Figure 3 shows part of the curve C with parametric equations $x = tan \theta$, $y = sin \theta$, $0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates $(\sqrt{3}, \frac{1}{2}, \sqrt{3})$ (a) Find the value of \(\theta\) at the point P - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 6

To find the coordinates of Q, we first need the slope of the normal line at point P. Using the derivative:

1. We find (\frac{dx}{d\theta} = sec^2 \theta) and (\frac{dy}{d\theta} = cos \theta).

At point P: [ m = \frac{dy}{dx} = \frac{cos \theta}{sec^2 \theta} = cos^3 \theta = \frac{1}{8} \rightarrow \theta = \frac{\pi}{3} ]

Then, the y-coordinate at Q is:

1. Since the line l is normal to C at P, it will cut the x-axis, where (y = 0): [ x = \sqrt{3} - \left(x - \sqrt{3}\right) \frac{dy}{dx}] With some calculation, we find: [ (k \sqrt{3}, 0) ext{ where } k = \frac{1}{2} ]

To find the volume V of the solid of revolution formed by rotating region S:

1. We use the formula: [ V = \pi \int_{0}^{\sqrt{3}} y^2 dx = \pi \int_{0}^{\sqrt{3}} (sin \theta)^2 dx]

2. Transforming to parametric terms, we can express this integral as: [ V = \int (y)(\frac{dx}{d\theta}) d\theta = \int y(\frac{dx}{d\theta})] Evaluating this integral will lead to: [ V = \frac{3\sqrt{3}}{2} \pi +\pi \text{ constant terms} ].

Therefore, the final volume can be expressed in the form (p\pi \sqrt{3} + q r^2) where (p = 1,; q = g;.)