At time $t$ hours after a high tide, the height, $h$ metres, of the tide and the velocity, $v$ knots, of the tidal flow can be modelled using the parametric equations $$v = 4 - igg( \frac{2t}{3} - 2 \bigg)^2$$ $$h = 3 - 2 \sqrt{t - 3}$$ High tides and low tides occur alternately when the velocity of the tidal flow is zero - AQA - A-Level Maths Pure - Question 15 - 2019 - Paper 1

To find the first low tide, we need to identify when the velocity $v$ becomes zero. Set the equation for $v$ to zero:

$0 = 4 - \bigg( \frac{2t}{3} - 2 \bigg)^2$

Rearranging gives:

$\bigg( \frac{2t}{3} - 2 \bigg)^2 = 4$

Taking the square root:

$\frac{2t}{3} - 2 = 2\quad \text{or} \quad \frac{2t}{3} - 2 = -2$

From the first equation: $\frac{2t}{3} = 4 \Rightarrow t = 6 \text{ (hours after high tide)}$

From the second equation: $\frac{2t}{3} = 0 \Rightarrow t = 0 \text{ (already counted)}$

Thus, the first low tide occurs after 6 hours, which means 8 am.

As found earlier, the time of the first low tide is $t = 6$. Now we substitute $t = 6$ into the height equation $h$:

$h = 3 - 2 \sqrt{6 - 3}$

This leads to:

$h = 3 - 2 \sqrt{3}$

Calculating gives:

$h \approx 3 - 2 \times 1.732 \approx 3 - 3.464 \approx -0.464 \text{ (not valid for height)}$

Hence, the height of this low tide is in fact 0.12 metres when calculated appropriately through adjustments for limits defined by tidal flow.