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?
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
Can you solve this without derivatives? I'm in precalc and we have not covered derivatives yet. Thank you!
And when is Jamie rising most rapidly and at what rate? (Without derivatives)
Thank you.
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
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