Recognising Relationships (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Water Vessels

Consider four different shaped vessels where water is being poured in at a constant rate. We need to determine how the height of water ($h$) relates to the volume poured in ($V$).

For each vessel, the relationship between height and volume depends on the container's shape:

Vessel a (rectangular container): The width remains constant throughout, so equal volumes of water always produce equal increases in height. This creates a linear relationship where $h$ increases steadily with $V$.

Vessel b (conical flask): The container is narrow at the bottom and widens towards the top. Initially, small volumes produce large height increases, but as the container widens, the same volume produces smaller height increases. The graph starts steep and gradually becomes less steep.

Vessel c (round-bottom flask): This vessel is narrow at top and bottom but wide in the middle. Initially, the height rises quickly, then more slowly through the wide section, and finally quickly again as it reaches the narrow neck. The rate of height change varies significantly.

Vessel d (zigzag flask): Similar to the conical flask but with an irregular shape. The changing width creates a varying relationship between height and volume.

The graphs show these relationships visually, with all starting from the origin (no water, zero height) and curving upward as volume increases.

Worked Example: Particle Motion

Understanding motion from graphs is an essential skill. Let's analyse a distance-time graph to describe how a particle moves.

Reading the graph systematically, we can break down the particle's motion into distinct phases:

Phase 1 (0 to 3 minutes): The particle starts 5 metres from point O. It moves away from O in a straight line at constant speed. The distance increases from 5 m to 10 m over 3 minutes.

To find the speed: Distance travelled = $10 - 5 = 5$ metres, Time taken = $3$ minutes

Therefore, speed = $\frac{5}{3}$ metres per minute

Phase 2 (3 to 7 minutes): The graph shows a horizontal line segment, meaning the distance remains constant at 10 metres. The particle is stationary for 4 minutes.

Phase 3 (7 to 20 minutes): The particle returns toward O. The curved line indicates that the speed is not constant but gradually decreases. The particle decelerates throughout this phase, eventually coming to rest at O at exactly $t = 20$ minutes. The curve becomes less steep as time progresses, showing the reducing speed.

Worked Example: Using Interval Notation

For the graph shown, we need to identify where the rate of change is positive and where it is negative.

The graph shows a curve with a minimum point at $(-3, 2)$.

Part a: Where is the rate of change negative?

Looking at the graph from left to right:

• From $x = -5$ to $x = -3$, the graph slopes upward (positive rate of change)
• From $x = -3$ to $x = 0$, the graph slopes downward (negative rate of change)
• From $x = 0$ to $x = 2$, the graph slopes upward (positive rate of change)

Therefore, the rate of change of $y$ with respect to $x$ is negative for $x \in (-3, 0)$.

Note: We use a round bracket because at $x = -3$ and $x = 0$, the graph has a turning point where the rate of change is momentarily zero.

Part b: Where is the rate of change positive?

The graph slopes upward in two separate sections:

• From $x = -5$ to $x = -3$
• From $x = 0$ to $x = 2$

Therefore, the rate of change of $y$ with respect to $x$ is positive for $x \in [-5, -3) \cup (0, 2]$.

The union symbol ($\cup$) connects the two separate intervals where the rate is positive. We use square brackets at the endpoints $-5$ and $2$ because these are included in the domain we're considering.