Worked examples — Gradient perpendicular to level curves - surfaces

Before anything, three symbols must be earned so nobody trips on line one.

Everything below is: compute $\nabla f$, apply the $90^{\circ }$ trick for the tangent, and confirm the dot product is $0$.

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The scenario matrix

Each row is a class of situation. "Quadrant" here always refers to where the gradient arrow $\nabla f$ points (its sign pattern), not to where the point sits. The last column names the worked example that lands on it.

Cell	What makes it different	Worst-case danger	Example
A $\nabla f$ in Quadrant I (both entries $>0$)	both partials $>0$	none — warm-up	Ex 1
B $\nabla f$ has a sign mix (one entry $<0$)	some partial $<0$	wrong tangent sign	Ex 2
C Axis / one partial $=0$	gradient points purely along one axis	tangent is a coordinate axis	Ex 3
D Degenerate: $\nabla f=0$	gradient vanishes	"perpendicular to WHAT?" breaks	Ex 4
E 3D level surface	tangent plane, not line	must use point-normal form	Ex 5
F Limiting behaviour	curve running to infinity	does the normal keep pointing "outward"?	Ex 6
G Real-world word problem	contour map / temperature	reading units, direction of ascent	Ex 7
H Exam twist: graph vs level set	normal to $z=f(x,y)$ not the level curve	the classic trap	Ex 8

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Example 1 — Cell A: gradient in Quadrant I (warm-up)

Forecast: guess before reading — will the gradient point into the ellipse or out of it, and roughly which way (up-right, up-left…)?

1. Compute the gradient. $\nabla f=(2x,6y)=(4,12)$. Why this step? The gradient is our normal vector; nothing happens until we have it. ${\partial }_x(x^2+3y^2)=2x$, ${\partial }_y(x^2+3y^2)=6y$. Both entries are positive, so $\nabla f$ lies in Quadrant I — this is Cell A.
2. Get a tangent direction. Apply the $90^{\circ }$ trick $(a,b)\mapsto (-b,a)$ to $(4,12)$: that gives $(-12,4)$ (or scaled, $(-3,1)$). Why this step? This is the mechanical way to produce "along the level set" from "across it."
3. Dot-product check. $(4,12)\cdot (-12,4)=-48+48=0$. Why this step? Zero dot product is the definition of perpendicular here — this is the whole theorem in one line.

Verify: Both partials are positive at $(2,2)$, so $\nabla f=(4,12)$ points up-and-to-the-right, out of the ellipse toward higher $f$ (steepest ascent).

Figure 1 — The blue ellipse $x^2+3y^2=16$. At $(2,2)$ the orange gradient arrow $(4,12)$ exits the curve; the green tangent arrow lies along the curve. The two arrows meet at a clean right angle (Cell A).

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Example 2 — Cell B: gradient with a sign mix

Forecast: one component of $\nabla f$ will be negative. Which one, and does the gradient still point toward increasing $f$?

1. Gradient. $\nabla f=(2x,-2y)=(2,-4)$. Why this step? Notice ${\partial }_y(-y^2)=-2y$: the minus sign is exactly where students drop it. The gradient $(2,-4)$ has a positive and a negative entry — a sign mix, so this is Cell B.
2. Sign sanity of the direction. At $(1,2)$, $f=1-4=-3$ ✅ (we are on the right curve). Moving in $+x$ raises $f$; moving in $+y$ lowers $f$. So the uphill arrow leans right and down — matching $(2,-4)$. Why this step? Cell B's danger is a flipped sign; check the physics against the formula.
3. Tangent + dot check. Apply the $90^{\circ }$ trick $(a,b)\mapsto (-b,a)$ to $(2,-4)$: $(4,2)$. Dot: $(2,-4)\cdot (4,2)=8-8=0$. ✅ Why this step? Confirms perpendicularity despite the sign mix.

Verify: Implicit slope of $x^2-y^2=-3$ is $\frac{dy}{dx}=-\frac{f_x}{f_y}=-\frac{2}{-4}=\frac{1}{2}$. Tangent $(4,2)$ has slope $2/4=\frac{1}{2}$. ✅ Match.

Figure 2 — The blue level curve of the saddle $x^2-y^2=-3$. At $(1,2)$ the orange gradient $(2,-4)$ points right-and-down (its $y$-entry is negative); the green tangent $(4,2)$ runs along the curve, perpendicular to it (Cell B).

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Example 3 — Cell C: one partial is zero (gradient along an axis)

Forecast: at the very top of a circle, which way must the tangent point? Horizontally or vertically?

1. Gradient. $\nabla f=(2x,2y)=(0,10)$. Why this step? The $x$-component is exactly $0$ — this is the "one partial vanishes" case. The gradient points straight up the $y$-axis.
2. Tangent. Apply the $90^{\circ }$ trick to $(0,10)$: $(-10,0)$, i.e. direction $(1,0)$ (or $(-1,0)$): a purely horizontal direction. Why this step? When the normal is vertical, the tangent must be horizontal — Cell C forces the tangent onto a coordinate axis.
3. Dot check. $(0,10)\cdot (1,0)=0+0=0$. ✅ Why this step? Even in this special case the theorem holds identically.

