Asked by Kelslie
A city averages 14 hours of daylight in June, 10 in December, 12 in both March and September. Assume that the number of hours of daylight varies sinusoidally over a period of one year. Write an expression for n, the number of hours of daylight, as a cosine function of t. Let t be in months and t=0 correspond to the month of January.
My work:
y=2cos((pi/6(x-d))+12
Ok how would I get the phase shift? Thanks in advance
My work:
y=2cos((pi/6(x-d))+12
Ok how would I get the phase shift? Thanks in advance
Answers
Answered by
Reiny
treat one of the data values as input into your partial equation.
e.g. 14 hours in June --- y = 14, x = 5
14 = 2cos((pi/6(5-d))+12
2 = 2cos((pi/6(5-d))
1 = cos((pi/6(5-d))
so pi/6(5-d) = 2pi , because cos 2pi = 1
which solves for d = -7
your equation is
<b>y=2cos(pi/6)(x+7) + 12 </b>
test it for one other given value.
e.g. Dec --> x = 11
y = 2cos(pi/6)18 + 12
= 2cos(3pi) + 12
= 2(-1) + 12
= 10 , as given
e.g. 14 hours in June --- y = 14, x = 5
14 = 2cos((pi/6(5-d))+12
2 = 2cos((pi/6(5-d))
1 = cos((pi/6(5-d))
so pi/6(5-d) = 2pi , because cos 2pi = 1
which solves for d = -7
your equation is
<b>y=2cos(pi/6)(x+7) + 12 </b>
test it for one other given value.
e.g. Dec --> x = 11
y = 2cos(pi/6)18 + 12
= 2cos(3pi) + 12
= 2(-1) + 12
= 10 , as given
Answered by
fondle
dad
Answered by
Joe
how r u getting pi?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.