Mathematics • Year 9 • Unit 2 • Lesson 17
Applications of Non-Linear Relationships
Drill the family-to-scenario matching habit and the "read features in context" routine: projectile parabolas, inverse-proportion hyperbolas, exponential growth, circular paths. Always with units, always rejecting impossible answers.
1. I do — fully worked example (projectile)
Read every line. Each step explains why we extract that piece of information, not just what the answer is.
Problem. A ball is thrown up with height $h = -5t^2 + 20t$ (m, s). Find (a) the time of maximum height, (b) the max height, (c) when the ball lands.
Step 1 — Identify the family and direction.
$t^2$ is the highest power $\Rightarrow$ parabola. $a = -5 < 0 \Rightarrow$ opens DOWN.
Reason: opens down $\Rightarrow$ the vertex is a MAXIMUM, which makes sense for a ball thrown up.
Step 2 — Find the $x$-intercepts.
Set $h = 0$: $-5t^2 + 20t = 0 \Rightarrow -5t(t - 4) = 0 \Rightarrow t = 0$ or $t = 4$.
Reason: $t = 0$ is the launch (height $= 0$); $t = 4$ s is the landing (height $= 0$ again).
Step 3 — Axis of symmetry $\to$ time of max.
Axis is midway between the two roots: $t = \dfrac{0 + 4}{2} = 2$ s.
Reason: parabolas are symmetric about their axis; the vertex sits on it.
Step 4 — Substitute to get max height.
$h(2) = -5(4) + 20(2) = -20 + 40 = 20$ m.
Reason: always state UNITS in word problems — naked numbers lose marks.
Step 5 — Reject any impossible solution.
Both $t$-values from Step 2 are $\ge 0$, so both are physical. Landing time: $t = 4$ s.
Reason: time can't be negative — always check, even if it's fine here.
Answer: (a) max at $t = 2$ s; (b) max height $= 20$ m; (c) lands at $t = 4$ s.
2. We do — fill in the missing steps
Same routine but with the working faded. Fill in each blank. 4 marks
Problem. A driver travels $120$ km. Time $T$ (hours) vs average speed $v$ (km/h). (a) Write the relation. (b) Find $T$ at $v = 60$ and $v = 30$. (c) Describe the shape of the graph.
Step 1 — Identify the relationship: distance $= $ speed $\times$ time. So $120 = v \cdot T$, which means $T = \_\_\_\_$.
Step 2 — Family: $T = \dfrac{120}{v}$ has $v$ in the __________________ $\Rightarrow$ family is __________________.
Step 3 — Domain: only $v > 0$ makes sense because __________________. So ONE branch in Quadrant __.
Step 4 — Substitute: at $v = 60$: $T = \dfrac{120}{60} = \_\_\_\_$ hours. At $v = 30$: $T = \dfrac{120}{30} = \_\_\_\_$ hours.
Step 5 — Pattern check: halving the speed (from $60$ to $30$) __________________ the time. That's the classic inverse-proportion signature.
3. You do — independent practice
Show your working under each problem. The first four are foundation (identify the family). The middle two are standard (substitute and interpret). The last two are extension (full reasoning with reject-impossible).
Foundation — name the family
3.1 Name the family that models the height of a kicked football over time. 1 mark
3.2 Name the family for "trip time vs average speed for a fixed distance". 1 mark
3.3 Name the family for "bacteria doubling every hour". 1 mark
3.4 Name the curve described by "path of a runner on a circular track of radius $100$ m". 1 mark
Standard — substitute and interpret
3.5 A stone is thrown so $h = -5t^2 + 30t$ (m, s). (a) Find the height at $t = 1$ s. (b) Find the time of landing ($h = 0$, $t > 0$). 2 marks
3.6 Six workers do a job in $10$ days. Assuming inverse proportion (worker-days = constant): (a) find $k$, (b) compute the time for $3$ workers. 2 marks
Extension — reject impossibles, include units
3.7 A projectile equation factorises so $h = 0$ when $t = -2$ or $t = 6$. (a) Which value is the landing time? (b) State, with reasons, why the other value is rejected. (c) Find the time of maximum height (use the axis of symmetry). 3 marks
3.8 A colony of $50$ bacteria doubles every hour: $N = 50 \cdot 2^t$. (a) Compute $N$ at $t = 0, 1, 2, 3, 4$ (with units — "cells"). (b) After how many WHOLE hours does the population first exceed $1000$? Show your working. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $T = \dfrac{120}{v}$)
Step 1: $T = \mathbf{\dfrac{120}{v}}$.
Step 2: $v$ in the denominator; family is hyperbola.
Step 3: $v > 0$ because speeds are positive — you can't drive at a negative speed. One branch in Quadrant 1.
Step 4: $T(60) = \mathbf{2}$ hours. $T(30) = \mathbf{4}$ hours.
Step 5: halving the speed doubles the time. Classic inverse proportion: $vT = 120$ (constant).
3.1 — Kicked football
Parabola (projectile under gravity).
3.2 — Trip time vs speed
Hyperbola (inverse proportion: $T \cdot v = $ constant).
3.3 — Bacteria doubling
Exponential growth (each step multiplies by $2$).
3.4 — Circular track
Circle (every point on the track is the same distance from the centre).
3.5 — Stone, $h = -5t^2 + 30t$
(a) $h(1) = -5(1) + 30(1) = -5 + 30 = 25$ m.
(b) Set $h = 0$: $-5t^2 + 30t = 0 \Rightarrow -5t(t - 6) = 0 \Rightarrow t = 0$ (launch) or $t = 6$ (landing). Lands at $t = 6$ s.
3.6 — Inverse proportion of workers
(a) $k = $ worker-days $= 6 \times 10 = 60$ worker-days.
(b) For $3$ workers: $T = \dfrac{60}{3} = 20$ days.
3.7 — Projectile with $t = -2$ or $t = 6$
(a) Landing time is $t = 6$ s.
(b) Reject $t = -2$ because time cannot be negative — it would imply the projectile was "on the ground" $2$ seconds before launch, which makes no physical sense.
(c) Axis of symmetry: $t = \dfrac{-2 + 6}{2} = 2$ s. So maximum height occurs at $t = 2$ s.
3.8 — Bacteria $N = 50 \cdot 2^t$
(a) $t = 0$: $50$ cells. $t = 1$: $100$ cells. $t = 2$: $200$ cells. $t = 3$: $400$ cells. $t = 4$: $800$ cells.
(b) Check $t = 4$: $N = 800$ cells (still below $1000$). Check $t = 5$: $N = 50 \times 32 = 1600$ cells (above $1000$). So the population first exceeds $1000$ cells at $t = 5$ hours.