Wednesday

March 4, 2015

March 4, 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

**Answer this Question**

**Related Questions**

Math - Outside temperature over a day can be modelled as a sinusoidal function. ...

math - The temperature outside was 3 degrees Fahrenheit at midnight. During the ...

math - The temperature in your house is controlled by a thermostat. The ...

math - At 9:00 pm the outside temperature was 20 celcius, by midnight the ...

algebra - The overnight low temperature was -5 centigrade. If the temperature ...

math - overnight the low temperature dropped to -6 degrees Fahrenheit. If the ...

algebra - How would this problem be solved? The low temperature on Monday was -5...

Algebra - Stella is recording temperatures every day for 5 days. On the first ...

Calculus - How would you begin this problem? On a standard summer day in upstate...

algebra1 - at midnight, the temperature was-12degrees F. by noon, the ...