How many seconds of the ride (1 full revolution) are spent higher than 12 meters above the ground?

Question

A Ferris wheel is 10 meters in diameter and boarded from a platform that is 4 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform.

The function H(t)=−5cos(π36t)+9 gives your height in meters above the ground t seconds after the wheel begins to turn.

How many seconds of the ride (1 full revolution) are spent higher than 12 meters above the ground?

Answer 1

To find out how many seconds of the ride are spent higher than 12 meters above the ground, we need to determine when the function H(t) = −5cos((π/36)t)+9 yields values greater than 12.

Set (H(t) > 12) and solve for t:

−5cos((π/36)t)+9 > 12

Subtract by 9 on both sides of the inequality:

−5cos((π/36)t) > 3

Divide by -5 and flip the sign (since we are dividing by a negative number):

cos(π/(36t)) < (-3/5)

The cosine function yields -3/5 when the angle is in the second and third quadrants of the unit circle. Since cos(π/36 * t) determines the height at any time t, we're interested in the interval of t within one period of the cosine function where its value is less than -3/5. The period of the cosine function here is 2π / (π/36) = 72 seconds, meaning a full revolution of the Ferris wheel takes 72 seconds.

Now, we need to find the arcs whose cosine values are -3/5. Since the cosine function is periodic and symmetric, we specifically look for these values in the interval [0, 72] seconds, which corresponds to the interval [0, 2π] radians in the context of the trigonometric function's period:

cos^-1(-3/5) ≈ 2.214 radians

Since we're dealing with the modified cosine wave, we need to map these radians back to seconds and apply them to the inequality within the 72-second period of the Ferris wheel ride.

The critical points within one period where the rider's height exceeds 12 meters correspond to phases before and after the apex of the cosine curve, considering the shift and scaling applied by the function. To find these times t, we convert the radians back into the Ferris wheel's timeframe using the relation given by the function's period:

(π/36 * t) = 2.214

t = (2.214 * 36) / π

Solving this gives us two points in time reflective of the unit circle's symmetry (second and third quadrants), but now we need to calculate their exact times during the 72-second cycle. These calculations provide the initial and final times when the height exceeds 12 meters, splitting into two intervals due to the function's periodic nature.

t ≈ (2.214 * 36) / π ≈ 25.36 seconds

and for the second critical point,

t ≈ ((2π - 2.214) * 36) / π ≈ 72 - 25.36 ≈ 46.64 seconds

Notice we've considered the conversion for both critical points on the unit circle to match the timeline of one revolution of the Ferris wheel. The ride's height exceeds 12 meters between these timestamps: (≈ 25.36) seconds and (≈ 46.64) seconds.

Hence, the duration spent above 12 meters is (46.64 - 25.36 = 21.28) seconds.

Answer 2

h(t) = 12 = -5cos(36pi/t) + 9

cos(36pi/t) = -3/-5 =.6

36pi/t = arccos.6 = about 53.15

36pi = 53.15t

t = 36pi/53.15 = about about 2.13 seconds when h=12 meters

it's 4 feet high at time t=0

a full revolution takes from h= min 4 to h= mid high = 9 to max high = 13 and back down to h=4 again

h=13 = -5cos(36pi/t) + 9

cos(36pi/t) = -4/-5 = .8

36pi/t = arccos.8

t = 36pi/arccos.8 = about 3.07 seconds

for 2(3.07- 2.13) seconds it's above 12 meters

= 2(.94)

= 1.88 seconds, it's above 12 meters

that's if by "pi36t" you mean (36pi)t