- Look for a quantity that returns to the same pattern after a fixed amount of time. - Distinguish repeating behavior from quantities that mainly keep increasing.

1. A periodic quantity repeats the same pattern after a fixed interval. 2. A Ferris-wheel rider returns to the same heights in the same order once every revolution. 3. The savings balance and total distance generally increase rather than repeat.

b) The rider's height is periodic because the same height pattern repeats every Ferris-wheel revolution.

- The midline lies halfway between the maximum and minimum values. - The amplitude measures the vertical distance from the midline to either extreme. - Check that midline plus and minus amplitude reproduce the given extremes.

1. The amplitude is half the distance from the minimum to the maximum: \(\frac{8.0-2.4}{2}=2.8\). 2. The midline is the average of the maximum and minimum: \(\frac{8.0+2.4}{2}=5.2\).

Amplitude: \(2.8\,\text{ft}\) Midline: \(5.2\,\text{ft}\)

- The period is the time required for one complete repeating cycle. - Frequency tells how many cycles occur in one unit of time. - Use the reciprocal relationship between period and frequency.

1. The height pattern repeats after one complete revolution, so the period is \(48\) seconds. 2. Frequency is the reciprocal of the period: \(\frac{1}{48}\) cycle per second.

a) \(48\,\text{s}\) b) \(\frac{1}{48}\) cycle per second

- Compare the values of sine and cosine at input \(0\). Which starts at its largest value? - Check how each basic graph moves immediately after input \(0\).

1. A positive cosine function starts at its maximum when its input is \(0\). 2. The described quantity is also at its maximum at \(t=0\) and then decreases. 3. Therefore, a cosine-based model is the natural choice.

Cosine, because a positive cosine model starts at a maximum and then decreases.

- Compare the values of both basic functions at input zero. - Which basic graph crosses its midline while rising, rather than starting at a high point?

1. A positive sine function starts at its midline when its input is \(0\). 2. It initially increases from that midline. 3. Therefore, a sine-based model matches the stated starting behavior naturally.

Sine, because a positive sine model starts on the midline and increases.

- Picture the direction of the beam at each recorded time. - Find how much time passes between the two observations.

1. The beam has returned to the same direction on its next pass, so it has completed one full rotation. 2. The period is \(10-2=8\,\text{s}\).

\(8\,\text{s}\)

- Read the maximum and minimum y-values from the graph. - The midline is halfway between those values. - Measure the horizontal distance between matching points on consecutive cycles.

1. The graph has maximum \(2\) and minimum \(-2\), so its midline is \(y=0\) and its amplitude is \(2\). 2. One full wave repeats every \(2\pi\) units along the x-axis. 3. Therefore, the period is \(2\pi\).

Amplitude: \(2\) Midline: \(y=0\) Period: \(2\pi\)

- Identify the highest and lowest y-values first. - Use the average and half-difference of the extremes for midline and amplitude. - Find two matching locations one full cycle apart to measure the period.

1. The graph ranges from \(-2\) to \(4\), so the maximum is \(4\) and the minimum is \(-2\). 2. The midline is \(\frac{4+(-2)}{2}=1\), and the amplitude is \(\frac{4-(-2)}{2}=3\). 3. The displayed minima occur at \(x=-2\pi\) and \(x=2\pi\). They are \(4\pi\) units apart, so the period is \(4\pi\).

Maximum: \(4\) Minimum: \(-2\) Midline: \(y=1\) Amplitude: \(3\) Period: \(4\pi\)

- Compare vertical distances from each graph's midline to its peaks. - Check whether the peaks occur at the same x-values. - Describe the change in outputs without changing the x-values.

1. Both functions have midline \(y=0\) and period \(2\pi\). 2. The amplitude of \(f\) is \(1\), while the amplitude of \(g\) is \(2\). 3. Multiplying every output of \(f\) by \(2\) produces \(g\), so \(g\) is a vertical stretch of \(f\) by a factor of \(2\).

\(f\): amplitude \(1\), midline \(y=0\), period \(2\pi\) \(g\): amplitude \(2\), midline \(y=0\), period \(2\pi\) The transformation is a vertical stretch by a factor of \(2\).

- Read the vertical distance from the midline to a peak for each graph. - Count how many complete repeating patterns fit in the displayed interval. - Relate the length of one repeating pattern to the period.

