Wave Dynamics

\((\frac{3\pi}{4}, 1)\) \(\rightarrow\) Midline

\((\frac{4\pi}{4}, 4)\) \(\rightarrow\) Peak (\(1 + 3\))

\((\frac{5\pi}{4}, 1)\) \(\rightarrow\) Midline

\((\frac{6\pi}{4}, -2)\) \(\rightarrow\) Trough

Key Takeaway: The cycle spans from \(x = \frac{\pi}{2}\) to \(x = \frac{3\pi}{2}\), which equals period \(\pi\).

Worked Walkthrough Slide 3 of 4

SYNTHESIS Reverse Engineering

Graph \(\rightarrow\) Equation

Extracting Equations From Wave Geometry

Given a continuous physical or graphical wave, use the four boundary equations:

1. Midline \(D = \frac{y_{\max} + y_{\min}}{2}\)

2. Amplitude \(|A| = \frac{y_{\max} - y_{\min}}{2}\)

3. Frequency Factor \(B = \frac{2\pi}{\text{Period}}\)

4. Phase Shift Sine: Midline rising | Cosine: Peak

Check for Understanding

A buoy bobs in the ocean. Its maximum height is 14 meters, minimum is 6 meters, and it takes 8 seconds between consecutive crests. At \(t=0\), it sits at maximum crest.

Target Equation:

\(h(t) = 4\cos\left(\frac{\pi}{4}t\right) + 10\)

\(A = \frac{14-6}{2}=4, \; D = \frac{14+6}{2}=10, \; B = \frac{2\pi}{8}=\frac{\pi}{4}, \; C=0\).

Classroom Mastery Slide 4 of 4

Human core body temperature follows a predictable 24-hour sinusoidal circadian cycle. In a clinical trial, a subject's minimum temperature was measured at 36.4°C at 4:00 AM (\(t=4\)), and the maximum temperature was measured at 37.6°C at 4:00 PM (\(t=16\)), where \(t\) represents hours after midnight.

(a) Find Amplitude \(A\), Midline \(D\), and Period \(T\):

(b) Write a cosine model \(T(t) = A\cos(B(t-C)) + D\):

(c) Predict the subject's temperature at 10:00 AM (\(t=10\)):

5 Part 5: Error Diagnostics

A student claims \(y = \sin(4x - \pi)\) has a phase shift of \(\pi\) units right. Identify and correct their error:

Wave Dynamics • Module 1 Page 2 of 2

Pt 2: \((\frac{\pi}{2}, 3)\) \(\rightarrow\) Midline

Pt 3: \((\frac{3\pi}{2}, 5)\) \(\rightarrow\) Peak \((3+2)\)

Pt 4: \((\frac{5\pi}{2}, 3)\) \(\rightarrow\) Midline

Pt 5: \((\frac{7\pi}{2}, 1)\) \(\rightarrow\) Trough

Curve: Inverted Cosine (starts at trough)

Domain plotted: \([-\pi/2, 7\pi/2]\), Range: \([1, 5]\).

4 Part 4: Circadian Biology Solutions

(a) Parameter Derivations:

Amplitude \(A = \frac{37.6 - 36.4}{2} = 0.6^\circ\text{C}\)
Midline \(D = \frac{37.6 + 36.4}{2} = 37.0^\circ\text{C}\)
Period \(T = 24\text{ h} \implies B = \frac{2\pi}{24} = \frac{\pi}{12}\)

(b) Cosine Model Equations:

Peak is at \(t = 16\) (4 PM):
\(T(t) = 0.6\cos\left(\frac{\pi}{12}(t - 16)\right) + 37.0\)
Alt (min at \(t=4\)): \(-0.6\cos(\frac{\pi}{12}(t - 4)) + 37\)

(c) Evaluation at \(t = 10\) (10:00 AM):

\(T(10) = 0.6\cos\left(\frac{\pi}{12}(10 - 16)\right) + 37.0 = 0.6\cos\left(-\frac{\pi}{2}\right) + 37.0 = 0.6(0) + 37.0 = \mathbf{37.0^\circ\text{C}}\) (Exact midline).

5 Part 5: Error Diagnostics Solution

Error: The student failed to factor out the frequency coefficient \(B=4\). The expression inside is \(4x - \pi = 4(x - \frac{\pi}{4})\). Therefore, the actual horizontal phase shift is \(\frac{\pi}{4}\) units to the right, not \(\pi\).

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Range: \((-\frac{\pi}{2}, \frac{\pi}{2})\)

Quadrants I & IV (open)

Has horizontal asymptotes at \(y = \pm \frac{\pi}{2}\).

