Yefim S. answered • 6d

Math Tutor with Experience

T(t) = Asin(Bt) + D; D = 80°; A = 98° - 80° = 18°; T(t) = 18sin(Bt) + 80;

Let we megere time from midnight, then at 3 PM t = 15 h, sin(15B) = 1; 15B = π/2; and B = π/30;

T(t) = 18sin((π/30)t) + 80;

At 7 AM t = 7 and T(7) = 18sin(7π/30) + 80 = 92°;

D(t) = Asin(Bt) + D; A = (93 - 67)/2 = 13°; D = (93 + 67)/2 = 80°; D(t) = 13sin(Bt) + 80;

If time is 5 AM then t = 5 hours and D(5) = 13(sin5B) + 80 = 67; sin(5B) = - 1; 5B = 3π/2, B = 3π/10

So, D(t) = 13sin(3πt/10) + 80

h(t)= 5 + 17.5(1 - cos(ωt + φ)); ω = 2π/10 = π/5; h(t) = 5 + 17.5cos(πt/5 + φ). At 6 PM we have t = 0 and cosφ = 0, and φ = π/2; So, h(t) = 5 + 17.5(1 - cos(πt/5 + π/2)) = 22.5 + 17.5sin(πt/5) > 29; sin(πt/5) > (29 -

- 22.5)/(-17.5); sin(πt/5) > 0.37143; sin^{-1}(0.37143) < πt/5 < π - sin^{-1}(0.37143);

0.38055 < πt/5 < 2.76104; 0.61 min < t < 4.39 min from 6 ocklock