temperatures by month in juneau html

Temperatures by Month in Juneau, Alaska

Michelle Harvey

Gabriel Zimmer

Ree’l Street

Our first problem deals with temperature which is by large determined by the position of the earth in its orbit. Now the earth’s orbit is a periodic function, so we assume that temperature as a function of time will also take the general form for periodic motion:

f (t) = a sin (w t )

where w is frequency = 2p /l , l = period of the function.

Fourier’s Theorem states that "any function of spatial period l can be represented by a sum of harmonic functions whose wavelengths (periods) are integral sub-multiples of l ". That is , {l , l /2, l /3,….l /n}

f (t) = c0 + c1 cos [( 2p /l ) t + f 1 ] + c2 cos [( 2p /l /2) t + f 2 ] + …. Cn cos [( 2p /l /n) t + f n ]

f (t) = c0 + S cn cos (w *t/n + f n)

where cn = constants of amplitude and f n = constant of phase. Note that c0 is the equilibrium point, f (0) = c0. It is convenient to use the trigonometric identity

cos (A + B) = cos A cos B – sin A sin B

to transform the Fourier expression, we let An = w t/n, Bn = f n, so that

f (t) = c0 + S [ cn cos An cos Bn – cn sin An sin Bn]

note that since Bn = f n = constant, the trig functions of Bn are also constant. We let an = cn Bn and bn = -cn sin Bn , so that

f (t) = co + S [ an cos An + bn sin An]

for our purposes, it is sufficient to take only the first term of the summation we have:

f (t) = c0 + a cos (w t) + b sin (w t)

we assume that the period for temperature is one year, so l = 1 year = 12 months = 365 days and, with our data taken monthly w = 2p /l = 2p /12 = p /6. In Juneau, the average temperature = c0 = 40.1. Our approximating function, then is of the form:

f (t) = 40.1 + acos (p t/6) + b sin (p t/6).

When dealing with a process that repeats itself, like seasonal temperatures, we can use the formula for oscillation as a guide:

y(t) = a cos(w t) + b sin(w t)

where t is time and w angular frequency.

With a small adjustment, we have a formula that will model the seasonal process:

y = M + a cos[(2p / 365)t] + b sin[(2p /365)t]

where M is our average yearly temperature and t is time(in days). This equation can also be written as

y = M + a cos[(2p /12)t] + b sin[(2p /12)t]

if t is in months.

Keeping in mind the formula of a line, y = a + bt, we can subtract M from y and divide by cos [(2p /12)t]:

___y-M ___ = a + b sin[(2p /12)t]

cos[(2p /12)t] cos[(2p /12)t]

(y-M) sec[(2p /12)t] = a + b tan[(2p /12)t]

Let us then assign:

Y = (y-M) sec[(2p /12)t] and T = tan [(2p /12)t]

So our format is linear:

Y = a + bt

Our goal is then to find a and b.

The following is our raw data collection:

Month	t = # of months	Y = temp ° F
January	1	22
February	2	28
March	3	31
April	4	39
May	5	46
June	6	53
July	7	56
August	8	55
September	9	49
October	10	42
November	11	33
December	12	27

Where the average yearly temperature M = 40.1° F. This data gives the following graph

Now we must transform our data. Note that with this oscillation, or season process, not only must we trasform our temperature(Y), but also our time (T).

	T = tan[(2p /12)t]	Y = (y-40.1) sec[(2p /12)t]
1	.57735	-20.90008
2	1.73205	-24.2000
3	not defined	.00000
4	-1.73205	2.20000
5	-.57735	-6.812733
6	not defined	-12.90000
7	.57735	-18.35974
8	1.73205	-29.80000
9	not defined	0.00000
10	-1.73205	3.80000
11	-0.57735	-8.19837
12	0.0000	-13.1000

When using Matlab’s function polyfit on our transformed data, we obtain the equation of the regression line:

Y = -8.8442 – 12.87271 T

Plugging in these values a = -6.8272 and b = -16.6052, into out oscillation formula we get:

y = 40.1 – 6.8272 cos[(2p /12)t] – 16.6052 sin[(2p /12)t]

After graphing this function, we can see that it approximates our original raw data very closely.