A curve C has parametric equations $$ x = 2 ext{sin}t, \ y = 1 - ext{cos}2t \\ rac{-rac{ ext{π}}{2} ext{}}{ ext{}} ext{ } rac{≤ t ≤ rac{ ext{π}}{2}}{ }$$ (a) Find $\frac{dy}{dx}$ at the point where $t = \frac{\pi}{6}$ (b) Find a cartesian equation for C in the form $y = f(x), \\ -k ≤ x ≤ k,$ stating the value of the constant $k$ - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 9

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

1. Differentiate $x$ and $y$:
   The equations are: $x = 2 \sin t, \quad y = 1 - \cos 2t$ Differentiating with respect to $t$ gives: $\frac{dx}{dt} = 2 \cos t$ $\frac{dy}{dt} = 2 \sin 2t$

2. Apply the chain rule: To find $\frac{dy}{dx}$, use: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \sin 2t}{2 \cos t} = \frac{\sin 2t}{\cos t}$

3. Evaluate at $t = \frac{\pi}{6}$:
   Substitute $t = \frac{\pi}{6}$: $\frac{dy}{dx} = \frac{\sin\left(2\cdot\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \frac{\sin\left(\frac{\pi}{3}\right)}{\frac{\sqrt{3}}{2}}$ This simplifies to: $\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1$ Therefore, $\frac{dy}{dx} = 1$.