MCR3U1 – Applying Sine & Cosine Functions to Model Behaviour in Word Problems
Sine & Cosine functions range from –1 to +1 → Amplitude is said to be 1.
The curves take 360° to complete ONE cycle or pattern.
We say that there is ONE cycle in 360°.
The Midpoint or the Axis of the function occurs at zero (0).
A Sine curve
• starts at the Midpoint (0),
• goes up to the maximum(+1),
• goes back down to the midpoint (0),
• continues down to the minimum (–1),
• then climbs back up to the midpoint (0).
( 0, 0)
( 90, 1)
(180, 0)
(270, –1)
(360, 1)
A Cosine curve
• starts at the maximum (+1)
• goes down to the midpoint (0),
• continues down to the minimum (–1)
• climbs up to the midpoint (0),
• then goes back up to the maximum (+1)
( 0, 1)
( 90, 0)
(180, –1)
(270, 0)
(360, 1)
To determine the equation of a situation that is modelled by a Sine or Cosine curve, you
need to find the following information:
i.
Amplitude [(Maximum – Minimum) ÷ 2] OR the Radius of the Circle
ii.
Midpoint [(Maximum + Minimum) ÷ 2] (What is the Half-way point of the range of
positions?)
iii.
Compression (How many cycles fit into 360? – That is, 360 ÷ Cycle Length)
iv. Phase Shift (Where is the initial position on the curve (Sine or Cosine) and where
does it go from there. On a “Normal” curve, how much to you have to shift the
curve to ensure that the curve starts at the right point?)
Function will be:
y = (Amplitude) sin [(Compression)x + (Phase Shift)] + (Midpoint)
MCR3UB Unit 09 Day 05
– 1 –
Modelling with Trig Functions
Example 1
Water Depth in a harbour is 21m at high tide and 11 m at low tide. One cycle is
completed every 12 hours. Find an equation for the water depth as a function of time
(t), after low tide.
Amplitude = [(Maximum – Minimum) ÷ 2]
= [(21 – 11) ÷ 2]
= 10 ÷ 2
=5
Midpoint = [(Maximum + Minimum) ÷ 2]
= [(21 + 11) ÷ 2]
= 32 ÷ 2
= 16
Compression (Number of Cycles in 360)
Compression = 360 ÷ 12 = 30
Phase Shift
Starts at Low Tide (Lowest Point of Sine curve)
Lowest point on the curve occurs at –90°.
That point must happen when t = 0
Phase Shift is 90° to the right. → (t – 90)
depth = 5 sin (30t – 90) + 16
depth = 5 sin (30(t – 3)) + 16
Check
At t = 0, tide will be 11 m (Low tide)
At t = 6, tide will be 21 m (High tide)
At t = 12, the tide will be at 11 (Low tide)
At t = 3, tide will be half way or 16 m.
Substitute into the equation.
t=0
t=3
t=6
t = 12
depth = 5 sin (30(0) – 90) + 16 → 5 sin (-90) + 16 → 5 (-1) + 16 → -5 + 16 → 11
depth = 5 sin (30(3) – 90) + 16 → 5 sin (0) + 16 → 5 (0) + 16 → 0 + 16 → 16
depth = 5 sin (30(6) – 90) + 16 → 5 sin (90) + 16 → 5 (1) + 16 → 5 + 16 → 21
depth = 5 sin (30(12) – 90) + 16 → 5 sin (270) + 16 → 5 (-1) + 16 → -5 + 16 → 11
MCR3UB Unit 09 Day 05
– 2 –
Modelling with Trig Functions
Example 2
An object suspended from a spring is oscillating up and down. The distance from the
high point to the low point is 30 cm, and the object takes 4s to complete 5 cycles. For
the first few cycles, the distance from the mean position, d(t) centimetres, with respect
to the time, t seconds, is modelled by a sine function.
Write an equation that describes the distance of the object from its mean position as a
function of time.
Amplitude = [(Maximum – Minimum) ÷ 2]
= [(30) ÷ 2]
(30 is given as the distance from high to low)
= 15
Midpoint = is given as the mean position
Assume that it is zero.
Compression (Number of Cycles in 360)
There are 5 cycles in each 4 second block
A cycle length will be 0.80 seconds.
Compression = 360 ÷ 0.80 = 450
Phase Shift
Starts at the Midpoint position
Assume that it goes UP first
That means there is NO Phase Shift
d(t) = 15 sin (450t)
If the object on the spring went down first, then the equation would be:
d(t) = -15 sin (450t)
Check
At t = 0, Object will be at 0 cm
To effectively measure the distance, you would have to ask where the object is at 0.20,
0.40,, 0.60, and 0.80 seconds. (One complete cycle.)