1. Both graphs reach \(1\) and \(-1\), so both amplitudes are \(1\). 2. From the graph, \(f\) completes one full cycle from \(0\) to \(2\pi\), so its period is \(2\pi\). 3. The graph of \(g\) completes two full cycles over the same interval, so one cycle has length \(\pi\). 4. Therefore, \(f\) has period \(2\pi\) and one displayed cycle, while \(g\) has period \(\pi\) and two displayed cycles.

a) Both amplitudes are \(1\). The period of \(f\) is \(2\pi\), and the period of \(g\) is \(\pi\). b) \(f\) completes \(1\) cycle; \(g\) completes \(2\) cycles.

- Read the highest and lowest y-values from the graph. - Average the extremes for the midline and take half their difference for amplitude. - Measure the horizontal distance between consecutive peaks.

1. The graph has maximum \(3\) and minimum \(1\). 2. The midline is \(\frac{3+1}{2}=2\), and the amplitude is \(\frac{3-1}{2}=1\). 3. The pattern repeats every \(2\pi\), so the period is \(2\pi\).

Maximum: \(3\) Minimum: \(1\) Midline: \(y=2\) Amplitude: \(1\) Period: \(2\pi\)

- Substitute the given time into the model before simplifying. - Recognize the resulting special-angle sine value. - Keep the exact form until the final decimal approximation.

1. Substitute \(t=2\): \(y(2)=10+4\sin\left(\frac{\pi}{3}\right)\). 2. Since \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), \(y(2)=10+2\sqrt{3}\). 3. Numerically, \(10+2\sqrt{3}\approx13.46\).

\(y(2)=10+2\sqrt{3}\approx13.46\). The height is about \(13.46\,\text{ft}\) at \(t=2\,\text{s}\).

- Use the extremes to locate the center and vertical distance to an extreme. - Compare the starting minimum with the basic cosine starting point. - Leave the horizontal coefficient unspecified because no period is given.

1. The amplitude is \(\frac{17-5}{2}=6\). 2. The midline is \(\frac{17+5}{2}=11\). 3. A positive cosine starts at a maximum, so starting at a minimum requires a negative amplitude coefficient. 4. Thus a suitable form is \(y=11-6\cos(Bt)\), where \(B\) depends on the period.

a) Amplitude \(6\); midline \(y=11\) b) \(y=11-6\cos(Bt)\)

- Focus on the coefficient multiplying \(t\) inside the sine function. - Relate that coefficient to the horizontal length of one full cycle. - Divide the total time span by the period only after finding the period.

1. For \(\sin(Bt)\), the period is \(\frac{2\pi}{|B|}\). 2. Here \(B=\frac{\pi}{4}\), so the period is \(\frac{2\pi}{\pi/4}=8\). 3. The interval from \(0\) to \(24\) has length \(24\), so it contains \(\frac{24}{8}=3\) complete cycles.

Period: \(8\) Complete cycles from \(0\) to \(24\): \(3\)

- Measure the horizontal distance between two matching peaks. - Read the highest and lowest y-values from the graph. - Use the extremes to find amplitude and midline.

1. Consecutive peaks occur \(6\) units apart, so the period is \(6\). 2. The maximum is \(3\) and the minimum is \(-1\). 3. The amplitude is \(\frac{3-(-1)}{2}=2\), and the midline is \(\frac{3+(-1)}{2}=1\).

Period: \(6\) Amplitude: \(2\) Midline: \(y=1\)

- Distinguish the total peak-to-trough distance from the distance from the midline to a peak. - The midline must be equally far from the maximum and minimum. - Check that midline plus and minus amplitude reproduce both extremes.

1. The difference \(14-6=8\) is the full vertical distance from minimum to maximum. 2. Amplitude is half that distance, so the amplitude is \(\frac{8}{2}=4\). 3. The midline is the average of the extremes: \(\frac{14+6}{2}=10\).

Linnea used the full range instead of half the range. The amplitude is \(4\), and the midline is \(10\).

- Distinguish the period from the coefficient multiplying the input. - Use the relationship between \(B\) and one full cycle of \(2\pi\). - Check what period Rafael's coefficient would actually produce.

1. For \(\sin(Bt)\), the period is \(\frac{2\pi}{|B|}\), so the inside coefficient is not the period itself. 2. Setting \(\frac{2\pi}{B}=10\) gives \(B=\frac{\pi}{5}\). 3. Rafael's coefficient \(10\) would instead give period \(\frac{2\pi}{10}=\frac{\pi}{5}\).

The coefficient should be \(\frac{\pi}{5}\), not \(10\). A suitable model is \(y=2+3\sin\left(\frac{\pi}{5}t\right)\).

