Gradient vector ∇f — definition, properties: Question bank

Reminders so every symbol here is earned:

$\nabla f$ (say "grad f") is the vector of partial derivatives, e.g. $(f_{x}, f_{y})$ in 2D.
$f_{x} = \partial f / \partial x$ is the slope of $f$ when you wiggle only $x$ and freeze $y$ .
$D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^}$ is the rate of change when you step in the unit direction $\overset{u}{^}$ (length exactly $1$ ).
$∣\nabla f ∣$ is the length of the gradient arrow; $θ$ is the angle between $\overset{u}{^}$ and $\nabla f$ .

True or false — justify

The gradient of a two-variable function is a vector, not a number.

True. A single number can only encode one slope, but there are infinitely many directions to walk in; you need a whole vector

(f_{x}, f_{y})

to reconstruct the slope in any direction via

\nabla f \cdot \overset{u}{^}

\nabla f

points in the direction the level curve

f = c

runs.

False. It points perpendicular (normal) to the level curve. Moving along the curve keeps

f

constant, so

D_{\overset{u}{^}} f = \nabla f \cdot \overset{u}{^} = 0

there — a zero dot product means the vectors are at right angles.

The maximum rate of increase of

f

at a point equals

\nabla f

False (type error). The maximum rate is the number

∣\nabla f ∣

; the direction achieving it is the vector

\nabla f

. A rate is a scalar, a direction is a vector — don't equate them.

\nabla f = 0

at a point, then

f

is flat in every direction there.

True. Every directional derivative is

\nabla f \cdot \overset{u}{^} = 0 \cdot \overset{u}{^} = 0

, so no direction goes up or down to first order — this is exactly the condition for a critical point (max, min, or saddle).

Doubling every value of

f

doubles the gradient.

True, by linearity:

\nabla (2 f) = 2\nabla f

. The hill gets twice as steep everywhere, so every slope, and hence the whole arrow, scales by

2

The gradient is always tangent to the graph surface

z = f (x, y)

False.

\nabla f

lives in the input plane (the

x y

-domain), lying flat on the ground pointing uphill — it is not a vector on the 3D surface at all. Confusing the domain arrow with a surface direction is a common slip.

Along the direction of steepest descent,

D_{\overset{u}{^}} f = ∣\nabla f ∣

False — it's

- ∣\nabla f ∣

. Steepest descent is

\overset{u}{^} = - \nabla f /∣\nabla f ∣

, giving

θ = π

and

cos π = - 1

, so the rate is negative: you go down as fast as ascent goes up.

If two functions have the same gradient everywhere, they are the same function.

False. They can differ by a constant, e.g.

f

and

f + 7

share every partial derivative. The gradient sees slopes, and adding a constant shifts height without changing any slope.

Spot the error

D_{\overset{u}{^}} f

toward

(4, 6)

from

(1, 2)

: just dot

\nabla f

with

(3, 4)

Error:

(3, 4)

has length

5

, not

1

. Dotting with a non-unit vector scales the true rate by

∣ v ∣ = 5

. Fix: normalise to

\overset{u}{^} = (3/5, 4/5)

first.

"The steepest direction is

- \nabla f

because we minimise the dot product."

Error: steepest ascent maximises

\nabla f \cdot \overset{u}{^}

, achieved at

θ = 0

, i.e. along

+ \nabla f

- \nabla f

is steepest descent. The sign got flipped.

\nabla f \cdot \overset{u}{^} = ∣\nabla f ∣ ∣ \overset{u}{^} ∣ cos θ = ∣\nabla f ∣ cos θ

— so the

∣ \overset{u}{^} ∣

just disappears for free."

Not for free — it disappears because

∣ \overset{u}{^} ∣ = 1

by assumption. If you feed a non-unit vector the

∣ \overset{u}{^} ∣

factor stays and multiplies your answer.

"On the circle

x^{2} + y^{2} = 25

(3, 4)

, the gradient

(6, 8)

is tangent, since the tangent is

(- 4, 3)

and

(6, 8)

'runs alongside' it."

Error:

(6, 8) \cdot (- 4, 3) = - 24 + 24 = 0

, so

(6, 8)

is perpendicular to the tangent, i.e. it is the normal (radial) direction — not tangent.

f_{x} = 2 x y

(1, 2)

2 \cdot 1 \cdot 2 = 4

, and since we want the gradient we take

∣\nabla f ∣ = 4

before finding

f_{y}

Error: the magnitude uses all components. You must compute

f_{y} = x^{2} = 1

too, then

∣\nabla f ∣ = 4^{2} + 1^{2} = 17

. Never take magnitude of a half-built vector.

