Question 8 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 8 - 2008 - Paper 1

To find the area of the surface $S$, we sum the areas of the individual sections $A_k$:

$A = \sum_{k=1}^{n} A_k = \sum_{k=1}^{n} \frac{2\pi R^2 \sin\delta \cos(2k - 1)\delta}{2}$

Simplifying this, we can use results from part (a) and the properties of summation to write:

$A = \pi R^2 \sin \delta \sum_{k=1}^{n} \cos(2k - 1)\delta$

By using the equation from part (a), we can substitute: $\sum_{k=1}^{n} \cos(2k - 1)\theta = \frac{\sin 2n\theta}{2 \sin \theta}.$

As $n$ approaches infinity, we analyze the expression for $A$. The summands tend to oscillate, but their influence is constrained as rac{\sin 2n\theta}{2 \sin \theta} becomes a bounded quantity, hence:

$\lim_{n \to \infty} A = \lim_{n \to \infty} \pi R^2 \sin \delta \cdot \frac{\sin 2n\theta}{2 \sin \theta}$

This approaches: $\frac{\pi R^2 \sin \delta}{\sin \theta}$. Thus, the limiting value of $A$ conforms according to the defined parametric relationships within the surface defined.

To find the second derivative:

First, differentiate $f(t)$: $f'(t) = a \cos(a + nt) \sin b - n \sin(a + nt) \sin b + n \sin a \cos(b - nt).$

Differentiating again gives: $f''(t) = -n^2 f(t).$

To show $f(0)=0$, substitute $t=0$: $f(0) = \sin(a) \sin b - \sin a \sin b = 0.$

Using the result of $f''(t)$ gives a second-order linear ordinary differential equation:

The solutions take the form: $f(t) = c_1 \sin(nt) + c_2 \cos(nt),$

Applying initial conditions determines $c_1, c_2$. Ultimately, this yields: $f(t) = \sin(a + b) \sin(nt).$

To solve:

Rearranging gives: $\sin(a + nt) \sin b - \sin a \sin(b - nt) = 0.$

This can be solved using product-to-sum identities or numerical methods depending on constraints on $a$, $b$, and $n$.

The periodic nature of sine indicates solutions will recur, hence values of $t$ will be situated at intervals based on these fundamental frequencies.