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junior college 2 | H2 Maths
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Alina
Alina

junior college 2 chevron_right H2 Maths chevron_right Singapore

Please help. Don't seem to be able to show.

Date Posted: 4 years ago
Views: 292
J
J
4 years ago
ABP is a right angled triangle. So we use trigo to try to find θ and dθ/dt. PAB remains at 90° except when at
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)

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Jiayang
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Alina
Alina
4 years ago
Oh thank you very much. I forgot to differentiate the inside value again my bad. No wonder the numerator was different. Gracias.