Integration 2

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2014-05-14 1:47 pm

A spherical tank of radius 2 m is initially half full of water. The water is drained through a hole at the bottom of the tank, and the depth of water in the tank at time t (in minutes) is denoted by h(t). If h = h(t) satisﬁes the diﬀerential equation

(4h - h^2)dh/dt = -k sqrt(2gh)
i.e (4h - h^2)dh/dt = -k √(2gh)

where k is a positive constant and g is the acceleration due to gravity, how long (in minutes) will it take for the water to drain completely? [Answer: 45/(15ksqrt(g))]

回答 (1)

用戶10012

2014-05-14 9:44 pm

✔ 最佳答案

(4h - h^2)dh/sqrt(2gh) = - kdt
1/sqrt (2g) [ 4sqrt h - h^(3/2)] dh = - k dt
1/sqrt(2g) ∫ [ 4 sqrt h - h^(3/2)] dh = ∫ - k dt
1/sqrt (2g) [ 8 h^(3/2)/3 - 2h^(5/2)/5] = - kt + C
when t = 0, h = 2
1/sqrt (2g) [ (8 sqrt 8)/3 - (2 sqrt 32)/5 ] = C
that is
1/sqrt(2g) [ 8h^(3/2)/3 - 2h^(5/2)/5 ] = - kt + 1/sqrt(2g) [ (8 sqrt 8)/3 - (2 sqrt 32)/5]
When h = 0, left hand side is zero,
so t = 1/ksqrt(2g) [ (16 sqrt 2)/3 - (8 sqrt 2)/5]
= 1/ksqrt g [ 16/3 - 8/5]
= 1/ksqrt g [ (80 - 24)/15]
= 1/ksqrt g ( 56/15).

2014-05-14 14:44:05 補充：
Remark : The answer should not be 45/15, if really 45/15, it will be written as 3, not 45/15.

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