Restricted Principal Branches Slide 3 of 4

APPLICATION Exact Evaluation

Reference Triangle Models

Evaluating Composite Trigonometric Expressions

To evaluate composite expressions like \(\cot\left(\arccos\left(-\frac{5}{13}\right)\right)\), construct a reference right triangle:

Model Walkthrough: \(\cot(\theta)\)

Step 1: Let \(\theta = \arccos\left(-\frac{5}{13}\right)\).

Step 2: Since input is negative, \(\theta\) lies in Quadrant II.

Step 3: \(\cos(\theta) = -\frac{5}{13} = \frac{x}{r}\), so \(x = -5, \; r = 13\).

Step 4: Solve \(y = \sqrt{13^2 - (-5)^2} = \sqrt{144} = 12\).

\(\cot(\theta) = \frac{x}{y} = -\frac{5}{12}\)

Algebraic Form: \(\sin(\arccos(u))\)

Let \(\theta = \arccos(u) \implies \cos\theta = u = \frac{u}{1}\).
Adjacent \(= u\), Hypotenuse \(= 1\).
Opposite \(= \sqrt{1 - u^2}\).

General Algebraic Identity:

\(\sin\bigl(\arccos(u)\bigr) = \sqrt{1 - u^2}\)

Algebraic & Numeric Evaluation Slide 4 of 4

4 Part 4: Reference Triangle Compositions

(a) Evaluate Numeric Value: \(\cot\left(\arccos\left(-\frac{7}{25}\right)\right)\)

Draw reference triangle in appropriate quadrant and label all sides.

(b) Algebraic Expression: \(\sin\left(\arctan(3x)\right)\)

Assume \(x > 0\). Construct a right triangle and express without trig functions.

5 Part 5: Maritime Navigation Application

A coastal patrol aircraft flies at a constant altitude of 4,000 meters. The radar tracking operator monitors a stranded vessel with line-of-sight angle of depression \(\theta = \arctan\left(\frac{4000}{x}\right)\), where \(x\) is the horizontal ground distance in meters.

(a) Find \(\theta\) when \(x = 4000\sqrt{3}\) meters:

(b) Determine exact line-of-sight distance using \(\csc\theta\):

Wave Dynamics • Module 2 Page 2 of 2

*Note on (a): \(\frac{5\pi}{3}\) is incorrect because it lies outside the required range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

4 Part 4: Reference Triangle Solutions

(a) Value: \(\cot\left(\arccos\left(-\frac{7}{25}\right)\right)\)

Let \(\theta = \arccos(-\frac{7}{25})\) (Quadrant II).

\(x = -7, \; r = 25\).

\(y = \sqrt{25^2 - (-7)^2} = \sqrt{576} = 24\).

\(\cot\theta = \frac{x}{y} = \mathbf{-\frac{7}{24}}\)

(b) Algebraic: \(\sin\left(\arctan(3x)\right)\)

Let \(\theta = \arctan(3x) \implies \tan\theta = \frac{3x}{1}\).

\(\text{Opp} = 3x, \; \text{Adj} = 1\).

\(\text{Hyp} = \sqrt{1^2 + (3x)^2} = \sqrt{1 + 9x^2}\).

\(\sin\theta = \frac{\text{Opp}}{\text{Hyp}} = \mathbf{\frac{3x}{\sqrt{1 + 9x^2}}}\)

5 Part 5: Maritime Navigation Solutions

(a) Angle of Depression \(\theta\):

\(\theta = \arctan\left(\frac{4000}{4000\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right)\)
\(\theta = \frac{\pi}{6}\text{ radians (or } 30^\circ)\)

(b) Line-of-Sight Distance:

\(\text{Distance } d = \frac{4000}{\sin\theta} = 4000\csc\left(\frac{\pi}{6}\right)\)
Since \(\csc\left(\frac{\pi}{6}\right) = 2\):
\(d = 4000 \times 2 = \mathbf{8,000\text{ meters}}\)

Wave Dynamics • Module 2 Page 2 of 2

(a) Channel max/min depth & tidal period:

(b) Set up & solve inequality \(D(t) \ge 12\):

(c) State the exact safe morning entry time window \([t_1, t_2]\):

Wave Dynamics • Module 3 Page 2 of 2

Wave Dynamics • Mastery Assessment Page 2 of 2

Students scoring at Level 1 or 2 complete targeted 5-point frame intervention prior to reassessing via alternate modular prompts.

