9. at one time, maple leaf village (which no longer exists) had north americas largest ferris wheel. the…

Explanation:

Step1: Determine the radius and center - height

The diameter of the Ferris - wheel is $d = 56$ m, so the radius $r=28$ m. The center of the Ferris - wheel is at a height of $h = 28 + 0.5=28.5$ m above the ground.

Step2: Write the general form of the sinusoidal function

The general form of a sinusoidal function for height $y$ as a function of time $t$ is $y = A\sin(B(t - C))+D$. Here, $A$ is the amplitude, $B$ is related to the period, $C$ is the phase - shift, and $D$ is the vertical shift. The amplitude $A = 28$ (radius of the Ferris - wheel), the period $T = 2.5$ min. Since $B=\frac{2\pi}{T}$, then $B=\frac{2\pi}{2.5}=\frac{4\pi}{5}$. The rider starts at $y = 0.5$ m at $t = 0$, so the phase - shift $C = 0$ and the vertical shift $D = 28.5$. The function is $y=28\sin(\frac{4\pi}{5}t)+28.5$.

Step3: Find the time intervals when $y>50$

Set $y>50$, so $28\sin(\frac{4\pi}{5}t)+28.5>50$. First, isolate the sine function: $28\sin(\frac{4\pi}{5}t)>50 - 28.5=21.5$. Then $\sin(\frac{4\pi}{5}t)>\frac{21.5}{28}\approx0.7679$. We know that $\sin^{-1}(0.7679)\approx0.879$ and $\pi-\sin^{-1}(0.7679)\approx2.263$. So, $\frac{4\pi}{5}t\in(0.879 + 2k\pi,2.263+2k\pi)$ for $k = 0,1,2,\cdots$. Solve for $t$: $t\in(\frac{0.879\times5}{4\pi}+ \frac{2k\pi\times5}{4\pi},\frac{2.263\times5}{4\pi}+\frac{2k\pi\times5}{4\pi})$. For $k = 0$: $t\in(\frac{0.879\times5}{4\pi},\frac{2.263\times5}{4\pi})\approx(\frac{4.395}{4\pi},\frac{11.315}{4\pi})\approx(0.35,0.90)$. Since the period $T = 2.5$ min, for $k = 1$: $t\in(\frac{4.395}{4\pi}+2.5,\frac{11.315}{4\pi}+2.5)\approx(0.35 + 2.5,0.90+2.5)=(2.85,3.40)$. For $k = 2$: $t\in(\frac{4.395}{4\pi}+5,\frac{11.315}{4\pi}+5)\approx(0.35 + 5,0.90+5)=(5.35,5.90)$.

To sketch the graph:

• The amplitude is 28, the vertical shift is 28.5, and the period is 2.5 min. The function starts at $(0,0.5)$. We can plot key points such as the maximum ($y = 28 + 28.5=56.5$), minimum ($y=-28 + 28.5 = 0.5$), and mid - line $y = 28.5$ for each period.

Answer:

The time intervals when the rider could see Niagara Falls are approximately $(0.35,0.90)$ min, $(2.85,3.40)$ min, $(5.35,5.90)$ min for the first three cycles.