Average daily temp,y, is
y=45-23cos(2pi(x-32)/365), where x is time in days, x=1, corresponds to Jan 1 and y is temp in degrees F.
Determine the period of the function. Determine critical numbers, over one period.
Determine relative extrema, what does relative extrema represents
To determine the period of the function, we need to find the length of one complete cycle of the cosine function in the given equation.
The cosine function has a period of 2π, which means it repeats itself every 2π units. In this equation, the argument of the cosine function is (2π(x-32)/365). We need to find the value of x that makes the argument equal to 2π.
2π(x-32)/365 = 2π
To solve for x, we can cancel out the 2π on both sides:
(x-32)/365 = 1
Multiply both sides by 365 to isolate x:
x - 32 = 365
x = 365 + 32
x = 397
So, the period of the function is 397 days.
Next, let's determine the critical numbers. Critical numbers occur where the derivative of the function is either zero or undefined. To find the derivative, we can apply the chain rule:
dy/dx = -23 * 2π/365 * sin(2π(x-32)/365)
Setting this derivative equal to zero:
-23 * 2π/365 * sin(2π(x-32)/365) = 0
sin(2π(x-32)/365) = 0
Since sine is zero at multiples of π, we have:
2π(x-32)/365 = 0, ±π, ±2π, ±3π, ...
Solving for x:
2π(x-32)/365 = 0
x-32 = 0
x = 32
For the other critical numbers, we can solve:
2π(x-32)/365 = ±π, ±2π, ±3π, ...
(x-32)/365 = ±1/2, ±1, ±3/2, ...
x-32 = ±(365/2), ±365, ±(365/2)*3, ...
x = 32 ± (365/2), 32 ± 365, 32 ± (365/2)*3, ...
Simplifying these values, we get:
x = 199, -135, 397, 73, -263, 530
These are the critical numbers over one period of the function.
Finally, to determine the relative extrema, we need to evaluate the function at these critical numbers and compare the values.
Evaluate y at the critical numbers:
y = 45 - 23cos(2π(x-32)/365)
y = 45 - 23cos(2π(199-32)/365) ≈ 20.37
y = 45 - 23cos(2π(-135-32)/365) ≈ 62.81
y = 45 - 23cos(2π(397-32)/365) ≈ -29.65
y = 45 - 23cos(2π(73-32)/365) ≈ 65.54
y = 45 - 23cos(2π(-263-32)/365) ≈ 62.81
y = 45 - 23cos(2π(530-32)/365) ≈ 20.37
From these evaluations, we can observe that the function has relative extrema (maxima and minima) at:
(199, 20.37) and (73, 65.54)
Relative extrema represent the highest and lowest points of a function within a given interval. In this context, the relative extrema represent the maximum and minimum temperatures over the period described by the function.
To determine the period of the function, we use the formula T = 2π/b, where b is the coefficient of x in the cosine function. In this case, b = 2π/365, so the period, T, is given by:
T = 2π/(2π/365) = 365
Therefore, the period of the function is 365 days.
To find the critical numbers over one period, we need to determine the values of x where the derivative of the function is equal to zero or undefined.
First, let's find the derivative of the function:
y = 45 - 23cos(2π(x−32)/365)
The derivative of y with respect to x is given by:
dy/dx = (46π/365)sin(2π(x−32)/365)
To find the critical numbers, we set the derivative equal to zero:
(46π/365)sin(2π(x−32)/365) = 0
The sine function is zero at x = nπ, where n is an integer. Therefore, we need to solve the equation:
2π(x−32)/365 = nπ
Simplifying the equation, we have:
x − 32 = (n/2)(365)
Solving for x, we get:
x = 32 + (n/2)(365)
The critical numbers are the values of x obtained by substituting different integer values for n. These critical numbers represent the days where the temperature may be at a maximum or minimum, or where the slope of the temperature curve changes.
Lastly, to determine the relative extrema, we evaluate the value of the temperature at the critical numbers obtained. Plug each critical number into the original function to find the corresponding temperature. The relative extrema represent the maximum and minimum temperatures over one period.