Question

Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. at which time the water is 16 feet above the height of low tide. Low tides occur 6 hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally.

(a) Find a formula for the function y = h(t) that computes the height of the tide above low tide at time t, where t indicates the number of hours after midnight. (In other words, y = 0 corresponds to low tide.)


h(t) = _____________________


(b) What is the tide height at 11:00 a.m.?


__________ ft

Answer 1

(a)

`\displaystyle h(t)=y(t)=8\left [ 1+sin\left ( (\pi)/(6)t+(\pi)/(3) \right ) \right ]`

(b) Height at 11:00 a.m. = 12 feet

Step-by-step explanation:

Sinusoidal Function

It refers to a mathematical curve with a smooth and periodic oscillation. Its name comes from the sine function but it can be a cosine function too. They only differ by the phase angle of 90 degrees or
`\pi/2` radians.

The sine function is characterized by

The minimum value is -1 when the argument is
`3\pi/2` radians or 270 degrees

The maximum value is 1 when the argument is
`\pi/2` or 90 degrees

It completes a full cycle in
`2\pi` radians (or 360 degrees)

It's zero at 0 and
`\pi`

It repeats itself along infinite cycles with the same characteristics

We need to model the height of the tide above low tide at time t, t expressed in hours from midnight

We know the following data

At 1:00 and 13:00, the tide is high at 16 feet above the low tide, assumed to be 0 m

At 7:00 and 19:00, the tide is low at 0 m.

(a) The general sine function is expressed as

`y(t)=Asin(wt+\phi)+B\ \ .........[1]`

Where A is the amplitude, w is the angular frequency,
`\phi` is the phase, B is the vertical shift

We know that at t=1, the sine must be at max (value =1) and at t=7 it must be at min (value=-1). Replacing in [1]

`y(1)=A(1)+B=A+B=16`

`y(7)=A(-1)+B=A-B=0`

We get these equations

`A+B=16`

`A=B`

`Solving, A=8,\ B=8`

Using the same data as before, when t=1 the argument of the sine must be
`\pi/2`, so it reaches it max, and when t=7, the argument must be
`3\pi/2`. Then:

`w+\phi=\pi/2`

`7w+\phi=3\pi/2`

Solving, we get

`w=\pi/6`

`\phi=\pi/3`

The function is complete now

`\displaystyle y(t)=8sin\left ( (\pi)/(6)t+(\pi)/(3) \right )+8`

Factoring

`\displaystyle h(t)=y(t)=8\left [ 1+sin\left ( (\pi)/(6)t+(\pi)/(3) \right ) \right ]`

(b) We need to find h when t=11

`\displaystyle h(11)=8\left [ 1+sin\left ( (\pi)/(6)11+(\pi)/(3) \right ) \right ]`

`\displaystyle h(11)=8\left [ 1+sin\left ( (13\pi)/(6) \right ) \right ]`

`h(11)=8\left (1+0.5 \right)`

`h(11)=12\ feet`