Derivatives, Taylor series, and local approximation

Start with a quantity, a reference and a domain

A function assigns one output to each allowed input. For this lesson, track one chosen point on a hypothetical body along a straight coordinate line. Fix the origin and take right as positive. Let t be elapsed time from a declared start, measured in seconds, and x(t) the point's signed position, measured in metres.

Supply the exact teaching rule x(t) = 1.00 m + (0.50 m/s)t + (0.20 m/s²)t², only for 0 ≤ t ≤ 4 s. The squared time t² means t multiplied by itself. Each term has metre units: (m/s)s = m and (m/s²)s² = m. The numbers specify a hypothetical model; their decimal places are not claims about instrument precision.

Our expansion point will be t₀ = 2 s, where x(t₀) = 1.00 m + 1.00 m + 0.80 m = 2.80 m. An increment h = t − t₀ has time units. Positive h selects a later time and negative h an earlier one. Keep t₀ + h inside the supplied domain.

A finite interval gives an average rate

For a nonzero time increment h, the difference quotient is [x(t₀ + h) − x(t₀)]/h. It divides a signed position change by a signed time change, giving units m/s. In this one-dimensional model it is the average signed velocity between the two times.

If h is negative, write the interval chronologically as [t₀ + h, t₀]. Reversing both differences leaves the quotient unchanged. A negative h therefore does not by itself mean negative velocity. For h = −0.50 s the point advances from 2.20 m at 1.50 s to 2.80 m at 2.00 s, giving +1.20 m/s.

Expanding the supplied quadratic gives x(t₀ + h) = 2.80 m + (1.30 m/s)h + (0.20 m/s²)h². For h ≠ 0, subtraction and division give the exact quotient 1.30 m/s + (0.20 m/s²)h. The table contains finite averages. Its rows illustrate a trend; the formula is what establishes the limit.

A limit defines the instantaneous derivative

A finite limit L describes values that get and stay arbitrarily close to L as a nonzero increment approaches zero. It does not require the expression to be defined at zero. At an interior point, the approaches from negative and positive increments must agree.

The ordinary derivative is x′(t₀) = lim as h → 0 of [x(t₀ + h) − x(t₀)]/h, when that two-sided limit exists as a finite value. The prime denotes differentiation; dx/dt is another notation. Here the extra term (0.20 m/s²)h tends to zero from either side, so x′(2 s) = +1.30 m/s.

The derivative is the slope of the tangent to the position-versus-time graph at the chosen point. In this declared frame it is the instantaneous signed velocity. Setting h = 0 in the original quotient would give the undefined division 0/0; taking a limit through nonzero h gives the derivative.

All these rates follow from the supplied continuous-time rule. A finite table of instrument readings would not, on its own, establish that rule or its derivative. The mathematical limiting process is not a claim that a physical timer can resolve arbitrarily small intervals.

Check that the derivative exists, and retain its units

A defined function need not have a derivative. Continuity at a point means that nearby outputs approach the output at that point; this alone is insufficient. For an original mathematical counterexample, supply q(t) = (1.00 m/s)|t − 2 s| on 0 ≤ t ≤ 4 s, where | | takes the nonnegative magnitude. At 2 s its left difference quotients are −1.00 m/s and its right ones +1.00 m/s. They disagree, so q has no ordinary derivative there despite being continuous.

We use ordinary two-sided derivatives at interior times 0 < t < 4 s. At 0 s only later inputs are supplied; at 4 s only earlier inputs are supplied. A derivative based only on the available side must be identified as one-sided. Do not invent times beyond the model to make a two-sided physical claim.

At a general interior time, expansion gives [x(t + h) − x(t)]/h = 0.50 m/s + (0.40 m/s²)t + (0.20 m/s²)h. Its limit is x′(t) = 0.50 m/s + (0.40 m/s²)t. Differentiating that rate by the same quotient process gives x″(t) = 0.40 m/s²; the double prime means the second derivative.

For this position model x″ is instantaneous signed acceleration, the time derivative of velocity. Its units are (m/s)/s = m/s². The original t² coefficient, 0.20 m/s², is half this acceleration. Position, velocity and acceleration are different quantity kinds even when they occur in the same equation.

Build an approximation about the chosen point

A Taylor polynomial uses derivatives at a stated expansion point to build a finite polynomial in the increment. The first Taylor polynomial is P₁(t₀ + h) = x(t₀) + x′(t₀)h. It matches position and slope at t₀. For our model, P₁ = 2.80 m + (1.30 m/s)h.

The second Taylor polynomial adds x″(t₀)h²/2: P₂ = 2.80 m + (1.30 m/s)h + (0.20 m/s²)h². Here the factor 1/2 gives the coefficient that matches the second derivative. Each added term still has position units.

For this supplied quadratic, P₂ exactly equals x throughout its stated domain. P₁ omits a term: define the signed remainder R₁ = x − P₁ = (0.20 m/s²)h². The absolute error is |R₁|. At h = +0.50 s, P₁ gives 3.45 m and x gives 3.50 m, so R₁ = +0.05 m. Matching a slope at one point does not make P₁ the body's full position rule.

State conditions before claiming an error bound

For a general position function, one sufficient condition is that it has a continuous second derivative on an open interval containing the entire segment between t₀ and the target. If a nonnegative bound M satisfies |x″(t)| ≤ M throughout that segment, Taylor's remainder theorem gives |x(t₀ + h) − P₁(t₀ + h)| ≤ (M/2)|h|². The bound uses the whole segment, not just the derivative at its centre.

