Multivariable-Calculus · Unit 11 · Video 2 · Interactive Practice

One Normal Vector, One Tangent Plane: The Gradient on a Hyperboloid

IKey Formulas

Formula	Name	At $P = (2, 1, 1)$ on $x^2 + y^2 - z^2 = 4$
$\nabla w(P) \perp \{\, w = c \,\}$ at $P$	Gradient is normal to the level surface	$\nabla w = \langle 2x, 2y, -2z \rangle$ , so $\nabla w(P) = \langle 4, 2, -2 \rangle$
$\nabla w(P) \cdot \langle x, y, z \rangle = \nabla w(P) \cdot \langle x_0, y_0, z_0 \rangle$	Tangent plane (gradient route)	$4x + 2y - 2z = 8$
$\Delta w \approx w_x\,\Delta x + w_y\,\Delta y + w_z\,\Delta z$	Linear approximation	$\Delta w \approx 4\,\Delta x + 2\,\Delta y - 2\,\Delta z$
$w_x(P)\,\Delta x + w_y(P)\,\Delta y + w_z(P)\,\Delta z = 0$	Tangent plane (differential route)	$4(x - 2) + 2(y - 1) - 2(z - 1) = 0$

Key Insight: The coefficients $4, 2, -2$ of $dw = 4\,dx + 2\,dy - 2\,dz$ are exactly the components of $\nabla w(2,1,1)$ — both routes use the same three numbers and the same point, so they give the same plane.

IIVisualization 1 — The Gradient as a Normal Vector

At every point of the level surface $w = x^2 + y^2 - z^2 = 4$ , $\nabla w$ is normal to the surface and fixes its tangent plane.

💡 The plane meets this surface along two straight lines through $P$ — at $(2, 1, 1)$ they are $(2,1,1) + t\langle 0, 1, 1 \rangle$ and $(2,1,1) + t\langle 4, -3, 5 \rangle$ — so near $P$ the surface lies above the plane in two opposite sectors and below it in the other two, like a saddle.

IIIVisualization 2 — Zooming In: the Linear Part Vanishes on the Plane

Along the tangent plane the linear part of $\Delta w$ is exactly zero; the true $\Delta w$ at distance $s$ shrinks like $s^2$ .

Slice through P Zoom ×1

💡 $\Delta w < 0$ along the $y = 1$ slice but $\Delta w > 0$ along the $z = 1$ slice: the tangent plane leaves the surface on opposite sides in different directions, and along $x = 2$ it never leaves at all.

IVQuiz Questions

Problem 1 · Tangent Plane from the Gradient

Given: the level surface $x^2 + 3y + z^2 = 11$ and the point $P = (1, 2, 2)$ on it — find the tangent plane at $P$ .

✅ Correct! The normal is

\nabla w(1,2,2) = \langle 2, 3, 4 \rangle

, and plugging in

P

fixes the constant:

2 + 6 + 8 = 16

❌ Not quite. The right-hand side is not the level value

11

; it comes from plugging

P

into

2x + 3y + 4z

. Check:

2(1) + 3(2) + 4(2) = 16 \ne 11

, so your plane misses

P

❌ Not quite. That plane passes through the origin, not through

P

. The tangent plane must contain

P

: plug it in to fix the constant.

❌ Not quite. You used the point

\langle 1, 2, 2 \rangle

as the normal vector. The normal is the gradient at the point,

\nabla w(P) = \langle 2x, 3, 2z \rangle\big|_P

❌ Not quite. Compute

\nabla w

, evaluate it at

P

, then plug

P

into the plane.

Show solution

Write the surface as the level set $w = 11$ of $w = x^2 + 3y + z^2$ ; $P$ is on it since $1 + 6 + 4 = 11$ .

\nabla w = \langle 2x,\ 3,\ 2z \rangle, \qquad \nabla w(1,2,2) = \langle 2, 3, 4 \rangle

A plane with normal $\langle 2, 3, 4 \rangle$ is $2x + 3y + 4z = \text{const}$ . Plug in $P$ :

2(1) + 3(2) + 4(2) = 16 \quad\Longrightarrow\quad 2x + 3y + 4z = 16

The level value $11$ never appears in the plane's equation: it only told us which level surface we were on.

Problem 2 · Watch the Sign of $-2z$

Given: the hyperboloid $x^2 + y^2 - z^2 = 4$ and the point $P = (1, 2, -1)$ on it — find the tangent plane at $P$ .

✅ Correct!

\nabla w(1,2,-1) = \langle 2, 4, 2 \rangle

— the third component is

-2(-1) = +2

— and

2 + 8 - 2 = 8

gives

x + 2y + z = 4

❌ Close, but check the sign. This plane does contain

P

, but its normal

\langle 1, 2, -1 \rangle

is the gradient direction at

(1, 2, 1)

. At

z = -1

, the component

-2z

equals

+2

❌ Close — right normal, wrong constant. Plug the actual

z

-coordinate

-1

1 + 2(2) + (-1) = 4

, not

6

. Your plane misses

P

❌ Not quite. Evaluate

\nabla w = \langle 2x, 2y, -2z \rangle

(1, 2, -1)

carefully, then plug

P

in.

