Unit 3 · Topic 3.7 · about 30 minutes · 7 parts · 14 scored checks, 1 of them multiple choice

AP Precalculus 3.7: Sinusoidal Function Context and Data Modeling

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

The graph of an invented lake temperature model, L(m) = 10 cos((pi/6)(m - 7)) + 14, over 20 months. It peaks at 24 degrees at months 7 and 19 and bottoms out at 4 degrees at months 1 and 13. A dashed midline at 14 is halfway between them, a red segment from the midline to the peak at month 7 is labeled amplitude 10, and a bracket between the peaks at months 7 and 19 is labeled period 12 months. The peak at (7, 24) is marked as the actual pair used to place the cosine.

Part 1 Think first When can the boat get in?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A small harbor's tide table (invented) shows low water, 0.4 meters, at 11:00 a.m. and the next high water, 3.2 meters, at 5:12 p.m. A fishing boat needs 1.1 meters of water at the harbor entrance to get in. That is a quarter of the way up from low to high. At about what time does the water reach 1.1 meters?

In short. A sinusoidal function model describes a periodic situation, and you build it from the context's maximum, minimum and period plus one known input-output pair, or by running a sinusoidal regression on data.

The midline is the average of the maximum and minimum, the amplitude is half their difference, and the time between consecutive maxima is the period, so $b = \frac{2 π}{period}$ .
The phase shift comes from comparing an actual input-output pair with the model: cosine is at its maximum where its input is 0, and sine is on its midline going up there.
Inside its contextual domain, the model predicts an output from an input, and it gives the inputs where the output reaches a stated value.

Part 2 Learn it Four short ideas, each checked as you go

2a. Midline, amplitude and period from the context

A sinusoidal model $f (t) = a \sin (b (t + c)) + d$ , or the same with cosine, has four numbers to fill in. A context usually hands you three of them.

The maximum and minimum give the vertical numbers. The midline is their average, $d = \frac{max + min}{2}$ , and the amplitude is half their difference, $\frac{max - min}{2}$ .

The time from one maximum to the next, or one minimum to the next, is the period. Since the period is $\frac{2 π}{| b |}$ , $b = \frac{2 π}{period}$ when $b$ is positive. The frequency is $\frac{1}{period}$ , in cycles per unit of input.

A lake (invented) is coldest, 4 degrees Celsius, in late January and warmest, 24 degrees, in late July, every year. Count months so that $m = 1$ is late January and $m = 7$ is late July. Then $d = 14$ and the amplitude is 10. The warmest point returns every 12 months, so $b = \frac{2 π}{12} = \frac{π}{6}$ and the frequency is $\frac{1}{12}$ of a cycle per month.

Keep the period and $b$ apart. The period is a time, 12 months; $b$ is what makes the input to cosine or sine run through $2 π$ in that time.

Quick check · Fill the statement

No calculator. In a desert town (invented), the air temperature follows the same cycle every day. It is highest, 36 degrees Celsius, at 3 p.m., and lowest, 16 degrees Celsius, at 3 a.m. Tap a blank, then tap an answer.

A sinusoidal model of the temperature has midline $y =$ and amplitude . The highs are hours apart, so that is the period, and the frequency is cycle per hour.

2b. The phase shift, from one actual pair

Midline, amplitude and $b$ fix the shape. The phase shift slides it into line with the context, and you find it by comparing an actual input-output pair with the pair the model gives.

Choose a pair the basic graph makes easy. Cosine is at its maximum where its input is 0. Sine is on its midline, going up, where its input is 0. So if you know when the context peaks, use cosine and make the inside 0 at that time.

The lake peaks at $m = 7$ , so

L (m) = 10 \cos (\frac{π}{6} (m - 7)) + 14

At $m = 7$ the inside is 0, $\cos 0 = 1$ , and $L (7) = 24$ , the actual maximum. In the form $a \cos (b (m + c)) + d$ , that is $c = - 7$ , a phase shift of 7 to the right.

