Trigonometry (Part II) 5. Modelling Periodic Function (Day 2)

Modeling Periodic Functions

Example 1: Pool Practice (Back and Forth Motion)

• Scenario Overview:     * A swimmer is training for a triathlon by swimming laps in a pool.     * Pool Dimensions: The pool is $50\,m$ in length.     * Lap Definition: One lap is equal to $100\,m$, comprised of two lengths ($50\,m \times 2$).     * Coach Position: The coach is seated halfway between the two ends of the pool. To represent this mathematically, the coach's location is at the $25\,m$ mark if measuring from one end of the pool.     * Swimming Session: The swimmer starts in the middle of the pool, next to the coach ($t = 0$). It takes the swimmer $40\,s$ to swim to one end of the pool and then return to the coach. This implies it takes $20\,s$ to reach the end and $20\,s$ to return to the center.     * Speed: The swimmer maintains a constant speed for the duration of three laps.

• Mathematical Parameters for the Sinusoidal Model:     * Amplitude ($|a|$): Since the coach is at the center ($25\,m$) and the ends are at $0\,m$ and $50\,m$, the maximum displacement from the center is $25\,m$. Therefore, $|a| = 25$.     * Period: A full cycle (one lap) involves swimming from the center to one end, back to the center, then to the opposite end, and back to the center once more. If the first half-cycle (to one end and back) takes $40\,s$, the full period is $80\,s$.     * Interval: The key points of the cycle occur every $\frac{80}{4} = 20\,s$.     * Vertical Shift ($c$): The midline of the motion is the coach's position at $25\,m$. Therefore, $c = 25$.     * Frequency Coefficient ($k$): Calculated using the formula $k = \frac{360}{\text{period}}$.         * $k = \frac{360}{80} = 4.5$     * Phase Shift ($d$): Because the swimmer starts at the midline ($c = 25$) at $t = 0$ and is moving toward a peak, a sine function with no phase shift ($d = 0$) can be used.

• Sinusoidal Model Equation:     * $d(t) = 25 \sin(4.5t) + 25$     * Variables: $d(t)$ is the distance from a reference end of the pool in metres ($m$); $t$ is the time in seconds ($sec$).

• Graph Sketching Details (representing three laps):     * The graph tracks distance against time.     * Total Duration: Since one lap is $80\,s$, three laps take $3 \times 80 = 240\,s$.     * Key Points (Time in seconds, Distance in metres):         * $(0, 25)$: Swimmer starts at coach.         * $(20, 50)$: Swimmer at first end.         * $(40, 25)$: Swimmer back at coach.         * $(60, 0)$: Swimmer at second end.         * $(80, 25)$: Swimmer back at coach (end of lap 1).         * $(100, 50)$, $(140, 0)$, $(180, 50)$, $(220, 0)$, $(240, 25)$.

Example 2: Bay of Fundy Tides (Up and Down Motion)

• Contextual Data:     * Source: Information collected from http://bayoffundytourism.com/.     * Temporal Data: The time taken to transition from low tide to high tide is approximately $6.25\,hours$ (expressed as $6\,hours$ and $13\,min$ in the transcript).     * Height Data: The height of the water at high tide is approximately $15\,metres$ ($50\,feet$).

• Assumptions and Setup:     * Low tide height is $0\,metres$.     * Low tide occurs at $2:00\,AM$.     * $t = 0$ is defined as $12:00\,AM$ (midnight).

• Mathematical Parameters for the Tidal Model:     * Amplitude ($|a|$): $\frac{15 - 0}{2} = 7.5\,m$.     * Vertical Shift ($c$): $\frac{15 + 0}{2} = 7.5\,m$.     * Period: Since low to high tide is a half-cycle ($6.25\,hours$), the full cycle is $6.25 \times 2 = 12.5\,hours$.     * Frequency Coefficient ($k$): $k = \frac{360}{12.5} = 28.8$.     * Phase Shift ($d$): A negative cosine function starts at its minimum. Since low tide (minimum) occurs at $2:00\,AM$, the graph is shifted to the right by $2$. Thus, $d = 2$.

• Sinusoidal Model Equation:     * $h(t) = -7.5 \cos[28.8(t - 2)] + 7.5$     * Variables: $h(t)$ is the height in metres ($m$); $t$ is the time in hours ($hours$).

• Calculations: Height of water at 6:45 PM     * Step 1: Convert time to hours since midnight.         * $6:45\,PM = 12\,hours + 6.75\,hours = 18.75\,hours$ (because $45/60 = 0.75$).     * Step 2: Substitute into the model.         * $h(18.75) = -7.5 \cos[28.8(18.75 - 2)] + 7.5$         * $h(18.75) = -7.5 \cos[28.8(16.75)] + 7.5$         * $h(18.75) = 11.52\,m$

Practice & Review Assignments

• Textbook Problems:     * Page 370 #8     * Page 393 #13, 14     * Page 399 #3, 5, 9
• Review Materials:     * Part 1 of the review is a separate handout.     * All solutions for the practice and review have been posted.