Introduction to Natural Science Trigonometric Modeling
1. Complete Ex 12, 14, 15 and 18 from Section 6.5 pg 264 in your Trigonometry Handout. 12.
This is an upright cosine graph with amplitude 3 so a = 3. It has period 4 so b = 2π/4 = π/2.
So y = 3 cos (πt/2)
14. This is an upright sine graph with amplitude 4 so a = 4. It has period 1 so b = 2π/1 = 2π.
So y = 4 sin (2πt).
15. This is an inverted cosine graph with amplitude 2 so a = −2 the midline is y = 2 so d = 2.
The period is 12 (since 3/4 of a period is 9) so b = 2π/12 = π/6. So y = −2 cos (πθ/6) + 2
18. This is an upright sine graph with amplitude 2 so a = 2. The period is π so b = 2π/π = 2.
The midline is y = −1 so d = −1. So y = 2 sin 2t − 1.
2. The number of hours of daylight in Olympia during different times of the year is modeled by
the function
πt π
y = 12 − 3 cos +
6 18
where y is hours of of daylight and t is the number of months since January 1st.
(a) What are the amplitude, period and midline of this function?
a = −3 ⇒ Amplitude is 3 hours. b = π/6 ⇒ period = 2π/b = 2π/(pi/6) = 12
months. d = 12 so the midline is y = 12.
(b) By considering your answers in (a) how many hours are there in the longest day and in
the shortest day of the year?
The longest day will be 12+3=15 hours. The shortest day will be 12-3=9 hours.
(c) According to the model at what time of the year does the longest day of the year occur?
πt π
The longest day is 15 hours so we solve the equation 15 = 12 − 3 cos + for t.
6 18
πt π πt π πt π
So 3 = −3 cos + ⇒ −1 = cos + ⇒ + = cos−1 (−1) = π
6 18 6 18 6 18
πt 17π 17
⇒ = ⇒ t= = 5 2 months since January 1st.
3
6 18 3
So the model predicts correctly that the longest day of the year is June 20th.
3. Some years ago an oil tanker called the Sea Empress ran a ground in a bay called Milford
Haven in the west of Wales because the captain misread the tide tables. Suppose the depth
of water in Milford Haven varies with the tide according to the equation
d = 7 + 2 sin (πt/6)
where d is the depth measured in metres and t is the number of hours since midnight.
(a) State the period of this function, and sketch a graph of it.
The period is 2π/b = 2π/(π/6) = 12 hours.
d
9
8
7
6
5
0 3 6 9 12 15 18 21 24 t
(b) What is the depth at high tide, and at what times during a 24 hour period does high
tide occur?
At high tide the depth is 9 m and this occurs 1/4 of a period after midnight. ie at 3:00
am (and again at 3:00 pm).
(c) What is the depth at low tide and at what times during a 24 hour period does low tide
occur?
At low tide the depth is 5 m and this occurs 3/4 of a period after midnight. ie at 9:00
am (and again at 9:00 pm).
(d) At what time in the morning is the depth of water 8 metres?
To find this time solve the equation 8 = 7 + 2 sin (πt/6) ⇒ 1 = 2 sin (πt/6) ⇒
sin (πt/6) = 1/2 = 0.5. But sin θ = 0.5 when θ = 30o = π/6 radians. So πt/6 = π/6 ⇒
t = 1 hour. The tide is at 8 m at 1:00 am. From the symmetry of the graph it is 8 m
again at 5:00 am (i.e. 1 hour before 6:00 am)
(e) The draught of a ship is the depth of water needed for it to float. The draught of the Sea
Empress is 6 metres. During which times may it enter Milford Haven without causing
an oil spill?
To find this time solve the equation 6 = 7 + 2 sin (πt/6) ⇒ −1 = 2 sin (πt/6) ⇒
sin (πt/6) = −1/2 = −0.5. But sin θ = −0.5 when θ = −30o = −π/6 radians. So
πt/6 = −π/6 ⇒ t = −1 hour. The tide is at 6 m at 1 hour before midnight. From the
symmetry of the graph it is 6 m again also at 11:00 am and 7:00 am (i.e. 1 hour before
noon and 1 hour after 6:00 am).
4. A spring with a mass hanging on the end is 12 cm in length. The mass is pulled down until
the spring is 15 cm in length and is then released. The spring oscillates in simple harmonic
motion with frequency 0.5 Hz.
(a) What are the amplitude and period of this motion?
The amplitude is 3 cm and the period is 1/0.5 = 2 seconds.
(b) Sketch a graph of the length of the spring in cm as a function of time in seconds since it
was released.
x
15
12
9
0 1 2 3 t
(c) Model the oscillations of the spring using an appropriate sinusoidal function.
This is an upright cosine function with amplitude 3 so a = 3. The midline is y = 12 so
d = 12 and the period 2 seconds so b = 2π/b = π. So a good model is y = 3 cos πt + 12
(d) Use your model to find how long it takes the spring to first reach 10 cm in length.
Solve the equation 10 = 3 cos πt + 12 ⇒ −2 = 3 cos πt ⇒ cos πt = −2/3 ⇒ πt =
cos−1 (−2/3) = 132o = 2.3 radians. So t = 2.3/π = 0.73 seconds
(e) Use your graph to help you find out how long after that it reaches 10 cm again?
From the symmetry of the graph the next time it reaches 10 cm is t = 2 − 0.73 = 1.27
seconds after it is released.