The sign is where models go wrong. $m - 7$ is 0 at $m = 7$ , but $m + 7$ is 0 at $m = - 7$ . Put the actual pair back in every time.

Sine works too, from a different pair. The lake crosses its midline going up halfway between January and July, at $m = 4$ , so $10 \sin (\frac{π}{6} (m - 4)) + 14$ draws the same curve.

Quick check · Decide and justify

No calculator. The water depth at a pier (invented) is greatest, 5 meters, at $t = 2$ hours after midnight, and least, 1 meter, at $t = 8$ . Which model fits these facts?

Which reason supports your answer?

2c. With technology: estimate, or run a regression

With data instead of a story, technology gives you two ways in.

Estimate the key values. Plot the data on a graphing calculator, read off the largest and smallest outputs and the inputs of two maxima in a row, and build the model by hand as above. Real data are noisy, so these are estimates; compare the model with a few actual pairs and adjust.

Run a sinusoidal regression. Enter the data as two lists and choose SinReg. It finds the four parameters that make the squared residuals smallest, like the earlier regressions. It needs data covering at least one full cycle, with several points in each cycle.

The result can look different from your model and still be the same curve. Some calculators print $y = a \sin (b x + c) + d$ ; rewrite $b x + c$ as $b (x + \frac{c}{b})$ for the form used here. A regression may also give a negative $a$ , a $c$ that differs from yours by a whole period, or a sine where you used a cosine. To compare two models, compare midline, amplitude, period and where each reaches its maximum.

Quick check · Sort it

No calculator. A student's model for a data set is $y = 2 \sin (\frac{π}{6} (x - 1)) + 5$ , which has its maximum at $x = 4$ . Each card is a model a classmate got for the same data, by regression or by hand. Tap a card, then whether it draws the same curve.

Same curve as $2 \sin (\frac{π}{6} (x - 1)) + 5$

A different curve

2d. Using the model: predict, solve, and stay in the domain

Once you have a model, it answers two kinds of question.

Predict an output. Substitute the input, and check what the input means first. If $h$ counts hours after 6 a.m., then 11 a.m. is $h = 5$ , not 11.

Find an input. Set the model equal to the output and solve. Without a calculator, the value will lead to a sine or cosine you know, such as $\frac{1}{2}$ . A sine equals $\frac{1}{2}$ twice in each cycle, once going up and once going down, so pick the one the question asks for. With a graphing calculator, graph the model and the horizontal line at that output, and read off the intersection you want.

Then ask whether the input makes sense. A model is useful over its contextual domain, the inputs the situation actually covers. A Ferris wheel model describes a rider only while the ride runs. A model fit to one week of greenhouse readings says nothing about a different season. Outside that domain the formula still produces a number, but the number is not about anything.

Quick check · Calculate

No calculator. On a school day, the temperature in a classroom is modeled by $T (h) = 4 \sin (\frac{π}{12} (h - 3)) + 20$ degrees Celsius, where $h$ is the number of hours after 6 a.m., for $0 \leq h \leq 12$ . What temperature does the model predict at 11 a.m.?

Your answer degrees Celsius

Part 3 Watch it done A Ferris wheel, from its numbers to a time

Watch each step, then do the last one yourself.

A Ferris wheel (invented) carries its seats from 3 meters above the ground at the bottom to 43 meters at the top, and makes one turn every 12 minutes. A rider boards at the bottom at $t = 0$ . Write a model $h (t) = a \sin (b (t + c)) + d$ for the rider's height in meters after $t$ minutes, then find the first time after boarding that the rider is 33 meters above the ground.

