A tank is filled with water up to the height H0. At the bottom of the tank, there is a tap which is opened at t=0. How does the height of liquid change with time?(Hint: dH/dt is proportional to -H)
This is a typical C4 differential equation question. The same algorithm could be used for a lot of other problems.
First of all, from the hint we have that dH/dt = -kH (1).
We need to separate the variables and then integrate both sides : dH/H = -kdt (2).
Now integrate both sides and dont forget about the integration constant!
lnH = -kt + c (3). In order to get rid of c, we have to use the boundary conditions given in the question, that is at t=0 , H=H0, and plug them into (3)
lnH0 = c (4) . Now plug (4) into (3) => lnH = -kt + lnH0 => lnH - ln H0 = -kt , using the subtraction of logs formula gives us: ln(H/H0) = -kt.
Taking e to the power of both sides : H/H0 = e-kt which gives our final answer : H = H0 e-kt.
Note that this is an exponential decrease where the value of k depends on various factors such as the viscosity, density of the fluid used or the diameter of the tap.
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