Cylindrical Water Tank - Differential Equations

[Morganita]

hi i need some help asap. please. ok

here's the question
Water is flowing out of a cylindrical tank. The rate at which the volume Vcm^3 of water is decresing with time is directly proportional to swrt(x) where x cm is the depth of water in the tank. A tank of radius 20cm and height 36cm takes 60 seconds to empty from full.
a)For a differenctial equation in x and time t
b)solve this differential equation and hence find the volume of water in the tank 30 seconds after it starts to drain.

What I've managed so far:
V= pi(r^2)*x
V=14400pi

dV/dt=-k(sqrt(x))

dV/dt=(dV/dx)*dx/dt

i know it isn't much. but i missed class for one day and now i have no idea what to do. please point me in the right direction.
thank you

 

[arthur ohlsten]

the volume of water in the tank when water is at height x, and radius of cylindar is r:
V=pi r^2 x
take derivative with respect to time ,t
dV/dt=400pi dx/dt

but dV/dt= k sqrt x [k can be either +/- at this point of the derivation]
substitute
k x^1/2 = 400 pi dx/dt
[x^1/2 k]dt=400 pi dx
k dt= 400 pi x^-1/2 dx
integrate
kt+c=400pi x^1/2 /[1/2]
kt + c = 800 pi x^1/2

at t=0 x=36 tank is full
substitute
c=800 pi 6
c=4800 pi
substitute

kt+4800pi =800pi x^1/2

at t = 60 x=0
60k+4800 pi =0
k=- 80

4800 pi -80t = 800 pi sqrt x
60 pi -t=10pi sqrt x answer
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I leave part b to you

Arthur