18. the sea ice area around the south pole fluctuates between about 18 million square kilometers in…

Explanation:

Step1: Set up the sinusoidal - function

Let (t) be the number of months since January. The general form of a sinusoidal function is (y = A\sin(B(t - C))+D). The amplitude (A=\frac{18 - 3}{2}=\frac{15}{2}=7.5), the vertical shift (D=\frac{18 + 3}{2}=10.5), and the period (T = 12), so (B=\frac{2\pi}{T}=\frac{\pi}{6}). Since the maximum occurs at (t = 8) (September), we can write the function as (y = 7.5\sin(\frac{\pi}{6}(t - 8))+10.5).

Step2: Set up the inequality

We want to find when (y>15), so (7.5\sin(\frac{\pi}{6}(t - 8))+10.5>15). First, subtract 10.5 from both sides: (7.5\sin(\frac{\pi}{6}(t - 8))>4.5). Then divide both sides by 7.5: (\sin(\frac{\pi}{6}(t - 8))>\frac{4.5}{7.5}=0.6).

Step3: Solve the trigonometric - inequality

We know that if (\sin\theta>0.6), then (\theta\in(\arcsin(0.6)+2k\pi,\pi-\arcsin(0.6)+2k\pi)) for (k\in\mathbb{Z}). Let (\theta=\frac{\pi}{6}(t - 8)). (\arcsin(0.6)\approx0.6435) and (\pi-\arcsin(0.6)\approx3.1416 - 0.6435 = 2.4981). So (\frac{\pi}{6}(t - 8)\in(0.6435 + 2k\pi,2.4981+2k\pi)). For (k = 0): (0.6435<\frac{\pi}{6}(t - 8)<2.4981). Multiply through by (\frac{6}{\pi}): (\frac{6\times0.6435}{\pi}<t - 8<\frac{6\times2.4981}{\pi}). (\frac{3.861}{\pi}<t - 8<\frac{14.9886}{\pi}). (1.23<t - 8<4.77). (9.23<t<12.77) and for the next - cycle ((k = 1)): (0.6435 + 2\pi<\frac{\pi}{6}(t - 8)<2.4981+2\pi). Multiply through by (\frac{6}{\pi}): (\frac{6\times(0.6435 + 2\pi)}{\pi}<t - 8<\frac{6\times(2.4981+2\pi)}{\pi}). (\frac{3.861}{\pi}+12<t - 8<\frac{14.9886}{\pi}+12). (13.23<t<16.77) (but we are considering (t\in[0,12]) for the first - year cycle). The time interval for (y>15) in the first year is from (t\approx9.23) to (t = 12) and then from (t = 0) to (t\approx4.77). The length of this interval is ((12 - 9.23)+4.77 = 7.54\approx8) months.

Answer:

8 months