Question

Jamie is riding a Ferris wheel that takes 15 seconds for each complete revolution. The diameter of the wheel is 10 meters and its center is 6 meters off the ground. When Jamie is 9 meters above the ground and rising, at what rate is Jamie gaining altitude?

Answer 1

period = 2π/k
15 = 2π/k
k = 2π/15
amplitude = 5
so we could start with y = 5sin(2πt/15)
centre is 6 m above ground, so...
y = 5sin(2πt/15) + 6

so when y = 9
5sin(2πt/15) + 6 = 9
sin(2πt/15) = 3/5 = .6
2πt/15 = arcsin .6 = .6435
t = 1.536...

dy/dt = 5(2π/15)cos(2πt/15) = appr 1.68 m/s

check my arithmetic

Answer 2

Can you solve this without derivatives? I'm in precalc and we have not covered derivatives yet. Thank you!

Answer 3

And when is Jamie rising most rapidly and at what rate? (Without derivatives)
Thank you.

Answer 4

ok, try this:
Finding that at t = 1.536 sec to reach 9 m did not require calculus, right?
when t = 1.536 , y = 5sin(2π(1.536))/15) + 6 = 9
let's look at a time just a tiny bit later ...
when t = 1.54, y = 9.00629

velocity of y = change in y / change in time
= (9.00629 - 9)/(1.54-1.536)
= 1.5725 , a bit off from my other answer.

redoing it while keeping all decimals in my calculator
i got 1.6745 m/s, much closer to my calculus answer

Answer 5

magnitude of speed = pi D/T = pi * 10/15 = 2.09 meters/second
vertical component of speed = 2.09 cos of angle to vertical
9 meters from ground is 3 meters above center of wheel
draw and see cos = 4/5
so vertical component = 2.09 * 4/5

max when at height of center cos angle = 1 so verrtical speed = 2.09 there

Answer 6

Physics is easier than math :)

Answer 7

Thank you to both of you. Huge help.