- Compare the vertical peak distances before considering horizontal repetition. - Count complete cycles for each labeled graph across the same x-interval. - A function with a longer period progresses through its cycle more slowly for the same change in x.

1. Both graphs range from \(-2\) to \(2\), so both amplitudes are \(2\). 2. The graph of \(f\) completes two cycles from \(0\) to \(4\pi\), so its period is \(2\pi\). 3. The graph of \(g\) completes one cycle on the same interval, so its period is \(4\pi\). 4. Therefore, \(g\) takes more horizontal distance to complete one cycle and changes through its cycle more slowly.

a) Both amplitudes are \(2\). The period of \(f\) is \(2\pi\), and the period of \(g\) is \(4\pi\). b) \(g\) changes through its cycle more slowly because it completes only one cycle while \(f\) completes two over the same interval.

- Read the vertical shift and outside coefficient first. - Use the inside coefficient to determine the period. - Combine midline and amplitude to find the extremes.

1. The vertical shift gives midline \(y=12\), and the cosine coefficient gives amplitude \(3\). 2. The period is \(\frac{2\pi}{\pi/6}=12\). 3. The maximum is \(12+3=15\), and the minimum is \(12-3=9\). 4. A period of \(12\) hours means the full pattern repeats every \(12\) hours.

Midline: \(y=12\) Amplitude: \(3\) Period: \(12\) hours Maximum: \(15\) Minimum: \(9\)

- Find the angle inside sine at each of the two times. - Compare the change in that angle with a complete turn around the unit circle.

1. The angle inside sine changes from \(\frac{\pi}{2}\cdot0=0\) to \(\frac{\pi}{2}\cdot4=2\pi\), a change of \(2\pi\) radians. 2. Sine completes one full cycle when its input increases by \(2\pi\), so the height completes one cycle in \(4\) seconds.

\(2\pi\) radians; one full cycle (in \(4\,\text{s}\)).

- Match peaks on the two curves rather than comparing their heights. - A horizontal shift changes peak locations but not the period. - Which curve reaches its peak later as x increases?

1. The first maximum of \(f\) is at \(x=\frac{\pi}{2}\); the first maximum of \(g\) is at \(x=\pi\). 2. Corresponding peaks are \(\frac{\pi}{2}\) apart, with \(g\) farther right. Thus \(g\) is \(f\) shifted right by \(\frac{\pi}{2}\).

Shift right by \(\frac{\pi}{2}\); first maximum of \(g\) at \(x=\pi\).

- Use the starting position and direction to select a basic sine or cosine form. - Relate the coefficient inside the trigonometric function to the desired period. - Place the amplitude outside the function and the midline as a vertical shift.

1. Midline-and-increasing behavior at \(t=0\) suggests a positive sine model. 2. For period \(6\), the angular frequency is \(B=\frac{2\pi}{6}=\frac{\pi}{3}\). 3. Using amplitude \(4\) and vertical shift \(10\) gives \(y(t)=10+4\sin\left(\frac{\pi}{3}t\right)\).

\(y(t)=10+4\sin\left(\frac{\pi}{3}t\right)\)

- Match the starting maximum with a basic trigonometric graph. - Convert the desired period into the coefficient multiplying \(t\). - Use the amplitude and midline as the outside coefficient and vertical shift.

1. Starting at a maximum suggests a positive cosine model. 2. For period \(8\), the angular frequency is \(B=\frac{2\pi}{8}=\frac{\pi}{4}\). 3. Using amplitude \(3\) and vertical shift \(5\) gives \(y(t)=5+3\cos\left(\frac{\pi}{4}t\right)\).

\(y(t)=5+3\cos\left(\frac{\pi}{4}t\right)\)

- Find amplitude and midline from the maximum and minimum depths. - Relate the period to the coefficient inside the trigonometric function. - Use the fact that the cycle starts at a maximum to select the phase.

1. The amplitude is \(\frac{8-2}{2}=3\), and the midline is \(\frac{8+2}{2}=5\). 2. The period is \(12\), so \(B=\frac{2\pi}{12}=\frac{\pi}{6}\). 3. Since the depth starts at a maximum, use cosine: \(d(t)=5+3\cos\left(\frac{\pi}{6}t\right)\).

\(d(t)=5+3\cos\left(\frac{\pi}{6}t\right)\)

- Connect the wheel's radius with amplitude and the center height with midline. - Use the revolution time as the period. - Decide how to make the model start at a minimum rather than a maximum.

