max of 50 and min of 4 ---> ampl = (50-4)/2 = 23
period is 30 min ---> 2π/k = 30 or k = π/15
You have:
a. If a child gets on the Ferris wheel when ,
how high will he be after riding for 15 minutes?
seems to be something missing after "wheel when, ..."
but after 15 minutes, which is half a period, the child would be at the max of 50 m
so far we have height = 23sin (π/15(t + d)) + 23
Just looking at my sketch, I would move the curve horizontally to the right 1/4 of a period so
that the height is zero when time is zero
so height = 23sin (π/15(t - 7.5)) + 23
let's see if that works
when t = 0, we want height = 0
0 = 23sin (π/15(0 + d)) + 23
sin (π/15(0 + d)) = -23/23 = -1
I know sin(3π/2) = -1 and also sin(-π/2) = -1
π/15(0 + d) = -π/2
π/15 d = -π/2
d = -7.5 , YES
https://www.wolframalpha.com/input/?i=y+%3D+23sin%28%CF%80%2F15%28x+-+7.5%29%29+%2B+23+from+0+to+32
looks like we got it.
height = 23sin (π/15(t - 7.5)) + 23
c) when t = 5
evaluate height, be replacing t with 5, watch the order of operation
looking at the graph, you should get about 11 m
d) set height = 45 and solve, you will get two answers ....
23sin (π/15(t - 7.5)) + 23 = 45
sin (π/15(t - 7.5)) = .956521.... , I stored it in calculator's memory
π/15(t - 7.5) = 1.2748... or π/15(t - 7.5) = π-1.2748... = 1.8667...
t = 13.586.. or t = 16.413..
So the child is higher than 45 m for about 16.412 - 13.586 or appr 2.8 minutes
check my arithmetic, this was fun!
A Ferris wheel at an amusement park reaches a maximum height of 50
metres and a minimum height of 4 metres. It takes 30 minutes for the wheel
to make one full rotation.
a. If a child gets on the Ferris wheel when ,
how high will he be after riding for 15 minutes?
b. Write a sine function to model the height of
the child minutes after boarding the Ferris wheel.
c. How high is the child if he has been riding
for 5 minutes
d. For what length to time will the child be
higher than 45 metres?
1 answer