Tuesday

January 27, 2015

January 27, 2015

Posted by **marie darling** on Tuesday, March 26, 2013 at 12:13am.

- precalculus -
**Reiny**, Tuesday, March 26, 2013 at 9:01amlet's pick a sine function

the range is 9-70 = 20

so a = 10

we know that the period is 24 hrs

so 2π/k=24

24k = 2π

k = π/12

so far we have:

D = 10 sin(π/12)(t) + 80

giving us a range from 70 to 90

But obviously the temp after midnight would decrease, whereas our function has it increasing to 90 when t = 6 (6:00 am)

We could do a phase shift, or more simply, just flip the function to**D = -10sin(π/12)t + 80**

check some values

t = 0 , D = -10sin0 + 80 = 80 , ok

t=6 , (6:00 am) D = -10 sin (π/2) + 80 = 70 , ok

t = 12 (noon), D = -10 sin π + 80 = 80 , ok

t = 18 , (6:00 pm) , D = -10sin 3π/2 + 80 = 90

all looks good

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