outside temperature over a day can be modeled as a sinusoidal function. suppose you know the temperature is…

outside temperature over a day can be modeled as a sinusoidal function. suppose you know the temperature is 45 degrees at midnight and the high and low temperature during the day are 63 and 27 degrees, respectively. assuming t is the number of hours since midnight, find an equation for the temperature, d, in terms of t.\nd(t) = \nquestion help: video 1 video 2\nsubmit question

outside temperature over a day can be modeled as a sinusoidal function. suppose you know the temperature is 45 degrees at midnight and the high and low temperature during the day are 63 and 27 degrees, respectively. assuming t is the number of hours since midnight, find an equation for the temperature, d, in terms of t.\nd(t) = \nquestion help: video 1 video 2\nsubmit question

Answer

Explanation:

Step1: Find the amplitude (A)

The amplitude (A=\frac{\text{High}-\text{Low}}{2}). Given high ( = 63) and low (=27), then (A=\frac{63 - 27}{2}=\frac{36}{2}=18).

Step2: Find the vertical shift (B)

The vertical shift (B=\frac{\text{High}+\text{Low}}{2}). So (B=\frac{63 + 27}{2}=\frac{90}{2}=45).

Step3: Find the period (T)

The period of a day - based sinusoidal function (where (t) is hours since midnight) is (T = 24). The formula for the angular frequency (\omega=\frac{2\pi}{T}), so (\omega=\frac{2\pi}{24}=\frac{\pi}{12}).

Step4: Determine the phase shift (C)

Since the function has a value of (45) at (t = 0) (midnight) and (B = 45), and we can use the cosine function (because (y = A\cos(\omega t)+B) has a maximum at (t = 0) when (C = 0) for the form (y=A\cos(\omega(t - C))+B)). The general form of a sinusoidal function is (D(t)=A\cos(\omega t)+B).

Substituting (A = 18), (\omega=\frac{\pi}{12}), and (B = 45) into the formula, we get (D(t)=18\cos\left(\frac{\pi}{12}t\right)+45).

Answer:

(D(t)=18\cos\left(\frac{\pi}{12}t\right)+45)