Applications of Circular Functions (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Rotating Wheel Motion

Problem: A wheel is mounted on a wall and rotates. The distance $d$ cm of a particular point $P$ on the wheel from the ground is given by:

where $t$ is the time in seconds.

Finding the initial position

Question: How far is point $P$ above the ground when $t = 0$?

Solution: Substitute $t = 0$ into the equation:

The point is 40 cm above the ground initially.

Calculating the period

Question: How long does it take for the wheel to rotate once?

Solution: Use the period formula with $n = \frac{4\pi}{3}$:

The wheel takes $\frac{3}{2}$ seconds (or $1.5$ seconds) to complete one rotation.

Finding maximum and minimum distances

Question: What are the maximum and minimum distances of point $P$ above the ground?

Solution: Since we have $d = 100 - 60\cos\left(\frac{4\pi}{3}t\right)$, the value depends on $\cos\left(\frac{4\pi}{3}t\right)$.

Minimum distance: This occurs when $\cos\left(\frac{4\pi}{3}t\right) = 1$ (making the subtracted term largest):

The minimum distance is 40 cm.

Maximum distance: This occurs when $\cos\left(\frac{4\pi}{3}t\right) = -1$ (making the subtracted term negative):

The maximum distance is 160 cm.

Sketching the graph

Question: Sketch the graph of $d$ against $t$.

Solution: The graph shows:

• Period of $\frac{3}{2}$ seconds
• Oscillates between minimum $40$ cm and maximum $160$ cm
• Starts at $40$ cm when $t = 0$

Solving inequalities

Question: In the first rotation, find the intervals when point $P$ is less than $70$ cm above the ground.

Solution: We need to solve $d < 70$:

The general solutions where $\cos(\theta) = \frac{1}{2}$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$ (in the first rotation).

Therefore:

From the graph, the distance is less than $70$ cm for:

• $t \in \left(0, \frac{1}{4}\right)$ and
• $t \in \left(\frac{5}{4}, \frac{3}{2}\right)$

Worked Example: Modelling Tide Heights

Problem: The height $h(t)$ metres of the tide above mean sea level at Warnung on 1 January is given approximately by:

where $t$ is the number of hours after midnight.

Finding the period

First, note that the period is:

This means the tide completes one full cycle every 12 hours.

Finding when high tide occurs

Question: When was high tide?

Solution: High tide occurs when $h(t) = 4$ (the maximum value):

This occurs when:

High tide occurs at 03:00 and 15:00 (3 p.m.).

Finding the high tide height

Question: What was the height of the high tide?

Solution: The high tide has height 4 metres above the mean height.

Evaluating at a specific time

Question: What was the height of the tide at 8 a.m.?

Solution: Substitute $t = 8$:

At 8 a.m., the water is $2\sqrt{3}$ metres below the mean height (approximately $3.46$ metres below mean level).