# Thread: How would you solve this problem?

1. ## How would you solve this problem?

Because of the tide, the depth of the water in a harbor is modeled by the equation d=-3cos(pi/6)t + 6, where d represents the depth of the water in meters and t represents the number of hours after midnight/ (i.e. t=0 means midnight, t=1 means 1 A.M., and so on.)

b) Surfing is allowed between 8 A.M. (08:00 hrs) and 7 P.M. (19:00 hrs), but only when the depth of the water is 6m or more. For how many hours is surfing allowed in one day?
The answer is 5 hours and I had to graph 2 cycles of the function to find that out. Are there any other approaches or methods to solving this other than drawing?

Thank you in advance.

2. ## Re: How would you solve this problem?

Determine when $d = 6$.
$6 = -3\cos\left(\dfrac{\pi}{6} t\right)+6$
$\cos \left(\dfrac{\pi}{6} t\right) = 0$

Cosine is zero at $\dfrac{(2n+1)\pi}{2}$ for all integers $n$. So, $\dfrac{\pi}{6} t = \dfrac{(2n+1)\pi}{2}$. Solving for $t$, we find $t = 3(2n+1) = 6n+3$. So, in a single day, $d=6$ at $t=3, 9, 15, 21$. Next, consider intervals when $d\ge 6$. We don't care about $(0,3)$ or $(21,24)$. We just care about $(3,9),(9,15),(15,21)$. Plug in one value for $t$ in each interval and if $d>6$ at that value, that will be the case for the whole interval. Similarly for $d<6$.

So, at $t=6$, $d = -3\cos(\pi)+6 = 9>6$, so $d>6$ on the whole interval $(3,9)$.

Next, try $t = 12$: $d = -3\cos(2\pi)+6 = 3<6$, so $d<6$ on the whole interval $(9,15)$.

Next, try $t=18$: $d = -3\cos(3\pi)+6 = 9>6$, so $d>6$ on the whole interval $(15,21)$.

So, among the intervals when surfing is allowed, $[8,9]\cup [15,19]$ are the only intervals that allow for surfing. So, it is allowed for 5 hours, as you found.