# Trig application problems.

#### spoc21

Hi,
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?

Any helpful tips will be greatly appreciated.

Thanks! Last edited by a moderator:

#### hollywood

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