1. **Problem 1: Ferris Wheel Altitude Modeling** We are given Nikki's seat altitude at various times on a Ferris wheel and need to model it with a sine function. 2. **Formula and Rules:** The general sine function for periodic motion is: $$h(t) = A \sin(B(t - C)) + D$$ where: - $A$ is the amplitude (half the distance between max and min heights), - $B = \frac{2\pi}{T}$ where $T$ is the period (time for one full cycle), - $C$ is the horizontal shift (phase shift), - $D$ is the vertical shift (midline or average height). 3. **Find parameters:** - Max altitude $= 14$ m, min altitude $= 2$ m - Amplitude $A = \frac{14 - 2}{2} = 6$ - Vertical shift $D = \frac{14 + 2}{2} = 8$ - Period $T = 60$ seconds (one full cycle) - Calculate $B = \frac{2\pi}{60} = \frac{\pi}{30}$ 4. **Phase shift $C$:** At $t=0$, altitude is minimum (2 m), sine normally starts at 0, so shift sine to start at minimum. Sine minimum occurs at $\frac{3T}{4} = 45$ seconds if no shift, so shift by $C = 45$ seconds. 5. **Equation:** $$h(t) = 6 \sin\left(\frac{\pi}{30}(t - 45)\right) + 8$$ --- 6. **Problem 2: Xtreme Skyflyer Height Modeling** Given max height 22 m, min height 3 m, time from max to min is 2.5 s. 7. **Parameters:** - Amplitude $A = \frac{22 - 3}{2} = 9.5$ - Vertical shift $D = \frac{22 + 3}{2} = 12.5$ - Half period $= 2.5$ s, so full period $T = 5$ s - $B = \frac{2\pi}{5}$ 8. **Equation starting at max height:** Use cosine since cosine starts at max: $$h(t) = 9.5 \cos\left(\frac{2\pi}{5} t\right) + 12.5$$ 9. **Find time when height is halfway between max and min:** Halfway height $= \frac{22 + 3}{2} = 12.5$ m Set $h(t) = 12.5$: $$9.5 \cos\left(\frac{2\pi}{5} t\right) + 12.5 = 12.5$$ $$\cos\left(\frac{2\pi}{5} t\right) = 0$$ Cosine is zero at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$: $$\frac{2\pi}{5} t = \frac{\pi}{2} \Rightarrow t = \frac{5}{4} = 1.25$$ $$\frac{2\pi}{5} t = \frac{3\pi}{2} \Rightarrow t = \frac{15}{4} = 3.75$$ So, halfway points occur at $t=1.25$ s and $t=3.75$ s. 10. **Effect of faster rotation:** Faster rotation means shorter period $T$, so $B = \frac{2\pi}{T}$ increases, making the function oscillate faster. --- 11. **Problem 3: Windmill Blade Height** Given: $$h(t) = 6.5 \cos(45 t) + 10.5$$ where $t$ in minutes. 12. **Height at $t=1.5$ minutes:** Calculate: $$h(1.5) = 6.5 \cos(45 \times 1.5) + 10.5 = 6.5 \cos(67.5) + 10.5$$ Using degrees, $\cos(67.5) \approx 0.3827$: $$h(1.5) \approx 6.5 \times 0.3827 + 10.5 = 2.4875 + 10.5 = 12.9875$$ Rounded to nearest centimeter: 13.0 m 13. **Times blade is 9 m during first cycle:** Set $h(t) = 9$: $$6.5 \cos(45 t) + 10.5 = 9$$ $$\cos(45 t) = \frac{9 - 10.5}{6.5} = -\frac{1.5}{6.5} = -0.2308$$ Find $\theta = 45 t$ where $\cos \theta = -0.2308$. First cycle $\theta$ from 0 to 360 degrees. $\cos \theta = -0.2308$ at approximately $103.4^\circ$ and $256.6^\circ$. Solve for $t$: $$t = \frac{103.4}{45} = 2.3 \text{ minutes}, \quad t = \frac{256.6}{45} = 5.7 \text{ minutes}$$ --- 14. **Problem 4: Sine Function from Graph** Given graph with x-axis from 4 to 24 and y-axis 0 to 60, sine wave starting near 60 at x=4, minimum near 10 at x=8, max near 55 at x=16. 15. **Estimate parameters:** - Amplitude $A = \frac{60 - 10}{2} = 25$ - Vertical shift $D = \frac{60 + 10}{2} = 35$ - Period $T = 24 - 4 = 20$ (one full cycle) - $B = \frac{2\pi}{20} = \frac{\pi}{10}$ 16. **Phase shift $C$:** Sine normally starts at 0, but here max near 60 at $x=4$, so shift sine left by 5 units to start near max. 17. **Equation:** Use cosine form for max start: $$y = 25 \cos\left(\frac{\pi}{10}(x - 4)\right) + 35$$ --- **Final answers:** 1) $$h(t) = 6 \sin\left(\frac{\pi}{30}(t - 45)\right) + 8$$ 2a) $$h(t) = 9.5 \cos\left(\frac{2\pi}{5} t\right) + 12.5$$ 2b) Halfway at $t=1.25$ s and $t=3.75$ s 2c) Faster rotation means smaller $T$, larger $B$, faster oscillations. 3a) Height at 1.5 min is approximately 13.0 m 3b) Blade is 9 m at $t=2.3$ min and $t=5.7$ min 4) $$y = 25 \cos\left(\frac{\pi}{10}(x - 4)\right) + 35$$