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 85 degrees at midnight and the high and low temperature during the day are 98 and 72 degrees, respectively. assuming t is the number of hours since midnight, find an equation for the temperature, d, in terms of t. d(t)=13 sin(πt/12)+85 × question help: video submit question

outside temperature over a day can be modeled as a sinusoidal function. suppose you know the temperature is 85 degrees at midnight and the high and low temperature during the day are 98 and 72 degrees, respectively. assuming t is the number of hours since midnight, find an equation for the temperature, d, in terms of t. d(t)=13 sin(πt/12)+85 × question help: video submit question

Answer

Explanation:

Step1: Find the amplitude

The amplitude $A$ is half the difference between the high - and low - temperatures. $A=\frac{98 - 72}{2}=\frac{26}{2}=13$.

Step2: Find the vertical shift

The vertical shift $D_0$ is the average of the high - and low - temperatures. $D_0=\frac{98 + 72}{2}=\frac{170}{2}=85$.

Step3: Find the period and angular frequency

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

Step4: Write the sinusoidal function

Since the temperature at $t = 0$ (midnight) is the average temperature, the general form of a sinusoidal function is $D(t)=A\sin(\omega t)+D_0$. Substituting $A = 13$, $\omega=\frac{\pi}{12}$, and $D_0 = 85$ we get $D(t)=13\sin(\frac{\pi t}{12})+85$. But we need to adjust the phase. Since the temperature is at the average value at $t = 0$, and we want a sine - type function that starts at the average value, the correct form considering the standard sine function properties is $D(t)=13\cos(\frac{\pi t}{12})+85$.

Answer:

$D(t)=13\cos(\frac{\pi t}{12})+85$