Question

A clock with an hour hand that is 15 inches long is hanging on a wall. At noon, the distance between the tip of the hour hand and the floor is 96 inches. At 3 p.m., the distance is 86 inches; at 6 p.m., 76 inches; at 9 p.m., 86 inches; and at midnight, the distance is again at 96 inches. If y represents the distance between the tip of the hour hand and the ceiling x hours after noon, write a sinusoidal function to model the relationship between these two variable quantities.

Answer 1

First of all, it is important to know that the tip of the hour hand and the ceiling varies sinusoidally, which means it has to be represented by a sinusoidal function

`y=A\sin (kx+r)+C`

According to the problem, the amplitude is 15 because that's the length of the hour hand. So, A = 15.

Given that it's about a clock, the period is 12 hours because each lap takes that time. So,

`(2\pi)/(k)=12`

We solve for k.

`\begin{gathered} 2\pi=12k \\ k=(2\pi)/(12) \\ k=(\pi)/(6) \end{gathered}`

We found the constant.

On the other hand, the maximum value takes place when x = 6. So, the maximum value is y = 53. Using the information we have at the moment, we form the following.

`\begin{gathered} 15\sin (\pi+r)+C=53 \\ -15\sin (r)+C=53 \end{gathered}`

Then, we evaluate the function when x = 0, and y = 23 to get another equation and form a system.

`15\sin (r)+C=23`

If we combine the equations, we get

`C=(76)/(2)=38`

Therefore, the equation is

`y=-15\cos ((\pi x)/(6))+38`

Observe that the function is a cosine because the term r is equal to pi/2.