For our exact model, x″ = 0.40 m/s² everywhere in the interior, so M = 0.40 m/s² works for all segments within 1.50 ≤ t ≤ 2.50 s. Choose the teaching requirement that absolute position error be at most 0.05 m. Whenever |h| ≤ 0.50 s, the bound is at most (0.40 m/s²)/2 × (0.50 s)² = 0.05 m. Equality at the two ends is allowed.

The time window follows from this model, expansion point and chosen error requirement. At t = 3 s, h = 1 s, P₁ gives 4.10 m while x gives 4.30 m: the 0.20 m error fails that requirement even though 3 s remains inside the model's domain. No single meaning of 'small' or universal acceptable window follows.

The remainder measures truncation error relative to the supplied mathematical rule. It does not include instrument uncertainty or disagreement between that rule and an actual body. Those need separate evidence and accounting.

A Taylor series requires another check

A finite Taylor polynomial stops after a chosen power of h. If derivatives of every order are available at t₀, the associated Taylor series has terms x(t₀) + x′(t₀)h + x″(t₀)h²/2! + x‴(t₀)h³/3! + ⋯. The triple prime means the third derivative. A factorial is the product of the positive integers up to that order: 2! = 2 and 3! = 6. Each term in the expansion is divided by the corresponding factorial.

An infinite series is interpreted through the limit of its finite partial sums. Calling those sums Pₙ, with n the highest retained power, equality to x at a target requires Rₙ = x − Pₙ to tend to zero there as n increases. The existence of derivatives of every order does not alone guarantee that equality. This lesson claims no universal convergence interval.

Our quadratic has zero derivatives above the second, so its Taylor expansion already stops at P₂ and reproduces the supplied rule. Nevertheless, evaluating that polynomial at 5 s would extend beyond the stated 0–4 s model. Exact algebra does not establish an actual position there. The ±0.50 s window is a finite linear-approximation error window, not a claimed radius of convergence of an infinite series.

Use the derivative to describe a local rate, then choose an approximation order and check its error on the allowed interval. The next lesson introduces integrals, including how a supplied velocity function can describe an accumulated signed displacement.

Worked examples

Try it yourself

1. For the supplied model, calculate the average signed velocity over 0–2 s and compare it with x′(2 s).

   Show the answer and reasoning

   The average is +0.90 m/s; the instantaneous value at 2 s is +1.30 m/s.

   x(0 s) = 1.00 m and x(2 s) = 2.80 m. The finite quotient is (2.80 m − 1.00 m)/(2 s) = 0.90 m/s. The derivative uses a limit about 2 s, where the changing model rate is 1.30 m/s.

2. Express x′(2 s) and x″(2 s) in millimetre-based units instead of metre-based units. Does this change the physical quantities?

   Show the answer and reasoning

   x′(2 s) = +1300 mm/s and x″(2 s) = +400 mm/s². The quantities are unchanged.

   1 m = 1000 mm. Multiplying the numerical metre values 1.30 and 0.40 by 1000 gives 1300 and 400. Time stays in seconds. Changing units changes the numerical descriptions, while velocity and acceleration remain distinct quantity kinds.

3. Why can the original quotient be undefined at h = 0 while x′(2 s) exists?

   Show the answer and reasoning

   The derivative is a limit through nonzero increments, not the value of the quotient at zero.

   At h = 0 the numerator and denominator are both zero, so division is undefined. For nonzero h the quotient equals 1.30 m/s + (0.20 m/s²)h. That expression approaches the same finite 1.30 m/s from both sides. Cancelling h was justified only for h ≠ 0.

4. For q(t) = (1.00 m/s)|t − 2 s| on 0–4 s, calculate the one-sided difference quotients about 2 s. Does the ordinary derivative exist?

   Show the answer and reasoning

   The left quotients are −1.00 m/s and the right quotients +1.00 m/s. No ordinary derivative exists at 2 s.

   q(2 s) = 0 m and q(2 s + h) = (1.00 m/s)|h|. For h < 0, |h|/h = −1; for h > 0 it is +1. The two finite limits disagree. Continuity of this mathematical rule does not remove its corner.

5. For P₁ about 2 s, does the target 2.25 s meet an absolute truncation-error requirement of at most 0.01 m?

   Show the answer and reasoning

   No. Its exact model error is 0.0125 m, which exceeds 0.01 m.

   h = +0.25 s. The remainder is (0.20 m/s²)(0.25 s)² = 0.0125 m. Directly, P₁ = 3.125 m and x = 3.1375 m. The target is inside the model's domain; it fails this stricter approximation requirement.

6. A learner says: 'Every function with derivatives of all orders equals its Taylor series, so the exact P₂ also predicts this body's position at 5 s.' Which claims require correction?

   Show the answer and reasoning

   Both claims. Series equality needs a convergence-to-the-function check, and this position model is supplied only on 0–4 s.

   For series equality, the remainders of the increasing partial sums must tend to zero at the target; having all derivatives is insufficient. For our quadratic, P₂ is already exact on the supplied domain. Substitution at 5 s would give 8.50 m algebraically, but the task provides no physical position rule there.

What to carry forward

• A nonzero finite interval gives an average rate; a finite two-sided quotient limit gives an ordinary instantaneous derivative at an interior point.
• Position differentiated with respect to time has velocity units; differentiating again gives acceleration units. The reference, direction and domain remain part of the model.
• Being defined or continuous at a point does not guarantee that a derivative exists there.
• Taylor polynomials use increments from a stated expansion point; a first polynomial matches value and slope, while its omitted terms create a remainder.
• A useful approximation window follows from explicit derivative conditions, a bound over the whole segment and a chosen error requirement.
• Finite Taylor error, infinite-series convergence, measurement uncertainty and a model's physical validity are separate checks.