Substitute into the equation.
t = 0.2
t = 0.4
t = 0.6
t = 0.8
d(0.2) = 15 sin (450(0.2)) → 15 sin (90) → 15(1) → 15
d(0.4) = 15 sin (450(0.4)) → 15 sin (180) → 15 (0) → 0
d(0.6) = 15 sin (450(0.6)) → 15 sin (270) → 15 (-1) → –15
d(0.8) = 15 sin (450(0.8)) → 15 sin (360) → 15 (0) → 0
MCR3UB Unit 09 Day 05
– 3 –
Modelling with Trig Functions
Example 3
A carnival Ferris wheel with a radius of 7m makes one complete revolution every 16
seconds. The bottom of the wheel is 1.5 metres above the ground. Find the equation
that models a person’s height while riding the Ferris wheel.
From the equation, you know that the lowest point is 1.5 metres and the highest is 1.5 +
7 + 7 = 15.5 metres.
Amplitude = [(Maximum – Minimum) ÷ 2]
= [(15.5 – 1.5) ÷ 2]
= [14 ÷ 2]
=7
Midpoint = [(Maximum + Minimum) ÷ 2]
= [(15.5 + 1.5) ÷ 2]
= 17 ÷ 2
= 8.5
(It can also be calculated as low point + Amplitude → 1.5 + 7 = 8.5)
Compression (Number of Cycles in 360)
There is one revolution every 16 seconds. (Cycle length)
Compression = 360 ÷ 16 = 22.5
Phase Shift
Starts at the Low point
Low point on the Sine curve is at -90°. This point needs to be at zero seconds.
Need a Phase Shift of 90° to the Right. (t – 90)
d(t) = 7 sin (22.5t - 90) + 8.5
d(t) = 7 sin [22.5(t – 4)] + 8.5
With a Cosine function, the curve starts at the highest point. It can be inverted
with a minus sign in front of the amplitude value.
d(t) = -7 cos (22.5t) + 8.5
Check
At t = 0, Object will be at 1.5 m. Since one revolution takes 16 seconds, also check at
4, 8, 12 & 16. (One complete cycle.)
Substitute into the equation.
t=0
d(0) = 7 sin (22.5(0) – 90) + 8.5→ 7 sin (–90) + 8.5 → 7(-1) + 8.5 → -7 + 8.5 = 1.5
t=4
d(4) = 7 sin (22.5(4) – 90) + 8.5→ 7 sin (0) + 8.5 → 7(0) + 8.5 → 8.5
t=8
d(8) = 7 sin (22.5(8) – 90) + 8.5→ 7 sin (90) + 8.5 → 7(1) + 8.5 → 7 + 8.5 = 15.5
t = 12 d(12) = 7 sin (22.5(12) – 90) + 8.5→ 7 sin (180) + 8.5 → 7(0) + 8.5 → 8.5
t = 16 d(16) = 7 sin (22.5(16) – 90) + 8.5→ 7 sin (270) + 8.5 → 7(-1) + 8.5 → -7 + 8.5 = 1.5
MCR3UB Unit 09 Day 05
– 4 –
Modelling with Trig Functions
Example 4
A clock pendulum completes one back-and-forth cycle every 2 seconds. The maximum
measure of the angle of the pendulum with the vertical is 12°. Express the measure of
the angle of the pendulum and the vertical as a function of time. Assume the
relationship is a sine function.
From the question, you know that the angle ranges from a –12 to +12°
Amplitude = [(Maximum – Minimum) ÷ 2]
Midpoint = [(Maximum + Minimum) ÷ 2]
Compression (Number of Cycles in 360)
Compression =
Phase Shift??
Starts where on a normal Sine Curve?
Equation
y = (Amplitude) sin [(Compression)x + (Phase Shift)] + (Midpoint)
Check
At t = 0, Object will be at 1.5 m. Since one revolution takes 16 seconds, also check at
4, 8, 12 & 16. (One complete cycle.)
Substitute into the equation.
t=0
t = 0.5
t=1
t = 1.5
t=2
MCR3UB Unit 09 Day 05
– 5 –
Modelling with Trig Functions
Example 5
A carnival Ferris wheel with a radius of 9.5 metres rotates once every 10 seconds. The
bottom of the Ferris wheel is 1.2 metres above the ground. Find the equation of the
sine function that gives a rider’s height above the ground, in metres, as a function of the
time, in seconds, with the rider starting at the bottom of the wheel.
Example 6
The water depth in a harbour is 8m at low tide and 20m at high tide. One cycle is
completed approximately every 12 hours. Find an equation for the water depth, d(t)
metres, as a function of the time, t hours, after high tide, which occurred at 03:00.
MCR3UB Unit 09 Day 05
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Modelling with Trig Functions