The Gradient, Tangent Planes and Normal Lines

Main idea

The gradient of $F(x, y, z)$ always sticks straight out of the level surface $F = k$ . So it's a ready-made normal vector, and that gives you tangent planes to any surface, even ones like spheres that aren't the graph of a function.

{\htmlData{sym=0}{F_x}}(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0

What the symbols mean

$F_x$
Gradient components
∇F at the point is the normal vector to the surface F = k.

Normal line

\vec r(t) = \langle x_0, y_0, z_0\rangle + t\,\nabla F(x_0, y_0, z_0)

Explanation

A normal line is simply the line that sticks straight out of a surface at a point, like a flagpole planted perpendicular to the ground, and the tangent plane is the flat sheet lying against the surface there. The gradient hands you the direction of that flagpole. That's useful for things like shading in 3D graphics, which needs to know which way every bit of surface faces. The tangent plane formula $F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$ looks long, but it's just $\langle F_x, F_y, F_z\rangle$ (the gradient, the direction sticking out) dotted with $\langle x - x_0, y - y_0, z - z_0\rangle$ (the arrow from the point to anywhere on the plane), set to zero. It's the plane equation from the first unit with the gradient as its normal.

The last lesson only handled surfaces of the form $z = f(x, y)$ . A sphere like $x^2 + y^2 + z^2 = 14$ isn't one of those, since each $(x, y)$ has two heights. But it is a level surface: the set where $F(x, y, z) = x^2 + y^2 + z^2$ equals $14$ . We already know that in 2D, the gradient is perpendicular to level curves. The same thing is true one dimension up.

Why it's perpendicular. Take any curve $\vec r(t)$ lying on the surface $F(x, y, z) = k$ . Along it, $F$ never changes, so its rate of change is zero. By the chain rule that rate is $\nabla F\cdot\vec r\,'(t)$ , so $\nabla F\cdot\vec r\,'(t) = 0$ . Think about what that says: $\vec r\,'$ is a tangent direction to the surface, and the curve could be any curve on the surface, heading any way. So the gradient is perpendicular to every tangent direction at once. That's exactly what a normal vector is.

Graphs are a special case. Write $z = f(x, y)$ as the level surface $F = f(x, y) - z = 0$ . Then $\nabla F = \langle f_x, f_y, -1\rangle$ , which is exactly the normal we found in the last lesson. Good, the two methods agree.

The normal line is then just a line through the point in the direction of $\nabla F$ , using the point-plus-direction form from the first unit.

See it

The dome z = 4 − x² − y², thought of as the level surface x² + y² + z = 4, with ∇F = ⟨2x, 2y, 1⟩ drawn at a point circling it at height 3. As you can see, the gradient always sticks straight out of the surface, leaning outward because the dome slopes away from the top.

Worked examples

Example 1

Find the tangent plane to the sphere $x^2 + y^2 + z^2 = 14$ at $(1, 2, 3)$ .

1
The sphere is the level surface of $F = x^2 + y^2 + z^2$ , so the normal is $\nabla F = \langle 2x, 2y, 2z\rangle = \langle 2, 4, 6\rangle$ at the point.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\nabla F$ for $F = x^2 + y^2 + z^2$ at $(1, 2, 3)$ .

Problem 2+5 XP

The tangent plane to $x^2 + y^2 + z^2 = 14$ at $(1, 2, 3)$ is $x + 2y + 3z = d$ . Find $d$ .

Problem 3+5 XP

Find the gradient of $f = xy$ at $(2, 3)$ (it's perpendicular to the level curve $xy = 6$ ).

Problem 4+5 XP

The normal line to the sphere at $(1, 2, 3)$ is $\langle 1, 2, 3\rangle + t\langle 1, 2, 3\rangle$ . Where is it at $t = 1$ ?

Problem 5+5 XP

The tangent plane to $z = x^2 + y^2$ at $(1, 1, 2)$ is $2x + 2y - z = d$ . Find $d$ .

Problem 6+5 XP

A rover rests on terrain $z = 50 - 0.01x^2 - 0.02y^2$ (meters) at the point above $(10, 10)$ . Using the normal vector $\langle -h_x, -h_y, 1\rangle$ , find how far the ground there is tilted from horizontal, in degrees.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the point with positive coordinates on the ellipsoid $x^2 + 2y^2 + 3z^2 = 21$ where the tangent plane is parallel to the plane $x + 4y + 6z = 0$ .

Where it's used

Computer graphics shades curved surfaces using normals from gradients of implicit surfaces; robots compute surface normals to know which way to push.

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