Intersect Prep

Building sinusoidal models for AP Precalculus

Updated

Tides, wheels, and daily temperatures all repeat. On AP Precalculus you model them with a sinusoidal function, and the third free-response question asks you to do it by hand, with no calculator.

This guide shows how to turn a highest value, a lowest value, and their times into a model, how to use the model to answer a question, and how to describe where it increases, decreases, and bends.

The four numbers in a sinusoidal model

A sinusoidal model has the form f(t)=acos⁡(b(t−c))+df(t) = a\cos(b(t - c)) + d or f(t)=asin⁡(b(t−c))+df(t) = a\sin(b(t - c)) + d. Each letter comes from one fact about the situation.

  • Midline, dd: halfway between the highest and lowest values, d=max+min2d = \frac{\text{max} + \text{min}}{2}.
  • Amplitude, ∣a∣|a|: half the distance from the lowest value to the highest, ∣a∣=max−min2|a| = \frac{\text{max} - \text{min}}{2}.
  • Period: the time for one full cycle. Then b=2πperiodb = \frac{2\pi}{\text{period}}.
  • Phase shift, cc: where the cycle starts. Cosine starts at a maximum, −cos⁡-\cos starts at a minimum, and sine starts on the midline going up.

A worked example from a high and a low

Here's the whole process on one context. The same steps work for any description that gives a highest value, a lowest value, and when they happen.

Water depth in a harbor

The water at a harbor entrance is 1414 feet deep at high tide, at t=4t = 4 hours after midnight. The next low tide is 44 feet deep, at t=10t = 10. The depth, in feet, is modeled by a sinusoidal function DD of tt. (a) Find the amplitude, the midline, and the period, and write an expression for D(t)D(t). (b) Find the depth at midnight. (c) A boat needs at least 11.511.5 feet of water to enter. For 0≤t≤120 \le t \le 12, when can it enter?
  1. 1Midline: d=14+42=9d = \frac{14 + 4}{2} = 9, so the midline is D=9D = 9.
  2. 2Amplitude: 14−42=5\frac{14 - 4}{2} = 5 feet.
  3. 3Period: high to low takes 10−4=610 - 4 = 6 hours, which is half a cycle, so the period is 1212 hours and b=2π12=π6b = \frac{2\pi}{12} = \frac{\pi}{6}.
  4. 4Phase shift: the maximum is at t=4t = 4, and cosine starts at a maximum, so use cos⁡\cos with c=4c = 4.

(a) Amplitude 55, midline D=9D = 9, period 1212 hours, and D(t)=5cos⁡(π6(t−4))+9D(t) = 5\cos\left(\frac{\pi}{6}(t - 4)\right) + 9.

51015202468101214(4, 14)(10, 4)(16, 14)(22, 4)
D(t)D(t)y=9y = 9The tide model over one day. High tides are 12 hours apart, and the dashed midline sits halfway between high and low.

(b) Midnight is t=0t = 0. D(0)=5cos⁡(−2π3)+9=5(−12)+9=6.5D(0) = 5\cos\left(-\frac{2\pi}{3}\right) + 9 = 5\left(-\frac{1}{2}\right) + 9 = 6.5 feet. Cosine is even, so cos⁡(−2π3)=cos⁡2π3\cos\left(-\frac{2\pi}{3}\right) = \cos\frac{2\pi}{3}.

(c) Solve 5cos⁡(π6(t−4))+9=11.55\cos\left(\frac{\pi}{6}(t - 4)\right) + 9 = 11.5. That gives cos⁡(π6(t−4))=12\cos\left(\frac{\pi}{6}(t - 4)\right) = \frac{1}{2}, so π6(t−4)=−π3\frac{\pi}{6}(t - 4) = -\frac{\pi}{3} or π3\frac{\pi}{3}. Multiply by 6π\frac{6}{\pi}: t−4=−2t - 4 = -2 or 22, so t=2t = 2 or t=6t = 6.

The depth is above 11.511.5 around the high tide at t=4t = 4, between those two times. The next time the angle reaches ±π3\pm\frac{\pi}{3} again is t=14t = 14, outside the interval.

246810122468101214(2, 11.5)(6, 11.5)
D(t)D(t)y=11.5y = 11.5The curve is above 11.5 feet from 2 a.m. to 6 a.m.

Answer: (a) D(t)=5cos⁡(π6(t−4))+9D(t) = 5\cos\left(\frac{\pi}{6}(t - 4)\right) + 9. (b) 6.56.5 feet. (c) From t=2t = 2 to t=6t = 6, 2 a.m. to 6 a.m.

On a calculator question you'd find the times in (c) by graphing y=D(x)y = D(x) and y=11.5y = 11.5 in Desmos and clicking the intersections. A sinusoid crosses most horizontal lines twice per period, so check the interval for every crossing.

Sine or cosine, and the sign of the shift

The same curve has many correct formulas. In the harbor model, the depth crosses the midline going up a quarter period before high tide, at t=4−3=1t = 4 - 3 = 1. So D(t)=5sin⁡(π6(t−1))+9D(t) = 5\sin\left(\frac{\pi}{6}(t - 1)\right) + 9 is the same function. Either one is correct, as long as the maximum lands at the right time.

  • Start at a maximum: acos⁡(b(t−c))+da\cos(b(t - c)) + d with a>0a > 0 and cc the time of the maximum.
  • Start at a minimum: −acos⁡(b(t−c))+d-a\cos(b(t - c)) + d, with cc the time of the minimum.
  • Start on the midline going up: asin⁡(b(t−c))+da\sin(b(t - c)) + d, with cc the time of the crossing.
  • Check your model by plugging in the time of the maximum. You should get the highest value.

