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Section 4.7

Mathematical Modeling
with Sine and Cosine

Precalculus

Concept

Sine & Cosine Are Everywhere

These functions model any phenomenon that repeats in a predictable cycle.

🌊
Tidal Height
🎵
Sound Waves
🌡️
Yearly Temperatures
🌅
Sunset Times
⚡
Electricity / Voltage
🕰️
Pendulum
🔩
Spring
🎡
Ferris Wheel
❤️
Heartbeat / EKG
Example

Tidal Depth at a Dock

Throughout the day, the depth of the water at the end of a dock varies with the tides. The table shows the depths \(y\) (in feet) at various times during the morning.
TimeDepth, \(y\)
Midnight3.4
2 a.m.8.7
4 a.m.11.3
6 a.m.9.1
8 a.m.3.8
10 a.m.0.1
Noon1.2

a. Use a cosine function to model the data. Let \(t\) be the time, with \(t = 0\) at midnight.

b. A boat needs at least 10 feet of water to moor at the dock. During what times in the afternoon can it safely dock?

c. Find a sine model for this data.

Example

Step 1: Make a Scatterplot

Plot the data points to visualize the pattern before building a model.

0 2 4 6 8 10 12 hours since midnight 0 2 4 6 8 10 12 depth (feet) MAX = 11.3 MIN = 0.1
Key Observations
The maximum depth is 11.3 ft at \(t = 4\) (4 a.m.) and the minimum is 0.1 ft at \(t = 10\) (10 a.m.). This looks like a cosine curve!
Part A

Building the Cosine Model

Find \(a\), \(b\), \(c\), and \(d\) in: \(y = a\cos\!\bigl(b(t - c)\bigr) + d\)
Step 1 — Vertical Translation (d)
\[d = \dfrac{\text{max} + \text{min}}{2} = \dfrac{11.3 + 0.1}{2} = 5.7\] Average of max and min values
Step 2 — Amplitude (a)
\[a = \dfrac{\text{max} - \text{min}}{2} = \dfrac{11.3 - 0.1}{2} = 5.6\] Half the distance between max and min
Step 3 — Period
\[\text{period} = 2(t_{\min} - t_{\max}) = 2(10 - 4) = 12\] Cosine completes half a cycle between consecutive max and min
Step 4 — Find b
\[\text{period} = \dfrac{2\pi}{b} \;\Longrightarrow\; 12 = \dfrac{2\pi}{b} \;\Longrightarrow\; b = \dfrac{\pi}{6}\]
Step 5 — Phase Shift (c)
High tide (max) occurs at \(t = 4\). Since cosine starts at its max, the phase shift is \(c = 4\).
Answer
\[y = 5.6\cos\!\left(\dfrac{\pi}{6}(t - 4)\right) + 5.7\]
Part A

Let's Confirm in the Calculator

The model \(y = 5.6\cos\!\left(\dfrac{\pi}{6}(t - 4)\right) + 5.7\) should fit the scatter data.

0 2 4 6 8 10 12 hours since midnight 0 2 4 6 8 10 12 depth (feet)
Midline
\(d = 5.7\) — the horizontal midline of the curve
Part B

When Can the Boat Safely Dock?

The boat needs at least 10 feet of water. Find where our model is above the line \(y = 10\).
0 4 8 12 16 20 24 hours since midnight noon 4 pm 8 pm 0 2 4 6 8 10 12 depth (feet)
The boat needs depth \(\ge 10\) ft, so we draw the line y = 10. Where the curve is above this line, the water is deep enough.
The curve crosses \(y = 10\) at four points over 24 hours -- because each full cycle of the cosine hits this height twice.
Why the Afternoon Pair?
The problem asks for afternoon docking times. The first two intersections (\(t \approx 2.67\) and \(5.33\)) are in the morning. The second pair (\(t \approx 14.67\) and \(17.33\)) falls after noon -- those are our answer window.
Reading the Graph
Between the two red points, the blue curve stays above the green line. That shaded region is when the depth \(\ge 10\) ft and the boat can safely dock.
Part B

Converting to Clock Time

We found \(t \approx 14.67\) and \(t \approx 17.33\). But what times are those?
Why Not Radians?
In this problem, \(t\) is measured in hours, not radians. The value 14.67 means 14.67 hours after midnight -- the decimal part (0.67) is a fraction of an hour, not an angle.
What Does the Decimal Mean?
The whole-number part gives us the hour. The decimal part is a fraction of an hour that we need to convert to minutes.
\[t = 14.67 \;\longrightarrow\; 14 \text{ hours} + 0.67 \times 60 \text{ min} = 14 \text{ hours } 40 \text{ min}\]
Left boundary
\(t = 14.67\) hours
↓
14 h 40 min = 2:40 PM
Right boundary
\(t = 17.33\) hours
↓
17 h 20 min = 5:20 PM
Answer
\[\text{The boat can safely dock between } \mathbf{2{:}40 \text{ PM}} \text{ and } \mathbf{5{:}20 \text{ PM}}\]
Part C

Finding the Sine Model

Convert from cosine to sine. We know: \(y = 5.6\cos\!\left(\dfrac{\pi}{6}(t - 4)\right) + 5.7\)
0 4 8 12 0 2 4 6 8 10 12 time y = 5.7
Step 1 — Graph the cosine model
Graph the cosine model and the horizontal line \(y = 5.7\) (the vertical translation / midline).
Step 2 — Find the first intersection
Find where the cosine curve crosses the midline while increasing. Using the calculator's Intersect feature: the first rising crossing is at (1, 5.7).
Step 3 — Set the phase shift
Set \(c = 1\). Since sine starts at the midline going up, this is exactly where our sine function begins its cycle. All other parameters stay the same (amplitude = 5.6, period = 12, vertical translation = 5.7).
Answer
\[y = 5.6\sin\!\left(\dfrac{\pi}{6}(t - 1)\right) + 5.7\]
What are your questions?
Section 4.7 -- Modeling with Sine and Cosine
Homework
Modeling with Sine and Cosine WORKSHEET