The depth of water, in metres, at a certain point in a harbour varies with the tide and can be modelled by a function of the form $$f(t) = a + b \cos ct$$ where $t$ is the time in hours from the first high tide on a particular Saturday and $a$, $b$, and $c$ are constants - Leaving Cert Mathematics - Question 9 - 2017

From the high tide of 5.5 m and low tide of 1.7 m:

1. Average depth (value of $a$): $a = \frac{5.5 + 1.7}{2} = 3.6$
2. Amplitude (value of $b$): $b = \frac{5.5 - 1.7}{2} = 1.9$

The period of the tide is from one high tide to the next which is 12 hours.

The formula for the period is given by: $\text{Period} = \frac{2\pi}{c}$

Setting the period to 12 hours: $12 = \frac{2\pi}{c}$

Rearranging this gives: $c = \frac{2\pi}{12} = \frac{\pi}{6} \approx 0.5$ (correct to 1 decimal place).

We set the equation equal to 5.2 m:

$5.2 = 3.6 + 1.9 \cos(0.5t)$

To find $t$:

1. Rearranging gives: $1.9 \cos(0.5t) = 5.2 - 3.6$ $1.9 \cos(0.5t) = 1.6$ $\cos(0.5t) = \frac{1.6}{1.9} \approx 0.8421$

2. Taking arccosine: $0.5t = \cos^{-1}(0.8421) \approx 0.579$ $t \approx 1.158 \text{ hours} = 1 ext{ hour } 9 ext{ minutes}$

For other possible solutions on the cosine function, repeat for $t$:

• Solutions would yield:
  • $14:34 - 1:{09}$
  • $13:26$ and $15:42$