Question

e is
The temperature, throughout the year, in a particular town can be graphed using
trigonometric functions. On July 24th, the hottest day of the year, the temperature
27 degrees Celsius. On Jan. 24th, the coldest day of the year, the temperature is - 11
degrees Celsius.
a) Determine the function for the temperature with respect to the number of
days since Jan. 1st. [Hint: Jan. 1st is day 1]
b) Determine what the temperature will be on July 6th and Nov. 16th​

Answer 1

Answer:

the range is from -11 to 27 or 38°

so in our trig function we have

T = 19sin(.....) + 8 as a start

the period is 365 days, so 2π/k = 365

and k = 2π/365

so we have T = 19 sin (2π/365)d + 8

where d is the number of the day in our year

(there are charts available that will give you the day number for a given date, e.g March 3 = 63)

In its current form our trig function will give up a max of 27 and a min of -11 , but for the wrong dates, so we need a phase shift

how about R = 19 sin (2π/365)(d +k)) + 8

check: for Jan 24 , d = 24, T = -11

-11 = 19sin(2π/365)(24+k) + 8

-19 = 19sin(2π/365)(24+k)

sin(2π/365)(24+k) = -1

I know sin (3π/2) = -1

(2π/365)(24+k) = 3π/2

24 + k = (3π/2)(365/2π) = 1095/4

k = 999/4

so T = 19 sin(2π/365)(d+999/4) + 8

checking it for jan24 or d = 24

my calculator gives me: T = -11 , YEAHHHH

checkin go July 24 , d = 205 ?? (check that)

T = 26.9936° , well, how about that

so for july 6, I get d = 187

T = 19 sin(2π/365)(187+999/4) + 8

= 25.94°

You do the one for Nov 16, happy button-pushing

(make sure you start with the innermost brackets, and follow the order of operation. Repeat my calculations first to make sure you get the same results. )