Question

the hours of daylight in atlanta oscillate according to this model where t is months from january 1st and h(t) is hours of daylight in a day. when should you plant a garden if you want to do it during a month where there are 11 hours of daylight?
h(t)=-2.25\cos(\frac{\pi}{6}t)+12
may t = 5.09
march t = 3.09
february t = 2.12
april t = 3.91

Explanation:

Step1: Set up the equation

Set \(h(t) = 11\), so \(11=-2.25\cos(\frac{\pi}{6}t)+12\).

Step2: Solve for \(\cos(\frac{\pi}{6}t)\)

Subtract 12 from both sides: \(11 - 12=-2.25\cos(\frac{\pi}{6}t)\), which gives \(- 1=-2.25\cos(\frac{\pi}{6}t)\). Then \(\cos(\frac{\pi}{6}t)=\frac{1}{2.25}=\frac{4}{9}\).

Step3: Solve for \(t\)

Using the inverse - cosine function, \(\frac{\pi}{6}t=\pm\cos^{-1}(\frac{4}{9}) + 2k\pi\), \(k\in\mathbb{Z}\).
First, \(\cos^{-1}(\frac{4}{9})\approx1.11\) (in radians).
Case 1: \(\frac{\pi}{6}t=\cos^{-1}(\frac{4}{9})+2k\pi\)
\(t=\frac{6}{\pi}\cos^{-1}(\frac{4}{9})+12k\). When \(k = 0\), \(t=\frac{6\times1.11}{\pi}\approx2.12\) (not a valid month for the answer options).
Case 2: \(\frac{\pi}{6}t = 2\pi-\cos^{-1}(\frac{4}{9})+2k\pi\)
\(t=\frac{6}{\pi}(2\pi-\cos^{-1}(\frac{4}{9}))+12k\).
\(t = 12-\frac{6}{\pi}\cos^{-1}(\frac{4}{9})+12k\). When \(k = 0\), \(t=12 - 2.12=9.88\) (not in the options).
Another way:
We can also use the general form of the cosine function \(y = A\cos(Bx - C)+D\) (here \(B=\frac{\pi}{6}\), \(A=-2.25\), \(C = 0\), \(D = 12\)).
We know that \(h(t)=11\), so \(11=-2.25\cos(\frac{\pi}{6}t)+12\)
\(\cos(\frac{\pi}{6}t)=\frac{1}{2.25}\)
Using a calculator:
If we assume \(t\) is in the range of \(0\leq t\leq12\) (months in a year)
\(\frac{\pi}{6}t=\cos^{-1}(\frac{1}{2.25})\) or \(\frac{\pi}{6}t = 2\pi-\cos^{-1}(\frac{1}{2.25})\)
\(t=\frac{6}{\pi}\cos^{-1}(\frac{1}{2.25})\approx\frac{6\times1.11}{\pi}\approx2.12\) (not in the options) or \(t = 12-\frac{6}{\pi}\cos^{-1}(\frac{1}{2.25})\)
Let's check the options by substituting:
For \(t = 3.91\)
\(h(3.91)=-2.25\cos(\frac{\pi}{6}\times3.91)+12\)
\(\frac{\pi}{6}\times3.91\approx2.04\) radians
\(\cos(2.04)\approx - 0.43\)
\(h(3.91)=-2.25\times(-0.43)+12\approx1 + 12=11\)

Answer:

April \(t = 3.91\)