Consider a body of water in a tank. The rate at which the water evaporates is pr

Question

Consider a body of water in a tank. The rate at which the water evaporates is

proportional to the exposed surface area. A particular tank has a cross section that is in

the shape of an isosceles triangle, with dimensions as depicted in the following picture.

Initially, the tank is full. But after one minute, the height decreases by 0.1 ft. Write an

appropriate differential equation and solve it to determine the volume, V, of the resulting

body of water as a function in time, t.

Please show clear work so I can understand what is going on.

Consider a body of water in a tank. The rate at which the water evaporates is proportional to the exposed surface area. A particular tank has a cross section that is in the shape of an isosceles triangle, with dimensions as depicted in the following picture. Initially, the tank is full. But after one minute, the height decreases by 0.1 ft. Write an appropriate differential equation and solve it to determine the volume, V, of the resulting body of water as a function in time, t.

Explanation / Answer

Initial volume = 0.5*6*3*10 = 90ft^3

dV/dt = -kA

A = 60*h/3

dV/dt = -60kh/3

V = 0.5*(6*h/3)*h*10 = 10h^2, h = sqrt(V/10)

dV/dt = 20h*dh/dt

dV/dt = 20h*dh/dt = -60kh/3

20dh/dt = -20k

20dh = -20kdt

dh = -kdt

-0.1 = -k*60

k = 1/600

dV/dt = -60k*h/3 = -20kh = -20*(1/600)*h = -h/30 = -sqrt(V/10)/30

2sqrt(90) - 2sqrt(V) = (-t/sqrt10)/30

60 - 2sqrt(10V) = t/30

sqrt(10V) = 30 - t/60

V = ((30 - t/60)^2)/10

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