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Quiz & Performance Task Solutions

Complete Scoring Solutions & Point Breakdowns

Answer Key • Page 2

1 Trig Mastery Quiz Solutions (40 Pts Total)

Section 1: Sinusoid \(f(x) = -3\sin(2x - \pi) + 2\) (15 Pts)

• Factored: \(-3\sin[2(x - \frac{\pi}{2})] + 2\). Amp: 3 (reflected), Period: \(\pi\), Shift: \(\frac{\pi}{2}\) right, Midline: \(y = 2\). (5 pts)

• 5 Points: \((\frac{\pi}{2}, 2) \rightarrow\) Midline, \((\frac{3\pi}{4}, -1) \rightarrow\) Trough (\(2-3\)), \((\pi, 2) \rightarrow\) Midline, \((\frac{5\pi}{4}, 5) \rightarrow\) Peak (\(2+3\)), \((\frac{3\pi}{2}, 2) \rightarrow\) Midline. (5 pts)

• Graph: Wave inverted starting at \(x = \frac{\pi}{2}\), passes through \((3\pi/4, -1)\) and \((5\pi/4, 5)\). (5 pts)

Section 2: Reciprocal \(g(x) = 2\sec(3x) - 1\) (10 Pts)

• (a) Companion Guide: \(y = 2\cos(3x) - 1\). Period: \(\frac{2\pi}{3}\). (3 pts)

• (b) Asymptotes: \(\cos(3x) = 0 \implies 3x = \frac{\pi}{2} + k\pi \implies \mathbf{x = \frac{\pi}{6} + \frac{k\pi}{3}, \; k \in \mathbb{Z}}\). (4 pts)

• (c) Range: Guide crest \(= 1\), guide trough \(= -3\). Range is \(\mathbf{(-\infty, -3] \cup [1, \infty)}\). (3 pts)

Section 3: Inverses & Triangles (10 Pts)

• Q3: (a) \(\arcsin(-1/2) = \mathbf{-\frac{\pi}{6}}\) (2 pts). (b) \(\arccos(-\frac{\sqrt{3}}{2}) = \mathbf{\frac{5\pi}{6}}\) (2 pts).

• Q4: Let \(\theta = \arccos(2x)\). Right triangle: \(\text{Adj} = 2x, \; \text{Hyp} = 1\). \(\text{Opp} = \sqrt{1 - 4x^2}\).

  \(\csc(\theta) = \frac{\text{Hyp}}{\text{Opp}} = \mathbf{\frac{1}{\sqrt{1 - 4x^2}}}\). (6 pts)

Section 4: Acoustic Modeling (5 Pts)

• (a) Amplitude \(= \mathbf{2.4\text{ mm}}\), Period \(T = \frac{2\pi}{440\pi} = \mathbf{\frac{1}{220}\text{ s}}\). (2 pts)

• (b) Frequency \(f = \frac{1}{T} = \mathbf{220\text{ Hz}}\). (1 pt)

• (c) \(d(\frac{1}{880}) = 2.4\cos(440\pi \cdot \frac{1}{880}) = 2.4\cos(\frac{\pi}{2}) = \mathbf{0\text{ mm}}\) (passing through equilibrium). (2 pts)

2 Harmonic Systems Task Quick Scoring Guide

Context 1 (Engine): \(f = \frac{3000}{60} = 50\text{ Hz}\), \(T = 0.02\text{ s}\), \(A = 5\text{ cm}\), \(D = 7\text{ cm}\), \(B = 100\pi\). Model: \(y(t) = 5\cos(100\pi t) + 7\). At \(t=0.005\text{ s}\), \(y = 5\cos(\pi/2)+7 = \mathbf{7\text{ cm}}\).

Context 2 (Spring): \(T = 0.4\text{ s}\), \(B = 5\pi\), \(y(t) = -8\cos(5\pi t)\). Passes equilibrium moving up at \(t = \mathbf{0.1\text{ s}, \; 0.5\text{ s}, \; 0.9\text{ s}}\).

Context 3 (Ecology): Hare: \(H(t) = 40000\cos(\frac{\pi}{5}(t-2)) + 50000\). Lynx: \(L(t) = 4000\cos(\frac{\pi}{5}(t-4.5)) + 6000\). At \(t=4.5\), \(H(4.5) = 40000\cos(\pi/2)+50000 = \mathbf{50,000\text{ hares}}\).

Context 4 (Tides): \(D(t) \ge 12 \implies \cos(\frac{\pi}{6}(t-3)) \ge 0.5 \implies -\frac{\pi}{3} \le \frac{\pi}{6}(t-3) \le \frac{\pi}{3} \implies \mathbf{1 \le t \le 5}\). Safe window: 1:00 AM to 5:00 AM.

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