**Question (8):**

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 *tÂ *seconds, the height *h *cm of liquid in the cylinder satisfies the differential equation:

, where *kÂ *is a positive constant.

**Solution:**

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 Â 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:

.

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b) When *h*=25, water is leaking out of the hole at 400 cm^3 per second. Show that *k*=0.02.

**Solution:**

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:

**Solution:**

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d) Using the substitution:

, or otherwise, find the exact value of:

**Solution:**

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.

**Solution:**

Remember that:

.

As this is the case,