Step 1 · Midline and amplitude
From the top and bottom heights: $d = \frac{43 + 3}{2} = 23 a = \frac{43 - 3}{2} = 20$ The 23 is the height of the wheel's center.
Step 2 · Period, then b
One turn takes 12 minutes, so the period is 12 and $b = \frac{2 π}{12} = \frac{π}{6}$ .
Step 3 · Phase shift from an actual pair
Sine is on its midline, going up, where its input is 0. The rider starts at the bottom and reaches the height of the center, going up, a quarter turn later, at $t = 3$ . So the inside should be 0 at $t = 3$ : $h (t) = 20 \sin (\frac{π}{6} (t - 3)) + 23$ Check against the actual pair $(0, 3)$ : $h (0) = 20 \sin (- \frac{π}{2}) + 23 = - 20 + 23 = 3$ .
Why this step?
Why $t - 3$ and not $t + 3$ ?
The inside has to be 0 at $t = 3$ , and $t - 3$ is 0 there. With $t + 3$ the inside is 0 at $t = - 3$ , three minutes before boarding, and $h (0) = 20 \sin \frac{π}{2} + 23 = 43$ : the model would start the rider at the top.
Your step · Calculate

No calculator. Finish it. A rider's height on a Ferris wheel is modeled by $h (t) = 20 \sin (\frac{π}{6} (t - 3)) + 23$ meters, $t$ minutes after boarding at the bottom. What is the first time after boarding that the rider is 33 meters above the ground?

Your answer minutes

Part 4 Try it Fit the tide and use the model

Here are water heights at a harbor (invented data), in meters, every 2 hours from midnight, $x = 0$ , to midnight the next night, $x = 24$ : 1, 2.07, 2.97, 2.79, 1.65, 0.72, 0.76, 1.92, 2.88, 2.9, 1.9, 0.79 and 0.74. Before you open the lab, find the highest and lowest readings and the hours of two high readings in a row, and estimate the midline, amplitude and period yourself.

The lab below opens on these data with the sinusoidal family selected. It prints the model in the form $y = a \sin (b (x + c)) + d$ with five significant figures, and under it the period $\frac{2 π}{b}$ , the midline and the amplitude. Compare them with your estimates. Then answer the two questions with the model, using the lab's graph and its Predict y at x = box or a graphing calculator.

Lab

Model Builder

Warm up first: type 5 into the Predict y at x = box and see that the model's height at 5 a.m. is a little higher than the readings at hours 4 and 6, because the model's high tide falls between them. If your calculator's SinReg prints $y = a \sin (b x + c) + d$ , divide its $c$ by its $b$ to compare it with the lab's form.

Open the full Model Builder lab

Try it · Calculate

Graphing calculator required. The water height at a harbor, in meters, was recorded every 2 hours from $x = 0$ (midnight) to $x = 24$ hours: 1, 2.07, 2.97, 2.79, 1.65, 0.72, 0.76, 1.92, 2.88, 2.9, 1.9, 0.79 and 0.74 (invented data). Run a sinusoidal regression on these data. The harbor closes its inner channel whenever the water height is below 1.2 meters. According to the model, at what time $x$ between $x = 4$ and $x = 12$ does the water height fall below 1.2 meters? Give your answer to the nearest tenth of an hour.

Your answer hours after midnight

Try it · Calculate

Graphing calculator required. The water height at a harbor, in meters, was recorded every 2 hours from $x = 0$ (midnight) to $x = 24$ hours: 1, 2.07, 2.97, 2.79, 1.65, 0.72, 0.76, 1.92, 2.88, 2.9, 1.9, 0.79 and 0.74 (invented data). Run a sinusoidal regression on these data. Large boats can use the inner channel only while the water height is above 2.6 meters. According to the model, how many hours does each of those stretches last? Give your answer to the nearest hundredth of an hour.

Your answer hours

Part 5 Explain it Write it the way the exam asks

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 37.5% of the exam score. This one is shaped like Question 3, Modeling a Periodic Context.

Free-response practice · Question 3 style

No calculator. A weight hangs from a spring and bobs up and down after it is pulled and let go. For the first few seconds, before the bouncing dies down, its height above a table, in centimeters, is modeled by a sinusoidal function $h$ of time $t$ seconds. The weight is at its highest point, 32 centimeters, at $t = 1$ , and it next reaches its lowest point, 8 centimeters, at $t = 2$ . (A) The function $h$ is given by $h (t) = a \sin (b (t + c)) + d$ . Find values of the constants $a$ , $b$ , $c$ and $d$ . (B) Use the model to find the height of the weight at $t = \frac{4}{3}$ seconds. (C) Find all times $t$ , for $0 \leq t \leq 2$ , when the model gives a height of 14 centimeters. Show the work that leads to your answers.

