The number of hours of​ daylight, H, on day t of any given year​ (on January​ 1, tequals​1) in a particular city can be modeled by the function :

H(t)=11+8.5 sin⁡[2π/365 (t-83)]​

Required:
a. On March 24, the 83rd day of the year, is the spring equinox. Find the number of hours of daylight in the city on this day. (Round to one decimal place as needed.)
b. On June 24, the 175th day of the year, is the summer solstice, the day with the maximum number of hours of daylight. Find the number of hours of daylight in the city on this day. (Round to one decimal place as needed.)
c. On December 24, the 358 day of the year, is the winter solstice, the day with a minimum number of hours of daylight. Find the number of hours of daylight in the city on this day. (Round to one decimal place as needed.)

Respuesta :

Answer:

(a)11 hours

(b)19.5 hours

(c)2.5 hours

Step-by-step explanation:

The number of hours of​ daylight, H, on day t of any given year in a particular city can be modeled by the function :

[tex]H(t)=11+8.5\sin\left(\dfrac{2\pi}{365}(t-83)\right)[/tex]

(a)When t=83

[tex]H(83)=11+8.5\sin\left(\dfrac{2\pi}{365}(83-83)\right)\\=11+8.5\sin 0\\=11$ hours[/tex]

(b)When t=175

[tex]H(175)=11+8.5\sin\left(\dfrac{2\pi}{365}(175-83)\right)\\=11+8.5\sin\left(\dfrac{2\pi}{365}\times 92 \right)\\=11+8.5\\=19.5$ hours[/tex]

(b)When t=358

[tex]H(175)=11+8.5\sin\left(\dfrac{2\pi}{365}(358-83)\right)\\=11+8.5\sin\left(\dfrac{2\pi}{365}\times 275 \right)\\=11-8.5\\=2.5$ hours[/tex]