Question

Ocean tides can be modeled by a sinusoidal function. Suppose that there is a low and high tide every 12 hours, and that high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. with the low tides 6 hours after high tides. Also suppose that the water level at high tide is 10 ft above the water level at low tide. a. Find a formula for the function that computes the height of the tide above low tide at time . (In other words, corresponds to low tide.) b. What is the height of the tide at 11:00 a.m.?

Answer 1

(a)
`y=5sin((\pi)/(6)(x+2))+5`

(b) 7.5ft above the low tide.

Explanation:

(a) To find the function that computes the height of the tide, you need to select the form of the sinusoidal function. For example, use the form:

`y=Asin(B(x-C))+D`

Where A is the amplitude, B the frequency, C the phase shift and D the vertical shift.

The amplitude is half the distance between the highest and the lowest tide:

`A=10/2=5ft`

The frequency is related to the period T by:

`B=(2\pi)/(T)`

The period is 12 hours, then

`B=(2\pi)/(12)=(\pi)/(6)`

The high tide is at 1:00 a.m. and 1:00 p.m. , this is the moment when
`sin(B(x-C))=1`, if
`sin((\pi)/(2))=1` then
`B(x-C)` must be equal to
`(\pi)/(2)` when
`x=1`:

`B(x-C)=(\pi)/(2)\\(\pi)/(6)(1-C)=(\pi)/(2)\\(1)/(6)(1-C)=(1)/(2)\\(1-C)=(6)/(2)\\-C=3-1\\C=-2`

The vertical shift is the sum of the lowest value, the height of the low tide (
`lt`) and the amplitude:

`D=5+lt`

The function is:

`y=5sin((\pi)/(6) (x+2))+5+lt`

Because the function must be the height above low tide height, subtract this heigh from the function:

`y=5sin((\pi)/(6) (x+2))+5+lt-lt`

`y=5sin((\pi)/(6) (x+2))+5`

(b) Use x=11 in the function

`y=5sin((\pi)/(6) (11+2))+5=2.5+5=7.5ft` above the low tide.