# 微分方程 應用問題part2

Consider a cylindrical hot tub with a 5-foot radius and height of 4 feet,placed on oneof its circular ends. Water is draining from the tub through a circular hole 5/8 inchesin diameter located in the base of the tub.

Assume a value k=0.6 to determine the rate at which the depth of the water is changing. Here it is useful to write dh/dt = dh/dV * dV/dt

Update:

(b)

Determine how much longer it takes to drain the lower half than the upper half of the tub.

Update 2:

5/8 inches in diameter 其他檢查過 題目 k=0.6 沒錯

(a) Assume a value k=0.6 to determine the rate at which the depth of the water is changing. Here it is useful to write dh/dt = dh/dV * dV/dt

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Consider a cylindrical hot tub with a 5-foot radius and height of 4 feet,placed on oneof its circular ends. Water is draining from the tub through a circular hole 5/8 inchesin diameter located in the base of the tub.

Assume a value k=0.6 to determine the rate at which the depth of the water is changing. Here it is useful to write dh/dt = dh/dV * dV/dt

設圓柱桶底端為O點,向上(高度)為y軸正向,

設水流出後第t秒,水位高度y(t),水體積V(t),水流出速率v(t),則

V(t)=π*5^2*y=25πy

v(t)=0.6√(2gy)=0.6*√(64y)=0.48√y

dV/dt=每秒水流出量= -v(t)*(水流口截面積)= -0.48√y*π*(5/8)*(1/12)=-π√y/40

(註:負號表V漸減)

又dV/dt=(d/dt)(25πy)=25π dy/dt

則 25π(dy/dt)= -π√y/40, y(0)=4

y'/√y = - 0.001

積分(對t): 2√y= -t/1000+C

y(0)=4, 則 C=4, y= (2- t/2000)^2

(上半流出) y(t)=2 = (2-t/2000)^2, t=2000(2-√2)

(全部流出) y(T)=0 = (2-T/2000)^2, T=2000

T-t=2000(√2 -1)

Ans:(a) dy/dt= -0.001/√y

(b)上半流出時間=2000(2-√2) 秒, 下半流出時間=2000(√2 -1)秒

註: t/(T-t)= √2

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