Question

Question #f8d27

Answer 1

Given that the water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 a.m. and high tide occurs at 3:30 p.m

So duration of half cycle is

`9:15amto3:30p m=6hr15min=6.25hr`

And duration of full cycle is `=2xx6.25hr=12.5hr`

At low tide height of water level `=12"in"`

At high tide height of water level `=52"in"`

So equilibrium height of water level `(12+52)/2 "in"=32"in"`

The amplitude or the maximum change from equilibrium level will be `=52-32=20" in"`

So the cosine function that models the variation in inches above and below the water level as a function of time is given by

`f(t)=32-20cos ((2pit)/12.5)` in
The following is the graphical representation of the variation of water level taking `9:15`am as zero hour

Answer 2

`f(t) = 32cos(t*(4pi)/25-pi)+20`

Explanation:

We know that the cosine function has a range of `[-1;1]`. By looking at the function we can see that this means a difference of 2 on the y-axis (from highest to lowest) over period of `pi` on the x-axis.

I'm asuming that in order to have anything below a water level, 12 inches at low tide refers to 12 inches below water level, and similarly 52 inches at high tide refers to above water level.
This means the water varies a total of `12-(-52) = 64" inches"`.

If a standard cosine function has a difference of 2, then 32 times that cosine function must have a difference of 64.
Our new cosine function `32*cos(t)` has a range of `[-32;32]`, we can alter this to get:
`f(t) = 32*cos(t)+20` , has a range of `[-12;52]`

The next step would be to get the timing right. A standard cosine function varies between its highest and lowest point over a period of `pi` on the x-axis. We want this function to do the same, but over a period of 6 hours and 15 min, that is (6+`1/4`)hours. (The time will be measured in hours).

When we look at `cos(u) = -1 , u = pi` , we want to find a constant (a), that makes our period of 6 hours and 15 min. equal to `pi`.
So, `pi = (6+1/4)a = (25/4)*a iff a = (4pi)/25`

Pluging this into our cosine function, we get this:
`f(t) = 32cos(t*(4pi)/25)+20`

Let's try it:
`t = 0` will be the starting point at 9:15 am, `t = 6+1/4` will be 6 hours and 15 min. later at 3:30 pm.
`f(0) = 32*cos(0*(4pi)/25)+20 = 52`

`f(6+1/4) = 32*cos((6+1/4)*(4pi)/25)+20 = -12`

So it's not quite there yet, as it starts at its high point, which is not what we want. To counteract this, we need to shift the whole function `pi` lenght on the x-axis. To do this, we just add `(-pi)` inside of `cos(u)`. This will shift it backwards so that its minimum point will be on `t = 0`

The final function will be:
`f(t) = 32cos(t*(4pi)/25-pi)+20`

`f(0) = -12`
`f(6+1/4) = 52`

Here's a graph of the function `f(t) = 32cos(t*(4pi)/25-pi)+20` graph{32cos(x*(4pi)/25-pi)+20 [-69.4, 90.6, -18.2, 61.8]}