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Rate of change in context

Find it and interpret it

  • In 7 of 20 recent exams
  • Paper 1 or 2
  • 1 to 4 marks
How it is set in the exam

Paper 2 (calculator) mostly (GDC), Paper 1 (no calculator) with exact values; mostly 1–4 marks per part, often inside a Section B context question. Common (7 of 20 exam versions).

Spot it in the exam
Exam phrases to spot (5)

  • “find the rate of change of VV when t=3t = 3”

  • “find dVdr\frac{dV}{dr} when r=2r = 2”

  • “how fast is … increasing”

  • “at what rate is … decreasing”

  • “interpret your answer in context”

In disguise (7)
  • the value of the quantity is given instead of the time: "when the temperature is 5050 °C" (solve for tt first)
  • a different quantity is given: dVdr\frac{dV}{dr} for a sphere "when the surface area is 36π36\pi" (find rr first)
  • a sign question, with no "rate" in it: "is the water level rising or falling at t=3t = 3? Justify your answer"
  • the long term: "describe what happens to the rate of change as tt increases" (the limit of the rate)
  • the long-run level of the quantity itself (the value the model settles at: a limit, which for a rational model is the horizontal asymptote, Asymptotes)
  • "show that the rate of growth is proportional to NN"
  • for how long the rate is above (or below) a level, often a total over several intervals: find the rate, then use Above or below a level on the rate instead of the model (solve rate =k= k with the GDC's intersections, then add the lengths of the intervals on the side asked for)
Variant

compare two rates ("which population is increasing faster at t=10t = 10?"). Find both rates, then write a sentence in context that compares them: that sentence is the reasoning mark.

Looks like it, but isn't (9)
  • the average rate over an interval: a chord gradient, no derivative
  • "the rate of change of yy with respect to xx is 23\frac{2}{3}", with no context and xx unknown: Find the point of contact. ("Find the rate of change of yy with respect to xx when x=2x = 2" is just the gradient at a given point.)
  • a rate is given and a total is asked: integrate (Total change from a rate)
  • "when is it decreasing fastest?": the greatest size of the rate, in Concavity and points of inflexion
  • a rate found from another rate (dVdt\frac{dV}{dt} from drdt\frac{dr}{dt}): related rates, HL only
  • the velocity or acceleration of a particle at a given time, or when it is at rest or turns: Velocity and acceleration: at rest, direction, speed. (Its velocity in the long run is a limit, and belongs here.)
  • no rate is asked: a model's initial value, a constant from a half-life or a data point, a value at a time, or the time it takes to reach a level: Exponential and log models
  • the horizontal asymptote of a function with no context: Rational functions: asymptotes, intercepts, sketch, range, or Exponential and log graphs: intercepts, asymptotes, domain, sign, range
  • the value of a sinusoidal model at a time, or its greatest value and when it happens, with no rate asked for: Read a sinusoid's features

Decide first: is a rate asked for (differentiate), or is a rate given and a total asked for (integrate)? See.

Worked example

Differentiate, evaluate, interpret

Paper 1 (no calculator)
Question

The concentration of a drug in a patient's blood, tt hours after it is taken, is C(t)=5te−0.4tC(t) = 5te^{-0.4t} mg L−1^{-1}, for t≥0t \ge 0.
(a) Show that C′(t)=e−0.4t(5−2t)C'(t) = e^{-0.4t}(5 - 2t). [3]
(b) Find the rate of change of CC when t=4t = 4, and interpret your answer in context. [3]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Steps

  1. Name the quantity and the variable.

    V(r)V(r) gives dVdr\frac{dV}{dr}. The units of the rate are the quantity's units per unit of the variable.

  2. If a value of the quantity (or of another quantity) is given

    , solve for the variable first.

  3. Find the rate:

    differentiate (), or use the GDC (gradient at a point) on .

  4. Evaluate and keep the sign:

    positive means increasing, negative means decreasing.

  5. Interpret

    with three things: when (the time, in context), the direction (increasing or decreasing: "changing" scores nothing), and the value with units. The interpretation's two marks need all three.

  6. Long term:

    use e−kt→0e^{-kt} \to 0 as t→∞t \to \infty (for k>0k > 0) to find the limit asked for (Limits).

Common mistakes

  • Giving the value instead of the rate

    D(10)D(10) is a depth; the rate is D′(10)D'(10).

  • An interpretation with a part missing

    say when, which direction (increasing or decreasing: "changing" is not accepted) and the value with units.

  • Using a given value of the quantity as tt

    "when T=50T = 50" means solve T(t)=50T(t) = 50 first.

More mistakes (3)

  • A double negative

    "decreasing at −1.17-1.17 m per hour" contradicts itself. Say "decreasing at 1.171.17 m per hour", or "the rate of change is −1.17-1.17 m per hour".

  • Degree mode

    with a trig model: the answer scores ().

  • Multiplying by a rate the question never gave

    related rates are HL.

Keep going

Easy to confuse with

Next typeFind the point of contact

More in Differentiation

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