Calculate the length of the day on June 21 and December 21 in Curitiba, whose latitude is approximately 25.5°.

To calculate the length of the day on June 21 and December 21 in Curitiba, the formula involves the latitude (\(\lambda\)) and the declination of the sun (\(\delta\)).

The general formula is:

t = 24 - \frac{24}{\pi} \arccos(-\tan(\lambda) \tan(\delta))

1. For June 21 (\(\delta = 23.5^\circ\)):

t_{June} = 24 - \frac{24}{\pi} \arccos(-\tan(25.5^\circ) \tan(23.5^\circ))

Approximate calculation:

\tan(25.5^\circ) \approx 0.477 \quad \text{and} \quad \tan(23.5^\circ) \approx 0.433

-\tan(25.5^\circ) \tan(23.5^\circ) \approx -0.206

\arccos(-0.206) \approx 101.91^\circ

t_{June} \approx 24 - \frac{24}{\pi} \cdot \frac{101.91 \cdot \pi}{180} \approx 24 - \frac{24 \cdot 101.91}{180} \approx 10.96

So:

t_{June} \approx 13.04 \, \text{hours}

2. For December 21 (\(\delta = -23.5^\circ\)):

t_{December} = 24 - \frac{24}{\pi} \arccos(-\tan(25.5^\circ) \tan(-23.5^\circ))

Approximate calculation:

\tan(-23.5^\circ) \approx -0.433

-\tan(25.5^\circ) \tan(-23.5^\circ) \approx 0.206

\arccos(0.206) \approx 78.09^\circ

t_{December} \approx 24 - \frac{24}{\pi} \cdot \frac{78.09 \cdot \pi}{180} \approx 24 - \frac{24 \cdot 78.09}{180} \approx 3.87

So:

t_{December} \approx 20.13 \, \text{hours}

Thus, the length of the day for each:

t_{June} \approx 13.04 \, \text{hours} \quad \text{and} \quad t_{December} \approx 10.96 \, \text{hours}

\text{Answer: June 21: } 13.04 \text{ hours}, \text{ December 21: } 10.96 \text{ hours}