The height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. Which of the following equations can be used to model the height as a function of time, t, in hours? Assume that the time at t = 0 is 12:00 a.m

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y=Acos(p)+m, A=amplitude, p=period, m=midline, in this case:

A=1/2, p=360(t/12)=30t, m=(10-9)/2+9=9.5 so

h(t)=(1/2)cos(30t)+9.5

Answer:

The required equation is [tex]y=0.5\cos (\frac{\pi t}{6})+9.5[/tex].

Step-by-step explanation:

It is given that the height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. It means the minimum height is 9 feet and the maximum height is 10.

At x=0 is 12:00 a.m, it means the function is maximum at x=0. So, the general equation is defined as

[tex]y=a\cos(\omega t)+b[/tex]

Where, a is amplitude, w is period, t is time in hours and b is midline.

Midline is the average of minimum and maximum height.

[tex]b=\frac{9+10}{2}=9.5[/tex]

Amplitude is the distance of maximum of minimum value from midline.

[tex]a=10-9.5=0.5[/tex]

Now, the equation can be written as

[tex]y=0.5\cos(wt)+9.5[/tex]                      .... (1)

At t = 0 is 12:00 a.m, so t=3 is 3:00 a.m and the y=9.5 at x=3.

[tex]9.5=0.5\cos(3w)+9.5[/tex]

[tex]\cos(3w)=0[/tex]

[tex]\cos(3w)=\cos(\frac{\pi}{2})[/tex]

[tex]3w=\frac{\pi}{2}[/tex]

[tex]w=\frac{\pi}{6}[/tex]

Substitute this value in equation (1).

[tex]y=0.5\cos (\frac{\pi t}{6})+9.5[/tex]

Therefore the required equation is [tex]y=0.5\cos (\frac{\pi t}{6})+9.5[/tex].

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