1. ## Application of trig

The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a
particular day is given by D(t) = 10 + 3 sin(π t/6), 0 ≤ t ≤ 24.
a) Find the value of t for which D(t) ≥ 8.5.
b) Boats which need a depth of w metres are permitted to enter the harbour only if the
depth of the water at the entrance is at least w metres for a continuous period of 1 hour.
Find, correct to 1 decimal place, the largest value of w which satisfies this condition.

i was able to sketch the graph, just having trouble with how you figure out D(t) ≥ 8.5

the answer in the book is {t : D(t) ≥ 8.5} = {t : 0 ≤ t ≤ 7} ∪
{t : 11 ≤ t ≤ 19} ∪ {t : 23 ≤ t ≤ 24}
this makes sense, i just dont know how to figure it out algebraically

The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a
particular day is given by D(t) = 10 + 3 sin(π t/6), 0 ≤ t ≤ 24.
a) Find the value of t for which D(t) ≥ 8.5.
b) Boats which need a depth of w metres are permitted to enter the harbour only if the
depth of the water at the entrance is at least w metres for a continuous period of 1 hour.
Find, correct to 1 decimal place, the largest value of w which satisfies this condition.

i was able to sketch the graph, just having trouble with how you figure out D(t) ≥ 8.5

the answer in the book is {t : D(t) ≥ 8.5} = {t : 0 ≤ t ≤ 7} ∪
{t : 11 ≤ t ≤ 19} ∪ {t : 23 ≤ t ≤ 24}
this makes sense, i just dont know how to figure it out algebraically
Here's my quick sketch of the graph (see attachment). Clearly when $t = 0, d>8.5$, and we can find the values for which $t = 8.5$ by solving the equation $8.5 = 10+3\sin\Big(\frac{\pi t}{6}\Big)$.

So: $8.5 = 10+3\sin\Big(\frac{\pi t}{6}\Big)$

$\Rightarrow 8.5-10 = 3\sin\Big(\frac{\pi t}{6}\Big)$

$\Rightarrow 3\sin\Big(\frac{\pi t}{6}\Big)=-1.5$

$\Rightarrow \sin\Big(\frac{\pi t}{6}\Big)=-0.5$

$\Rightarrow \frac{\pi t}{6}= \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{19\pi}{6},\frac{23\pi}{6} ...$

(Is this where you were getting stuck? Look at the point moving around the unit circle, noting that $\sin\Big(\frac{\pi}{6}\Big)=0.5$
. So the first value of $\frac{\pi t}{6}$ for which $\sin\Big(\frac{\pi t}{6}\Big) = -0.5$ is $\frac{\pi t}{6}=\frac{7\pi}{6,}$, etc)

$\Rightarrow t = 7, 11, 19,23$

... and using the sketch-graph, the answers then follow.