The high Roller Ferris wheel was completed in 2014 in Las Vegas. It
is the largest in the world, with a height of 550 ft and a 520 ft diameter. Riders enter a cabin at the bottom of the wheel and are taken on a rotation that lasts 30 min. Write a sine function that models the height of a cabin for the duration of the ride.
4 years ago
Answered By Soroosh M
First of all, a general sine equation is as follows:
$y=Asin\left(Bx+C\right)+D$y=Asin(Bx+C)+D
Where A is the amplitude, how much the height changes relative to the midpoint of the circle, which in this case, would be half of the diameter.
$\frac{2\pi}{B}$2πB is the period, meaning the the time it takes for the function to complete one full cycle.
$-\frac{C}{B}$−CB is the phase shift, or "horizontal translation" of the sine curve
D is the amount of "vertical translation" of a sine curve from its midpoint
Amplitude, A, for a full circle rotation would be half of the diameter, which is 260ft. The period of the graph would be $\frac{2\pi}{B}=30$2πB=30 . Rearranging for B, we get $B=\frac{2\pi}{30}=\frac{\pi}{15}$B=2π30=π15 . Phase shift does not occur because we assume that the sine curve starts at zero. Vertical translation would be the distance from the midpoint of the sine curve (in this case, center of the circle) to the imaginary x-axis (in this case, the ground), this would be equal to: $\left(550ft-520ft\right)+\frac{520ft}{2}=30ft+260ft=290ft$(550ƒt−520ƒt)+520ƒt2=30ƒt+260ƒt=290ƒt .
Rewriting the equation using the values for A,B,C,D and the proper variables (h, height in ft, as a function of t, time in minutes) yields:
4 years ago
Answered By Soroosh M
First of all, a general sine equation is as follows:
$y=Asin\left(Bx+C\right)+D$y=Asin(Bx+C)+D
Where A is the amplitude, how much the height changes relative to the midpoint of the circle, which in this case, would be half of the diameter.
$\frac{2\pi}{B}$2πB is the period, meaning the the time it takes for the function to complete one full cycle.
$-\frac{C}{B}$−CB is the phase shift, or "horizontal translation" of the sine curve
D is the amount of "vertical translation" of a sine curve from its midpoint
Amplitude, A, for a full circle rotation would be half of the diameter, which is 260ft. The period of the graph would be $\frac{2\pi}{B}=30$2πB =30 . Rearranging for B, we get $B=\frac{2\pi}{30}=\frac{\pi}{15}$B=2π30 =π15 . Phase shift does not occur because we assume that the sine curve starts at zero. Vertical translation would be the distance from the midpoint of the sine curve (in this case, center of the circle) to the imaginary x-axis (in this case, the ground), this would be equal to: $\left(550ft-520ft\right)+\frac{520ft}{2}=30ft+260ft=290ft$(550ƒ t−520ƒ t)+520ƒ t2 =30ƒ t+260ƒ t=290ƒ t .
Rewriting the equation using the values for A,B,C,D and the proper variables (h, height in ft, as a function of t, time in minutes) yields:
$h\left(t\right)=260sin\left(\frac{\pi}{15}t\right)+290$h(t)=260sin(π15 t)+290
Hope this helps.
4 years ago
Answered By Majid B
Information Given:
$h_{max}=550ft$hmax=550ƒ t , $Diameter=D=520ft$Diameter=D=520ƒ t => , $t=0s$t=0s => $h\left(0\right)=h_{min}=550ft-520ft=30ft$h(0)=hmin=550ƒ t−520ƒ t=30ƒ t , $Period=T=30min$Period=T=30min
General Sinsoidal Function:
$h=a.\sin\left(b\left(t-c\right)\right)+d$h=a.sin(b(t−c))+d
$a=amplitude=\left(\frac{h_{max}-h_{min}}{2}\right)=\left(\frac{550ft-30ft}{2}\right)=260ft$a=amplitude=(hmax−hmin2 )=(550ƒ t−30ƒ t2 )=260ƒ t
$d=vertical$d=vertical $displacement=\left(\frac{h_{max}+h_{min}}{2}\right)=\left(\frac{550ft+30ft}{2}\right)=290ft$displacement=(hmax+hmin2 )=(550ƒ t+30ƒ t2 )=290ƒ t
$b=\frac{2\pi}{Period}=\frac{2\pi}{T}=\frac{2\pi}{30}=\frac{\pi}{15}$b=2πPeriod =2πT =2π30 =π15
$c=Phase$c=Phase $Shift=Horizontal$Shiƒ t=Horizontal $Transation:$Transation: $h\left(0\right)=30ft$h(0)=30ƒ t => $30=260.\sin\left(\frac{\pi}{15}\left(0-c\right)\right)+290$30=260.sin(π15 (0−c))+290 => $\sin\left(-\frac{\pi c}{15}\right)=\frac{30-290}{260}=\frac{-260}{260}=-1$sin(−πc15 )=30−290260 =−260260 =−1 => $-\frac{\pi c}{15}=\frac{3\pi}{2}$−πc15 =3π2 => $c=-\frac{45}{2}$c=−452
Final Answer:
$h\left(t\right)=260.\sin\left(\frac{\pi}{15}\left(t-\frac{\left(-45\right)}{2}\right)\right)+290=260.\sin\left(\frac{\pi}{15}t+\frac{3\pi}{2}\right)+290$h(t)=260.sin(π15 (t−(−45)2 ))+290=260.sin(π15 t+3π2 )+290