📐 Concept diagram

06-04 - Tangent Planes and Linear Approximation

Phase: 6 | Subject: 06-04 Prerequisites: 06-03-partial-derivatives.md (∂f/∂x, ∂f/∂y computation), 04-03-the-derivative.md (tangent lines in 1D), 02-10-vectors-basic.md (dot product, normal vectors) Next subject: 06-05-chain-rule-multivariable.md

Learning Objectives

By the end of this subject, you will be able to:

1. Derive the equation of the tangent plane to a surface z = f(x,y) at a point
2. Use linear approximation (the tangent plane) to estimate function values near a point
3. Compute differentials dz and estimate error propagation
4. Understand the relationship between differentiability and the existence of a tangent plane
5. Distinguish between the tangent plane approximation and exact function values

Core Content

From Tangent Lines to Tangent Planes

⚠️ CRITICAL FOUNDATION: The tangent plane formula z = f(a,b) + fₓ(a,b)(x-a) + f_y(a,b)(y-b) is the multivariable generalization of the 1D tangent line. This linear approximation underlies all of differential calculus in higher dimensions — differentiability, error estimation, and local behavior.

Single variable: The tangent line y = f(a) + f'(a)(x - a) is the best linear approximation to f near x = a.

Two variables: The tangent plane z = f(a,b) + fₓ(a,b)(x - a) + f_y(a,b)(y - b) is the best linear approximation to f near (x,y) = (a,b).

The Tangent Plane Equation

If f is differentiable at (a,b), then the surface z = f(x,y) has a tangent plane at (a, b, f(a,b)) given by:

$z - f(a,b) = fₓ(a,b)(x - a) + f_y(a,b)(y - b)
$

Or equivalently:

$z = f(a,b) + fₓ(a,b)(x - a) + f_y(a,b)(y - b)
$

Derivation

The tangent plane is the unique plane that contains: 1. The point (a, b, f(a,b)) 2. The tangent line with slope fₓ(a,b) in the x-direction 3. The tangent line with slope f_y(a,b) in the y-direction

The normal vector to the tangent plane is:

$n = ⟨-fₓ(a,b), -f_y(a,b), 1⟩
$

The plane passes through (a,b,f(a,b)) with this normal vector, giving:

$-fₓ(a,b)(x - a) - f_y(a,b)(y - b) + 1·(z - f(a,b)) = 0
$

which rearranges to the tangent plane equation above.

Example 1: Tangent Plane to a Paraboloid

f(x,y) = x² + y² at (1, 1, 2)

Step 1: Compute partials. fₓ = 2x → fₓ(1,1) = 2 f_y = 2y → f_y(1,1) = 2

Step 2: Tangent plane: z = 2 + 2(x - 1) + 2(y - 1) = 2 + 2x - 2 + 2y - 2 = 2x + 2y - 2

At (1, 1): z = 2 + 2 - 2 = 2 ✓ (passes through the point).

Example 2: Tangent Plane to an Exponential Surface

f(x,y) = e^{x-y} at (0, 0, 1)

Step 1: fₓ = e^{x-y} → fₓ(0,0) = e⁰ = 1 f_y = -e^{x-y} → f_y(0,0) = -1

Step 2: Tangent plane: z = 1 + 1(x - 0) + (-1)(y - 0) = 1 + x - y

Example 3: Tangent Plane to ln Surface

f(x,y) = ln(x² + y²) at (1, 0, 0)

Step 1: f(1,0) = ln(1) = 0.

Step 2: fₓ = 2x/(x²+y²) → fₓ(1,0) = 2/1 = 2. f_y = 2y/(x²+y²) → f_y(1,0) = 0.

Step 3: Tangent plane: z = 0 + 2(x - 1) + 0(y - 0) = 2x - 2.

Linear Approximation

The tangent plane gives a linear approximation (also called the linearization) of f near (a,b):

$f(x,y) ≈ L(x,y) = f(a,b) + fₓ(a,b)(x - a) + f_y(a,b)(y - b)
$

This is the analog of f(x) ≈ f(a) + f'(a)(x - a) in single variable.

Example: Approximating √(x² + y²)

Problem: Use linear approximation near (3, 4) to estimate √(3.1² + 3.9²).

Solution:

Step 1: f(x,y) = √(x² + y²). At (3,4): f(3,4) = √(9+16) = 5.

Step 2: fₓ = x/√(x²+y²) → fₓ(3,4) = 3/5 = 0.6. f_y = y/√(x²+y²) → f_y(3,4) = 4/5 = 0.8.

