ce amath---rate of change

2007-09-17 2:02 am

18. A ball is thrown vertically upwards from the roof of a building 40 metres in height. After t seconds, the height of the ball above the roof is h metres, where h=20t-5t^2. At this instant, the angle of elevation of the ball from a point O, which is at a horizontal distance of 55 meters from the building, is θ.
a)find
i)tanθ in terms of t.
ii)the value of θ when t=3
b) Find the rate of change of θ with respect to time when t=3.

回答 (1)

wmheric

2007-09-17 2:29 am

✔ 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.]

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