Expert: Ahmed Salami - 11/23/2010

Question
QUESTION: Dear Mr.Ahmed,
      Mr.Scotto gave me the following answer which I have uploaded and right now i want to ask him about C - the phase shift, but he is on vacation.So, can you please help me with phase shift.He says that phase shift needs to be changed according to 2pi/365.But i am unable to change it according to 2pi/365.
         THANKS A LOT!!!
         Bulbul.

ANSWER: Hi Bulbul,
First of all, your question didnt say anything about 09:17 and 04:35 being maximum and minimum rise times of the sun. I'll of course assume that to be the case looking at your answer.
Now, some background on the sine function.
The most basic form of the sine function is
y = Asin(t)
where A is the amplitude (the maximum value, also the minimum value is -A) and it has a period of 2π
The y-axis distance between the maximum and minimum values is 2A and the distance on the t-axis is π (half of the period)
The graph also starts at the origin.

For the sine function of the form
y = Asin(t-c)
everything is still the same except that the graph doesnt intercept the origin. It seems to be moved c units to the right.

For the sine function of the form
y = Asin(Bt)
A is the amplitude but now the period is 2π/B
The y-axis distance between the maximum and minimum values is still 2A but the distance on the t-axis is π/B
The graph also starts at the origin.

For the sine function of the form
y = Asin[B(t-c)]
A is the amplitude but now the period is 2π/B
The y-axis distance between the maximum and minimum values is still 2A and the distance on the t-axis is π/B
The graph seems to be moved Bc units to the right of the origin.

The sinusoidal function
y = Asin[B(t-c)] + D    (which is the complete form that takes into account every type of phase shift and amplification)
is simply the graph of y = Asin[B(t-c)] moved D units upwards. Notice that the graph only actually moves upwards if D is positive, otherwise it moves downwards.
Also, the maximum value is now A+D and the minimum value is -A+D

We can now begin our modelling of sunrise times.
From the information provided,
557 minutes after midnight is the maximum rise time and it occurs at t = 355 days
275 minutes after midnight is the minimum rise time and it occurs at t = 173 days
And from our knowledge of sinusoidal functions, we know that the difference in minutes at those values of t is equal to 2A
557 - 275 = 2A
282 = 2A
A = 141

A+D = 557
D = 557 - 141 = 416
OR
-A+D = 8.3
D = 275 + 141 = 416
OR
We could just say that D is midway between the two extreme values i.e
D = (557 + 275)/2
 = 832/2 = 416

The difference between the t values is π/B
355 - 173 = π/B
182 = π/B
B = π/182

To get c, we could go back to the equation
y = Asin[B(t-c)] + D
y-D = Asin[B(t-c)]
(y-D)/A = sin[B(t-c)]
B(t-c) = arcsin[(y-D)/A]
t - c = arcsin[(y-D)/A] / B
Inserting the value y = 275 at t = 173
173 - c = arcsin[(275 - 416)/141] / (π/182)
173 - c = arcsin[(-141)/141] / (π/182)
173 - c = arcsin(-1) / (π/182)
173 - c = (-π/2) / (π/182)
173 - c = (-π/2)(182/π)
173 - c = -182/2
173 - c = -91
c = 264
OR (just to make sure)
Inserting the value y = 557 at t = 355
355 - c = arcsin[(557 - 416)/141] / (π/182)
355 - c = arcsin[(141)/141] / (π/182)
355 - c = arcsin(1) / (π/182)
355 - c = (π/2) / (π/182)
355 - c = (π/2)(182/π)
355 - c = 182/2
355 - c = 91
c = 264

Of course we could just have said c = (173 + 355)/2 = 528/2 = 264 but i just wanted you to see that c doesnt need any further adjustments.

The complete modelling equation is therefore,
y = 141sin[(π/182)(t - 264)] + 416
where y is the number of minutes after midnight and t is the number of days from the beginning of the year.

This solution might seem elaborate but its to give you more understanding and clear any doubts.

Regards and best of luck.

---------- FOLLOW-UP ----------

QUESTION: Dear Dr.Ahmed,
         I am sorry to disturb you again but can you please explain me why "B = pi/182. And The difference between the t values is π/B
355 - 173 = π/B
182 = π/B
B = π/182."
If this is a sine function than shouldnt the formula of period be 2pi/B , as the formula pi/B is only designated for tangent functions? Shouldnt B = 2pi/365 ? because,
In the question it is written that period is 365 days.
       So, Period = 2pi/B
         365  = 2pi/B
         B   = 2pi/365.
Isnt this right? Why is this wrong,Mr.Ahmed? My most of the friends have done that. Can you please explain me and clear my doubt, I am in a great confusion! I told my friends the thing which you taught but they say that the period is already given as 365 in the question.
        I am very sorry to disturb you once again!!!          Thanks a LOT!!!
         Bulbul.  

Answer
Hi Bulbul,
I totally understand your concern. The period of the sinusoidal function i've written is 364 and not 365 as the question required.
Note that B = 2π/365 = π/182.5
which is approximately equal to π/182 to a few decimal places.
The point is this, strictly speaking, the difference between the t values at the maximum and minimum sunrise times should be EXACTLY half the period of the sinusoidal function. That is a definite property of a sinusoidal function and this is obviously not the case for a period of 365. Actually, we do not need the period to be stated if we've been given the t values at the maximum and minimum points as these on their own will determine the period.
Ask your friends these, what if the question asked the period to be 1000? How does a sinusoidal function with a period of 1000 have maximum and minimum points at 173 and 355? This is definitely not possible as it doesnt conform with symmetry because the distance between them can only be 500. What i'm saying is that a period of 365 contradicts the placement of the two t values. It is of course close but if you want exactness, B has to be π/182.
As a check, use the equation i provided to find the sunrise times at t = 173 and 355 and you'll get exactly 275 and 557. If you use B = 2π/365 = π/182.5, you'll only get close, but not exact, values.

You can always get back to me.

Regards