Question 457399
<pre>
{{{drawing(400,400,-2,24,-5,70) graph(400,400,-2,24,-5,70,

(sqrt(x)/sqrt(x))*(35-28cos((pi/6.2)x)),3.5 )}}}

We find the period of that function as that is the number of
hours from the time the ocean is a certain number of feet
deep until it is that same number of feet deep again. 

The period of that function is {{{2pi}}}{{{"÷"}}}{{{pi/6.2}}} or 
{{{2pi*expr(6.2/pi)}}} = 12.4 hours.

Low tide is when t=0 or midnight or 12:00AM. So there will
be another low tide again at 12.4 hours later or t=12.4 at 
12:24PM.  Half-way between those two low tides there will be a 
high tide and that will be at t=6.2 hours after midninght, at 
6:12AM.  12.4 hours later there will be another high tide at 
t=18.6 hours after midnight and that will be at 6:36PM.

The low tide is when t=0, substituting t=0 in the equation

d = 35 - 28cos(pi/6.2)t
d = 35-28cos(pi/6.2)0
d = 35-28cos(0)
d = 35-28(1)
d - 35-28
d = 7

So the lowest possible depth is 7 feet, so it will never be
3.5 feet there.  The green line on the graph is at 3.5 feet
and we see that the ocean there never gets that shallow.

So there is no solution to the b part.

Edwin</pre>