Mathematics Advanced • Year 11 • Module 3 • Lesson 7
The Second Derivative
Apply second derivatives to motion problems, curve shape, and rate-of-change interpretation in real contexts.
Problem 1 — Lift in a building (kinematics)
An express lift in a Sydney CBD building rises with displacement
s(t) = t³ − 6t² + 9t metres, 0 ≤ t ≤ 5 s
measured upward from its starting floor.
Set up: What are we solving for?
(i) Write down expressions for v(t) and a(t) (velocity and acceleration). 2 marks
(ii) Find the time at which the acceleration is zero, and state the velocity at that instant. Interpret physically. 3 marks
(iii) Between t = 0 s and t = 2 s, is the lift accelerating or decelerating (in the upward sense)? Justify with the sign of a(t). 2 marks
Stuck? Revisit lesson § Worked Example 3 (kinematics).Problem 2 — Cooling tea curve (concavity)
The temperature (in °C) of a cup of tea, t minutes after pouring, is modelled approximately by
T(t) = 20 + 70 · e−0.1t, t ≥ 0
You may use d/dt[ekt] = k · ekt.
Set up: What are we solving for?
(i) Find T′(t) and T″(t). 3 marks
(ii) State the sign of T′(t) and the sign of T″(t) for all t > 0. Interpret each in plain English (one sentence each). 2 marks
(iii) A drinker says "the tea is cooling, but it's cooling more slowly as time passes." Match this statement to the signs of T′ and T″, and explain why this corresponds to a curve that is decreasing and concave up. 2 marks
Problem 3 — Drone height (projectile-like)
A drone's height above a tennis court (in metres) at time t seconds is
h(t) = −5t² + 20t + 2, 0 ≤ t ≤ 4
Set up: What are we solving for?
(i) Find h′(t) and h″(t). What does h″(t) tell you about the drone's acceleration? 2 marks
(ii) Find the time at which h′(t) = 0 and the corresponding maximum height. Use the sign of h″(t) to confirm it is a maximum. 3 marks
(iii) Compare the constant negative value of h″(t) to a real physical setting: under what assumption is −10 m/s² approximately the "Earth-surface gravity" value? Comment on whether this model is a reasonable representation of free-fall. 2 marks
Stuck? Revisit lesson § Concept (sign of f″ and concavity).Problem 4 — Population growth slowing (inflection)
A new ant colony's size N (thousands of ants) is modelled by
N(t) = t³ − 9t² + 24t, 0 ≤ t ≤ 6 weeks
Set up: What are we solving for?
(i) Find N′(t) and N″(t). 2 marks
(ii) Find the time t at which the growth rate N′(t) is changing most slowly (i.e. the point of inflection of N). Find N at that time. 3 marks
(iii) Interpret the inflection point in plain English: before this time the growth was ____________; after this time the growth is ____________ . 2 marks
Problem 5 — Reading signs from a graph (interpretive)
The graph of y = f(x) is shown schematically with the following observations:
• f(0) = 2, f′(0) = 0 (horizontal tangent at the y-intercept).
• f(2) = 6 (a local maximum).
• f(4) = 2 (a point of inflection where the curve flattens then steepens downward).
• f(6) = −2 (a local minimum).
Set up: What are we solving for?
(i) State the sign (positive, negative, or zero) of f′(x) at each of x = 0, 2, 4, 6. 2 marks
(ii) State the sign of f″(x) at each of x = 2, 4, 6. (Hint: relate concavity to maximum / inflection / minimum.) 2 marks
(iii) In one sentence each, justify your sign for f″(2) and f″(4). 2 marks
Stuck? At a local max, the curve must bend downward (concave down); at a local min, it bends upward.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Lift
Set up. We are using the second derivative to find acceleration of a lift, and interpreting its sign as "speeding up" vs "slowing down".
(i) v(t) = s′(t) = 3t² − 12t + 9. a(t) = s″(t) = 6t − 12.
(ii) a(t) = 0 ⇒ t = 2 s. v(2) = 3(4) − 24 + 9 = −3 m/s. Physically: at t = 2 s the acceleration vanishes momentarily; the lift is moving downward at 3 m/s at that instant (it has already overshot and is on the return part of its motion within this model).
(iii) For 0 ≤ t < 2: a(t) = 6t − 12 < 0. So in the upward sense the lift is decelerating (its upward velocity is decreasing).
Problem 2 — Cooling tea
Set up. We are reading the rate (T′) and the rate-of-the-rate (T″) to describe how quickly the tea is cooling and whether the cooling itself is slowing.
(i) T′(t) = 70 · (−0.1) · e−0.1t = −7 e−0.1t. T″(t) = (−7)(−0.1) e−0.1t = 0.7 e−0.1t.
(ii) T′(t) < 0 for all t > 0 (e−0.1t > 0): the tea is cooling (temperature falling). T″(t) > 0 for all t > 0: the cooling rate is becoming less negative, i.e. the tea is cooling more slowly as time passes.
(iii) "Cooling" ↔ T′ < 0. "Cooling more slowly" ↔ T′ is increasing (becoming less negative) ↔ T″ > 0. A function that is decreasing (T′ < 0) and concave up (T″ > 0) has exactly this shape: a downward curve that flattens out as it approaches an asymptote (here the room temperature of 20°C).
Problem 3 — Drone
Set up. We are using the constant second derivative of a quadratic position model as a stand-in for gravitational acceleration.
(i) h′(t) = −10t + 20. h″(t) = −10 m/s² (constant). h″ < 0 means the upward velocity is decreasing throughout the flight — the drone is decelerating upward / accelerating downward.
(ii) h′(t) = 0 ⇒ 10t = 20 ⇒ t = 2 s. h(2) = −20 + 40 + 2 = 22 m. Confirm: h″(2) = −10 < 0, so t = 2 s is a maximum.
(iii) The value −10 m/s² is approximately the magnitude of g (acceleration due to gravity) near the Earth's surface, used in this constant-acceleration model. The model is reasonable for short-duration motion near the ground but ignores air resistance and any thrust the drone produces — a free-falling object (no thrust, no drag) would follow exactly this kind of quadratic h(t).
Problem 4 — Ant colony
Set up. We are locating an inflection point — the instant when growth slows from "accelerating" to "decelerating".
(i) N′(t) = 3t² − 18t + 24. N″(t) = 6t − 18.
(ii) N″(t) = 0 ⇒ t = 3 weeks. Sign change: at t = 2, N″ = −6 (−); at t = 4, N″ = 6 (+). ✓ N(3) = 27 − 81 + 72 = 18 thousand ants.
(iii) Before t = 3 weeks the growth rate N′(t) was decreasing (N″ < 0, decelerating growth). After t = 3 weeks the growth rate is increasing (N″ > 0, accelerating growth). The colony hits its slowest growth-rate at the inflection point.
Problem 5 — Reading f′ and f″ from a graph
Set up. We are deducing the signs of first and second derivatives at named features (max, min, inflection) without an algebraic formula.
(i) f′(0) = 0 (horizontal tangent). f′(2) = 0 (local max). f′(4) ≠ 0 in general — since this is a non-stationary inflection where the curve is still falling, f′(4) is negative. f′(6) = 0 (local min).
(ii) f″(2) < 0 (concave down at a local maximum). f″(4) = 0 (point of inflection). f″(6) > 0 (concave up at a local minimum).
(iii) f″(2) < 0 because at a local maximum the curve must bend downward; otherwise the point would be a minimum or a horizontal inflection. f″(4) = 0 because a point of inflection is where concavity changes sign — the bend transitions from one direction to the other, so the second derivative passes through zero.