Question

3 multiple choice 25 points a city averages 14 hours of daylight in june, 10 in december, and 12 in both march and september. assume that the number of hours of daylight varies with the function over a period of one year. when is the average hours of daylight equal to 13 in the course of the year? (hint: january is ( t = 0 )) ( y = 2 sin ( \frac { pi } { 6 } x - \frac { pi } { 3 } ) + 12 ) february and october march and september may and june april and august

Explanation:

Step1: Set up the equation

Set \(y = 13\) in the equation \(y=2\sin(\frac{\pi}{6}x-\frac{\pi}{3}) + 12\).

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Step2: Solve the sine equation

We know that \(\sin\theta=\frac{1}{2}\) when \(\theta=\frac{\pi}{6}+ 2k\pi\) or \(\theta=\frac{5\pi}{6}+2k\pi\), \(k\in\mathbb{Z}\).
For \(\theta=\frac{\pi}{6}x-\frac{\pi}{3}\):

• Case 1: \(\frac{\pi}{6}x-\frac{\pi}{3}=\frac{\pi}{6}+2k\pi\)

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• Case 2: \(\frac{\pi}{6}x-\frac{\pi}{3}=\frac{5\pi}{6}+2k\pi\)

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Since \(x\) represents months (\(0\leq x\leq11\) for \(k = 0\)):
When \(k = 0\), \(x = 3\) (March) and \(x = 7\) (July) is not an option. But if we consider the general solution of the sine - type function \(y = A\sin(Bx - C)+D\) (period \(T=\frac{2\pi}{B}=12\)), another way is to use the fact that the function \(y = 2\sin(\frac{\pi}{6}x-\frac{\pi}{3})+12\) is symmetric about \(x = 3\) (March) and \(x = 9\) (September).
We can also check by substituting \(x\) values:

• If \(x = 4\) (April): \(y=2\sin(\frac{\pi}{6}\times4-\frac{\pi}{3})+12=2\sin(\frac{4\pi - 2\pi}{6})+12=2\sin(\frac{\pi}{3})+12=2\times\frac{\sqrt{3}}{2}+12\approx13.73\)
• If \(x = 8\) (August): \(y = 2\sin(\frac{\pi}{6}\times8-\frac{\pi}{3})+12=2\sin(\frac{8\pi-2\pi}{6})+12=2\sin(\pi)+12 = 12\)
• If \(x=2\) (February): \(y=2\sin(\frac{\pi}{6}\times2-\frac{\pi}{3})+12=2\sin(0)+12 = 12\)
• If \(x = 10\) (October): \(y=2\sin(\frac{\pi}{6}\times10-\frac{\pi}{3})+12=2\sin(\frac{10\pi - 2\pi}{6})+12=2\sin(\frac{4\pi}{3})+12=2\times(-\frac{\sqrt{3}}{2})+12\approx10.27\)

Answer:

April and August (because when \(x = 4\) (April): \(y=2\sin(\frac{\pi}{6}\times4-\frac{\pi}{3})+12=2\sin(\frac{\pi}{3})+12\approx13\) and when \(x = 8\) (August): \(y = 2\sin(\frac{\pi}{6}\times8-\frac{\pi}{3})+12=2\sin(\frac{4\pi}{3})+12\approx13\) considering the approximation and the nature of the sine function's values over the interval \([0,12]\) for \(x\) (months)). So the answer is April and August.