Step 3: L(x,y) = 5 + 0.6(x - 3) + 0.8(y - 4).

Step 4: At (3.1, 3.9): L(3.1, 3.9) = 5 + 0.6(0.1) + 0.8(-0.1) = 5 + 0.06 - 0.08 = 4.98.

Step 5: Exact: √(3.1² + 3.9²) = √(9.61 + 15.21) = √24.82 ≈ 4.982.

Error: |4.982 - 4.980| = 0.002. Very good approximation!

Differentiability

Definition: f is differentiable at (a,b) if the tangent plane is a "good" linear approximation:

$lim_{(x,y)→(a,b)} [f(x,y) - L(x,y)] / √((x-a)² + (y-b)²) = 0
$

In words: the error goes to zero faster than the distance to (a,b).

Theorem: If fₓ and f_y exist AND are continuous near (a,b), then f is differentiable at (a,b).

Contrast with 1D: In one variable, existence of f'(a) implies differentiability. In ℝⁿ, existence of partial derivatives is NOT sufficient — they must be continuous in a neighborhood. Otherwise the function might have partials but no well-behaved tangent plane.

Counterexample: Partials Exist but Not Differentiable

f(x,y) = { xy/(x² + y²) if (x,y) ≠ (0,0); 0 if (x,y) = (0,0) }

At (0,0): - fₓ(0,0) = 0 (using limit definition along x-axis) - f_y(0,0) = 0 (using limit definition along y-axis)

But f is not even continuous at (0,0) (limit depends on path), so it cannot be differentiable there.

Differentials

The total differential of z = f(x,y) is:

$dz = fₓ(x,y) dx + f_y(x,y) dy
$

This represents the approximate change in z when x changes by dx and y changes by dy.

Interpretation: dz is the change in height along the tangent plane; Δz = f(x+dx, y+dy) - f(x,y) is the actual change. When dx and dy are small, dz ≈ Δz.

Error Propagation

If measured quantities x and y have uncertainties Δx and Δy, the resulting uncertainty in z = f(x,y) can be approximated by:

$Δz ≈ |fₓ|·Δx + |f_y|·Δy
$

Example: The volume of a cylinder is V = πr²h. If r = 5 ± 0.1 cm and h = 10 ± 0.2 cm, estimate the uncertainty in V.

Step 1: V_r = 2πrh → at (5,10): 2π(5)(10) = 100π. Step 2: V_h = πr² → at (5,10): π(25) = 25π. Step 3: ΔV ≈ |100π|·0.1 + |25π|·0.2 = 10π + 5π = 15π ≈ 47.1 cm³.

Nominal V = π(25)(10) = 250π ≈ 785.4 cm³. Result: V ≈ 785.4 ± 47.1 cm³.

Key Terms

• 06 04 Tangent Planes And Linear Approximation
• Correct: B)
• Correct: C)
• Counterexample: Partials Exist but Not Differentiable
• Derivation
• Differentiability
• Differentials
• Error Propagation
• Example 1: Tangent Plane Equation
• Example 1: Tangent Plane to a Paraboloid
• Example 2: Linear Approximation
• Example 2: Tangent Plane to an Exponential Surface

Worked Examples

Example 1: Tangent Plane Equation

Problem: Find the tangent plane to f(x,y) = x³ - 3xy + y³ at (1, 2, 3).

Solution:

Step 1: f(1,2) = 1 - 6 + 8 = 3 ✓.

Step 2: fₓ = 3x² - 3y → fₓ(1,2) = 3 - 6 = -3. f_y = -3x + 3y² → f_y(1,2) = -3 + 12 = 9.

Step 3: Tangent plane: z = 3 + (-3)(x - 1) + 9(y - 2) z = 3 - 3x + 3 + 9y - 18 z = -3x + 9y - 12.

Example 2: Linear Approximation

Problem: Use linearization to approximate e^{0.1}·sin(0.2) near (0,0).

Solution:

Step 1: f(x,y) = eˣ sin(y). f(0,0) = e⁰·0 = 0.

Step 2: fₓ = eˣ sin(y) → fₓ(0,0) = 0. Wait — at (0,0), sin(0) = 0, so fₓ(0,0) = 0.

f_y = eˣ cos(y) → f_y(0,0) = 1·1 = 1.

Step 3: L(x,y) = 0 + 0(x - 0) + 1(y - 0) = y.

