Sinusoidal Function Context and Data Modeling

Worked example

1. 01

   Max and min: the lowest point is 3 m, and the highest point is one diameter higher, $3 + 50 = 53$ m. Midline $(53 + 3) \div 2 = 28$. Amplitude $(53 - 3) \div 2 = 25$, which is the radius.

2. 02

   Period 20 minutes, so $b = 2\pi \div 20 = \pi /10$.

3. 03

   At $t = 0$ the rider is at the minimum, so use cosine with $a < 0$. No shift is needed, because the cycle starts at $t = 0$.

   $$H(t)=-25\,\cos\left(\frac{\pi}{10}t\right)+28$$
4. 04

   $H(12) = -25 \cos (6\pi /5) + 28$. Since $\cos (6\pi /5) = -\cos (\pi /5)$, this is $25 \cos (\pi /5) + 28 \approx 48.2$ m.

5. 05

   Graph $y = H(x)$ and $y = 40$ and find the two intersections in the first rotation: $t \approx 6.59$ and $t \approx 13.41$. The rider is above 40 m between them, for about $13.41 - 6.59 \approx 6.81$ minutes.

   The highlighted part of the curve is where the rider is above 40 m.

   Notice the symmetry: the two intersections are equally far from the top of the ride at $t = 10$ $(10 - 6.59 = 13.41 - 10)$. A sinusoid is symmetric about every maximum and minimum, which gives you a quick check on calculator answers.

Worked example

1. 01

   (a) The lowest value is at $t = 1$ and the highest at $t = 7$. Minimum to maximum is half a cycle, so half the period is about 6 months, and the period is about 12 months. The frequency is about $1/12$ cycle per month. That matches the context: the seasons repeat once a year.

2. 02

   (b) Use the extreme data values: midline ≈ $(27.9 + 4.9) \div 2 = 16.4$ and amplitude ≈ $(27.9 - 4.9) \div 2 = 11.5$.

3. 03
   $> (c) b = 2\pi \div 12 = \pi /6$. The highest point is at $t = 7$, so use cosine with $a > 0$, shifted right 7:

   $$T(t)\approx 11.5\,\cos\left(\frac{\pi}{6}(t-7)\right)+16.4$$
4. 04

   (d) Enter the months in L1 and the temperatures in L2, set the calculator to RADIAN mode, and run sinusoidal regression (SinReg). Rounded to three decimal places, the result is:

   $$T(t)\approx 12.347\,\sin(0.522t-2.163)+16.812$$

   Our estimate (dashed) and the regression model (solid) nearly overlap.

   The regression looks different on paper, but it describes almost the same curve. Its period is $2\pi \div 0.522 \approx 12.04$ months, and its midline is 16.8 instead of 16.4. To find where its cycle starts, factor out $b$ as in 3.6B: $0.522t - 2.163 = 0.522(t - 4.14)$, so the sine curve crosses its midline going up at $t \approx 4.14$. A sine curve reaches its maximum a quarter period later, at about $4.14 + 12.04 \div 4 \approx 7.15$, which is mid-July, right next to our estimate of $t = 7$.

   Why is the regression amplitude (12.3) bigger than our estimate (11.5)? Our estimate used only the two most extreme readings. The regression balances all twelve points, and the flat stretch from June to August pulls the curve a little higher. The regression's squared errors add up to about 6.3, compared with about 18.5 for our estimate, so it fits this data better.

Worked example

1. 01

   (a) 7:30 a.m. is $t = 7.5$. $D(7.5) = 1.8 \cos ((5\pi /31)(4.5)) + 2.8 \approx 1.63$ m. That is less than 2.0 m, so the boat cannot leave.

2. 02

   (b) Graph $y = D(x)$ and $y = 2.0$, and find the intersections on either side of low tide ($t = 9.2$): $t \approx 7.01$ and $t \approx 11.39$.

3. 03

   Convert to clock times: 0.01 hour is less than a minute, and 0.39 hour ≈ 23 minutes. The water is too shallow from about 7:01 a.m. to about 11:23 a.m., roughly 4 hours 23 minutes.

4. 04

   Check with symmetry: low tide at $t = 9.2$ should sit exactly in the middle. $(7.01 + 11.39) \div 2 = 9.2$ ✓.

   The highlighted part of the curve is where the water is shallower than 2.0 m.