Your answer stays on this page. It is never sent or saved.

Model answer

(A) $d = \frac{32 + 8}{2} = 20$ and $a = \frac{32 - 8}{2} = 12$ . Highest to lowest takes 1 second, which is half a cycle, so the period is 2 and $b = \frac{2 π}{2} = π$ . Sine is at its maximum where its input is $\frac{π}{2}$ , so I need $π (1 + c) = \frac{π}{2}$ , which gives $c = - \frac{1}{2}$ . So $h (t) = 12 \sin (π (t - \frac{1}{2})) + 20$ . Check: $h (1) = 12 \sin \frac{π}{2} + 20 = 32$ and $h (2) = 12 \sin \frac{3 π}{2} + 20 = 8$ . (B) $h (\frac{4}{3}) = 12 \sin \frac{5 π}{6} + 20 = 12 \cdot \frac{1}{2} + 20 = 26$ centimeters. (C) $12 \sin (π (t - \frac{1}{2})) + 20 = 14$ gives $\sin (π (t - \frac{1}{2})) = - \frac{1}{2}$ . For $0 \leq t \leq 2$ the input $π (t - \frac{1}{2})$ runs from $- \frac{π}{2}$ to $\frac{3 π}{2}$ , where sine is $- \frac{1}{2}$ only at $- \frac{π}{6}$ and $\frac{7 π}{6}$ . So $t - \frac{1}{2} = - \frac{1}{6}$ or $\frac{7}{6}$ , and $t = \frac{1}{3}$ or $t = \frac{5}{3}$ seconds.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds $a = 12$ and $d = 20$ from the maximum 32 and the minimum 8, or $a = - 12$ with a value of $c$ that matches it.
Finds $b = π$ from a period of 2 seconds, twice the 1 second from highest to lowest.
Finds $c = - \frac{1}{2}$ , or a value differing from it by a whole period, and checks that $h (1) = 32$ .
Finds $h (\frac{4}{3}) = 26$ centimeters.
Gives both times, $t = \frac{1}{3}$ and $t = \frac{5}{3}$ , and no others in $0 \leq t \leq 2$ .

Part 6 Check yourself A greenhouse over two days

The two case questions use the same table. Answer them from it.

A grower records the air temperature in a greenhouse every 3 hours, starting at 3 a.m. on Monday. In the table, $t$ is the number of hours after midnight at the start of Monday, so $t = 27$ is 3 a.m. on Tuesday. The daily pattern repeats, and the grower wants a sinusoidal model $T (t)$ for the temperature.

Greenhouse air temperature (invented data)
$t$ (hours)	Temperature (degrees Celsius)
3	15.0
6	17.3
9	21.8
12	27.1
15	29.0
18	26.7
21	22.2
24	16.9
27	15.0
30	17.2
33	21.9
36	27.0
39	29.0

Case question 1 · Calculate

No calculator. A model of the greenhouse table has the form $T (t) = a \cos (b (t + c)) + d$ , with $b > 0$ , and $b = k π$ . Use the greenhouse table to find $k$ . Give the exact value; a fraction such as 5/3 is fine.

Your answer

Case question 2 · Multiple choice

No calculator. A grower models the greenhouse table with $T (t) = 7 \sin (b (t + c)) + 22$ , where $b > 0$ comes from the time between consecutive maxima. Which value of $c$ makes the model's maximum fall at $t = 15$ , the time of an actual maximum in the table?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A bicycle pedal's height above the road is greatest, 46 centimeters, at $t = 0$ seconds and least, 10 centimeters, at $t = 0.5$ seconds, and the rider pedals at a steady pace. A student builds a model $h (t) = a \cos (b (t + c)) + d$ . Tap the first line that goes wrong.

