Can someone please explain this to me?? I'm really stuck on it. First of all, i don't know if I should use cosine or sine for the function because it only says to make a sinusoidal function that models the data, and a sinusoidal function could be either sine or cosine. PLEASE HELP!!!
Trigonometric Stuff: It is possible to use a sinusoidal function to model the amount of perceived daylight in a certain location over the course of a year. For Portland, Oregon, there is a minimum of 9 hours of “daylight” on the 1st day of winter and a maximum of 17 hours of “daylight” on the 1st day of summer. Let D represent the number of hours of “daylight” in Portland, Oregon, T days after the 1st day of spring (assume that T = 0 corresponds to March 20th). You may assume that 1 year has 365 days.
Find a formula for such a function, being sure to explain the practical meanings of any important pieces of the formula (amplitude,midline, and period). Use your formula to determine on what days of the year (month and day, not just T’s value) Portland has about 11hours of “daylight” and about 15 hours of “daylight”. Please round to the nearest day, if not exact.
Please can someone help!?!!?
well, you could say
D = .5(9+17)+ A sin wt + B cos wt
that is the midline , 13
if you start at the minimum with t = 0
and D = 9
then to get a minimum at t = 0
say D = 13 + A sin 0 + B cos 0
D = 13 + 0 + B
but we know D = 9
so B = -4
so we have
D = 13 - 4 cos w t
now when t = 365
w t = 2 pi, a whole circle
so
w * 365 = 2 pi
w = 2 pi/365
d = 13 - 4 cos (2 pi t/365)
sine function
it starts at the median (average) day length and increases to the summer solstice, then decreases to the winter solstice, then increases back to the average at the vernal equinox
the midline is the average ... (9 + 17) / 2
the amplitude is max minus average
the period is 365 (days)
whoops I took t = 0 at Jan 1
should be at March 20 or 1/4 year after Jan 1
so
alter that to get min at 1/4 year
D = 13 - 4 cos (2pi t/365 - x)
when t = (1/4)(365) or 91.25 days
2 pi * .25 - x = 0
x = 2 pi * .25
so in the end
D = 13 - 4 cos (2 pi t/365 - pi/2)
which is
D = 13 + 4 sin (2 pi t/365)
Thank you so much!!! It makes more sense now!!! Thanks again!!
To model the amount of perceived daylight in Portland over the course of a year, we can use a sinusoidal function. In this case, the function can be either a sine or cosine function since both are sinusoidal functions. The choice between sine and cosine depends on the starting point and phase of the function.
Given that there are 9 hours of daylight on the 1st day of winter and 17 hours on the 1st day of summer, we can determine the amplitude, midline, and period of the function.
The amplitude is the distance from the midline to the highest or lowest point of the function. In this case, the amplitude is (17 - 9)/2 = 4. This means that the function will oscillate between 9 + 4 = 13 hours and 17 - 4 = 13 hours.
The midline is the horizontal line that represents the average value of the function. In this case, the midline is at (9 + 17)/2 = 13 hours.
The period represents the length of one complete cycle of the function. In this case, the period is the number of days it takes to go from the 1st day of winter to the 1st day of summer, which is 365 days.
Now let's set up the formula for the sinusoidal function to model the data. Since T = 0 corresponds to March 20th, we can use T = 0 as the starting point for our function.
Using the cosine function, the equation can be written as:
D = A cos[(2π/P)T] + M
Where:
D is the number of hours of daylight
A is the amplitude (4 in this case)
P is the period (365 in this case)
T is the number of days after the 1st day of spring (starting from March 20th)
M is the midline (13 in this case)
D = 4 cos[(2π/365)T] + 13
To find the days when Portland has about 11 hours of daylight, we can set D to 11 and solve the equation for T:
11 = 4 cos[(2π/365)T] + 13
Solving for T will give you the day of the year when Portland has about 11 hours of daylight.
Similarly, to find the days when Portland has about 15 hours of daylight, we can set D to 15 and solve the equation:
15 = 4 cos[(2π/365)T] + 13
Again, solving for T will give you the day of the year when Portland has about 15 hours of daylight.
Remember to round the results to the nearest day since the given values are in hours.
Now you have a formula to model the data and can use it to determine the specific days of the year when Portland has about 11 hours and 15 hours of daylight.