Question

Model equation for tide: h=2 cos⁡(π/6 t-2π/3)+4
Given the above, a large boat needs at least 4 meters of water to secure it at the end of the pier. Determine what span of time after noon, including both a starting and ending time, the boat can first safely be secured, justifying your answer.
So far, I have gotten this, but I'm stuck!
4=2 cos⁡(π/6 t-2π/3)+4
0=2 cos⁡(π/6 t-2π/3)
0=cos⁡(π/6 t-2π/3)
0=cos⁡(π/6 t) cos⁡(2π/3)+sin⁡(π/6 t) sin⁡(2π/3)

Am I even remotely on the right track??? Can someone PLEASE help?!?

Answer 1

Answer on Question #44762 – Math - Trigonometry

Model equation for tide: $h = 2 \cos \left( \frac{\pi}{6} t - \frac{2\pi}{3} \right) + 4$

Given the above, a large boat needs at least 4 meters of water to secure it at the end of the pier. Determine what span of time after noon, including both a starting and ending time, the boat can first safely be secured, justifying your answer.

So far, I have gotten this, but I'm stuck!

$$\begin{array}{l} 4 = 2 \cos \left( \frac{\pi}{6} t - \frac{2\pi}{3} \right) + 4 \\ 0 = 2 \cos \left( \frac{\pi}{6} t - \frac{2\pi}{3} \right) \\ 0 = \cos \left( \frac{\pi}{6} t - \frac{2\pi}{3} \right) \\ 0 = \cos \left( \frac{\pi}{6} t \right) \cos \left( \frac{2\pi}{3} \right) + \sin \left( \frac{\pi}{6} t \right) \sin \left( \frac{2\pi}{3} \right) \\ \end{array}$$

Answer.

A large boat needs at least 4 meters of water, so h must be not less than 4.

So our first goal is to solve $2 \cos \left( \frac{\pi}{6} t - \frac{2\pi}{3} \right) + 4 = 4$ or

$$\cos \left( \frac{\pi}{6} t - \frac{2\pi}{3} \right) = 0 \text{ for 't'}$$ $$\frac{\pi}{6} t - \frac{2\pi}{3} = \frac{\pi}{2} + k\pi, \text{ is integer};$$ $$\frac{\pi}{6} t = \frac{3\pi}{6} + \frac{4\pi}{6} + k\pi; \quad \frac{\pi}{6} t = \frac{7\pi}{6} + k\pi; \quad t = 7 + 6k, \text{ is integer}.$$

We should get the solutions $t = 1$, $t = 7$, $t = 13$, $t = 19$ etc. (i.e. add 6 hours to each solution to get another one).

Since we only care about hours after noon, this means that $t \geq 12$.

So at $t = 13$ (1 pm), the height is 4 meters. As time goes on, the water rises. When it reaches $t = 19$ (7 pm), the water comes back to 4 meters, which then starts to dip below that mark. So the water is at least 4 meters high from the hours of 1 pm to 7 pm.

The water will eventually come back up, but it will reach the 4 meter mark at 1 am the next day.

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