Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm³/s 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 - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 8

At the point when h = 25, the rate at which water is leaking out is given as 400 cm³/s. We substitute h into the derived equation:

$\frac{dh}{dt} = 0.4 - k\sqrt{25} = 0.4 - 5k.$ Setting this equal to the leakage rate of -400, we have:

$400 = 0.4 - 5k.$

Solving for k gives:

$5k = 0.4 - 400 \Rightarrow k = \frac{0.4 - 400}{5} = 0.02.$ Thus, we have shown k = 0.02.

Starting from the equation:

$\frac{dh}{dt} = 0.4 - 0.02\sqrt{h},$ we can rewrite it in the separable form:

$\frac{dh}{0.4 - 0.02\sqrt{h}} = dt.$

Integrating both sides will give the time taken to fill the cylinder.

Using the substitution ( h = (20 - x)^2 ), we find that:

1. When h = 0, x = 20.
2. When h = 100, x = 10.

Then, the integral transforms to:

$\int_{20}^{10} \frac{50}{20 - (20 - x)} (-2(20 - x))dx \text{ which simplifies to } \int_{10}^{20} \frac{100}{x} dx.$

Evaluating gives:

$= 100 \ln(x) |_{10}^{20} = 100(\ln(20) - \ln(10)) = 100 \ln(2) \text{ in exact form.}$

From the evaluated integral, we have:

$\text{Time} = 200 \ln(2) \text{ seconds.}$

Calculating yields approximately:

$\text{Time} \approx 386.2934611 \text{ seconds which is } 6 \text{ minutes and } 26 \text{ seconds.}$