The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function $$f(t) = 12 - 25 + 4 imes 75 imes ext{sin} \left( \frac{2 \pi t}{365} \right)$$, where $t$ is the number of days after March 21$^{st}$ (76 days after March 21$^{st}$) - Leaving Cert Mathematics - Question 9 - 2015

To find when the length of the day is approximately 15 hours, we set:

$f(t) = 15$

This gives:

$12 - 25 + 4 \times 75 \times \text{sin} \left( \frac{2 \pi t}{365} \right) = 15$

Solving for the sine part:

$\text{sin} \left( \frac{2 \pi t}{365} \right) = \frac{15 + 25 - 12}{300} = 0.1$

Consequently:

$\frac{2 \pi t}{365} = 0.5789471$

Then:

$t \approx 35.87$

Thus, the date is approximately April 26$^{th}$.

To find the derivative $f^{'}(t)$, we differentiate the function:

$f(t) = 12 - 25 + 4 imes 75 \text{sin} \left( \frac{2 \pi t}{365} \right)$

Thus,

$f^{'}(t) = 0 + 0 + 4 \times 75 \times \text{cos} \left( \frac{2 \pi t}{365} \right) \times \frac{2 \pi}{365}$

Simplifying gives:

$f^{'}(t) = \frac{9.5\pi}{365} \text{cos} \left( \frac{2 \pi t}{365} \right)$

The maximum value of $f(t)$ occurs when:

$\text{sin} \left( \frac{2 \pi t}{365} \right) = 1$

So,

$t = 12 - 25 + 4 \times 75 \times 1$

= 12 - 25 + 300 = 17 \text{ hours.}$$

To find the average length of the day, we set up the integral:

$\text{Average} = \frac{1}{b - a} \int_{a}^{b} f(x) dx$

where $a = 0, b = 184$:

$= \frac{1}{184} \int_{0}^{184} \left[ 12 - 25 + 4 \times 75 \times \text{sin} \left( \frac{2 \pi x}{365} \right) \right] dx$

Evaluating gives:

$= \frac{1}{184} \left[ 2254 - 275 \times (0) - 777 \right] = 15 \text{ hours } 15 ext{ minutes.}$