MATHEMATICS · LESSON 07 OF 10
Linear approximation
Near a point, a differentiable function is almost a straight line: its tangent line gives quick estimates and error sizes.
What this lesson explains
Engineers constantly replace complicated functions by simpler ones that are accurate enough nearby: small-angle approximations, linearised models of reactors and circuits, and error estimates for measurements. The tangent line is the simplest such replacement, and differentials tell you how an error in a measured input spreads to a calculated output.
Before you begin
Tangent line
The tangent to y = f(x) at x = a is y = f(a) + f′(a)(x − a). See “Derivative as a rate”.
Relative and percentage error
If a quantity Q has error ΔQ, the relative error is ΔQ/Q and the percentage error is 100·ΔQ/Q %.
The idea, made visible
If you zoom in far enough on a smooth curve, it looks like a straight line — its tangent line. So near x = a we can approximate f(x) by the tangent line L(x) = f(a) + f′(a)(x − a). This is the linear approximation (or linearisation) of f at a.
Choose a as a point where f is easy to evaluate exactly and close to the input you care about. To estimate √4.08, use a = 4 because √4 = 2 is known exactly.
The approximation is only good near a. The error grows as x moves away, and its size depends on how strongly the graph bends (the second derivative). If the graph is concave down (bending below its tangents), the tangent line lies above the curve and the estimate is too large; if concave up, the estimate is too small.
The same idea in “change” language: a small change dx in the input produces a change in output of about dy = f′(x) dx. This differential is the change along the tangent line. It is how measurement errors propagate: if a radius is measured with a small error, the resulting error in a computed area is about A′(r) × (error in r).
| x | L(x) = 2 + (x − 4)/4 | true √x | error L(x) − √x |
|---|---|---|---|
| 4.08 | 2.02 | 2.019901 | 0.000099 |
| 4.5 | 2.125 | 2.121320 | 0.00368 |
| 5 | 2.25 | 2.236068 | 0.0139 |
| 9 | 3.25 | 3 | 0.25 |
The error is always positive here because √x is concave down: the tangent lies above the curve. The error grows roughly like (x − 4)² as x moves away from 4.
Key terms
- Linearisation
- L(x) = f(a) + f′(a)(x − a), the tangent-line function at a. For x near a, f(x) ≈ L(x).
- Differential
- dx is an independent small change in x; dy = f′(x) dx is the corresponding change along the tangent line.
- Actual change
- Δy = f(x + dx) − f(x), the true change on the curve. For small dx, Δy ≈ dy.
- Error propagation
- If an input is measured with error dx, the computed output has error about |f′(x)| |dx|; the relative error is about |dy/y|.
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| a | the base point where f and f′ are known exactly | input units |
| L(x) | linear approximation (tangent line) | output units |
| dx, dy | differentials: small input change and the tangent-line output change | input units, output units |
| Δy | actual change in output | output units |
Linear approximation
f(x) ≈ L(x) = f(a) + f′(a)(x − a)
Conditions and limits: f differentiable at a, and x close to a. Accuracy depends on how close x is and how much the graph curves.
Differential
dy = f′(x) dx, Δy ≈ dy
Conditions and limits: dx small. The approximation Δy ≈ dy improves as dx → 0.
Relative error of a power
Q = kxn ⇒ dQ / Q = ndx / x
Conditions and limits: Follows from dQ = nkxn−1dx. A 1% error in a radius gives about 2% error in an area and 3% in a volume.
Common small-value approximations
sin x ≈ x; cos x ≈ 1; ex ≈ 1 + x; ln(1 + x) ≈ x; (1 + x)n ≈ 1 + nx
Conditions and limits: Linearisations at a = 0, valid for |x| small (x in radians for sin and cos).
Why it works: Where (1 + x)ⁿ ≈ 1 + nx comes from
Linearise f(x) = (1 + x)n at a = 0.
f(0) = 1.
Base value.
f′(x) = n(1 + x)n−1, so f′(0) = n.
Chain rule.
L(x) = 1 + n(x − 0) = 1 + nx.
Tangent line at 0.
Example: √1.02 = (1 + 0.02)1/2 ≈ 1 + 0.01 = 1.01 (the true value is 1.00995…).
A first worked example
How to make a linear approximation
- Identify the function f and the value you want, f(x).
- Choose a nearby base point a where f(a) and f′(a) are easy to compute exactly.
- Compute f(a) and f′(a), and write L(x) = f(a) + f′(a)(x − a).
- Evaluate L at the required x. Decide whether the estimate is too high or too low from the concavity.
