Question
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
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
dad
how r u getting pi?
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