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Nov 2002 p3 q7
1870
The diagram shows a curved rod AB of length 100 cm which forms an arc of a circle. The end points A and B of the rod are 99 cm apart. The circle has radius r cm and the arc AB subtends an angle of 2α radians at O, the centre of the circle.
(i) Show that α satisfies the equation \(\frac{99}{100}x = \sin x\).
(ii) Given that this equation has exactly one root in the interval \(0 < x < \frac{1}{2} \pi\), verify by calculation that this root lies between 0.1 and 0.5.
(iii) Show that if the sequence of values given by the iterative formula \(x_{n+1} = 50 \sin x_n - 48.5 x_n\) converges, then it converges to a root of the equation in part (i).
(iv) Use this iterative formula, with initial value \(x_1 = 0.25\), to find α correct to 3 decimal places, showing the result of each iteration.
Solution
(i) The arc length is given by \(2r\alpha = 100\). The chord length is given by \(2r \sin \alpha = 99\). Dividing these equations gives \(\frac{99}{100} = \frac{\sin \alpha}{\alpha}\), leading to \(\frac{99}{100}x = \sin x\).
(ii) Calculate \(f(0.1) = \frac{99}{100} \times 0.1 - \sin(0.1) \approx 0.099 - 0.0998 = -0.0008\) and \(f(0.5) = \frac{99}{100} \times 0.5 - \sin(0.5) \approx 0.495 - 0.4794 = 0.0156\). Since \(f(0.1) < 0\) and \(f(0.5) > 0\), there is a root between 0.1 and 0.5.
(iii) The iterative formula \(x_{n+1} = 50 \sin x_n - 48.5 x_n\) rearranges to \(x = 50 \sin x - 48.5 x\), which is equivalent to \(\frac{99}{100}x = \sin x\).
(iv) Using the iterative formula with \(x_1 = 0.25\):