- For error propagation, write dy = f′(x) dx with dx = measurement error; report absolute and relative errors.
Estimate a square root
Problem. Use a linear approximation to estimate √4.08.
f(x) = √x, a = 4: f(4) = 2.
4 is close to 4.08 and has an exact root.
f′(x) = 1 / 2√x, so f′(4) = 1 / 4.
Power rule with exponent 1/2.
L(x) = 2 + 1 / 4(x − 4).
Tangent line at 4.
L(4.08) = 2 + 0.08 / 4 = 2.02.
Substitute x = 4.08.
Result: √4.08 ≈ 2.02.
What it means: The true value is 2.019901…, so the error is about 0.0001. The estimate is slightly high because √x is concave down.
A different case
Estimate a square
Problem. Use the linearisation of x2 at x = 3 to estimate (3.02)2.
f(3) = 9 and f′(x) = 2x, so f′(3) = 6.
Base value and slope.
L(x) = 9 + 6(x − 3); L(3.02) = 9 + 6(0.02) = 9.12.
Evaluate the tangent line.
Result: (3.02)2 ≈ 9.12.
What it means: The exact value is 9.1204; the error 0.0004 = (0.02)² is exactly the curvature term that the line ignores. x² is concave up, so the estimate is low.
More worked cases
Each case below uses a different skill. Every step and result is shown.
Error in a computed area
Problem. The radius of a circular plate is measured as 10 cm with a possible error of ±0.1 cm. Estimate the maximum error in the computed area and the percentage error.
A = πr2, so dA = 2πr dr.
Differential of the area.
dA = 2π(10)(0.1) = 2π ≈ 6.28 cm2.
Substitute r = 10 and dr = 0.1.
A = π(10)2 = 100π ≈ 314.2 cm2; relative error dA / A = 2π / 100π = 0.02.
Compare the error with the value.
Result: Maximum error ≈ ±6.3 cm2, which is about 2%.
What it means: A 1% error in the radius becomes about a 2% error in the area, as the power rule for relative errors predicts (n = 2).
Small-angle approximation
Problem. Estimate sin(0.1) (radians) and compare with the true value.
Linearise sin x at a = 0: sin 0 = 0 and cos 0 = 1, so L(x) = x.
Tangent line of sin x at the origin.
sin(0.1) ≈ 0.1.
Substitute.
Result: sin(0.1) ≈ 0.1; the true value is 0.099833…, an error of about 0.17%.
What it means: This is the approximation used for a pendulum swinging through small angles. It only works in radians.
Common misunderstandings
Misunderstanding: Choosing a base point that is far away or hard to evaluate.
Correct idea: Pick a close to x with f(a) and f′(a) exact, such as a perfect square for roots.
Misunderstanding: Using the approximation far from a.
Correct idea: The error grows quickly with distance: √9 ≈ 3.25 from a = 4 is poor.
Misunderstanding: Confusing dy with Δy.
Correct idea: dy is the change along the tangent; Δy is the true change on the curve. They are close only for small dx.
Misunderstanding: Using degrees in sin x ≈ x.
Correct idea: The approximation requires radians: sin(5°) ≈ 0.0873, not 5.
Going further (optional): Newton’s method (check whether your outline includes it)
Newton’s method uses linearisation to solve f(x) = 0. From a guess xn, follow the tangent line to where it crosses the x-axis: xn+1 = xn − f(xn) / f′(xn).
Example: for f(x) = x2 − 2 (root √2), start at x0 = 1. Then x1 = 1 − −1 / 2 = 1.5, x2 = 1.5 − 0.25 / 3 ≈ 1.41667, x3 ≈ 1.414216. The true value is 1.414214. The method fails if f′(xn) = 0 or if the starting guess is poor.
Keep in mind
- f(x) ≈ L(x) = f(a) + f′(a)(x − a)Linear approximation
- dy = f′(x) dx, Δy ≈ dyDifferential
- Q = kxn ⇒ dQ / Q = ndx / xRelative error of a power
- sin x ≈ x; cos x ≈ 1; ex ≈ 1 + x; ln(1 + x) ≈ x; (1 + x)n ≈ 1 + nxCommon small-value approximations
Scope of this lesson
- Only first-order (linear) approximation is covered. Quadratic and Taylor approximations belong to later calculus.
- Error estimates here are first-order; for large errors compute the exact change.
Next: Rolle and Mean Value Theorems. If a smooth function goes from one value to another, somewhere its instantaneous rate equals its average rate.
Original study text. Sources and credits.