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Home » #### \( \frac10\pi3 \) cm or approximately 10.47 cm - RoadRUNNER Motorcycle Touring & Travel Magazine

#### \( \frac10\pi3 \) cm or approximately 10.47 cm - RoadRUNNER Motorcycle Touring & Travel Magazine

📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 Question: A museum curator uses a digital model where the position $ s(t) $ of a rotating exhibit arm is given by $ s(t) = t - \frac{t^5}{5} $. Define $ c_n $ by $ c_1 = \frac{1}{2} $ and $ c_{n+1} = s(c_n) $. Compute $ c_3 $.

📌 Solution: Given $ s(t) = t - \frac{t^5}{5} $, $ c_1 = \frac{1}{2} $. Compute recursively:

📌 $ c_2 = s\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{5} \left(\frac{1}{2}\right)^5 = \frac{1}{2} - \frac{1}{5} \cdot \frac{1}{32} = \frac{1}{2} - \frac{1}{160} = \frac{80 - 1}{160} = \frac{79}{160} $

📌 $ c_3 = s\left(\frac{79}{160}\right) = \frac{79}{160} - \frac{1}{5} \left(\frac{79}{160}\right)^5 $

📌 First compute $ \left(\frac{79}{160}\right)^5 $. Since this is tedious by hand, we compute step-by-step:

📌 Let $ a = \frac{79}{160} $. Then $ a^2 = \frac{6241}{25600} $,

📌 $ a^3 = \frac{79 \cdot 6241}{160 \cdot 25600} = \frac{493039}{4096000} $,

📌 $ a^5 = a^3 \cdot a^2 = \frac{493039}{4096000} \cdot \frac{6241}{25600} = \frac{493039 \cdot 6241}{4096000 \cdot 25600} $

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