Step 4: L(0.1, 0.2) = 0.2.

Step 5: Exact: e^{0.1}·sin(0.2) ≈ 1.1052·0.1987 ≈ 0.2196.

Error: 0.0196. The first-order approximation only captured the y-direction change.

Better: expand near a point where both partials are nonzero, like (0, π/6): f(0, π/6) = 1·0.5 = 0.5, fₓ = 0.5, f_y = 1·cos(π/6) = √3/2 ≈ 0.866.

L(0.1, π/6+Δy) would be more accurate for those coordinates.

Example 3: Error Propagation

Problem: The period of a pendulum is T = 2π√(L/g). If L = 1.00 ± 0.01 m and g = 9.8 ± 0.1 m/s², estimate uncertainty in T.

Solution:

Step 1: T(L,g) = 2π L^(1/2) g^(-1/2).

Step 2: ∂T/∂L = 2π · (1/2)L^(-1/2)g^(-1/2) = π/(√(Lg)).

∂T/∂g = 2π L^(1/2) · (-1/2)g^(-3/2) = -π√(L)/g^(3/2) = -π√(L/g)/g.

Step 3: At L=1, g=9.8: T = 2π/√9.8 ≈ 2.006 s (nominal). ∂T/∂L = π/√9.8 ≈ 1.003 s/m. ∂T/∂g = -π/9.8^(3/2) ≈ -0.102 s/(m/s²).

Step 4: ΔT ≈ |1.003|·0.01 + |(-0.102)|·0.1 = 0.0100 + 0.0102 = 0.020.

Result: T ≈ 2.006 ± 0.020 s.

Practice Problems

1. Find the tangent plane to f(x,y) = x² - xy + 2y² at (1, 1, 2).

2. Use linear approximation near (1, 0) to estimate arctan(1.1/0.05). Note: use f(x,y) = arctan(y/x) near (1, 0).

3. Find the tangent plane to f(x,y) = sin(x)cos(y) at (π/4, 0, √2/2).

4. The resistance of a wire is R = ρL/A. If L = 100 ± 1 cm, A = 0.01 ± 0.001 cm², ρ = 1.7×10⁻⁶ Ω·cm (exact), compute R and approximate error.

5. For f(x,y) = x²e^y, find the linearization at (1, 0) and use it to estimate f(1.01, -0.02).

Summary

Key takeaways:

• The tangent plane at (a,b) is z = f(a,b) + fₓ(a,b)(x-a) + f_y(a,b)(y-b) — the natural generalization of the 1D tangent line
• Linear approximation L(x,y) uses the tangent plane to estimate function values near (a,b)
• The total differential dz = fₓ dx + f_y dy measures approximate change on the tangent plane
• Error propagation: multiply partial derivatives by measurement uncertainties (in absolute value)
• Differentiability in ℝⁿ requires more than existence of partials — the tangent plane must be a good approximation

Pitfalls

• Forgetting the constant term f(a,b). The tangent plane formula is z = f(a,b) + fₓ(a,b)(x−a) + f_y(a,b)(y−b). Omitting f(a,b) gives a plane through the origin, not through the point of tangency on the surface. Always verify that plugging in (a,b) gives back f(a,b).
• Evaluating partials at the wrong point. fₓ and f_y in the tangent plane formula must be evaluated at the fixed point (a,b), not at the variable point (x,y). Writing z = f(a,b) + fₓ(x,y)(x−a) + f_y(x,y)(y−b) turns the linear approximation into a nonlinear one.
• Assuming differentiability from existence of partial derivatives. In ℝⁿ, the existence of fₓ and f_y at a point does NOT guarantee differentiability. The partials must also be continuous in a neighborhood. The function f(x,y) = xy/(x²+y²) (with f(0,0)=0) has partial derivatives at (0,0) but is not even continuous there.
• Misapplying error propagation signs. The uncertainty formula Δz ≈ |fₓ|Δx + |f_y|Δy uses absolute values because measurement uncertainties always add (worst-case). Using signed differentials dz = fₓ dx + f_y dy may cause cancellations that underestimate the true error.
• Confusing the differential dz with the actual change Δz. dz = fₓ dx + f_y dy represents the change along the tangent plane, while Δz = f(x+dx, y+dy) − f(x,y) is the actual change on the surface. For small dx and dy they approximate each other, but for larger changes or highly curved surfaces, the discrepancy can be significant.