Check yourself · Put it in order

No calculator. At a harbor (invented), high tides of 2.8 meters come at 2 a.m. and 2 p.m., low tides are 0.4 meters, and a student has already found $d = 1.6$ and $a = 1.2$ from them. Put the rest of the work, building $W (t) = a \cos (b (t + c)) + d$ with $t$ in hours after midnight and then using it, in order. Tap the steps in sequence.

Check yourself · Decide and justify

No calculator. A Ferris wheel ride lasts three turns, 36 minutes in all, and a rider's height is modeled by $h (t) = 20 \sin (\frac{π}{6} (t - 3)) + 23$ meters, $t$ minutes after boarding. Should the operator use $h (50)$ to report the rider's height 50 minutes after boarding?

Which reason supports your answer?

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

The function p is sinusoidal. The table gives p at the whole numbers from 0 to 8, and the chart plots more points on its graph. The maximum value of p occurs at $x = 0$ and next at $x = 8$ , and the minimum value between them occurs at $x = 4$ .

Values of the sinusoidal function $p$
$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$p (x)$	$3$	$1 + \sqrt{2}$	$1$	$1 - \sqrt{2}$	$- 1$	$1 - \sqrt{2}$	$1$	$1 + \sqrt{2}$	$3$

Points on the graph of $p$

Each dot is $(x, p (x))$ , at every half unit from 0 to 8, rounded to two decimal places.

From 3.5 Sinusoidal Functions · Decide and justify

No calculator. Is the graph of p concave up or concave down on the interval $2.5 < x < 3.5$ ?

Which reason supports your answer?

From 3.6 Sinusoidal Function Transformations · Sort it

No calculator. Each card is a transformation of $y = \sin θ$ that changes exactly one of four features of its graph: the amplitude, the period, the midline or the phase shift. Tap a card, then the feature it changes.

Amplitude

Period

Midline

Phase shift

In three lines

The midline is the average of the maximum and minimum, the amplitude half their difference, and $b = \frac{2 π}{period}$ , with the period measured between consecutive maxima.
Place the phase shift with an actual pair: cosine peaks where its input is 0, sine crosses its midline going up there, and the sign of $c$ is checked by putting the pair back in.
Build from key values or a sinusoidal regression, then predict and solve only inside the contextual domain.

Vocabulary

Period: The input-value interval between consecutive maxima or consecutive minima, $\frac{2 π}{| b |}$ for $a \sin (b (θ + c)) + d$ .
Frequency: The reciprocal of the period: the number of cycles per unit of input.
Amplitude: Half the difference between the maximum and minimum output values, $| a |$ in $a \sin (b (θ + c)) + d$ .
Vertical shift: The value $d$ , the average of the maximum and minimum, which moves the midline from $y = 0$ to $y = d$ .
Phase shift: The horizontal translation of $- c$ units in $a \sin (b (θ + c)) + d$ , found by comparing an actual input-output pair with the model's.
Sinusoidal regression: A fit by technology, such as SinReg on a graphing calculator, of a sinusoidal function model to a data set.
Contextual domain: The input values the situation actually covers, over which a model's predictions mean something.

Course alignment, for teachers

AP Precalculus topic 3.7, Unit 3: Trigonometric and Polar Functions.

Skill 1.C: Construct new functions, using transformations, compositions, inverses, or regressions, that may be useful in modeling contexts, criteria, or data, with and without technology.
Skill 3.C: Support conclusions or choices with a logical rationale or appropriate data.

This lesson maps all 5 essential knowledge statements of the topic to the part that teaches each one: 5 are checked by a scored question. The map is in the lesson file.

Part 1 Think first When can the boat get in?

Part 2 Learn it Four short ideas, each checked as you go

2a. Midline, amplitude and period from the context

2b. The phase shift, from one actual pair

2c. With technology: estimate, or run a regression

2d. Using the model: predict, solve, and stay in the domain

Part 3 Watch it done A Ferris wheel, from its numbers to a time

Part 4 Try it Fit the tide and use the model

Model Builder

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A greenhouse over two days

Part 7 Keep it Bring back earlier ideas

Vocabulary

Labs that use this topic