Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
junior college 2 | H2 Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Please help. Don't seem to be able to show.
t = 0 where AB = 0, and when AP = 0
tan APB = opp/adj = AB/AP
tan θ = AB/AP
From the info we can deduce that :
AP = 1000 - 6t
(when t = 0, AP = AO = 1000m. So as the moving point P moves towards A at 6m/s, the distance AP decreases by 6m/s)
AB = 2t
(when t = 0, AB = 0m. B ascends at 2m/s so AB = 2t)
θ = arctan (AB/AP)
θ = arctan ( 2t/(1000 - 6t) )
θ = arctan ( t/(500 - 3t) )
Since arctan (u) = u'/(1 + u²) ,
dθ/dt = (d/dt (t/(500 - 3t)) ) / ( 1 + t²/(500 - 3t)²)
dθ/dt = ( (1)(500 - 3t) - (-3)(t) )/(500 -3t)² ÷ ( 1 + t²/(500 - 3t)²)
= 500/((1 + t²/(500 - 3t)²) x (500 - 3t)² )
= 500/(t² + (500 - 3t)²)
(Shown)
See 1 Answer