question 11 outside temperature over a day can be modeled as a sinusoidal function. suppose you know the…

Explanation:

Step1: Determine the amplitude

The amplitude $A$ is half the difference between the high and low temperatures. So $A=\frac{75 - 55}{2}=\frac{20}{2}=10$.

Step2: Determine the vertical - shift

The vertical - shift $k$ is the average of the high and low temperatures. So $k=\frac{75 + 55}{2}=\frac{130}{2}=65$.

Step3: Determine the period

The period $P$ of a daily temperature cycle is 24 hours. The formula for the angular frequency $\omega=\frac{2\pi}{P}$, so $\omega=\frac{2\pi}{24}=\frac{\pi}{12}$.

Step4: Determine the phase - shift

Since the temperature is 65 degrees at midnight ($t = 0$), and the general form of a sinusoidal function is $D(t)=A\sin(\omega(t - h))+k$ or $D(t)=A\cos(\omega(t - h))+k$. Using the cosine function (because at $t = 0$, a cosine function $y = A\cos(\omega t)+k$ starts at its mid - value when there is no phase - shift), and with no phase - shift ($h = 0$), the function is $D(t)=10\cos(\frac{\pi}{12}t)+65$.

Answer:

$D(t)=10\cos(\frac{\pi}{12}t)+65$