Mathematics Advanced • Year 11 • Module 3 • Lesson 1
Average & Instantaneous Rates of Change
HSC-style writing on the difference quotient — including an extended response that derives the instantaneous rate as a limit.
1. Short-answer questions
1.1 The displacement of a particle along a straight line at time t seconds is s(t) = t² + 3t metres. Find the average velocity of the particle over the interval [1, 4]. State the units. 2 marks Band 3
1.2 A water tank holds V(t) = 100 − 2t² litres after t minutes (0 ≤ t ≤ 5). By computing average rates of change of V over [2, 2.1] and [2, 2.01], estimate the instantaneous rate of change of volume at t = 2. State the sign and what it tells you about the tank. 3 marks Band 3-4
1.3 The graph of y = f(x) passes through the points P(1, 2) and Q(1 + h, 2 + 3h + h²), where h ≠ 0.
(a) Write the gradient of the secant PQ as a simplified expression in h.
(b) Hence determine the value the secant gradient approaches as h → 0, and state what this number represents geometrically. 4 marks Band 4
2. Extended response
2.1 A runner's position from the start line, in metres at time t seconds, is given by s(t) = t² for 0 ≤ t ≤ 6.
(a) Calculate the average velocity of the runner over the intervals [3, 4], [3, 3.5], [3, 3.1], [3, 3.01] and [3, 3.001]. Present your values in a table.
(b) State the value to which these average velocities are converging, and hence write a sentence stating the instantaneous velocity of the runner at t = 3 seconds.
(c) Derive the same answer algebraically by simplifying ( s(3 + h) − s(3) ) / h and considering what happens as h → 0.
(d) Generalise: by simplifying ( s(t + h) − s(t) ) / h, find a formula for the instantaneous velocity of the runner at any time t ∈ [0, 6]. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct values of s at each endpoint for all five intervals (s(3) = 9 used throughout).
• 1 mark — correct average velocities tabulated (7, 6.5, 6.1, 6.01, 6.001 m/s).
Part (b) — 1 mark
• 1 mark — states the values are converging to 6 m/s and writes that the instantaneous velocity at t = 3 is 6 m/s.
Part (c) — 2 marks
• 1 mark — expands (3 + h)² = 9 + 6h + h², subtracts 9, divides by h, simplifies to 6 + h.
• 1 mark — concludes that 6 + h → 6 as h → 0, hence the instantaneous velocity is 6 m/s, in agreement with (b).
Part (d) — 3 marks
• 1 mark — expands (t + h)² = t² + 2th + h², subtracts t², divides by h, simplifies to 2t + h.
• 1 mark — concludes 2t + h → 2t as h → 0; states the instantaneous velocity formula is 2t.
• 1 mark — checks the formula at t = 3 (gives 6 m/s), confirming it agrees with the earlier work, and notes correct units of m/s.
Your response:
Stuck on (d)? Use the same algebra as (c) but keep t general instead of substituting 3.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Average velocity of s(t) = t² + 3t on [1, 4] (2 marks)
Sample response. s(1) = 1 + 3 = 4; s(4) = 16 + 12 = 28. Average velocity = (s(4) − s(1)) / (4 − 1) = (28 − 4) / 3 = 24 / 3 = 8 m/s.
Marking notes. 1 mark — correct values of s(1) and s(4) with substitution shown. 1 mark — correct application of the difference quotient with units. Bare answer "8" with no working scores 0.5; answer without units loses the unit half.
1.2 — Estimate instantaneous rate of V(t) = 100 − 2t² at t = 2 (3 marks)
Sample response. V(2) = 100 − 8 = 92. V(2.1) = 100 − 2(4.41) = 91.18. Average rate over [2, 2.1] = (91.18 − 92) / 0.1 = −8.2 L/min. V(2.01) = 100 − 2(4.0401) = 91.9198. Average rate over [2, 2.01] = (91.9198 − 92) / 0.01 = −8.02 L/min. The values are converging to about −8 L/min. The negative sign means the tank is losing water (volume is decreasing) at an instantaneous rate of approximately 8 L/min when t = 2.
