Mathematics Advanced • Year 11 • Module 4 • Lesson 10
Differentiating ax
Apply the ln a factor to real growth/decay problems in finance, biology, technology and physics.
Problem 1 — Compound interest growth rate (financial)
An investment of $1000 grows according to
A(t) = 1000 × 1.06t, t in years
Set up: What are we solving for?
(i) Find dA/dt as a function of t, including the ln 1.06 factor. 2 marks
(ii) Evaluate dA/dt at t = 0 and at t = 10 years. Give answers in dollars per year (to the nearest cent). 3 marks
(iii) Show that dA/dt = (ln 1.06) × A. What does the constant ln 1.06 represent (think about the "continuously compounded" growth rate)? 2 marks
Stuck on (iii)? d/dt(at) = at ln a; rewrite the right side using 1000·1.06t.Problem 2 — Drug elimination expressed in base ½ (medical)
A drug clears from the blood with half-life 4 hours, so the concentration is
C(t) = 100 × (½)t/4, t in hours, C in mg/L
Set up: What are we solving for?
(i) Use d/dt(akt) = k akt ln a (with a = ½, k = 1/4) to find dC/dt. Note ln(½) = −ln 2. 3 marks
(ii) Evaluate the rate of clearance at t = 0 and at t = 4. Round to 3 decimal places. 2 marks
(iii) Show that dC/dt = −(ln 2 / 4) × C. Hence verify that the model satisfies dC/dt = −ke C with ke = ln 2 / 4 (the elimination constant for first-order pharmacokinetics). 2 marks
Problem 3 — Moore's Law (technology doubling)
The number of transistors per chip is modelled by
N(t) = N0 × 2t/2, t in years from a chosen baseline
(this says transistor density doubles every 2 years).
Set up: What are we solving for?
(i) Differentiate N(t) with respect to t, simplifying the result in the form (constant) × N(t). 3 marks
(ii) Express your "constant" in (i) as ln 2 divided by the doubling time. Hence write down the analogous "constant" for a model that doubles every 18 months. 2 marks
(iii) If a chip currently has N = 1010 transistors, find the instantaneous rate of growth (transistors per year). Use ln 2 ≈ 0.6931. 2 marks
Problem 4 — Pressure-amplitude growth (acoustics)
The sound-pressure level L (dB) corresponds to a pressure amplitude p (Pa) by
p(L) = p0 × 10L/20, p0 = 20 µPa (reference pressure)
Set up: What are we solving for?
(i) Find dp/dL using the chain rule. 3 marks
(ii) Show that dp/dL = (ln 10 / 20) × p. Interpret in one sentence: how does the rate of change of pressure scale with the current pressure? 2 marks
(iii) For L = 60 dB, compute the pressure amplitude p and the rate dp/dL. Give p in Pa and dp/dL in Pa/dB, to 3 s.f. (use p0 = 20 × 10−6 Pa). 2 marks
Problem 5 — Stationary point with a general base
The function g(x) = (1 − x) · 2x models a quantity that depends on an exponentially-growing factor scaled by a decreasing linear factor.
Set up: What are we solving for?
(i) Find g'(x), factorising the result. 3 marks
(ii) Find the stationary point of g, giving the x-coordinate in exact form (in terms of ln 2) and to 2 d.p. 3 marks
(iii) Use the sign of g' immediately to the left and right of the critical x to classify the stationary point. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Compound interest rate
Set up. We differentiate an at model to obtain a growth rate, and reinterpret the ln a factor as a continuously compounded rate.
(i) dA/dt = 1000 × 1.06t × ln 1.06 = 1000 ln 1.06 × 1.06t.
(ii) ln 1.06 ≈ 0.05827. At t = 0: dA/dt ≈ 1000 × 0.05827 = $58.27 per year. At t = 10: 1.0610 ≈ 1.7908, so dA/dt ≈ 1000 × 0.05827 × 1.7908 ≈ $104.36 per year.
