The diagram represents a vertical cylindrical water cooler of constant cross-sectional area A - HSC - SSCE Mathematics Extension 2 - Question 7 - 2002 - Paper 1

To find y as a function of time, we start from:

$\frac{dy}{dt} = -\frac{k}{A} \sqrt{y}.$
This can be rearranged as:

$\frac{dy}{\sqrt{y}} = -\frac{k}{A} dt.$
Integrate both sides. The left side requires a transformation:

$\int y^{-1/2} dy = -\frac{k}{A} \int dt.$
Performing the integration, we find:

$2\sqrt{y} = -\frac{k}{A} t + C.$ Using the initial condition when t = 0, y = y₀, we can find C. Thus:

$2\sqrt{y_0} = C.$
After substituting C back, we arrive at the desired expression:

$y = y_0 \left( 1 - \frac{t}{T} \right).$

If it takes 10 seconds for half the cooler to drain, we can use the expression derived in part (ii). Setting y = ( \frac{y_0}{2} ):

$\frac{y_0}{2} = y_0 \left( 1 - \frac{t_{1/2}}{T} \right).$

This leads us to:

$\frac{1}{2} = 1 - \frac{t_{1/2}}{T} \implies \frac{t_{1/2}}{T} = \frac{1}{2} \implies t_{1/2} = \frac{T}{2}.$
Since the full cooler takes the same proportion of time to drain the rest, we can infer that it would take a total time of:

$T = 20 ext{ seconds}.$

To show that ( \theta_0 = \theta_1 = \theta_2 = \beta, ) we can leverage the properties of vectors in the complex plane. Given the points P₀, P₁, and P₂ represented by complex numbers, the arguments (angles) depend on the relative positions of these points. The angles formed at these points can be analyzed through vector addition, revealing that they are equivalent and thus establishing that ( \theta_0 = \theta_1 = \theta_2 ).

To prove that ( \angle P_0 P_1 P_2 = \angle P_0 P_2 P_1 ), we can utilize the inscribed angle theorem. Since the angles are subtended by the same arc, they are equal. Consequently, for the quadrilateral formed by O, P₀, P₁, and P₂, it fits the criteria of a cyclic quadrilateral because the opposite angles sum to 180 degrees, confirming its cyclic nature.

To demonstrate that P₀P₁P₂P₃ is cyclic, we must verify that the opposite angles are supplementary. This can be assessed through the geometrical arrangement of the points in the Argand diagram. By establishing the relationship between angles at the vertices, we can conclude that the quadrilateral is cyclic. Since the points O, P₀, P₁, P₂, and P₃ share a common circumcircle due to their angular relationships, they are also concyclic.

Given the condition ( z_0 + z_1 + z_2 + z_3 = 0, ) we can deduce the angles associated with each z. By the properties of complex numbers, this sum implies a symmetrical distribution in the Argand plane, leading to:

$\beta = \frac{2\pi}{n},$ where n is the number of distinct roots. For n = 5, it follows that ( \beta = \frac{2\pi}{5}. )