✔ 最佳答案
18.a)i) tan θ = (h + 40)/55 = (20t - 5t² + 40)/55
ii) When t = 3,
tan θ = [20(3) - 5(3)² + 40]/55 = 1
θ = π/4
b)
Differentiating both sides of the expression in (a)(i) with respect to t, we have
(sec² θ)(dθ/dt) = (20 - 10t)/55
From (a)(ii), when t = 3, sec² θ = sec² (π/4) = 2
So, (2)(dθ/dt) = [20 - 10(3)]/55 = -2/11
dθ/dt = -1/11
Therefore, θ is decreasing (changing) at a rate of 1/11 s¯¹ (-1/11 s¯¹) when t = 3.
[N.B. In physics, we sometimes write rad. s¯¹, but since angles in radians actually possess NO unit in mathematics, s¯¹ is more appropriate than rad. s¯¹.]