Mathematics, 29.07.2019 10:00 theojw

Aferris wheel has a radius of 25 m. the centre of the wheel is 26 m above the ground. the wheel rotates twice in 22 min. a) determine the equation of the sinusoidal function, h(x), that models the height of a passenger on the ride as a function of time. b) if a passenger gets on at the bottom of the wheel, when is the passenger 30 m above the ground? express the answer to the nearest tenth of a minute.

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Part aanswer: $h(x) = -25\cos\left(\dfrac{2\pi}{11}x\right)+26$ explanation: part b involves the passenger getting on at the bottom of the wheel. so let us make everything easy for us. we can model this with a cosine curve with x = 0 as a minimum point for the cosine curve. therefore, the cosine curve will have the form h(x) = a cos( bx ) + k, where "a" is a negative number(no phase shift because we want the minimum to be at x = 0)if the radius is 25 m and the centre is 26 m, then the maximum height is centre + radius = 26m + 25m = 51 m and the minimum height is centre - radius = 26m - 25m = 1 mthe amplitude of this sinusoidal function is the difference between the maximum and the centre (or centre and minimum), which is just the radius, 25. so this converts onto our equation as $a = -25$ $h(x) = a \cos( bx ) + k \implies -25 \cos (bx) + k$ the vertical displacement is the centre of the ferries is 26m. so $k = 26$ $h(x) = -25 \cos (bx) + 26$ if the wheel rotates twice in 22 minutes, then the wheel has one full revolution in 11 minutes. the period is therefore 11 minutes.b is related to the period by the equation $b = \dfrac{2 \pi}{\mathrm{period}}$ with period = 11, we have $b = 2\pi / 11$ $h(x) = -25\cos\left(\dfrac{2\pi}{11}x\right)+26$ ======part banswer: at 3.0 min and 8.0 minexplanationuse our equation from part a. since h(x) represents the height in metres, set h(x) to equal 30 and solve for x $\begin{aligned} 30 & = -25\cos\left(\tfrac{2\pi}{11}x\right)+26 \\ 4 & = -25\cos\left(\tfrac{2\pi}{11}x\right) \\ -\tfrac{4}{25} & = \cos\left(\tfrac{2\pi}{11}x\right) \\ \cos^{-1}\left(-\tfrac{4}{25}\right) & = \cos^{-1}\big( \cos\left(\tfrac{2\pi}{11}x\right)\big) \\ \cos^{-1}\left(-\tfrac{4}{25}\right) & = \tfrac{2\pi}{11}x \end{aligned}$ the reference angle is $\cos^{-1}\left(\tfrac{4}{25}\right)$ and $\cos^{-1}\left(\tfrac{4}{25}\right) \approx 1.4101 + 2\pi n,\ \text{$n$ an integer}$ since cosine is negative in quadrants ii and iii: in quadrant ii: $\cos^{-1}\left(-\tfrac{4}{25}\right) = \pi - \cos^{-1}\left(\tfrac{4}{25}\right)$ so $\begin{aligned} \pi - \cos^{-1}\left(\tfrac{4}{25}\right) & = \tfrac{2\pi}{11}x \\ 1.610807 + 2\pi n & = \tfrac{2\pi}{11}x \\ x & = \tfrac{11}{2\pi}(1.610807 + 2\pi n) \\ x & \approx 3.03 + 11n,\ \text{$n$ an integer} \end{aligned}$ so after 3.0 minutes of passenger getting on, he is 30 m above the ground.in quadrant iii: $\cos^{-1}\left(-\tfrac{4}{25}\right) = \cos^{-1}\left(\tfrac{4}{25}\right) + \pi$ so $\begin{aligned} \cos^{-1}\left(\tfrac{4}{25}\right) + \pi & = \tfrac{2\pi}{11}x \\ 4.55169 + 2\pi n & = \tfrac{2\pi}{11}x \\ x & = \tfrac{11}{2\pi}(4.55169 + 2\pi n) \\ x & \approx 7.96 + 11n,\ \text{$n$ an integer} \\ & \approx 8.0 + 11n \end{aligned}$ so after 8.0 minutes of passenger getting on, he is 30 m above the ground.

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