The height $x$ metres, of a column of water in a fountain display satisfies the differential equation \[ \frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}\n\] where $i$ is the time in seconds after the display begins - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Next, we integrate both sides:

[ \int \sqrt{x} , dx = \int \frac{8\sin 2t}{3} , dr ] This leads to: [ \frac{2}{3}x^{\frac{3}{2}} = -\frac{4}{3}cos(2t) + C ]

Given that initially the column of water has zero height (i.e., when $t = 0$, $x = 0$), we can substitute these values to find $C$: [ \frac{2}{3}(0)^{\frac{3}{2}} = -\frac{4}{3}cos(0) + C\ 0 = -\frac{4}{3} + C \Rightarrow C = \frac{4}{3} ]

Now substituting $C$ back into the equation, we have: [ \frac{2}{3}x^{\frac{3}{2}} = -\frac{4}{3}cos(2t) + \frac{4}{3} \Rightarrow x^{\frac{3}{2}} = 2 - 2cos(2t) ]

Finally, we express $x$ as: [ x = \left(2 - 2cos(2t)\right)^{\frac{2}{3}}\n]

To find the maximum height of the column of water, we observe the function: [ \cos(2t) \text{ oscillates between -1 and 1. Therefore, the maximum value occurs when } cos(2t) = -1. ] This gives: [ x_{ ext{max}} = \left(2 - 2(-1)\right)^{\frac{2}{3}} = (4)^{\frac{2}{3}} = 4^{\frac{2}{3}} = 4^{\frac{2}{3}} = 252 cm ]