Figure 2 shows a sketch of the curve C with parametric equations $x = 4 \tan t, \quad y = 5\sqrt{3}\sin 2t, \quad 0 \leq t \leq \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 6 - 2016 - Paper 4

To find $\frac{dy}{dx}$, we use the chain rule for parametric equations:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

First, we compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

1. The derivative of $x$: $x = 4 \tan t \implies \frac{dx}{dt} = 4 \sec^2 t$

2. The derivative of $y$: $y = 5\sqrt{3}\sin 2t \implies \frac{dy}{dt} = 10\sqrt{3}\cos 2t$

Now substituting for $t = \frac{\pi}{3}$ (since $P$ corresponds to this value):

• For $\frac{dx}{dt}$: $\frac{dx}{dt} \bigg|_{t = \frac{\pi}{3}} = 4 \sec^2\left(\frac{\pi}{3}\right) = 4 \cdot 4 = 16$

• For $\frac{dy}{dt}$: $\frac{dy}{dt} \bigg|_{t = \frac{\pi}{3}} = 10\sqrt{3}\cos\left(2 \cdot \frac{\pi}{3}\right) = 10\sqrt{3} \cdot (-\frac{1}{2}) = -5\sqrt{3}$

Putting it together:

$\frac{dy}{dx} = \frac{-5\sqrt{3}}{16}$

Thus, the exact value of $\frac{dy}{dx}$ at the point P is $\frac{-5\sqrt{3}}{16}$.