conversion question with liters A tank has a square base of 3 m by 3 m.  Water is poured into the tank at a constant rate.
The mass of water in the tank is observed to increase at a rate of dm/dt
= 500 kg/s.
You may assume throughout that 1 liter of water weighs 1 kg.
Show that the mass of the water increases with height at a rate of 9000 kg/m.
Hence, find dh/dm.
Why does 9 m$^3$= 9000 liters?
 A: You asked why $9$ m$^3$ = $9000$ L.  Differentiation is not needed for that.  A liter is $1000$ cm$^3.$  A meter is $100$ cm.  Can you conclude from there that $1$ m$^3$=$1000$ L, or $9$ m$^3$ = $9000$ L?
A: This is a related rates problem.
The volume of the water in the tank as of a function of time:
$$V(t)=3\cdot3\cdot h(t)=9\cdot h(t)$$
Notice that $h(t)$, the height of the water in the tank, is also as a function of time.
Since 1 liter of water weighs 1 kg, the rate at which the volume of the water in the tank grows is equivalent to the rate at which the water is being poured into the tank which is $500$ kg/s:
$$\frac{dV}{dt}=500\ L/s=500\cdot 0.001\ m^3/s$$
1 liter, by the way, is equal to $0.001\ m^3$.
The rate at which the height of the water in the tank grows:
$$\frac{dV}{dt}=9\frac{dh}{dt}\implies
\frac{dh}{dt}=\frac{500\cdot 0.001}{9}\ m/s
$$
Given the fact that $\frac{dm}{dt}=500$ kg/s and $\frac{dh}{dt}=\frac{500\cdot 0.001}{9}$ m/s, we can conclude that each second for every $500$ kg of water we get a $\frac{500\cdot 0.001}{9}$ m increase in height:
$$
\frac{500\ kg/s}{\frac{500\cdot 0.001}{9}\ m/s}=
\frac{500}{1}\div\frac{500\cdot 0.001}{9}=
\frac{500}{1}\cdot\frac{9}{500\cdot 0.001}=9000\ kg/m.
$$
I hope I understood the problem correctly.
