# math

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?

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1. period = 2π/k
15 = 2π/k
k = 2π/15
amplitude = 5
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

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

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

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posted by will
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

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posted by Reiny
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

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posted by Damon
6. Physics is easier than math :)

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posted by Damon
7. Thank you to both of you. Huge help.

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posted by will

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