Trig application problems.

May 2010
I`m having a lot of trouble solving the following question:

The water at a local beach has an average depth of 1 metre at low tide. The average depth of the water at high tide is 8 m. If one cycle takes 12 hours:
    1. Determine the equation of this periodic function using cosine as the base function where 0 time is the beginning of high tide.
    2. What is the depth of the water at 2 am?
    3. Many people dive into the beach from the nearby dock. If the water must be at least 3 m deep to dive safely, between what daylight hours should people dive?
I am really confused about this problem, and dont even know how to start

Any helpful tips will be greatly appreciated.

Thanks! :)
Last edited by a moderator:
Mar 2010
1. The height is given by \(\displaystyle h(t)=h_{avg}+A\cos{\omega{t}}\). Normally there would be a phase variable, but we are given that there is a maximum at t=0, so the phase must be zero. The difference between high and low tide is \(\displaystyle (h_{avg}+A)-(h_{avg}-A)=2A=8-1=7\), so \(\displaystyle A=3.5\) and since the max is 8, \(\displaystyle h_{avg}=8-3.5=4.5\) (I could have also used the min). Since one cycle is 12 hours, \(\displaystyle \omega=\frac{2\pi}{12}=\frac{\pi}{6}\). So the equation is:

\(\displaystyle h(t)=4.5+3.5\cos{\frac{\pi}{6}t}\)

The other two problems just work with this equation. For (2), you set t=2 and find h(t). For (3), you need to know what range of t makes h(t) is greater than or equal to 3.

Post again in this thread if you're still having problems.

- Hollywood