Mathematics Advanced • Year 11 • Module 3 • Lesson 1
Average & Instantaneous Rates of Change
Apply the difference quotient to multi-step problems involving sprints, drug clearance, water tanks, traffic flow and population growth.
Problem 1 — Usain Bolt's 100 m world record
Bolt's position (in metres) along the track during his 9.58 s 100 m record is modelled below. Split times at five checkpoints were recorded.
| Time t (s) | 0 | 1.89 | 4.64 | 6.31 | 9.58 |
|---|---|---|---|---|---|
| Position s (m) | 0 | 20 | 60 | 80 | 100 |
Set up: What are we solving for?
(i) Compute Bolt's average speed for the whole race in m/s, to 3 sig fig. 1 mark
(ii) Compute the average speed in each 20 m segment: 0→20 m, 20→60 m, 60→80 m, 80→100 m. 3 marks
(iii) Identify the segment with the highest average speed. Explain in one sentence why his fastest instantaneous speed during the race is greater than the answer to part (i). 2 marks
Stuck? Revisit lesson § Real-world link.Problem 2 — Drug clearance from the bloodstream
The concentration of a painkiller in a patient's blood (in mg/L) at time t hours after the dose is modelled by
C(t) = 80 / (t + 4), for 0 ≤ t ≤ 12
Set up: What are we solving for?
(i) Find the average rate of change of C over [0, 4]. State the sign and explain what it means physically. 2 marks
(ii) Estimate the instantaneous rate of change of C at t = 2 by computing the average rate over [2, 2.1] and over [2, 2.01]. Round to 3 sig fig. 3 marks
(iii) The next dose is due when the rate of clearance slows to about −1 mg/L per hour. Without solving formally, state whether this happens before, at, or after t = 4, justifying in one sentence using the trend of your answers. 2 marks
Problem 3 — Draining a water tank
A water tank holds V(t) = 500 − 5t² litres at time t minutes after the outlet is opened, until the tank empties.
Set up: What are we solving for?
(i) Find the time at which the tank is empty (V = 0). 1 mark
(ii) Compute the average rate of change of V over [0, 5] and over [5, 10]. Comment on the comparison in one sentence — what does the difference tell you about the flow rate? 3 marks
(iii) Estimate the instantaneous flow rate (in L/min) at t = 6 by computing the average rate of change over [6, 6.01]. State the sign and interpret physically. 2 marks
Stuck? Revisit lesson § Trap 03 — instantaneous rate still needs two points (very close together).Problem 4 — Counting cars at a tunnel entrance
A camera at a tunnel entrance counts the cumulative number of cars N(t) that have entered by time t minutes after 8:00 am. Recorded values:
| t (min) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| N (cars) | 0 | 140 | 320 | 520 | 700 | 820 | 880 |
Set up: What are we solving for?
(i) Find the average rate of cars per minute over the full half-hour. 1 mark
(ii) Find the average rate (cars/min) in each 5-minute interval. 3 marks
(iii) Identify the 5-minute interval with the highest average flow. Explain in one sentence how a single average over the half-hour can hide the morning rush peak. 2 marks
Problem 5 — Bacterial population
A bacterial population (in thousands) at time t hours is modelled by P(t) = 200 + 30t − t².
Set up: What are we solving for?
(i) Find the average growth rate over [0, 10] and over [10, 20]. 2 marks
(ii) Using the difference quotient with h = 0.001, estimate the instantaneous growth rate at t = 10 to 1 decimal place. 2 marks
(iii) Simplify ( P(t + h) − P(t) ) / h algebraically, then state what value the expression approaches as h → 0. Use this to identify the time at which the instantaneous growth rate is exactly zero. 3 marks
Stuck on (iii)? Expand (t + h)², subtract t², divide every term by h. The expression should be in the form (something) + (something with h in it).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bolt
Set up. Average speed = change in position ÷ change in time, applied to the whole race and then to each 20 m segment.
(i) (100 − 0) / 9.58 = 10.44 m/s (3 sig fig).
(ii) 0→20 m over 1.89 s: 20 / 1.89 ≈ 10.58 m/s. 20→60 m over (4.64 − 1.89) = 2.75 s: 40 / 2.75 ≈ 14.55 m/s. 60→80 m over (6.31 − 4.64) = 1.67 s: 20 / 1.67 ≈ 11.98 m/s. 80→100 m over (9.58 − 6.31) = 3.27 s: 20 / 3.27 ≈ 6.12 m/s.
