Respuesta :
Answer:
The function is [tex]D = 2sin ( \frac{\pi}{2} - \frac{\pi}{12} t) + 6[/tex]
Step-by-step explanation:
From the question we are told that
The maximum depth is [tex]d = 8 \ ft[/tex]
The minimum depth is [tex]d_i = 4 \ ft[/tex]
Generally the average depth is mathematically represented as
[tex]d_a = \frac{8 + 4}{2}[/tex]
=> [tex]d_a = 6 \ ft[/tex]
Generally the amplitude is mathematically represented as
[tex]A = d - d_a[/tex]
=> [tex]A = 8 - 6[/tex]
=> [tex]A = 2[/tex]
Generally the period is 24 hours given that the the interval between the maximum depth and the minimum depth is half a day
Generally the period is mathematically represented as
[tex]T = \frac{2 \pi }{w}[/tex]
here w is the angular frequency
So
[tex]w = \frac{2 \pi}{24}[/tex]
[tex]w = \frac{\pi}{12}[/tex]
Generally the depth can be modeled with a sin function as follows
[tex]D = Acos (wt) + d_a[/tex]
Now from co-function identity we have that [tex]for \ cos (z) = sin (\frac{\pi}{2} - z)[/tex]
So
[tex]D = Asin ( \frac{\pi}{2} - wt) + d_a[/tex]
[tex]D = 2sin ( \frac{\pi}{2} - \frac{\pi}{12} t) + 6[/tex]
The function that models the depth, in feet, of the water at the pier yesterday, as a function of time t in minutes past high tide is; D = 2 sin((π/2) - (π/12)t) + 6
We are given;
- Maximum depth; d2 = 8 ft
- Minimum depth; d1 = 4 ft
Thus;
Average depth; d = (d1 + d2)/2
d = (4 + 8)/2
d = 6 ft
Now, to find the amplitude, we will just subtract the minimum depth from the maximum one to get; A = d2 - d1
A = 8 - 6
A = 2 ft
Now, the period T is a whole day which is 24 hours and so we can find the angular frequency ω from the formula;
ω = 2π/T
Thus;
ω = 2π/24
ω = π/12
- Now, the general formula for the depth function is given as; D = A sin(π/2 - ωt) + d
Where;
d_i is average depth
Thus;
D = 2 sin((π/2) - (π/12)t) + 6
Read more about sinusoidal functions at; https://brainly.com/question/2410297
