# The height, in metres, of a nail in water wheel above the surface of the water, as a function of time, can be?

modeled by the function h(t) = -4sin pi/4(t-1) + 2.5, where t is the time in seconds. during what periods of time is the nail below the water in the first 24 s that the wheel is rotating?

how do i solve this? thank you

You want to know when h(t) is less than 0. So:

-4sin (Pi/4 (t-1)) + 2.5 < 0

-4sin(Pi/4 (t-1)) < -2.5

sin(Pi/4 (t-1)) > 5/8

Now - the period of the function is 2Pi = (Pi/4 t); t = 8 seconds (ignore the offset here). So you're going through three full rotations of the wheel. And the answers will be:

Pi/4 (t-1) = sin^-1 (5/8) or Pi - sin^-1(5/8) + 2Pi or 4Pi

t-1 = 4/Pi sin^-1 (5/8) or 4 - 4/Pi sin^-1(5/8) + 8 or 16

And the six solutions are t = :

1 + 4Pi sin^-1 (5/8)

9 + 4Pi sin^-1 (5/8)

17 + 4Pi sin^-1 (5/8)

5 - 4/Pi sin^-1 (5/8)

13 - 4/Pi sin^-1 (5/8)

21 - 4/Pi sin^-1 (5/8)

#1

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