(iii) The 20→60 m segment has the highest average speed (≈ 14.55 m/s). His peak instantaneous speed must be at least as fast as this segment average, so it must exceed the overall race average of 10.44 m/s. Average over the whole race smooths out the fast middle and slow start/finish.
Problem 2 — Drug clearance C(t) = 80 / (t + 4)
Set up. The drug concentration is falling; we want the difference quotient over the given interval, then over shrinking intervals near t = 2.
(i) C(0) = 80/4 = 20; C(4) = 80/8 = 10. Average rate = (10 − 20) / 4 = −2.5 mg/L per hour. The negative sign means the drug is being cleared (concentration is falling).
(ii) C(2) = 80/6 ≈ 13.333. C(2.1) = 80/6.1 ≈ 13.115. Rate over [2, 2.1] ≈ (13.115 − 13.333) / 0.1 ≈ −2.18 mg/L per h. C(2.01) = 80/6.01 ≈ 13.311. Rate over [2, 2.01] ≈ (13.311 − 13.333) / 0.01 ≈ −2.22 mg/L per h. The values are converging to approximately −2.22 mg/L per h.
(iii) Magnitudes are getting smaller as t increases (compare −2.5 over [0, 4] with −2.22 at t ≈ 2 then expected smaller still at larger t). The rate of −1 mg/L per h is shallower than the rates we have found, so it happens after t = 4.
Problem 3 — Tank V(t) = 500 − 5t²
Set up. Find when V = 0 first, then apply the difference quotient over [0, 5] and [5, 10], then estimate at t = 6.
(i) 500 − 5t² = 0 ⇒ t² = 100 ⇒ t = 10 minutes (taking the positive root).
(ii) V(0) = 500; V(5) = 500 − 125 = 375; V(10) = 0. Over [0, 5]: (375 − 500)/5 = −25 L/min. Over [5, 10]: (0 − 375)/5 = −75 L/min. The tank empties faster in the second half — the outlet flow rate is increasing in magnitude as the tank drains (the model is not constant flow).
(iii) V(6) = 500 − 180 = 320; V(6.01) = 500 − 5(36.1201) = 500 − 180.6005 = 319.3995. Average rate ≈ (319.3995 − 320) / 0.01 = −60.05 L/min ≈ −60 L/min. The negative sign confirms water is leaving the tank; magnitude is 60 L/min at that instant.
Problem 4 — Tunnel traffic flow
Set up. Total cars ÷ total time gives the overall average; difference quotient on each 5-minute slice gives the segment flows; identify the peak segment and compare.
(i) 880 cars / 30 min = ≈ 29.3 cars/min.
(ii) 0→5: 140/5 = 28 cars/min. 5→10: 180/5 = 36 cars/min. 10→15: 200/5 = 40 cars/min. 15→20: 180/5 = 36 cars/min. 20→25: 120/5 = 24 cars/min. 25→30: 60/5 = 12 cars/min.
(iii) The peak 5-minute average is 40 cars/min in the 10→15 interval. The half-hour average of 29.3 cars/min combines the rush peak (40) with the much quieter 12 cars/min at the end, so the rush peak is hidden inside the bulk average — exactly the problem average rates have when speeds are uneven.
Problem 5 — Bacterial population P(t) = 200 + 30t − t²
Set up. Apply the difference quotient over given intervals; numerically estimate the instantaneous rate at t = 10; finally derive the algebraic expression to find where it equals zero.
(i) P(0) = 200; P(10) = 200 + 300 − 100 = 400; P(20) = 200 + 600 − 400 = 400. Over [0, 10]: (400 − 200)/10 = 20 thousand/h. Over [10, 20]: (400 − 400)/10 = 0 thousand/h.
(ii) P(10) = 400; P(10.001) = 200 + 30(10.001) − (10.001)² = 200 + 300.030 − 100.020001 = 400.009999. Rate ≈ (400.009999 − 400)/0.001 ≈ 10.0 thousand/h (1 dp).
(iii) P(t + h) = 200 + 30(t + h) − (t + h)² = 200 + 30t + 30h − t² − 2th − h². P(t + h) − P(t) = 30h − 2th − h². Divide by h: 30 − 2t − h. As h → 0 the expression approaches 30 − 2t. Setting 30 − 2t = 0 gives t = 15 hours: at this moment, the instantaneous growth rate is zero (the population is at its maximum).