an inverted right circular cone with a base radius of 40 cm and a height of 100 cm. the cone is full of liquid .suppose the liquid is flowing out of the cone at a rate of 5 cm^3 per minute.
(a) let the height of the liquid be h cm and the radius of the surface be r cm.
(i)prove that r= 2/5 h
(ii)prove that the volume V cm^3 of the solution can be found by V= 4/75兀h
(b) if the height of the liquid inside the container is 50 cm, what is the rate of decrease in the liquid level?
點解AND 點做?
maths m1
2011-12-11 9:30 pm
回答 (1)
2011-12-11 9:39 pm
✔ 最佳答案
(i) right circular cone, radius = 40 and height = 10So, using similar triangle
40/r = 100/h => 100r = 40h => r = (2/5)h
(ii) V = (1/3)πr^2h => V = (1/3)π(2h/5)^2h = (4π/75)h^3
(b) dV/dt = -5
(4π/75)(3h^2)(dh/dt) = -5
(4π/25)(h^2)(dh/dt) = -5
Sub. h = 50
dh/dt = (-125/4)(π/2500) = -π/80 cm/s
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