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# differential equations C4 watch

1. Ok, I am doing questions on forming differential equations where you need to use the chain rule, but I can't get the right answer for this one:

Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t>0, the volume of fluid remaining in the tank is Vm^3. The rate at which the fluid flows out in m^3min^-1 is proportional to the square root of V.
Show that the depth h metres of fluid in the tank satisfies the differential equation dh/dt=-k(h)^1/2, where k is a constant

so I let t=time; V=volume; A=area and h=height

Vol of cylinder= pi.r^2.h; dV/dh= 2pi.r.h Is this correct? this may be where I am going wrong? then dh/dV=1/2pi.r.h
area of cross section- pi.r^2

so dV/dt=-k(V)^1/2
We are looking for dh/dt
By the chain rule: dh/dt= dh/dV . dV/dt
therefore: dh/dt=(1/2pi.r.h) . -k(v)^1/2
dh/dt=(1/2pi.r.h) . (-k(pi.(r)^2.h))

This takes me to the answer dh/dt= -k.r

Where am I going wrong? Please ask me if you do not understand some of the notation, but please help me
2. (Original post by Gabriela)
Vol of cylinder= pi.r^2.h; dV/dh= 2pi.r.h Is this correct? this may be where I am going wrong? then dh/dV=1/2pi.r.h
area of cross section- pi.r^2
That is dV/dr, not dV/dh
3. (Original post by Rickelton)
That is dV/dr, not dV/dh
Oh yes so dV/dh= pi.r^2 ?

and thank you for the quick reply
4. (Original post by Gabriela)
Oh yes so dV/dh= pi.r^2 ?

and thank you for the quick reply
thats right

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