outside temperature over a day can be modelled using a sine or cosine function. suppose you know the high…

Explanation:

Step1: Find the amplitude

The amplitude $A$ is half the difference between the high and low temperatures. So $A=\frac{64 - 30}{2}=\frac{34}{2}=17$.

Step2: Find the vertical shift

The vertical - shift $k$ is the average of the high and low temperatures. So $k=\frac{64 + 30}{2}=\frac{94}{2}=47$.

Step3: Determine the period

The period of a day is $T = 24$ hours. For a sine or cosine function $y = A\sin(B(t - h))+k$ or $y=A\cos(B(t - h))+k$, the formula for the period is $T=\frac{2\pi}{B}$. Since $T = 24$, then $B=\frac{2\pi}{24}=\frac{\pi}{12}$.

Step4: Determine the phase - shift

The low temperature occurs at $t = 6$. For a cosine function $y = A\cos(B(t - h))+k$, the minimum of the cosine function $y=\cos(x)$ occurs at $x=\pi$. We want $B(t - h)=\pi$ when $t = 6$. Substituting $B=\frac{\pi}{12}$, we have $\frac{\pi}{12}(6 - h)=\pi$. Solving for $h$: First, divide both sides of the equation $\frac{\pi}{12}(6 - h)=\pi$ by $\pi$ to get $\frac{1}{12}(6 - h)=1$. Then multiply both sides by 12 to obtain $6 - h=12$, and $h=- 6$.

Step5: Write the equation

Using the cosine function form $D(t)=A\cos(B(t - h))+k$, substituting $A = 17$, $B=\frac{\pi}{12}$, $h=-6$, and $k = 47$, we get $D(t)=17\cos\left(\frac{\pi}{12}(t + 6)\right)+47$.

Answer:

$17\cos\left(\frac{\pi}{12}(t + 6)\right)+47$