Tangent Planes
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.4: “Tangent Planes and Linear Approximations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-4-tangent-planes-and-linear-approximations |
| Textbook used in class | Stewart, Calculus, Section 14.4: “Tangent Planes and Linear Approximations” (Examples 1, 2) |
Quick Reference
Tangent plane to $z = f(x,y)$ at the point $(a, b, f(a,b))$: $$z - f(a,b) = f_x(a,b)(x-a) + f_y(a,b)(y-b).$$
Equivalently, define $F(x,y,z) = f(x,y) - z$; then the normal vector to the surface at the point is $\nabla F = \langle f_x, f_y, -1\rangle$.
Motivation
In single-variable calculus, the tangent line at a point is the best linear approximation to the curve there. For a surface $z = f(x,y)$, the analogous object is the tangent plane: the plane that touches the surface at one point and matches the surface to first order in all directions.
The tangent plane equation follows directly from the two partial derivatives: $f_x$ gives the slope of the surface in the $x$-direction, and $f_y$ gives the slope in the $y$-direction. Together they determine a unique plane.
Key Concepts
1. Deriving the Tangent Plane Equation
The surface $z = f(x,y)$ has two sets of tangent lines at any point $(a,b,f(a,b))$:
- The line tangent to the trace $y = b$, with slope $f_x(a,b)$ in the $xz$-plane.
- The line tangent to the trace $x = a$, with slope $f_y(a,b)$ in the $yz$-plane.
The unique plane containing both tangent lines is the tangent plane: $$z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b).$$
2. Normal Vector
Rewrite as $f_x(x-a) + f_y(y-b) - (z - f(a,b)) = 0$. The coefficient vector $\langle f_x(a,b), f_y(a,b), -1\rangle$ is a normal to the tangent plane (and hence to the surface at that point).
Worked Example
Find the tangent plane to $z = x^2 + xy + y^3$ at $(1, 2, f(1,2))$. (Stewart 14.4, Example 1.)
Step 1. Compute $f(1,2) = 1 + 2 + 8 = 11$.
Step 2. Compute partial derivatives. $$f_x = 2x + y \implies f_x(1,2) = 2+2 = 4.$$ $$f_y = x + 3y^2 \implies f_y(1,2) = 1+12 = 13.$$
Step 3. Write the tangent plane equation. $$z - 11 = 4(x-1) + 13(y-2).$$ $$z = 4x + 13y - 19.$$
the tangent plane equation uses the point $(a,b)$, not the point $(a,b,f(a,b))$. The full point of tangency is the three-dimensional point $(a, b, f(a,b))$ on the surface. The equation $z - f(a,b) = f_x(a,b)(x-a) + f_y(a,b)(y-b)$ must include $f(a,b)$ on the left side. Forgetting this gives a plane through the origin shifted incorrectly.
Leveled Practice
Problem 1. Find the tangent plane to $z = e^{x-y}$ at the point $(1, 1, 1)$.
Show answer
$f_x = e^{x-y}$, $f_x(1,1) = 1$.
$f_y = -e^{x-y}$, $f_y(1,1) = -1$.
Tangent plane: $z - 1 = 1(x-1) + (-1)(y-1) = x - y$.
So $z = x - y + 1$.