Answer:
The temperature first reach 63 degrees at 9:09 AM.
Step-by-step explanation:
We can write a model for the temperature during the day as:
[tex]T(t)=Asin(\omega t+\phi)+B[/tex]
The minimum temperature is 56:
[tex]T=A*(-1)+B=56\\\\B=56+A[/tex]
The maximum temperature is 74:
[tex]T=A*1+B=A+(56+A)=74\\\\2A=(74-56)=18\\\\A=9\\\\B=56+A=65[/tex]
The cicle repeats daily, so T(0)=T(24). Other way to calculate it is that:
[tex]\omega\cdot 24=2\pi\\\\\omega=\pi/12[/tex]
(2 pi is one cycle for the sin function).
The average temperature occurs when
[tex]sin( \frac{\pi}{12} t+\phi)=0[/tex]
Then we, we calculate this for the 10 AM (t=10)
[tex]sin( \frac{\pi}{12} \cdot 10+\phi)=0\\\\\frac{\pi}{12} \cdot 10+\phi=2\pi\\\\ \phi=\frac{7}{6} \pi[/tex]
Then, we have all parameters calculated and the model is:
[tex]T(t)=9sin(\frac{\pi}{12}\cdot t+\frac{7\pi}{6} )+65[/tex]
We hace to calculate how many hours after midnight, to two decimal places, does the temperature first reach 63 degrees
[tex]T(t)=9sin(\frac{\pi}{12}\cdot t+\frac{7\pi}{6} )+65 = 63\\\\9sin(\frac{\pi}{12}\cdot t+\frac{7\pi}{6} ) = 63-65\\\\sin(\frac{\pi}{12}\cdot t+\frac{7\pi}{6} ) = -2/9\\\\\frac{\pi}{12}\cdot t+\frac{7\pi}{6}=arcsin(-2/9)=6.06\\\\t=(6.06-7\pi/6)/(\pi/12)\\\\t= 9.144[/tex]
The value t=9.144 is equal to 9:09 AM.
