Linear Approximations in 2D
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 3, 4) |
Quick Reference
Linearization of $f$ near $(a,b)$: $$L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b).$$
Approximation: $f(x,y) \approx L(x,y)$ for $(x,y)$ near $(a,b)$.
$L$ is exactly the $z$-coordinate on the tangent plane at $(a,b,f(a,b))$.
Motivation
The tangent plane serves double duty: it is a geometric object (the plane touching the surface) AND an analytic tool (the best linear approximation to the function). When you need to estimate $f(1.02, 0.97)$ and computing the exact value is hard, the linearization around $(1,1)$ gives a quick, accurate approximation for small displacements.
This is the multivariable analogue of the single-variable tangent line approximation $f(x) \approx f(a) + f'(a)(x-a)$.
Key Concept
The linearization $L(x,y)$ is identical to the tangent plane equation. The two concepts are the same formula viewed from different angles: the tangent plane is a geometric surface, the linearization is a function approximation. Using $L(x,y) \approx f(x,y)$ is valid when $(x,y)$ is close to $(a,b)$ and $f$ is differentiable there.
Worked Example
Use the linearization of $f(x,y) = \sqrt{x}\,e^{y^2}$ at $(1, 0)$ to approximate $f(1.1, -0.1)$. (Adapted from Stewart 14.4.)
Step 1. Evaluate $f$ and its partials at $(1,0)$.
$f(1,0) = \sqrt{1}\,e^0 = 1$.
$f_x = \dfrac{1}{2\sqrt{x}}\,e^{y^2} \implies f_x(1,0) = \dfrac{1}{2}$.
$f_y = \sqrt{x}\,e^{y^2}\cdot 2y \implies f_y(1,0) = 0$.
Step 2. Write the linearization. $$L(x,y) = 1 + \tfrac{1}{2}(x-1) + 0(y-0) = 1 + \tfrac{1}{2}(x-1).$$
Step 3. Evaluate at $(1.1, -0.1)$. $$L(1.1, -0.1) = 1 + \tfrac{1}{2}(0.1) = 1.05.$$
The exact value is $\sqrt{1.1}\,e^{0.01} \approx 1.0488 \times 1.01005 \approx 1.059$, so the approximation is good for this small displacement.
the linearization only accounts for the $x$-change, not the $y$-change. The linearization uses BOTH partial derivatives: $L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$. Both the $x$-displacement and the $y$-displacement contribute. Students sometimes include only the term corresponding to whichever variable they are changing, forgetting that both affect $f$.
Leveled Practice
Problem 1. Use linearization to approximate $f(2.1, 0.9)$ for $f(x,y) = x^2 y - 3y$.
Show answer
At $(a,b) = (2,1)$: $f(2,1) = 4 - 3 = 1$.
$f_x = 2xy \implies f_x(2,1) = 4$.
$f_y = x^2 - 3 \implies f_y(2,1) = 1$.
$L(x,y) = 1 + 4(x-2) + 1(y-1)$.
$L(2.1, 0.9) = 1 + 4(0.1) + 1(-0.1) = 1 + 0.4 - 0.1 = 1.3$.
Exact: $f(2.1, 0.9) = (4.41)(0.9) - 2.7 = 3.969 - 2.7 = 1.269$. Close.