"To move at constant height, walk in the direction

\nabla f

Error: walking along

\nabla f

climbs fastest. Constant height needs

D_{\overset{u}{^}} f = 0

, i.e.

\overset{u}{^} ⊥ \nabla f

— walk across the arrow, along the level curve.

"Since

\nabla (f + g) = \nabla f + \nabla g

, also

\nabla (f g) = \nabla f \nabla g

Error: gradient is linear (sum rule holds) but not multiplicative. The product rule gives

\nabla (f g) = f \nabla g + g \nabla f

, a mix of both, using the product rule on each partial.

Why questions

Why does the chain rule force the directional derivative to be a dot product?

Stepping in direction

\overset{u}{^} = (u_{1}, u_{2})

moves

x

at rate

u_{1}

and

y

at rate

u_{2}

; the chain rule sums each partial slope times its rate:

f_{x} u_{1} + f_{y} u_{2}

, which is exactly

(f_{x}, f_{y}) \cdot (u_{1}, u_{2})

. The dot product isn't chosen — it's the shape of the chain rule.

Why is the steepest-ascent direction guaranteed to be

\nabla f

and not something else?

Because

D_{\overset{u}{^}} f = ∣\nabla f ∣ cos θ

and

cos θ

is largest (

= 1

) when

θ = 0

, i.e. when

\overset{u}{^}

points the same way as

\nabla f

. No other unit direction can beat pointing straight along the gradient.

Why must

\overset{u}{^}

be a unit vector in

D_{\overset{u}{^}} f

We want a rate per unit length travelled. A longer vector would count more ground per "step" and inflate the rate, so we fix the step length to exactly

1

to make the number a fair, comparable slope.

Why is the gradient perpendicular to level sets rather than parallel?

Along a level set

f

is constant, so the change is zero:

\nabla f \cdot \overset{u}{^}_{tangent} = 0

. A zero dot product means orthogonal, so

\nabla f

sticks out at right angles to the curve, not along it.

Why does a smooth function look linear near a point, making one vector enough to describe it?

Zooming in on any smooth hill flattens it into a tilted plane, and a plane's tilt is fully captured by its two axis slopes

(f_{x}, f_{y})

— a single vector. That local linearity is why the gradient alone determines all nearby behaviour.

Why can the same gradient value occur at many different points of a function?

The gradient is a vector field — a separate arrow at every point. Two points with the same local steepness and direction of climb (like two identical slopes on a hill) share the same gradient value even though they are different locations.

Edge cases

At a local maximum of

f

, what is

\nabla f

and why?

It is

0

. At the very top no direction climbs, so every directional derivative is zero, which forces both partials — and hence the whole gradient — to vanish.

At a saddle point, is

\nabla f = 0

, and does that mean

f

is a max or min?

Yes,

\nabla f = 0

(it's a critical point), but it is neither a max nor min:

f

rises in some directions and falls in others. A zero gradient signals "flat to first order," not which shape.

\nabla f = 0

, what is the "direction of steepest ascent"?

Undefined. Steepest ascent is

\nabla f /∣\nabla f ∣

, but dividing by

∣\nabla f ∣ = 0

is illegal — geometrically there is no uphill because the ground is level in all directions to first order.

For the constant function

f (x, y) = 5

, what is

\nabla f

everywhere?

The zero vector

(0, 0)

at every point. A flat sheet has no slope in any direction, so all partials are zero and every level set is the whole plane.

θ = π /2

between

\overset{u}{^}

and

\nabla f

, what is

D_{\overset{u}{^}} f

and what are you doing physically?

D_{\overset{u}{^}} f = ∣\nabla f ∣ cos (π /2) = 0

: you are walking along the level curve, staying at exactly the same height, neither climbing nor descending.

Can

∣\nabla f ∣

be negative?

No. It is a length, computed as

f_{x}^{2} + f_{y}^{2} \geq 0

. The rate in a chosen direction can be negative (going downhill), but the gradient's magnitude never is.

If you scale

f

- 1

(flip the hill into a valley), what happens to the gradient?

It reverses:

\nabla (- f) = - \nabla f

. Every uphill becomes downhill, so the steepest-ascent arrow flips to point the opposite way — this is precisely why gradient descent follows

- \nabla f

Connections

Partial derivatives — the components each trap above manipulates.
Directional derivative — the unit-vector rule lives here.
Chain rule (multivariable) — source of the dot-product form.
Level curves and surfaces — the perpendicularity traps.
Gradient descent — why the sign of $\nabla f$ matters.
Tangent plane and linearisation — the local-linearity "why".
Lagrange multipliers — normals of level sets, tested above.