(iii) dA/dt = 1000 × ln 1.06 × 1.06t = (ln 1.06)(1000 × 1.06t) = (ln 1.06) × A. ✓ ln 1.06 ≈ 0.0583 is the "continuously compounded" equivalent of the 6% annual rate — the instantaneous growth rate as a proportion of the current balance.
Problem 2 — Drug half-life
Set up. We differentiate an exponential decay in base ½ and verify the standard pharmacokinetic differential equation.
(i) dC/dt = 100 × (¼) × (½)t/4 × ln(½). Wait — d/dt(akt) = k akt ln a, with k = 1/4, a = ½: dC/dt = 100 × (1/4) × (½)t/4 × ln(½) = 25 (½)t/4 × (−ln 2) = −25 ln 2 × (½)t/4.
(ii) At t = 0: dC/dt = −25 ln 2 × 1 ≈ −25 × 0.6931 ≈ −17.329 mg/L per hour. At t = 4 (one half-life): (½)1 = ½, so dC/dt ≈ −25 × 0.6931 × 0.5 ≈ −8.664 mg/L per hour.
(iii) dC/dt = −25 ln 2 × (½)t/4 = −(ln 2 / 4) × [100 × (½)t/4] = −(ln 2 / 4) × C. ✓ So ke = ln 2 / 4 ≈ 0.173 per hour, the standard "first-order elimination constant" — the half-life divided into ln 2.
Problem 3 — Moore's Law
Set up. We differentiate a doubling-time model and identify the "growth constant" as ln 2 / (doubling time).
(i) dN/dt = N0 × (1/2) × 2t/2 × ln 2 = (ln 2 / 2) × N0 × 2t/2 = (ln 2 / 2) × N(t).
(ii) The constant is ln 2 / (doubling time). For doubling every 1.5 years: constant = ln 2 / 1.5 ≈ 0.4621 per year.
(iii) dN/dt = (ln 2 / 2) × 1010 = (0.6931 / 2) × 1010 = 0.3466 × 1010 = ≈ 3.47 × 109 transistors per year.
Problem 4 — Sound pressure
Set up. We differentiate p(L) = p0 · 10L/20 using the chain rule with the ln 10 factor.
(i) dp/dL = p0 × (1/20) × 10L/20 × ln 10 = (p0 ln 10 / 20) × 10L/20.
(ii) dp/dL = (ln 10 / 20) × [p0 × 10L/20] = (ln 10 / 20) × p. ✓ The rate of change of pressure with L is proportional to the current pressure — louder sounds change pressure faster per dB.
(iii) 1060/20 = 103 = 1000. p = (20 × 10−6) × 1000 = 0.0200 Pa. dp/dL = (ln 10 / 20) × 0.0200 ≈ (2.303/20) × 0.0200 ≈ 2.30 × 10−3 Pa/dB.
Problem 5 — Stationary point of g(x) = (1 − x) · 2x
Set up. We use the product rule with d/dx(2x) = 2x ln 2, factor, then find and classify the critical point.
(i) g'(x) = (−1) · 2x + (1 − x) · 2x ln 2 = 2x[−1 + (1 − x) ln 2].
(ii) Set g'(x) = 0: 2x > 0 always, so −1 + (1 − x) ln 2 = 0 ⇒ (1 − x) ln 2 = 1 ⇒ 1 − x = 1/ln 2 ⇒ x = 1 − 1/ln 2. Numerical: 1/ln 2 ≈ 1.4427, so x ≈ 1 − 1.4427 ≈ −0.44.
(iii) Sign of g'(x) = 2x[(1 − x) ln 2 − 1]. For x slightly less than −0.44, (1 − x) is slightly larger, so (1 − x) ln 2 > 1, bracket positive ⇒ g' > 0 (increasing). For x slightly more than −0.44, bracket negative ⇒ g' < 0 (decreasing). Increasing → decreasing means local maximum at x = 1 − 1/ln 2.