1. The radius gives amplitude \(20\), and the center height gives midline \(25\). 2. The period is \(30\), so \(B=\frac{2\pi}{30}=\frac{\pi}{15}\). 3. Starting at the lowest point is modeled by subtracting cosine: \(h(t)=25-20\cos\left(\frac{\pi}{15}t\right)\).

\(h(t)=25-20\cos\left(\frac{\pi}{15}t\right)\)

- Use the maximum and minimum to determine amplitude and midline. - Convert the twelve-month period to the inside coefficient. - A cosine maximum occurs when its inside angle is \(0\), so place that at the stated month.

1. The amplitude is \(\frac{80-50}{2}=15\), and the midline is \(\frac{80+50}{2}=65\). 2. A period of \(12\) gives \(B=\frac{2\pi}{12}=\frac{\pi}{6}\). 3. To place a cosine maximum at \(m=7\), shift the input by \(7\): \(T(m)=65+15\cos\left(\frac{\pi}{6}(m-7)\right)\).

\(T(m)=65+15\cos\left(\frac{\pi}{6}(m-7)\right)\)

- Identify the model's midline from the vertical shift. - Ask when the cosine contribution is zero. - Restrict the solutions to the stated single-period interval.

1. The midline is \(y=5\), so set \(5+3\cos\left(\frac{\pi}{4}t\right)=5\). 2. This requires \(\cos\left(\frac{\pi}{4}t\right)=0\). 3. In one cycle, cosine is zero at inside angles \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). 4. Solving gives \(t=2\) and \(t=6\).

\(t=2\) and \(t=6\)

- Read the midline and amplitude from the maximum and minimum. - Check the graph's value at \(x=0\). - Use the spacing between repeated minima or maxima to verify the period.

1. The graph has midline \(y=1\), amplitude \(3\), and period \(4\). 2. At \(x=0\), the graph is at its minimum value \(-2\). 3. A negative cosine term with vertical shift \(1\) starts at that minimum, so choice b) matches.

b) \(q(x)=1-3\cos\left(\frac{\pi}{2}x\right)\)

- Substitute the shift point into the inside expression. - Determine the period before reasoning about quarter-cycle timing. - Locate where a positive sine cycle reaches its maximum relative to an upward midline crossing.

1. At \(t=1\), the sine input is \(0\), so \(y=5\), the midline, and the positive sine pattern is increasing there. 2. The period is \(\frac{2\pi}{\pi/2}=4\). 3. A positive sine function reaches its first maximum one quarter-period after the upward midline crossing. 4. One quarter of \(4\) is \(1\), so the first maximum after \(t=1\) occurs at \(t=2\).

a) At the midline \(y=5\) and increasing b) Period \(4\) c) \(t=2\)

- Use the highest and lowest table values to determine vertical parameters. - Track the sequence midline, maximum, midline, minimum, midline to identify one full cycle. - Match the starting pattern with a basic sine or cosine cycle before writing the equation.

1. The maximum is \(14\) and the minimum is \(6\), so the amplitude is \(4\) and the midline is \(10\). 2. The values return to the same midline position and direction after \(8\) time units, so the period is \(8\). 3. The table starts on the midline, rises to a maximum, returns to the midline, falls to a minimum, and returns again, matching a positive sine cycle. 4. With period \(8\), \(B=\frac{2\pi}{8}=\frac{\pi}{4}\), so a model is \(y=10+4\sin\left(\frac{\pi}{4}t\right)\).

Amplitude: \(4\) Midline: \(y=10\) Period: \(8\) Model: \(y=10+4\sin\left(\frac{\pi}{4}t\right)\)

- Keep the given amplitude and midline in their usual positions in a sinusoidal model. - Convert cycles per second into the angular coefficient multiplying \(t\). - Choose a sine or cosine form that naturally matches a maximum at \(t=0\).

1. A frequency of \(0.25\) cycles per second gives angular frequency \(B=2\pi(0.25)=\frac{\pi}{2}\). 2. Since the platform begins at a maximum, cosine needs no phase shift. 3. Using amplitude \(4\) and midline \(6\), \(h(t)=6+4\cos\left(\frac{\pi}{2}t\right)\).

\(h(t)=6+4\cos\left(\frac{\pi}{2}t\right)\)

- Check the midline, amplitude, starting position, and starting direction separately from the rate of repetition. - Distinguish cycles per unit time from radians per unit time. - Relate the inside coefficient of a sinusoid to the number of full cycles completed per unit time.