Verify: The top of a circle of radius $5$ is $(0,5)$; its tangent line is horizontal ($y=5$), matching direction $(1,0)$. ✅

Figure 3 — The blue circle $x^2+y^2=25$. At the top point $(0,5)$ the orange gradient $(0,10)$ points straight up; the green tangent arrows run horizontally left and right, perpendicular to it (Cell C).

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Example 4 — Cell D: the degenerate case $\nabla f=0$

Forecast: does the theorem even apply here? Guess "yes / no / undefined."

1. Gradient. $\nabla f=(2x,2y)=(0,0)$ at the origin. Why this step? The gradient is the zero vector. And $0\cdot \mathbf{u}=0$ for every $\mathbf{u}$.
2. Interpret. Perpendicularity means "the only direction that dots to zero is the normal." But every direction dots to zero with $0$. So "the perpendicular direction" is not well-defined. Why this step? This is the honest limit of the theorem — it needs $\nabla f\ne 0$. Points where $\nabla f=0$ are critical points; the level set can pinch to a single point (here $f=0$ is just the origin, not a curve at all).
3. The rule to remember. The perpendicularity result holds wherever $\nabla f\ne 0$. At $\nabla f=0$ there is no unique normal. Why this step? Cell D is the one every exam hopes you forget.

Verify: $\nabla f(0,0)=(0,0)$, so its magnitude is $0$; the normal is undefined.

Figure 4 — Dashed gray circles are the level curves $f=1,4,9$. The level set $f=0$ is the single red dot at the origin. The faded orange arrows fan out in every direction: since $\nabla f=0$, no single one is "the" perpendicular (Cell D).

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Example 5 — Cell E: a 3D level surface (tangent plane)

Forecast: the normal now has three components. Guess whether the $z$-component is positive or negative.

1. Gradient. $\nabla f=(2x,2y,-1)=(2,2,-1)$. Why this step? In 3D the gradient is still the normal; ${\partial }_z(-z)=-1$ gives the negative third entry.
2. Point-normal plane. Write $\Delta \mathbf{x}=(x,y,z)-(1,1,2)$ for the displacement from our point to a general point on the plane (so $\Delta x=x-1$, $\Delta y=y-1$, $\Delta z=z-2$). The plane is all displacements perpendicular to the normal, $\nabla f\cdot \Delta \mathbf{x}=0$: $2(x-1)+2(y-1)-1(z-2)=0\Rightarrow 2x+2y-z=2.$ Why this step? A plane is "all displacements from the point that are perpendicular to the normal" — exactly $\nabla f\cdot \Delta \mathbf{x}=0$ (see Tangent Plane to a Surface).
3. Tangent-vector spot check. Take a curve $\mathbf{r}(t)$ (a moving point, as defined at the top) that stays on the surface and passes through $(1,1,2)$: let $x=1+t$, $y=1$, then $z=(1+t{)}^2+1$, so its velocity at $t=0$ is ${\mathbf{r}}^{\prime }(0)=(1,0,2)$. Dot with normal: $(2,2,-1)\cdot (1,0,2)=2+0-2=0$. ✅ Why this step? Confirms the gradient is perpendicular to an actual curve lying in the surface — the parent's Step-1 argument, made concrete.

Verify: Plane $2x+2y-z=2$ passes through $(1,1,2)$: $2+2-2=2$ ✅.

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Example 6 — Cell F: limiting behaviour (curve to infinity)

Forecast: as the curve flattens toward the $x$-axis far out, which way does the normal tilt?

1. Gradient formula. $\nabla f=(y,x)$. Why this step? ${\partial }_x(xy)=y$, ${\partial }_y(xy)=x$ — the normal literally is the swapped coordinates.
2. Evaluate the three points.
   • At $(2,3)$: $\nabla f=(3,2)$ — points up-right, moderately.
   • At $(6,1)$: $\nabla f=(1,6)$ — now steeply up.
   • At $(100,0.06)$: $\nabla f=(0.06,100)$ — almost straight up. Why this step? Far out along the $x$-axis the curve is nearly horizontal, so its normal must be nearly vertical — and it is.
3. Perpendicular in the limit. Tangent slope of $y=6/x$ is $-6/x^2\to 0^{-}$ (curve flattens). The gradient $(y,x)$ as an arrow has slope $x/y=x/(6/x)=x^2/6\to \infty$ (vertical). Their product $=(-6/x^2)(x^2/6)=-1$ for every $x$. ✅ Why this step? Two lines are perpendicular exactly when slopes multiply to $-1$; here it holds for all $x$, including the limit.

Verify: At $(2,3)$: tangent slope $-6/4=-1.5$, gradient-arrow slope $x/y=2/3\approx 0.667$; product $=(-1.5)(2/3)=-1$ ✅.

Figure 5 — The blue hyperbola $xy=6$. Orange gradient arrows $(y,x)$ are drawn at $(2,3)$ and $(6,1)$. As the point runs out along the $x$-axis the curve flattens, so the normal tilts toward vertical (Cell F).

