Sine function-word problem

The tide in a local coastal community can be modeled using a sine function. Starting at noon, the tide is at its "average" height of 3 meters measured on a pole located off of the shore. 5 hours later is high tide with the tide at a height of 5 meters measured at the same pole. 15 hours after noon is low tide with the tide at a height of 1 meter measured at the same pole. Use a Sine function to describe the tides motion.

1. The tide in a local costal community can be modelled using a sine function. Starting at noon, the tide is at its "average" height of 3 metres measured on a pole located off of the shore. 5 hours later is high tide with the tide at a height of 5 metres measured at the same pole. 15 hours after noon is low tide with the tide at a height of 1 metre measured at the same pole. Use this information to model the tide motion using a sine function. Show all work. [4]

The equation is of the form y = A sin(B(x - C)) + D, where A is the amplitude, B affects the period, C is the horizontal shift, and D is the vertical shift.

The vertical shift is 3 metres, so D = 3.

Since the water went from an "average" depth of 3 metres to its high tide of 5 metres, the amplitude is 2 metres. So, A = 2.

The normal period of a sine graph is 360˚ or 2π radians. Since it took 10 hours to go from high tide to low tide, the entire period is 20 hours. This means the new period is 20. Since B = 360˚/new period, then b = 360/20 = 18.

There is no horizontal shift, so C = 0.


y = 2 sin(18x) + 3 <==ANSWER

sinusoidal axis: 3 meters
period/4=5hours (it took 5 hours to go from sinusoidal axis to max)
period=20hours
amp=2

2π/20=HS
π/10=HS
if noon = 0 hours

h(t)=2sin(π/10(t))+3

t y
-5 1
0 3
5 5
10 3
15 1
20 3
25 5
Period = 2pi / 20 = pi/10
Amplitude = 2
Vertical translation = 3
y = 2 sin (pi/10 *t ) +3 where t = 0 @ noon , t = 1 @ 5 pm , t = 2 @ 10pm etc