NapkinCalc

Calculus 1C — Applications of the Derivative

Linear approximation

continues from lesson 1 — values defined earlier in the course stay live here

ELI5: near a point, a curve looks just like its tangent line — so use the easy line to estimate the hard curve: f(x) ≈ f(a) + f′(a)·(x − a). It's how you do calculator-free estimates.

Estimate √4.1 by hand. Take f(x) = √x at a = 4: f(4) = 2 and f′(4) = 1/(2√4) = 1/4. So √4.1 ≈ 2 + ¼·(0.1) = 2.025 — and the true value is 2.0248. Spot on.

approxroot=2+14⋅0.1approx_{root} = 2 + \frac{1}{4} \cdot 0.1 = 2.02502.0250 tangent-line estimate: 2.025
trueroot=4.1true_{root} = \sqrt{4.1} = 2.02482.0248 the honest value
✓ pass abs(approxroot−trueroot)<0.001\mathrm{abs}\left(approx_{root} - true_{root}\right) < 0.001 the line nails it to three decimals nearby

Real-world hook: this is error propagation — a 1 % slip in a measured radius becomes about a 2 % slip in the computed area. Engineers carry the derivative through to see how input wobble amplifies into output wobble.

Try it yourself: estimate √9.1 using a = 9 (where f′ = 1/(2·3) = 1/6).

estroot=3+16⋅0.1est_{root} = 3 + \frac{1}{6} \cdot 0.1 = 3.01673.0167 ✏️ Your turn: replace 3.5 with the tangent-line estimate 3 + (1/6)·(0.1). It should land within 0.01 of the true √9.1.
✓ pass abs(estroot−9.1)<0.01\mathrm{abs}\left(est_{root} - \sqrt{9.1}\right) < 0.01 green when your estimate is close to the real √9.1