Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = an(2x)$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 5 - 2005 - Paper 1

A cyclic quadrilateral is defined as a quadrilateral whose vertices all lie on the circumference of a circle. In this case, quadrilateral DP AE is cyclic because the points D, P, A, and E all lie on the circumference of the circle defined by points A and B, thus satisfying the condition of a cyclic quadrilateral.

To prove that the angles are equal, we can use the property of angles subtended by the same arc in a circle. Since points A, P, E, and B lie on the arc AB, it follows that: $\angle APE = \angle ABC$ This holds because both angles subtend the same chord AB of the circle.

Since we have established that $\angle APE = \angle ABC$ and both angles are subtended from points A and E, we can deduce that lines AP and BC are perpendicular to one another at point P. Thus, it follows that the line PQ is perpendicular to BC, given that it extends vertically from point P down to line BC.

We express the linear combination of sine and cosine in the form $R \sin(3t - \alpha)$. To find R and α, we compare:

1. Find R using the formula: $R = \sqrt{A^2 + B^2}$ where $A = \sqrt{3}$ and $B = -1$. Thus: $R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$
2. To find α, we use the relationship: $\tan(\alpha) = \frac{-1}{\sqrt{3}}$ This gives $\alpha = \frac{\pi}{3}$, thus: $\sqrt{3}\sin(3t) - \cos(3t) = 2\sin(3t - \frac{\pi}{3})$

The amplitude of the simple harmonic motion is given by R, which we found to be 2. The period T can be calculated from the angular frequency ω: Since the function $x = 5 + 2 \sin(3t - \frac{\pi}{3})$, the angular frequency is 3: $T = \frac{2\pi}{\omega} = \frac{2\pi}{3}$

The speed of the particle can be derived from the position function by taking the derivative: $v = \frac{dx}{dt}$ Calculating the derivative and setting it to zero allows us to find the maximum speed:

1. Differentiate $x = 5 + 2\sin(3t - \frac{\pi}{3})$: $v(t) = 6\cos(3t - \frac{\pi}{3})$
2. To find when $v = 0$ we solve: $6\cos(3t - \frac{\pi}{3}) = 0$ This occurs when: $3t - \frac{\pi}{3} = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$ From this we can determine the times at which the speed reaches its maximum.