the height h meters of a seat on a Ferris wheel after t minutes is given by

h(t)= -23.5cos(0.4t) +25

For t ≥ 0

a) find the initial height of the seat

To find the initial height, t=0, so;

h(0)= -23.5cos(0.4(0)) +25

__h(0)=1.5m__

once a passenger's seat is more than 30m above the ground, there are no trees in view and they can take unobstructed photographs of the nearby city.

b) Given that passengers only complete one rotation on the Ferris wheel, calculate how long they will have to take unobstructed photographs of the nearby city

I'm unsure of this question. I first tried making h(t)=30m which ended up with me having two graphs

y=-23.5cos(0.4t) +25,

and y=30

so I decided to find the intersections between these for the first wave.

t=4.46, and t=11.3

but I end up hitting a wall, I don't know what to do now. Any ideas?

Guest Feb 4, 2020

#1**+2 **

h(t)= -23.5cos(0.4t) +25

Your answer to (a) is correct !!!

(b) we have this inequality

h(t)= -23.5cos(0.4t) +25 > 30

Look at the graph here : https://www.desmos.com/calculator/utxpmoarvo

Notice that the graph is above 30m between t ≈ 4.463 sec and t ≈ 11.245 sec

Subtracting.....

11.245 - 4.463 =

6.782 sec = the time the ride is above 30m

Your approach was correct !!!

CPhill Feb 4, 2020