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

Explanation:

Step1: Determine the amplitude and mid - line

The amplitude $A$ of the sinusoidal function is given by $\frac{\text{max}-\text{min}}{2}$. Here, $\text{max} = 64$ and $\text{min}=36$, so $A=\frac{64 - 36}{2}=\frac{28}{2}=14$. The mid - line $D$ (average value) is $\frac{64 + 36}{2}=50$.

Step2: Write the general form of the sinusoidal function

The general form of a sinusoidal function is $y = A\sin(B(x - C))+D$. The period of a daily temperature function is $T = 24$ hours, and since $T=\frac{2\pi}{B}$, then $B=\frac{2\pi}{24}=\frac{\pi}{12}$. The average daily temperature first occurs at 10 AM (10 hours after midnight), so $C = 10$. The function is $y=14\sin\left(\frac{\pi}{12}(x - 10)\right)+50$.

Step3: Set up the equation to solve for $x$

We want to find $x$ when $y = 46$. So we set up the equation $46=14\sin\left(\frac{\pi}{12}(x - 10)\right)+50$. First, subtract 50 from both sides: $46−50=14\sin\left(\frac{\pi}{12}(x - 10)\right)$, which gives $- 4=14\sin\left(\frac{\pi}{12}(x - 10)\right)$. Then $\sin\left(\frac{\pi}{12}(x - 10)\right)=-\frac{4}{14}=-\frac{2}{7}$.

Step4: Solve for $x$

We know that if $\sin\theta=-\frac{2}{7}$, then $\theta=\sin^{-1}\left(-\frac{2}{7}\right)+2k\pi$ or $\theta=\pi-\sin^{-1}\left(-\frac{2}{7}\right)+2k\pi$. We want the first non - negative solution for $x$. $\frac{\pi}{12}(x - 10)=\sin^{-1}\left(-\frac{2}{7}\right)$. $x-10=\frac{12}{\pi}\sin^{-1}\left(-\frac{2}{7}\right)$. $x = 10+\frac{12}{\pi}\sin^{-1}\left(-\frac{2}{7}\right)$. Using a calculator, $\sin^{-1}\left(-\frac{2}{7}\right)\approx - 0.28975$. $x=10+\frac{12}{\pi}\times(- 0.28975)\approx10 - 1.10=8.90$.

Answer:

$8.90$