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.