Show solution

$P$ is on the surface: $1 + 4 - 1 = 4$ . With $w = x^2 + y^2 - z^2$ :

\nabla w(1, 2, -1) = \langle 2(1),\ 2(2),\ -2(-1) \rangle = \langle 2, 4, 2 \rangle

2x + 4y + 2z = 2(1) + 4(2) + 2(-1) = 8

Dividing by $2$ : $x + 2y + z = 4$ .

Common mistakes: copying $-2$ for the third component from the video's point $(2, 1, 1)$ (there $z = 1$ ), or plugging $z = +1$ when fixing the constant.

Problem 3 · The Differential Route

Given: $w = x^2 + y^2 - z^2$ near $P = (2, 1, 1)$ , where $w(P) = 4$ — find the linear approximation of $\Delta w$ for the move $(\Delta x, \Delta y, \Delta z) = (0.1, -0.1, 0.2)$ , and the $\Delta z$ that puts $(2.1,\ 0.9,\ 1 + \Delta z)$ on the tangent plane.

Linear approximation of $\Delta w$ for $(0.1, -0.1, 0.2)$ ?

Which $\Delta z$ puts $(2.1,\ 0.9,\ 1 + \Delta z)$ on the tangent plane?

✅ Correct! The linear part is

-0.2

; setting it exactly to zero gives

\Delta z = 0.1

, and indeed

w(2.1, 0.9, 1.1) = 4.01

— almost on the surface.

❌ Check the sign of the $z$ -term. The coefficient is

-2

, so the

z

-term is

-2(0.2) = -0.4

, not

+0.4

❌ That is the exact change.

w(2.1, 0.9, 1.2) - 4 = -0.22

; the linear approximation keeps only

4\,\Delta x + 2\,\Delta y - 2\,\Delta z

❌ Not quite. Substitute into

4\,\Delta x + 2\,\Delta y - 2\,\Delta z

term by term.

❌ Check $\Delta z$ . The tangent plane is

4\,\Delta x + 2\,\Delta y - 2\,\Delta z = 0

exactly; with

\Delta x = 0.1

\Delta y = -0.1

solve for

\Delta z

Show solution

Step 1: Linear approximation. At $P$ , $\nabla w = \langle 4, 2, -2 \rangle$ , so

\Delta w \approx 4(0.1) + 2(-0.1) - 2(0.2) = 0.4 - 0.2 - 0.4 = -0.2

(The exact change is $4.41 + 0.81 - 1.44 - 4 = -0.22$ ; the difference is the second-order part.)

Step 2: Tangent plane. Set the linear formula exactly to zero:

4(0.1) + 2(-0.1) - 2\,\Delta z = 0 \quad\Longrightarrow\quad 0.2 = 2\,\Delta z \quad\Longrightarrow\quad \Delta z = 0.1

Check: $4(2.1) + 2(0.9) - 2(1.1) = 8.4 + 1.8 - 2.2 = 8$ ✓, and $w(2.1, 0.9, 1.1) = 4.41 + 0.81 - 1.21 = 4.01 \approx 4$ .

Problem 4 · Every Point at Once

Given: any point $P_0 = (x_0, y_0, z_0)$ on $x^2 + y^2 - z^2 = 4$ . Its tangent plane can be written $x_0\,x + y_0\,y - z_0\,z = c$ — find $c$ .

✅ Correct! Plugging

P_0

in gives

x_0^2 + y_0^2 - z_0^2

, which is

4

because

P_0

is on the surface. At

(2, 1, 1)

2x + y - z = 4

, the video's plane divided by

2

❌ Not quite.

c = 0

is a plane through the origin, and no point of this surface is the origin (

0 \ne 4

), so it cannot contain

P_0

❌ Close.

8

is the constant for the normal

\langle 2x_0, 2y_0, -2z_0 \rangle

; the equation here has already been divided by

2

❌ Watch the sign. Plugging

P_0

into

x_0\,x + y_0\,y - z_0\,z

gives

x_0^2 + y_0^2 - z_0^2

, with a minus on the

z

-term — and that has a fixed value on this surface.

❌ Not quite. Plug

P_0

itself into the plane and use the surface equation.

Show solution

$\nabla w(P_0) = \langle 2x_0, 2y_0, -2z_0 \rangle$ , so the plane is $2x_0\,x + 2y_0\,y - 2z_0\,z = \text{const}$ . Plug in $P_0$ :

\text{const} = 2x_0^2 + 2y_0^2 - 2z_0^2 = 2\,(x_0^2 + y_0^2 - z_0^2) = 2 \cdot 4 = 8

Divide by $2$ : $x_0\,x + y_0\,y - z_0\,z = 4$ , so $c = 4$ at every point of the surface.

Check at $(2, 1, 1)$ : $2x + y - z = 4$ , which is $4x + 2y - 2z = 8$ divided by $2$ ✓.

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