1. The midline, amplitude, and negative sine sign match the stated conditions. 2. The error is the inside coefficient. A frequency of \(2\) cycles per minute requires angular frequency \(B=2\pi(2)=4\pi\), not \(2\). 3. Therefore, the corrected model is \(d(t)=1-4\sin(4\pi t)\).

The coefficient \(2\) incorrectly treats the frequency as the angular coefficient. Since \(B=2\pi f=4\pi\), the corrected model is \(d(t)=1-4\sin(4\pi t)\).

- Try simplifying the sine and cosine inputs at the two times before comparing models. - A useful model must match the observations, not just produce heights in the right range. - How is the input coefficient related to the time for one full cycle?

1. At \(t=0\), \(h_1(0)=h_3(0)=7\), but \(h_2(0)=9\), so only \(h_1\) and \(h_3\) match the first observation. 2. At \(t=2\), \(h_1(2)=9\), while \(h_2(2)=h_3(2)=5\). Only \(h_3\) matches both observations. 3. Its period is \(\frac{2\pi}{\pi/4}=8\) seconds.

\(h_3(t)=7-2\sin\left(\frac{\pi}{4}t\right)\); period \(8\,\text{s}\).

- Which inputs make cosine easy to evaluate exactly? - When finding the stopped height, which time should go into the formula? - Consider whether a formula for motion should still describe the rider when there is no motion.

1. At \(t=3\), the input is \(\frac{\pi}{2}\), so \(h(3)=8\,\text{ft}\). 2. At \(t=6\), the input is \(\pi\), so \(h(6)=8-6(-1)=14\,\text{ft}\). 3. At \(t=30\), continuing the formula gives \(8-6\cos(5\pi)=14\,\text{ft}\). 4. The stopped wheel holds the height it had at \(t=24\): \(h(24)=8-6\cos(4\pi)=2\,\text{ft}\). The continued sinusoid assumes motion beyond the time the wheel stopped.

a) \(h(3)=8\,\text{ft}\); \(h(6)=14\,\text{ft}\) b) Continuing the formula predicts \(14\,\text{ft}\), but the actual stopped height is \(2\,\text{ft}\). The moving-wheel model no longer applies after \(24\,\text{s}\).

- What constant height lies halfway between the greatest and least depths? - How does the time from a maximum to the next minimum compare with a full cycle? - Which input makes basic cosine reach a maximum?

1. The midline is \(\frac{11+3}{2}=7\) and the amplitude is \(\frac{11-3}{2}=4\). 2. Successive maxima are \(10-2=8\) seconds apart, giving period \(8\) and coefficient \(B=\frac{2\pi}{8}=\frac{\pi}{4}\). 3. A positive cosine shifted to start at its maximum at \(t=2\) is \(d(t)=7+4\cos\left(\frac{\pi}{4}(t-2)\right)\).

\(d(t)=7+4\cos\left(\frac{\pi}{4}(t-2)\right)\)

- The inside angle controls where the basic cosine graph starts. - Think about how to recognize a time shift when the inside expression is not factored. - Which angle values make cosine reach its maximum?

1. Rewrite \(\frac{\pi}{4}t-\frac{\pi}{2}=\frac{\pi}{4}(t-2)\). Thus \(g\) is \(f\) shifted right by \(2\) seconds. 2. At \(t=0\), \(g(0)=\cos\left(-\frac{\pi}{2}\right)=0\). 3. The earliest nonnegative time with inside angle \(0\) is \(t=2\), when \(g(2)=\cos0=1\).

Shift right by \(2\,\text{s}\); \(g(0)=0\); first maximum at \(t=2\,\text{s}\).

- Use the highest and lowest points to find the midline and amplitude. - Measure the horizontal distance between consecutive maxima to find the period. - A positive cosine model reaches a maximum when its inside angle is \(0\). - Place the horizontal shift at a visible maximum.

1. The graph has maximum \(3\) and minimum \(-1\), so the amplitude is \(2\) and the midline is \(D=1\). 2. Consecutive maxima are \(4\) units apart, so the period is \(4\). Thus \(B=\frac{2\pi}{4}=\frac{\pi}{2}\). 3. A maximum occurs at \(x=1\), so choose a cosine shift of \(C=1\). 4. One matching equation is \(r(x)=1+2\cos\left(\frac{\pi}{2}(x-1)\right)\).

One matching equation is \(r(x)=1+2\cos\left(\frac{\pi}{2}(x-1)\right)\).