A merry-go-round ride has outer horse figures that go up and down exactly three times during one rotation of the carousel. One of the high-points for each horse occurs where a rider is just close enough to reach out and try to grab a metal ring from a mechanical dispenser. If the rider succeeds in grabbing a brass ring instead of an iron ring, the rider has won a free ride on the merry-go-round. Hence the expression, "reaching for the brass ring." Write a sinusoidal function that models the height of one of the horse figures as a function of the rotation of the main carousel with theta = 0 at the ring dispenser. The amplitude of the horses vertical motion is 1.1 meters around an average height of 1.6 meters.

Question

A merry-go-round ride has outer horse figures that go up and down exactly three times during one rotation of the carousel.  One of the high-points for each horse occurs where a rider is just close enough to reach out and try to grab a metal ring from a mechanical dispenser.  If the rider succeeds in grabbing a brass ring instead of an iron ring, the rider has won a free ride on the merry-go-round.  Hence the expression, "reaching for the brass ring."  Write a sinusoidal function that models the height of one of the horse figures as a function of the rotation of the main carousel with theta = 0 at the ring dispenser.  The amplitude of the horses vertical motion is 1.1 meters around an average height of 1.6 meters.

phi · Answer

it sounds like we want to use cosine
because cos(0) (at the "ring dispenser") is at a max and we want
"One of the high-points for each horse occurs where a rider is just close enough to reach out and try to grab a metal ring from a mechanical dispenser"

haleyelizabeth2017 · Answer

but we have to use sin...

phi · Answer

if the carousel rotates at theta, the horse is going up and down 3 times faster so 3theta
so you want
$$ A \cos (3 \theta) + B$$
you need A and B

if you have to use sin, shift the sin pi/2 to the left
i.e use
$$ A \sin\left(3 \theta + \frac{\pi}{2}\right) +B$$

haleyelizabeth2017 · Answer

Ah okay

haleyelizabeth2017 · Answer

$$1.1 \sin\left(3 \theta + \frac{\pi}{2}\right) +1.6$$How did you get $$3\theta ?$$

haleyelizabeth2017 · Answer

Oh wait....I misread at first x'D

phi · Answer

outer horse figures that go up and down exactly three times during one rotation

phi · Answer

up and down means (to me) one full period

haleyelizabeth2017 · Answer

Okay :)

haleyelizabeth2017 · Answer

So if we had to use sin like this time...and it's similar to this, would it be $\frac{\pi}{2}$ most of the time?

phi · Answer

in this problem, you want the sin to peak at theta =0
However, you (should) know that sin peaks at pi/2, so shift the sin to the left by pi/2
shift to the left means add pi/2  (shift to the right would be - pi/2)

If they problem is to align the peak at say theta=pi/4 (45 degrees) rather than 0, you would shift the sin from pi/2 to pi/4  which means pi/4 to the left, i.e. add pi/4

haleyelizabeth2017 · Answer

ah

haleyelizabeth2017 · Answer

Thank you very much @phi :)