The curve shown in Figure 3 has parametric equations $x = t - 4 \, ext{sin} \, t, \quad y = 1 - 2 \, ext{cos} \, t, \quad \frac{2\, ext{π}}{3} \leq t \leq \frac{2\text{π}}{3}$ The point $A$, with coordinates $(k, 1)$, lies on the curve - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 9

To find the gradient at the point $A(k, 1)$, we need to calculate rac{dy}{dx} using the parametric equations. First, we find rac{dx}{dt} and rac{dy}{dt}:

$\frac{dx}{dt} = 1 - 4\text{cos} \, t$
$\frac{dy}{dt} = 2 \text{sin} \, t$
Now we calculate rac{dy}{dx}:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\text{sin} \, t}{1 - 4\text{cos} \, t}$
Substituting $t = \frac{3\text{π}}{2}$:

$\frac{dy}{dx} = \frac{2\text{sin} \left( \frac{3\text{π}}{2} \right)}{1 - 4\text{cos} \left( \frac{3\text{π}}{2} \right)}$
We know that: $\text{sin} \left( \frac{3\text{π}}{2} \right) = -1 \text{ and } \text{cos} \left( \frac{3\text{π}}{2} \right) = 0$
Hence,

$\frac{dy}{dx} = \frac{2(-1)}{1 - 4(0)} = \frac{-2}{1} = -2$
Therefore, the gradient of the curve at point $A$ is:

$\frac{dy}{dx} = -2$

To find where the gradient is $-\frac{1}{2}$, we set:

$\frac{dy}{dx} = -\frac{1}{2}$
Thus,

$\frac{2\text{sin} \, t}{1 - 4\text{cos} \, t} = -\frac{1}{2}$
Cross-multiplying gives:

$4\text{sin} \, t = - (1 - 4\text{cos} \, t)$
This simplifies to:

$4\text{sin} \, t + 1 - 4\text{cos} \, t = 0$
Rearranging yields:

$4\text{sin} \, t - 4\text{cos} \, t + 1 = 0$
This can be rewritten as:

$\text{sin} \, t - \text{cos} \, t = -\frac{1}{4}$
Using the identity:

$\text{tan} \, t = \frac{\text{sin} \, t}{\text{cos} \, t}$
is a useful approach, leading to:

$\text{sin} \, t = -\frac{1}{4}\text{cos} \, t$
Then substituting into the Pythagorean identity gives:

$\left(-\frac{1}{4}\text{cos} \, t\right)^2 + \text{cos}^2 \, t = 1$
Expanding and solving:

$\frac{1}{16}\text{cos}^2 \, t + \text{cos}^2 \, t = 1 \Rightarrow \frac{17}{16}\text{cos}^2 \, t = 1$
Thus:

$\text{cos}^2 \, t = \frac{16}{17} \Rightarrow \text{cos} \, t = \pm \sqrt{\frac{16}{17}} = \pm\frac{4}{\sqrt{17}}$
Substituting back to find $ext{sin} \, t$ gives:

If $\text{cos} \, t = \frac{4}{\sqrt{17}}$, then:

$\text{sin} \, t = -\frac{1}{4} \cdot \frac{4}{\sqrt{17}} = -\frac{1}{\sqrt{17}}$
Using the values we can find the angles:

$t = \text{tan}^{-1}\left( \frac{-\frac{1}{\sqrt{17}}}{\frac{4}{\sqrt{17}}} \right) = \text{tan}^{-1}\left(-\frac{1}{4}\right)$
By evaluating we find the angle in radians, leading to:

$t \approx 0.2448 \; \text{radians (to 4 decimal places)}$