The average thickness of the ice covering on arctic lake can be modeled by the function T(x) =9cos(Pi/6 x)+15 where T(x) is the average thickness in month x (x=1 -> January)

Question

The average thickness of the ice covering on arctic lake can be modeled by the function T(x) =9cos(Pi/6 x)+15 where T(x) is the average thickness in month x (x=1 -> January) 

A) how thick is the ice in mid-March?

B) for what months of the year is the ice 10.5 in. thick?

Answer 1

A) Mid March would be x = 3, assuming that the average thickness occurs in the middle of the month.

T(3) = 9 cos(π/2) + 15 = 0 + 15 = 15 in.

B) 10.5 = 9 cos(π x/6) + 15, subtract 15 and divide by 9 -> -4.5/9 = cos(π x/6) = -1/2. This occurs at π x/6 = 2π/3 and 4π/3, so x = 4 (April) and 8 (August).