ and the max rate of increase there.

Recall Solution

$f_{x} = e^{x} sin y$ (the $sin y$ is constant w.r.t. $x$ ; $\frac{d}{d x} e^{x} = e^{x}$ ).
$f_{y} = e^{x} cos y$ (the $e^{x}$ is constant w.r.t. $y$ ; $\frac{d}{d y} sin y = cos y$ ). At $(0, π /2)$ : $e^{0} = 1$ , $sin (π /2) = 1$ , $cos (π /2) = 0$ . $\nabla f = (1 \cdot 1, 1 \cdot 0) = (1, 0) .$ Max rate $= ∣\nabla f ∣ = 1$ , in direction $(1, 0)$ (the $+ x$ direction).

L4.4

Two functions: $f = x^{2}$ , $g = y^{3}$ . Verify $\nabla (2 f + 3 g) = 2\nabla f + 3\nabla g$ at $(1, 1)$ .

Recall Solution

Left side. $2 f + 3 g = 2 x^{2} + 3 y^{3}$ , so $\nabla (2 f + 3 g) = (4 x, 9 y^{2}) = (4, 9)$ at $(1, 1)$ . Right side. $\nabla f = (2 x, 0) = (2, 0)$ ; $\nabla g = (0, 3 y^{2}) = (0, 3)$ . $2 (2, 0) + 3 (0, 3) = (4, 0) + (0, 9) = (4, 9)$ ✓. Why it works: each partial derivative is linear, and stacking linear operations into a vector keeps them linear — this is the same linearity that makes the tangent plane exact for linear parts.

Level 5 — Mastery

L5.1

On the ellipse-like surface, find the point(s) on the circle $x^{2} + y^{2} = 25$ where $f (x, y) = x + 2 y$ is largest, using gradients. (A taste of Lagrange multipliers.)

Recall Solution

Idea. At the maximum along the constraint circle, moving along the circle can't increase $f$ , so $\nabla f$ must be parallel to the circle's normal $\nabla g$ , where $g = x^{2} + y^{2}$ .

$\nabla f = (1, 2)$ (constant).
$\nabla g = (2 x, 2 y)$ . Set $\nabla f = λ \nabla g$ : $1 = 2 λ x$ , $2 = 2 λ y \Rightarrow y = 2 x$ . Substitute into $x^{2} + y^{2} = 25$ : $x^{2} + 4 x^{2} = 25 \Rightarrow x^{2} = 5 \Rightarrow x = \pm 5$ , $y = \pm 25$ . The maximum picks the $+$ branch (aligned with $\nabla f$ ): $(5, 25)$ , giving $f = 5 + 2 (25) = 55 \approx 11.180.$ The minimum is at $(- 5, - 25)$ with $f = - 55$ .

L5.2

$f (x, y) = x^{2} y$ . Prove the directional derivative at $(2, 3)$ can never exceed $410$ , whatever unit direction you pick.

Recall Solution

$D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = ∣\nabla f ∣ ∣ \overset{u}{^} ∣ cos θ = ∣\nabla f ∣ cos θ$ (since $∣ \overset{u}{^} ∣ = 1$ ). Because $cos θ \leq 1$ for every angle, $D_{\overset{u}{^}} f \leq ∣\nabla f ∣ = ∣ (12, 4) ∣ = 160 = 410$ . Equality holds only when $θ = 0$ , i.e. $\overset{u}{^}$ points along $\nabla f$ . So $410$ is a hard ceiling — the max rate. This is the Cauchy–Schwarz bound in disguise.

L5.3

The temperature is $T (x, y) = 100 - x^{2} - 2 y^{2}$ . A bug at $(1, 1)$ wants to cool down fastest. Give its unit direction of motion and the cooling rate. Then find a direction along which its temperature stays constant.

Recall Solution

$\nabla T = (- 2 x, - 4 y) = (- 2, - 4)$ at $(1, 1)$ . Cool fastest = steepest descent = $- \nabla T = (2, 4)$ . Unit: $\frac{( 2 , 4 )}{20} = (\frac{1}{5}, \frac{2}{5})$ . Cooling rate $= - ∣\nabla T ∣ = - 4 + 16 = - 20 = - 25 \approx - 4.472$ (temperature drops at $25$ per unit length). Constant-temperature direction: perpendicular to $\nabla T = (- 2, - 4)$ . Solve $- 2 u_{1} - 4 u_{2} = 0 \Rightarrow u_{1} = - 2 u_{2}$ ; take $(- 2, 1)$ , unit $(- \frac{2}{5}, \frac{1}{5})$ . This is tangent to the isotherm (level curve) through the bug.

L5.4

$f (x, y) = ln (x^{2} + y^{2})$ . Show $∣\nabla f ∣$ depends only on the distance $r = x^{2} + y^{2}$ from the origin, and state that dependence.

Recall Solution

Let $r^{2} = x^{2} + y^{2}$ . Using $\frac{d}{d x} ln (u) = \frac{u _{x}}{u}$ :

$f_{x} = \frac{2 x}{x ^{2} + y ^{2}}$ , $f_{y} = \frac{2 y}{x ^{2} + y ^{2}}$ . $∣\nabla f ∣ = \frac{( 2 x ) ^{2} + ( 2 y ) ^{2}}{x ^{2} + y ^{2}} = \frac{2 x ^{2} + y ^{2}}{x ^{2} + y ^{2}} = \frac{2}{x ^{2} + y ^{2}} = \frac{2}{r} .$ So $∣\nabla f ∣ = 2/ r$ : steepness grows as you approach the origin and fades to $0$ far away — matching the level curves (circles) being crowded near the centre. (Undefined at $r = 0$ , where $ln$ blows up.)

Recall Self-test checklist

Can compute $\nabla f$ in 2D and 3D? ::: L1–L2, L4.3 Can turn a target point into a unit $\overset{u}{^}$ and get $D_{\overset{u}{^}} f$ ? ::: L2.3 Can find steepest ascent/descent directions and rates? ::: L3.2, L4.2, L5.3 Can find directions of zero change (tangent to level set)? ::: L3.3, L5.3 Can build a tangent plane from $\nabla f$ ? ::: L4.1 Can bound $D_{\overset{u}{^}} f$ and use $\nabla f ∥ \nabla g$ ? ::: L5.2, L5.1

Connections

Partial derivatives — every gradient component is one.
Directional derivative — the $\nabla f \cdot \overset{u}{^}$ engine used throughout.
Chain rule (multivariable) — behind the linearity checks.
Tangent plane and linearisation — L4.1.
Level curves and surfaces — L3.1, L3.3, L5.3.
Gradient descent — L4.2, L5.3 (moving along $- \nabla f$ ).
Lagrange multipliers — L5.1.