Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm^3 per second and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm^2.
a) Show that at time t seconds, the height h cm of liquid in the cylinder satisfies the differential equation:
, where k is a positive constant.
Well firstly let’s draw a little diagram to depict what is actually happening:
As you can see, dV/dt is the rate at which the volume of liquid in the cylinder is changing over time. You have to remember that liquid is being poured into the large vertical cylinder at a constant rate of 1600 cm^3 per second. What you also have to note is that the vertical cylinder is losing liquid at a rate proportional to the square root of the height already in the cylinder. Therefore we have .
Now the area of the circular cross section should be: . However, we are told that A=4000. Knowing that the Volume of the cylinder is , we can transform the volume formula to V=4000h.
Ok, so how can we get dh/dt? This is the formula we can use to derive it:
We know that:
We also know that:
So what we get is:
b) When h=25, water is leaking out of the hole at 400 cm^3 per second. Show that k=0.02.
Ok, with this information we can say that:
Moving on, we can determine that:
c) Separate the variables of the differential equation:
to show that the time taken to fill the cylinder from empty to a height of 100cm is given by:
d) Using the substitution:
, or otherwise, find the exact value of:
Therefore, we need to integrate:
e) Hence find the time taken to fill the cylinder from empty to a height of 100cm, giving your answer in minutes and seconds to the nearest second.
As this is the case,