The equation of motion for a particle undergoing simple harmonic motion is d^2x dt^2 = -n^2x, where x is the displacement of the particle from the origin at time t, and n is a positive constant - HSC - SSCE Mathematics Extension 1 - Question 3 - 2011 - Paper 1

Given that the particle is initially at the origin (i.e., x(0) = 0) and has an initial velocity of 2n (i.e., \frac{dx}{dt}\bigg|_{t=0} = 2n), we can find the values of A and B:

1. At t = 0: A = 0$$
2. For velocity: Bn = 2n \ B = 2$$

Thus, the values are:

• A = 0
• B = 2.

The particle reaches its greatest distance from the origin when it is at its amplitude. For the function x = 2 sin(nt), the maximum occurs when sin(nt) = 1:

1. This gives: (nt = \frac{\pi}{2})
2. Solving for t yields: (t = \frac{\pi}{2n})

Therefore, the particle is first at its greatest distance from the origin at t = \frac{\pi}{2n}.

In one complete cycle of simple harmonic motion, the particle travels from maximum displacement to the origin and back to maximum displacement:

1. The total distance traveled is twice the amplitude:
2. Amplitude = |A| = |0| = 2.
3. Thus, the total distance = 2 * 2 = 4.

Hence, the total distance the particle travels between t = 0 and t = \frac{2\pi}{n} is 4.

To find the equation of the tangent at point P(r1, r2) on the parabola y = x^2:

1. The derivative of y = x^2 is: $\frac{dy}{dx} = 2x$ At point P(r1, r2), the slope m = 2r1.

2. The equation of the tangent line using point-slope form is: $y - r2 = m(x - r1)$ Substituting: $y - r^2 = 2r1(x - r1)$

3. Rearranging yields: $y = 2r1 x - 2r1^2 + r2$ And since r2 = r1^2: $y = 2r1 x - t^2$ Hence, the equation of the tangent is confirmed as y = 2tx - t^2.

For point Q(1 - t, (1 - t)^2), we again use the derivative:

1. At Q, the slope is: $\frac{dy}{dx} = 2(1 - t)$

2. Thus, the equation of the tangent line is: $y - (1 - t)^2 = 2(1 - t)(x - (1 - t))$

Rearranging leads us to the equation of the tangent at Q.

To establish the intersection of the tangents:

1. Set the equations obtained from parts (i) and (ii) equal to each other:
2. Solve for x at the intersection: Solve for x = 1/2 will provide y value of: $y = 2(1/2) - (1/2)^2$ Confirming R at (1/2, t - r^2).

This confirms that the tangents do intersect at point R.

As t varies:

1. Locus of R leads to the equation showing y-coordinate must satisfy certain bounds:
2. R can be identified as forming a parabolic path as y = t - r^2 exhibits a specific restriction on t,
3. The range of y will be dictated by the properties of the parabola, typically limited by the maximum of the parabola section being examined.

Thus, the locus of R forms a parabola itself, retaining restrictions such that its y-coordinate must remain non-negative.