Ferris Wheel

[nycmathguy]

Section 5.3

We end this Section with a thrill ride application.

Do not go. Stop here.

1. Is part (a) correct?

2. I don't understand the set up for part (b).

 

[MathLover1]

part (a) is correct, just substitute k=0,1 and you have t=8 and t=24 seconds
Hence, the at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground.

b.
h(t)=53+50sin((pi/16)*t-pi/2)........max height will be where sin((pi/16)*t-pi/2)=1
h(t)=53+50
h(t)=103
103=53+50sin((pi/16)*t-pi/2) ........solve for t
103-53=50sin((pi/16)*t-pi/2)

50/50=sin((pi/16)*t-pi/2)
sin((pi/16)*t-pi/2)=1

t = 16 (2 n - 1), n element Z
so, one cycle last 32s, so at the top of the ride at, 16s, 48s, 80s,.....

t=16 seconds -to the top of the wheel

 

[nycmathguy]

part (a) is correct, just substitute k=0,1 and you have t=8 and t=24 seconds
Hence, the at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground.

b.
h(t)=53+50sin((pi/16)*t-pi/2)........max height will be where sin((pi/16)*t-pi/2)=1
h(t)=53+50
h(t)=103
103=53+50sin((pi/16)*t-pi/2) ........solve for t
103-53=50sin((pi/16)*t-pi/2)

50/50=sin((pi/16)*t-pi/2)
sin((pi/16)*t-pi/2)=1

t = 16 (2 n - 1), n element Z
so, one cycle last 32s, so at the top of the ride at, 16s, 48s, 80s,.....

t=16 seconds -to the top of the wheel

At least I got part (a). I did not know about plugging values for k in my answer for t. Thank you for explaining part (b).