Linear approximation, explained with examples
Zoom in far enough on a smooth curve and it looks like a straight line. Linear approximation uses that tangent line to estimate values of a function near a known point. It is a short but frequently tested topic in AP Calculus AB and BC.
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Local linearity
If a function is differentiable at a point, its graph looks almost straight when you zoom in close to that point. The straight line it resembles is the tangent line. Near the point of tangency, the tangent line and the curve are very close, so the tangent line's easy-to-calculate values are good estimates for the curve's values. This idea is called local linearity.
The formula
The tangent line to y = f(x) at x = a is
L(x) = f(a) + f'(a)(x − a)
For x close to a, f(x) ≈ L(x). The recipe has three steps: choose a nearby point a where f(a) and f'(a) are easy to find; write the tangent line; substitute the x-value you want.
Worked examples
Example 1: a square root
Estimate √26 using a linear approximation.
- Use f(x) = √x at a = 25: f(25) = 5 and f'(x) = 1/(2√x), so f'(25) = 1/10.
- L(x) = 5 + (1/10)(x − 25). L(26) = 5 + 0.1 = 5.1.
- The true value is 5.0990…, so the estimate is very close.
Example 2: another root
Estimate √4.1.
- a = 4: f(4) = 2, f'(4) = 1/4.
- L(4.1) = 2 + 0.25 × 0.1 = 2.025. True value: 2.02485…
Example 3: a trig value
Estimate sin(0.1), with 0.1 in radians.
- a = 0: sin 0 = 0 and cos 0 = 1, so L(x) = x.
- sin(0.1) ≈ 0.1. True value: 0.0998…. This is the small-angle approximation used in physics for pendulums.
Example 4: an exponential
Estimate e0.05.
- a = 0: e0 = 1 and the derivative is also 1, so L(x) = 1 + x.
- e0.05 ≈ 1.05. True value: 1.05127…
Example 5: from a table (AP style)
A function f has f(3) = 7 and f'(3) = −2. Estimate f(3.2).
- L(x) = 7 − 2(x − 3).
- f(3.2) ≈ 7 − 2(0.2) = 6.6. AP questions often give only f(a) and f'(a), sometimes in a table or from a differential equation.
Example 6: a cube root
Estimate ∛26.
- a = 27: f(27) = 3, f'(x) = (1/3)x−2/3, so f'(27) = 1/27.
- L(26) = 3 + (1/27)(−1) = 3 − 0.0370 ≈ 2.963. True value: 2.9625…
Example 9: from a differential equation
y = f(x) satisfies dy/dx = x + y, with f(1) = 2. Use the tangent line at x = 1 to estimate f(1.1).
- The slope at (1, 2) is 1 + 2 = 3, so L(x) = 2 + 3(x − 1).
- f(1.1) ≈ 2 + 0.3 = 2.3. This is one step of Euler's method, which simply repeats linear approximation many times.
Linear approximation and Euler's method
Euler's method is linear approximation used again and again: take a small step along the tangent, find the new slope, and step again. If you understand why one tangent-line step works, and why concavity decides whether it overshoots or undershoots, Euler's method follows naturally. AP Calculus BC free-response questions often ask for both in the same problem, so practise them together.
Overestimate or underestimate?
The tangent line lies on one side of the curve near the point, and concavity tells you which side.
- If f is concave up (f'' > 0), the curve bends above the tangent line, so the tangent line gives an underestimate.
- If f is concave down (f'' < 0), the curve bends below the tangent line, so the tangent line gives an overestimate.
In Example 1, √x is concave down (its second derivative is negative), so 5.1 is an overestimate, which matches the true value of 5.099. For ex, which is concave up, 1.05 is an underestimate of e0.05.
Example 7: justifying the direction
f(2) = 4, f'(2) = 3 and f''(x) > 0 for all x. Estimate f(2.1) and say whether it is an over- or underestimate.
- L(2.1) = 4 + 3(0.1) = 4.3.
- Since f'' > 0, the graph is concave up, so the tangent line lies below the curve and 4.3 is an underestimate. Name f'' and concavity in your justification.
Differentials
Linear approximation can also be written with differentials. If x changes by a small amount dx, the change in y is approximately dy = f'(x) dx. This is handy for estimating errors: if a measurement has a small error, dy estimates the resulting error in a calculated quantity.
Example 8: propagating an error
The radius of a sphere is measured as 10 cm with a possible error of 0.1 cm. Estimate the possible error in its volume.
- V = (4/3)πr³, so dV = 4πr² dr = 4π × 100 × 0.1.
- dV ≈ 40π ≈ 126 cm³, about 3% of the volume of roughly 4189 cm³.
Why accuracy falls further away
A tangent line matches the curve's value and slope at x = a, but not its bending. The further you move from a, the more the curve bends away from the line, so the error grows. Choose a as close as possible to the value you want, and expect linear approximations to be reliable only for small changes. Taylor polynomials, in AP Calculus BC, improve on this by also matching the curvature.
Common mistakes
- Choosing a point a where f(a) or f'(a) is hard to evaluate.
- Forgetting the (x − a) factor and writing f(a) + f'(a)x.
- Using the first derivative instead of concavity to decide over- or underestimate.
- Using degrees in trigonometric approximations; calculus needs radians.
Practice questions
Try each one before opening the answer.
Q1. Estimate √50 using a = 49.
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Q2. f(1) = 2, f'(1) = 5. Estimate f(0.9).
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Q3. Estimate ln(1.1) using a = 1.
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Q4. If f'' < 0 near a, is a tangent-line estimate too high or too low?
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Frequently asked questions
Is linear approximation on AP Calculus AB?
Yes. Local linearity and linearisation are in Unit 4 of both AB and BC, and they often appear in free-response questions with a table or differential equation.
Is linear approximation in A level Maths?
The idea appears in A level Maths through tangents and small changes, and more formally in Further Maths with Maclaurin series.
Key takeaways
- Near a point, a differentiable curve looks like its tangent line.
- L(x) = f(a) + f'(a)(x − a).
- Choose a close to x with easy values of f(a) and f'(a).
- Concave up gives an underestimate; concave down gives an overestimate.
- Differentials: dy = f'(x) dx estimates small changes and errors.