Mathematics Advanced • Year 11 • Module 4 • Lesson 9
Differentiating ex
Apply differentiation of ex to real growth/decay problems: tangent lines, stationary points, drug clearance, charging capacitors.
Problem 1 — Tangent line to an exponential curve (geometric)
Consider the curve y = ex.
Set up: What are we solving for?
(i) Find dy/dx and evaluate the gradient at the point (1, e). 2 marks
(ii) Hence find the equation of the tangent line to y = ex at (1, e). Give your answer in the form y = mx + c. 2 marks
(iii) Show that the tangent crosses the x-axis at x = 0. Explain in one sentence what is special about this — what does it reveal about the geometry of y = ex? 2 marks
Stuck? Tangent equation: y − y1 = m(x − x1).Problem 2 — Drug clearance from the bloodstream (medical)
The concentration of a drug in the bloodstream t hours after a single dose is
C(t) = 50 e−0.2 t, t ≥ 0, C in mg/L
Set up: What are we solving for?
(i) Find dC/dt and evaluate the rate of change at t = 0 and t = 5 hours. State the units. 3 marks
(ii) Show that dC/dt = −0.2 C, and explain in one sentence what this differential equation says physically about how drug clearance depends on the amount currently in the body. 2 marks
(iii) Find the time at which the drug concentration is 10 mg/L and the rate at which it is decreasing at that moment. 3 marks
Problem 3 — Charging a capacitor (electrical)
A capacitor in an RC circuit charges according to
V(t) = 12 (1 − e−t/3), t in seconds, V in volts
Set up: What are we solving for?
(i) Find dV/dt. State the initial rate of charging dV/dt at t = 0. 3 marks
(ii) Show that dV/dt is always positive and that it approaches 0 as t → ∞. Hence describe in one sentence how the rate of charging changes with time. 2 marks
(iii) Find the time at which V reaches 90% of its limiting value (i.e. 10.8 V), giving your answer to 2 d.p. 2 marks
Stuck on (iii)? Solve 12(1 − e−t/3) = 10.8 ⇒ e−t/3 = 0.1.Problem 4 — Stationary point of a product (curve sketching)
The function f(x) = x² e−x arises in radial distribution calculations in chemistry.
Set up: What are we solving for?
(i) Find f'(x), factorising the result. 3 marks
(ii) Find all stationary points of f and use the sign of f' on either side of each critical x to classify them. 3 marks
(iii) State the behaviour of f as x → ∞ and as x → −∞. Hence give a one-line description of the overall shape of the graph. 2 marks
Problem 5 — Bacterial growth and verification (biology)
A bacterial colony's population is modelled by
P(t) = 100 e0.4 t, t in hours, P = number of bacteria
Set up: What are we solving for?
(i) Compute dP/dt and verify by direct substitution that dP/dt = 0.4 P. 2 marks
(ii) Find the rate of growth (bacteria per hour) when the population has just reached 1000. 2 marks
(iii) Explain in one sentence why the proportionality dP/dt = k P (here k = 0.4) is what makes the exponential the right model — what does it say about how new bacteria are produced? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Tangent to y = ex at (1, e)
Set up. We use the derivative to get the slope, then write the tangent line.
(i) dy/dx = ex. At x = 1: gradient = e ≈ 2.718.
(ii) y − e = e(x − 1) ⇒ y = ex − e + e = y = e x (i.e. y = ex with c = 0).
(iii) Set y = 0: 0 = ex ⇒ x = 0. The tangent passes through the origin. Geometric meaning: the tangent to y = ex at (1, e) extended back hits the x-axis exactly 1 unit to the left of the point of tangency — a special property of the function y = ex that holds at every point because slope and y-coordinate are equal.
Problem 2 — Drug clearance
Set up. We differentiate an exponential decay to obtain the rate of clearance, then interpret it.
(i) dC/dt = 50 · (−0.2) e−0.2 t = −10 e−0.2 t. At t = 0: dC/dt = −10 mg/L per hour. At t = 5: dC/dt = −10 e−1 ≈ −3.68 mg/L per hour.
(ii) dC/dt = −10 e−0.2 t = −0.2 · (50 e−0.2 t) = −0.2 C. ✓ Physically: the body clears the drug at a rate proportional to how much is currently present — when there is twice as much, it clears twice as fast.
(iii) Solve 50 e−0.2 t = 10 ⇒ e−0.2 t = 0.2 ⇒ −0.2 t = ln(0.2) ⇒ t = −ln(0.2)/0.2 = ln 5 / 0.2 ≈ 1.6094 / 0.2 ≈ 8.05 hours. Rate then: dC/dt = −0.2 × 10 = −2 mg/L per hour.
Problem 3 — Capacitor charging
Set up. We are differentiating the exponential approach to a limit and interpreting the rate.
(i) dV/dt = 12 · (−(−1/3)) e−t/3 = 4 e−t/3. At t = 0: dV/dt = 4 V/s (initial rate of charging is fastest).
(ii) e−t/3 > 0 for all t, so dV/dt = 4 e−t/3 > 0 always (V is always increasing). As t → ∞, e−t/3 → 0, so dV/dt → 0. The capacitor charges fastest at the start, then slows as V approaches the limiting value of 12 V.
(iii) 12(1 − e−t/3) = 10.8 ⇒ 1 − e−t/3 = 0.9 ⇒ e−t/3 = 0.1 ⇒ −t/3 = ln(0.1) = −ln 10 ⇒ t = 3 ln 10 ≈ 3 × 2.3026 ≈ 6.91 seconds.
Problem 4 — Stationary points of f(x) = x² e−x
Set up. We use the product rule, find critical points, and use sign analysis to classify them.
(i) f'(x) = 2x · e−x + x² · (−e−x) = e−x(2x − x²) = x e−x(2 − x).
(ii) Set f'(x) = 0: e−x > 0 always, so x(2 − x) = 0 ⇒ x = 0 or x = 2. Stationary points (0, f(0)) = (0, 0) and (2, f(2)) = (2, 4e−2) ≈ (2, 0.541).
Sign of f' = x(2 − x) e−x: for x < 0, x < 0 and (2 − x) > 0, so f' < 0 (decreasing). For 0 < x < 2, both factors positive ⇒ f' > 0 (increasing). For x > 2, x > 0 but (2 − x) < 0 ⇒ f' < 0 (decreasing). (0, 0) is a local minimum; (2, 4e−2) is a local maximum.
(iii) As x → ∞: x² grows polynomially but e−x decays faster, so f(x) → 0+. As x → −∞: x² → ∞ and e−x → ∞, so f(x) → ∞. Shape: from ∞ on the left, decreasing through a local minimum at the origin, rising to a local maximum near x = 2, then decaying back to 0 along the positive x-axis.
Problem 5 — Bacterial growth
Set up. We are differentiating an exponential growth model to verify the differential equation it satisfies, then interpreting the result.
(i) dP/dt = 100 · 0.4 e0.4 t = 40 e0.4 t. Also 0.4 P = 0.4 × 100 e0.4 t = 40 e0.4 t. Equal. ✓ So dP/dt = 0.4 P.
(ii) When P = 1000, dP/dt = 0.4 × 1000 = 400 bacteria per hour.
(iii) dP/dt = k P says the rate of producing new bacteria is proportional to the number currently present — each bacterium contributes a fixed amount to the reproductive rate, so a population that doubles produces new bacteria twice as fast. Exponentials are precisely the functions with this self-similar growth pattern.