How to find the normal line to a surface using the gradient

A normal line passes through a point on a surface and points perpendicular to the tangent plane there. The gradient gives us that direction directly. Once we have the gradient at the point, we can write both the normal line and the tangent plane.

Use the gradient as a direction vector

For a level surface $F (x, y, z) = c$ , suppose $P = (x_{0}, y_{0}, z_{0})$ lies on the surface and the gradient there is nonzero. Evaluate the partial derivatives at that point.

\nabla F (P) = (F_{x} (P), F_{y} (P), F_{z} (P))

A normal line through P has these parametric equations.

\begin{array}{l} x = x_{0} + t F_{x} (P) \\ y = y_{0} + t F_{y} (P) \\ z = z_{0} + t F_{z} (P) \end{array}

Parametric form also works when a direction component is zero: that coordinate stays constant. When all three direction components are nonzero, we can also use symmetric form.

\frac{x - x_{0}}{F_{x} (P)} = \frac{y - y_{0}}{F_{y} (P)} = \frac{z - z_{0}}{F_{z} (P)}

Schematic: the normal line passes through the point of tangency and is perpendicular to the tangent plane.

Check the point and evaluate the gradient

Use the surface $x^{2} + y^{2} + 2 z^{2} = 37$ at $P = (- 1, 2, 4)$ . The point lies on this surface because $(- 1)^{2} + 2^{2} + 2 (4^{2}) = 37$ .

\begin{array}{l} F_{x} = 2 x \\ F_{y} = 2 y \\ F_{z} = 4 z \\ \nabla F (- 1, 2, 4) = (- 2, 4, 16) \end{array}

Write the normal line

\begin{array}{l} x = - 1 - 2 t \\ y = 2 + 4 t \\ z = 4 + 16 t \end{array}

\frac{x + 1}{- 2} = \frac{y - 2}{4} = \frac{z - 4}{16}

At t = 0, the line passes through the required point. Its direction is the gradient, so it is perpendicular to the tangent plane.

Find the tangent plane from the same gradient

\begin{array}{l} - 2 (x + 1) + 4 (y - 2) + 16 (z - 4) = 0 \\ - x + 2 y + 8 z = 37 \end{array}

Apply the method to the video example

For $2 (x - 2)^{2} + (y - 1)^{2} + (z - 3)^{2} = 10$ at $(3, 3, 5)$ , the partial derivatives are $4 (x - 2)$ , $2 (y - 1)$ , and $2 (z - 3)$ . Their values at the point are $(4, 4, 4)$ . A normal direction can therefore be simplified to $(1, 1, 1)$ .

\begin{array}{l} x = 3 + t \\ y = 3 + t \\ z = 5 + t \end{array}

The tangent plane is $4 (x - 3) + 4 (y - 3) + 4 (z - 5) = 0$ , or $x + y + z = 11$ . Both examples use the same process: check the point, evaluate the gradient, then use it as the normal direction.

Learn mathKrista KingFebruary 19, 2024math, learn online, online course, online math, equilibrium solutions