Factor before you read the shift

What is the period of f(t)=3cos⁡(2t−π)+1f(t) = 3\cos(2t - \pi) + 1, and at what value of tt in 0≤t<π0 \le t < \pi does ff reach its maximum?

Factor out bb: 2t−π=2(t−π2)2t - \pi = 2\left(t - \frac{\pi}{2}\right). The phase shift is π2\frac{\pi}{2}, not π\pi.

The period is 2π2=π\frac{2\pi}{2} = \pi. Cosine is at its maximum when its input is 00, so ff is at its maximum when t=π2t = \frac{\pi}{2}. Check: f(π2)=3cos⁡0+1=4f\left(\frac{\pi}{2}\right) = 3\cos 0 + 1 = 4, which is d+ad + a.

Answer: Period π\pi, maximum of 44 at t=π2t = \frac{\pi}{2}

How a sinusoid increases, decreases, and bends

  • Increasing from each minimum to the next maximum, and decreasing from each maximum to the next minimum.
  • Concave down above the midline, and concave up below it.
  • Changing fastest as it crosses the midline, and slowest near a maximum or minimum.

Free-response questions usually ask you to back these up with average rates of change. Over consecutive intervals of the same length, increasing rates of change mean concave up, and decreasing rates of change mean concave down.

Falling faster and faster

For the harbor model D(t)=5cos⁡(π6(t−4))+9D(t) = 5\cos\left(\frac{\pi}{6}(t - 4)\right) + 9, find the average rates of change over [4,5][4, 5], [5,6][5, 6], and [6,7][6, 7]. Is the depth decreasing at an increasing rate or a decreasing rate on 4<t<74 < t < 7? Give a reason.

D(4)=14D(4) = 14, D(5)=5cos⁡π6+9=9+532D(5) = 5\cos\frac{\pi}{6} + 9 = 9 + \frac{5\sqrt{3}}{2}, D(6)=5cos⁡π3+9=11.5D(6) = 5\cos\frac{\pi}{3} + 9 = 11.5, and D(7)=5cos⁡π2+9=9D(7) = 5\cos\frac{\pi}{2} + 9 = 9.

The average rates of change are 532−5≈−0.670\frac{5\sqrt{3}}{2} - 5 \approx -0.670, then 2.5−532≈−1.8302.5 - \frac{5\sqrt{3}}{2} \approx -1.830, then −2.5-2.5 feet per hour.

Write: "The average rates of change over consecutive one-hour intervals are negative and getting more negative, so the depth is decreasing at an increasing rate on 4<t<74 < t < 7." That matches the picture: from high tide down to the midline, the water drops faster and faster.

Answer: About −0.670-0.670, −1.830-1.830, and −2.5-2.5 feet per hour; decreasing at an increasing rate

Fitting a sinusoid to data in Desmos

When you're given a table instead of a description, fit the model in Desmos. This comes up on calculator multiple-choice questions.

  1. 1Enter the data in a table, or as lists x1=[…]x_1 = [\ldots] and y1=[…]y_1 = [\ldots].
  2. 2If the period is known, type bb in. For a 12-month cycle, type y1∼asin⁡(π6(x1−c))+dy_1 \sim a\sin\left(\frac{\pi}{6}(x_1 - c)\right) + d.
  3. 3Read aa, cc, and dd from the regression. A negative aa, or a cc that differs from yours by a full period, describes the same curve.
  4. 4Graph the model over the points to check the fit, then evaluate it or intersect it with a horizontal line.

Typing the known period keeps Desmos from settling on a curve that wiggles through the points with a period that makes no sense. More on choosing and checking models is in the regression guide.

A quick check

Reading the maximum from a formula

Let g(t)=−4cos⁡(π5t)+7g(t) = -4\cos\left(\frac{\pi}{5}t\right) + 7. What is the maximum value of gg, and what is the smallest positive tt where it occurs?
  1. A. Maximum 77 at t=5t = 5
  2. B. Maximum 1111 at t=10t = 10
  3. C. Maximum 1111 at t=5t = 5
  4. D. Maximum 33 at t=0t = 0

The midline is 77 and the amplitude is 44, so the maximum is 1111. The negative sign means gg starts at its minimum, 33, at t=0t = 0. The period is 2ππ/5=10\frac{2\pi}{\pi/5} = 10, so the maximum is half a period later, at t=5t = 5.

Answer: C

For how this fits into the third free-response question, see the free-response guide.

Keep going

Practice on exam-style questions

Intersect Prep has 117 multiple-choice questions for Unit 3, with a full solution for each one. Create a free account to try a full lesson with the calculator. One payment of $29 unlocks the course for 3 months, or $39 for 6 months.

Questions

How do I find the period from a maximum and a minimum?
The time from a maximum to the next minimum is half the period, so double it. Then b is 2π divided by the period.
Should I use sine or cosine?
Either works. Cosine is easiest when you know when the maximum happens, negative cosine when you know the minimum, and sine when you know a midline crossing going up.
Where is a sinusoidal function changing fastest?
As it crosses the midline. It changes slowest near its maximum and minimum values.
Is the sinusoidal free-response question done with a calculator?
No. FRQ 3, Modeling a Periodic Context, is in the no-calculator part, so values come from the unit circle.

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