I have previously posted on question on here, and @CPhill was kind enough to help me out with this, and I really appreciate that. However, he answered the question in degrees. I was wondering if anyone could do it in radians...
To form a sin or cos equation, Ive only gotten as far as to find the amplitude and the midpoint. I'm not sure how to find the period of horizontal shift. Any help would be appreciated.
Question :
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The hours of daylight, throughout the year, in a particular town can be graphed using trigonometric functions. On June 21, the longest day of the year, there are 15.3 daylight hours. On Dec. 21, the shortest day of the year, there are 8.7 daylight hours.
a) Determine the function for the number of daylight hours with respect to the number of days since Jan. 1st.
(Hint: Jan. 1st is day 1)
b) Determine the number of daylight hours the town will have on March 27th and October 2nd.
Let the number of days = 365
Then each day must be 2pi/365 rads
The midline is [ as before ] 12 and the amplitude here is 15. 8 - 12 = 3.3
I'm going to use the sine function here
I'm letting January 1st = "Day 0"...that is......where x = 0
Using the basic function y = 3.3sin [ (2pi* x / 365 ) ] we need to find the phase shift
Using this function.....one minimum occurs when x ≈ -91.25
And we need this minimum to occur on Dec 21st, i.e., 11 units (days) before x = 0 ( Jan 1st) ....so....the phase shift, P, can be found thusly :
-91.25 + P = -11 ⇒ P = 80.25....so.....this means that we need to shift the graph 80.25 units (days) to the right ........
So......the phase shift can be expressed as 80.25 *2 *pi / 365
So....... the [approximate] function is y = 3.3sin [ (2 pi * x / 365) - (80.25 * 2pi / 365) ] + 12
March 27th is the 86th day of the year..so x = 85 days after Jan 1st...so the hours of daylight when x = 85 are ≈ 12.27 hrs
And October 2nd is the 275th day of the year.....x = 274 days after Jan 1st....so the hours of daylight when = 274 ≈ 11.365 hrs
Here is the graph : https://www.desmos.com/calculator/kyp7tucydf