Asked by joe
To test the resistance of a new product to changes in static pressure, the product is placed in a
controlled environment. The static pressure in this environment as a function of time can be
described by the cosine function. The maximum static pressure is 5000 pascals, the minimum is 100
pascals, and at 𝒕 = 𝟎, the static pressure is at its maximum. In 40 hours the maximum static
pressure is reached 5 times (including the time at 𝒕 = 𝟎hours and 𝒕 = 𝟒𝟎hours.) What is the
equation of the cosine function that describes the static pressure in the environment?
Here is what I got so far: f(x) = 2450 sin (k x) + 2550
How do I find the value of k?
controlled environment. The static pressure in this environment as a function of time can be
described by the cosine function. The maximum static pressure is 5000 pascals, the minimum is 100
pascals, and at 𝒕 = 𝟎, the static pressure is at its maximum. In 40 hours the maximum static
pressure is reached 5 times (including the time at 𝒕 = 𝟎hours and 𝒕 = 𝟒𝟎hours.) What is the
equation of the cosine function that describes the static pressure in the environment?
Here is what I got so far: f(x) = 2450 sin (k x) + 2550
How do I find the value of k?
Answers
Answered by
mathhelper
From the statement:
In 40 hours the maximum static
pressure is reached 5 times
we can see that the period is 8 hours,
period = 2π/k
k = 2π/period = 2π/8 = π/4
Why did you choose a sine function, when they recommended the cosine function?
The cosine curve has a maximum when x = 0, so it would be the obvious choice:
f(x) = 2450cos((π/4)x) + 2550
checking:
We should have a max when x = 0
a min when x = 4
when x = 0, f(0) = 2450cos0 + 2550 = 2450(1) + 2450 = 5000 , check!
when x = 4, f(4) = 2450cos(π) + 2550 = 2450(-1)+2450 = 100 , check!
when x = 8 , f(8) = 2450cos(2π)+2550 = 5000, check
In 40 hours the maximum static
pressure is reached 5 times
we can see that the period is 8 hours,
period = 2π/k
k = 2π/period = 2π/8 = π/4
Why did you choose a sine function, when they recommended the cosine function?
The cosine curve has a maximum when x = 0, so it would be the obvious choice:
f(x) = 2450cos((π/4)x) + 2550
checking:
We should have a max when x = 0
a min when x = 4
when x = 0, f(0) = 2450cos0 + 2550 = 2450(1) + 2450 = 5000 , check!
when x = 4, f(4) = 2450cos(π) + 2550 = 2450(-1)+2450 = 100 , check!
when x = 8 , f(8) = 2450cos(2π)+2550 = 5000, check
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