The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function $$f(t) = 12.25 + 4.75 ext{sin} \left(\frac{2\pi}{365} t\right),$$ where $t$ is the number of days after March 21$^{st}$ and $\left(\frac{2\pi}{365}\right)$ is expressed in radians - Leaving Cert Mathematics - Question 9 - 2015

To find a date with a day length of 15 hours, we set:

$15 = 12.25 + 4.75 \sin \left(\frac{2\pi}{365} t\right)$

Rearranging gives:

$4.75 \sin \left(\frac{2\pi}{365} t\right) = 15 - 12.25 = 2.75$

Thus,

$\sin \left(\frac{2\pi}{365} t\right) = \frac{2.75}{4.75} \approx 0.578947$

Now, find the angle:

$\frac{2\pi}{365} t = \arcsin(0.578947)$

After calculating and converting to a day, we find that it occurs around April 26$^{th}$.

Let's differentiate $f(t)$ with respect to $t$:

$f'(t) = 0 + 4.75 \cdot \cos \left(\frac{2\pi}{365} t\right) \cdot \frac{2\pi}{365}$

This yields:

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

The longest day occurs when (\sin \left(\frac{2\pi}{365} t \right) = 1):

$f(t) = 12.25 + 4.75 (1) = 17 \text{ hours}.$

The average length of the day can be found using the formula:

$\text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt$

Where $a = 0$ and $b = 184$:

$= \frac{1}{184} \int_{0}^{184} \left[12.25 + 4.75 \sin \left(\frac{2\pi}{365} t \right)\right] dt$

After computing this integral, we find:

$\approx 15 \, \text{hours} \, 15 \, \text{minutes}.$