F4 Maths M1 Differentiation(1)

2014-06-18 5:48 pm

The figure shows a rectangular table placed with its corners A and B against two walls PQ and RQ respectively. The diagonal AC =1m and angle CAB=20°. The corner A is pulled away from Q along the wall PQ at a rate of 0.5m/s so that B moves towards Q along RQ.

a) let BQ =y m and AQ =x m. Show that the distance of C from the wall PQ is
(y+x tan20°) m.
b) when AQ = 0.8m, find the rate at which the distance of C from PQ decreases.

Thank you very much!

回答 (1)

用戶10012

2014-06-18 10:26 pm

✔ 最佳答案

(a) BC = 1 x sin 20 = sin 20.
Let M be the point on RBQ such that CM is perpendicular to RBQ.
Triangle AQB is similar to triangle CMB.
So BM/CB = x/AB
BM/sin 20 = x/cos 20
so BM = x tan 20
so distance of C from wall (ground) PQ, L = BM + BQ = y + x tan 20.
(b)
L = y + x tan 20 .............(1)
BQ^2 + AQ^2 = x^2 + y^2 = AB^2 = cos^2 20..............(2)
From (1)
dL/dt = dy/dt + tan 20 dx/dt........(3)
From (2)
2x dx/dt + 2y dy/dt = 0
so dy/dt = - x/y ( dx/dt)
Sub into (3)
dL/dt = (-x/y) dx/dt + tan 20 (dx/dt) = ( - x/y + tan 20) dx/dt
Now AQ = x = 0.8,
from (2)
y = sqrt [ cos^2 20 - 0.8^2] = 0.493
so dL/dt = [ - 0.8/0.493 + tan 20] ( 0.5) = - 0.6294 m/s
that is C is decreasing at a rate of 0.6294 m/s when AQ = 0.8m

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