The curve shown in Figure 3 has parametric equations $x = t - 4 ext{ sin } t, \, y = 1 - 2 ext{ cos } t, \, \frac{2 ext{π}}{3} \leq t \leq \frac{2 ext{π}}{3}$ The point $A$, with coordinates $(k, 1)$, lies on the curve - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 8

To find the gradient, we first need to find $\frac{dy}{dx}$. We compute the derivatives:

$\frac{dx}{dt} = 1 - 4\text{ cos } t$ $\frac{dy}{dt} = 2\text{ sin } t$

Thus, the gradient is given by:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\text{ sin } t}{1 - 4\text{ cos } t}$

Substituting $t = \frac{\text{π}}{2}$ into the equation:

$\frac{dy}{dx} = \frac{2(1)}{1 - 4(0)} = 2$

Hence, the gradient at point $A$ is $2$.

To find $t$ where the gradient is equal to $-\frac{1}{2}$, we set up the equation:

$\frac{2\text{ sin } t}{1 - 4\text{ cos } t} = -\frac{1}{2}$

Cross-multiplying gives:

$4\text{ sin } t = - (1 - 4\text{ cos } t)$

Rearranging leads to:

$4\text{ sin } t + 1 = 4\text{ cos } t$

Squaring both sides leads to:

$(4\text{ sin } t + 1)^2 = (4\text{ cos } t)^2$

Using the identity $\text{ sin }^2 t + \text{ cos }^2 t = 1$, we can solve this polynomial equation and find:

$t \approx 3.0310$

Thus the value of $t$ is approximately 3.0310 to four decimal places.