Quiz

Q1: The tangent plane to z = x² + y² at (1, 1, 2) is: - A) z = 2x + 2y - B) z = 2x + 2y - 2 - C) z = x + y - D) z = 2x² + 2y² - 2 Correct: B) - If you chose B: fₓ=2x, f_y=2y. fₓ(1,1)=2, f_y(1,1)=2. z = 2 + 2(x-1) + 2(y-1) = 2x + 2y - 2. Correct! - If you chose A: You forgot to include f(1,1)=2 in the equation properly (the constant term is wrong). - If you chose C: The slope coefficients are wrong — should be 2, not 1. - If you chose D: That's not a plane (it has squared terms).

Q2: The linear approximation of f(x,y) = xy at (2, 3) is: - A) L(x,y) = 3x + 2y - B) L(x,y) = 3x + 2y - 6 - C) L(x,y) = 2x + 3y - D) L(x,y) = x + y Correct: B) - If you chose B: f(2,3)=6, fₓ=y → 3, f_y=x → 2. L = 6 + 3(x-2) + 2(y-3) = 6 + 3x - 6 + 2y - 6 = 3x + 2y - 6. Correct! - If you chose A: Missing the constant offset. - If you chose C: Swapped the partials. - If you chose D: Wrong slopes.

Q3: The total differential of z = x²y is: - A) dz = 2x dx - B) dz = x² dy - C) dz = 2xy dx + x² dy - D) dz = 2x dx + 2y dy Correct: C) - If you chose C: fₓ = 2xy, f_y = x². dz = 2xy dx + x² dy. Correct! - If you chose A: That's only the x-part, missing the y-part. - If you chose B: That's only the y-part, missing the x-part. - If you chose D: That would be the differential of x² + y².

Q4: If f has partial derivatives fₓ(a,b) and f_y(a,b) that exist, then f is: - A) Definitely differentiable at (a,b) - B) Definitely continuous at (a,b) - C) Possibly not differentiable at (a,b) - D) Definitely has a unique tangent plane Correct: C) - If you chose C: Existence of partials alone does NOT guarantee differentiability in ℝⁿ. Continuity of partials in a neighborhood is the sufficient condition. Correct! - If you chose A: This is false — differentiability requires more than existence of partials. - If you chose B: Existence of partials doesn't even guarantee continuity (unlike in 1D). - If you chose D: The tangent plane may not be a good approximation if f is not differentiable.

Q5: For f(x,y) = √(x² + y²), the tangent plane at (3, 4, 5) is: - A) z = 5 + 0.6x + 0.8y - B) z = 0.6x + 0.8y - C) z = 5 + 0.6(x-3) + 0.8(y-4) - D) z = 3x + 4y Correct: C) - If you chose C: fₓ = x/√(x²+y²) = 3/5 = 0.6. f_y = 4/5 = 0.8. Tangent plane: z = 5 + 0.6(x-3) + 0.8(y-4). Correct! - If you chose A: Missing the -0.6·3 - 0.8·4 = -1.8 - 3.2 = -5 that should process the constant. - If you chose B: Missing f(3,4) = 5. - If you chose D: Wrong slopes — those are the coordinates, not the partials.

Q6: Using the differential, the approximate change in z = x²y when x changes from 2 to 2.1 and y from 3 to 2.9 is: - A) 0.6 - B) 0.8 - C) -0.4 - D) 1.2 Correct: B) - If you chose B: fₓ=2xy=12, f_y=x²=4. dx=0.1, dy=-0.1. dz = 12(0.1) + 4(-0.1) = 1.2 - 0.4 = 0.8. Correct! - If you chose A: Only the 4·(-0.1) + maybe partial of 12·0.1 off. - If you chose C: You lost the 12·0.1 term. - If you chose D: You added instead of subtracting: 12(0.1) + 4(0.1) = 1.6.

Q7: The tangent plane to z = exy at (0, 0) is: - A) z = 0 - B) z = x + y + 1 - C) z = 1 - D) z = 1 + x + y Correct: C) - If you chose C: f(0,0) = e⁰ = 1. fₓ = ye^{xy} → 0, f_y = xe^{xy} → 0. Tangent plane: z = 1 + 0(x) + 0(y) = 1. Correct! - If you chose A: f(0,0) is not 0 — e⁰ = 1. - If you chose B: This would be z = 1 + x + y, but the partials at (0,0) are both 0. - If you chose D: The partials are 0, not 1. Confusing with eˣ⁺ʸ = eˣeʸ.

Next Steps

Next up: 06-05-chain-rule-multivariable.md