Marking notes. 1 mark — correct V values and one correct average rate (either interval). 1 mark — both rates correct and the convergence noted (close to −8). 1 mark — correct sign-interpretation (volume decreasing/water leaving). Common errors: subtracting V(2) − V(2.1) instead of V(2.1) − V(2) (flips the sign); leaving off the negative sign in the conclusion.
1.3 — Secant gradient through P(1, 2) and Q(1 + h, 2 + 3h + h²) (4 marks)
(a) Sample response. Gradient PQ = ( (2 + 3h + h²) − 2 ) / ( (1 + h) − 1 ) = (3h + h²) / h. Factor h in numerator: = h(3 + h) / h = 3 + h (valid for h ≠ 0).
(b) Sample response. As h → 0, the expression 3 + h approaches 3. Geometrically this is the gradient of the tangent to the curve at the point P(1, 2) — i.e., the instantaneous rate of change of f at x = 1.
Marking notes. (a) 1 mark — correct numerator and denominator; 1 mark — correctly factor and cancel h with the note "h ≠ 0". (b) 1 mark — correct limit value 3; 1 mark — geometric interpretation (tangent gradient / instantaneous rate). Failing to note "h ≠ 0" before cancelling loses 0.5; concluding "average rate" instead of "instantaneous rate / tangent" loses the geometric mark.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). s(3) = 9. Compute s at each right-endpoint and form the difference quotient:
| Interval | s at right end | Δs | Δt | Average velocity (m/s) |
|---|---|---|---|---|
| [3, 4] | 16 | 7 | 1 | 7 |
| [3, 3.5] | 12.25 | 3.25 | 0.5 | 6.5 |
| [3, 3.1] | 9.61 | 0.61 | 0.1 | 6.1 |
| [3, 3.01] | 9.0601 | 0.0601 | 0.01 | 6.01 |
| [3, 3.001] | 9.006001 | 0.006001 | 0.001 | 6.001 |
[1 mark — correct s values; 1 mark — correct average velocities tabulated.]
Part (b). The average velocities 7, 6.5, 6.1, 6.01, 6.001 are converging to 6 m/s. Hence the instantaneous velocity of the runner at t = 3 s is 6 m/s. [1 mark.]
Part (c). ( s(3 + h) − s(3) ) / h = ( (3 + h)² − 9 ) / h = ( 9 + 6h + h² − 9 ) / h = ( 6h + h² ) / h = 6 + h (h ≠ 0). [1 mark — algebraic simplification.] As h → 0, the expression 6 + h approaches 6. Hence the instantaneous velocity at t = 3 s is exactly 6 m/s, in agreement with (b). [1 mark — limit and conclusion.]
Part (d). Generalising the same algebra:
( s(t + h) − s(t) ) / h = ( (t + h)² − t² ) / h = ( t² + 2th + h² − t² ) / h = ( 2th + h² ) / h = 2t + h.
[1 mark — generalised simplification.] As h → 0, the expression 2t + h approaches 2t. Hence the instantaneous velocity of the runner at time t is given by the formula:
v(t) = 2t m/s.
[1 mark — limit and statement of formula.] Check at t = 3: v(3) = 6 m/s, matching (b) and (c). Units are m/s, consistent with the units of s(t). [1 mark — check, units, conclusion.]
Total: 8/8.
Band descriptors for marker.
Band 3: Computes average velocities over the given intervals but does not notice convergence; cannot complete the algebraic limit. Often confuses position with velocity (uses s instead of Δs). ≈ 3-4 marks.
Band 4: Completes (a) and (b) correctly with units; attempts (c) but stops at the algebraic simplification without taking the limit. ≈ 5-6 marks.
Band 5: Completes (a)-(c). Attempts (d) but does not check at t = 3 or leaves out the limit step. ≈ 6-7 marks.
Band 6: All four parts complete, units carried through, algebra labelled with "h ≠ 0" before cancelling, the limit explicitly stated and checked, and v(t) = 2t identified as the instantaneous velocity formula. 8/8.