Asked by adiba
suppose that the water level varies 70 inches between low tide at 8:.40 AM and high tide at 2:55PM .what he cosine function that models the variation in inches above and below the average water level as a function of the number of hours since 8:40AM .at what point in the cycle does the function cross the midline.what does the midline represent.
Answers
Answered by
Reiny
recall that the most general equation of a cosine curve is
y = a cos k(x - h) + d
I assume you know what each of these parameters represent
"water level varies 70 inches between low tide at 8:.40 AM and high tide at 2:55PM"
----> a = 35
low = 8:40 am, high = 2:55 pm, so 1/2 of a period = 14:55 - 8:40 = 6:15
so my period is 12:30 or 12.5 hrs
period = 2π/k
k = 2π/12.5 = 4π/25
so far we would have y = 35 cos (4π/25)(t ) + d
I think I will let d = 0, so the midline would be the midpoint between high and low tide.
The standard cosine has a max of 1, when t = 0 , and drops as we move to the right.
we want our equation to have a value of 35 when t = 8:40 or t = 26/3
so we have to move our normal cosine 26/3 to the right
---> y = 35 cos (4π/25)(t - 26/3)
let's test it:
when t = 26/3 , y = 35cos(4π/25)(0) = 35(1) = 35 , check!
when t = 14:55 = 179/12 , y = 35cos(4π/25)(25/4) = 35(-1) = -35 , check!
how about half way:
t = 283/24
y = 35cos(4π/25)(25/8) = 35cos(1.57079...) = 35(0) = 0 , yeahhh!!
My equation is correct.
further proof:
http://www.wolframalpha.com/input/?i=plot+y+%3D+35+cos+((4%CF%80%2F25)(x+-+26%2F3)),+for+0%3Cx%3C24
y = a cos k(x - h) + d
I assume you know what each of these parameters represent
"water level varies 70 inches between low tide at 8:.40 AM and high tide at 2:55PM"
----> a = 35
low = 8:40 am, high = 2:55 pm, so 1/2 of a period = 14:55 - 8:40 = 6:15
so my period is 12:30 or 12.5 hrs
period = 2π/k
k = 2π/12.5 = 4π/25
so far we would have y = 35 cos (4π/25)(t ) + d
I think I will let d = 0, so the midline would be the midpoint between high and low tide.
The standard cosine has a max of 1, when t = 0 , and drops as we move to the right.
we want our equation to have a value of 35 when t = 8:40 or t = 26/3
so we have to move our normal cosine 26/3 to the right
---> y = 35 cos (4π/25)(t - 26/3)
let's test it:
when t = 26/3 , y = 35cos(4π/25)(0) = 35(1) = 35 , check!
when t = 14:55 = 179/12 , y = 35cos(4π/25)(25/4) = 35(-1) = -35 , check!
how about half way:
t = 283/24
y = 35cos(4π/25)(25/8) = 35cos(1.57079...) = 35(0) = 0 , yeahhh!!
My equation is correct.
further proof:
http://www.wolframalpha.com/input/?i=plot+y+%3D+35+cos+((4%CF%80%2F25)(x+-+26%2F3)),+for+0%3Cx%3C24
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.