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Example 7 — Cell G: real-world word problem (temperature map)

Forecast: guess whether "warm up fastest" points toward or away from the origin.

1. Current value. $T(3,2)=100-9-8=83^{\circ }$C. The isotherm is $100-x^2-2y^2=83$, i.e. $x^2+2y^2=17$ (an ellipse). Why this step? Establishes the level curve the bug is standing on.
2. Gradient = steepest-ascent direction. $\nabla T=(-2x,-4y)=(-6,-8)$ °C/cm. Why this step? The gradient always points toward the fastest increase — here toward higher temperature (see Directional Derivative and Steepest Descent / Gradient Descent). Since both entries are negative, "warmer" is toward the origin (down-left), which makes sense: $T$ peaks at the center.
3. Rate of fastest warming. $\Vert \nabla T\Vert =\sqrt{(-6{)}^2+(-8{)}^2}=\sqrt{36+64}=10$ °C/cm. Why this step? The magnitude of the gradient is the maximum rate of change per unit distance — units $={}^{\circ }$C per cm.
4. Constant-temperature direction. Apply the $90^{\circ }$ trick to $(-6,-8)$: e.g. $(8,-6)$ or $(-8,6)$. Why this step? Walking along the isotherm means zero change, i.e. tangent to the level curve = perpendicular to the gradient. This answers part (c).

Verify: Part (a): $T(3,2)=100-9-8=83$ ✅, isotherm $9+2(4)=17$ ✅. Part (b): $\Vert \nabla T\Vert =\sqrt{36+64}=10$ °C/cm, direction $(-6,-8)$ toward origin ✅. Part (c): dot check $\nabla T\cdot (8,-6)=(-6)(8)+(-8)(-6)=-48+48=0$ ✅ — the constant-temperature direction is genuinely perpendicular to the gradient.

Figure 6 — The blue isotherm $x^2+2y^2=17$ on the plate. At the bug's spot $(3,2)$ the orange gradient $(-6,-8)$ points down-left toward the hot center (steepest warming); the green tangent $(8,-6)$ runs along the isotherm — the direction of constant temperature, perpendicular to the gradient (Cell G).

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Example 8 — Cell H: the exam twist (graph vs level set)

Forecast: true or false? And if the graph's normal has three components, what is the third?

1. Separate the two objects. $\nabla f=(6,8)$ is a 2D vector in the input plane; it is normal to the level curve $x^2+y^2=25$, not to the 3D surface $z=f(x,y)$. Why this step? The trap is conflating the level curve (in the map) with the graph (a bowl in 3D).
2. Build the graph's normal properly. Write the graph as a level surface of a new function $g(x,y,z)=f(x,y)-z=x^2+y^2-z$. Its level set $g=0$ is the graph. Why this step? Now we can use the same theorem: $\nabla g$ is normal to $g=0$.
3. Compute. $\nabla g=(f_x,f_y,-1)=(6,8,-1)$. Why this step? ${\partial }_z(-z)=-1$ supplies the crucial third component the student forgot.

Verify: Tangent to the graph at $(3,4,25)$: use the moving point $\mathbf{r}(t)=(3+t,4,(3+t{)}^2+16)$, whose velocity at $t=0$ is ${\mathbf{r}}^{\prime }(0)=(1,0,6)$. Dot with $(6,8,-1)$: $6+0-6=0$ ✅ — the 3D normal is genuinely perpendicular to the surface.

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The whole matrix in one glance

Read the flowchart top-to-bottom as the decision you make at each point: first compute the gradient, then ask whether it is zero, then ask how many variables you have. Each leaf names the cell(s) it handles.

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Recall Self-test (reveal after guessing)

At $(0,5)$ on $x^2+y^2=25$, what is the tangent direction? ::: Horizontal, $(1,0)$ — because $\nabla f=(0,10)$ is vertical. Why does the perpendicularity theorem fail at the origin of $x^2+y^2$? ::: $\nabla f=0$; every direction dots to zero, so no unique normal. Real-world: max rate of temperature change at a point equals ::: $\Vert \nabla T\Vert$, in degrees per unit distance. Normal to the graph $z=x^2+y^2$ at $(3,4,25)$? ::: $(6,8,-1)$, from $g=f-z$. For $f=xy$, why is the gradient nearly vertical far out along the x-axis? ::: The curve flattens ($\text{slope}\to 0$), so its normal $(y,x)$ tilts toward vertical.

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Connections

• Directional Derivative — Ex 7: rate of steepest ascent is $\Vert \nabla T\Vert$.
• Multivariable Chain Rule — Ex 5's curve-through-surface check is the parent derivation made concrete.
• Tangent Plane to a Surface — Ex 5, Ex 8 use the point-normal form.
• Lagrange Multipliers — Ex 4's $\nabla f=0$ exception matters when constraint gradients vanish.
• Steepest Descent / Gradient Descent — Ex 7: walking against $\nabla T$ cools fastest.